heat convection by latif m. jiji - solutions
TRANSCRIPT
PROBLEM 1.1
Heat is removed from a rectangular surface by
convection to an ambient fluid at T . The heat transfer
coefficient is h. Surface temperature is given by
sT = 2/1x
A
where A is constant. Determine the steady state heat transfer rate from the plate.
(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable. (ii) Ambient temperature and heat transfer coefficient are uniform. (iii) Surface temperature varies along the rectangle.
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface temperature is not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oCqs = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
d qs = h (Ts - T ) dAs (b)
The next step is to express )(xTs in terms of distance x along the triangle. )(xTs is specified as
sT = 2/1x
A (c)
L
x0 W
x0
L
Wsdq
dx
PROBLEM 1.1 (continued)
The infinitesimal area dAs is given by
dAs = W dx (d) where
x = axial distance, m W = width, m
Substituting (c) and into (b)
d qs = h(2/1x
A - T ) Wdx (e)
Integration of (f) gives qs
qs = sdq =
L
dxTAxhW0
2/1 )( (f)
Evaluating the integral in (f)
LTALhWqs2/12
Rewrite the above
TALhWLqs2/12 (g)
Note that at x = L surface temperature )(LTs is given by (c) as
2/1)( ALLTs (h)
(h) into (g)
TLThWLq ss )(2 (i)
(iii) Checking. Dimensional check: According to (c) units of C are 2/1o C/m . Therefore units
sq in (g) are W.
Limiting checks: If h = 0 then qs = 0. Similarly, if W = 0 or L = 0 then qs = 0. Equation (i)
satisfies these limiting cases.
(5) Comments. Integration is necessary because surface temperature is variable.. The same procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
PROBLEM 1.2
A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to
an ambient fluid at T . The heat transfer coefficient h varies along the surface according to
h = C
x1 2/
where C is constant and x is distance along the base measured from the apex. Determine the
total heat transfer rate from the triangle.
(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling may be helpful. (ii) Ambient temperature and surface temperature are uniform. (iii) Surface area and heat transfer coefficient vary along the triangle.
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface area and heat transfer coefficient are not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs
and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation and (3) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oCqs = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Applying (a) to an infinitesimal area dAs
d qs = h (Ts - T ) dAs (b)
The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer coefficient h is given by
h = C
x1 2/ (c)
The infinitesimal area dAs is given by
W
x
L
dx
dqs
PROBLEM 1.2 (continued)
dAs = y(x) dx (d) where
x = distance along base of triangle, m y(x) = height of the element dAs, m
Similarity of triangles give
y(x) = W
Lx (e)
where
L = base of triangle, m W = height of triangle, m
Substituting (c), (d) and (e) into (b)
d qs = C
x1 2/(Ts - T )
W
Lx dx (f)
Integration of (f) gives qs. Keeping in mind that C, L, W, Ts and T are constants, (f) gives
qs = sdq = )( TTL
WCs
L
x
x
02/1
dx (g)
Evaluating the integral in (g)
qs =2
3C W L
1/2 (Ts - T ) (h)
(iii) Checking. Dimensional check: According to (c) units of C are W/m3/2-oC. Therefore units of qs in (h) are
qs = C(W/m3/2-oC) W(m) L1/2(m1/2) (Ts - T )(oC) = W
Limiting checks: If h = 0 (that is C = 0) then qs = 0. Similarly, if W = 0 or L = 0 or Ts = T
then qs = 0. Equation (h) satisfies these limiting cases.
(5) Comments. Integration was necessary because both area and heat transfer coefficient vary with distance along the triangle. The same procedure can be followed if the ambient temperature or surface temperature is non-uniform.
PROBLEM 1.3
A high intensity light bulb with surface heat flux sAq )/( is cooled by a fluid at T . Sketch the
fluid temperature profiles for three values of the heat transfer coefficients: h1, h2, and h3, where
h1 < h2 < h3.
(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling. (iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature gradient is described by Fourier’s law.(vii) Ambient temperature is constant.
(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and surface gradient..
(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply Fourier’s law to determine temperature gradient at the surface.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature and (4) constant properties.
(ii) Analysis. Newton’s law of cooling
)(/ TThAq ss (a)
Solve for sT
h
AqTT s
s
)/( (b)
This result shows that for constant sAq )/( surface temperature decreases as h is increased.
Apply Fourier’s law
0
/
y
sy
TkAq (c)
where y is the distance normal to the surface. Rewrite (c)
k
Aq
y
T w
y
/
0
(d)
This shows that temperature gradient at the surface remains constant independent of h. Based on (b) and (d) the temperature profiles corresponding to three values of h are shown in the sketch.
(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature
C)Cw/m(
)w/m()/()C()C( o
o2
2oo
h
AqTT s
s
T
sTT
y
2h
1h
3hsAq )/(
PROBLEM 1.3 (continued)
(2) Each term in (d) has units of C/mo
C/mC)W/m-(
)C/m(/C/m)( o
o
2oo
0 k
Aq
y
T w
y
Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0
in (b) gives .sT
(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the thermal conductivity of the fluid is constant and radiation is neglected.
PROBLEM 1.4
Explain why fanning gives a cool sensation.
(1) Observations. (i) Metabolic heat leaves body at the skin by convection and radiation. (ii) Convection is described by Newton's law of cooling. (iii). Fanning increases the heat transfer coefficient and affects temperature distribution, including surface temperature. (iv). Surface temperature decreases as the heat transfer coefficient is increased. (v) Surface temperature is described by Newton’s law of cooling. (vi) Ambient temperature is constant.
(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature.
(3) Solution Plan. Apply Newton's law of cooling to examine surface temperature.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature, (4) constant surface heat flux and (5) constant properties.
(ii) Analysis. Newton’s law of cooling
)( TThq ss (a)
where
h = heat transfer coefficient, CW/m o2
sq surface heat flux, 2W/m
sT = surface temperature, Co
T =ambient temperature, Co
Solve (a) for sT
h
qTT s
s (b)
This result shows that for constant sq , surface temperature decreases as h is increased. Since
fanning increases h it follows that it lowers surface temperature and gives a cooling sensation.
(iii) Checking. Dimensional check: Each term in (b) has units of temperature
C)Cw/m(
)w/m()C()C( o
o2
2oo
h
qTT s
s
T
y
TsT
sq
fan
fanno
skin
PROBLEM 1.4 (continued)
Limiting check: for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0 in
(b) gives .sT
(5) Comments. (i) The analysis is based on the assumption that surface heat flux remains constant. (ii) Although surface temperature decreases with fanning, temperature gradient at the surface remains constant. This follows from the application of Fourier’s law at the surface
s
sy
Tkq
Solving for syT )/(
k
q
y
T s
s
constant
PROBLEM 1.5
A block of ice is submerged in water at the melting temperature. Explain why stirring the water
accelerates the melting rate.
(1) Observations. (i) Melting rate of ice depends on the rate of heat added at the surface. (ii) Heat is added to the ice from the water by convection. (iii) Newton's law of cooling is applicable. (iv). Stirring increases surface temperature gradient and the heat transfer coefficient. An increase in gradient or h increases the rate of heat transfer. (v) Surface temperature remains constant equal to the melting temperature of ice. (vi) water temperature is constant.
(2) Problem Definition. Determine effect of stirring on surface heat flux.
(3) Solution Plan. Apply Newton's law of cooling to examine surface heat flux.
(4) Plan Execution.
(i) Assumptions. (1) no radiation ,(2) uniform water temperature, (3) constant melting (surface) temperature.
(ii) Analysis. Newton’s law of cooling
)( TThq ss (a)
where
h = heat transfer coefficient, CW/m o2
sq surface heat flux, 2W/m
sT = surface temperature, Co
T =ambient water temperature, Co
Stirring increases h . Thus, according to (a) surface heat flux increases with stirring. This will accelerate melting.
(iii) Checking. Dimensional check: Each term in (a) has units of heat flux.
Limiting check: For sTT (water and ice are at the same temperature), no heat will be added to
the ice. Set sTT in (a) gives .0sq
(5) Comments. An increase in h is a consequence of an increase in surface temperature gradient. Application of Fourier’s law at the surface gives
s
sy
Tkq (b)
y
Tsq
0sT
stirringstirringno
ice
ice
water
PROBLEM 1.5 (continued)
Combining (a) and (b)
TT
y
Tk
hs
s (c)
According to (c), for constant sT and T , increasing surface temperature gradient increases h.
PROBLEM 1.6
Consider steady state, incompressible, axisymmetric parallel flow in a tube of radius or . The
axial velocity distribution for this flow is given by
)(2
2
1or
ruu
where u is the mean or average axial velocity. Determine the three components of the total
acceleration for this flow.
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) For parallel
streamlines 0vv r . (iii) Axial velocity is independent of axial and angular distance.
(2) Problem Definition. Determine the total acceleration in the r, and z directions.
(3) Solution Plan. Apply total derivative in cylindrical coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) Constant radius tube, (2) constant density and (3) streamlines are parallel to surface.
(ii) Analysis. Total acceleration in cylindrical coordinates is given by
tzrrr
Dd rrz
rr
rr vvv
vvvvv
Dt
v
dt
v2
(1.23a)
tzrrr
Ddz
rr
vvv
vvvvvv
Dt
v
dt
v (1.23b)
tzrr
Dd zzz
zzr
zz vvv
vvvv
Dt
v
dt
v (1.23c)
For streamlines parallel to surface
0vv r (a)
The axial velocity zu v is given by
)(2
2
1o
zr
ruuv (b)
From (b) it follows that
0tz
zz vv (c)
Substituting into (1.23a), (1.23b) and (1.23c)
Radial acceleration: 0Dt
v
dt
v rr Dd
Angular acceleration;Dt
v
dt
v Dd0
PROBLEM 1.6 (continued)
Axial acceleration: 0Dt
v
dt
v zz Dd
(5) Comments. All three acceleration components vanish for this flow.
PROBLEM 1.7
Consider transient flow in the neighborhood of a vortex line where the
velocity is in the tangential direction given by
t
r
rtrV o
4exp1
2),(
2
Here r is the radial coordinate, t is time, o is circulation
(constant) is kinematic viscosity. Determine the three components
of total acceleration.
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) streamlines are
concentric circles. Thus the velocity component in the radial direction vanishes ( 0rv ). (iii)
For one-dimensional flow there is no motion in the z-direction ( 0zv ). (iv) The -velocity
component, v , depends on distance r and time t.
(2) Problem Definition. Determine the total acceleration in the r, and z directions.
(3) Solution Plan. Apply total derivative in cylindrical coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) streamlines are concentric circles (2) no motion in the z-direction.
(ii) Analysis. Total acceleration in cylindrical coordinates is given by
The three components of the total acceleration in the cylindrical coordinates zr ,, are
tzrrr
Dd rrz
rrr
rr vvv
vvvvv
Dt
v
dt
v2
(1.23a)
tzrrr
Ddz
rr
vvv
vvvvvv
Dt
v
dt
v (1.23b)
tzrr
Dd zzz
zzr
zz vvv
vvvv
Dt
v
dt
v (1.23c)
For the flow under consideration the three velocity component, ,rv v and zv are
0rv (a)
t
r
rtrv o
4exp1
2),(
2
(b)
0zv (c)
Radial acceleration: (a) and (c) into (1.23a)
Vr
PROBLEM 1.7 (continued)
r
D r2v
Dt
v (d)
(b) into (d) 2
2
32
2
4exp1
4
)
t
r
r
D or
Dt
v(e)
Tangential acceleration: (a) and (c) into (1.23b)
t
D v
Dt
v (f)
(b) into (f)
t
r
tt
r
r
vD o
4exp
1
42
22
Dt (g)
Axial acceleration: (a) and into (1.23c)
0Dt
v zD (h)
(iii) Checking. Dimensional check: Units of acceleration in (e) and (g) are .m/s2 Note that
according to (b), units of o are /sm2 and the exponent of the exponential is dimensionless.
Thus units of (e) are
22
332
242
4exp1
)m(4
)/sm()
t
r
r
vD or
Dt= 2m/s
Units of (g) are
t
r
tt
r
rDt
Dv o
4exp
)s(
1
)s()/sm(4
)m(
)m(2
)/sm( 2
2
222
= 2m/s
Limiting check: (1) For 0o , all acceleration components vanish. Setting 0o in (e) and
(g) gives 0Dt
v
Dt
v DD r
(2) According to (b) at t , the tangential velocity vanishes ( v = 0). Thus all acceleration
components should vanish. Setting t in (e) and (g) gives 0Dt
v
Dt
v DD r .
(5) Comments. The three velocity components must be known to determine the three acceleration components.
PROBLEM 1.8
An infinitely large plate is suddenly moved parallel to its surface with a velocity oU . The
resulting transient velocity distribution of the surrounding fluid is given by
0
2 )exp()/2(1 dUu o
where the variable is defined as
t
ytx
2),(
Here t is time, y is the vertical coordinate and is kinematic viscosity. Note that streamlines
for this flow are parallel to the plate. Determine the three components of total acceleration.
(1) Observations. (i) This problem is described by Cartesian coordinates. (ii) For parallel streamlines the y-velocity component 0v . (iii) For one-dimensional flow there is no motion in
the z-direction (w = 0). The x-velocity component depends on distance y and time t.
(2) Problem Definition. Determine the total acceleration in the x, y and z directions.
(3) Solution Plan. Apply total derivative in Cartesian coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) streamlines are parallel to surface and (2) no motion in the z-direction.
(ii) Analysis. Total acceleration in Cartesian coordinates is given by
t
f
z
fw
y
f
x
fu
Dt
Df
dt
dfv (1.21)
where f represents any of the three velocity components u, v or w. The x-velocity component u is given by
0
2 )exp()/2(1 dUu o (a)
where
t
ytx
2),( (b)
Note that u depends on y and t only. For one-dimensional parallel flow
0wv (c)
Total acceleration in the x-direction, xa . Set f = u in (1.21)
t
u
z
uw
y
u
x
uu
Dt
Du
dt
duax v (d)
oUx
y
plate 0
PROBLEM 1.8 (continued)
Since u depends on y and t only, it follows that
0x
u (e)
Substitute (c) and (e) into (d)
t
uax (f)
This derivative is obtained using the chain rule
td
du
t
uax (g)
Using (a)
)exp(2 2oU
d
du(h)
Using (b)
ttt
yt
y
t 4
1
42
2/3 (i)
Substitute (h) and (i) into (g)
t
U
t
ua o
x
)exp(
2
2
(g)
Total acceleration in the y-direction, ya . Set f = v in (1.21)
tzw
yxu
Dt
D
dt
day
vvvv
vvv (h)
Apply (c) to (h)
0ya (i)
Total acceleration in the z-direction, za . Set f = w in (1.21)
t
w
z
ww
y
w
x
wu
Dt
Dw
dt
dwaw v (j)
Apply (c) to (h)
0za (k)
(iii) Checking. Dimensional check: Units of acceleration in (g) are /s.m2 Note that is
dimensionless. Thus units of (g) are
/sm)s(
)exp(
2
)m/s( 22
t
Ua o
x
Limiting check: (1) For 0oU , the acceleration .0xa Setting 0oU in (g) gives .0xa
(2) According to (b) at t , .0),(y Evaluation (a) at 0),(y gives
PROBLEM 1.8 (continued)
oUyu ),( (l)
Since u is constant every where it follows that the xa must be zero. Setting 0 and t in
(g) gives .0xa
(5) Comments. The three velocity components must be known to determine the three acceleration components.
PROBLEM 1.9
Consider two parallel plates with the lower plate stationary and the upper plate moving with a
velocity .oU The lower plate is maintained at temperature 1T and the upper plate at .oT The
axial velocity of the fluid for steady state and parallel streamlines is given by
H
yUu o
where H is the distance between the two plates.
Temperature distribution is given by
11
22
)(2
TH
yTT
H
yy
kH
UT o
o
where k is thermal conductivity and is viscosity. Determine the total temperature derivative.
(1) Observations. (i) This problem is described by Cartesian coordinates. (ii) For parallel streamlines the y-velocity component 0v . (iii) For one-dimensional flow there is no motion in
the z-direction (w = 0). The x-velocity component depends on distance y only.
(2) Problem Definition. Determine the total temperature derivative.
(3) Solution Plan. Apply total derivative in Cartesian coordinates.
(4) Plan Execution.
(ii) Assumptions. (1) streamlines are parallel to surface, (2) no motion in the z-direction and (3) temperature distribution s one dimensional, ).(yTT
(ii) Analysis. Total acceleration in Cartesian coordinates is given by
t
f
z
fw
y
f
x
fu
Dt
Df
dt
dfv (1.21)
where f represents temperature. Let f = T in (1.21)
t
T
z
Tw
y
T
x
Tu
Dt
DT
dt
dTv (a)
where
H
yUu o (b)
and
v = w = 0 (c)
Temperature distribution is given by
11
22
)(2
TH
yTT
H
yy
kH
UT o
o (d)
Using (d)
x
y
oU
0
oT
1T
PROBLEM 1.9 (continued)
0t
T
x
T (e)
Substituting (b), (c) and (e) into
0Dt
DT
dt
dT (f)
(iii) Checking. Dimensional check: Each term in (d) has units of temperature.
(5) Comments. Velocity and temperature distribution must be know in order to determine the total derivative of temperature.
PROBLEM 1.10
One side of a thin plate is heated electrically such
that surface heat flux is uniform. The opposite side
of the plate is cooled by convection. The upstream
velocity is V and temperature is T . Experiments
were carried out at two upstream velocities, 1V
and 2V where .12 VV All other conditions were unchanged. The heat transfer coefficient
was found to increase as the free stream velocity is increased. Sketch the temperature profile T(y) of the fluid corresponding to the two velocities.
(1) Observations. (i) Heat flux leaving the surface is specified (fixed). (ii) Heat loss from the surface is by convection and radiation. (iii) Convection is described by Newton's law of cooling. (iv). Changing the heat transfer coefficient affects temperature distribution. (v). Surface temperature decreases as the heat transfer coefficient is increased. (vi) Surface temperature gradient is described by Fourier’s law.(vii) Ambient temperature is constant.
(2) Problem Definition. Determine effect of heat transfer coefficient on surface temperature and surface gradient..
(3) Solution Plan. (i) Apply Newton's law of cooling to examine surface temperature. (ii) Apply Fourier’s law to determine temperature gradient at the surface.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) no radiation ,(3) uniform ambient fluid temperature and (4) constant properties.
(ii) Analysis. Newton’s law of cooling
)( TThq so (a)
Solve for sT
h
qTT o
s (b)
This result shows that for constant oq , surface temperature decreases as h is increased. Apply
Fourier’s law
0y
oy
Tkq (c)
where y is the distance normal to the surface. Rewrite (c)
k
q
y
T o
y 0
(d)
This shows that temperature gradient at the surface remains constant independent of h. Based on (b) and
oq
x
y
T
V
T
T
y
2h
1h
oq2sT
PROBLEM 1.10 (continued)
(d) the temperature profiles corresponding to two values of h are shown in the sketch.
(iii) Checking. Dimensional check: (1) Each term in (b) has units of temperature
C)Cw/m(
)w/m()/()C()C( o
o2
2oo
h
AqTT w
s
(2) Each term in (d) has units of C/mo
C/mC)W/m-(
)C/m(/C/m)( o
o
2oo
0 k
Aq
y
T w
y
Limiting check: (i) for h = 0 (no heat leaves the surface), surface temperature is infinite. Set h = 0
in (b) gives .sT
(5) Comments. Temperature gradient at the surface is the same for all values of h as long as the thermal conductivity of the fluid is constant and radiation is neglected.
PROBLEM 1.11
Heat is removed from an L-shaped area by convection. The heat
transfer coefficient is h and the ambient temperature is .T Surface
temperature varies according to
xceos TxT )(
where c and oT are constants. Determine the rate of heat transfer
from the area.
(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable. (ii) Ambient temperature and heat transfer coefficient are uniform. (iii) Surface temperature varies along the area. (iv) The area varies with distance x.
(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient.
(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface temperature is not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature.
(ii) Analysis. Newton's law of cooling states that
qs = h As (Ts - T ) (a)
where
As = surface area, m2
h = heat transfer coefficient, W/m2-oCqs = rate of surface heat transfer by convection, W
Ts = surface temperature, oC
T = ambient temperature, oC
Since the L-shaped area varies with distance x, it is divided into two parts, 1 and 2, each having constant width. Applying
(a) to an infinitesimal area 1sdA
d 11 )( sss dATThq (b)
Integration of (b) from x = 0 to x = a gives the total heat from area 1
a2
0
a
a
a
a
a2x
x0
21
a
a
a2
a
sdq
dx dx
PROBLEM 1.11 (continued)
a
sss dATThq0
11 )( (c)
Similarly, for area 2
d 2)(2 sss dATThq (d)
Integration form x = a to x = 2a gives the total heat from area 2
a
asss dATThq
2
2)(2 (e)
1sdA and 2sdA are given by
dxadAs1 (f)
dxadAs 22 (g)
Surface temperature )(xTs is specified as
xceos TxT )( (h)
Substitute (f) and (h) into (c)
dxTTahq
axceos
0
1 )(
Evaluate the integral
aTc
Tahq caeo
s )1(1 (i)
Similarly, (g) and (h) into (e)
dxTTahq
a
a
xceos
2
)(22
Evaluate the integral
aTc
Tahq caca eeo
s )(2 22 (j)
The total heat transfer from the L-shaped area is
21 sss qqq aTc
TahaT
c
Tahq cacaca eee oo
s )(2)1( 21
Rearrange
12 2
oos
T
TacTh
c
aq caca ee (k)
(iii) Checking. Dimensional check: According to (a) units of c are1/m . Therefore units sq
each term in the bracket of (k) is dimensionless and the coefficient has units of W.
PROBLEM 1.11 (continued)
Limiting checks: (1) If h = 0 then qs = 0. Similarly, if a = 0 the area vanishes and qs = 0.
Equation (i) satisfies these limiting cases.
(2) If 0oT , the entire surface is at uniform temperature .0sT Application of Newton’s law
of cooling (a) gives
hTaqs23 (l)
Setting 0oT in (k) gives same result.
(5) Comments. Integration is necessary because surface temperature is variable. The same procedure can be followed if the ambient temperature or heat transfer coefficient is non-uniform.
PROBLEM 2.1
[a] Consider transient (unsteady), incompressible, three dimensional flow. Write the continuity
equation in Cartesian coordinates for this flow.
[b] Repeat [a] for steady state.
The continuity equation in Cartesian coordinates is
0wzy
uxt
v (2.2a)
[a] For incompressible flow the density is constant. Thus the can be taken out of the
differentiation sign. In addition, for constant density
0t
Equation (2.2a) becomes
0z
w
yx
u v (a)
[b] Equation (a) holds for steady state as well.
PROBLEM 2.2
Far away from the inlet of a tube, entrance
effects diminish and stream lines become
parallel and the flow is referred to as fully
developed. Write the continuity equation in
the fully developed region for incompressible
fluid.
(1) Observations. (i) The fluid is incompressible. (ii) Radial and tangential velocity components are zero. (iii) Streamlines are parallel. (iv) Cylindrical geometry.
(2) Problem Definition. Simplify the continuity equation for this flow.
(3) Solution plan. Apply continuity equation in cylindrical coordinates.
(4) Plan Execution.
(i) Assumptions. (1) Incompressible fluid and (2) radial and tangential velocity components are zero.
(ii) Analysis. The continuity equation in cylindrical coordinates is given by (2.4)
011
zrzr
rrrt
vvv (2.4)
This equation is simplified for:
Incompressible fluid: is constant, 0t
Parallel streamlines (no radial velocity): 0rv
Parallel streamlines (no tangential velocity): 0v
Introducing the above simplifications into (2.4), gives
0z
zv (a)
this result shows that the axial velocity component is invariant with z.
(iii) Checking. Dimensional check: Each term in (2.4) has units of density per unit time.
(5) Comments. (i) The axial velocity varies with radial distance only. (ii) Equation (a) holds for unsteady state as well. The reason is because for incompressible flow steady or unsteady the following applies
0t
(b)
r r
z
ullyf developed
PROBLEM 2.3
Consider incompressible flow between parallel
plates. Far away from the entrance the axial
velocity component does not vary with the axial
distance.
[a] Determine the velocity component in the y-
direction.
[b] Does your result in [a] hold for steady as well as unsteady flow? Explain.
(1) Observations. (i) The fluid is incompressible. (ii) axial velocity is invariant with axial distance. (iii) Plates are parallel. (iv) Cartesian geometry.
(2) Problem Definition. Determine the velocity component v in the y-direction.
(3) Solution plan. Apply continuity equation.
(4) Plan Execution.
(i) Assumptions. (1) Incompressible fluid, (2) axial velocity is invariant with axial distance and (3) two-dimensional flow.
(ii) Analysis. The continuity equation in Cartesian coordinates is
0wzy
uxt
v (2.2a)
For incompressible flow the density is constant. Thus the can be taken out of the
differentiation sign. In addition, for constant density
0t
(a)
Since the axial velocity u is invariant with axial distance x, it follows that
0x
u (b)
For two-dimensional flow
0z
(c)
(a)-(c) into (2.2a)
0y
v (d)
Integrating (d)
),( txfv (e)
PROBLEM 2.3 (continued)
At the wall the velocity must vanish. Thus
0v everywhere in the flow field (f)
[b] Equation (f) holds for steady state as well.
PROBLEM 2.4
The radial and tangential velocity components for
incompressible flow through a tube are zero. Show
that the axial velocity does not change in the flow
direction. Is this valid for steady as well as
transient flow?
(1) Observations. (i) The fluid is incompressible. (ii) Radial and tangential velocity components are zero. (iii) Streamlines are parallel. (iv) Cylindrical geometry.
(2) Problem Definition. Show that 0z
v z .
(3) Solution plan. Apply continuity equation in cylindrical coordinates.
(4) Plan Execution.
(i) Assumptions. (1) Incompressible fluid and (2) radial and tangential velocity components are zero.
(ii) Analysis. The continuity equation in cylindrical coordinates is given by (2.4)
011
zrzr
rrrt
vvv (2.4)
This equation is simplified for:
Incompressible fluid: is constant, 0t
No radial velocity: 0rv
No tangential velocity 0v
Introducing the above simplifications into (2.4), gives
0z
zv (a)
this result shows that the axial velocity component is invariant with z.
Equation (a) holds for unsteady state as well. The reason is because for incompressible flow steady or unsteady the following applies
0t
(b)
(iii) Checking. Dimensional check: Each term in (2.4) has units of density per unit time.
(5) Comments. (i) Since the radial and tangential velocity component vanishes everywhere in the flow field, it follows that the streamlines are parallel to the surface. (ii) The axial velocity varies with radial distance only.
PROBLEM 2.5
Show that yxxy
(1) Observations. (i) Shearing stresses are tangential surface forces. (ii) xy and yx are
shearing stresses in a Cartesian coordinate system.(iii) Tangential forces on an element result in angular rotation of the element. (iv) If the net external torque on an element is zero its angular acceleration will vanish.
(2) Problem Definition. Find the relationship between xy and yx acting on an element.
(3) Solution Plan. Apply Newton’s law of angular motion to an element dydx .
(4) Plan Execution.
(i) Assumptions. Continuum,
(ii) Analysis. Consider an element dydx with tangential shearing stresses acting on its four
sides. The depth of the element is unity. Apply Newton’s law of angular motion
I0 (a)
where
I = moment of inertia about 0, 2mKg
= angular acceleration, 2rad/s
0 = torque about center 0, m-N
Note that normal forces (pressure and normal stress, not shown) exert no torque on the element since their resultants pass through the center 0. The moment of inertia of the element dydx about the
center 0 is
dxdydydx
I12
)()( 22
(b)
The sum of all external torques acting on the element due to shearing stresses is
2222][][0
dxdydx
x
dxdy
dydxdy
y
dydx
xyxyxy
yxyxyx (c)
Note that in the above each shearing stress is multiplied by area to obtain force and by the arm to give torque. Equation (c) is simplified by neglecting third order
dxdydx
dydx
dydy
dxdy
dx xyyxxyxyyxyx )(2222
0 (d)
dx
dy
y
x
yx
dyy
yxyx
dxx
xyxy
xy
0
PROBLEM 2.5 (continued)
Substituting (b) and (d) into (a)
dxdydydx
dxdyxyyx12
)()()(
22
Simplify
12
)()()(
22 dydxxyyx (e)
The right hand side of (e) is of higher order and thus can be neglected to give
0)( xyyx
Thus
xyyx (f)
(iii) Checking. Dimensional check: Units of (a);
)s
rad()mKg()mN(
22
0 I
This gives
2s
mKgN
which is the correct units for Newton.
(5) Comments. It is incorrect to conclude that xyyx because the element is in static
equilibrium.
PROBLEM 2.6
A fluid flows axially between parallel plates.
Assume: Newtonian fluid, steady state, constant
density, constant viscosity, negligible gravity and
parallel streamlines. Write the three components of
the momentum equations for this flow.
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines: no velocity component in the y-direction. (iv) Axial flow: no velocity component in the z-direction. (v) The Navier-Stokes equations give the three momentum equations.
(2) Problem Definition. Determining the three momentum equations for the flow under consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in Cartesian coordinates. Simplify according to the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction) and (6) axial flow (no motion in the z-direction).
(ii) Analysis. The Navier-Stokes for constant properties are
x-direction:2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
y-direction:2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
z-direction:2
2
2
2
2
2
z
w
y
w
x
w
z
pg
z
ww
y
w
x
wu
t
wzv (2.10z)
These equations are simplfied as follows:
Steady state: 0t
No gravity: g = 0
Parallel streamlines: 0v
Axial flow: w = 0
Substituting these simplifications inot (2.10)
x-direction:2
2
2
2
y
u
x
u
x
p
x
uu (a)
PROBLEM 2.6 (continued)
However, continuity equation gives
0x
w
xx
u
zw
yxu
t
vv (2.2b)
For incompressible flow this simplifies to
0z
w
yx
u v(b)
This simplifies to
0x
u(c)
Substituting (c) into (a)
x-direction:2
2
y
u
x
p (d)
Equations (2.10y) and (2,10z) simplify to
y-direction: 0y
p (e)
z-direction: 0z
p (f)
(iii) Checking. Dimensional check: units of (d)
)sm
m()
ms
kg()
mm
N(
22
2
2 y
u
x
p
2s
m-kgN
Units of (d) are correct.
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes equations.
PROBLEM 2.7
A fluid flows axially (z-direction) through a tube.
Assume: Newtonian fluid, steady state, constant
density, constant viscosity, negligible gravity and
parallel streamlines. Write the three components of
the momentum equations for this flow.
(1) Observations. (i) Properties are constant. (ii) Cylindrical coordinates. (iii) Parallel streamlines: no velocity component in the r-direction. (iv) Axial flow: no velocity component in the -direction. (v) No variation in the -direction. The Navier-Stokes equations give the three
momentum equations.
(2) Problem Definition. Determining the three momentum equations for the flow under consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in cylindrical coordinates. Simplify according to the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to surface (no motion in the r-direction) and (6) axial flow (no motion in the -direction).
(ii) Analysis. The Navier-Stokes for constant properties are
r-direction:
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
-direction:
2
2
22
2
2
21)(
11
zrrr
rrr
p
rg
tzrrr
r
r
vvvv
vvv
vvvvvv zr
(2.11 )
z-direction:
2
2
2
2
2
11
zrrr
rrz
pg
tzrr
zzzz
zzzz
vvv
vvv
vvvv zr
(2.11z)
These equations are simplfied as follows:
Steady state: 0t
Parallel streamlines: 0r
v
PROBLEM 2.7 (continued)
Axial flow: v = 0
No gravity: 0zr ggg
No variation in the -direction, 0
Substituting these simplifications into (2.11)
r-direction 0r
p (a)
-direction: 0p
(b)
z-direction:2
21
zrr
rrz
p
z
zzz vvvv z
(c)
Equation (c) is simplified further using the continuity equation in cylindrical coordinates
011
zrzr
rrrt
vvv (2.4)
This equation is simplified to
0z
zv (d)
(d) into (c)
z-direction:r
rrrz
p zv (e)
(iii) Checking. Dimensional check: units of (e)
322223 N/mmkg/s)s/mm())m(
1)mkg/s()N/m(
r
v zrrrz
p
Thus units of (e) are correct.
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes equations.
PROBLEM 2.8
Consider two-dimensional flow (x,y) between parallel
plates. Assume: Newtonian fluid, constant density and
viscosity. Write the two components of the momentum
equations for this flow. How many unknown do the
equations have? Can they be solved for the unknowns? If not what other equation(s) is needed to obtain a solution?
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Two dimensional flow (no velocity component in the z-direction. (iv) The Navier-Stokes equations give two momentum equations.
(2) Problem Definition. Determining the two momentum equations for the flow under consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in Cartesian coordinates. Simplify according to the conditions of the problem and count the unknown dependent variables.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state and (4) two-dimensional flow (no motion in the z-direction).
(ii) Analysis. The Navier-Stokes for constant properties are
x-direction:2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
y-direction:2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
These equations are simplfied as follows:
Steady state: 0t
Two dimensional flow: z
w = 0
Substituting these simplifications inot (2.10)
x-direction:2
2
2
2
y
u
x
u
x
pg
y
u
x
uu xv
(a)
y-direction:2
2
2
2
yxy
pg
yxu y
vvvv
v(b)
These two equations contains three unknowns: ,u v and p. A third equation is needed to obtain a
solution. This equation is continuity
PROBLEM 2.8 (continued)
0x
w
xx
u
zw
yxu
t
vv (2.2b)
For incompressible two-dimensional flow this simplifies to
0xx
u v(c)
(iii) Checking. Dimensional check: Each term in (a) and (b) has units of 22 ms
kg.
(5) Comments. It is not surprising that continuity is needed to obtain a solution to the flow field. Conservation of mass (continuity) and momentum (Navier-Stokes equations) must be satisfied.
PROBLEM 2.9
Consider Two-dimensional (r,z) flow through a tube. Assume: Newtonian, constant density and viscosity.
Write the two components of the momentum equations for this flow. How many unknowns do the equations have?Can the equations be solved for the unknowns? If not what other equation(s) is needed to obtain a solution?
(1) Observations. (i) Properties are constant. (ii) Cylindrical coordinates. (iii) Two dimensional flow (no velocity component in the -direction. (iv) The Navier-Stokes equations give two
momentum equations.
(2) Problem Definition. Determining the two momentum equations for the flow under consideration.
(3) Solution Plan. Apply the Navier-Stokes equations of motion in cylindrical coordinates. Simplify according to the conditions of the problem and count the unknown dependent variables.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density and viscosity, (3) steady state and (4) two-dimensional flow (no motion in the -direction).
(ii) Analysis. The Navier-Stokes in the r and z directions for constant properties are
r-direction:
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
z-direction:
2
2
2
2
2
11
zrrr
rrz
pg
tzrr
zzzz
zzzz
vvv
vvv
vvvv zr
(2.11z)
These equations are simplfied as follows:
Steady state: 0t
Two dimensional flow: v = 0
Substituting these simplifications into (2.11)
r r
z
PROBLEM 2.9 (continued)
r-direction:2
2
)(1
zr
rrrr
pg
zrr
rr
rz
rr
vv
vv
vv (a)
z-direction:2
21
zrr
rrz
pg
zr
zzz
zz vvvv
vv zr (b)
These two equations contains three unknowns: ,rv zv and p. A third equation is needed to obtain
a solution. This equation is continuity in cylindrical coordinates
The continuity equation in cylindrical coordinates is given by (2.4)
011
zrzr
rrrt
vvv (2.4)
This equation is simplified for:
Incompressible fluid: is constant, 0t
Equation (2.4) becomes
01
zr
rr
zr
vv (c)
(iii) Checking. Dimensional check: Each term in (a) and (b) has units of 22 ms
kg.
(5) Comments. It is not surprising that continuity is needed to obtain a solution to the flow field. Conservation of mass (continuity) and momentum (Navier-Stokes equations) must be satisfied.
PROBLEM 2.10
In Chapter 1 it is stated that fluid motion and fluid nature play a role in convection heat transfer. Does the energy equation substantiate this observation? Explain.
(1) Observations. (i) Motion in energy consideration is represented by velocity components. (ii) Fluid nature is represented by fluid properties.
(2) Problem Definition. Determine if the energy equation depends on velocity and fluid properties.
(3) Solution Plan. Write the energy equation and determine if it depends on velocity and properties.
(4) Plan Execution.
(i) Assumptions. (1) Continuum, (2) Newtonian fluid and (3) negligible nuclear, radiation and electromagnetic energy transfer.
(ii) Analysis. The energy equation in given by
Dt
DpTTk
Dt
DTc p (2.15)
where
pc specific heat at constant pressure
k thermal conductivity
p pressure
coefficient of thermal expansion or compressibility
= dissipation function
The coefficient of thermal expansion is a property of material defined as
pT
1 (2.16)
The dissipation function is associated with energy dissipation due to friction. It is important in high speed flow and for very viscous fluids. In Cartesian coordinates is given by
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Note that the total temperature derivative in (2.15) is defined as
t
T
z
Tw
y
T
x
Tu
Dt
DTv (a)
PROBLEM 2.10 (continued)
Examination of the above equations shows that energy equation (2.15) depends on the velocity
components u, v and w. In addition, (2.15) depends on , ,pc , k and . These are properties
of fluid.
(5) Comments. To determine temperature distribution it is necessary to know the velocity distribution.
PROBLEM 2.11
A fluid flows axially (x-direction) between parallel plates. Assume: Newtonian fluid, steady state, constant density,constant viscosity, constant conductivity, negligible
gravity and parallel streamlines. Write the energy
equation for this flow.
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines: no velocity component in the y-direction. (iv) Axial flow: no velocity component in the z-direction.
(2) Problem Definition. Determining the energy equations for the flow under consideration.
(3) Solution Plan. Apply the energy equations in Cartesian coordinates. Simplify according to the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density, viscosity and conductivity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction), (6) axial flow (no motion in the z-direction) and (7) negligible nuclear, radiation and electromagnetic energy transfer.
((4) Plan Execution.
(i) Assumptions. (1) Continuum, (2) Newtonian fluid, (3) no energy generation and (4) constant properties.
(ii) Analysis. The energy equation for this case is given by
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc p v (2.19b)
where
pc specific heat at constant pressure
k thermal conductivity
p pressure
T temperature density
= dissipation function
The dissipation function in Cartesian coordinates is given by
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
These equations are simplfied as follows:
PROBLEM 2.11 (continued)
Steady state: 0t
No gravity: g = 0
Parallel streamlines: 0v
Axial flow: 0z
w
Substituting these simplifications into (2.19b)
2
2
2
2
y
T
x
Tk
x
Tuc p (a)
Similarly (2.17) simplifies to
2
3
222
2x
u
y
u
x
u (b)
Further simplifications are obtained using continuity equation (2.2b)
0x
w
xx
u
zw
yxu
t
vv (2.2b)
For incompressible parallel flow this becomes
0x
u(c)
(c) into (b)
2
y
u (d)
Substitute (d) into (a) gives the energy equation for this flow
2
2
2
2
2
y
u
y
T
x
Tk
x
Tuc p (e)
(iii) Checking. Dimensional check: Each term in (e) has units of 3W/m .
(5) Comments. The continuity equation provides additional simplification of the dissipation function.
PROBLEM 2.12
An ideal gas flows axially (x-direction) between parallel
plates. Assume: Newtonian fluid, steady state, constant
viscosity, constant conductivity, negligible gravity and
parallel streamlines. Write the energy equation for this
flow.
(1) Observations. (i) Properties are constant. (ii) Cartesian coordinates. (iii) Parallel streamlines: no velocity component in the y-direction. (iv) Axial flow: no velocity component in the z-direction. (v) The fluid is an ideal gas.
(2) Problem Definition. Determining the energy equations for the flow under consideration.
(3) Solution Plan. Apply the energy equations in Cartesian coordinates. Simplify according to the conditions of the problem.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant density, viscosity and conductivity, (3) steady state, (4) negligible gravity, (5) streamlines are parallel to the plates (no motion in the y-direction), (6) axial flow (no motion in the z-direction), (7) negligible nuclear, radiation and electromagnetic energy transfer and (8) Ideal gas.
(ii) Analysis. The energy equation for this case in given by
VpTkDt
DTcv (2.23)
Rewriting and noting that k is constant
Vpz
T
y
T
x
Tk
t
T
z
Tw
y
T
x
Tuc
2
2
2
2
2
2
vv (a)
where
vc specific heat at constant pressure
k thermal conductivity
p pressure
T temperature wu ,,v velocity components in x, y and z directions
density
= dissipation function
The dissipation function in Cartesian coordinates is given by
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
PROBLEM 2.12 (continued)
These equations are simplfied as follows:
Constant k
Steady state: 0t
Parallel streamlines: 0v
Axial flow: 0z
w
Incompressible fluid: 0V
Substituting these simplifications into (a)
2
2
2
2
y
T
x
Tk
x
Tucv (b)
Similarly (2.17) simplifies to
2
3
222
2x
u
y
u
x
u (b)
Further simplifications are obtained using continuity equation (2.2b)
0x
w
xx
u
zw
yxu
t
vv (2.2b)
For incompressible parallel flow this becomes
0x
u(c)
(c) into (b)
2
y
u (d)
Substitute (d) into (a) gives the energy equation for this flow
2
2
2
2
2
y
u
y
T
x
Tk
x
Tucv (e)
(iii) Checking. Dimensional check: Each term in (e) has units of 3W/m .
(5) Comments. The continuity equation provides additional simplification of the dissipation function.
PROBLEM 2.13
Consider two-dimensional free convection over a vertical plate. Assume:Newtonian fluid, steady state, constant viscosity, Boussinesq approximation and
negligible dissipation. Write the governing equations for this case. Can the flow field be determined independently of the temperature field?
(1) Observations. (i) This is a two-dimensional free convection problem. (ii) The flow is due to gravity. (iii) The flow is governed by the momentum and energy equations. Thus the governing equations are the Navier-Stokes equations of motion and the energy equation. (iv) The geometry is Cartesian.
(2) Problem Definition. Determine: the x and y components of the Navier-Stokes equations of motion, and the energy equation for the flow under consideration .
(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) Boussinesq approximations and (5) negligible nuclear, electromagnetic and radiation energy transfer.
(ii) Analysis. Momentum equations. The Navier Stokes equations of motion for free convection are given in (2.29)
VvppTTgDt
VD 21 (2.29)
This vector equation gives the x and y components
2
2
2
2
2
21)(
z
u
y
u
x
u
x
pTTg
z
uw
y
u
x
uu
t
uv (a)
2
2
2
2
2
2
zyxy
p
zw
yxu
t
vvvvvv
vv (b)
Gravity is assumed to point in the negative x-direction. The Cartesian coordinates energy equation for incompressible constant conductivity fluid is given by equation (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
where is the dissipation function. These equations are simplified based on the following assumptions
Steady state: 0t
Axial flow: 0z
w
y
x
g
u
PROBLEM 2.13 (continued)
No dissipation: 0
(a), (b) and (2.19b) become
2
2
2
21)(
y
u
x
u
x
pTTg
y
u
x
uu v (c)
2
2
2
21
yxy
p
yxu
vvvv
v (d)
2
2
2
2
y
T
x
Tk
y
T
x
Tuc p v (e)
Equations (c), (d) and (e) are the governing equations for this flow. Examination of momentum equations (c) and (d) shows that they contain the unknown temperature variable T. Thus these equations can not be solved for the flow field without invoking the energy equation. Note that the three equations contain four unknowns: ,u ,v p and T. Continuity provides the fourth
equation.
(iii) Checking. Dimensional check: Each term of momentum equations (c) and (d) has units of 2m/s . Each term in (e) has units of 3W/m .
Limiting check: If the fluid is not moving, the energy equation should reduce to pure conduction. Setting 0u in (e) gives
02
2
2
2
y
T
x
T
This is the correct equation for this limiting case.
(5) Comments. (i) Governing equations (c), (d) and (e) are coupled. Thus they must be solved, together with continuity, for the flow field and temperature field. (ii) In energy equation (e),
properties kc p , and represent fluid nature. Velocity components u and v represent fluid
motion. This confirms the observation made in Chapter 1 that fluid motion and nature play a role in convection heat transfer (temperature distribution).
PROBLEM 2.14
Discuss the condition(s) under which the Navier-Stokes equations of motion can be solved
independently of the energy equation.
Solution
Examination of the smallest rectangle in Table 2.1 shows that for constant properties (density and viscosity), continuity and momentum (4 equations) contain the four flow field unknowns u,
v, w and p. Thus for constant properties the Navier-Stokes equations and continuity can be
solved for the flow field independently of the energy equation.
PROBLEM 2.15
Consider a thin film of liquid condensate which is falling over a flat surface by
virtue of gravity. Neglecting variations in the z-direction and assuming Newtonian fluid, steady state, constant properties and parallel streamlines.
[a] Write the momentum equation(s) for this flow.
[b] Write the energy equation including dissipation effect
(1) Observations. (i) The flow is due to gravity. (ii) For parallel streamlines the
velocity component v = 0 in the y-direction. (iii) Pressure at the free surface is
uniform (atmospheric). (iv) Properties are constant. (v) The geometry is Cartesian.
(2) Problem Definition. Determine: [a] the x and y components of the Navier-Stokes equations of motion, and [b] the energy equation for the flow under consideration .
(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the x-direction only, (4) constant properties, (5) uniform ambient pressure, (6) parallel streamlines. (7) negligible nuclear, electromagnetic and radiation energy transfer.
(ii) Analysis. [a] Momentum equations. The Navier Stokes equations of motion in Cartesian coordinates for constant properties are given in equations (2.10x ) and (2.10y)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
The gravitational components are
,gg x 0yg (a)
Based on the above assumptions, equations (2.10) are simplified as follows:
Steady state: 0tt
u v (b)
Axial flow (x-direction only): 0z
w (c)
Parallel flow: 0v (d)
Substituting (a)-(d) into (2.10x) and (2.10y), gives
gx
uu
2
2
2
2
y
u
x
u
x
p (e)
and
y
x
g
Problem 2.15 (continued)
0y
p (f)
The x-component (e) can be simplified further using the continuity equation for incompressible flow, equation (2.3)
0zyx
uV
wv (g)
Substituting (c) and (d) into (g), gives
0x
u (h)
Using (h) into (e) gives the x-component
g2
2
y
u
x
p= 0 (i)
Integrating (f) with respect to y)(xfp (j)
where f(x) is the constant of integration. At the free surface, ,Hy the pressure is uniform
equal to .p Therefore, setting Hy in (j) gives
pxf )( (k)
Substituting (k) into (j) gives the pressure solution
pp (l)
Differentiating (k) with respect to x givesx
p = 0 (m)
Substituting (m) into (i) gives the x-component of the Navier-Stokes equations
02
2
yd
udg (n)
[b] Energy equation. The Cartesian coordinates energy equation for incompressible constant conductivity fluid is given by equation (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
where the dissipation function in Cartesian coordinates is given by equation (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Problem 2.15 (continued)
Based on the above assumptions, these equations are simplified as follows:
Steady state: 0t
T (o)
Substituting (c), (d) and (o) into (2.19b), gives
2
2
2
2
y
T
x
Tk
x
Tuc p (p)
The dissipation function (2.17) is simplified using (c), (d) and (h)
2
y
u (q)
Substituting (q) into (p) gives the energy equation
2
2
2
2
2
y
u
y
T
x
Tk
x
Tuc (r)
(iii) Checking. Dimensional check: Each term of the x-component equation (n) must have the same units
g = kg/m2-s2
2
2
yd
ud = (kg/m-s)
m s
m
/2
= kg/m2-s2
Each term in (r) has the same units of 3W/m .
Limiting check: If the fluid is not moving, the energy equation should reduce to pure conduction. Setting 0u in (i) gives
02
2
2
2
y
T
x
T
This is the correct equation for this limiting case.
(5) Comments. (i) For two-dimensional incompressible parallel flow, the momentum equations
are considerably simplified because the vertical velocity component v = 0.
(ii) The flow is in fact one-dimensional since u does not change with x and is a function of yonly.
(iii) In energy equation (r), properties ,,kc p and represent fluid nature. The velocity u
represents fluid motion. This confirms the observation made in Chapter 1 that fluid motion and nature play a role in convection heat transfer (temperature distribution).
(ii) The last term in energy equation (r) represents dissipation.
PROBLEM 2.16
A wedge is maintained at T1 along one side and T2 along
the opposite side. A solution for the flow field is obtained
based on Newtonian fluid and constant properties. The
fluid approaches the wedge with uniform velocity and
temperature. Examination of the solution shows that the
velocity distribution is not symmetrical with respect to the
x-axis. You are asked to support the argument that the
solution is incorrect.
(1) Observations. (i) This is a forced convection problem. (ii) Flow properties (density and viscosity) are constant. (iii) Upstream conditions are uniform (symmetrical) (iv) The velocity vanishes at both wedge surfaces (symmetrical). (v) Surface temperature is asymmetric. (vi) Flow field for constant property fluids is governed by the Navier-Stokes and continuity equations. (vii) If the governing equations are independent of temperature, the velocity distribution over the wedge should be symmetrical with respect to x. (viii) The geometry is Cartesian.
(2) Problem Definition. Determine if the governing equations for the velocity distribution is independent of temperature.
(3) Solution Plan. Examine the Navier-Stokes and continuity equations in Cartesian coordinates for dependency on temperature.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) two-dimensional (x and y), (3) constant properties and (4) uniform upstream conditions.
(ii) Analysis. The Navier Stokes equations of motion in Cartesian coordinates for constant properties are given in equations (2.10x ) and (2.10y)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
For two-dimensional conditions (w = 0) these equations become
2
2
2
2
y
u
x
u
x
pg
y
u
x
uu
t
uxv (a)
2
2
2
2
yxy
pg
yxu
ty
vvvv
vv (b)
These equations contain three unknowns: u, v and p. Continuity provides the fourth equation
0wzy
uxt
v (2.2a)
For two-dimensional constant density (2.2a) simplifies to
1T
2TT
V
y
x
PROBLEM 2.16 (continued)
0yx
u v (b)
Since properties are constant, and are constant. Thus (a), (b) and (c) are independent of
temperature. It follows that the solution to these equations for u, v and p is independent of the
asymmetry of the boundary temperature. A solution based on the assumption of constant property that give asymmetrical velocity distribution must be incorrect.
(iii) Checking. Dimensional check: Each term in (a) and (b) has units of kg/m2-s2
(5) Comments. (i) Although the flow was assumed two-dimensional, the same conclusion applies to three-dimensional flow as long as the geometry is symmetrical about the x-axis and upstream conditions are uniform. (ii) Examination of the smallest rectangle in Table 2.1 shows that for constant properties (density and viscosity), continuity and momentum (4 equations)
contain the four flow field unknowns u, v, w and p. Thus for constant properties the Navier-
Stokes equations and continuity can be solved for the flow field independently of the energy equation. This is valid for steady as well as transient flow.
PROBLEM 2.18
Consider two-dimensional (x and y), steady, constant
properties, parallel flow between two plates
separated by a distance H. The lower plate is
stationary while the upper plate moves axially with a
velocity oU . The upper plate is maintained at
uniform temperature oT and the lower plate is cooled
with a flux oq . Taking into consideration dissipation, write the Navier-Stokes equations of
motion, energy equation and boundary conditions at the two plates.
(1) Observations. (i) The geometry is Cartesian. (ii) Properties are constant. (ii) Axial flow (no
motion in the z-direction), (iv) Parallel streamlines means that the normal velocity component is
zero. (v) Specified flux at the lower plate and specified temperature at the upper plate.
(2) Problem Definition. Determine: [a] the x and y components of the Navier-Stokes equations of motion, [b] the energy equation for the flow under consideration, and [c] velocity and temperature boundary conditions at the lower and upper plates.
(3) Solution Plan. Start with the Cartesian coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) flow is in the x -direction, (3) constant properties
and (4) negligible nuclear, electromagnetic and radiation energy transfer.
(ii) Analysis. The Navier Stokes equations of motion in Cartesian coordinates for constant properties are given in equations (2.10x ) and (2.10y)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y
The energy equation given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc p v (2.19b)
The dissipation function in Cartesian coordinates is given by (2.17)
2
3
2222222
2z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u vvvv(2.17)
These equations are simplfied as follows:
Parallel streamlines: 0v
PROBLEM 2.18 (continued)
Axial flow: 0z
w
Further simplifications are obtained using continuity equation (2.2b)
0z
w
yx
u
zw
yxu
t
vv (2.2b)
For incompressible parallel flow this reduces to
0x
u (a)
Substituting these simplifications into (2.10x), (2.10y), (2.19b) and (2.17)
2
2
y
u
x
pg
t
ux
(b)
0y
pg y (c)
2
2
2
2
y
T
x
Tk
x
Tu
t
Tc p (d)
2
y
u(e)
The boundary conditions on velocity components are
(1) 0)0,(xu
(2) 0)0,(xv
(3) oUHxu ),(
(4) 0),( Hxv
The boundary conditions on temperature are
(1) oqy
xTk
)0,(
(2) oTHxT ),(
(iii) Checking. Each term in momentum equations (b) and (c) has units of 22 skg/m . Each
term in energy equation (d) has units of .W/m3
Limiting check: If the upper plate is stationary and there is no axial pressure gradient, the problem reduces to one-dimensional transient conduction. Since 0vu , (d) becomes
2
2
2
2
y
T
x
Tk
t
Tc p (f)
However, since the boundary conditions on temperature are independent of x, axial temperature
PROBLEM 2.18 (continued)
gradient vanishes and (f) simplifies further to
2
2
y
Tk
t
Tc p (g)
This is the one-dimensional transient conduction equation.
(5) Comments. (i) The continuity equation provides important simplifications in the momentum and energy equations. (ii) For steady state set tT / is set equal to zero.
PROBLEM 2.19
A shaft or radius 1r rotates concentrically inside a sleeve of inner
radius 2r . Lubrication oil fills the clearance between the shaft and
the sleeve. The sleeve is maintained at uniform temperature oT .
Neglecting axial variation and taking into consideration dissipation,
write the Navier-Stokes equations of motion, energy equation and
boundary conditions for this flow. Assume constant properties.
(1) Observations. (i) The geometry is cylindrical. (ii) No variation in
the axial and angular directions. (iii) Properties are constant.
(2) Problem Definition. Determine: [a] the r and components of the Navier-Stokes equations
of motion, [b] the energy equation and [c] boundary conditions for the flow under consideration.
(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the -direction, (4)
constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6) negligible gravity.
(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for constant properties are given in equations (2.11r) and (2.11 )
r-direction:
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
-direction:
2
2
22
2
2
21)(
11
zrrr
rrr
p
rg
tzrrr
r
r
vvvv
vvv
vvvvvv zr
(2.11 )
These equations are simplfied as follows:
Steady state: 0t
No gravity: 0ggr
No axial flow:z
0z
v
r
1r2r
shaft
oT
PROBLEM 2.19 (continued)
Symmetry: 0
Substituting these simplifications inot (2.11r)
)(1
2
rr
r vvv
v rrrrr
p
rr (a)
Similarly, (2.11 ) become
)(1
vvvv
v r rrrrrr
r (b)
However, continuity equation gives
011
zrzr
rrrt
vvv (2.4)
For incompressible fluid and 0z
, this simplifies to
0rrr
v (c)
Integrating
Cr rv
where C is constant of integration. Since 0)( 1rrv it follows that C = 0. Therefore
0rv (d)
Introducing (d) into (a) and (b)
r-direction:r
p
r
2v
(e)
-direction 0)(1
vrrrr
(f)
[b] Energy equation. For constant properties the energy equation is given by
2
2
2
2
2 0
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrp v
vv (2.24)
where the dissipation function is
PROBLEM 2.19 (continued)
22
2222
0
1
0
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
Equations (2.24) and (2.25) are simplified for the conditions of this problem
r
Tr
rrk
10 (g)
2
rr
0 vv (h)
[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are
(1) 11)( rrv
(2) 0)( 2rv
The boundary conditions on energy equation (g) are
(1) Insulated surface at :1r 0)( 1
r
rT
(2) Specified temperature at :2r oTrT )( 2
(iii) Checking. Dimensional check: Units of (e)
)mm
N()
ms
m()
m
kg(
22
22
3 r
p
r
v
22222 sm
kg
mm
N
sm
kg
Units of each term in (g)
ms
kg
m
W)C/m()m(
)m()m(
1)
Cm
W(
33
o
o r
Tr
rrk
ms
kg)s/1()
ms
kg(
3
2
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes and energy equations.
PROBLEM 2.20
A rod of radius ir moves axially with velocity
oU inside a concentric tube of radius or . A
fluid having constant properties fills the space
between the shaft and tube. The tube surface is
maintained at uniform temperature .oT Write
the Navier-Stokes equations of motion, energy
equation and surface boundary conditions
Taking into consideration dissipation. Assume that the streamlines are parallel to the surface.
(1) Observations. (i) The geometry is cylindrical. (ii) No variation in the angular direction. (iii)
Properties are constant. (iv) Parallel streamlines means that the radial velocity component is
zero.
(2) Problem Definition. Determine: [a] the r and z components of the Navier-Stokes equations of motion, [b] the energy equation for the flow under consideration and [c] boundary conditions.
(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the z -direction, (4) constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6) negligible gravity.
(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for constant properties are given in equations (2.11r) and (2.11 z )
r-direction:
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
z-direction:
2
2
2
2
2
11
zrrr
rrz
pg
tzrr
zzzz
zzzz
vvv
vvv
vvvv zr
(2.11z)
These equations are simplfied as follows:
Steady state: 0t
No gravity: 0zr gg
No tangential flow: 0v
PROBLEM 2.20 (continued)
Symmetry: 0
Parallel streamlines: 0r
v
Substituting these simplifications inot (2.11r)
r-directionr
p0 (a)
Similarly, (2.11 z ) become
:2
21
zrr
rrz
p
z
zzz vvvv z (b)
However, continuity equation gives
011
zrzr
rrrt
vvv (2.4)
For incompressible fluid and 0rv , this simplifies to
0z
zv (c)
Introducing (d) into (b)
z-directionr
rrrz
p zv10 (d)
[b] Energy equation. For constant properties the energy equation is given by
2
2
2
2
2
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrp v
vv (2.24)
where the dissipation function is
22
2222
1
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
Equations (2.24) and (2.25) are simplified for the conditions of this problems
2
21
z
T
r
Tr
rrk
z
Tc zpv (g)
2
r
zv (h)
PROBLEM 2.20 (continued)
[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are
(1) oiz Ur )(v
(2) 0)( oz rv
The boundary conditions on energy equation (g) are
(1) Equality of temperature at :ir ),()( , zrTzrT iri
(2) Equality of flux at :irr
zrTk
r
zrTk ir
ri ),(),(
(3) Specified temperature at :or oo TzrT )),(
where the subscript r refers to the rod.
(iii) Checking. Dimensional check: United of each term in (d)
32 m
N)
mm
N(
z
p
322 m
N
ms
kg
m)(
m/s)(m)(
m)(m)(
1)
ms
kg(
rr
rr
zv
Units of each term in (h)
3
o
o3 m
W
m
C
s
m
Ckg
J
m
kg
z
Tc zpv
32
o
o2
2
m
W
m
C
Cm
W
z
Tk
ms
kg)s/1()
ms
kg(
3
2
(5) Comments. (i) The continuity equation provides additional simplification of the Navier-Stokes and energy equations. (ii) The temperature distribution is two-dimensional. (iii) To solve for the temperature distribution it is necessary to write a heat equation for the rod as well as thermal boundary conditions at two axial locations.
PROBLEM 2.21
A rod or radius ir rotates concentrically inside a tube of inner radius or .
Lubrication oil fills the clearance between the shaft and the tube. Tube
surface is maintained at uniform temperature oT . The rod generates heat
volumetrically at uniform rate q . Neglecting axial variation and taking
into consideration dissipation, write the Navier-Stokes equations of
motion, energy equation and boundary conditions for this flow. Assume
constant properties.
(1) Observations. (i) The geometry is cylindrical. (ii) No variation in
the axial and angular directions. (iii) Properties are constant.
(2) Problem Definition. Determine: [a] the r and components of the Navier-Stokes equations
of motion, [b] the energy equation and [c] boundary conditions for the flow under consideration.
(3) Solution Plan. Start with the cylindrical coordinates Navier-Stokes equations of motion and energy equation for constant properties. Simplify them for this special case.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) flow is in the -direction, (4)
constant properties, (5) negligible nuclear, electromagnetic and radiation energy transfer and (6) negligible gravity.
(ii) Analysis. [a]. The Navier Stokes equations of motion in cylindrical coordinates for constant properties are given in equations (2.11r) and (2.11 )
r-direction:
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
-direction:
2
2
22
2
2
21)(
11
zrrr
rrr
p
rg
tzrrr
r
r
vvvv
vvv
vvvvvv zr
(2.11 )
These equations are simplfied as follows:
Steady state: 0t
No gravity: 0ggr
No axial flow:z
0z
v
oT
0
oq
ir
or
PROBLEM 2.21 (continued)
Symmetry: 0
Substituting these simplifications inot (2.11r)
)(1
2
rr
r vvv
v rrrrr
p
rr (a)
Similarly, (2.11 ) become
)(1
vvvv
v r rrrrrr
r (b)
However, continuity equation gives
011
zrzr
rrrt
vvv (2.4)
For incompressible fluid and 0z
, this simplifies to
0rrr
v (c)
Integrating
Cr rv
where C is constant of integration. Since 0)( or rv it follows that C = 0. Therefore
0rv (d)
Introducing (d) into to (a) and (b)
r-direction:r
p
r
2v
(e)
-direction 0)(1
vrrrr
(f)
[b] Energy equation. For constant properties the energy equation is given by
2
2
2
2
2 0
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrp v
vv (2.24)
where the dissipation function is
22
2222
0
1
0
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
Equations (2.24) and (2.25) are simplified for the conditions of this problems
PROBLEM 2.21 (continued)
r
Tr
rrk
10 (g)
2
rr
0 vv (h)
[c] Boundary conditions. The boundary conditions for Navier-Stokes equations (e) and (f) are
(1) ii rr )(v
(2) 0)( orv
The boundary conditions on energy equation (g) are
(1) Specified flux at :ir ii q
r
rTk
)(
(2) Specified temperature at :or oo TrT )(
Conservation of energy for the rod gives the flux iq :
Energy generated in rod = energy leaving surface at ir
iii qrrq 22
2
ii
rqq (i)
(iii) Checking. Dimensional check: United of (e)
)mm
N()
ms
m()
m
kg(
22
22
3 r
p
r
v
22222 sm
kg
mm
N
sm
kg
Units of each term in (g)
ms
kg
m
W)C/m()m(
)m()m(
1)
Cm
W(
33
o
o r
Tr
rrk
ms
kg)s/1()
ms
kg(
3
2
(5) Comments. The continuity equation provides additional simplification of the Navier-Stokes and energy equations.
PROBLEM 2.22
Air flows over the two spheres. The radius of sphere 2 is double
that of sphere 1. However, the free stream velocity for sphere 1 is
double that for sphere 2. Determine the ratio of the average heat
transfer coefficients 21 / hh for the two spheres.
(1) Observations. (i) This is a forced convection problem. (ii) The same fluid flows over both spheres. (iii) Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient.
(2) Problem Definition. Since the average heat transfer coefficient h is correlated in terms
of the Nusselt number, the problem becomes one of determining the Nusselt number for each sphere and taking their ratio.
(3) Solution Plan. Use the results of dimensional analysis to obtain a relationship between the Nusselt number and the significant parameters in forced convection.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state and (3) constant properties.
(ii) Analysis. Non-dimensional form of the governing equations for convection gives
DNu =h D
k = f( DRe , Pr, DrG , E) (a)
where
D = diameter of sphere, m E = Eckert number
DrG = Grashof number
h = average heat transfer coefficient, W/m2-oCk = thermal conductivity of fluid, W/m-oC
DNu = average Nusselt number
Pr = Prandtl number
DRe = Reynolds number
Assume that free convection is negligible compared to forced convection. This eliminates the Grashof number. Furthermore, neglect dissipation effects. This eliminates the Eckert number. Thus (a) is simplified to
DNu =k
Dh = f( DRe , Pr) (b)
Since the same fluid flows over both spheres, it follows that the Prandtl number is the same for both. Thus (b) becomes
DNu =k
Dh = f( DRe ) (c)
The Reynolds number is defined as
2
1
1V
2V
PROBLEM 2.22 (continued)
DRe =DV
(d)
where
V free stream velocity, m/s
= kinematic viscosity, m2/s
Solving equation (c) for h
h =k
Df( DRe ) (e)
According to (e), to calculate h for each sphere it is necessary to determine: (1) the exact form
of the function f( DRe ), (2) the diameter, (3) the thermal conductivity and (4) the Reynolds
number. However, of interest is determining the ratio of the heat transfer coefficients for two spheres. Applying (e) to the two spheres and taking the ratio of the resulting equation
h
h
1
2
=21
12
D
D
RefD
RefD (f)
where the subscripts 1 and 2 refer to sphere 1 and 2, respectively. Using (d) to determine the two Reynolds numbers
111
DVReD (g)
and
222
DVReD (h)
HoweverD2 = 2 D1 (i)
and
2/12 VV (j)
Substituting (i) and (j) into (h)
2
2 112
DVReD =
111
DReDV
(k)
Thus the Reynolds number is the same for both spheres. It follows that
f(2DRe ) = f(
1DRe ) (l)
Substituting this result into (f) gives
h
h
1
2
=D
D
2
1
= 2 (m)
(iii) Checking. Qualitative check: From (e) one concludes that for the same fluid (same k)and Reynolds number, the heat transfer coefficient is inversely proportional to the diameter. This confirms the result in (m).
2DRe2DRe
PROBLEM 2.22 (continued)
(5) Comments. (i) For constant Reynolds and Prandtl numbers the heat transfer coefficient increases as the diameter decreases. (ii) The ratio of the total heat transfer rate from the two spheres is obtained from Newton's law of cooling
2
1
))((
))((
2
1
221
212
222
211
2
1
D
D
DD
DD
TTDh
TTDh
q
q
s
s
Thus, although the heat transfer coefficient for the small sphere is greater than that of the large sphere, its total heat transfer rate is smaller by a factor of two.
PROBLEM 2.23
The average Nusselt number for laminar free convection over an isothermal vertical plate is
determined analytically and is given by
)(43
44/1
PrfGr
k
LhNu L
L
where LGr is the Grashof number based on the length of the plate L and f(Pr) is a function of the
Prandtl number. Determine the percent change in the average heat transfer coefficient if the
length of the plate is doubled.
(1) Observations. (i) This is a free convection problem. (ii) The average heat transfer
coefficient h depends on the vertical length L of the plate. (iii) L appears in the Nusselt number as well as the Grashof number.
(2) Problem Definition. Derive a relationship between the average heat transfer coefficient h
and the length of a vertical plate L.
(3) Solution Plan. Solve the given Nusselt number correlation equation for h in terms of length L.
(4) Plan Execution.
(i) Assumptions. (1) Laminar flow and (2) given correlation equation for Nusselt number applies to both plates.
(ii) Analysis. The percent change in h is given by
% change in h = 100 11001
2
1
12
h
h
h
hh (a)
where the subscripts 1 and 2 refer to plates of length L and 2L, respectively and h is the average
heat transfer coefficient. The average Nusselt number LNu is given by
k
LhNu L
4/1
43
4 LGrf(Pr) (b)
where
f(Pr) = function of Prandtl numberGrL = Grashof number
h = average heat transfer coefficient, W/m2-oCk = thermal conductivity, W/m-oCL = plate length, m
NuL = average Nusselt number
Pr = Prandtl number
The Grashof number is defined as
GrL =2
3)( LTTg s (c)
where
PROBLEM 2.23 (continued)
g = gravitational acceleration, m/s2
sT = surface temperature, oC
T = ambient temperature, oC
= coefficient of thermal expansion, 1/K (or 1/oC)
= kinematic viscosity, m2/s
Substituting (c) into (b) and solving for h
h =
4/1
24
)(
3
4 TTgk s
4/1
1)(
LPrf (d)
Applying (d) to the two plates
h1 =
4/1
24
)(
3
4 TTgk s
4/11
1)(L
Prf (e)
and
h2 =
4/1
24
)(
3
4 TTgk s f
4/12
1Pr
L (f)
Taking the ratio of (e) and (f) 4/1
2
1
1
2
L
L
h
h (g)
Substituting (g) into (a)
% change in h = 100 1)( 4121
/L/L (h)
(iii) Computations. For the case where the length L2 = 2L1, equation (h) gives
% change in h = 100 [(1/2)1/4 1] = 9.15 %
(iv) Checking. Dimensional check: Units of h in (d) should be W/m2-oC:
h =
4/1
222
o2oo
)/sm(4
)C)()(m/s()C/1()CW/m(
3
4 TTgk s f(Pr)
4/14/1 )m(
1
L= W/m2-oC
Qualitative check: According to (d) the average heat transfer coefficient is inversely proportional to L
1/4. Thus increasing L, decreases h . This is consistent with the negative result obtained ( 9.15 %) which indicates a decrease in h .
(5) Comments. Although h decreases as the length of the plate is increased, the total heat transfer rate increases. Newton's law of cooling gives
4/3
1
2
1
2
4/1
2
1
11
22
11
22
1
2
L
L
L
L
L
L
Lh
Lh
TTLh
TTLh
q
q
s
s = (2)3/4 = 1.68
PROBLEM 2.24
An experiment was performed to determine the average heat transfer coefficient for forced convection over spheres. In the experiment a sphere of diameter 3.2 cm is maintained at uniform
surface temperature. The free stream velocity of the fluid is 23.4 m/s. Measurements showed that
the average heat transfer coefficient is 62 CW/m o2 .
[a] Predict the average heat transfer coefficient for the same fluid which is at the same free
stream temperature flowing over a sphere of diameter 6.4 cm which is maintained at the same
surface temperature. The free stream velocity is 11.7 m/s.
[b] Which sphere transfers more heat?
(1) Observations. (i) This is a forced convection problem. (ii) The same fluid flows over both spheres. (iii) Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient. (iv) Newton’s law of cooling gives the heat transfer rate.
(2) Problem Definition. Since the average heat transfer coefficient h is expressed in terms
of the Nusselt number, the problem becomes one of determining the Nusselt number for each sphere and taking their ratio.
(3) Solution Plan. Use the results of dimensional analysis to obtain a relationship between the Nusselt number and the significant parameters in forced convection. Apply Newton’s law of cooling to determine heat transfer rate.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state and (3) constant properties.
(ii) Analysis and computations. Non-dimensional form of the governing equations for forced convection gives
DNu =h D
k = f( DRe , Pr, E) (a)
where
D = diameter of sphere, m E = Eckert number
h = average heat transfer coefficient, W/m2-oCk = thermal conductivity of fluid, W/m-oC
DNu = average Nusselt number
Pr = Prandtl number
DRe = Reynolds number
Assume that dissipation is negligible, equation (a) is simplified to
DNu =h D
k = f( DRe , Pr) (b)
Since the same fluid flows over both spheres, it follows that the Prandtl number is the same for both. Thus (b) becomes
PROBLEM 2.24 (continued)
DNu =h D
k = f( DRe ) (c)
The Reynolds number is defined as
DVReD (d)
where
V free stream velocity, m/s
= kinematic viscosity, m2/s
Solving equation (c) for h
h =k
Df( DRe ) (e)
Applying (2) to the two spheres and taking the ratio to eliminate k
h
h
1
2
=)(
)(
21
12
D
D
RefD
RefD (f)
where the subscripts 1 and 2 refer to sphere 1 and 2, respectively. Using (d) to determine the two Reynolds numbers
1DRe = 11 DV (g)
and
222
DVReD (h)
The total heat transfer rate is determined using Newton’s law of cooling
)( TTAhq sT (i)
where A is surface area of sphere given by
2DA (j)
(j) into (i)
)( TTDhq s (k)
Applying (k) to the two spheres and taking their ratio
22
11
22
11
2
1
)(
)(
Dh
Dh
TTDh
TTDh
q
q
s
s (l)
(iii) Computations. Substituting numerical values in (g) and (h)
)/sm(
)/sm(88.74
)/sm(
)m(2.3)m/s(4.232
2
21DRe
and
PROBLEM 2.24 (continued)
)/sm(
)/sm(88.74
)/sm(
)m(4.6)m/s(7.112
2
22DRe
Thus the two Reynolds numbers are identical. It follows that
)()(21 DD RefRef (i)
Substituting (i) into (f)
h
h
1
2
=D
D
2
1
(j)
Solving (j) for 2h
CW/m31m)(4.6
m)(2.3)CW/m(62 o2o2
2
112
D
Dhh
Substitute (j) into (l)
121
12
2
1
DD
DD
q
q
Checking. Dimensional check: computations showed that the Reynolds number is dimensionless and that units of h are correct.
Qualitative check: From (e) one concludes that for the same fluid (same k) and Reynolds number, the heat transfer coefficient is inversely proportional to the diameter. Results show that increasing the diameter by a factor of 2 reduces the heat transfer coefficient by the same factor.
(5) Comments. For constant Reynolds and Prandtl numbers the heat transfer coefficient increases as the diameter decreases.
PROBLEM 2.25
Atmospheric air flows between parallel plates with a mean velocity of m/s10 . One plate is
maintained at C25 o while the other at C.115 o
[a] Calculate the Eckert number. Can dissipation be neglected?
[b] Use scale analysis to compare the magnitude of normal conduction, ,/ 22 yTk with
dissipation, .)/( 2yu Is dissipation negligible compared to conduction?
(1) Observations. (i) Dissipation is important when the Eckert number is high compared to unity. (ii) If the ratio of dissipation to conduction is small compared to unity, it can be neglected.
(2) Problem Definition. [a] Compute the Eckert number. [b] Estimate normal conduction and dissipation using scaling.
(3) Solution Plan. [a] Using the definition of the Eckert number, compute its value for the given data. [b] Use scaling to estimate the ratio of dissipation to normal conduction.
(4) Plan Execution.
(i) Assumption. (1) Newtonian fluid and (2) continuum.
(ii) Analysis. The Eckert number is defined in equation (2.43) as
)( 21
2
ss TTc
u
p
(a)
where
pc specific heat, CJ/kg o
1sT temperature of plate 1 = 25 Co
2sT temperature of plate 2 = 115 Co
u mean axial velocity = 10 m/s
Dissipation is given by
dissipation =
2
y
u(b)
where
u axial velocity, m/s
y normal coordinate, m
viscosity, kg/s-m
Normal conduction is given by
conduction = 2
2
y
Tk (c)
To scale dissipation and conduction, the following scales are introduced
0
2sT
1sT
uH y
PROBLEM 2.25 (continued)
Scale for :T )( 21 ss TTT
Scale for :u uu
Scale for :y Hy
where H is spacing between plates. Rewrite (b) and (c)
dissipation =
2
y
u(d)
conduction = 2)(
)(
y
Tk (e)
Use the above scales to estimate (d) and (e)
dissipation =2
2
H
u(f)
conduction = 2
21 )(
H
TTk ss (g)
Taking the ratio of (f) and (g)
)(conduction
ndissipatio
21
2
ss TTk
u (h)
(iii) Computations. Properties of air are determined at the average temperatureT
C702
)C)(11525(
2
oo
21 ss TTT
7.1008pc CJ/kg o
02922.0k CW/m o
61047.20 mkg/s
Substitute into (a)
0011.0sJ
mkg0011.0
C)25)(-C)(115J/kg(7.1008
)/sm()10(2
2
oo
222
E
Substitute into (h)
conduction
ndissipatio00078.0
W
W00078.0
s-W
m-kg00078.0
C)25)(-C)(115m-0.02922(W/
m/s)(m)(10)kg/s(1047.203
2
oo
226
Computations show that the Eckert number is small compared to unity. Thus dissipation can be neglected. Computations also show that dissipation is small compared to normal conduction. Thus it can be neglected.
(iv) Checking. Dimensional check: Computations show that units for dissipation and conduction are correct.
PROBLEM 2.25 (continued)
(5) Comments. (i) The spacing between the two plates, H, need not be specified to compare dissipation with conduction. Their ratio in (h) is independent of H. (ii) The Eckert number is a measure of the importance of dissipation.
PROBLEM 2.26
An infinitely large plate is immersed in an infinite fluid. The plate is suddenly moved along its
plane with velocity .oU Neglect gravity and assume constant properties.
[a] Show that the axial Navier-Stokes equation is given by
2
2
y
u
t
u
[b] Due to viscous forces, the effect of plate motion
penetrates into the fluid. The penetration depth )(t
increases with time. Use scaling to derive an expression
for )(t .
(1) Observations. (i) The plate is infinite. (ii) No changes take place in the axial direction (infinite plate). (iii) This is a transient problem. (iv) Constant properties. (v) Cartesian coordinates.
(2) Problem Definition. [a] Determine the equation of motion for resulting from a suddenly accelerated plate. [b] Use scaling to estimate ).(t
(3) Solution Plan. [a] Apply the Navier-Stokes equations of motion. Introduce continuity to identify simplifying conditions. [b] Assign scales to each variable in the governing equation to estimate ).(t
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant properties, (3) no motion in the z-direction and (4) negligible gravity.
(ii) Analysis. [a] The Navier-Stokes for two-dimensional constant properties are
x-direction:2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
y-direction:2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
These equations are simplfied as follows:
No gravity: g = 0
No axial variation: 0x
No motion in the z-direction: 0z
w
Substituting these simplifications into (2.10)
PROBLEM 2.26 (continued)
x-direction:2
2
y
u
y
u
t
uv (a)
y-direction:2
2
yy
p
yt
vvv
v (b)
However, continuity equation gives
0z
w
yx
u
zw
yxu
t
vv (2.2b)
For two-dimensional incompressible flow this simplifies to
0yx
u v (c)
This simplifies to
0y
v (d)
Integration of (d) gives
)(tfv (e)
where f(t) is “constant” of integration. This time function is determined from the no slip boundary condition on v
0)0,(xv (f)
Applying (f) to (e) gives
0)(tf (g)
(g) into (e)
0v (h)
Substitute (h) into (a) and (b)
x-direction:2
2
y
u
t
u (i)
y-directiony
p0 (j)
[b] To obtain scaling estimate of )(t rewrite (i)
y
u
yt
u(k)
Scale for :u )0(Uu
Scale for :u )0(tt
Scale for :y )0(y
PROBLEM 2.26 (continued)
Substitute into (k)
2
U
t
Ul
Introduce the definition of kinematic viscosity
(m)
Substitute (m) into (l) and solve for
tt)( n
(iii) Checking. Dimensional check: (i) Each term in (i) has units of 22 mkg/s .
)m/sm()mkg/s()m/s()(kg/m 2
2
223
y
u
t
u= 22 mkg/s
(ii) Units of (n) should be length:
m(s)/s)m()( 2 tt
Limiting check: Initially the penetration thickness is zero. Setting t = 0 in (n) gives 0)0( ,
which is the correct result.
(5) Comments. (i) A major simplification of this problem is due to the assumption of infinite plate. Due to this assumption all derivatives with respect to x vanish. (ii) The same governing equation (i) applies to an oscillating plate moving in a plane normal to y. (iii) Scaling estimate of the penetration thickness )(t is independent of plate velocity U.
PROBLEM 2.27
An infinitely large plate is immersed in an infinite fluid at
uniform temperature iT . The plate is suddenly maintained at
temperature .oT Assume constant properties and neglect gravity.
[a] Show that the energy equation is given by
2
2
y
T
t
T
[b] Due to conduction, the effect of plate temperature propagates into the fluid. The penetration
depth )(t increases with time. Use scaling to derive an expression for )(t .
(1) Observations. (i) The plate is infinite. (ii) No changes take place in the axial direction (infinite plate). (iii) This is a transient problem. (iv) Constant properties. (v) Cartesian coordinates. (vi) Gravity is neglected. Thus there is no free convection. (vii) The fluid is stationary.
(2) Problem Definition. [a] Determine the energy equation resulting from a step change in surface temperature. [b] Use scaling to estimate the thermal penetration thickness ).(t
(3) Solution Plan. [a] Apply energy equation and simplify it for the conditions of the problem. [b] Assign scales to each variable in the governing equation for temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant properties, (3) stationary fluid, and (4) negligible gravity.
(ii) Analysis. (i) Assumptions. (1) Continuum, (2) Newtonian fluid, (3) constant properties and (4) negligible nuclear, radiation and electromagnetic energy transfer.
(ii) Analysis. The energy equation for this case is given by
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc p v (2.19b)
where
pc specific heat at constant pressure
k thermal conductivity
p pressure
T temperature density
= dissipation function
Stationary fluid: 0wu v
PROBLEM 2.27 (continued)
No axial variation: 0x
No variation in the z-direction: 0z
Substituting these simplifications into (2.19.b)
2
2
y
Tk
t
Tc p
(a)
Introduce the definition of thermal diffusivity
k
c p (b)
(b) into (a)
2
2
y
T
t
T (c)
[b] To obtain scaling estimate of )(t rewrite (i)
y
T
yt
T(d)
Scale for :T )( io TTT
Scale for :u )0(tt
Scale for :y )0(y
Substitute into (d)
2
ioio TT
t
TTe
Solve for
tt)( f
(iii) Checking. Dimensional check: (i) Each term in (i) has units of. C/so :
C/s)C/m(/s)m()C/s( o2o
2
22o
y
T
t
T
(ii) Units of (f) should be length:
m(s)/s)m()( 2 tt
Limiting check: Initially the penetration thickness is zero. Setting t = 0 in (f) gives 0)0( ,
which is the correct result.
(5) Comments. (i) A major simplification of this problem is due to the assumption of infinite plate. Consequently, all derivatives with respect to x vanish. (ii) Since the fluid is stationary, the problem is one of pure conduction.
PROBLEM 3.1
A large plate moves with constant velocity oU parallel to a stationary plate separated by a
distance H. An incompressible fluid fills the channel formed by the plates. The stationary plate
is at temperature 1T and the moving plate is at temperature oT . Taking into consideration
dissipation, determine the maximum temperature and
the heat flux at the moving plate. Assume laminar
flow and neglect gravity effect and pressure
variation in the channel.
(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The effect of pressure gradient is negligible. (iv) The fluid is incompressible (constant density). (v) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field. Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
where the dissipation function is given by (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates
0z
w
yx
u
zw
yxu
t
vv (2.2b)
For constant density
PROBLEM 3.1 (continued)
0zyxt
(a)
Since plates are infinite
0wzx
(b)
Substituting (a) and (b) into (2.2b), gives
0y
v (c)
Integrating (c) )(xfv (d)
To determine the “constant” of integration )(xf we apply the no-slip boundary condition at the
lower plate 0)0,(xv (e)
Equations (d) and (e) give 0)(xf
Substituting into (d) 0v (f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel. To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,(2.10x)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
This equation is simplified as follows: Steady state
0t
u (g)
Negligible gravity effect
0xg (h)
Negligible axial pressure variation
0x
p (i)
Substituting (b) and (f)-(i) into (2.10x) gives
02
2
dy
ud (j)
The solution to (j) is
21 CyCu (k)
where 1C and 2C are constants of integration. The two boundary conditions on u are:
0)0(u and oUHu )( (l)
These conditions give
PROBLEM 3.1 (continued)
H
UC o
1 and 02C (m)
Substituting (m) into (k)
H
y
U
u
o
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy equation. Substituting (b) and (f) into (2.17) gives
2
y
u (n)
Using solution (3.8) into (n) gives
2
2
H
U o (o)
Noting that for steady state 0/ tT and using (b), (f) and (o), the energy equation (2.10b)
simplifies to
02
2
2
2
H
U
dy
Tdk o (p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at uniform surface temperature. Equation (p) is solved by direct integration
432
2
2
2CyCy
kH
UT o (q)
where 3C and 4C are constants of integration. The two boundary conditions on (q) are
1)0( TT and oTHT )( (r)
These boundary conditions and solution (q) give
Hk
U
H
TTC oo
2
21
3 and 14 TC (s)
Substituting (s) into (q) and rearranging the result in dimensionless form, give
H
y
H
y
TTk
U
H
y
TT
TT
o
o
o
1)(2 1
2
1
1 (t)
This can be written in terms of the Eckert and Prandtl numbers as
H
y
H
y
k
c
TTc
U
H
y
TT
TT p
op
o
o
1)(2 1
2
1
1
H
y
H
yPrE
H
y
TT
TT
o
121
1 (u)
where
PROBLEM 3.1 (continued)
)( 1
2
TTc
UE
op
o andk
cPr
p (v)
The maximum temperature occurs where the temperature gradient is zero. Differentiating (u),
setting the result equal to zero and solving for position of maximum temperature my
PrEH
ym 1
2
1 (w)
Substituting (w) into (u) gives the maximum temperature mT
8
PrE
PrETT
TT
o
m
2
1
2
1
1
1 (x)
The heat flux at the moving surface is determined by applying Fourier’s law at Hy
dy
HdTkHq
)()(
Using (u) into the above
12
)()( 1 PrE
H
TTkHq o (y)
(iii) Checking. Dimensional check: Each term in (3.8), (t), (u), (w) and (x) is dimensionless.
Units of (y) should be 2W/m :
2
o1
o
m
W
)m(
)C)()(CW/m()(
H
TTkHq o
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature solution (t) satisfies (p).
Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and temperature solution (t) satisfies boundary conditions (r).
Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting 0oU
in (3.8) gives .0)(yu
(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be linear and surface flux at the upper plate will be due to conduction between the two surfaces. Setting
0oU in (v) gives E = 0. When this is substituted into (u) and (y) gives the anticipated linear
temperature distribution and a surface flux of
H
TTkHq o )(
)( 1
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be linear. Setting 0 in (v ) gives 0Pr . When this is substituted into (u) gives a linear temperature
distribution.
PROBLEM 3.1 (continued)
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. Alternatively, one could state that the streamline are parallel. This means that .0/ yvv
substituting this into the continuity equation for two-dimensional incompressible flow gives .0/ xu This is identical equation (b) which is based on assuming infinite plate.
(ii) According to (w), maximum temperature occurs in the upper half of the channel.
k
UTT o
o2
)0(2
PROBLEM 3.2
A large plate moves with constant velocity oU
parallel to a stationary plate separated by a
distance H. An incompressible fluid fills the
channel formed by the plates. The upper plate
is maintained at uniform temperature oT and
the stationary plate is insulated. A pressure gradient dxdp / is applied to the fluid. Taking into
consideration dissipation, determine the temperature of the insulated plate and the heat flux at
the upper plate. Assume laminar flow and neglect gravity effect.
(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The effect of pressure gradient must be included. (iv) The fluid is incompressible. (v) Using Fourier’s law, Temperature distribution gives surface heat flux of the moving plate. (vi) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field. Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no end effects and (v) negligible gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
where the dissipation function is given by (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates
0z
w
yx
u
zw
yxu
t
vv (2.2b)
For constant density
0zyxt
(a)
PROBLEM 3.2 (continued)
Since plates are infinite
0wzx
(b)
Substituting (a) and (b) into (2.2b), gives
0y
v (c)
Integrating (c) )(xfv (d)
To determine the “constant” of integration )(xf , we apply the no-slip boundary condition at the
lower plate 0)0,(xv (e)
Equations (d) and (e) give 0)(xf
Substituting into (d) 0v (f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel. To determine the horizontal component u we apply the Navier-Stokes equations (2.10)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
These equations are simplified as follows: Steady state
0t
u (g)
Negligible gravity effect
0yx gg (h)
Substituting (b) and (f)-(h) into (2.10x) and (2.10y) gives
2
2
dy
ud
x
p (i)
and
0y
p (j)
Equation (j) shows that pressure does not vary in the y-direction and thus it can either be a function of x or constant. Integrating (i) twice
212
2
1CyCy
dx
dpu (k)
where 1C and 2C are constants of integration. The two boundary conditions on u are:
PROBLEM 3.2 (continued)
(1) 0)0(u
(2) oUHu )(
These conditions give
dx
dpH
H
UC o
21 and 02C (l)
Substituting (l) into (k)
H
y
dx
dp
U
H
H
y
U
u
oo
12
12
(m)
With the velocity distribution determined, we return to the dissipation function and energy equation. Substituting (b) and (f) into (2.17) gives
2
y
u (n)
Using solution (m) into (n) gives 2
1
2y
dx
dp
dx
dpH
H
Uo (o)
Noting that for steady state 0/ tT and using (b), (f) and (o), the energy equation (2.10b)
simplifies to
01
2
2
2
2
ydx
dp
dx
dpH
H
U
dy
Tdk o (p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at uniform surface temperature. Equation (p) is solved by direct integration
214
2
32
21
12
1
2322ByBy
dx
dp
ky
dx
dp
dx
dpH
H
U
ky
dx
dpH
H
U
kT oo (q)
where 1B and 2B are constants of integration. The two boundary conditions on (q) are
(1) 0)0(
dy
dT
(2) oTHT )(
These boundary conditions and solution (q) give
01B (r)
4
2
32
2
2
1
12
1
2322H
dx
dp
kH
dx
dp
dx
dpH
H
U
kH
dx
dpH
H
U
kTB oo
o (s)
Substituting (r) into (q) and rearranging the result in dimensionless form, give
PROBLEM 3.2 (continued)
4
4
3
3
22
2
2
224
112
111
2
8
111
2
8
1
H
y
H
y
dx
dpH
U
H
y
dx
dpH
U
dx
dp
k
H
TT ooo (t)
Surface temperature of the insulated plate, T(0), is obtained by setting y = 0 in (t)
12
11
2
8
11
2
8
1)0(
2
2
224
dx
dpH
U
dx
dpH
U
dx
dp
k
H
TT ooo (u)
Surface heat flux at the upper plate is obtained by applying Fourier’s law at y = H
dy
HdTkHq
)()(
Using (t) into the above
3
11
2
2
11
2
4
1)(
2
2
223
dx
dpH
U
dx
dpH
U
dx
dpH
Hq oo (v)
(iii) Checking. Dimensional check: Each term in (m), (t), (u) and (v) is dimensionless. Each term in (q) has units of temperature.
Differential equation check: Velocity solution (m) satisfies equation (i) and temperature solution (t) satisfies (p).
Boundary conditions check: Velocity solution (m) and temperature solution (t) satisfy their respective boundary conditions.
Limiting check: (i) If the upper plate is stationary and there is no axial pressure gradient the fluid
will also be stationary. Setting 0/ dxdpUo in (m) gives .0)(yu
(ii) If the upper plate is stationary and there is no axial pressure gradient, there will be no fluid
motion and dissipation will vanish. The temperature distribution will be uniform equal to .oT
Setting 0/ dxdpUo in (t) gives .)( oTyT Similarly, surface heat flux will vanish. Setting
0/ dxdpUo in (v) gives .0)(Hq
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. (ii) According to (t), maximum temperature occurs at the insulated plate. (iii) According to the dimensionless form of solutions (u) and (v), the problems is characterized by the single dimensionless parameter
dx
dpH
Uo
2
2
PROBLEM 3.3
Incompressible fluid is set in motion between two
large parallel plates by moving the upper plate with
constant velocity oU and holding the lower plate
stationary. The clearance between the plates is H.
The lower plate is insulated while the upper plate
exchanges heat with the ambient by convection. The heat transfer coefficient is h and the
ambient temperature is .T Taking into consideration dissipation determine the temperature of
the insulated plate and the heat flux at the moving plate. Assume laminar flow and neglect
gravity effect.
(1) Observations. (i) Moving plate sets fluid in motion in the x-direction. (ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The fluid is incompressible (constant density). (iv) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field. Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible gravitational effect.
(ii) Analysis. Since the objective is the determination of temperature distribution and heat transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
where the dissipation function is given by (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates
0z
w
yx
u
zw
yxu
t
vv (2.2b)
For constant density
PROBLEM 3.3 (continued)
0zyxt
(a)
Since plates are infinite
0wzx
(b)
Substituting (a) and (b) into (2.2b), gives
0y
v (c)
Integrating (c) )(xfv (d)
To determine the “constant” of integration )(xf we apply the no-slip boundary condition at the
lower plate 0)0,(xv (e)
Equations (d) and (e) give 0)(xf
Substituting into (d) 0v (f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel. To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,(2.10x)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
This equation is simplified as follows: Steady state
0t
u (g)
Negligible gravity effect
0xg (h)
Negligible axial pressure variation
0x
p (i)
Substituting (b) and (f)-(i) into (2.10x) gives
02
2
dy
ud (j)
The solution to (j) is
21 CyCu (k)
where 1C and 2C are constants of integration. The two boundary conditions on u are:
0)0(u and oUHu )( (l)
These conditions give
PROBLEM 3.3 (continued)
H
UC o
1 and 02C (m)
Substituting (m) into (k)
H
y
U
u
o
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy equation. Substituting (b) and (f) into (2.17) gives
2
y
u (n)
Using solution (3.8) into (n) gives
2
2
H
U o (o)
Noting that for steady state 0/ tT and using (b), (f) and (o), the energy equation (2.10b)
simplifies to
02
2
2
2
H
U
dy
Tdk o (p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at uniform surface temperature. Equation (p) is solved by direct integration
432
2
2
2CyCy
kH
UT o (q)
where 3C and 4C are constants of integration. The two boundary conditions on (q) are
(1) 0)0(
dy
dT
(2) TyThdy
HdTk )(
)(
These boundary conditions and solution (q) give
03C (r)
k
U
khH
UTC oo
2
22
4 (s)
Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give
2
2
2 2
1
2
1
H
y
hH
k
k
U
TT
o
(t)
The dimensionless parameter khH / is known as the Biot number, Bi . It is associated with
convection boundary conditions.
PROBLEM 3.3 (continued)
The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.
hH
k
k
U
TT
o2
1)0(2
(u)
The heat flux at the moving surface is determined by applying Fourier’s law at Hy
dy
HdTkHq
)()(
Substituting (t) into the above
H
UHq o
2
)( (v)
(iii) Checking. Dimensional check: Each term in (3.8), (t) and (u) is dimensionless. Each term
in solution (v) has units of 2W/m .
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature solution (t) satisfies (p). Boundary conditions check: Velocity solution (3.8) satisfies boundary conditions (l) and temperature solution (t) satisfies boundary conditions (r).
Limiting check: (i) If the upper plate is stationary the fluid will also be stationary. Setting 0oU
in (3.8) gives .0)(yu
(ii) If the upper plate is stationary, dissipation will vanish, temperature distribution will be
uniform equal to the ambient temperature .T Setting 0oU in (u) gives .)( TyT Similarly
the heat flux )(Hq vanishes. Substituting 0oU (v) gives .0)(Hq
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to
.T Setting 0 in (u) gives .)( TyT Similarly the heat flux )(Hq vanishes. Substituting
0 (v) gives .0)(Hq
Global energy balance: Energy leaving the channel must equal to the work done to move the plate. Consider the work done by the plate on the fluid
ooUW (w)
where
W work done per unit surface area by the plate on the fluid
o shearing stress at the moving plate
However, shearing stress is given by
y
Huo
)( (x)
(3.8) into (x)
H
Uoo (y)
(y) into (w)
PROBLEM 3.3 (continued)
H
UW o
2
(z)
This is identical to the heat removed from the upper plate given in equation (v).
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. Alternatively, one could state that the streamline are parallel. This means that .0/ yvv (ii)
The solutions is characterized by a single dimensionless parameter ,/ khH which is the Biot
number. (iii) The Nusselt number at moving plate, ),(HNu is defined as
k
hHHNu )( (w)
The heat transfer coefficient h is based on the overall temperature drop, defined as
)()0(
)(
HTT
Hqh (x)
Centerline temperature and moving plate temperature are obtained by evaluating (t) at y = 0 and y = H
hH
k
k
UTT o
2
1)0(
2
and
Hh
UTHT o
2
2
1)(
The above two equations give
k
UHTT o
2
2
1)()0( (y)
(v) and (y) into (x)
H
kh 2
Substituting into (w) gives the Nusselt number
2)(HNu (z)
PROBLEM 3.4
Two parallel plates are separated by a
distance 2H. The plates are moved in
opposite direction with constant velocity oU .
Each plate is maintained at uniform
temperature oT . Taking into consideration
dissipation determine the heat flux at the
plates. Assume laminar flow and neglect
gravity effect
(1) Observations. (i) Moving plates set fluid in motion in the positive and negative x-direction.(ii) Since plates are infinite the flow field does not vary in the axial direction x. (iii) The fluid is incompressible (constant density). (iv) The fluid is stationary at the center plane y = 0. (v) Symmetry dictates that no heat is conducted through the center plane. (vi) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field. Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no end effects, (vi) uniform pressure (vi) negligible gravitational effect.
(ii) Analysis. Taking advantage of symmetry only the upper half of the channel is analyzed. Since the objective is the determination of temperature distribution and heat transfer rate, it is logical to begin the analysis with the energy equation. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
where the dissipation function is given by (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Thus it is clear from (2.19b) and (2.17) that the determination of temperature distribution requires the determination of the velocity components u , v and w. This is accomplished by
applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates
0z
w
yx
u
zw
yxu
t
vv (2.2b)
PROBLEM 3.4 (continued)
For constant density
0zyxt
(a)
Since plates are infinite
0wzx
(b)
Substituting (a) and (b) into (2.2b), gives
0y
v (c)
Integrating (c) )(xfv (d)
To determine the “constant” of integration )(xf we apply the no-slip boundary condition at the
lower plate 0)0,(xv (e)
Equations (d) and (e) give 0)(xf
Substituting into (d) 0v (f)
Since the vertical component v vanishes everywhere, it follows that the streamlines are parallel. To determine the horizontal component u we apply the Navier-Stokes equation in the x-direction,(2.10x)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
This equation is simplified as follows: Steady state
0t
u (g)
Negligible gravity effect
0xg (h)
Negligible axial pressure variation
0x
p (i)
Substituting (b) and (f)-(i) into (2.10x) gives
02
2
dy
ud (j)
The solution to (j) is
21 CyCu (k)
where 1C and 2C are constants of integration. The two boundary conditions on u are:
0)0(u and oUHu )( (l)
PROBLEM 3.4 (continued)
These conditions give
H
UC o
1 and 02C (m)
Substituting (m) into (k)
H
y
U
u
o
(3.8)
With the velocity distribution determined, we return to the dissipation function and energy equation. Substituting (b) and (f) into (2.17) gives
2
y
u (n)
Using solution (3.8) into (n) gives
2
2
H
U o (o)
Noting that for steady state 0/ tT and using (b), (f) and (o), the energy equation (2.10b)
simplifies to
02
2
2
2
H
U
dy
Tdk o (p)
In arriving at (p) axial temperature variation was neglected. This is valid for infinite plates at uniform surface temperature. Equation (p) is solved by direct integration
432
2
2
2CyCy
kH
UT o (q)
where 3C and 4C are constants of integration. The two boundary conditions on (q) are
(1) 0)0(
dy
dT
(2) oTHT )(
These boundary conditions and solution (q) give
03C (r)
k
UTC o
o2
2
4 (s)
Substituting (r) and (s) into (q) and rearranging the result in dimensionless form, give
2
2
21
2
1
H
y
k
U
TT
o
o (t)
The temperature of the insulated surface, T(0), is obtained by evaluating (t) at y = 0.
PROBLEM 3.4 (continued)
The heat flux at the moving surface is determined by applying Fourier’s law at Hy
dy
HdTkHq
)()(
Substituting (t) into the above
H
UHq o
2
)( (u)
Similarly, the heat flux at the lower plate is
H
UHq o
2
)( (v)
(iii) Checking. Dimensional check: Each term in (3.8) and (t) is dimensionless. Each term in
solutions (u) and (v) has units of 2W/m .
Differential equation check: Velocity solution (3.8) satisfies equation (j) and temperature solution (t) satisfies (p).
Boundary conditions check: Velocity solution (3.8) and temperature solution (t) satisfy their respective boundary conditions.
Limiting check: (i) If the two plates are stationary, the fluid will also be stationary. Setting
0oU in (3.8) gives .0)(yu
(ii) If the upper and lower plates are stationary, dissipation will vanish, temperature distribution
will be uniform equal to the ambient temperature .oT Setting 0oU in (t) gives .)( oTyT
Similarly the heat flux )(Hq vanishes. Substituting 0oU (u) gives .0)(Hq
(iii) If the fluid is inviscid, dissipation will vanish and temperature should be uniform equal to
.oT Setting 0 in (t) gives .)( oTyT Similarly the heat flux )(Hq vanishes. Substituting
0 (u) gives .0)(Hq
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. Alternatively, one could state that the streamline are parallel. This means that .0/ yvv (ii)
Symmetry provides additional simplification. (iii) The Nusselt number at the upper moving plate, ),(HNu is defined as
k
hHHNu )( (w)
Defining the heat transfer coefficient is defined as
oTT
Hqh
)0(
)((x)
Centerline temperature is obtained by evaluating (t) at y = 0
k
UTT o
o
2
2
1)0( (y)
PROBLEM 3.4 (continued)
(u) and (y) into (x) gives h
H
kh 2
Substituting into (w) gives the Nusselt number
2)(HNu (z)
A more appropriate definition of the heat transfer coefficient is based on the mean temperature,
,mT rather than the temperature at the center. That is
om TT
Hqh
)(
PROBLEM 3.5
Incompressible fluid flows in a long tube of radius
or . Fluid motion is driven by an axial pressure
gradient ./ zp The tube exchanges heat by
convection with an ambient fluid. The heat transfer
coefficient is h and the ambient temperature is .T
Taking into consideration dissipation, assuming
laminar incompressible axisymmetric flow, and neglecting gravity, axial temperature variation
and end effects, determine:
[a] Surface temperature.
[b] Surface heat flux.
[c] Nusselt number based on [ )()0( orTT ].
(1) Observations. (i) Fluid motion is driven by axial pressure drop. (ii) For a very long tube the flow field does not vary in the axial direction z. (iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. It is removed from the fluid by convection at the surface. (v) The Nusselt number is a dimensionless heat transfer coefficient. (vi) To determine surface heat flux and heat transfer coefficient requires the determination of temperature distribution. (vii) Temperature distribution depends on the velocity distribution. (viii) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Apply the energy equation to determine temperature distribution. Fourier’s law gives surface heat flux. Equation (1.10) gives the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform ambient temperature, (vii) uniform heat transfer coefficient and (viii) negligible gravitational effect.
(ii) Analysis. [a] Since temperature distribution is obtained by solving the energy equation, we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)
2
2
2
2
2 0
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrP v
vv (2.24)
where the dissipation function is given by (2.25)
22
0
2
0
222
0
1
0
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
Equations (2.24) and (2.25) show that the determination of temperature distribution requires the
determination of the velocity components rv , v and .zv The flow field is determined by
PROBLEM 3.5 (continued)
solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates
011
zrzr
rrrt
vvv (2.4)
For constant density
0zrt
(a)
For axisymmetric flow
0v (b)
For a long tube with no end effects axial changes in velocity are negligible
0z
(c)
Substituting (a)-(c) into (2.4)
0rrdr
dv (d)
Integrating (d)
)(zfr rv (e)
To determine the “constant” of integration )(zf we apply the no-slip boundary condition at the
surface
0),( zrov (f)
Equations (e) and (f) give 0)(zf
Substituting into (e)
0rv (g)
Since the radial component rv vanishes everywhere, it follows that the streamlines are parallel
to the surface. To determine the axial component zv we apply the Navier-Stokes equation in the
z-direction, (2.11z)
2
2
2
2
2
11
zrrr
rrz
pg
tzrr
zzzz
zzzz
vvv
vvv
vvvv zr
(2.11z)
This equation is simplified as follows:
Steady state
0t
(h)
Negligible gravity effect
0zr gg (i)
Substituting (b), (c) and (g)-(i) into (2.11z) gives
PROBLEM 3.5 (continued)
01
dr
dr
dr
d
rz
p zv (3.11)
Since zv depends on r only, equation (3.11) can be written as
)(1
rgdr
dr
dr
d
rz
p zv (j)
Integrating (j) with respect to z
oCzrgp )( (k)
where oC is constant of integration. We turn our attention now to the radial component of
Navier-Stokes equation, (2.11r)
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
Substituting (b), (g) and (i) into (2.11r), gives
0r
p (l)
Integrating (l) )(zfp (m)
where )(zf is “constant” of integration. We now have two solutions for the pressure p: (k) and
(m). Equating the two, gives
)()( zfCzrgp o (n)
One side of (n) shows that the pressure depends on z only while the other side shows that it depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is by requiring that
g(r) = C (o)
where C is a constant. Substituting (o) into (j)
Cdr
dr
dr
d
rz
p zv1 (p)
Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to give the axial velocity distribution. Integrating once
12
2
1Cr
zd
pd
dr
dr zv
Separating variables and integrating again
PROBLEM 3.5 (continued)
212 ln
4
1CrCr
zd
pdzv (q)
where 1C and 2C are constants of integration. The two boundary conditions on zv are
,0)0(
dr
d zv 0)( oz rv (r)
Equations (q) and (r) give 1C and 2C
,01C 22
4
1or
zd
pdC
Substituting into (q)
)(4
1 22oz rr
zd
pdv (3.12)
With the velocity distribution determined we return to the energy equation (2.24) and the dissipation function (2.25). We note that for a long tube at uniform surface temperature with no end effects, axial temperature variation can be neglected. Thus
02
2
z
T
z
T (s)
Substituting (b), (c), (g), (h) and (s) into (2.24)
01
dr
dTr
dr
d
rk (t)
Using (b), (c) and (g) into (2.25) gives the dissipation function for this flow
2
dr
d zv
Substituting the velocity solution (3.11) into the above, gives
2
2
2
1r
zd
pd (u)
Using (u) to eliminate in (t) and rearranging, we obtain
dr
dTr
dr
d 3
2
4
1r
zd
pd
k
Integrating the above twice
434
2
ln64
1CrCr
zd
pd
kT (v)
Two boundary conditions are needed to evaluate the constants of integration 3C and 4C . They
are:
(1) 0)0(
dr
dT
PROBLEM 3.5 (continued)
(2) ])([)(
TrThdr
rdTk o
o
Equations (v) and the two boundary conditions give the two constants
,03C ]14[64
24
4o
o
hr
k
zd
pd
k
rTC
Substituting the above into (v)
2424
64]14[
64 zd
pd
k
r
hr
k
zd
pd
k
rTT
o
o (w)
This solution can be expressed in dimensionless form as
4
4
2414
64
ooo
r
r
hr
k
zd
pd
k
r
TT (x)
The dimensionless parameter k
hro in (x) is known as the Biot number.
Surface temperature is obtained by setting orr in (w)
hzd
pdrTrT o
o
1
16)(
23
(y)
[b] Surface heat flux )( orq is obtained by applying Fourier’s law
dr
rdTkrq o
o
)()(
Using (w) into the above 23
16)(
zd
pdrrq o
o (z)
[c] The Nusselt number is defined as
k
hr
k
hDNu o2
(z-1)
where D is tube diameter. The heat transfer coefficient h is determined using equation (1.10)
dr
rdT
rTT
kh o
o
)(
)]()0([ (z-2)
Substituting (w) into the above
or
kh
4 (z-3)
Introducing (z-3) into (z-1) 8Nu (z-4)
PROBLEM 3.5 (continued)
(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity. Each term in (w) has units of temperature. Each term in (x) is dimensionless.
Differential equation check: Velocity solution (3.12) satisfies equation (p) and temperature solution (w) satisfies (t).
Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r) and temperature solution (w) satisfies boundary conditions (1) and (2).
Limiting check: (i) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary. Setting
0/ dzdp in (3.12) gives .0zv
(ii) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary and no dissipation takes place
and thus surface heat transfer should vanish Setting 0/ dzdp in (z) gives .0)( orq
(iii) Global conservation of energy. Heat transfer rate leaving the tube must be equal to the work required to pump the fluid. Pump work for a tube section of length L is
QppW )( 21 (z-1)
Where
1p = upstream pressure
2p = downstream pressure
Q = volumetric flow rate, given by
or
v0
2 rdrQ z
Substituting (3.12) into the above and integrating
4
8or
dz
dpQ (z-2)
Combining (z-1) and (z-2))
)(8
21
4
ppdz
dprW o (z-3)
Work per unit area W is
Lr
WW
o2
Substituting (z-3) into the above
L
pp
dz
dprW o )(
16
213
(z-4)
However
dz
dp
L
pp )( 21
Combining this result with (z-4) gives 23
16 dz
dprW o
PROBLEM 3.5 (continued)
This result is identical to surface heat transfer rate given in (z)
(5) Comments. (i) The assumption of a long tube with negligible end effects is a key factor in simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting end effects since it leads to the same mathematical simplifications. (ii) Solution (w) shows that maximum temperature occurs at the center .0r
(iii) The Nusselt number is constant independent of Reynolds and Prandtl numbers.
PROBLEM 3.6
Fluid flows axially in the annular space
between a cylinder and a concentric rod . The
radius of the rod is ir and that of the cylinder
is .or Fluid motion in the annular space is
driven by an axial pressure gradient ./ zp
The cylinder is maintained at uniform
temperature .oT Assume incompressible laminar axisymmetric flow and neglect gravity and end
effects. Show that the axial velocity is given by
1)/ln()/ln(
)/(1)/(
4
22
2
o
io
oio
oz rr
rr
rrrr
dz
dprv
(1) Observations. (i) Fluid motion is driven by axial pressure drop. (ii) For a very long tube the flow field does not vary in the axial direction z. (iii) The fluid is incompressible (constant density). (iv) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density and viscosity), (v) no end effects and (vi) negligible gravitational effect.
(ii) Analysis. The flow field is determined by solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates
011
zrzr
rrrt
vvv (2.4)
For constant density
0zrt
(a)
For axisymmetric flow
0v (b)
For a long tube with no end effects axial changes in velocity are negligible
0z
(c)
Substituting (a)-(c) into (2.4)
0rrdr
dv (d)
Integrating (d)
PROBLEM 3.6 (continued)
)(zfr rv (e)
To determine the “constant” of integration )(zf we apply the no-slip boundary condition at the
surface
0),( zrov (f)
Equations (e) and (f) give 0)(zf
Substituting into (e)
0rv (g)
Since the radial component rv vanishes everywhere, it follows that the streamlines are parallel
to the surface. To determine the axial component zv we apply the Navier-Stokes equation in the
z-direction, (2.11z)
2
2
2
2
2
11
zrrr
rrz
pg
tzrr
zzzz
zzzz
vvv
vvv
vvvv zr
(2.11z)
This equation is simplified as follows:
Steady state
0t
(h)
Negligible gravity effect
0zr gg (i)
Substituting (b), (c) and (g)-(i) into (2.11z) gives
01
dr
dr
dr
d
rz
p zv (3.11)
Since zv depends on r only, equation (3.11) can be written as
)(1
rgdr
dr
dr
d
rz
p zv (j)
Integrating (j) with respect to z
oCzrgp )( (k)
where oC is constant of integration. We turn our attention now to the radial component of
Navier-Stokes equation, (2.11r)
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
Substituting (b), (g) and (i) into (2.11r), gives
PROBLEM 3.6 (continued)
0r
p (l)
Integrating (l) )(zfp (m)
where )(zf is “constant” of integration. We now have two solutions for the pressure p: (k) and
(m). Equating the two, gives
)()( zfCzrgp o (n)
One side of (n) shows that the pressure depends on z only while the other side shows that it depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is by requiring that
g(r) = C (o)
where C is a constant. Substituting (o) into (j)
Cdr
dr
dr
d
rz
p zv1 (p)
Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to give the axial velocity distribution. Integrating once
12
2
1Cr
zd
pd
dr
dr zv
Separating variables and integrating again
212 ln
4
1CrCr
zd
pdzv (q)
where 1C and 2C are constants of integration. The two boundary conditions on zv are
,0)( iz rv 0)( oz rv (r)
Equations (q) and (r) give 1C and 2C1
221 ln)(
4
1
i
ooi
r
rrr
zd
pdC
2
1
222
4
1lnln)(
4
1ii
i
ooi r
zd
pdr
r
rrr
zd
pdC
Substituting into (q) and rearranging
1)/ln()/ln(
)/(1)/(
4
22
2
o
io
oio
oz rr
rr
rrrr
dz
dprv (3.12)
(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity.
Differential equation check: Velocity solution (3.12) satisfies equation (p).
PROBLEM 3.6 (continued)
Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r).
Limiting check: (i) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary. Setting
0/ dzdp in (3.12) gives .0zv
(5) Comments. The assumption of a long tube with negligible end effects is a key factor in simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting end effects since it leads to the same mathematical simplifications.
PROBLEM 3.7
A rod of radius ir is placed concentrically inside a cylinder of radius .or The rod moves axially
with constant velocity oU and sets the
fluid in the annular space in motion. The
cylinder is maintained at uniform
temperature .oT Neglect gravity and end
effects, and assume incompressible
laminar axisymmetric flow
[a] Show that the axial velocity is given by
)/ln()/ln(
ooi
oz rr
rr
Uv
[b] Taking into consideration dissipation, determine the heat flux at the outer surface and the
Nusselt number based on [ oi TrT )( ]. Neglect axial temperature variation.
(1) Observations. (i) Fluid motion is driven by axial motion of the rod. Thus motion is not due to pressure gradient. (ii) For a very long tube the flow field does not vary in the axial direction z.(iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. It is removed from the fluid by conduction at the surface. (v) The Nusselt number is a dimensionless heat transfer coefficient. (vi) To determine the heat transfer coefficient require the determination of temperature distribution. (vii) Temperature distribution depends on the velocity distribution. (viii) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Apply the energy equation to determine temperature distribution. Fourier’s law gives surface heat flux. Equation (1.10) gives the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform surface temperature and (vii) constant rod velocity, (viii) negligible axial pressure gradient and (ix) negligible gravitational effect.
(ii) Analysis. [a] Velocity distribution is governed by the continuity equation and Navier-Stokes equations of motion. The flow field is determined by solving the continuity and the Navier-Stokes equations. We begin with the continuity equation in cylindrical coordinates
011
zrzr
rrrt
vvv (2.4)
For constant density
0zrt
(a)
For axisymmetric flow
PROBLEM 3.7 (continued)
0v (b)
For a long tube with no end effects axial changes in velocity are negligible
0z
(c)
Substituting (a)-(c) into (2.4)
0rrdr
dv (d)
Integrating (d)
)(zfr rv (e)
To determine the “constant” of integration )(zf we apply the no-slip boundary condition at the
surface
0),( zrov (f)
Equations (e) and (f) give 0)(zf
Substituting into (e)
0rv (g)
Since the radial component rv vanishes everywhere, it follows that the streamlines are parallel
to the surface. To determine the axial component zv we apply the Navier-Stokes equation in the
z-direction, (2.11z)
2
2
2
2
2
11
zrrr
rrz
pg
tzrr
zzzz
zzzz
vvv
vvv
vvvv zr
(2.11z)
This equation is simplified as follows:
Steady state
0t
(h)
Negligible gravity effect
0zr gg (i)
Substituting (b), (c) and (g)-(i) into (2.11z) gives
01
dr
dr
dr
d
rz
p zv (3.11)
Since zv depends on r only, equation (3.11) can be written as
)(1
rgdr
dr
dr
d
rz
p zv (j)
Integrating (j) with respect to z
PROBLEM 3.7 (continued)
oCzrgp )( (k)
where oC is constant of integration. We turn our attention now to the radial component of
Navier-Stokes equation, (2.11r)
2
2
22
2
2
2
21)(
1
zrrr
rrrr
pg
tzrrr
rrr
r
rrz
rrr
vvvv
vvv
vvvvv
(2.11r)
Substituting (b), (g) and (i) into (2.11r), gives
0r
p (l)
Integrating (l) )(zfp (m)
where )(zf is “constant” of integration. We now have two solutions for the pressure p: (k) and
(m). Equating the two, gives
)()( zfCzrgp o (n)
One side of (n) shows that the pressure depends on z only while the other side shows that it depends on r and z. This, of course, is a contradiction. The only possibility for reconciling this is by requiring that
g(r) = C (o)
where C is a constant. Substituting (o) into (j)
Cdr
dr
dr
d
rz
p zv1 (p)
Thus the axial pressure gradient in the tube is constant. Equation (p) can now be integrated to give the axial velocity distribution. Integrating once
12
2
1Cr
zd
pd
dr
dr zv
Separating variables and integrating again
212 ln
4
1CrCr
zd
pdzv (q)
where 1C and 2C are constants of integration. The two boundary conditions on zv are
,0)( iz rv 0)( oz rv (r)
Equations (q) and (r) give 1C and 2C
,01C 22
4
1or
zd
pdC
Substituting into (q)
PROBLEM 3.7 (continued)
)(4
1 22oz rr
zd
pdv (3.12)
Since temperature distribution is obtained by solving the energy equation, we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)
2
2
2
2
2 0
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrP v
vv (2.24)
where the dissipation function is given by (2.25)
22
0
2
0
222
0
1
0
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
Equations (2.24) and (2.25) show that the determination of temperature distribution requires the
determination of the velocity components rv , v and .zv
With the velocity distribution determined we return to the energy equation (2.24) and the dissipation function (2.25). We note that for a long tube at uniform surface temperature with no end effects, axial temperature variation can be neglected. Thus
02
2
z
T
z
T (s)
Substituting (b), (c), (g), (h) and (s) into (2.24)
01
dr
dTr
dr
d
rk (t)
Using (b), (c) and (g) into (2.25) gives the dissipation function for this flow
2
dr
d zv
Substituting the velocity solution (3.11) into the above, gives 2
2
2
1r
zd
pd
(u)
Using (u) to eliminate in (t) and rearranging, we obtain
dr
dTr
dr
d 3
2
4
1r
zd
pd
k (3.13)
Integrating the above twice
434
2
ln64
1CrCr
zd
pd
kT (v)
PROBLEM 3.7 (continued)
Two boundary conditions are needed to evaluate the constants of integration 3C and 4C . They
are:
0)0(
dr
dT and oo TrT )( (w)
Equations (v) and (w) give the two constants
,03C 4
2
464
1oo r
zd
pd
kTC
Substituting the above into (v)
4
424
164
o
oo
r
r
zd
pd
k
rTT (3.14a)
This solution can be expressed in dimensionless form as
4
4
241
64
oo
o
r
r
zd
pd
k
r
TT (3.14b)
[b] Surface heat flux )( orq is obtained by applying Fourier’s law
dr
rdTkrq o
o
)()(
Using (3.14) into the above 23
16)(
zd
pdrrq o
o (3.15)
[c] The Nusselt number is defined as
k
hr
k
hDNu o2
(x)
where D is tube diameter. The heat transfer coefficient h is determined using equation (1.10)
dr
rdT
TT
kh o
o
)(
])0([ (y)
Substituting (3.14a) into (y) or
kh
4 (z)
Substituting (z) into (x) 8Nu (3.16)
(iii) Checking. Dimensional check: Each term in (3.12) has units of velocity. Each term in
)14.3( a as units of temperature. Each term in (3.15) has units of .W/m2
Differential equation check: Velocity solution (3.12) satisfies equation (p) and temperature solution (3.14) satisfies (3.13).
PROBLEM 3.7 (continued)
Boundary conditions check: Velocity solution (3.12) satisfies boundary conditions (r) and temperature solution (3.14) satisfies boundary conditions (w).
Limiting check: (i) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary. Setting
0/ dzdp in (3.12) gives .0zv
(ii) If pressure is uniform ( 0/ dzdp ) the fluid will be stationary and no dissipation takes place
and thus surface heat transfer should vanish Setting 0/ dzdp in (3.15) gives .0)( orq
(iii) Global conservation of energy. Heat transfer rate leaving the tube must be equal to the rate of work required to pump the fluid. Pump work for a tube section of length L is
QppW )( 21 (z-1)
Where
1p = upstream pressure
2p = downstream pressure
Q = volumetric flow rate, given by
or
v0
2 rdrQ z
Substituting (3.12) into the above and integrating
4
8or
dz
dpQ (z-2)
Combining (z-1) and (z-2))
)(8
21
4
ppdz
dprW o (z-3)
Work per unit area W is
Lr
WW
o2
Substituting (z-3) into the aboveL
pp
dz
dprW o )(
16
213
(z-4)
However
dz
dp
L
pp )( 21
Combining this result with (z-4) gives 23
16 dz
dprW o
This result is identical to surface heat transfer rate given in (3.15)
PROBLEM 3.7 (continued)
(5) Comments. (i) The assumption of a long tube with negligible end effects is a key factor in simplifying the problem. This assumption eliminates the z-coordinate as a variable and results in governing equations that are ordinary. Assuming parallel streamlines is equivalent to neglecting end effects since it leads to the same mathematical simplifications.
(ii) Solution (3.14) shows that maximum temperature occurs at the center .0r
(iii) The Nusselt number is constant independent of Reynolds and Prandtl numbers.
PROBLEM 3.8
A liquid film of thickness H flows down an inclined plane due to
gravity. The plane is maintained at uniform temperature oT and
the free film surface is insulated. Assume incompressible laminar
flow and neglect axial variation of velocity and temperature and
end effects.
[a] Show that the axial velocity is given by
2
22
2
1sin
H
y
H
ygHu
[b] Taking into consideration dissipation, determine the heat flux at the inclined plane.
(1) Observations. (i) Fluid motion is driven by gravity. (ii) No velocity and temperature variation in the axial direction. (iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. (v) Temperature distribution depends on the velocity distribution. (vi) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply Navier-Stokes and continuity equations in Cartesian coordinates to determine the flow field. Apply the energy equation to determine temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) no end effects, (v) no motion in the z-direction and (vi) velocity and temperature do not vary in the axial direction. .
(ii) Analysis. [a] The Navier-Stokes equation for constant properties are
x-direction:2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
y-direction:2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
Note that the sign of the gravity force in (2.10y) is negative since this component of gravity points in the negative y-direction. That is
singg x and cosgg y
These equations are simplified as follows:
0zx
u
tt
u v
PROBLEM 3.8 (continued)
2
2
y
u
x
pg
y
uxv (a)
2
2
2
2
yxy
pg
yxu y
vvvv
v (b)
The continuity equation introduces additional simplifications. For two-dimensional constant properties, the continuity equations gives
0yx
u v
Since 0x
u, it follows that
0y
v (c)
IntegratingCv (d)
However, the no slip condition at the wall gives
0(0)v
Applying this condition to (d) gives
C = 0 Thus
0v (d) Substituting (d) into (a) and (b)
2
2
0y
u
x
pg x (e)
y
pg y0 (f)
Integrating (f)
1Cygp y (g)
The pressure at the free surface is atmospheric. Thus
apHp )(
(g) gives
1CHgp ya
HgpC ya1
Substituting into (g)
ay pyHgp )( (h)
This result shows that pressure is independent of x. thus
PROBLEM 3.8 (continued)
0x
p (i)
Substituting into (e)
xgy
u2
2
(j)
Integrating twice
32
2
2CyC
ygu x (k)
The two boundary conditions are
(1) 0)0(u
(2) 0)(
dy
Hdu
These boundary conditions give
Hg
C x2 and 03C
Substituting into (k)
2
22
2
1
H
y
H
yHgu x (l)
However
singg x (m)
Introducing (m) into (l)
2
22
2
1sin
H
y
H
ygHu (n)
[b] The energy equation is the starting point for determining the temperature distribution. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc p v (2.19b)
where the dissipation function is given by (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
Equation (2.19b) simplifies to
02
2
y
Tk (o)
PROBLEM 3.8 (continued)
The dissipation function simplifies to
2
y
u (p)
Using (n) into (p) 2
2
2
1)sin(
H
ygH (q)
Introducing (q) into (o)
01)sin(
22
2
2
H
ygH
dy
Tdk (r)
Integrating (r) twice
212
4322
1232
)sin(AyA
H
y
H
yy
k
gHT (s)
The two boundary conditions are
(1) oTT )0(
(2) 0)(
dy
HdT
The two boundary conditions give the constants 1A and 2A
23
1 )sin(3
gk
HA
oTA2
Substituting into (s) and rearranging the result in dimensionless form
4
4
3
3
2
2
24 12
1
3
1
2
1
3
1
)sin( H
y
H
y
H
y
H
y
k
gH
TT o (t)
Surface heat flux is determined using Fourier’s law
dy
dTkq
)0()0(
3
)sin()0(
23 gHq (u)
(iii) Checking. Dimensional check: Equation (n) has units of velocity. Each term in (t) is dimensionless. Each term in (u) has units of heat flux.
Differential equation check: Velocity solution (n) satisfies equation (j) and temperature solution (t) satisfies (r).
Boundary conditions check: Velocity solution (n) satisfies the two boundary conditions following equation (k) and temperature solution (t) satisfies the boundary conditions following equation (s).
PROBLEM 3.8 (continued)
Limiting check: (i) If gravity or inclination angle vanishes the fluid will be stationary. Setting 0g or 0 in (n) gives .0u
(ii) If gravity or inclination angle vanishes the fluid will be stationary, dissipation will also
vanish and the temperature will be uniform throughout equal to .oT Setting 0g or 0 in
(t) gives .oTT
(5) Comments. (i) Neglecting axial variation of u and T are key simplifying assumptions in this problem. (ii) Surface heat flux is negative since all energy generated due to friction must leave through the inclined plane.
PROBLEM 3.9
A liquid film of thickness H flows down an inclined plane due to gravity. The plane exchanges
heat by convection with an ambient fluid. The heat transfer coefficient is h and the ambient
temperature is .T The inclined surface is insulated. Assume incompressible laminar flow and
neglect axial variation of velocity and temperature and end effects.
[a] Show that the axial velocity is given by
2
22
2
1sin
H
y
H
ygHu
[b] Taking into consideration dissipation, determine the heat flux
at the free surface.
(1) Observations. (i) Fluid motion is driven by gravity. (ii) No velocity and temperature variation in the axial direction. (iii) The fluid is incompressible (constant density). (iv) Heat is generated due to viscous dissipation. (v) Temperature distribution depends on the velocity distribution. (vi) the inclined surface is at specified temperature and the free surface exchanges heat by convection with the ambient. (vii) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply Navier-Stokes and continuity equations in Cartesian coordinates to determine the flow field. Apply the energy equation to determine temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) no end effects, (v) no motion in the z-direction, (vi) uniform heat transfer coefficient and ambient temperature and (vii) velocity and temperature do not vary in the axial direction.
(ii) Analysis. [a] The Navier-Stokes equati on for constant properties are
x-direction:2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
y-direction:2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
Note that the sign of the gravity force in (2.10y) is negative since this component of gravity points in the negative y-direction. That is
singg x and cosgg y
These equations are simplified as follows:
0zx
u
tt
u v
PROBLEM 3.9 (continued)
2
2
y
u
x
pg
y
uxv (a)
2
2
2
2
yxy
pg
yxu y
vvvv
v (b)
The continuity equation introduces additional simplifications. For two-dimensional constant properties, the continuity equations gives
0yx
u v
Since 0x
u, it follows that
0y
v (c)
Integrating
Cv
However, the no slip condition at the wall gives
0(0)v
Applying this condition gives
C = 0 Thus
0v (d) Substituting (d) into (a) and (b)
2
2
0y
u
x
pg x (e)
y
pg y0 (f)
Integrating (f)
1Cygp y (g)
The pressure at the free surface is atmospheric. Thus
apHp )(
(g) gives
1CHgp ya
HgpC ya1
Substituting into (g)
ay pyHgp )( (h)
This result shows that pressure is independent of x. thus
PROBLEM 3.9 (continued)
0x
p (i)
Substituting into (e)
xgy
u2
2
(j)
Integrating twice
32
2
2CyC
ygu x (k)
The two boundary conditions are
(1) 0)0(u
(2) 0)(
dy
Hdu
These boundary conditions give
Hg
C x2 and 03C
Substituting into (k)
2
22
2
1
H
y
H
yHgu x (l)
However
singg x (m)
Introducing (m) into (l)
2
22
2
1sin
H
y
H
ygHu (n)
[b] The energy equation is the starting point for determining the temperature distribution. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc p v (2.19b)
where the dissipation function is given by (2.17)
2
3
2
2222222
z
w
yx
u
z
u
x
w
y
w
zxy
u
z
w
yx
u
v
vvv
(2.17)
PROBLEM 3.9 (continued)
Equation (2.19b) simplifies to
02
2
y
Tk (o)
The dissipation function simplifies to
2
y
u (p)
Using (n) into (p) 2
2
2
1)sin(
H
ygH (q)
Introducing (q) into (o)
01)sin(
22
2
2
H
ygH
dy
Tdk (r)
Integrating (r) twice
232
4322
1232
)sin(CyC
H
y
H
yy
k
gHT (s)
The two boundary conditions are
(1) 0)0(
dy
dT
(2) THThdy
HdTk )(
)(
The two boundary conditions give the constants 3C and 4C
03C
43
)sin( 32
4
H
h
k
k
HgTC
Substituting into (s) and rearranging
4
4
3
3
2
224
12
1
3
1
2
1
4
1
3
1)sin(
H
y
H
y
H
y
hH
k
k
gHTT (t)
Rewriting (t) in dimensionless form
4
4
3
3
2
2
24 12
1
3
1
2
1
4
1
3
1
)sin( H
y
H
y
H
y
hH
k
k
gH
TT (v)
Surface heat flux is determined using Fourier’s law
dy
HdTkHq
)()(
(t) into the above
PROBLEM 3.9 (continued)
3
)sin()(
23 gHHq (w)
(iii) Checking. Dimensional check: Equation (n) has units of velocity. Each term in (t) has units of temperature. Each term in (w) has units of flux. Each term in (v ) and (y) is dimensionless.
Differential equation check: Velocity solution (n) satisfies e quation (j) and temperature solution (t) satisfies (r).
Boundary conditions check: Velocity solution (n) and temperature solution (t) satisfy their respective boundary conditions.
Limiting check: (i) If gravity or inclination angle vanishes the fluid will be stationary. Setting 0g or 0 in (n) gives .0u
(ii) If gravity or inclination angle vanishes the fluid will be stationary, dissipation will also
vanish and the temperature will be uniform throughout equal to .T Setting 0g or 0 in
(t) gives .TT
Qualitative check: All dissipation heat must leave the free surface. Equation (w) shows that surface heat flux is positive (leaving the fluid).
(5) Comments. (i) Neglecting axial variation in u and T are key simplifying assumptions in this problem. (ii) Distinction should be made between the ambient heat transfer coefficient and the liquid film heat transfer coefficient. They are not identical.
PROBLEM 3.10
Lubricating oil fills the clearance space of between a rotating
shaft and its housing. The shaft radius is cm6ir and housing
radius is cm.1.6ir The angular velocity of the shaft is
RPM3000 and the housing temperature is C.40ooT Taking
into consideration dissipation, determine the maximum oil
temperature and the heat flux at the housing. Neglect end effects
and assume incompressible laminar flow. Properties of lubricating
oil are: CW/m138.0 ok and skg/m0356.0 .
Solution
This problem is identical to Example 3.3. The maximum temperature occurs at the shaft surface
irr . This temperature is given in equation (3.21)
1)/ln(2)/()/(1
2
4)( 2
2
2 iooi
oi
ioi rrrr
rr
r
kTrT (3.21)
where
k thermal conductivity = CW/m138.0 o
ir shaft radius = 0.06 m
or housing radius = 0.061 m
oT housing temperature = 40 Co
viscosity = skg/m0356.0
angular velocity = 3000 RPM = 100 rad/s
The heat flux at the housing surface per unit length is given in (3.22)
2
2
)/(1
)(4)(
oi
io
rr
rrq (3.22)
Computation.
Substituting into (3.21)
1m)(06.0
m)(061.0ln2(m/m))061.0/06.0(
(m/m))061.0/06.0(1
)(0.06(m)s/rad)(100(2
C)W/m138.0(4
skg/m0356.0C)(40)( 22
2
22oo
max irTT
C8.86 omaxT
Substituting into (3.22)
22
2
(m/m)1)(0.06/0.061
(0.06)(m))s/rad)(100(m)/s0.0356)(kg(4)( orq = 4,888
3s
mkg = 4,888
m
W
Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes.
PROBLEM 3.10 (continued)
(ii) Temperature rise of the lubricating oil and energy dissipation increase as the clearance between the shaft and the housing is decreased. This is evident from equations (3.22) which
show that in the limit as 1)/( oi rr , .q
(iii) The energy dissipated due to friction is considerable. The heat dissipated for a housing radius of 1.0 cm is 48.9 W.
PROBLEM 3.11
Consider lubrication oil in the clearance between a shaft and its
housing. The radius of the shaft is ir and that of the housing is or .
The shaft rotates with an angular velocity and its housing
exchanges heat by convection with the ambient fluid. The heat
transfer coefficient is h and the ambient temperature is .T Taking
into consideration dissipation, determine the maximum
temperature of the oil and surface heat flux at the housing. Assume
incompressible laminar flow and neglect end effects.
(1) Observations. (i) Fluid motion is driven by shaft rotation (ii) The housing is stationary. (iii) Axial variation in velocity and temperature are negligible for a very long shaft. (iv) Velocity and temperature do not vary with angular position. (v) The fluid is incompressible (constant density). (vi) Heat ge nerated by viscous dissipation is removed from the oil at the housing. (vii) No heat is conducted through the shaft. (viii) The maximum temperature occurs at the shaft. (ix) Heat fl ux at the housing is determined from temperature distribution and Fourier’s law of conduction. (x) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution in the oil.
(3) Solution Plan. This problem is similar to Example 3.3. The flow field is given by (3.18). The energy equation is given in (l) of Example (3.18). The only difference between this problem and Example 3.3 is the boundary condition at the housing. Housing heat flux can be determined using the solution to temperature distribution and Fourier’s law.
(4) Plan Execution
(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform ambient temperature and (vii) negligible gravitational effect.
(ii) Analysis. The velocity distribution is given by
1)/(
)/()/()/()(2
2
io
iiio
i rr
rrrrrr
r
rv (3.18)
Following Example 3.3, the energy equation is
01
dr
dTr
dr
d
rk (a)
and the solution is
432
2
2
2
ln1
)/(1
2
4)( CrC
rrr
r
krT
oi
i (b)
where 3C and 4C are the integration constants. Two boundary conditions are needed to
determine 3C and 4C . They are:
PROBLEM 3.11 (continued)
(1) 0)(
dr
rdT i
(2) TrThdr
rdTk o
o )()(
These boundary conditions give the two constants
2
2
2
2
3
1
)/(1
2
2ioi
i
rrr
r
kC
and
o
o
i
o
ooi
i
ohr
kr
r
r
hr
k
rr
r
krTC ln212
)/(1
2
4 2
22
2
2
24
Substituting the above into (b) and rearranging
1)/(12)/ln()/(2)/()/(1
2
4)( 222
2
22
2
io
o
oioo
oi
i
o
i rrhr
krrrrrr
rr
r
r
r
kTrT (c)
This solution can be expressed in dimensionless form as
112ln24
1
)/(1
2
)(2
2
2
2
2
2
2
2
2
2
i
o
o
oo
i
o
o
i
oi
ir
r
hr
k
r
r
r
r
r
r
r
r
rr
r
k
TrT (d)
The maximum temperature is at the shaft’s surface. Setting irr in (c) gives
22
2
2max )/(1)/(2)/ln(21)/(1
2
4)( oioi
o
oi
oi
ii rrrr
hr
krr
rr
r
kTrTT (e)
Energy generated due to dissipation per unit shaft length, ),( orq is determined by applying
Fourier’s law at the housing. Thus
dr
rdTkrrq o
oo
)(2)(
Using (c), the above gives
2
2
)/(1
)(4)(
oi
io
rr
rrq (f)
(iii) Checking. Dimensional check: each term in solutions (c) has units of temperature. (f) has the correct units of W/m. and (3.20b) is dimensionless. Equation (3.22) has the correct units of W/m.
Boundary conditions check: Temperature solution (c) satisfies the two boundary conditions on temperature.
Limiting check: (i) If the shaft does not rotate no dissipation takes place and thus surface heat
transfer should vanish. Setting 0 in (f) gives .0)( orq
PROBLEM 3.11 (continued)
(ii) If the fluid is inviscid no dissipation takes place and thus surface heat transfer should vanish.
Setting 0 in (f) gives .0)( orq
Global conservation of energy. Heat transfer rate from the hous ing must equal to work required to overcome friction at the shaft’s surface. The rate of shaft work per unit length is given by
iii rrrW )(2 (p)
where
W = work done on the fluid per unit shaft length
)( ir = shearing stress at the shaft’s surface, given by
irrrdr
dri
vv 0)( (q)
Substituting (3.18) into the above
2)/(12)(
oi
irr
r (r)
Combining (p) and (r) and rearranging, gives
2
2
)/(1
)(4
oi
i
rr
rW (s)
This result is identical to surface heat transfer rate given in (f)
(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes.
(ii) Temperature rise of the lubricatig oil and energy dissipation increase as the clearance between the shaft and the housing is decreased.
(iii) Velocity distributions are governed by a single parameter )./( oi rr Temperature distribution
is governed by two parameters: )/( oi rr and the Biot number ./ khro
(iv) Heat transfer rate at the housing, equation (f), is identical to that of Example 3.3 given in equation (3.22). This is not surprising since dissipation energy for constant property fluids is a function of flow field. Thus, dissipation energy is the same for problems with identical flow fields even if they have different temperature boundary conditions.
PROBLEM 3.12
A rod of radius ir is placed concentrically inside a sleeve of radius
or . Incompressible fluid fills the clearance between the rod and the
sleeve. The sleeve is maintained at uniform temperature oT while
rotating with constant angular velocity . Taking into consideration
dissipation, determine the maximum fluid temperature and surface
heat flux at the sleeve. Assume incompressible laminar flow and
neglect end effects.
(1) Observations. (i) Fluid motion is driven by sleeve rotation (ii) The shaft is stationary. (iii) Axial variation in velocity and temperature are negligible for a very long shaft. (iv) Velocity and temperature do not vary with angular position. (v) The fluid is incompressible (constant density). (vi) Heat ge nerated by viscous dissipation is removed from the oil at the housing. (vii) No heat is conducted through the shaft. (viii) The maximum temperature occurs at the shaft. (ix) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution in the oil.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Use the energy equation to determine temperature distribution. Apply Fourier’s law at the housing to determine the rate of energy generated by dissipation.
(4) Plan Execution
(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) uniform surface temperature and (vii) negligible gravitational effect.
(ii) Analysis. Temperature distribution is obtained by solving the energy equation. Thus we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)
2
2
2
2
2 0
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrP v
vv (2.24)
where the dissipation function is given by (2.25)
22
0
2
0
222
0
1
0
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
The solution to (2.24) requires the determination of the velocity components rv , v and .zv
These are determined by solving the continuity and the Navier-Stokes equations in cylindrical coordinates. The continuity equation is given by equation (2.4)
011
zrzr
rrrt
vvv (2.4)
For constant density
PROBLEM 3.12 (continued)
0zrt
(a)
For axisymmetric flow
0 (b)
For a long shaft with no end effects axial changes are negligible
0z
zv (c)
Substituting (a)-(c) into (2.4)
0rrdr
dv (d)
Integrating (d)
Cr rv (e)
To determine the constant of integration C we apply the no-slip boundary condition at the housing surface
0)( orrv (f)
Equations (e) and (f) give 0C
Substituting into (e)
0rv (g)
Since the radial component rv vanishes everywhere, it follows that the streamlines are
concentric circles. To determine the tangential velocity v we apply the Navier-Stokes equation
in the -directions, equation (2.11 )
2
2
22
2
2
21)(
11
zrrr
rrr
p
rg
tzrrr
r
r
vvvv
vvv
vvvvvv zr
(2.11 )
For steady state
0t
(h)
Neglecting gravity and applying (b),(c), (g) and (h), equation (2.11 ) simplifies to
0)(1
vrdr
d
rdr
d (i)
Integrating (i) twice
r
Cr
C 21
2v (j)
where 1C and 2C are constants of integration. The two boundary conditions on v are
PROBLEM 3.12 (continued)
0)( irv , oo rr )(v (j)
Boundary conditions (j) give 1C and 2C
22
2
1
2
io
o
rr
rC ,
22
22
2
io
oi
rr
rrC (k)
Substituting (k) into (j) and rearranging in dimensionless form, gives
r
r
r
r
rr
rr
r
r i
ioi
oi
o2)/(1
)/()(v (l)
We now return to the energy equation (2.24) and the dissipation function (2.25). Using (b), (c), (g) and (h), equation (2.24) simplifies to
01
dr
dTr
dr
d
rk (m)
The dissipation function (2.25) is simplified using (b), (c) and (g)
2
0
rdr
d vv
Substituting the velocity solution (l) into the above, gives
4
2
2
21
)/(1
2
rrr
r
oi
i (n)
Combining (n) and (m) and rearranging, we obtain
dr
dTr
dr
d3
2
2
21
)/(1
2
rrr
r
koi
i (o)
Integrating (o) twice
432
2
2
2
ln1
)/(1
2
4)( CrC
rrr
r
krT
oi
i (n)
where 3C and 4C are the integration constants. Two boundary conditions are needed to
determine 3C and 4C . They are:
0)(
dr
rdT i and oo TrT )( (o)
Equations (n) and (o) give the two constants
2
2
2
2
3
1
)/(1
2
2ioi
i
rrr
r
kC
and
o
iooi
io r
rrrr
r
kTC ln
21
)/(1
2
4 22
2
2
2
4
PROBLEM 3.12 (continued)
Substituting the above into (o)
)/ln(2)/()/()/(1
2
4)( 22
2
2rrrrrr
rr
r
kTrT oioi
oi
io
(p)
This solution can be expressed in dimensionless form as
)/ln(2)/()/(
)/(1
2
4
)( 22
2
2
rrrrrr
rr
r
k
TrToioi
oi
i
o (q)
The maximum temperature is at the shaft’s surface. Setting irr in (p) gives
)/ln(2)/(1)/(1
2
4)( 2
2
2 iooi
oi
ioi rrrr
rr
r
kTrT (r)
Energy generated due to dissipation per unit shaft length, ),( orq is determined by applying
Fourier’s law at the housing. Thus
dr
rdTkrrq o
oo
)(2)(
Using (q), the above gives
2
2
)/(1
)(4)(
oi
io
rr
rrq (s)
(iii) Checking. Dimensional check: each term in solutions (l) and (q) is dimensionless. Equation (s) has the correct units of W/m.
Differential equation check: Velocity solution (l) satisfies equation (i) and temperature solution (p) satisfies (o).
Boundary conditions check: Velocity solution l) satisfies boundary conditions (j) and temperature solution (p) satisfies boundary conditions (o).
Limiting check: (i) If sleeve does not rotate the fluid will be stationary. Setting 0 in (l)
gives .0v
(ii) If the sleeve does not rotate no dissipation takes place and thus surface heat transfer should
vanish. Setting 0 in (s) gives .0)( orq
(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes.
(ii) Temperature rise of the lubricating oil and energy dissipation increase as the clearance between the shaft and the housing is decreased. This is evident from equation (s) which show
that in the limit as 1)/( oi rr , .q
(iii) Velocity and temperature distribu tion are governed by a single parameter )./( oi rr
PROBLEM 3.13
A hollow shaft of outer radius or rotates with constant angular
velocity while immersed in an infinite fluid at uniform
temperature .T Taking into consideration dissipation, determine
surface temperature and heat flux. Assume incompressible laminar
flow and neglect end effects.
(1) Observations. (i) Fluid motion is driven by shaft rotation (ii) Axial variation in velocity and temperature are negligible for a very long shaft. (iii) Velocity, pressure and temperature do not vary with angular position. (iv) The fluid is incompressible (constant density). (v) Heat ge nerated by viscous dissipation is conducted radially. (vi) The determination of surface temperature and heat flux requires the determination of temperature distribution in the rotating fluid. (vii) Use cylindrical coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution in the rotating fluid.
(3) Solution Plan. Apply continuity and Navier-Stokes equations in cylindrical coordinates to determine the flow field. Use the energy equation to determine temperature distribution.
(4) Plan Execution.
(i) Assumptions. (I) Steady state, (ii) laminar flow, (iii) axisymmetric flow, (iv) constant properties (density, viscosity and conductivity), (v) no end effects, (vi) no angular and axial variation of velocity, pressure and temperature and (vii) negligible gravitational effect.
(ii) Analysis. Temperature distribution is obtained by solving the energy equation. Thus we begin the analysis with the energy equation. The energy equation in cylindrical coordinates for constant properties is given by (2.24)
2
2
2
2
2 0
11
z
TT
rr
Tr
rrk
z
TT
rr
T
t
Tc zrP v
vv (2.24)
where the dissipation function is given by (2.25)
22
0
2
0
222
0
1
0
12
122
rzzr
rrrzrrr
zrz
rzrr
vvvv
vvvvvvv
(2.25)
The solution to (2.24) requires the determination of the velocity components rv , v and .zv
These are determined by solving the continuity and the Navier-Stokes equations in cylindrical coordinates. The continuity equation is given by equation (2.4)
011
zrzr
rrrt
vvv (2.4)
For constant density
PROBLEM 3.13 (continued)
0zrt
(a)
For axisymmetric flow
0 (b)
For a long shaft with no end effects axial changes are negligible
0z
zv (c)
Substituting (a)-(c) into (2.4)
0rrdr
dv (d)
Integrating (d)
Cr rv (e)
To determine the constant of integration C we apply the no-slip boundary condition at the housing surface
0)( irrv (f)
Equations (e) and (f) give 0C
Substituting into (e)
0rv (g)
Since the radial component rv vanishes everywhere, it follows that the streamlines are
concentric circles. To determine the tangential velocity v we apply the Navier-Stokes equation
in the -direction, equation (2.11 )
2
2
22
2
2
21)(
11
zrrr
rrr
p
rg
tzrrr
r
r
vvvv
vvv
vvvvvv zr
(2.11 )
For steady state
0t
(h)
Neglecting gravity and applying (b),(c), (g) and (h), equation (2.11 ) simplifies to
0)(1
vrdr
d
rdr
d (i)
Integrating (3.17) twice
r
Cr
C 21
2v (j)
where 1C and 2C are constants of integration. The two boundary conditions on v are
PROBLEM 3.13 (continued)
oo rr )(v , 0)(v
These boundary conditions give 1C and 2C
01C , 22 orC (k)
Substituting (k) into (j) and rearranging in dimensionless form, gives
r
r
r
r o
o
)(v (l)
We now return to the energy equation (2.24) and the dissipation function (2.25). Using (b), (c), (g) and (h), equation (2.24) simplifies to
01
dr
dTr
dr
d
rk (m)
The dissipation function (2.25) is simplified using (b), (c) and (g)
2
0
rdr
d vv
Substituting the velocity solution (l) into the above, gives
4
22 1
2r
ro (n)
Combining (m) and (n) and rearranging
dr
dTr
dr
d3
22 1
2r
rk
o (o)
Integrating (o) twice
432
22 ln
12
4)( CrC
rr
krT o (p)
where 3C and 4C are the integration constants. Two boundary conditions are needed to
determine 3C and 4C . They are:
(1) )(T finite
(2) TT )(
Boundary condition (1) gives
03C
Boundary condition (2) gives
TC4
Substituting the above into (n)
PROBLEM 3.13 (continued)
2
22 1
24
)(r
rk
TrT o (q)
This solution can be expressed in dimensionless form as
2
2
2)(
)(
r
r
rk
TrT o
o
(r)
Surface temperature obtained by setting orr in (q)
2)()( oo rk
TrT (s)
Surface heat flux per unit shaft length, ),( orq is determined by applying Fourier’s law at orr
dr
rdTkrrq o
oo
)(2)(
Using (q) the above gives 2)(4)( oo rrq (t)
(iii) Checking. Dimensional check: each term in solutions (l) and (r) is dimensionless.
Equation (q) has the correct units of Co and equation (t) has units of W/m.
Differential equation check: Velocity solution (l) satisfies equation (i) and temperature solution (q) satisfies (o).
Boundary conditions check: Velocity solution (l) and temperature solution (q) satisfy their respective boundary conditions.
Limiting check: (i) If shaft does not rotate the fluid will be stationary. Setting 0 in (l) gives
.0v
(ii) If the shaft does not rotate no dissipation takes place and fluid temperature should be uniform
equal to T and surface heat transfer should vanish. Setting 0 in (q) gives TrT )( .
Setting 0 in (t) gives .0)( orq
Global conservation of energy: Surface heat transfer rate must equal to work required to overcome friction at the shaft’s surface. The rate of shaft work per unit length is given by
ooo rrrW )(2 (u)
where
W = work done on the fluid per unit shaft length
)( or = shearing stress at the shaft’s surface, given by
orrrdr
dro
vv0)( (v)
Substituting (l) into the above
2)( ir (w)
PROBLEM 3.13 (continued)
Combining (u) and (w) 2)(4 orW (y)
This result is identical to surface heat transfer rate given in (t).
(5) Comments. (i) The key simplifying assumption is axisymmetry. This resulted in concentric streamlines with vanishing normal velocity and angular changes. (ii) Surface temperature is lowest in the entire region. (iii) Heat flow direct ion is negative. (iii) This problem was solved by specifying two conditions at infinity. If surface temperature is specified instead of fluid temperature at infinity, the solution determines ).(T
PROBLEM 3.14
Two large porous plates are separated by a distance H. An incompressible fluid fills the channel
formed by the plates. The lower plate is maintained at temperature 1T and the upper plate at 2T .
An axial pressure gradient dxdp / is applied to the
fluid to set it in motion. A fluid at temperature 1T is
injected through the lower plate with a normal
velocity .ov Fluid is removed along the upper plate
at velocity .ov The injected fluid is identical to the
channel fluid. Neglect gravity, dissipation and
axial variation of temperature.
[a] Show that the axial velocity is given by
H
y
Hv
yv
dx
dp
v
Hu
o
o
o )/exp(1
)/exp(11
[b] Determine surface heat flux at each plate.
(1) Observations. (i) Axial pressure gradient sets fluid in motion. (ii) The fluid is incompressible. (iii) The flow field is determined by solving the continuity and Navier-Stokes equations. (iv) Energy equation gives the temperature distribution. (v) Fourier’s law and temperature distribution give surface heat flux. (vi) Axial variation of temperature is neglected. (viii) Use Cartesian coordinates.
(2) Problem Definition. Determine the velocity and temperature distribution.
(3) Solution Plan. Apply continuity and Navier-Stokes equations to determine the flow field. Apply the energy equation to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (i) Steady state, (ii) laminar flow, (iii) constant properties (density, viscosity and conductivity), (iv) infinite plates, (v) no axial variation of temperature, (vi) negligible gravitational and (vii) negligible dissipation.
(ii) Analysis. [a] Velocity distributi on. Applying continuity and the Navier-Stokes equations. We begin with the continuity equation in Cartesian coordinates
0z
w
yx
u
zw
yxu
t
vv (2.2b)
For constant density
0zyxt
(a)
Since plates are infinite
0wzx
(b)
Substituting (a) and (b) into (2.2b), gives
PROBLEM 3.14 (continued)
0y
v (c)
Integrating (c) Cv (d)
where C is constant of integration. The boundary condition on v is
ovv )0( (e)
Equations (d) and (e) give
ovC
Substituting into (d)
ovv (f)
To determine the horizontal component u we apply the Navier-Stokes equations (2.10)
2
2
2
2
2
2
z
u
y
u
x
u
x
pg
z
uw
y
u
x
uu
t
uxv (2.10x)
2
2
2
2
2
2
zyxy
pg
zw
yxu
ty
vvvvvv
vv (2.10y)
These equations are simplified as follows: Steady state
0t
u (g)
Negligible gravity effect
0yx gg (h)
Substituting (b) and (f)-(h) into (2.10x) and (2.10y) gives
2
21
dy
ud
x
p
dy
duvo (i)
and
0y
p (j)
where ./ Equation (j) shows that pressure does not vary in the y-direction and thus it can
be a function of x or constant. Integrating (i) once
dy
duy
xd
pdCu
vo 11 (k)
To solve this equation it is rewritten first as
)( yQuPdy
du (l)
where
ovP , 1
1)( Cy
dx
dpyQ (m)
PROBLEM 3.14 (continued)
The solution to (l) is
2)( CdyyQeeuPdyPdy
(n)
substituting (m) into (n) and evaluating the integrals
yo
o
o
o
veCC
vvy
vdx
dpu
)/(21)/(
1 (o)
The constants 1C and 2C are determined form the boundary conditions on u
(1) 0)0(u
(2) 0)(Hu
These conditions give
Hvedx
dpHC
oHov
1
1/1 (p)
Substituting (l) into (k)
1
1/2 Ho
ov
edx
dpH
vC (q)
(p) and (q) into (o)
H
y
e
e
dx
dp
v
Hu
Ho
yo
ov
v
1
11/
/
(r)
[b] Temperature distribution and Nusselt number. With the velocity distribution determined, the energy equation is applied to determine temperature distribution. The energy equation for constant properties is given by (2.19b)
2
2
2
2
2
2
z
T
y
T
x
Tk
z
Tw
y
T
x
Tu
t
Tc v (2.19b)
Neglecting dissipation and using (a) and (b) this equation is simplified
2
2
dy
Td
dy
dTov (s)
where pck / is thermal diffusivity. To Integrate (s) it is rewritten as
dy
dy
dT
dy
dTd
ov
Integrating
3lnln Cydy
dT ov
Rewriting
PROBLEM 3.14 (continued)
ydy
dT
C
ov
3
1ln
/3
yoeCdy
dT v
Integrating again
4/
3 CeCTy
o
ov
v (t)
where 3C and 4C are constants of integration. The two boundary conditions on (q) are
(1) 1)0( TT
(2) 2)( THT
These boundary conditions and solution (t) give
1/12
3 H
o
oe
TTC
v
v (u)
1
)(/
/12
24 H
H
o
o
e
eTTTC
v
v
(w)
Substituting (u) into (w) into (t) and rearranging the result in dimensionless form, give
1]/exp[
)]/)(/exp[(]/exp[)( 212
H
HyHHTTTT
o
oo
v
vv (x)
This result in now expressed in terms of the Prandtl number. Note that
Pr
Substituting into (x)
1])/(exp[
)]/()/(exp[])/(exp[)( 212
PrHv
HyPrHvPrHvTTTT
o
oo (y)
This result can be rearranged in dimensionless form as
1])/(exp[
)]/()/(exp[])/(exp[
21
2
PrHv
HyPrHPrHv
TT
TT
o
oo v (z)
Surface heat flux is determined by applying Fourier’s law at each plate
dy
dTkq
)0()0( (z-1)
dy
HdTkHq
)()( (z-2)
Substituting (y) into (z-1) and (z-2)
PROBLEM 3.14 (continued)
1])/(exp[)0( 21
PrHv
TTkPr
vq
o
o (z-3)
1])/(exp[
])/(exp[)()( 21
PrHv
PrHvTTkPr
vHq
o
oo (z-4)
Expressed in dimensionless form, (K) and (L) become
1])/(exp[
1
)(
)0(
21
PrHvTTPrk
v
q
oo
(z-5)
1])/(exp[
])/(exp[
)(
)(
21
PrHv
PrHv
TTPrkv
Hq
o
o
o
(z-6)
(iii) Checking. Dimensional check: Each term in (z), (z-5), and (z-6) is dimensionless. The exponents of all exponentials are dimensionless.
Differential equation check: Velocity solution (r) satisfies e quation (i) and temperature solution (x) satisfies (s).
Boundary conditions check: Velocity solution (r) and temperature solution (x) satisfy their respective boundary conditions.
Limiting check: (i) If there is no axial pressure gradient, the fluid will be stationary. Set 0/ dxdp in (r) gives .0)(yu
(ii) If 21 TT , surface heat flux will vanish. Set 21 TT in (z-3) and (z-4) gives
.0)()0( Hqq
(5) Comments. (i) Treating the plate as infinite is one of the key simplifying assumptions. This eliminates the x-coordinate as a variable and results in governing equations that are ordinary. (ii) Contrary to expectation, the axial velocity plays no role in the temperature distribution This is evident from energy equation (s) and temperature solutions (y).
(iii) According to the dimensionless form of solutions (z), (z -5) and (z-6), the problems is characterized by the following single dimensionless parameter
PrHvo
Note that this parameter is a combination of Prandtl number, geometry, injection velocity, and kinematic viscosity , a property.
(iv) Taking the ratio of (N) to (M) provides a co mparison of surface heat flux at the two plates
])/(exp[)0(
)(PrHv
q
Hqo
This result indicates that heat flux at the upper plate is higher than that at the lower plate.
PROBLEM 4.1
Put a check mark in the appropriate column for each of the following statements.
Statement true false may be
(a) 0)/()/( yxu v is valid for transient flow. x
(b) The y-momentum equation is neglected in boundary layer flow.
x
(c) Boundary layer equatio ns are valid for all Reynolds numbers.
x
(d) Pressure gradient is zero outside the boundary layer.
x
(e)
2
2
2
2
y
u
x
u for a streamlined body. x
(f) In boundary layer flow fluid velocity upstream of an object is undisturbed.
x
(g) Axial pressure gradient is neglected in boundary layer flow. x
(i) Axial conduction is neglected in boundary layer flow.
x
PROBLEM 4.2
Examine the three governing equations, (2.2), (4.13) and (4.18) for two-dimensional, constant
properties, laminar boundary layer flow.
[a] How many dependent variables do these equations have?
[b] How is the pressure p determined?
[c] If streamlines are parallel in the boundary layer what terms will vanish?
[d] Can (2.2) and (4.13) be solved for the velocity field u and v independently of the energy
equation (4.18)?
Solution: The three equations are:
0yx
u v (2.2)
:2
21
y
u
dx
dp
y
u
x
uu v (4.13)
2
2
y
T
y
T
x
Tu v (4.18)
[a] There are four dependent variables: ,u ,v ,p and T.
[b] The pressure p is determined form the inviscid external flow. The solution to the Navier-
Stokes equations with 0 (Euler’s equations of motion) for the flow over the same object
gives p .
[c] The following terms will vanish if streamlines are parallel:
If 0v then 0y
v. When this is substituted into (2.2) gives 0
x
u. Equations (4.13) and
(4.18) become:
2
210
y
u
dx
dp
and
2
2
y
T
x
Tu
[d] Equations (2.2) and (4.13) can be solved for u and v independently of energy equation (4.18)
since temperature does not enter in (2.2) and (4.13).
PROBLEM 4.3
Air flows over a semi-infinite plate with a free stream velocity V = 0.4 m/s and a free stream
temperature C.20oT The plate is maintained at C.60osT Can boundary layer
approximations for the flow and temperature fields be applied at:
[a ] location x = 1.5 mm?[b] location x = 15 mm?
Note: Evaluate air properties at the average film temperature .2/)( TTT sf
(1) Observations. (i) This is forced convection flow over a streamlined body. (ii) Viscous (velocity) boundary layer approximations can be made if the Reynolds number Rex > 100. (iii) Thermal (temperature) boundary layer approximations can be made if the Peclet number Pex = Rex Pr > 100. (iv) The Reynolds number decreases as the distance along the plate is decreased.
(2) Problem Definition. Determine the local Reynolds and Peclet numbers at the locations of interest.
(3) Solution Plan. Write the definitions of Rex and Pex and calculate their values at x = 1.5 mm and x = 15 mm.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional flow and (5) streamlined body.
(ii) Analysis. The local Reynolds and Peclet numbers are defined as number
Rex =V x
(a)
andPex = Rex Pr (b)
where
Pex = local Peclet number Pr = Prandtl number Rex = local Reynolds number
V = upstream velocity = 0.4 m/s x = distance from the leading edge of the plate, m
= kinematic viscosity, m2 /s
Properties are evaluated at the film temperature Tf defined as
Tf = T Ts
2 (c)
where
Tf = film temperature, oCTs = surface temperature = 60oC
T = free stream temperature = 20oC
t
y
Ts
T V,
x
PROBLEM 4.3 (continued)
(iii) Computations. Use (c) to calculate Tf
Tf = (60 + 20)( oC)/2 = 40oC
At this temperature Appendix C gives
Pr = 0.71
= 16.96 10-6 m2/s
[a] At x = 1.5 mm = 0.0015 m, equation (a) gives
Rex =)s/m(1096.16
)m(0015.0)s/m(4.026
= 35.4
Since this value is less than 100, it follows that velocity boundary layer approximations can not be made. Using (b) to calculate Pex
Pex = 35.4 x 0.71 = 25.1
Since this is smaller than 100, it follows that temperature boundary layer approximations can not be made.
[b] At x = 15 mm = 0.015 m the Reynolds number and Peclet number of part [a] will increase by a factor of 10. Thus
Rex = 354
and
Pex = 251
Since both Rex and Pex are larger than 100, it follows that both velocity and temperature boundary layer approximations can be made at this location.
(iv) Checking. Dimensional check: Computations showed that equation (a) is dimensionally consistent.
(5) Comments. The Reynolds and Peclet numbers should be calculated to establish if boundary layer approximations can be made.
PROBLEM 4.4
Water at C25o flows with uniform velocity V = 2 m/s over a streamlined object. The object is 8
cm long and its surface is maintained at C.85osT Use scaling to: [a] show that 1/ L , [b]
evaluate the inertia termsx
uu and
y
uv , and [c] evaluate the viscous terms
2
2
x
uand .
2
2
y
u
(1) Observations. (i) The surface is streamlined. (ii) The fluid is water. (iii) Inertia and viscous effects can be estimated using scaling. (iv) If a viscous term is small compared to inertia, it can
be neglected. (v) Properties should be evaluated at the film temperature .2/)( TTT sf
(2) Problem Definition. Estimate the magnitudes ,/ Lx
uu ,
y
uv ,
2
2
x
uand .
2
2
y
u
(3) Solution Plan. Use the scaling to estimate the magnitudes of the above terms.
(4) Plan Execution.
(i) Assumptions. (1) Continuum and (2) streamlined surface.
(ii) Analysis. Scales:
u V (a)
Lx (b)
y (c)
Scaling of continuity equation gives a scale for v
LVv (4.7d)
[a] A balance between inertia term x
uu and viscous term
2
2
y
u gives
LReL
1 (4.14b)
where LRe is the Reynolds number defined as
LVReL (d)
where
08.0L m
2V m/s
kinematic viscosity, /sm2
[b] Scales for inertia terms are:
L
VV
x
uu (e)
and
PROBLEM 4.4 (continued)
V
y
uvv
Using (4.7d) to eliminate v in the above, gives
L
VV
y
uv (f)
Thus the two inertia terms are of the same magnitude.
[c] The two viscous terms are scaled as: First term:
2
2
x
u2L
V (g)
Second term:
2
2
y
u2
V (h)
For 1/ L , comparing (g) with (h) shows that
2
2
x
u<<
2
2
y
u (4.2)
(iii) Computation. Properties are evaluated at the film temperature fT
C552
)C)(2585(
2
ooTT
T sf
s
m105116.0
26
[a] Substituting into (d)
5
2610127.3
/s)m(105116.0
)m(08.0)m/s(2Re
Equation (4.14b) gives
00179.010127.3
1
5L
m000143.0)m(08.000179.0
Thus 1/ L .
[b] The two inertia terms are of the same order of magnitude, given by (e) or (f)
)m/s(50)m(08.0
)m/s(2)m/s(2 2
x
uu
[c] The first viscous term is given by (g)
PROBLEM 4.4 (continued)
2
2
x
u 2m/s00016.0)m()08.0(
)m/s(2)/sm(105116.0
22
26
The second viscous term is given by (h)
2
2
y
u 2m/s50)m()000143.0(
)m/s(2)/sm(105116.0
22
26
Thus the first viscous term can be neglected since it is much smaller than the second term.
(iv) Checking: Dimensional check: Inertia and viscous terms have the same units.
(5) Comments. Computation showed that the second viscous term is identical to the inertia term. This is consequence of equating the two terms to derive (4.14b).
PROBLEM 4.5
Water at C25o flows with uniform velocity V = 2 m/s over a streamlined object. The object is 8
cm long and its surface is maintained at C.85osT Use scaling to: [a] show that 1/ Lt , [b]
evaluate the convection terms x
Tu and
y
Tv , and [c] evaluate the conduction terms
2
2
x
T
and .2
2
y
T
(1) Observations. (i) The surface is streamlined. (ii) The fluid is water. (iii) Convection and conduction effects can be estimated using scaling. (iv) If a conduction term is small compared to
convection, it can be neglected. (v) The scale for Lt / depends on whether t or .t
(vi) Properties should be evaluated at the film temperature .2/)( TTT sf
(2) Problem Definition. Estimate the magnitudes ,/ Ltx
Tu ,
y
Tv ,
2
2
x
Tand .
2
2
y
T
(3) Solution Plan. Use the scaling to estimate the magnitudes of the above terms.
(4) Plan Execution.
(i) Assumptions. (1) Continuum and (2) streamlined surface.
(ii) Analysis. Scales: Two scales are used for u depending on whether t or .t
Case (1): t .
u V (a)
Lx (b)
ty (c)
TTT s (d)
Scaling of continuity equation gives a scale for v
vL
V t (4.23)
Case (2): t .
u tV (4.29)
Scaling of the continuity equation gives
vL
V t2
(4.30)
[a] Scaling of Lt / depends on whether t or .t The two cases are considered.
Case (1): t . A balance between convection x
Tu and normal conduction
2
2
y
T gives
PROBLEM 4.5 (continued)
LePrRL
t 1 (4.24)
where Pr is the Prandtl number and LRe is the Reynolds number defined as
LVReL (f)
where
08.0L m
2V m/s
kinematic viscosity, /sm2
Case (2): t . A balance between convection x
Tu and normal conduction
2
2
y
T gives
LeRPrL
t1/3
1 (4.31)
[b] Scales for convection terms.
Case (1): t .
L
TTV
x
Tu s (g)
and
t
s TT
y
Tvv
Using (4.23) to eliminate v in the above, gives
L
TTV
y
T sv (h)
Thus the two convection terms are of the same magnitude.
Case (2): t . Using (d) and (4.29)
L
TTV
x
Tu st (i)
where is scaled as
LRe
L (4.14b)
Substituting (4.14b) into (i)
2L
TTReV
x
Tu s
Lt (j)
PROBLEM 4.5 (continued)
Similarly, using (c), (d), (4.14b) and (4.30) give the same result of (j).
[c] Scale for conduction terms are: Axial conduction is scaled as
2
2
x
T2L
TTs (i)
normal conduction is scaled as
2
2
y
T2t
s TT (j)
For 1/ Lt , comparing (i) with (j) shows that
2
2
x
T<<
2
2
y
T (4.2)
(iii) Computation. Properties are evaluated at the film temperature fT
C552
)C)(2585(
2
ooTT
T sf
Pr = 3.27
s
m10566.1
27
s
m105116.0
26
[a] Substituting into (f)
5
2610127.3
/s)m(105116.0
)m(08.0)m/s(2Re
Case (1): t . Equation (4.24) gives
4
51089.9
10127.3)(27.3(
1
L
t
m1091.7)m(08.01089.9 54t
Thus 1/1 L for this case.
Case (2): t . Equation (4.31) gives
3
53/110205.1
10127.3)27.3(
1
L
t
m1064.9)m(08.010205.1 53t
[b] The two convection terms are of the same order of magnitude for both cases.
PROBLEM 4.5 (continued)
Case (1): t . Using (g)
s
C1500
)m(08.0
)C)(2585()m/s(2
oo
x
Tu
Case (2): t . Noting that m1064.9 5t , and using (j)
1011)m()08.0(
)C)(2085(10127.3)m(1064.9)m/s(2
22
o55
x
Tu
[c] Axial conduction is given by (i)
2
2
x
T
s
C10175.1
)m()08.0(
)C)(2585()/sm(10566.1
o4
22
o27
Normal conduction is given by(j). Two cases are considered:
Case (1): t .
m1091.7 5t
2
2
y
T2s
m1502
)m()1091.7(
)C)(2585()/sm(10566.1
227
o27
Case (2): t .
m1064.9 5t
2
2
y
T2s
m1011
)m()1064.9(
)C)(2585()/sm(10566.1
227
o27
Thus axial conduction can be neglected since it is much smaller than normal conduction.
(iv) Checking: Dimensional check: convection and conduction terms have the same units.
(5) Comments. (i) 1/ Lt . (ii) Axial conduction is small compared to normal conduction.
(iii) Computation showed that the normal conduction is identical to the convection term for both cases. This is a consequence of equating the two terms to derive (4.24) and (4.31).
PROBLEM 4.6
Atmospheric air at 25 Co flows over a surface at 115 Co . The free stream velocity is 10 m/s.
[a] Calculate the Eckert number.
[b] Use scale analysis to show that the dissipation term 2)/( yu is small compared to the
conduction term )./( 22 yTk
(1) Observations. (i) The fluid is air. (ii) Dissipation and conduction can be estimated using scaling. (iii) Dissipation is negligible if the Eckert number is small compared to unity.
(2) Problem Definition. Compute the Eckert number. Estimate the magnitudes of dissipation and conduction terms.
(3) Solution Plan. Use the definition of Eckert number to compute its value. Apply scaling to estimate the dissipation and conduction terms.
(4) Plan Execution.
(i) Assumptions. Continuum.
(ii) Analysis. [a] The Eckert number is defined as
)(
2
TTc
VE
sp
(a)
where
pc specific heat, Ckg-
Jo
E Eckert number
sT = surface temperature = 115 Co
T = free stream temperature = 25 Co
[b] The ratio of dissipation to condu ction is estimated using scaling.
Dissipation =
2
y
u
Conduction =2
2
y
Tk
where
k thermal conductivity, Cm-
Wo
viscosity,s-m
kg
Scales:
PROBLEM 4.6 (continued)
u V (a)
Lx (b)
y (c)
TTT s (d)
The dissipation and conduction terms are estimated using scales (a)-(d).
Dissipation
2V
(e)
Conduction2
TTk s (f)
where
sT surface temperature = 115 Co
T free stream temperature = 25 Co
Taking the ratio of (e) to (f)
Conduction
nDissipatio
)(
2
TTk
V
s
(g)
(iii) Computation. Properties are evaluated at the film temperature fT
C702
)C)(25115(
2
ooTT
T sf
pc 1008.7Ckg-
Jo
02922.0kCm-
Wo
61047.20s-m
kg
[a] Substituting into (a)
0011.0-sJ
-mkg0011.0
C))(25115(C)J/kg(7.1008
)m/s()10(2
2
oo
22
E
[b]Conduction
nDissipatio00078.0
)C)(25115)(CW/m(02922.0
)s/m()10()skg/m(1047.20oo
2226
(iv) Checking: Dimensional check: The ratio of dissipation to conduction in (g) must be dimensionless:
PROBLEM 4.6 (continued)
1s-W
m-kg
)C)()(CW/m(
)s/m()skg/m(3
2
oo
222
TTk
V
s
Limiting check: If 0V there is no dissipation. Setting 0V in (e) gives the correct result.
(5) Comments. (i) Since the Eckert number is small compared to unity dissipation is negligible. (ii) Dissipation is negligible compared to conduction.
PROBLEM 4.7
Air at C20o flows over a streamlined surface with a free stream velocity of m/s10 . Use scale
analysis to determine the boundary layer thickness at a distance of 80 cm from the leading edge.
(1) Observations. (i) The surface is streamlined. (ii) The fluid is air.
(2) Problem Definition. Estimate the magnitude of boundary layer thickness at a specified
distance from the leading edge.
(3) Solution Plan. Use the scaling to estimate the magnitudes of .
(4) Plan Execution.
(i) Assumptions. (1) Continuum and (2) streamlined surface.
(ii) Analysis. Scale analysis gives as
LReL
1 (4.14b)
where LRe is the Reynolds number defined as
LVReL (a)
where
8.0L m
10V m/s
61009.15 /sm2
(iii) Computation. Substituting into (a)
5
2610302.5
/s)m(1009.15
)m(8.0)m/s(10Re
Equation (4.14b) gives
00137.010302.5
1
5L
m0011.0)m(8.000137.0
(iv) Checking: Dimensional check: The Reynolds number is dimesionles.
(5) Comments. .1/ L
PROBLEM 4.8
In boundary layer flow, pressure gradient normal to the
flow direction is assumed zero. That is .0/ yp If this is
correct, how do you explain lift on the wing of an airplane
in flight?
Solution
Although 0/ yp in boundary layer flow, .0/ xp Thus, pressure distribution around the
surface is accounted for. Lift is th e net force acting on a surface in the y-direction.
PROBLEM 4.9
Derive an equation describing the vertical velocity component v at the edge of the boundary
layer for two-dimensional incompressible flow over a semi-infinite flat plate. Assume laminar
flow. Compare your result with scaling estimate.
(1) Observations. (i) This is a forced convection problem over a flat plate. (ii) At the edge of the
thermal boundary layer, the axial velocity is Vu . (iii) Blasius solution gives the distribution
of the velocity components u(x,y) and v(x,y). (iv) Scaling gives an estimate of v(x,y).
(2) Problem Definition. Determine the vertical velocity at the edge of the viscous boundary layer, ).,(xv
(3) Solution Plan. Use Blasius solution for ),( yxv . Evaluate ),( yxv at .y
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Blasius so lution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex < 5 105), (6) uniform upstream velocity and temperature, (7) flat plate and (8) negligible changes in kinetic and potential energy.
(ii) Analysis. Blasius solution for ),( yxv is given by
fd
df
xVV 2
1v (4.43)
where V free steam velocity and is kinematic viscosity. The variable and f are defined as
x
Vyyx ),( (4.41)
d
df
V
u (4.42)
At the edge of the thermal boundary layer, y
1d
df
V
u (a)
According to Blasius solution, at 1d
df
V
u, Table 4.1,
8),(x (b)
and27923.6)(f (c)
Substituting(b) and (c) into (4.43)
PROBLEM 4.9 (continued)
)27923.68(2
1
xVV
)v(x,
xReV
8604.0)v(x, (d)
where the Reynolds number is defined as xV
Rex .
Scaling estimate of the velocity component v in the boundary layer is given by
vx
V (4.7d)
where
xRex
1 (4.16)
substituting (4.16) into (4.7d) and rearranging
V
v
xRe
1 (e)
(iii) Checking. Dimensional check: Since the Reynolds number is dimensionless it follows that each term in (d) is dimensionless.
Qualitative check: The layer thickness increases with distance x. Solution (c) confirms this behavior.
(5) Comments. Recalling that Blasius solution gives
xRex
2.5 (4.46)
Comparing (d) with (4.46) shows that t for 8.9Pr . Examination of Fig. 4.6 shows that
t for all fluids with 0.1Pr and that t for 0.1Pr .
PROBLEM 4.10
Sketch the streamlines in boundary layer flow over a semi-infinite flat plate.
Since the flow within the boundary layer is two-dimensional the vertical velocity component does not vanish. Thus stream lines are not parallel.
PROBLEM 4.11
Define the thickness of the velocity boundary layer in Blasius solution as the distance y where
the velocity u = 0.988 V . Derive an expression for /x.
(1) Observations. (i) This is a laminar boundary layer flow problem. (ii) Blasius solution gives the velocity distribution for the flow over a semi-infinite flat plate. (iii) A solution for the boundary layer thickness depends on how the thickness is defined.
(2) Problem Definition. Determine the distance y from the surface of the plate to the location
where the velocity ratio u/V = 0.988.
(3) Solution Plan. Use Blasius solution, Table 4.1, to dete rmine the location of the edge of the
velocity boundary layer where u/V = 0.988.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4)
laminar flow (Rex < 5 105), (5) constant properties, (6) uniform upstream velocity, (7) flat plate and (8) boundary layer flow (Rex > 100).
(ii) Analysis. Blasius solution, Table 4.1, gives the axial velocity ratio u/V as a
function of the variable which is defined as
= yx
V (a)
where
V = free stream velocity, m/sx = axial coordinate, m y = normal coordinate, m
= dimensionless variable
= kinematic viscosity, m2/s
The viscous boundary layer thickness is the value of y where the velocity ratio u/V reaches
an arbitrarily selected value. At f ( ) = u/V = 0.988, Table 47.1 gives = 4.8. This value of
corresponds to the edge of the boundary layer where y = . Thus (a) gives
4.8 = = yx
V =
x
V(b)
Solving for
V
x8.4 (c)
Dividing both sided of (c) by x, rearranging and using the definition of Reynolds number gives
xxVx Re
8.48.4 (d)
(iii) Checking. Dimensional check: in equation (c) should have units of length
PROBLEM 4.11 (continued)
)sm/(
)m()s/m( 2
V
x = m
(5) Comments. (i) The thickness of the viscous boundary layer depends on how it is defined.
Thus, it is not uniquely determined. (ii) Regardless of how is defined, the solution takes the
form of equation (d). Only the cons tant in (d) changes according to how is defined.
PROBLEM 4.12
Water flows over a semi-infinite plate with an upstream velocity of 0.2
m/s. Blasius solution is used to calculate at three locations along the
plate. Results are tabulated. Are these results valid? Explain.
(1) Observations. (i) Blasius solution is valid for laminar boundary layer flow over a semi-infinite plate. (ii) The transition Reynolds number from laminar to
turbulent flow is 5105 . (iii) Boundary layer approximations are valid if the Reynolds number is
greater than 100.
(2) Problem Definition. Determine the Reynolds number at each location.
(3) Solution Plan. Determine the Reynolds number at each location to establish the applicability of Blasius solution. Where applicable, use Bl asius result to determine the boundary layer thickness.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Blasius solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex < 5 105), (6) viscous boundary layer flow (Rex > 100), (7) uniform upstream velocity, (8)
flat plate, (9) negligible changes in kinetic and potential energy and (10) no buoyancy ( = 0 or g= 0).
(ii) Analysis. The Reynolds number is computed to establish if the flow is laminar and if boundary layer approximations can be made. The Reynolds number is defined as
Rex =xV
(a)
where
Rex = Reynolds number
V = upstream velocity = 0.2 m/s x = distance from the leading edge of the plate, m
= kinematic viscosity = 0.5116 610 m2 /s
To determine if the flow is laminar or turbulent, compare the Reynolds number with the
transition Reynolds number. For flow over a flat plate the transition Reynolds number txRe is
5105txRe (b)
The flow is laminar if Rex < txRe . Viscous boundary layer approximations are valid for
100xRe (c)
Blasius solution for the boundary layer thickness is
xRex
2.5 (d)
x(cm) (cm)
300 1.441
40 0.526
0.01 0.0083
V
y
x0
PROBLEM 4.12 (continued)
(iii) Computations. Evaluating the Reynolds number at x = 300 cm, equation (a) gives
6
2610173.1
)/sm(105116.0
)m(3)m/s(2.0xRe
Since this value is larger than the transition Reynolds number given in (b), it follows that the flow is turbulent and thus Blasius solu tion (d) does not apply. Using (d) gives
cm441.1m01441.0)m(310173.1
2.5
6
Although this is the reported value for , it is incorrect.
Evaluating the Reynolds number at x = 40 cm, equation (a) gives
5
26105637.1
)/sm(105116.0
)m(4.0)m/s(2.0xRe
Since this value is smaller than the transition Reynolds number given in (b), and since it is larger than 100, it follows that the flow is laminar and thus Blasius solution (d) is applicable. Using (d) gives
cm526.0m00526.0)m(4.010173.1
2.5
6
Thus the reported value for is correct.
Evaluating the Reynolds number at x = 0.01 cm, equation (a) gives
39)/sm(105116.0
)m(0001.0)m/s(2.026xRe
Since this value is smaller than 100, it follows that boundary layer approximations are not valid and thus Blasius solution (d) doe s not apply. Using (d) gives
cm0083.0m000083.0)m(0001.039
2.5
Although this is the reported value for , it is incorrect
(iv) Checking. Dimensional check: Computations showed that equations (a) and (d) are dimensionally correct.
(5) Comments. In applying Blasius results it is important to verify that the conditions leading to Blasius solution are satisfied.
PROBLEM 4.13
Consider laminar boundary layer flow over a semi-infinite flat plate. Evaluate the wall shearing
stress at the leading edge. Comment on your answer. Is it valid? If not explain why.
(1) Observations. (i) This is an external flow problem over a flat plate. (ii) Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable. (iii) Of interest is the value of the local stress at the leading edge of the plate.
(2) Problem Definition. Determine the local wall shearing stress for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Use Blasius solution for the local wall stress. Evaluate wall stress at the leading edge.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Blasius so lution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex < 5 105), (6) viscous boundary layer flow (Rex > 100, (7) uniform upstream velocity and (8) flat plate.
(ii) Analysis. Blasius solution gives
o =x
VV33206.0 (4.47)
where
V = free stream velocity, m/s x = axial distance measured from the leading edge, m
viscosity, mkg/s
= kinematic viscosity, m2/s
o wall shearing stress, 2N/m
Noting that all quantities in equation (4.47) are constant except the variable x, (4.47) can be rewritten as
x
o
constant (a)
(iii) Computations. To determine the shearing stress at the leading edge, set x = 0 in (a)
0
constant)0(o =
(iv) Checking. Dimensional check: Units of in (4.47) should be N/m2:
222 m
N
ms
kg
m)(/s)m(
m/s)(m/s)()mkg/s(
x
VVo
PROBLEM 4.13 (continued)
(5) Comments. The shearing stress cannot be infinite. This suggests that Blasius solution is not valid at the leading edge. One of the assumptions leading to Blasius solution is Rex > 100. However, at the leading edge x = 0, the Reynolds number is given by
Rex =V x
= 0
Therefore, Blasius solution ca nnot be used to determine o at the leading edge. In fact the
solution breaks down at small values of x where the corresponding local Reynolds number and Peclet number are smaller than 100.
PROBLEM 4.14
Water at 20 Co flows over a m2m2 plate with a free stream velocity of 0.18 m/s. Determine
the force needed to hold the plate in place. Assume laminar boundary layer flow.
(1) Observations. (i) This is an external flow problem over a flat plate. (ii) The force needed to hold the plate in place is equal to the total shearing force by the fluid on the plate, (iii) Integration of wall shear over the surface gives the total shearing force. (iv) Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable.
(2) Problem Definition. Determine the local wall shearing stress for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Use Blasius solution for the local wall stress. Integrate shearing force over the total surface area of plate.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Blasius so lution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar flow
(Rex < 5 105), (6) viscous boundary layer flow (Rex > 100, (7) uniform upstream velocity and (8) flat plate.
(ii) Analysis. Since shearing stress is not uniform, shearing force must be integrated over the area to obtain the total restraining force. Thus
L
o WdxxF0
)( (a)
where
L plate length = 2 m W plate width = 2 m
x distance along surface, m
o wall shearing stress, 2N/m
The local shearing stress is given by Blasius solution
o =x
VV33206.0 (4.47)
where
V = free stream velocity = 0.18 m/s
viscosity = 310003.1 mkg/s
= kinematic viscosity = 610004.1 m2/s
Substituting (4.47) into (a)
L
dxxV
WVF0
2/133206.0
Evaluating the integral
x
dxF
L
W
V
PROBLEM 4.14 (continued)
LVWVF 66412.0 (b)
(iii) Computations. Equation (b) gives
s
mkg144.0
/s)m(10004.1
(m/s)2(m)18.0)2(m)m)0.18(m/skg/s)(10003.1(66412.0
26
3F
F = 0.144 newton
(iv) Checking. Dimensional check: Computations showed that units of force are correct.
(5) Comments. According to (4.47) shearing stress decreases with distance from the leading
edge. Thus doubling the length of the plate increases the total force by a factor of 2 . This is
evident in equation (b) which shows that LF .
PROBLEM 4.15
Consider Blasius solution for uniform flow over a semi-infinite plate. Put a check mark in the
appropriate column for each of the following statements.
Solution.
Statement true False Undetermined
(a)0
dx
dpbecause the flow is laminar. x (1)
(b) Wall shearing stress increases with distance from the leading edge of plate.
x (2)
(c) Solution is not valid for 100xRe . x
(d) Solution is not valid for 5105xRe . x
(e) The solution is valid for 100xRe . x(3)
(f) Boundary layer thickness is uniquely defined. x
(g) Solution is not valid for a curved plate. x(4)
(h) The solution for the wall shear at the leading edge (x = 0) is not valid.
x(5)
(i) The plate does not disturb upstream flow. x
(j) Solution is not valid for 5105xRe . x(3)
(1) In boundary layer flow 0dx
dp for flat plate only.
(2) See equation (7.12).
(3) Blasius solution is valid for 5105100 xRe .
(4) For a curved surface 0dx
dp.
(5) 0xRe at x = 0. Boundary layer approximation is not valid at 100xRe .
PROBLEM 4.16
Imagine a cold fluid flowing over a thin hot plate. Using your intuition, would you expect the
fluid just upstream of the plate to experience a temperature rise due to conduction from the hot plate? How do you explain the assumption in Pohlhausen's solution that fluid temperature is
unaffected by the plate and therefore T(0, y) = T ?
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Of interest is the region where the upstream fluid reaches the leading edge of the plate. (iii) The fluid is heated by the plate. (iv) Heat fr om the plate is conducted through the fluid in all directions. (v) Pohlhausen’s solution assumes that heat is not conducted upstream from the plate and therefore fluid temperature at the leading edge is the same as upstream temperature.
(2) Problem Definition. Determine the conditions under which axial conduction in force convection flow can be neglected.
(3) Solution. In reality heat from the plate is conducted in all directions and thus one would expect the temperature of the incoming fluid to be affected by the presence of the plate. This becomes more obvious if one imagines that fluid velocity is decreased and in the limit the fluid becomes stationary. Clearly, for a stationary fluid heat is conducted from the plate through the fluid in all directions. Howeve r, as the velocity of the fluid increases, energy conducted upstream is carried downstream by convection. This tends to minimize the effect of axial conduction. and eventually may be ignored when compared with normal conduction. The condition for this approximation is
Pe = Rex Pr > 100 (a)where
Pe = Peclet number Pr = Prandtl number Rex = local Reynolds number
Pohlhausen’s solution is based on the above boundary layer approximation. By neglecting axial conduction the temperature of the incoming fluid at the leading edge is assumed to be the same as the free stream temperature. That is
T(0,y) = Twhere
T = temperature distribution in the fluid, oC
T = free stream temperature, oCy = coordinate normal to plate, m
x
y
00 Ts
T
V T(0,y) = ?
t
PROBLEM 4.17
Consider laminar boundary layer flow over a semi-infinite flat plat. The plate is maintained at
uniform temperature .sT Assume constant properties and take into consideration dissipation.
[a] Does Blasius solution apply to this case? Explain.
[b] Does Pohlhausen’s solution apply to this case? Explain.
Solution
[a] Blasius solution is based on the assumption that properties are constant independent of temperature. Thus, Blasius solu tion applies to this problem.
[b] Pohlhausen’s solution neglects dissipati on. Thus it does not apply to this problem.
PROBLEM 4.18
A fluid with Prandtl number 9.8 flows over a semi-infinite flat plat. The plate is maintained at
uniform surface temperature. Derive an expression for the variation of the thermal boundary
layer thickness with distance along the plate. Assume steady state laminar boundary layer flow
with constant properties and neglect dissipation. Express your result in dimensionless form.
(1) Observations. (i) This is a forced convection problem over a flat plate. (ii) At the edge of the thermal boundary layer, fluid
temperature is TT . (iii) Pohlhausen’s
solution gives the temperature distribution in the boundary layer. (iv) The thermal boundary
layer thickness t increases with distance from the leading edge. (v) t depends on the Prandtl
number.
(2) Problem Definition. Determine the variation of the thermal boundary layer thickness with distance for a fluid with Prandtl number of 9.8.
(3) Solution Plan. Use Pohlhausen’s graphical solution for the temperature distribution of laminar flow over a flat plate to determine the thermal boundary layer thickness.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation.
(ii) Analysis. Pohlhausen’s solution is shown in graphical form in fig. 4.6
Fig. 4.6
0.1
6.0
4.0
2.0
2 4 6 8 12 1410
01.0Pr
1.0
)(7.0 air110100
0
8.0
xVy
s
s
TT
TT
PROBLEM 4.18 (continued)
At the edge of the thermal boundary layer, ty the dimensionless temperature in Fig. 4.6 is
equal to unity
1s
s
TT
TT (a)
For a fluid with 8.9Pr the corresponding value of ),( tx is obtained from Fig. 4.6 as
8.2x
Vt (b)
Solving (b) for t
V
xt 8.2 (c)
Expressing this result in dimensionless form
x
t
RexVx
8.28.2 (d)
(iii) Checking. Dimensional check: Since the Reynolds number is dimensionless it follows that each term in (d) is dimensionless.
Qualitative check: The thermal boundary layer thickness increases with distance x. Solution (c) confirms this behavior.
(5) Comments. Recalling that Blasius solution gives
xRex
2.5 (4.46)
Comparing (d) with (4.46) shows that t for 8.9Pr . Examination of Fig. 4.6 shows that
t for all fluids with 0.1Pr and that t for 0.1Pr .
PROBLEM 4.19
Use Pohlhausen’s solution to determine the heat flux at the leading edge of a plate. Comment on your answer. Is it valid? If not explain why.
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable. (iii) Of interest is the va lue of the local heat flux at the leading edge of the plate. (iv) Knowing the local transfer coefficient and using Newton’s law, gives the heat flux
(2) Problem Definition. Determine the local heat transfer coefficient for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Apply Newton’s law of cooling to determine the local heat flux. Use Pohlhausen’s solution for the local heat transfer coefficient. Apply the solution at the leading edge.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. Application of Newton’s law gives
][)( TThxq ss (a)
where
h = local heat transfer coefficient, W/m2-oC
sq = surface heat flux, W/m2
Ts = surface temperature, oC
T = free stream temperature, oC
Based on the above assumptions, Pohlhausen’s soluti on for the local heat transfer coefficient is
d
d
x
Vkh
)0( (b)
where
k = thermal conductivity, W/m- oC
V = free stream velocity, m/s T = temperature variable, oC
)/()( ss TTTT , dimensionless temperature
x = axial distance measured from the leading edge, m
= kinematic viscosity, m2/s
00
sTx
y
t
V
T
PROBLEM 4.19 (continued)
= yV
x, dimensionless variable
y = vertical coordinate, m
Noting that all quantities in equation (a) are constant except the variable x, (a) can be rewritten as
h = x
constant (c)
(iii) Computations. To determine the heat transfer coefficient at the leading edge, set x = 0 in (c)
h(0) = cons ttan
0 =
Substituting into (a)
][)( TTxq ss
(iv) Checking. Dimensional check: Units of h in (a) should be W/m2-oC:
h = k (W/m-oC)V
x
m s
m s m
/
/2d
d )0((1/1) = W/m2-oC
(5) Comments. Physically, the heat transfer coefficient cannot be infinite. This suggests that Pohlhausen’s solution is not valid at the leading edge. One of the assumptions leading to Pohlhausen’s solution is Rex > 100. However, at the leading edge x = 0, the Reynolds number is given by
Rex =V x
= 0
Therefore, Pohlhausen’s solution cannot be used to determine h at the leading edge. In fact the solution breaks down at small values of x where the corresponding local Reynolds number and Peclet number are smaller than 100.
PROBLEM 4.20
Consider laminar boundary layer flow over a semi-infinite flat plate at uniform surface
temperature .sT The free stream velocity is V and the Prandtl number is 0.1. Determine the
temperature gradient at the surface ./)0,( dyxdT
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution is assumed to be applicable. (iii) Of interest is the value of the normal temperature gradient at the surface.
(2) Problem Definition. Determine the normal temperature gradient at the surface, ,/)0,( dyxdT for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Apply Pohlhausen’s solution for the temperature distribution for laminar flow over a flat plate.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. Surface temperature gradient is given by
dy
d
d
d
d
dT
y
yxT ),( (a)
where
x
Vyyx ),( (4.41)
s
s
TT
TT (4.58)
Substituting (4.41) and (4.58) into (a)
d
d
x
VTT
y
yxTs )(
),( (b)
Evaluating (b) at 0y gives the temperature gradient at the surface
d
d
x
VTT
y
xTs
)0()(
)0,( (c)
The problem reduces to determining ./)0( dd This is given by Pohlhausen’s solution as
PROBLEM 4.20 (continued)
dd
fdd
d
Pr
Pr
2
2
332.0)0( (4.64)
where )(f is given by Blasius solution an d is listed in Table 4.1. The integral in (6.64) must be
evaluated numerically. An alternate approximate method for determining dd /)0( is using
Fig. 4.6 to determine the slope of the 098.0Pr curve.
(iii) Computations.
Numerical integration of (4.64) gives
1568.0)0(
d
d (d)
Substituting into (c)
x
VTT
y
xTs )(1568.0
)0,( (e)
Approximating the Prandtl number of 0.098 by 0.1 and using Fig. 4.6, gives the slope at the wall as
125.0)0(
d
d (f)
(iv) Checking. Dimensional check: Surface temperature gradient in (e) should have units ofC/m:
m
C
)(m/s)m(
)m/s()C/m)((
)0,( o
2
o
x
VTT
y
xTs
Limiting check: Surface temperature gradient should vanish for TTs . Setting TTs in (e)
gives .0/)0,( yxT
Qualitative check: Surface temperature gradient should increase as the free stream velocity is increased. This is confirmed by ©.
Comments. (1) Using numerical integration to evaluate the integral in (4.64) is necessary since this integral cannot be evaluated analytically. (2) As the thermal boundary layer thickness increases surface heat flux, and thus surface temperature gradient, should decrease. This follows from the observation that the thermal boundary layer acts as an insulation layer. Equation (e) show that surface temperature gradient decreases with distance x. (3) Fig. 4.6 provides a reasonable estimate surface temperature gradient.
PROBLEM 4.21
Fluid flows between two parallel plates. It enters with uniform velocity V and temperature .T
The plates are maintained at uniform surface temperature .sT Assume laminar boundary layer
flow at the entrance. Can Pohlhausen solution be applied to determine the heat transfer
coefficient? Explain.
Solution
The velocity distribution in Pohlhausen’s solution is based on Blasius solution. Blasius solution is limited to the flow over a single plate. For a single plate axial pressure gradient is set equal to zero. That is, Blasius solution is based on
0x
p
For flow between parallel plates axial pressure gradient does not vanish. That is
0x
p
In fact, pressure decreases in the flow direction. Thus, Blasius solution does not apply to the flow between parallel plates. It follows that Pohlhausen’s also does not apply.
PROBLEM 4.22
Two identical rectangles, A and B, of dimensions
L1 L2 are drawn on the surface of a semi-infinite
flat plate as shown. Rectangle A is oriented with
side L1 along the leading edge while rectangle B is
oriented with side L2 along the edge. The plate is
maintained at uniform surface temperature.
[a] If the flow over rectangle A is laminar, what is
it for B ?
[b] If the heat transfer rate from plate A is
435 W, what is the rate from plate B ?
(1) Observations. (i) This is an external forced convection problem over two flat plates. (ii) Both plates have the same surf ace area. (iii) For flow over a flat plate, the heat transfer coefficient h decreases with distance from the leading edge. (iv) Since the length in the flow direction is not the same for the two plates, the average heat transfer coefficient is not the same. It follows that the total heat transfer rate is not the same. (v) The flow over a flat plate
is laminar if the Reynolds number is less than 5 105.
(2) Problem Definition. Determine the Reynolds number at the trailing end of plate B.Obtain a solution for the average heat transf er coefficient for laminar forced convection over a flat plate.
(3) Solution Plan. Examine the Reynolds number at the trailing end of plate B to establish if the flow is laminar or turbulent. Use Newton’s law of cooling to determine the heat transfer from each plate. Use Pohlhausen’s solution to obtain a solution for the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream veloc ity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy,
(11) negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation.
(ii) Analysis. [a] To establish if the flow is laminar or turbulent, compare the Reynolds number with the transition Reynolds number. For the flow over a flat plate the transition Reynolds number is
txRe = 5 105 (a)
The flow is considered laminar if Rex < txRe . The Reynolds number for plate B is
1LRe = 1LV (b)
T
sTV B
A2L
1L
1L
2L
viewtop
PROBLEM 4.22 (continued)where
1LRe = Reynolds number at trailing end of plate B
L1 = length of plate B in the flow direction, m
V = upstream velocity, m/s
= kinematic viscosity, m2 /s
Similarly, for plate A
2LRe = 2LV (c)
where
2LRe = Reynolds number at trailing end of plate A
L2 = length of plate A in the flow direction, m
Taking the ratio of (b) and (c) and rearranging
1LRe =2
1
L
L2LRe (d)
Since L1 < L2, equation (d) gives
1LRe < 2LRe (e)
Thus,1LRe <
2LRe <txRe . Since the flow is laminar for plate A, it follows that it is also
laminar for plate B.
[b] Application of Newton' s law of cooling gives
q = h A (Ts - T ) (f) where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oCq = total surface heat transfer rate, W
Ts = surface temperature, oC
T = free stream temperature, oC
Applying (f) to the two plates and taking the ratio of the resulting equations
q
q
h
h
B
A
B
A
(g)
where the subscripts refer to plates A and B. The average heat transfer coefficient for laminar flow over a flat plate is given by Pohlhausen's solution, equations (4.67) and (4.71b)
3/1)(664.0 PrL
Vkh (h)
where
PROBLEM 4.22 (continued)
k = thermal conductivity, W/m-oC Pr = Prandtl number
Applying (h) to plates A and B, noting that L = L2 for A and L = L1 for B, and taking the ratio of the results
h
h
L
L
B
A
2
1
(i)
Substituting into (g) and solving for qB
q qB A
L
L
2
1
(j)
(iii) Computations. With qA = 435 W, equation (j) gives
qB = 435(W)L
L
2
1
Since L2 > L1, it follows that the heat transfer rate from B is greater than that from A.
(iv) Checking. Dimensional check: Units of h in equation (h) should be W/m2-oC
h = k(W/m-oC)V
L )m)(s/m(
s/m2
Pr1/3
= W/m2-oC
Limiting check: For the special case of L1 = L2 (square plate), the heat transfer rate from the two plates should be the same. Setting L1 = L2 in equation (j) gives q qB A .
(5) Learning and Generalizing. To maximize the rate of heat transfer from a flat rectangular plate under laminar flow conditions, the long side of the plate should face flow direction.
PROBLEM 4.23
A semi-infinite plate is divided into four equal sections of one centimeter long each. Free
stream temperature and velocity are uniform and the flow is laminar. The surface is maintained
at uniform temperature. Determine the ratio of the heat transfer rate from the third section to
that from the second section.
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable. (iii) Of interest is the value of the heat transfer rate from a section of the plate at a specified location and of a given width. (iv) Newton’s law of cooling gives the heat transfer rate.
(2) Problem Definition. Determine the local heat transfer coefficient for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Apply Newton’s law of cooling to determine the local heat flux. Use Pohlhausen’s solution for the local heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. Application of Newton’s law to the surface element Wdx gives
WdxTTxhdq s ][)( (a)
where
h = local heat transfer coefficient, W/m2-oC q = surface heat flux, W Ts = surface temperature, oC
T = free stream temperature, oC W = plate width, m
x = axial distance measured from the leading edge, m
The local heat transfer coefficient is given by
d
d
x
Vkxh
)0()( (4.66)
where
k = thermal conductivity, W/m- oC
V = free stream velocity, m/s
T
V
sT
4321
x
1x 2x 3x4x
dx
W
PROBLEM 4.23 (continued)
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= yV
x, dimensionless variable
y = vertical coordinate, m
Substituting (4.66) into (a)
x
dx
d
dVWkTTdq s
)0(][ (b)
Integration of (b) between two values of x gives the heat transfer rate for that section. Application of (b) to sections 3 and 4, give
)()0(
][2)0(
][ 232
3
2
xxd
dVWkTT
x
dx
d
dVWkTTq ss
x
x
(c)
)()0(
][)0(
][ 343
4
3
xxd
dVWkTT
x
dx
d
dVWkTTq ss
x
x
(d)
where
01.02x m
02.03x m
03.04x m
Taking the ratio of (c) and (d)
23
34
2
3
xx
xx
q
q (e)
(iii) Computations. Equation (e) gives
7673.0)m(01.0)m(02.0
)m(02.0)m(03.0
2
3
q
q
(iv) Checking. Dimensional check: Units of q in (c) should be W:
W)m(])[1/1()0(
/sm
(m/s)C)m/W()m()C]([ 342
oo xxd
dVkWTTq s
Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (c) and (d)
gives .032 qq
Qualitative check: The heat transfer rate should increase as the free stream velocity, plate width, or thermal conductivity are increased. This is confirmed by (c) and (d).
Comments. Although each section is rectangular in shape, the same procedure can be followed to determine the heat transfer rate from any configuration drawn on the plate.
PROBLEM 4.24
A fluid at a uniform velocity and temperature flows over a semi-infinite flat plate. The
surface temperature is uniform. Assume laminar boundary layer flow.
[a] What will be the percent change in the local heat transfer coefficient if the free
stream velocity is reduced by a factor of two?
[b] What will be the percent change in the local heat transfer coefficient if the distance from
the leading edge is reduced by a factor of two?
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Of interest is the variation of the local heat transfer coefficient with free stream velocity and distance from the leading edge. (iii) Pohlhausen's solution applies to this problem.
(2) Problem Definition. Obtain a solution for the local heat transfer coefficient and examine its dependency on the free stream velocity and distance from the leading edge.
(3) Solution Plan. Use Pohlhausen's solution for the loca l heat transfer coefficient over a semi-infinite flat plate.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional,
(5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and te mperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible
dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. Pohlhausen's solution for the local heat transfer coefficient gives
d
d
x
Vkh
)0( (a)
where
h = local heat transfer coefficient, W/m2-oCk = thermal conductivity, W/m-oC
V = free stream velocity, m/s T = temperature variable, oCTs = surface temperature, oC
T = free stream temperature, oC
)/()( ss TTTT , dimensionless temperature
x = axial coordinate, measured from the leading edge, m y = vertical coordinate, measured from the surface, m
= kinematic viscosity, m2/s
PROBLEM 4.24 (continued)
= yV
x, dimensionless variable
However, k, and d
)0( are constants. Thus, equation (a) can be rewritten as
x
VCh (b)
where C is a constant. The percent change in h is given by
Percent change = 100 ( 12 hh )/h1
= 100 (h2/h11) (c)
where subscripts 1 and 2 refer to the initial and changed conditions, respectively. Applying (b) to the two locations and substituting into (c)
Percent change = 100 121
12
xV
xV (d)
(iii) Computations.
[a] Percent change in h if the free stream velocity V is reduced by a factor of two:
Applying (d) for 2/12 VV and x1 = x2, gives
Percent change = 100 11
2
V
V100 )12/1( 3.29 %
[b] Percent change in h if the distance x from the leading edge is reduced by a factor of two:
Applying (d) for x2 = x1/2 and 21 VV , gives
Percent change = 100 %4.41)12(10012
1
x
x
(iv) Checking. Dimensional check: Units of h in (a) should be W/m2-oC:
h = k (W/m-oC)V
x
m s
m s m
/
/2d
dT )0(*
(1/1) = W/m2-oC
Qualitative check: As the free stream velocity decreases, the local heat transfer coefficient should decrease. Computations confirm this as indicated by the negative sign in the percent change when free stream velocity is reduced by a factor of two.
(5) Comments. (i) The local heat transfer coefficient increases as the distance from the leading edge is decreased and as the free stream velocity is increased. (ii) Since percent change involves taking ratios, the problem is solved without knowing the nature of the fluid
and the magnitudes of V and x.
PROBLEM 4.25
Use Pohlhausen's solution to derive an expression for the ratio of the thermal boundary layer
thickness for two fluids. The Prandtl number of one fluid is 1.0 and its kinematic viscosity is
012 10 6. m / s2 . The Prandtl number of the second fluid is 100 and its kinematic viscosity is
68 10 6. m / s2 .
(1) Observations. (i) This is an external flow problem. (ii) At the edge of the thermal
boundary layer, ty , fluid temperature
approaches free stream temperature. That is,
TT and 1)/()(*ss TTTTT . (iii)
According to Pohlhausen's solution, Fig. 4.6, the thermal boundary layer thickness depends
on the Prandtl number, free stream velocity V ,
kinematic viscosity and location x.
(2) Problem Definition. Derive an expression for the thermal boundary layer thickness for laminar flow over a semi-infinite flat plate.
(3) Solution Plan. Use Pohlhausen's solution, Fig. 4.6, to determine t at Pr = 1 and Pr = 100.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis and Computations. At the edge of the thermal boundary layer, ty , fluid
temperature is
TxT t),(
and therefore the dimensionless temperature *T is
1)/()()/()(*ssss TTTTTTTTT (a)
where
T = temperature variable, oCTs = surface temperature, oC
T = free stream temperature, oC*T = dimensionless temperature, oC
y = coordinate normal to the plate, m
t = thermal boundary layer thickness = value of y where *T = 1, m
Let the subscripts 1 and 2 de note conditions corresponding to Pr = 1 and Pr = 100, respectively.
0
0
2 4 6 8 10 12 14
0.2
0.4
0.6
0.8
1.0
T T
T T
s
s
y V x/
Pr = 0,01
0.1
1.01
00
0.71 (air)
Fig. 7.2
PROBLEM 4.25 (continued)
Using Fig. 4.6 at Pr = 1, the value of corresponding to *T = 1 is
2.51
11x
Vt (b)
Similarly, for Pr = 100
5.12
22x
Vt (c)
where
V = free stream velocity, m/s x = coordinate along the plate, m
= dimensionless variable = xVy /
1 = kinematic viscosity (of fluid with Pr = 1) = 0.12 610 m2/s
2 = kinematic viscosity (of fluid with Pr = 100) = 6.8 610 m2/s
Taking the ratio of (b) and (c) gives
46.0)s/m(108.6
)s/m(1012.0
5.1
2.5
5.1
2.526
26
2
1
2
1
t
t
(iii) Checking. Dimensional check: The right hand side of equation (b) should be dimensionless:
)m()/sm(
)(m/s)m(
21
11x
Vt = dimensionless
(5) Learning and Generalizing. (i) In calculating the ratio 21/ tt , the location x and the free
stream velocity V are assumed to be the same for both fluids. (ii) The ratio 21/ tt is constant
independent of location.
PROBLEM 4.26
Water at 25oC flows over a flat plate with a uniform velocity of 2 m/s. The plate is maintained at
85oC. Determine the following:
[a] The thermal boundary layer thickness at a distance of 8 cm from the leading edge.[b] The heat flux at this location.[c] The total heat transfer from the first 8 cm of the plate.
[d] Whether Pohlhausen's solution can be used to find the heat flux at a distance of 80 cm from
the leading edge.
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) The Reynolds number and Peclet number should be checked to determine if the flow is laminar and if boundary layer approximations are valid. (iii) Pohlhausen's solution is applicable
if 100 < Rex < 100 105 and Pex = Rex Pr > 100. (iv) Thermal boundary layer thickness and heat transfer coefficient vary along the plate. (v) Newton’s law of cooling gives local heat flux. (vi) The fluid is water.
(2) Problem Definition. The problem is determining the temperature distribution in the fluid. Knowing the temperature distribution for laminar flow (Pohlhausen's solu tion), the thermal boundary layer thickness, local and average heat transfer coefficients, local heat flux and total heat transfer rate can be determined.
(3) Solution Plan. Check the Reynolds and Peclet numbers to determine if boundary layer approximations can be made, the flow is laminar and if Pohlhausen’s solution is applicable. Use Pohlhausen’s solution to determine the local heat transfer coefficient and Newton’s law of cooling to determine heat flux and heat transfer rate.
(4) Plan Execution.
(i) Assumptions. Assume and verify that Pohlhausen’s solution is applicable. Assumptions leading to Pohlhausen's solution are: (1) Ne wtonian fluid, (2) steady state, (3) constant
properties, (4) two-dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential
energy, (11) negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) no radiation.
(ii) Analysis. The Reynolds and Peclet numbers are defined as
Rex =V x
(a)
andPex = Rex Pr (b)
where
Pex = Peclet number Pr = Prandtl number Rex = Reynolds number
V = upstream velocity = 2 m/s x = distance from the leading edge of the plate, m
00
sT x
yt
V
T
)(xqs
)(xh
PROBLEM 4.26 (continued)
= kinematic viscosity, m2 /s
Properties are evaluated at the film temperature Tf defined as
Tf = T Ts
2 (c)
where
Tf = film temperature, oCTs = surface temperature = 85oC
T = free stream temperature = 25oC
To determine if the flow is laminar or turbulent, compare the Reynolds number with the
transition Reynolds number. For the flow over a flat plate, transition Reynolds number tx
Re is
txRe = 5 105 (d)
The flow is laminar if Rex < txRe . Viscous boundary layer approximations are valid for
100xRe
Thermal boundary layer approximations are valid for
Pex > 100Substituting into (c)
Tf = (85 + 25)( oC)/2 = 55oC
Properties of water at this temperature are
k = 0.6458 W/m-oCPr = 3.27
= 0.5116 10-6 m2/s
The Reynolds and Peclet numbers are determined at x = 0.08 m
Rex = 2 0 08
0 5116 10312 744
6 2
( / ) . ( )
. /,
m s m
m s
and
Pex = 312,744 3.27 = 1.0227 106
Therefore, boundary layer approximations can be made and the flow is laminar. Pohlhausen’s solution is applicable.
[a] Boundary layer thickness t . Pohlhausen’s solution, Fig. 4.6, is used to determine the
thermal boundary layer thickness. At the edge of the thermal boundary layer, y = t , fluid
temperature is approximately the same as ambient temperature. That is
T(x, t ) T (e)
Or, in terms of the dimensionless temperature *T
*T =s
st
TT
TxT ),( 1 (f)
PROBLEM 4.26 (continued)where
T = temperature variable, oC*T = dimensionless temperature
y = distance from the plate, m
t = thermal boundary layer thickness, m
At T = 1 and Pr = 3.27, Fig. 4.6 gives the thermal boundary layer thickness in terms of the
dimensionless variable
3 = yx
V=
x
Vt (g)
Solving (g) for t
t = 3V
x (h)
[b] Local surface heat flux qs . This is the local heat transfer per unit area. Applying Newton's
law of cooling at location x gives
qs = h (Ts - T ) (i)
where
h = local heat transfer coefficient, W/m2-oCqs = local surface heat flux, W/m2
The local heat transfer coefficient is given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.24b) gives
h = 0.332 kV
xPr
1/3 (j)
Substituting (j) into (i) gives
qs = 0.332 k (Ts - T )V
xPr
1/3 (k)
[c] Total heat transfer rate qT. Applying Newton’s law of cooling and using the average heat transfer coefficient gives the total heat transfer from the plate
qT = )( TTLWh s (l)
where
h = average heat transfer coefficient, W/m2-oCL = length of plate = 0.08 m W = width of plate, m
The average heat transfer coefficient is given by equations (4.67) and (4.71b)
3/1664.0 Pr
vL
Vkh (m)
Substituting (m) into (l) and rearranging
1/3)()(664.0 PrReTTkW
qLs
T (n)
PROBLEM 4.26 (continued)
where ReL is the Reynolds number at x = L.
(iii) Computations.
[a] Boundary layer thickness t . Substituting into equation (h)
t = 3)s/m(2
)m(08.0)s/m(105116.0 26
= 0.00043 m = 0.43 mm
[b] Surface heat flux qs . Equation (k) gives
qs = 0.332 (0.6458)(W/m-oC) (85 -25)(oC)2
05116 10 0 086 2
( / )
. ( / ) . ( )
m s
m s m (3.27)1/3
=133,477 W/m2
[c] Total heat transfer rate qT . Equation (n) gives
q
WT = 0.664 (0.6458)(W/m-oC) (85 - 25)(oC) (312,744)1/2 (3.27)1/3 = 21,356 W/m
[d] Validity of Pohlhausen’s solution at x = 0.8 m. The Reynolds number at x = 0.8 m is
Rex = 2 0 8
0 5116 103127 10
6 2
6( / ) . ( )
. /.
m s m
m s
Since the Reynolds number is greater than the transition Reynolds number the flow is turbulent and therefore Pohlhausen’s solution does not apply.
(iv) Checking. Dimensional check: Computations showed that units for equations (a), (h), (k) and (n) are dimensionally consistent.
Quantitative check: Equations (4.66) and (4.67) show that for a plate of length L, the average heat transfer coefficient is twice the local coefficient at x = L. This means that the average heat flux is twice the flux at x = L. Thus, the average flux for this problem is 2(133,477)(W/m2) = 266,954(W/m2). Therefore
sT qLWq (o)
where sq is the average heat flux. Substituting into (o)
W
qT = 0.08(m)266,954(W/m2) = 21,356 W/m
Limiting check: If Ts = T , the local heat flux and the total heat transfer rate should vanish.
Setting Ts = T in equations(k) and (n) gives the expected results.
(5) Comments. (i) It is important to check the conditions for boundary layer flow and for laminar flow before proceeding with the solution (i.e. check the Reynolds and Peclet numbers). (ii) Typically, the thickness of the thermal boundary layer is relatively small. In this example, at a distance of 8 cm from the leading edge is 0.43 mm.
PROBLEM 4.27
The cap of an electronic package is cooled by forced convection. The free stream temperature is
25oC. The Reynolds number at the downstream end of the cap is 110,000. Surface temperature
was found to be 145oC. However, reliability requires that
surface temperature does not exceed 83oC. One possible
solution to this design problem is to increase the free stream
velocity by a factor of 3. You are asked to determine if
surface temperature under this plan will meet design
specification.
(1) Observations. (i) This is an external forced convection problem over a flat plate. (ii) Increasing the free stream velocity, increases the average heat transfer coefficient. This in turn causes surface temperature to drop. (iii) Based on this observation, it is possible that the proposed plan will meet design specification. (iv) Since the Reynolds number at the downstream end of the package is less than 500,000, it follows that the flow is laminar throughout. (v) Increasing the free stream velocity by a factor of 3, increases the Reynolds number by a factor of 3 to 330,000. At this Reynolds number the flow is still laminar. (vi) The power supplied to the package is dissipated into heat and transferred to the surroundings from the surface. (vii) Pohlhausen's solution can be applied to this problem. (viii) The ambient fluid is unknown.
(2) Problem Definition. Find the relationship between surface temperature, free stream velocity and heat transfer coefficient.
(3) Solution Plan. Surface temperature is related to heat transfer coefficient by Newton's law of cooling. Heat transfer coefficient is related to free stream velocity in Pohlhausen’s solution. Apply Newton’s law of cooling and Pohlhausen’s solution to the cap.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation and
(15) all power supplied to package leaves the cap as heat.
(ii) Analysis. Application of Newton's law of cooling gives
Tq = h A (Ts - T ) (a)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
Tq = total surface heat transfer rate, W
Ts = surface temperature, oC
T = free stream temperature = 25oC
V T,cap at Ts
PROBLEM 4.27 (continued)
The heat transfer rate Tq is determined by the power supplied to the package which is assumed
constant. Changing the free stream velocity results in a change in the average heat transfer
coefficient and surface temperature. However, Tq , A and T remain the same. Let subscripts 1
and 2 refer to conditions corresponding to the two velocities V 1 and V 2, where
V 2 = 3 V 1 (b)
Applying (a) to the two cases gives
Tq = h1A (Ts1 - T ) (c)
and
Tq = h2 A (Ts2 - T ) (d)
Taking the ratio of (c) and (d)
11 1
2 2
h T T
h T T
s
s
( )
( )
Solving the above for Ts2
Ts2 = T + )( 1
2
1 TTh
hs (e)
The average heat transfer coefficient for laminar flow over a flat plate is given by Pohlhausen's solution (7.25)
3/1)(664.0 PrL
Vkh (f)
where
k = thermal conductivity, W/m-oC L = length of cap, m
Pr = Prandtl number
V = free stream velocity, m/s
= kinematic viscosity, m2/s
Properties in Pohlhausen’s solution are evaluated at the film temperature, 2/)( TTT sf .
Since surface temperature is expected to change, it follows that properties will also change. However, this effect cannot be considered in th e solution since the cooling fluid is not known. As a first approximation, changes in properties due to changes in surface temperature will be
neglected. Thus, k, Pr and are assumed to remain constant and equation (f) simplifies to
VCh (g)
where C is a constant. Applying (g) to the two cases and taking the ratio of the resulting equations
2
1
2
1
V
V
h
h (h)
Substituting (h) into (e) gives
PROBLEM 4.27 (continued)
Ts2 = T +2
1
V
V (Ts1 - T ) (i)
(iii) Computations. Setting V 2 = 3 V 1 and Ts1 = 145oC in (i) gives
Ts2 = 25 (oC) + )C)(25145(3/1 o = 94.3oC
Since this temperature is above the design limit of 83oC, the proposed plan will not work.
(iv) Checking. Dimensional check: Since the ratio of velocities is dimensionless, it follows that equation (i) is dimensionally consistent.
Qualitative check: As V is increased, surface temperature Ts should decrease. This is confirmed by equation (i).
Limiting check: There should be no change in surface temperature for the special case of V 2 =
V 1. Setting V
V
1
2
= 1 in (i) gives the expected result of Ts2 = Ts1.
(5) Comments. (i) This problem is solved without knowing the size of the cap, the magnitude of the free stream velocity and the nature of the fluid. (ii) Neglecting radiation is a conservative assumption since it overestimates surface temperature. Taking radiation into consideration results in a lower temperature than 94.3oC. Thus, the proposed design modification should not be rejected without first examining the effect of radiation. (iii) If the cooling fluid is known, changes in properties due to changes in surface temperature can be accounted for using a trial
and error procedure. A value for Ts2 is assumed, properties at Tf2 are determined and 21 / hh is
calculated and used in equation (e) to calculate Ts2. If the calculated Ts2 is not close to the assumed value, the procedure is repeated until a satisfactory agreement is obtained between
assumed and calculated values. (iv) For V 2 = 4.3V 1, the resulting surface temperature will be 83oC.This meets design specification.
PROBLEM 4.28
The back of the dinosaur Stegosaurus has two rows
of fins. Each row is made up of several fins arranged
in line and separated by a space. One theory
suggests that providing a space between neighboring
fins reduces the weight on the back of the dinosaur
when compared with a single long fin along the back.
On the other hand, having a space between
neighboring fins reduces the total surface area. This
may result in a reduction in the total heat loss.
Model the fins as rectangular plates positioned in line as shown. The length of each plate is L
and its height is H. Consider two fins separated by a distance L. Compare the heat loss from the
two fins with that of a single fin of length 3L and height H. Does your result support the
argument that spaced fins lead to a reduction in heat loss? To simplify the analysis assume
laminar flow.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. The total heat transfer rate can be determined using the average heat transfer coefficient. (iv) For laminar flow, Pohlhausen's solution gives the heat transfer coefficient. (v) For two in-line fins heat transfer from the down stream fin is influenced by the upstream fin. The further the two fins are apart the less the interference will be.
(2) Problem Definition. Determine the average heat transfer coefficient for plates of length Land 3L.
(3) Solution Plan. Use Pohlhausen’s solution to determine the average heat transfer coefficient. Apply Newton's law of cooling to determine the h eat transfer rate from two plates of length L each and compare with the heat transfer from a single plate of length 3L.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect
radiation and (15) neglect interference between in-line plates.
(ii) Analysis. Of interest is the ratio of the total heat transfer rate from two single plates of length L each and a single plate of length 3L. Newton’s law of cooling applied to a single plate of length L gives
)(1 TTAhq s (a)
PROBLEM 4.28 (continued)
where
A plate area, 2m
h average heat transfer coefficient, CW/m o2
1q heat transfer rate from a single plate of length L, W
sT surface temperature, Co
T free stream temperature, Co
The area of a single plate of length L and height H is
A = HL (b)
The average Nusselt number is obtained from Pohlhausen's solution
LL RePrk
LhNu 3/1664.0 (c)
where
k = thermal conductivity, W/m-oC
L = plate length, m Pr Prandtl number
LRe Reynolds number
The Reynolds number is defined as
LVRe (d)
where
V free stream velocity, m/s
= kinematic viscosity, m2/s
Substituting (b)-(d) into (a)
LTTV
kHPrq s )(664.0 3/11 (e)
The total heat transfer rate from two in-line fins, 2q , is
LTTV
kHPrqq s )(664.022 3/112 (f)
The heat transfer rate 3q from a single plate of length 3L is obtained from (e) by replacing L
with 3L
LTTV
kHPrq s 3)(664.0 3/13 (g)
Taking the ratio of (f) to (g)
15.13
2
3
2
q
q (h)
PROBLEM 4.28 (continued)
(iii) Checking. Dimensional check: Units of q1 in equation (e) should be W
m)()C)(()s/m(
s)/m()m()CW/m(664.0 o
2
o3/11 LTT
VHkPrq s = W
Limiting check: If TTs the heat transfer rate should vanish regardless of plate arrangement.
Setting TTs in (f) and (g) gives
032 qq
(5) Comments. (i) Based on the assumption of no inte rference between neighboring in-line plates, the heat transfer rate from two in-line plates of length L each separated by a distance Lexceeds that of a single plate of length 3L by 15%. The weight of tw o in-line plates of length Leach is 2/3rd of that of a single plate of length 3L. Thus the two in-line plate arrangement has advantages in heat transfer rate and weight when compared to a single plate of length 3L. (iii) The analysis can be generalized to n in-line plates each of length L separated by spacing L as compared with a single plate of length (2n-1)L. The ratio of the heat transfer rates for this case is given by
1212 n
n
q
q
n
n (i)
For example, for n = 10
29.21102
10
19
10
q
q
Thus the heat transfer rate from 10 in-line plates of length L each separated by a distance L
exceeds that of a single plate of length 19L by 129%. The weight is reduced by approximately ½.
PROBLEM 4.29
A fluid with Prandtl number 0.098 flows over a semi-
infinite flat plate. The free stream temperature is T and
the free stream velocity is V . The surface of the plate is
maintained at uniform temperature .sT Assume laminar
flow.
[a] Derive an equation for the local Nusselt number.
[b] Determine the heat transfer rate from a section of the plate between 1x and 2x . The width of
the plate is W.
[c] Derive an equation for the thermal boundary layer thickness ).(xt
(1) Observations. (i) This is an external forced convection problem for flow over a flat plate. (ii) Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable. (iii) Knowing the heat transfer coefficient, the local Nusselt number can be determined. (iv) the Newton’s law of cooling gives the heat transfer rate. (iv) Pohlhausen’s solution gives the thermal boundary layer thickness.
(2) Problem Definition. Determine the local heat transfer coefficient and thermal boundary layer thickness for laminar boundary layer flow over a flat plate.
(3) Solution Plan. Use Pohlhausen’s solution to determine the local heat transfer coefficient. Apply Newton’s law of cooling to determine the heat transfer rate. Use Fig. 4.6 to determine the thermal boundary layer thickness.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are:(1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. [a]: The local heat tran sfer coefficient is given by
d
d
x
Vkxh
)0()( (4.66)
where
k = thermal conductivity, W/m- oC
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= yV
x, dimensionless variable
y = vertical coordinate, m
T
V
x
1x 2x
W0
dx
PROBLEM 4.29 (continued)
The local Nusselt number is defined as
k
hxNux (a)
Substituting (4.66) into (a)
d
dReNu xx
)0( (b)
where xRe is the local Reynolds number defined as
xVRex (c)
The constant dd /)0( is given in Table 4.2
138.0)0(
d
d (d)
Substituting (d) into (b)
xx ReNu 138.0 (e)
[b] Application of Newton’s law to the surface element Wdx gives
WdxTTxhdq s ][)( (f)
where
h = local heat transfer coefficient, W/m2-oC q = surface heat flux, W Ts = surface temperature, oC
T = free stream temperature, oC W = plate width, m
x = axial distance measured from the leading edge, m
Substituting (4.66) into (f)
x
dx
d
dVWkTTdq s
)0(][ (g)
Integration of (g) from 1x to 2x
)()0(
][2)0(
][ 12
2
1
xxd
dVWkTT
x
dx
d
dVWkTTq ss
x
x
Introducing the definition of the local Reynolds number (c) and using (d), the above gives
)(][2276.012
xx ReReWkTTq s (h)
[c] At the edge of the thermal boundary layer, ty the dimensionless temperature in Fig. 4.6
is equal to unity
1s
s
TT
TT (i)
For a fluid with 098.0Pr the corresponding value of ),( tx is obtained from Fig. 4.6 as
PROBLEM 4.29 (continued)
12x
Vtt (j)
Solving (b) for t
V
xt 12 (k)
Expressing this result in dimensionless form
x
t
RexVx
1212 (l)
(iii) Checking. Dimensional check: (1) Each term in (e) and (l) is dimensionless.
(2) Units q in equation (h) are
WC)m/W()m()C]([ oo kWTTq s
Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (h) gives
.0q
Qualitative check: The thermal boundary layer thickness increases with distance x. Solution (k) confirms this behavior.
(5) Comments. (i) The value of dd /)0( was obtained by interpolation in Table 4.2. A more
accurate value can be obtained using equation (6.46)
dd
fdd
d
Pr
Pr
2
2
332.0)0( (4.64)
However, this requires numerical ev aluation of the integral in (6.46).
(ii)Recalling that Blasius solution gives
xRex
2.5 (4.46)
Comparing (l) with (4.46) shows that t for 098.0Pr . Examination of Fig. 4.6 shows that
t for all fluids with 0.1Pr and that t for 0.1Pr .
PROBLEM 4.30
Two identical triangles are drawn on the surface of a flat
plate as shown. The plate, which is maintained at uniform
surface temperature, is cooled by laminar forced
convection. Determine the ratio of the heat transfer rate
from the two triangles, q1/q2.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) For each triangle the area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of each triangle. (vi) Pohlhausen's solution may be a pplicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along each triangle.
(3) Solution Plan. Apply Newton's law of cooling to an element of each triangle, ydx, determine the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect
radiation.
(ii) Analysis. Of interest is the ratio of the total heat transfer rate from triangle 1 to triangle 2. Since both the heat transfer coefficient and area vary along each triangle, it follows that Newton's law of cooling s hould be applied to an element dA at a distance x from the leading edge:
dq = h(x) (Ts - T )dA (a) where
dA = area of element, m2
dq = rate of heat transfer from element, W h = local heat transfer coefficient, W/m2-oC
Ts = surface temperature, oC
T = free stream temperature, oCx = distance along plate, m
The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution (4.66)
d
d
x
Vkh
)0( (b)
where
TsT
V
viewtop
1 2H
Lx
1y
2y
PROBLEM 4.30 (continued)
k = thermal conductivity, W/m-oC
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= x
Vy , dimensionless variable
y = vertical coordinate, m
Noting that all quantities in equation (b) are constant except the variable x, (b) is rewritten as
h = x
constant (c)
The next step is to determine the infinitesimal area dA for each triangle. Using the subscripts 1 and 2 to refer to triangles 1 and 2, respectively,
dxxydA )(11 (d)
and
dxxydA )(22 (e)
where
)(1 xy = side of element in triangle 1, m
)(2 xy = side of element in triangle 2, m
Similarity of triangles gives
)()(1 xLL
Hxy (f)
xL
Hxy )(2 (g)
Substituting (f) into (d) and (g) into (e) gives
dA1 = (H/L)(L x )dx (h)
dA2 = (H/L) x dx (i)where
H = base of triangle, m L = length of triangle, m
Substituting (c) and (h) into (a) and integrating from x = 0 to x = L, gives
q1 =
L
s
L
s dxx
xL
L
HTTCdx
x
xL
L
HTTCdq
02/1
02/11 )()(
Carrying out the integration yields
2/11 )()3/4( HLTTCq s (j)
Similarly, substituting (c) and (i) into (a) and integration from x = 0 to x = L gives
PROBLEM 4.30 (continued)
q2 =
L
s
L
s dxxL
HTTCdx
x
x
L
HTTCdq
0
2/1
02/12 )()(
Carrying out the integration yields
q2 = (2/3)C(Ts - T )H L1/2 (k)
Taking the ratio of (j) and (k)
22
1
q
q (l)
(iii) Checking. Dimensional check: Units of q1 in equation (j) should be W. First, units of C are determined
C = k(W/m-oC)[V (m/s)/ (m2/s)]1/2
d
d )0((1/1) = W/m3/2-oC
Units of 1q are
1q = C(W/m3/2-oC)(Ts- T )(oC)H(m)L1/2 (m1/2) = W
Since 2q has the same form as 1q , it follows that the units of 2q in equation (k) are also correct.
Qualitative check: The result shows that the rate of heat transfer from triangle 1 is greater than that from triangle 2. This is expected since the heat transfer coefficient increases as the distance from the leading edge is decreased and triangle 1 has its base at x = 0 where h is maximum.
(5) Comments. (i) Although the two triangles have the same area, the rate of heat transfer from triangle 1 is double that from triangle 2. Thus, orientation and proximity to the leading edge of a flat plate play an important role in determining the rate of heat transfer. (ii) The same approach can be used to determine heat transfer for configurations other than rectangles, such as circles and ellipses.
PROBLEM 4.31
An isosceles triangle is drawn on a semi-infinite flat
plate at a uniform surface temperature .sT Consider
laminar uniform flow of constant properties fluid
over the plate. Determine the rate of heat transfer
between the triangular area and the fluid.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Heat transfer rate can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of the triangle. (vi) Pohlhausen's solu tion may be applicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.
(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect
radiation.
(ii) Analysis. Taking advantage of symmetry, we consider the right angle triangle representing half the isosceles triangle. Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives
dq = 2h(x) (Ts - T )ydx (a) where h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from 2 elements, W Ts = surface temperature, oC
T = free stream temperature, oCx = distance along plate, m
y = y(x) height of element
Note that the factor 2 is introduced to account for heat transfer from the two right angle triangles representing the isosceles triangles. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution
d
d
x
Vkh
)0( (4.66)
TsT
V
viewtop
H
L
dx
y
PROBLEM 4.31 (continued)
where
k = thermal conductivity, W/m-oC
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= x
Vy , dimensionless variable
The variable y(x) is given by
xL
Hxy )( (b)
where
H = base of triangle, m L = length of triangle, m
Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = L, gives
L
s dxxd
dVk
L
HTTq
0
2/1)0()(2
Carrying out the integration yields
d
dLVHTTkq s
)0()(
3
4 (c)
This result can be expressed in terms of the Reynolds number as
d
dReHTTkq Ls
)0()(
3
4 (d)
where
LVReL (e)
(iii) Checking. Dimensional check: Units of q in equation (d) should be W
q = k(W/m-oC) )C)(( oTTs H (m)d
d )0((1/1) = W
Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (d) gives
.0q
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is increased. Thus, the same triangle rotated 180 degrees, with its base at x = 0, will have a higher heat transfer rate.
PROBLEM 4.32
Determine the total heat transfer rate from a half circle
drawn on a semi-infinite plate as shown. Assume laminar
two-dimensional boundary layer flow over the plate.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Heat transfer rate can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along over the area of the semi-circle. (vi) Pohlhausen's solution gives the heat transfer coefficient.
(2) Problem Definition. Determine the local heat transfer coefficient along the semi-circle.
(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect
radiation.
(ii) Analysis. Taking advantage of symmetry, we consider the upper half the semi-circle. Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives
dq = 2h(x) (Ts - T )ydx (a) where
h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from upper and lower elements, W Ts = surface temperature, oC
T = free stream temperature, oCx = distance along plate, m
y = y(x) height of element
Note that the factor 2 is introduced to account for heat transfer from the two halves of the semi-circle. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution
d
d
x
Vkh
)0( (4.66)
where
k = thermal conductivity, W/m-oC
PROBLEM 4.32 (continued)
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= x
Vy , dimensionless variable
The variable y(x) is given by 22 )()( oo rxrxy
This can be simplified to 22)( xxrxy o (b)
where
or radius of semi-circle, m
Substituting (4.66) and (b into (a) and integrating from orx to orx 2 , gives
o
o
r
rdxxr
d
dVkTTq os
2
)2()0(
)(2
Carrying out the integration yields
d
dVrTTkq os
)0()(
3
4 2/3 (c)
This result can be expressed in terms of the Reynolds number as
d
dRerTTkq
oros
)0()(
3
4 (d)
where
orVRe
or (e)
(iii) Checking. Dimensional check: Units of q in equation (d) should be W
q = k(W/m-oC) )C)(( oTTs or (m)d
d )0((1/1) = W
Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (d) gives
.0q
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is
increased. Thus, the same semi-circle rotated 180 degrees and remaining at distance or from the
leading edge of the plate, will have a lower heat transfer rate.
PROBLEM 4.33
Consider steady, two-dimensional, laminar boundary
layer flow over a semi-infinite plate. The surface is
maintained at uniform temperature .sT Determine the
total heat transfer rate from the surface area described
by LxHxy /)( as shown.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Heat transfer rate can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient changes along the plate. (iv) The area changes with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of the triangle. (vi) Pohlhausen's solution may be applicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.
(3) Solution Plan. Apply Newton's law of cooling to an element ydx, determine the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect
radiation.
(ii) Analysis. Newton's law of cooling applied to an element ydx at a distance x from the leading edge gives
dq = h(x) (Ts - T )ydx (a) where h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from 2 elements, W Ts = surface temperature, oC
T = free stream temperature, oCx = distance along plate, m
y = y(x) height of element
The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution
d
d
x
Vkh
)0( (4.66)
where
k = thermal conductivity, W/m-oC
T
V
L
xHy
y
x L
H
sT viewtop
PROBLEM 4.33 (continued)
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= x
Vy , dimensionless variable
The variable y(x) is given by
L
xHxy )( (b)
where
H = height at x = L, m L = length, m
Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = L, gives
L
s dxd
dVk
L
HTTq
0
)0()(
Carrying out the integration yields
d
dLVHTTkq s
)0()( (c)
This result can be expressed in terms of the Reynolds number as
d
dReHTTkq Ls
)0()( (d)
where
LVReL (e)
(iii) Checking. Dimensional check: Units of q in equation (d) should be W
q = k(W/m-oC) )C)(( oTTs H (m)d
d )0((1/1) = W
Limiting check: The heat transfer rate should vanish for TTs . Setting TTs in (d) gives
.0q
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is increased. Thus, the same shape rotated 180 degrees, with its base H at x = 0, will have a higher heat transfer rate.
PROBLEM 4.34
Fluid flows over a semi-infinite flat plat which is maintained
at uniform surface temperature. It is desired to double the
rate of heat transfer from a circular area of radius 1R by
increasing its radius to 2R . Determine the percent increase
in radius need to accomplish this change. In both cases the
circle is tangent to the leading edge. Assume laminar
boundary layer flow with constant properties.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) this problem involves determining the heat transfer rate from a circle tangent to the leading edge of a plate. (iii) Heat transfer rate can be determined using Newton’s law of cooling. (iv) The local heat transfer coefficient changes along the plate. (v) The area changes with distance along the plate. (vi) The total heat transfer rate can be determined by integration along the length of the triangle. (vii) Pohlhausen's so lution may be applicable to this problem.
(2) Problem Definition. Determine the local heat transfer coefficient along the triangle.
(3) Solution Plan. Determine the heat transfer rate from a circle which is tangent to the leading edge of a plate. Apply Newton's law of cooling to an element ydx, determine the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5)
laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no
buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect
radiation.
(ii) Analysis. Consider a circle of radius R
which is tangent to the leading edge of the plate. Taking advantage of symmetry, we consider the upper half of the circle. Newton's la w of cooling applied to an element ydx at a distance x from the leading edge gives
dq = 2h(x) (Ts - T )ydx (a) where
h = local heat transfer coefficient, W/m2-oC dq = rate of heat transfer from 2 elements, W Ts = surface temperature, oC
T = free stream temperature, oCx = distance along plate, m
y = y(x) height of element
T sT
V
viewtop
1R 2R
T
Vx0
R
xR
y
PROBLEM 4.34 (continued)
Note that the factor 2 is introduced to account for the two haves of the circle. The local heat transfer coefficient h(x) is obtained from Pohlhausen's solution
d
d
x
Vkh
)0( (4.66)
where
k = thermal conductivity, W/m-oC
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= x
Vy , dimensionless variable
The variable y(x) is given by 222 2)()( xRxxRRxy (b)
Substituting (4.66) and (b) into (a) and integrating from x = 0 to x = 2R, gives
R
s dxxRd
dVkTTq
2
0
)2()0(
)(2
Carrying out the integration yields
2/32/3 )0(
)(3
)2(2R
d
dVTTkq s
(c)
Applying (c) to two circles of radii 1R and 2R and taking the ratio of the two results, gives
2/31
2/32
1
2
R
R
q
q (d)
Doubling the heat transfer rate gives
21
2
q
q (e)
Thus the radius of the circle needed to double the heat transfer rate is obtained by substituting (e)
into (d) and solving for 2R
113/2
2 587.1)2( RRR (f)
Thus percent increase in R is
% increase = 7.58587.1
1001
11
1
12
R
RR
R
RR
(iii) Checking. Dimensional check: Units of q in equation (c) should be W
PROBLEM 4.34 (continued)
q = k(W/m-oC) )m()/sm(
)m/s()C)(( 3/22/3
2
o RV
TTs (m)d
d )0((1/1) = W
Limiting check: (1) The heat transfer rate should vanish for TTs . Setting TTs in (c) gives
.0q
(2) The heat transfer rate must vanish for a circle of radius R = 0. Setting R = 0 in (c) gives .0q
(5) Comments. Examination of (4.66) shows that the heat transfer coefficient decreases as x is increased. Thus, moving a circle in the x-direction decreases the rate of heat transfer.
PROBLEM 4.35
Liquid potassium (Pr << 1) flows over a semi-
infinite plate. Assume laminar boundary layer
flow, suggest a simplified velocity profile for
solving the energy equation
Solution
For Pr << 1 the viscous boundary layer
thickness is much thinner than the thermal boundary layer thickness t . Thus, throughout
much of the thermal boundary layer the axial velocity is .Vu Therefore, the term convective
term is approximate as
x
TV
x
Tu .
PROBLEM 4.36
For very low Prandtl numbers the thermal boundary layer is much thinner than the viscous
boundary layer. Thus little error is introduced if the velocity everywhere in the thermal boundary
layer is assumed to be the free stream velocity .V Show that for laminar boundary layer flow
over a flat plate at low Prandtl numbers the local Nusselt number is given by
2/12/1564.0 RePrNux
How does this result compare with scaling prediction?
(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that the axial velocity is uniform throughout the thermal boundary layer. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution. (iii) The Nusselt number depends on the temperature gradient at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a flat plate using a uniform velocity throughout the thermal boundary layer.
(3) Solution Plan. Follow Pohlhausen’s solution replacing Bl asius solution to the flow field with uniform axial velocity equal to the free stream velocity.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation and
(15) assume uniform axial velocity equal to the free stream velocity.
(ii) Analysis. The local Nusselt number for laminar flow over a flat plate is given by
xx Red
dNu
)0( (4.68)
where
xVRex
V free stream velocity, m/s
x
Vyyx ),(
s
s
TT
TT
PROBLEM 4.36 (continued)
sT surface temperature, Co
T free stream temperature, Co
The temperature gradient at the surface d
d )0( is given in equation (b) of Appendix B
1
0 0
)(2
exp)0(
ddfPr
d
d (b)
where )(f is the solution to the velocity field. In Pohlhausen’s solution )(f is obtained from
Blasius solution to the flow fiel d for laminar boundary layer flow over a semi-infinite flat plate. Rather than use Blasius solution we now use a simp lified flow field of uniform velocity given by
1d
df
V
u (c)
Integrating (c) f (d)
Substituting (d) into (b) 1
0 02exp
)0(dd
Pr
d
d
Evaluating the integral 1
0
2
4exp
)0(d
Pr
d
d (e)
The definite integral in (e) is recognized as the error function. Thus
PrPr
dPr
d
d564.0
4exp
)0(1
0
2 (f)
Introducing (f) into (4.68)
xx RePrNu 564.0 (g)
Scaling results for the case of 1Pr is given by
xx eRPrNu , for Pr <<1 (4.55)
This compares favorably with the exact solution (g).
(iii) Checking. Dimensional check; Each term in (g) is dimensionless.
(5) Comments. Scaling prediction of the local Nusselt number gives the same dependency on the Prandtl and Reynolds numbers as the exact solution. The constant 0.564 in the exact solution is replaced by unity in the scaling prediction.
PROBLEM 4.37
Consider laminar boundary layer flow over a flat plate at a uniform temperature .sT When the
Prandtl number is very high the viscous boundary layer is much thicker than the thermal
boundary layer. Assume that the thermal boundary layer is entirely within the part of the
velocity boundary layer in which the velocity profile is approximately linear. Show that for such
approximation the Nusselt number is given by
2/13/1339.0 RePrNux
Note:0
3/13 )3/1(
3)exp(
cdxcx , where is the Gamma function.
(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that the axial velocity varies linearly in the y-direction. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution. (iii) The Nusselt number depends on the temperature gradient at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a flat plate using a uniform velocity throughout the thermal boundary layer.
(3) Solution Plan. Follow Pohlhausen’s solution replacing Bl asius solution to the flow field with linear axial velocity distribution.
(4) Plan Execution.
(i) Assumptions. All assumptions leading to Pohlhausen's solution are applicable. These are: (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-dimensional, (5) laminar
flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy
( = 0 or g = 0) and (13) no energy generation ( 0q ). In addition, (14) neglect radiation and
(15) assume uniform axial velocity equal to the free stream velocity.
(ii) Analysis. The local Nusselt number for laminar flow over a flat plate is given by
xx Red
dNu
)0( (4.68)
where
xVRex
V free stream velocity, m/s
x
Vyyx ),(
PROBLEM 4.37 (continued)
s
s
TT
TT
sT surface temperature, Co
T free stream temperature, Co
The temperature gradient at the surface d
d )0( is given in equation (b) of Appendix B
1
0 0
)(2
exp)0(
ddfPr
d
d (b)
where )(f is the solution to the velocity field. In Pohlhausen’s solution )(f is obtained from
Blasius solution to the flow fiel d for laminar boundary layer flow over a semi-infinite flat plate. Rather than use Blasius solution we now use a si mplified flow field of linear velocity given by
Ad
df
V
u (c)
where A is constant. Integrating (c)
BAf2
)(2
(d)
The constants A and B are determined from Blasius bound ary condition (4.45b) and solution. Boundary condition (4.45b) gives
0)0(f (4.45b)
Table 7.2 of Blasius solution gives
33206.0)0(
2
2
d
fd (e)
These two conditions give A and B
16603.0A , 0B (f)
Equation (d) becomes
216603.0)(f (g)
Substituting (g) into (b) 1
0 0
2
216603.0exp
)0(dd
Pr
d
d
Evaluating the indefinite integral1
0
3
616603.0exp
)0(d
Pr
d
d (h)
The definite integral in (h) is evaluated next. Let
PROBLEM 4.37 (continued)
3
6
16603.0Prz
It follows that
3/13/1
16603.0
6z
Pr
Differentiating
dzzPr
d 3/23/1
16603.0
6
3
1
Substituting into (h)
1
3/2
0
3/1
16603.0
6
3
1)0(dzze
Prd
d z (i)
The definite integral in (i) is recognized as the Gamma function given by
)(0
1 ndzze nz 1n (j)
Comparing the integral in (i) with (j) gives
3
1n (k)
Using (j) and (k), equation (i) becomes
13/1
)3/1(16603.0
6
3
1)0(
Prd
d (l)
The value of )3/1( is obtained from tables of Gamma function
679.2)3/1( (m)
1/3339.0)0(
Prd
d (n)
Substituting (n) into (4. 2/13/1339.0 xx RePrNu (o)
(iii) Checking. Dimensional check: (1) The exponent of the exponential in(h) is dimensionless. (2) Each term in (o) is dimensionless.
(5) Comments. Scaling prediction of the local Nusselt number gives the same dependency on the Prandtl and Reynolds numbers as the exact solution (o). The constant 0.339 in the exact solution is replaced by unity in the scaling prediction. Scaling results for the case of 1Pr isgiven by
2/11/3xx eRPrNu , for Pr >>1 (4.57)
PROBLEM 4.38
Consider steady, two-dimensional, laminar boundary
layer flow over a porous flat plate at uniform surface
temperature. The plate is subject to a uniform suction
ovv )0,(x . Far away downstream both the axial
velocity and the temperature may be assumed to be
functions of y only. Free stream velocity is V and
free stream temperature is .T Determine the heat transfer coefficient and Nusselt number in
this region.
(1) Observations. (i) The flow and temperature fields for this boundary layer problem are simplified by assuming that the axial velocity and temperature do not vary in the x-direction. (ii) The heat transfer coefficient depends on the temperature gradient at the surface. (iii) Temperature distribution depends on the flow field. (iv) The effect of wall suction must be taken into consideration.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a flat plate.
(3) Solution Plan. Introduce simplifying assumptions in the energy equation for boundary layer flow and solve for the temperature distribution for laminar flow over a plate with suction.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible
dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In
addition, (14) neglect radiation, (15) uniform suction velocity and (16) negligible axial variation of velocity and temperature.
(ii) Analysis. The local heat transfer coefficient h is given by
TT
y
xT
khs
)0,(
(1.10)
where
k thermal conductivity, CW/m o
T fluid temperature, Co
sT surface temperature, Co
T free stream temperature, Co
y normal coordinate, m
The local Nusselt number is defined as
ov
T
Vx
y
PROBLEM 4.38 (continued)
k
hxNux (a)
Temperature distribution in (1.10) is obtained from the solution to boundary layer energy equation
2
2
y
T
y
T
x
Tu v (4.18)
where
u axial velocity, m/s
v normal velocity, m/s
thermal diffusivity, /sm2
Since temperature variation in the axial direction is neglected, it follows that
0x
T (b)
(b) into (a)
2
2
y
T
y
Tv (c)
To solve (c) for the temperature distribution, velocity component v must be determined. The continuity equation for two-dimensional incompressible flow is given by
0yx
u v (2.2)
Since velocity variation in the axial direction is neglected, it follows that
0x
u (d)
(d) into (2.2)
0y
v (e)
Integration of (e) gives cv (f)
where c is constant of integration. Surface boundary condition on v is
ovv (g)
where
ov surface suction velocity, m/s
Substituting (g) into (c)
2
2
y
T
y
Tov-
Since T is independent of x, it follows that the above can be written as
PROBLEM 4.38 (continued)
2
2
dy
Td
dy
dTov- (h)
The solution to (h) requires two boundary conditions. They are:
sTxT )0,( (i)
TxT ),( (j)
To solve (h), it is rewritten as
dy
dy
dT
y
Td
ov (k)
Integrate (j)
1lnln Cydy
dT ov (l)
where 1C is constant of integration. To integrate (l) again, it is first rewritten as
ydy
dT
C
ov
1
1ln
or
)exp(1 yCdy
dT ov (m)
Integrate (m)
21 )exp( CyCT o
o
v
v (n)
where 2C is constant of integration. Application of boundary conditions (i) and (j) gives 1C and
2C
)(1 TTC sov
(o)
TC2 (p)
(o) and (p) into (n)
)exp()( yTTTT sov
(q)
Substituting (q) into (1.10) gives the heat transfer coefficient h
ovkh (r)
However, is defined as
pc
k (s)
PROBLEM 4.38 (continued)
where
pc specific heat, CJ/kg o
density, 3kg/m
(s) into (r)
ovpch (t)
Substituting (t) into (a) gives the local Nusselt number
k
xcNu
px
ov
The above can be written in a more revealing way as
x
k
cNu
px
ov
This is recognized as
xx RePrNu (u)
where the local Reynolds number is defined as
xRex
ov (v)
(iii) Checking. Dimensional check: (1) The exponent of the exponential in(q) is dimensionless. (2) Each term in (q) has units of temperature. (3) Equation (t) give the correct units for h.
Boundary conditions check: Solution (q) satisfies conditions (i) and (j).
Limiting check: For the special case of TTs , fluid temperature should be uniform equal to
.T . Setting TTs in (q) gives .TT
(5) Comments. (i) Neglecting axial variation of velocity and temperature are the key
simplifying assumptions in this problem. (ii) Free stream velocity V does not enter into the
solution for the temperature distribution and heat transfer coefficient. (iii) The Reynolds number in solution (u) depends on the suction velocity and not free stream velocity.
PROBLEM 4.39
A semi infinite plate is heated with uniform flux q along its length. The free stream temperature
is T and free stream velocity is .V Since the heat transfer coefficient varies with distance
along the plate, Newton’s law of cooling requires that surface temperature must also vary to
maintain uniform heat flux. Consider the case of laminar boundary layer flow over a plate whose
surface temperature varies according to
nCxTxTs )(
Working with the solution to this case, show that 2/1n corresponds to a plate with uniform
surface flux.
(1) Observations. (i) This is a forced convection flow over a plate with variable surface temperature. (ii) The local heat flux is determined by Newton’s law of cooling. (iii) The local heat transfer coefficient and surface temperature vary with distance along the plate. The variation of surface temperature and heat transfer coefficient must be such that Newton’s law gives uniform heat flux. (iv) The local heat transfer coefficient is obtained from the local Nusselt number.
(2) Problem Definition. Determine the value of the exponent n that results in uniform heat flux.
(3) Solution Plan. Apply Newton’s law of cooling to determine the local surface heat flux. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is laminar use the result of numerical solution to forced convection over a plate with variable surface temperature to determine the local Nusselt number and heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105, to be verified), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100, to be verified), (7) uniform upstream velocity and temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible
dissipation, (11) no buoyancy ( = 0 or g = 0), (12) no energy generation ( 0q ) and (14)
negligible radiation
(ii) Analysis. Newton’s law of cooling gives
))(( TTxhq s (a)
where
)(xh local heat transfer coefficient, CW/.m o2
q surface heat transfer rate, 2W/m
)(xTs surface temperature, Co
T free stream temperature, Co
Surface temperature variation is given by
nCxTxTs )( (b)
Substituting (b) into (a)
PROBLEM 4.39 (continued)
)(xhCxq n (c)
According to (c), surface flux is constant if
nxxh /1)( (d)
The local heat transfer coefficient is obtained from the local Nusselt number defined as
k
hxNux (e)
Solving for h(x)
x
kNuxh x)( (f)
The solution to the Nusselt number for a plate with surface temperature variation described in equation (b) is
xx Red
dNu
)0( (4.80)
where dd /)0( is the dimensionless surface temperature gradient. It is a constant which
depends on the Prandtl number and the exponent n. Substituting (4.80) into (f)
x
kRe
d
dxh x
)0()(
Using the definition of Reynolds number into the above
2/1)0()0()( x
V
d
dk
x
kxV
d
dxh (g)
Using (g) into (c)
2/1)0(x
V
d
dkCxq n (h)
Examination of (h) shows that for q to be constant independent of x, the exponent n must be
2
1n (i)
(iii) Checking. Dimensional check: (1) Each term in (4.80) is dimensionless. (2) Equations(g) and (h) are dimensionally correct.
Limiting check: (1) For the special case of C = 0, surface temperature according to (b) will be the
same as ambient temperature. That is .TTs The corresponding heat flux for this case should
vanish. Setting C = 0 in (h) gives .0q
(5) Comments. (i) The key to the solution to this problem is the determination of the variation of h with distance x.
PROBLEM 4.40
Water flows over a semi-infinite flat plate which is maintained at a variable surface
temperature sT given by
75.0)( CxTxTs
where
C = 54.27 oC / (m)0.75
T = free stream temperature = C.3o
x = distance from the leading edge, m
Determine the average heat transfer coefficient for a plate if length L = 0.3 m. Free stream
velocity is 1.2 m/s.
(1) Observations. (i) This is a forced convection flow over a plate with variable surface temperature. (ii) The Reynolds number should be computed to determine if the flow is laminar or turbulent. (iii) The local heat transfer coefficient and surface temperature vary with distance along the plate. (iv) The local heat transfer coefficient is obtained from the solution to the local Nusselt number. (v) The determination of the Nusselt number requires determining the temperature gradient at the surface.
(2) Problem Definition. Determine surface temperature gradient for a plate at variable surface temperature.
(3) Solution Plan. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is laminar use the result of numerical solution to forced convection over a plate with variable surface temperature to determine surface temperature gradient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105, to be verified), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100, to be verified), (7) uniform upstream velocity and temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible
dissipation, (11) no buoyancy ( = 0 or g = 0), (12) no energy generation( 0q ) and (13)
negligible radiation
(ii) Analysis.
The maximum Reynolds number is given by
LVReL (a)
where
L = plate length = 0.3 m
V free stream velocity = 1.2 m/s
kinematic
Properties are evaluated at the film temperature defined as
T
V
L
x
75.0xCTTs
W
PROBLEM 4.40 (continued)
2
sf
TTT (b)
where
T free stream temperature = 3 Co
sT average surface temperature, Co
The average surface temperature is defined as
2
)()0( LTTT ss
s (c)
Surface temperature is given by 75.0)( CxTxTs (d)
where
0.75
o
m
C27.54C
Evaluating sT at x = 0 and x = 0.3 m and substituting into (d) gives
C142
(m))(0.3)C/m(27.45C3C3
2
o0.750.750.75ooo75.0CLTT
Ts
Substituting into (b)
C5.82
C))(143( oo
fT
Properties of water at this temperature are
Cm
W5791.0
ok
942.9Pr
s
m103716.1
26
Substituting into (c)
)/sm(103716.1
)m(3.0)m/s(2.126LRe = 262,467
Thus this is laminar boundary layer flow. The local Nusselt number for laminar flow over a flat plate with variable surface temperature is given by
xx Red
dNu
)0( (4.80)
where
PROBLEM 4.40 (continued)
k
hxNux (e)
)(xh local heat transfer coefficient, CW.m o2
x = distance from the leading edge, m
Substitute (e) into (4.80) and solve for h
xRex
k
d
dh
)0( (f)
The dimensionless surface temperature gradient, ,/)0( dd is obtained from the
numerical solution for laminar flow over a flat plate at a surface temperature of the form
ns CxTxT )( (g)
The solution is presented in Fig. 4.8. The exponent n in (g) characterizes surface temperature variation.
(iii) Computations. For the problem under consideration 75.0n and 942.9Pr , Fig. 4.8
gives
1.1.
)0(
d
d (h)
Substituting into (f)
xRex
kh 1.1 (i)
The corresponding Nusselt number is
xx Rek
hxNu 1.1 (j)
(iv) Checking. Dimensional check: (1) Equation (f) has the correct units for heat transfer coefficient. (2) The Reynolds number in (a) and Nusselt number in (j) are dimensionless.
(5) Comments. (i) The key to the solution to this problem is the determination of the constant ./)0( dd This constant is determined through the use of Fig. 4.8. (ii) Fig. 4.8 is applicable to a
class of variable surface temperature described by equation (g) only
ns CxTxT )( (g)
(iii) According to (d) surface temperature varies from C3o at the leading edge to C25o . Clearly,
surface temperature is not uniform. For uniform surface temperature the Nusselt number is given by (4.72c)
10for,339.0 3/1 PrRePrNu xx (4.72c)
0 1.00.5 1.5
1.0
d
d )0(
0.7Pr
n
10
30
2.0
re, temperatusurfaceing with varrplatefor(0)
d
d4.8Fig.
ns CxT-T [4]
PROBLEM 4.40 (continued)
In this problem 942.9Pr . Using this value in (4.72c) gives
xx ReNu 729.0 (k)
This is 33% smaller than the variable surface temperature solution given in (j).
PROBLEM 4.41
Air flows over a plate which is heated non-uniformly such that its surface temperature increases linearly as the distance from the leading edge is increased according to
CxTxTs )(
where
C = 24 oC /cm
T = free stream temperature = C20o
x = distance from the leading edge, m
Determine the total heat transfer rate from a square
plate cm.10cm10 Free stream velocity is 3.2 m/s.
(1) Observations. (i) This is a forced convection flow over a plate with variable surface temperature. (ii) The Reynolds number should be computed to determine if the flow is laminar or turbulent. (iii) Newton’s law of cooling gives the heat transfer rate from the plate. (iv) The local heat transfer coefficient and surface temperature vary with distance along the plate. Thus determining the total heat transfer rate requires integration of Newton’s law along the plate. (v) The local heat transfer coefficient is obtained from the local Nusselt number.
(2) Problem Definition. Determine the average Nusselt number for forced convection over a flat plate with variable surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling to determine the total heat transfer rate form the plate. Compute the Reynolds number to establish if the flow is laminar or turbulent. If the flow is laminar use the result of numerical solution to forced convection over a plate with variable surface temperature to determine the average Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105, to be verified), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100, to be verified), (7) uniform upstream velocity and temperature, (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible
dissipation, (11) no buoyancy ( = 0 or g = 0), (12) no energy generation ( q ) and (13)
negligible radiation
(ii) Analysis. The total heat transfer from the plate is given by Newton’s law of cooling
L
s WdxTTxhq0
))(( (a)
where
)(xh local heat transfer coefficient, CW.m o2
L plate length = 0.1 m q heat transfer rate, W
)(xTs surface temperature, Co
CxTTs
T
V
L
x
PROBLEM 4.41 (continued)
T free stream temperature = 20 Co
W plate width = 0.1 m
x distance along plate, m
The local heat transfer coefficient is obtained from the local Nusselt number defined as
k
hxNux (b)
The maximum Reynolds number is given by
LVReL (c)
where
V free stream velocity = 3.2 m/s
kinematic
Properties are evaluated at the film temperature defined as
2
sf
TTT (d)
where
sT average surface temperature, Co
The average surface temperature is defined as
2
)()0( LTTT ss
s (e)
Surface temperature is given by
CxTxTs )( (f)
where C = 24 oC /cm. Evaluating sT at x = 0 and x = 0,1 m and substituting into (e) gives
C1402
)C/cm)10(cm24(C20C20
2
ooooCLTT
Ts
Substituting into (d)
C802
C))(14020( oo
fT
Properties of air at this temperature are
Cm
W02991.0
ok
706.0Pr
s
m1092.20
26 0 1.00.5 1.5
1.0
d
d )0(
0.7Pr
n
10
30
2.0
re, temperatusurfaceing with varrplatefor(0)
d
d4.8Fig.
ns CxT-T [4]
PROBLEM 4.41 (continued)Substituting into (c)
)/sm(1092.20
)m(1.0)m/s(2.326LRe =15,296
Thus this is laminar boundary layer flow. The local Nusselt number for laminar flow over a flat plate is given by
xx Red
dNu
)0( (4.80)
where ./)0( dd is the dimensionless surface temperature gradient. Numerical solution to
dd /)0( for laminar flow over a flat plate at a surface temperature of the form
ns CxTxT )( (g)
is given in Fig. 4.8. The exponent n in (g) characterizes surface te mperature variation.. Using the definition of local Nusselt number and Reynolds number, (4.80) is solved for the local heat transfer coefficient
x
V
d
dk
xV
x
k
d
dh
1)0()0( (h)
Substituting (f) and (h) into (a)
L
dxxV
d
dkWCq
0
)0(
Evaluating the integral
2/3)0(
3
2L
V
d
dkWCq
Expressing this result in terms of the Reynolds number
LV
d
dkWLCq
)0(
3
2
LRed
dkWLCq
)0(
3
2 (i)
Equation (i) is rewritten in dimensionless form as
LRed
d
kWLC
q )0(
3
2 (j)
(iii) Computations. For the problem under consideration 1n and 706.0Pr , Fig. 4.8 gives
48.0.
)0(
d
d (h)
Substituting into (i)
PROBLEM 4.41 (continued)
W4.28296,15.480m/m)C/cm)100(c4(2m)0.1(m))(1.0)(CW/m)(02991.0(3
2 ooq
(iv) Checking. Dimensional check: (1) Computations showed that (i) has the correct units for heat transfer rate. (2) Each term in (j) is dimensionless.
Limiting check: (1) If the width of plant is zero the heat transfer rate will vanish. Setting W = 0 in (i) gives q = 0. (2) For the special case of C = 0, surface temperature according to (f) will be the
same as ambient temperature. That is .TTs The corresponding heat transfer rate for this case
is q = 0. Setting C = 0 in (i) gives q = 0.
(5) Comments. (i) The key to the solution to this problem is the determination of the constant ./)0( dd This constant is determined through the use of Fig. 4.8. (ii) Fig. 4.8 is applicable to a
class of variable surface temperature described by equation (g) only
ns CxTxT )( (g)
PROBLEM 4.42
The surface temperature of a plate varies with distance
from the leading edge according to
8.0)( CxTxTs
Two identical triangles are drawn on the surface as
shown. Fluid at uniform upstream temperature T and
uniform upstream velocity V flows over the plate. Assume laminar boundary layer flow.
Determine the ratio of the heat transfer rate from the two triangles, q1/q2.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Convection heat transfer from a surface can be determined using Newton’s law of cooling. (iii) The local heat transfer coefficient and surface temperature vary along the plate. (iv) For each triangle the area varies with distance along the plate. (v) The total heat transfer rate can be determined by integration along the length of each triangle.
(2) Problem Definition. Determine the local heat transfer coefficient along each for laminar boundary layer flow over a plate with variable surface temperature.
(3) Solution Plan. Apply Newton's law of cooling to an element of each triangle, ydx, determine the local heat transfer coefficient along the plate and integrate over the area.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) flat plate, (10) negligible changes in kinetic and potential energy, (11) negligible
dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ). In
addition, (14) neglect radiation.
(ii) Analysis. Of interest is the ratio of the total heat transfer rate from triangle 1 to triangle 2. Since both the heat transfer coefficient and area vary along each triangle, it follows that Newton's law of cooling s hould be applied to an element dA = ydx at a distance x from the leading edge:
dq = h(x) (Ts - T ) y(x)dx (a) where
dq = rate of heat transfer from element, W h = local heat transfer coefficient, W/m2-oC
Ts = surface temperature, oC
T = free stream temperature, oCx = distance along plate, m
y = y(x) = width of element
T
V
viewtop
1 2 H
L
)(xTs
TsT
V
viewtop
1 2H
Lx
1y
2y
PROBLEM 4.42 (continued)
Integration of (a) gives the total heat transfer rate
dxxyTTxhq
L
s )())((0
(b)
To evaluate the integral we must specify ( TTs ), y(x) and h(x). Surface temperature variation
is given as8.0)( CxTxTs (c)
Similarity of triangles gives
)()(1 xLL
Hxy (d)
xL
Hxy )(2 (e)
The local heat transfer coefficient h(x) for a plate with variable surface temperature described by (c) is given by (4.78)
d
d
x
Vkh
)0( (4.78)
where
k = thermal conductivity, W/m-oC
V = free stream velocity, m/s
)/()( ss TTTT , dimensionless temperature
= kinematic viscosity, m2/s
= x
Vy , dimensionless variable
y = coordinate normal to plate, m
The dimensionless surface temperature gradient, ,/)0( dd is obtained from the numerical
solution for laminar flow over a flat plate at a surface temperature of the form
ns CxTxT )( (f)
Fig. 4.8 gives dd /)0( . As shown in Fig. 4.8, dd /)0( depends on the Prandtl number and
the exponent n in (f). Since both triangles have the same surface temperature distribution and the same fluid, it follows that dd /)0( is the same for both triangles.
To determine the heat transfer rate, ,1q from triangle 1, equations (c), (d) and (4.78) are
substituted into (b)
dxxx
xL
d
dV
L
HkCq
L
0
8.01
)()0(
Evaluating the integral
PROBLEM 4.42 (continued)
99.2
)0( 3.1
1
L
d
dVkCHq (g)
Similarly, for triangle 2, equations (c), (e) and (4.78) are substituted into (b)
dxxx
x
d
dV
L
HkCq
L
0
8.01
)0(
Evaluating the integral
3.2
)0( 3.1
2
L
d
dVkCHq (h)
Taking the ratio of (g) and (h)
769.099.2
3.2
2
1
q
q (i)
(iii) Checking. Dimensional check: (1) Each term in (d) and (e) has units of length. (2)
Noting that units of C are 8.0o C/m , equations (g) and (h) have the correct units for heat.
Limiting check: For the limiting case of a plate which is maintained at the free stream
temperature, that is TTs , the corresponding heat transfer rate should vanish for both
triangles. According to (f), TTs when C = 0. Setting C = 0 in (g) and (h) gives .021 qq
(5) Comments. (i) According to (i) the heat transfer ratio .1/ 21 qq By contrast, in Problem
4.30 in which surface temperature is uniform, 1/ 21 qq . The reason for this reversal is the
increase in surface temperature with x favors triangle 2 where the area also increases with x.
Although the two triangles have the same area, the rate of heat transfer from triangle 1 is double that from triangle 2. Thus, orientation and proximity to the leading edge of a flat plate play an important role in determining the rate of heat transfer. (ii) The same approach can be used to determine heat transfer for configurations other than rectangles, such as circles and ellipses.
PROBLEM 4.43
Construct a plot showing the variation of xx ReNu / with wedge angle. Where xNu is the local
Nusselt number and xRe is the local Reynolds number. Assume laminar boundary layer flow of
air.
(1) Observations. (i) This is a forced convection boundary layer flow over a wedge. (ii) Wedge surface is maintained at uniform temperature. (iii) The flow is laminar. (iv) The fluid is air. (v) Similarity solution for the local Nusselt number is presented in Section 4.4.3. (vi) The Nusselt number depends on the Reynolds number and the dimensionless temperature gradient at the surface ./)0( dd (vii) Surface temperature
gradient depends on wedge angle.
(2) Problem Definition. Determine the variation of dd /)0( with wedge angle.
(3) Solution Plan. Use the wedge solution of Section 4.4.3 to determine the variation of local Nusselt number with wedge angle.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. The local Nusselt number for laminar boundary layer flow over a wedge is given by
xx Red
dNu
)0( (4.96)
where
xVRex , local Reynolds number
)(xV external flow velocity over the wedge, m/s
x
xVyyx
)(),( , similarity variable
s
s
TT
TT, dimensionless temperature
PROBLEM 4.43 (continued)
Rewrite (496)
d
d
Re
Nu
x
x )0( (a)
Surface temperature gradient depends on wedge angle and Prandtl number. Table 4.3 lists dd /)0( corresponding to four wedge angles and five Prandtl numbers. For air with Prandtl
number 7.0Pr , Table 4.3 gives the values of dd /)0( used to construct a pot of
xx ReNu / vs. wedge angle .
(4) Comments. (i) The local heat transfer coefficient decreases with distance along the surface. (ii) Local Nusselt number and heat transfer coeffi cient increase with wedge angle. The increase is approximately linear.
Pr = 0.7
Wedgeangle
d
d )0(
0 0.292
5/ 0.331
2/ 0.384
0.496
x
x
Re
Nu
wedge angle
0.5
0.4
0.3
0 60 120 180
deg,
PROBLEM 4.44
Consider laminar boundary layer flow over a wedge. Show that the average Nusselt number Nu
for a wedge of length L is given by
LRed
d
mNu
)0(
1
2
where the Reynolds number is defined as .)(LLV
ReL
(1) Observations. (i) This is a forced convection boundary layer flow over a wedge. (ii) Wedge surface is maintained at uniform temperature. (iii) The flow is laminar. (iv) The average Nusselt number depends on the average heat transfer coefficient.. (v) Similarity solution for the local heat transfer coefficient is presented in Section 4.4.3.
(2) Problem Definition. Determine the average heat transfer coefficient for laminar flow over a wedge at uniform surface temperature.
(3) Solution Plan. Start with the definitions of the average Nusselt number. Use the wedge solution of Section 4.4.3 to determine the local heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. The average Nusselt number for a wedge of length L is defined as
k
LhNu L (a)
where
h = average heat transfer coefficient, CW/m o2
k = thermal conductivity, CW/m o
L = wedge length, m
The average heat transfer coefficient is defined as
L
dxxhL
h0
)(1
(b)
where the local heat transfer coefficient, h(x), is given by (4.95)
d
d
x
xVkxh
)0()()( (4.95)
PROBLEM 4.44 (continued)
where
)(xV external flow velocity over the wedge, m/s
x = distance along wedge surface form the leading end, m
x
xVyyx
)(),( , similarity variable
s
s
TT
TT, dimensionless temperature
= kinematic viscosity, sm2
(4.95) into (b)
dxx
xV
d
dkxh
L
0
)()0()( (c)
The external flow velocity, ),(xV is given by
mCxxV )( (4.82)
where C is constant and m is a measure of wedge angle , defined as
2m (4.83)
(4.82) into (c) L
m dxxd
dC
L
kh
0
2/)1()0(
Evaluating the integral
2/)1()0(
1
2 mLd
dC
L
k
mh (e)
(e) into (a)
2/)1()0(
1
2 mL L
d
dC
mNu
To express in terms of the Reynolds number, rewrite the above
LCL
d
d
mNu
m
L)()0(
1
2
(4.82) into the above
LLV
d
d
mNu L
)()0(
1
2 (f)
Introduce the definition of Reynolds number
LL Red
d
mNu
)0(
1
2 (g)
PROBLEM 4.44 (continued)
(iii) Checking: Dimensional check: (1) Units of the constant C is determined from (4.82) as
mms
mC . Using this shows that (e) has the correct units. (2) Equation (f) is dimensionless,
Limiting check: For a wedge with zero angle ( m = 0), the solutions should reduce to the flat
plate solution of Pohlhausen. Setting m = 0 in (g) gives
LL Red
dNu
)0(2 (h)
This agrees with Pohlhausen’s result of equation (4.69).
(4) Comments. In determining the average heat transfer coefficient for a wedge, the variation of the velocity outside the boundary layer with distance x must be taken into consideration. That is,
unlike Pohlhausen’s solution for the flat plate, for the wedge ).(xVV
PROBLEM 4.45
Compare the total heat transfer rate from a o90 wedge, ,wq with that from a flat plate, ,pq of
the same length. Construct a plot of pw qq / as a function of Prandtl number.
(1) Observations. (i) Newton’s law of cooling gives the heat transfer rate from a surface. (ii)
Total heat transfer from a surface depends on the average heat transfer coefficient h . (iii) Both
flat plate and wedge are maintained at uniform surface temperature. (iv) Pohlhausen’s solution
gives h for a flat plate. (v) Similarity solution for the local heat transfer coefficient for a wedge is presented in Section 4.4.3.
(2) Problem Definition. Determine the average heat transfer coefficient for laminar flow over a flat plate and a wedge.
(3) Solution Plan. Apply Newton’s law of cooling to flat plate and
wedge. Use Pohlhausen’s and wedge solutions to h .
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3)
constant properties, (4) two-dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstream velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow over wedge, (10) negligible changes in
kinetic and potential energy, (11) negligible dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) no energy generation ( 0q ).
(ii) Analysis. Newton’s law of cooling gives
)( TTAhq s (a)
where
A = surface area, 2m
h average heat transfer coefficient, CW/m o2
q heat transfer rate, W
sT surface temperature, Co
T free stream temperature, Co
Let the subscript p denote plate and w denote wedge. Apply (a) to the plate and wedge and take their ratio
w
p
w
p
h
h
q
q(b)
The problem reduces to determining ph and wh .
The average heat transfer coefficient for boundary layer laminar flow over a flat plate at uniform surface temperature is given by Pohlhausen’s solution (4.67)
PROBLEM 4.45 (continued)
p
pLp
d
dRe
L
kh
)0(2 (4.67)
where
k = thermal conductivity, CW/m o
L = wedge length, m
LReLV
V free stream velocity, m/s
x
Vyyxp ),( , dimensionless similarity variable
x = axial coordinates, m y = normal coordinates, m
s
spp
TT
TT, dimensionless temperature
),( yxTT pp , temperature distribution
= kinematic viscosity, sm2
Using the definition of Reynolds number, (4.67) us rewritten as
p
pp
d
dLV
L
kh
)0(2 (c)
The average heat transfer coefficient foe a wedge is defined as
L
ww dxxhL
h0
)(1
(d)
The local heat transfer coefficient for a wedge, )(xhw , is given by (4.95)
w
ww
d
d
x
xVkxh
)0()()( (4.95)
where
)(xV external flow velocity over the wedge, m/s
x
xVyyxw
)(),( , similarity variable
s
sww
TT
TT, dimensionless temperature
),( yxTT ww , temperature distribution
= kinematic viscosity, sm2
PROBLEM 4.45 (continued)
(4.95) into (d)
dxx
xV
d
dkxh
L
ww
0
)()0()( (e)
The external flow velocity, ),(xV is given by
mCxxV )( (4.82)
where C is constant and m is a measure of wedge angle , defined as
2m (4.83)
(4.82) into (e) L
mww dxx
d
dC
L
kh
0
2/)1()0(
Evaluating the integral
2/)1()0(
1
2 mww L
d
dC
L
k
mh (f)
Substitute (c) and (f) into (a)
p
p
wm
p
w
d
d
d
d
V
CL
mq
q
)0(
)0(
1
1 (g)
(iii) Checking: Dimensional check: (1) Equation (4.82) shows that mCL has units of velocity. It
follows that (g) is dimensionless. Similarly, units of (f) are correct.
Limiting check: For a wedge with zero angle ( m = 0), the wedge solution should reduce to
the flat plate solution and the heat ratio should be unity. For this case, according to (4.82),
.VC Setting m = 0 in (g) gives .1/ pw qq
(4) Comments. (i) The local heat transfer coefficient decreases with distance along the surface. (ii) Local Nusselt number and heat transfer coeffi cient increase with wedge angle. The increase is approximately linear.
PROBLEM 4.46
For very low Prandtl numbers the thermal boundary layer is much thinner than the viscous boundary layer. Thus little error is introduced if the velocity everywhere in the thermal boundary
layer is assumed to be the free stream velocity .V Show that for laminar boundary flow layer
flow over a wedge at low Prandtl numbers the local Nusselt number is given by
xx RePrm
Nu)1(
(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer is the same as that of the external flow. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3. (iii) The local Nusselt number depends the local heat transfer coefficient which depends on the temperature gradient at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a wedge for the simplified velocity field described above.
(3) Solution Plan. Follow analysis of section 4.4.3 for determining the local Nusselt number using a simplified flow field solution.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy ( = 0 or g = 0), (13) no energy generation ( 0q ) and
(14) Pr << 1. Thus assume that axial velocity with in the thermal boundary layer is the same as that of the external flow.
(ii) Analysis. The local Nusselt number, xNu , for a wedge is given by equation (4.96) of
Section 4.4.3
xx Red
dNu
)0( (4.68)
where
xxVRex
)(
)(xV external flow velocity over the wedge, m/s
x
xVyyx
)(),( , similarity variable
PROBLEM 4.46 (continued)
s
s
TT
TT, dimensionless temperature
= kinematic viscosity, sm2
The problem reduces to determining the dimensionless temperature gradient at the surface dd /)0( given by equation (4.94)
1
0 0
)(2
1)(mexp
)0(ddF
Pr
d
d (4.94)
where m is measure of wedge angle , defined as
2m (4.83)
The function )(F is obtained from the solution to the flow field over the wedge. It is defined in
(4.87) as
)(
),(
xV
yxu
d
dF (4.87)
where u(x,y) is the axial velocity within the thermal boundary layer. However, for Pr << 1 we assume that
)(),( xVyxu (a)
(a) into (4.87)
1d
dF (b)
Integration of (b) gives F (c)
(b) into (4.94) 1
0 02
1)(exp
)0(dd
Prm
d
d (d)
Evaluating the integral in the integrand
1
0
2
2
1)(exp
)0(d
Prm
d
d (e)
The definite integral in (e) is recognize d as the error function. To proceed, let
4
1)( Prmz (f)
Thus
dzPrm
d1)(
4 (g)
PROBLEM 4.46 (continued)
Substitute (f) and (g) into (e)
1
0
2 )exp(1)(
4)0(dzz
Prmd
d (h)
Howeverz
z dzz e0
22erf (i)
and
12
erf
0
2
dzze (j)
(i) into (h)
Prm
Prmd
d 1)(
1)(
)0(1
(k)
(k) into (4.68)
xx RePrm
Nu)1(
(l)
(iii) Checking. Dimensional check; All equations are dimensionless.
Limiting check: For a wedge with zero angle ( m = 0), the solutions should reduce to the flat
plate solution with Pr << 1 and .Vu . Setting m = 0 in (l) gives
xx PrReNu 564.0 (m)
This agrees with the solution to Problem 4.36.
(5) Comments. The assumption that axial velocity within the thermal boundary layer is the same as that of the external flow provided a major simplification in the solution. It made it possible to obtain a solution for the Nusselt number without the need for numerical integration.
PROBLEM 4.47
Consider laminar boundary layer flow over a wedge at a uniform temperature .sT When the
Prandtl number is very high the viscous boundary layer is much thicker than the thermal
boundary layer. Assume that the velocity profile within the thermal boundary layer is
approximately linear. Show that for such approximation the local Nusselt number is given by
2/11/3)0()1(489.0 RePrFmNux
Note: 0
3/13 )3/1(
3)exp(
cdxcx , where is the Gamma function.
(1) Observations. (i) The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer varies linearly with the normal distance. (ii) Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3. (iii) The local Nusselt number depends on the local heat transfer coefficient which depends on the temperature gradient at the surface.
(2) Problem Definition. Determine the temperature gradient at the surface for laminar flow over a wedge for the simplified velocity field described above.
(3) Solution Plan. Follow the analysis of section 4.4.3 for determining the local Nusselt number using a simplified flow field solution.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) two-
dimensional, (5) laminar flow (Rex < 5 105), (6) viscous and thermal boundary layer flow (Rex > 100 and Pex > 100), (7) uniform upstr eam velocity and temperature, (8) uniform surface temperature, (9) symmetrical flow, (10) negligible changes in kinetic and potential energy, (11)
negligible dissipation, (12) no buoyancy ( = 0 or g = 0), (13) no energy generation ( 0q ) and
(14) Pr >> 1. Thus assume that axia l velocity within the thermal boundary layer varies linearly with distance normal to wedge surface.
(ii) Analysis. The local Nusselt number, xNu , for a wedge is given by equation (4.96) of
Section 4.4.3
xx Red
dNu
)0( (4.68)
where
xxVRex
)(
)(xV external flow velocity over the wedge, m/s
PROBLEM 4.47 (continued)
x
xVyyx
)(),( , similarity variable
s
s
TT
TT, dimensionless temperature
= kinematic viscosity, sm2
The problem reduces to determining the dimensionless temperature gradient at the surface dd /)0( given by equation (4.94)
1
0 0
)(2
1)(mexp
)0(ddF
Pr
d
d (4.94)
where m is a measure of wedge angle , defined as
2m (4.83)
The function )(F is obtained from the solution to the flow field over the wedge. It is defined in
(4.87) as
)(
),(
xV
yxu
d
dF (4.87)
where u(x,y) is the axial velocity within the thermal boundary layer. However, for Pr >>1 we use a simplified flow field of linear velocity given by
Ad
dF
V
u (c)
where A is constant. Integrating (c)
BAF2
)(2
(d)
The constant B is determined from wedge flow boundary condition. (4.89b) and solution. Boundary condition (4.89b) gives
0)0(F (4.89b)
Apply (4.89b) to (d)B = 0 (e)
Differentiate (d) twice and apply at the surface
)0(2
2
Fd
FdA (f)
The constant )0(F depends on wedge angle. It is obtained from the solution to flow field over
the wedge. Values of )0(F for four angles are listed in Table 4.3. Thus (d) can be written as
PROBLEM 4.47 (continued)
2
2
)0()(
FF (g)
Substitute (g) into (4.94) 1
0 0
2
4
(0)1)(exp
)0(dd
PrFm
d
d \ (h)
Evaluating the integral in the integrand
1
0
3
12
(0)1)(exp
)0(d
PrFm
d
d (i)
The definite integral in (i) is evaluated next. Let
3
12
(0)1)( PrFmz
It follows that
3/13/1
(0)1)(
12z
PrFm
Differentiating
dzzPrFm
d 3/23/1
(0)1)(
12
3
1
Substituting into (i)
1
3/2
0
3/1
(0)1)(
12
3
1)0(dzze
PrFmd
d z (j)
The definite integral in (j) is recognized as the Gamma function given by
)(0
1 ndzze nz 1n (k)
Comparing the integral in (j) with (k) gives
3
1n (l)
Using (k) and (l), equation (j) becomes
13/1
)3/1((0)1)(
12
3
1)0(
PrFmd
d (m)
PROBLEM 4.47 (continued)
The value of )3/1( is obtained from tables of Gamma function
679.2)3/1( (n)
(n) into (m)
1/33/1(0)1)(489.0)0(
PrFmd
d (o)
Substitute (n) into (o)
2/11/33/1(0)1)(489.0 xx RePrFmNu (o)
(iii) Checking. Dimensional check: (1) The exponent of the exponential in (j) is dimensionless. (2) All equations are dimensionless.
Limiting check: For a wedge with zero angle ( m = 0), the solutions should reduce to the flat
plate solution with Pr >> 1 and linear axial velocity. Setting m = 0 in (o) gives
2/11/33/1(0)489.0 xx RePrFNu (p)
However, for m = 0 the flow field reduces to Blasius solution. Thus, using Table 4.1, gives
33206.0)0()0( fF (q)
(q) into (p) 2/11/3339.0 xx RePrNu
This agrees with the solution to Problem 4.37.
(5) Comments. The assumption that axial velocity within the thermal boundary layer varies linearly in the direction normal to the surface provided a major simplification in the solution. It made it possible to obtain a solution for the Nusselt number without the need for numerical integration.
PROBLEM 5.1
For fluids with 1Pr the thermal boundary
layer thickness is much larger than the viscous
boundary layer. That is .1/t It is
reasonable for such cases to assume that fluid
velocity within the thermal layer is uniform
equal to the free stream velocity. That is
Vu
Consider uniform laminar boundary layer flow over a flat plate. The surface is maintained at
uniform temperature sT and has an insulated leading section of length .ox Assume a third degree
polynomial temperature profile, show that the local Nusselt number is given by
2/12/1
2/1
153.0 xo
x eRPrx
xNu
where the local Reynolds number is ./Re xVx
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be uniform,
Vu . This represents a significant simplification.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) two-
dimensional, (4) laminar flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible
dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) uniform velocity within the thermal boundary layer ( 1Pr ).
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution
PROBLEM 5.1 (continued)
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
However, For 1Pr the velocity boundary layer thickness is much smaller than the thermal boundary layer thickness. Thus we assume
Vu (b)
For the temperature profile we assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (c)
The boundary conditions on the temperature are
(1) sTxT 0,
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (c) and the four boundary conditions give the coefficients )(xbn
,0 sTbt
sTTb1
)(2
31 , ,02b
33
1)(
2
1
t
sTTb
Substituting the above into (c)
3
3
2
1
2
3)(),(
ttss
yyTTTyxT (d)
Substituting (d) into (1.10)
t
kh
2
3 (e)
Combining (a) and (e)
tx
xNu
2
3 (f)
The problem reduces to finding t which is obtained using the energy equation. Substituting (b)
and (d) into (5.7) )(
0
3
3
22
31)(
2
)(3xt
tts
t
s dyyy
dx
dTTV
TT
Evaluating the integral in the above
dx
dV
dx
dV t
tttt 8
3)8/1()4/3(
2
3
Separating variables and rearranging
PROBLEM 5.1 (continued)
dxV
d tt
4 (f)
Integrating
CxV
t
82 (g)
where C is constant of integration. The boundary condition on t is
0)( ot x (h)
Applying (h) to (g) gives
oxV
C8
Substituting into (g) and solving for t
x
x
xVx
ot 18
(i)
Substituting (i) into (f)
)/(1
1
24
3
xx
xVNu
ox (j)
Noting that
p
p
ckk
cPr
/
/(k)
Using (k), equation (j) is expressed in terms of Prandtl and Reynolds numbers
)/(1530.0
xx
PrReNu
o
xx (l)
(iii) Checking. Dimensional check: All terms in equations (i)-(l) are dimensionless.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions on temperature.
Comparison with scaling results: For the special case of 0ox , scaling estimate of the Nusselt
number for Pr << 1 is given in equation (4.55)
xx ePrRNu , for Pr <<1 (4.55)
Thus, the dependency on the Prandtl and Reynolds number is the same in (l) and (4.55).
(5) Comments. (i) For the special case of no leading insulated section, equation (l) reduces to
xx PrReNu 530.0 (m
The exact solution to this case gives
xx PrReNu 564.0 (n)
Thus the error in the integral solution is 6%.
PROBLEM 5.2
For fluids with 1Pr the thermal boundary layer
thickness is much smaller than the viscous boundary
layer. That is .1/t It is reasonable for such
cases to assume that fluid velocity within the thermal
layer is linear given by
yVu
Consider uniform laminar boundary layer flow over a flat plate with an insulated leading section
of length .ox The plate is maintained at uniform surface temperature .sT Assume a third degree
polynomial temperature profile, show that the local Nusselt number is given by
2/13/13/1
4/3/1319.0 eRPrxxNu ox
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be linear,
)/(yVu .
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a linear velocity profile and a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant
properties, (4) laminar flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible
dissipation, (12) no buoyancy ( = 0 or g = 0) and (13) linear velocity within the thermal boundary layer ( 1Pr ).
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
PROBLEM 5.2 (continued)
However, For 1Pr the velocity boundar (5. (5. (5. (5. (5. (5. (5. (5.r than the thermal boundary layer thickness. Thus we assume
yVu (b)
where is the thickness of the velocity boundary layer. Application of the integral form of the
momentum equation, (5.5) gives ).(x The solution for )(x for this case is detailed in Example
5.1 and is given by
xRex
12 (5.26)
This gives
xV
12 (c)
For the temperature profile we assume a third degree polynomial
33
2210 )()()()(),( yxbyxbyxbxbyxT (d)
The boundary conditions on the temperature are
(1) sTxT 0,
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (d) and the four boundary conditions give the coefficients )(xbn
,0 sTbt
sTTb1
)(2
31 , ,02b
33
1)(
2
1
t
sTTb
Substituting the above into (d)
3
3
2
1
2
3)(),(
ttss
yyTTTyxT (e)
Substituting (e) into (1.10)
t
kh
2
3 (f)
Combining (a) and (f)
tx
xNu
2
3 (g)
The problem reduces to finding t which is obtained using the energy equation. Substituting (b)
and (e) into (5.7)
PROBLEM 5.2 (continued)
)(
0
3
3
22
31)(
2
)(3xt
tts
t
s dyyyy
dx
dTTV
TT
Evaluating the integral in the above2
102
3 t
t dx
dV
Multiplying and dividing by and rearranging the above
2
2115 tt
dx
d
V (h)
Define
tr (i)
substituting (i) into (h)
2115r
dx
dr
V (j)
However, )(x is given in (c). Substituting (c) into the above and rearranging
21
12
15rx
dx
dr
x (k)
Differentiating the right hand side and noting that /Pr
32
2
12
125.1r
xdx
drrx
xPr (l)
Rearranging and separating variables
3
2
-)/5.2(
4
rPr
drr
x
dx (m)
Integrating and using the boundary condition 0)( ot x
rx
x rPr
drr
x
dx
o 03
2
-)/5.2(
4 (n)
Evaluating the integrals
3/43)/5.2(lnln rPrx
x
o
The above can be written as
3/43)/5.2( rPrx
x
o
(o)
PROBLEM 5.2 (continued)
Using the definition of r in (i) and solving (o) for r
3/14/3]/[1
5.2xx
Prr o
t (p)
Substituting (c) for
3/14/3]/[1
5.212xx
Prx
Vot (q)
Using the definition of Reynolds number
3/14/3
1/21/3
3/1 )/(11
)5.2(12 xxRePrx
ot (r)
Substituting (r) into (g)
2/13/13/1
4/3/1319.0 eRPrxxNu ox (s)
(iii) Checking. Dimensional check: All terms in equations (g)-(j) and (m)-(p) are dimensionless.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions on temperature.
Comparison with scaling results: For the special case of 0ox , scaling estimate of the Nusselt
number for Pr >> 1 is given in equation (4.57)
xx eRPrNu 1/3)( , for Pr >>1 (4.57)
Thus, the dependency on the Prandtl and Reynolds number is the same in (s) and (4.57).
(5) Comments. (i) For the special case of no leading insulated section, equation (s) reduces to
xx RePrNu3/1
)(319.0 (t)
The exact solution to this case gives
xx RePrNu 3/1)(339.0 , for Pr >> 1 (n)
Thus the error in the integral solution is 5.9%.
PROBLEM 5.3
A square array of chips is mounted flush on a flat plate. The array measures cmcm LL . The
forward edge of the array is at a distance ox from the leading edge of the plate. The chips
dissipate uniform surface flux .sq The plate is cooled by forced convection with uniform
upstream velocity V and temperature T . Assume laminar boundary layer flow. Assume
further that the axial velocity within the thermal
boundary layer is equal to the free stream velocity.
Use a third degree polynomial temperature profile.
[a] Show that the local Nusselt number is given by
)/(175.0
xx
eRPrNu
o
xx
[b] Determine the maximum surface temperature.
(1) Observations. (i) The velocity is assumed to be uniform, ,Vu throughout the thermal
boundary layer. (ii) A leading section of length ox is unheated. (iii) at oxx , surface heat flux
is uniform. (iv) The determination of the Nusselt number requires the determination of the temperature distribution. (v) Surface temperature is unknown. (vi) The maximum surface temperature for a uniformly heated plate occurs at the trailing end.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate which heated with uniform surface flux. This reduces to determining the thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar
flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) flat surfa ce, (7) uniform surface heat flux, (8) negligible changes in kinetic and potential energy, (9) negligible axial conduction,
(10) negligible dissipation and (11) no buoyancy ( = 0).
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TxT
y
xTk
hs )(
)0,(
(1.10)
Substitute (1.10) into (a)
PROBLEM 5.3 (continued)
xTxT
y
xT
Nus
x)(
)0,(
(b)
Thus the Nusselt number depends on the temperature distribution ).,( yxT Note that surface
temperature varies with location x and is unknown. The integral form of the energy equation is used to determine the temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
The axial velocity u is assumed to be uniform equal to the free stream velocity. Thus
Vu (c)
(c) into (5.7) )(
0
)(0,
xt
dyTTdx
dV
y
xT (d)
Once temperature distribution is determine d, surface temperature and maximum surface temperature will be known.
We assume a third degree polynomial temperature profile
33
2210 )()()()(),( yxbyxbyxbxbyxT (e)
The boundary conditions on the temperature are
(1) sqy
xTk
)0,(
(2) TxT t ),(
(3) 0,
y
Tx t
(4) 0)0,(
2
2
y
xT
Equation (e) and the four boundary conditions give the coefficients )(xbn
ts
k
qTb
3
20
k
qb s
1 , ,02b23
1
3 t
s
k
qb
Substitute the above into (e)
2
3
33
2),(
t
ts y
yk
qTyxT (f)
PROBLEM 5.3 (continued)
Surface temperature is obtained by setting y = 0 in (f)
ts
sk
qTxTxT
3
2)0,()( (g)
The Nusselt number is obtained by substituting (f) and (g) into (b)
tx
xNu
2
3 (h)
The problem reduces to determining .t Substitute (f) into (d) and simplify
)(
033
22
3xt
t
t dyy
ydx
d
V
Evaluate the integral in the above
222
12
1
2
1
3
2ttt
dx
d
V
dx
d
V
t2
4
1 (i)
Separate variables and integrate
2
4
1tddx
V
tx
Evaluate the integrals
CxV
t24 (j)
where C is constant of integration determined from the boundary condition on t
0t at oxx (k)
Apply (k) to (j)
oxV
C 4
Substitute into (j) and solve for t
)(4 ot xxV
(l)
The Nusselt number is obtained by substituting (l) into (h)
)(42
3
o
x
xxV
xNu
PROBLEM 5.3 (continued)
This simplifies to
)/1(4
3
xx
xVNu
ox (m)
Noting that pck / , (m) is rewritten as
)/1(
1
4
3
)/1(4
3
xx
xV
k
c
xxk
xVcNu
o
p
o
px
Introduce the definitions of Prandtl and Reynolds numbers, the above gives
)/(175.0
xx
eRPrNu
o
xx (n)
Surface temperature is obtained by substituting (l) in (g)
)(3
4)( o
ss xx
Vk
qTxT (o)
According to (o), the maximum surface temperature occurs at the trailing end of the plate.
For the special case where heating begins at the leading edge, ,0ox equations (n) and (o) give
xx eRPrNu 75.0 (p)
Vk
qTxT s
s3
4)( (q)
(iii) Checking. Dimensional check: (1) All term in (b), (h), (m) and (n) are dimensionless. (2) Each term in (f) and (o) has units of temperature.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions on temperature.
Limiting check: If 0sq surface temperature will be the same as free stream temperature.
Setting 0sq in (0) gives .)( TxTs
(5) Comments. For the special case of 0ox , scaling estimate of the Nusselt number for
1Pr is given in equation (4.55)
xx eRPrNu , for Pr <<1 (4.55)
Thus, the dependency on the Prandtl and Reynolds number is the same in (p) and (4.55).
PROBLEM 5.4
A liquid film of thickness H flows by gravity down
an inclined surface. The axial velocity u is given by
2
2
2H
y
H
yuu o
where ou is the free surface velocity. At 0x the
surface is maintained at uniform temperature .sT
The fluid temperature upstream of this section is
.T Assume laminar boundary layer flow and that .1/ Ht Determine the local Nusselt
number and the total surface heat transfer from a section of width W and length L. Neglect heat
loss from the free surface. Use a third degree polynomial temperature profile.
(1) Observations. (i) The velocity distribution is known. (ii) Surface temperature is uniform. (iii) The determination of the Nusselt number requires the determination of the temperature distribution. (iv) Newton’s law of cooling gives the heat transfer rate. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at uniform surface temperature. This reduces to determining the thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution. Apply Newton’s law of cooling to determine the heat transfer rate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar
flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) flat surfa ce, (7) uniform surface temperature, (8) negligible changes in kinetic and potential energy, (9) negligible axial
conduction, (10) negligible dissipation and (11) no buoyancy ( = 0).
(ii) Analysis The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
Substitute (1.10) into (a)
PROBLEM 5.4 (continued)
xTT
y
xT
Nus
x
)0,(
(b)
Thus, the Nusselt number depends on the temperature distribution ).,( yxT Integration of
Newton’s law of cooling gives the total heat transfer rate
L
s dxxhWTTq0
)()( (c)
The integral form of the energy equation is used to determine the temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
The velocity distribution is given by
2
2
2H
y
H
yuu o (d)
We assume a third degree polynomial temperature profile
33
2210 )()()()(),( yxbyxbyxbxbyxT (e)
The boundary conditions on the temperature are
(1) sTxT )0,(
(2) TxT t ),(
(3) 0,
y
Tx t
(4) 0)0,(
2
2
y
xT
Equation (e) and the four boundary conditions give the coefficients )(xbn
,0 sTbt
sTTb1
)(2
31 , ,02b
33
1)(
2
1
t
sTTb
Substitute the above into (e)
3
3
2
1
2
3)(),(
ttss
yyTTTyxT (f)
The Nusselt number is obtained by substituting (f) into (b)
tx
xNu
2
3 (g)
PROBLEM 5.4 (continued)
The problem reduces to determining .t Substitute (d) and (f) into (5.7)
)(
0
12
1
2
32
1
2
33
3
2
2xt
tto
t
dyyy
H
y
H
y
dx
du
Expand the integrand
)(
0
5
2
13
2
324131
2
32
2
22
2
33
xt
tttto
t
dyH
yy
H
y
Hy
H
y
H
y
Hdx
du
Evaluate the integral
23
2 5
1
24
11
2
3tto
t HHdx
du (h)
To solve this differential equation for t it is first simplified by noting that
23
2 5
1
24
1tt
HH
Thus (h) is approximated by
2
5
11
2
3to
t Hdx
du (i)
Rewrite (i)
dx
dH
u
tt
to
22
15
Separate variables and integrate
t
tto
ddxu
Hx
2
4
15
Evaluate the integrals
Cxu
Ht
o
3
4
45 (j)
where C is constant of integration determined from the boundary condition on t
0t at 0x (k)
Apply (k) to (j) gives 0C
Substitute into (j) and solve for t
3/1
4
45x
u
H
ot (l)
The Nusselt number is obtained by substituting (l) into (g)
PROBLEM 5.4 (continued)
xxH
uNu o
x
3/11
45
4
2
3
Noting that pck / , (m) is rewritten as
3/1
2
2
3/1
3/1
2
23/1
10
1
45
4
4
3
H
x
k
c
k
Hu
H
x
k
HucNu
poopx
Introduce the definitions of Prandtl and Reynolds numbers, the above gives
3/1
2
2
3/110
1
H
xeRPrNu Hx (m)
where the Reynolds number is defined as
HueR o
x (n)
With the Nusselt number determined, the local heat transfer coefficient can be formulated and the total heat transfer rate computed. Equate (a) and (m)
3/1
2
2
3/110
1
H
xeRPr
k
hxH
Solve for h
3/13/1
3/23/1)()(
10xRePr
H
kh H (o)
Substitute (o) into (c)
L
Hs dxxPrReH
kWTTq
0
3/13/1
3/23/1)(
10)( (p)
Evaluate the integral
3/23/1
3/1)/()()(
10
1
2
3HLRePrWkTTq Hs (q)
Rewrite (q) in dimensionless form
3/23/1
3/1)/()(
10
1
2
3
)(HLRePr
WkTT
qH
s
(r)
(iii) Checking. Dimensional check: (1) Each term in (f) has units of temperature. (2) Each term is (g) has units of heat. (3) each term in (b), (g), (m) and (r) is dimensionless.
PROBLEM 5.4 (continued)
Limiting check: If surface temperature is the same as free stream temperature, ,TTs the heat
transfer rate will vanish. Setting TTs in (q) gives q = 0.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions on temperature.
Qualitative check: The total heat transfer rate is expected to increase with increasing length L.This is in agreement with result (q).
(5) Comments. (i) The Nusselt number in (m) depends on a single parameter PrRe . This
product is known as the Peclet number Pe. (ii) The dimensionless heat transfer rate depends on two parameters: the Peclet number, Pe, and the geometric parameter L/H. (iii) The solution is
not valid for .Ht Thus there is a maximum length for which the solution is valid. This
maximum length, ,maxL is determined from (l) by setting Ht and letting .maxLx The
result is 2
max45
4 HuL o (s)
Expressed in dimensionless form, (s) becomes
HRePrH
L
45
4max (t)
PROBLEM 5.5
A thin liquid film flows under gravity down an
inclined surface of width W. The film thickness
is H and the angle of inclination is . The
solution to the equations of motion gives the
axial velocity u of the film as
2
22
22
sin
H
y
H
ygHu
Heat is added to the film along the surface
beginning at 0x at uniform flux .sq Determine the total heat added from 0x to the section
where the thermal boundary layer penetrates half the film thickness. Assume laminar boundary
layer flow and that .1/ Ht Neglect heat loss from the free surface. Use a third degree
polynomial temperature profile.
(1) Observations. (i) The velocity distribution is known. (ii) Total heat transfer is equal to heat flux times surface area. (iii) Heat fl ux is given. However, the distance x = L at which
2/Ht is unknown.
(2) Problem Definition. Determine the thickness of the thermal boundary layer ).(xt
(3) Solution Plan. (i) Express total heat in terms of heat flux and surface area. (ii) Use the
integral form of the energy equation to determine ).(xt
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties, (4) laminar
flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) flat surfa ce, (7) uniform surface heat flux, (8) negligible changes in kinetic and potential energy, (9) negligible axial conduction,
(10) negligible dissipation and (11) no buoyancy ( = 0).
(ii) Analysis. The total heat transfer rate from the surface is
LWqAqq ssT (a)
where
A surface area, 2m
L = distance along surface where 2/Ht , m
Tq total heat transfer rate, W
W = width of surface, m
To determine )(xt , we apply the integral form of the energy equation
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
where the axial velocity u is given by
PROBLEM 5.5 (continued)
2
22
22
sin
H
y
H
ygHu (b)
We assume a third degree polynomial temperature profile
33
2210 )()()()(),( yxbyxbyxbxbyxT (c)
The boundary conditions on the temperature are
(1) sqy
xTk
)0,(
(2) TxT t ),(
(3) 0,
y
Tx t
(4) 0)0,(
2
2
y
xT
Equation (c) and the four boundary conditions give the coefficients )(xbn
ts
k
qTb
3
20
k
qb s
1 , ,02b23
1
3 t
s
k
qb
Substituting the above into (c)
2
3
33
2),(
t
ts y
yk
qTyxT (d)
Substituting (b) and (d) into (5.7) and recalling that pck /
)(
033
22
2
sin2
3
2
22xt
t
tss dy
yy
k
q
H
y
H
ygH
dx
d
c
q
p
)(
036
1
2
1
333
2sin 5
2
324
2
2
xt
t
t
t
t dyyyyyH
HyyHdx
dg
c
k
p
Evaluating the integral in the above
43
72
1
15
sintt
H
dx
dg
c
k
p
Separating variables and rearranging
43
72
1
15
sintt
Hd
gdx
PROBLEM 5.5 (continued)
Integrating and noting that 0)0(t
43
72
1
15
sintt
Hgx (e)
Evaluating the above at Lx , where L is the distance form the leading end to the location
where 2/Ht , gives
sin
5760
43)2/(
72
1)2/(
15
sin 443 gH
HHHg
L (f)
Substituting (f) into (a)
sT qWgH
qsin
5760
43 4
(g)
(iii) Checking. Dimensional check: (i) Each term in (d) has units of temperature. (ii) Each term is (g) has units of watts.
Boundary conditions check: Assumed temperature profile satisfies the four boundary conditions on temperature.
Qualitative check: The total heat transfer rate is expected to increase with increasing angle .
This is in agreement with result (g).
(5) Comments. (i) Equation (g) shows that Tq is proportional to .4H Thus film thickness has
significant effect on the total heat transfer rate.
PROBLEM 5.6
A plate is cooled by a fluid with Prandtl number 1Pr . Surface temperature varies with
distance from the leading edge according to
xCTxTs )0,(
where C is constant. For such a fluid it is
reasonable to assume that .Vu Use a
third degree polynomial temperature
profile to show that the local Nusselt
number is given by
2/12/175.0 eRPrNux
and that surface heat flux is uniform. Assume laminar boundary layer flow.
(1) Observations. (i) The determination of the Nusselt number requires the determination of the velocity and temperature distributions. (ii). Velocity is assumed uniform. (iii) Surface temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution. Apply Newton’s law of cooling to determine surface heat flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)
laminar flow (Rex < 5 105), (5) viscous boundary layer flow (Rex > 100), (6) thermal boundary layer (Pe > 100), (7) uniform upstream velocity and temperature , (8) flat plate, (9) (10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,
(12) negligible dissipation, (13) no buoyancy ( = 0 or g = 0) and (14) 1Pr .
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
The heat transfer coefficient h given by equation (1.10)
TxT
y
xTk
hs )(
)0,(
(1.10)
PROBLEM 5.6 (continued)
Thus the temperature distribution ),( yxT must be determined. Surface heat flux is obtained
using Newton’s law of cooling
)( TThq ss (b)
The integral form of the energy equation is used to determine temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
Axial velocity distribution u(x,y) for 1Pr is assume to be the same as free stream velocity. Thus
Vu (c)
We assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (d)
The temperature boundary conditions are:
(1) )(0, xTxT s
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
The four boundary conditions are used to determine the coefficient in (d). The assumed profile becomes
3
3
2
1
2
3)()(),(
ttss
yyxTTxTyxT (e)
Substitute (e) into (1.10)
t
kxh
2
3)( (f)
Introducing (f) into (a)
tx
xNu
2
3 (g)
Thus the problem reduces to determining the thermal boundary layer thickness t . This is
accomplished using the integral form of the energy equation (5.7). Substituting (c) and (e) into (5.7)
PROBLEM 5.6 (continued)
)(
0
3
3
2
1
2
31)(
)(
2
3xt
tts
t
s dyyy
TxTdx
dTxT
V
Evaluate the integral
tst
s TxTdx
dTxT
V)(
8
3)(
2
3 (h)
However, surface temperature is given by
xCTTs (i)
Substitute into (h)
tt
xdx
dx
V4 (j)
To solve (j) for )(xt we let
2zx t (k)
Solve (k) for t
x
zt
2
(l)
Substitute (k) and (l) into (j)
dx
dzz
dx
dzzx
V
3242
2
Separate variables
dzzdxxV
32
Integrate
oCzxV
4
4
12 (m)
where oC is a constant determined from the boundary condition on )(xt :
0)0(t (n)
Apply (n) to (l) gives z(0) = 0 (o)
Apply (o) to (m) gives oC = 0. Equation (m) becomes
424 zxV
(p)
Use (l) to eliminate z in (p)
PROBLEM 5.6 (continued)
224 txxV
Solve the above for t
xV
t 2 (q)
Substitute (q) into (g) gives the local Nusselt number
xVNux
4
3
Noting that pck / , the above becomes
xV
k
c
k
xVcNu
ppx
4
3
4
3 (r)
Expressing this result in terms of the Prandtl and local Reynolds number, gives
2/12/175.0 RePrNux (s)
The heat transfer coefficient is determined to examine surface heat flux. Substitute (q) into (f), gives h(x)
x
Vkxh
4
3)( (t)
Substitute (i) and (t) into (b), simplify
VkCqs
4
3 (u)
This result shows that surface heat flux is uniform.
(5) Checking. Dimensional check: (i) Equations (g), (r) and (s) are dimensionless. (ii) Equations (e), (f), (h), (q), (t) and (u) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (e) satisfies the four boundary conditions.
Limiting check: If surface temperature is the same as free stream temperature, the heat flux
will be zero. According to (i) TxTs )( when C = 0. Setting C = 0 in (u) gives 0sq .
(6) Comments. The solution to the case of a specified uniform surface flux is given in Section 5.7.3. The corresponding surface temperature is given in (5.35)
1/21/3396.2)(
x
ss
RerP
x
k
qTxT (5.35)
Note that the above can be rewritten as
xCTxTs )(
PROBLEM 5.6 (continued)
where C is constant. This is identical to equation (i) which gives the specified surface
temperature in this problem.
PROBLEM 5.7
A plate is cooled by a fluid with Prandtl number 1Pr . Surface temperature varies with
distance form the leading edge according to
xCTxTs )0,(
where C is constant. For such a fluid it
is reasonable to assume that axial
velocity within the thermal boundary
layer is linear given by
yVu
Determine the local Nusselt number and show that surface heat flux is uniform. Use a third
degree polynomial temperature profile and assume laminar boundary layer flow.
(1)Observations. (i) The determination of the Nusselt number requires the determination of the velocity and temperature distributions. (ii). Velocity is assumed linear. (iii) Surface temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the velocity and temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the viscous and thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the momentum and energy equations to determine the velocity and temperature distribution. Apply Newton’s law of cooling to determine surface heat flux. Since
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)
laminar flow (Rex < 5 105), (5) viscous boundary layer flow (Rex > 100), (6) thermal boundary layer (Pe > 100), (7) uniform upstream velocity and temperature , (8) flat plate, (9) (10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,
(12) negligible dissipation, (13) no buoyancy ( = 0 or g = 0) and (14) 1Pr .
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
The heat transfer coefficient h given by equation (1.10)
TxT
y
xTk
hs )(
)0,(
(1.10)
PROBLEM 5.7 (continued)
Thus the temperature distribution ),( yxT must be determined. Surface heat flux is obtained
using Newton’s law of cooling
)( TThq ss (b)
The integral form of the energy equation is used to determine temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
Axial velocity distribution u(x,y) for 1Pr is assume to vary linearly with normal distance y from the plate. Thus
yVu (c)
where is the viscous boundary layer thickness. Before proceeding with the determination
of temperature distribution, must be determined. We apply the integral form of the
momentum equation (5.5)
dyudx
dudy
dx
dV
y
xuv
xx
0
2
0
0, (5.5)
Substitute (c) into (5.5)
dyydx
dVydy
dx
dV
Vv
xx
0
2
2
2
0
2 11
Evaluate the integrals and simplify
dx
dVv
6
1
Separate variables and integrate
1
2
26 Cdx
V (d)
The constant of integration 1C is obtained from the boundary condition on
0)0(
This condition gives 1C = 0. Substituting into (d) and solving for
xV
12 (e)
Turning now to the temperature distribution, we assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (f)
PROBLEM 5.7 (continued)
The temperature boundary conditions are:
(1) )(0, xTxT s
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
The four boundary conditions are used to determine the coefficient in (f). The assumed profile becomes
3
3
2
1
2
3)()(),(
ttss
yyxTTxTyxT (g)
Substitute (g) into (1.10)
t
kxh
2
3)( (h)
Introducing (h) into (a)
tx
xNu
2
3 (i)
Thus the problem reduces to determining the thermal boundary layer thickness t . This is
accomplished using the integral form of the energy equation (5.7). Substituting (c) and (g) into (5.7)
)(
0
3
3
2
1
2
31)(
)(
2
3
xt
tts
t
s dyyy
TxTy
dx
dTxT
V
Evaluate the integral
2)(
10
1)(
2
3t
s
t
s TxT
dx
dTxT
V (j)
However, surface temperature is given by
xCTTs (k)
Substitute into (j)
215 tt
x
dx
dx
V (l)
To solve (l) for )(xt we first rewrite (l)
PROBLEM 5.7 (continued)
2
15 t
t
xdx
dx
V (m)
Use (e) to eliminate 1/ on the left side of (m) and on the right side
2
4
5 t
t
xdx
d (n)
To solve (n) for /t , let
2zx t (o)
substitute into (n)
dx
dzzx 2
8
5
Separate variables and integrate
232/3
3
1
24
10Czx (p)
where 2C is a constant determined from the boundary condition on )(xt
0)0(t (q)
Apply (q) to (o) z(0) = 0 (r)
Apply (r) to (p) gives 1C = 0. Equation (p) becomes
32/3
4
5zx (s)
Use (o) to eliminate z in (s) 3
4
5 t
Use (e) to eliminate and solve for x
t
xVx
t3/1
2
35 (t)
Note that
Pr (u)
and
PROBLEM 5.7 (continued)
xVRex (v)
Substitute (u) and (v) into (t)
2/13/1
2
35x
t RePrx
(w)
(w) into (i) gives the local Nusselt number
2/13/1
5
3xx RePrNu (x)
To examine surface heat flux we determine the heat transfer coefficient. Substitute (a) into (w), and solve for h(x)
2/13/1
5
3xRePr
x
kh
Use (v) to eliminate the Reynolds number in the above
xV
Prkh
2/13/1
5
3 (y)
Substitute (k) and (y) into (b), simplify 2/1
3/1
5
3 VPrkCqs (z)
This result shows that surface heat flux is uniform.
(5) Checking. Dimensional check: (i) Equations (i), (t), (w) and (x) are dimensionless. (ii) Equations (e), (g), (h), (y), and (z) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (g) satisfies the four boundary conditions.
Limiting check: If surface temperature is the same as free stream temperature, the heat flux
will be zero. According to (k) TxTs )( when C = 0. Setting C = 0 in (z) gives 0sq .
(6) Comments. The solution to the case of a specified uniform surface flux is given in Section 5.7.3. The corresponding surface temperature is given in (5.35)
1/21/3396.2)(
x
ss
RerP
x
k
qTxT (5.35)
Note that the above can be rewritten as
xCTxTs )(
where C is constant. This is identical to equation (k) which gives the specified surface
temperature in this problem.
PROBLEM 5.8
Surface temperature of a plate increases exponentially with distance from the leading edge
according to
)exp()0,( xCTxTs
where C and are constants. The plate is cooled
with a low Prandtl number fluid ( 1Pr ). Since
for such fluids t , it is reasonable to assume
uniform axial velocity within the thermal boundary
layer. That is
Vu
Assume laminar boundary layer flow and use a third degree polynomial temperature
profile.
[a] Show that the local Nusselt number is given by
2/12/12/1)exp(175.0 xx eRPrxxNu
[b] Determine surface flux distribution.
(1) Observations. (i) The determination of the Nusselt number requires the determination of the velocity and temperature distributions. (ii). Velocity is assumed uniform. (iii) Surface temperature is variable. (iv) Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile to determine the temperature distribution. Apply Newton’s law of cooling to determine surface heat flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4)
laminar flow (Rex < 5 105), (5) viscous boundary layer flow (Rex > 100), (6) thermal boundary layer (Pe > 100), (7) uniform upstream velocity and temperature , (8) flat plate, (9) (10) negligible changes in kinetic and potential energy, (11) negligible axial conduction,
(12) negligible dissipation, (13) no buoyancy ( = 0 or g = 0) and (14) 1Pr .
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
PROBLEM 5.8 (continued)
The heat transfer coefficient h given by equation (1.10)
TxT
y
xTk
hs )(
)0,(
(1.10)
Thus the temperature distribution ),( yxT must be determined. Surface heat flux is obtained
using Newton’s law of cooling
)( TThq ss (b)
The integral form of the energy equation is used to determine temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
Axial velocity distribution u(x,y) for 1Pr is assume to be the same as free stream velocity. Thus
Vu (c)
We assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (d)
The temperature boundary conditions are:
(1) )(0, xTxT s
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
The four boundary conditions are used to determine the coefficient in (d). The assumed profile becomes
3
3
2
1
2
3)()(),(
ttss
yyxTTxTyxT (e)
Substitute (e) into (1.10)
t
kxh
2
3)( (f)
Introducing (f) into (a)
tx
xNu
2
3 (g)
PROBLEM 5.8 (continued)
Thus the problem reduces to determining the thermal boundary layer thickness t . This is
accomplished using the integral form of the energy equation (5.7). Substituting (c) and (e) into (5.7)
)(
0
3
3
2
1
2
31)(
)(
2
3
xt
tts
t
s dyyy
TxTdx
dTxT
V
Evaluate the integral
tst
s TxTdx
dTxT
V)(
8
3)(
2
3 (h)
However, surface temperature is given by
)exp( xCTTs (i)
Substitute into (h)
)exp()exp(
4 xdx
dx
Vt
t
(j)
Differentiate the right side of (j), the above becomes
dx
dxx
x
V
tt
t
)exp()exp()exp(
4
The above is simplified to
dx
d
V
ttt
24
Rewrite as
dx
d
V
ttt
24
Separate variables
24 t
tt
V
ddx
Integrate
ot CV
x 24ln2
1 (k)
where oC is a constant determined from the boundary condition on )(xt
0)0(t (l)
Apply (l) to (k) gives
VCo 4ln
2
1 (m)
(m) into (k)
PROBLEM 5.8 (continued)
V
Vx
t
4
4
ln2
2
(n)
Rewrite
V
Vx
t
4
4
)2exp(
2
Solve the above for t
)2exp(14 xV
t (o)
Substitute (o) into (g) gives the local Nusselt number
)2exp(1
1
42
32
x
xVNux (p)
To express (p) in terms of the Prandtl and local Reynolds numbers, rewrite the above and
note that pck /
)2exp(1
)(
4
3
x
xxV
k
cNu
px
This is written as
xx PrRex
xNu
)2exp(1
)(75.0 (r)
The heat transfer coefficient is determined to examine surface heat flux. Substitute (o) into (f), gives h(x)
)2exp(1
1
4
3)(
x
Vkxh (s)
Substitute (i) and (s) into (b), simplify
)2exp(1
)2exp(
4
3
x
xVkCqs (t)
This result shows that surface heat flux is varies with distance along the surface.
Expressing the above in dimensionless form, gives
)2exp(1
)2exp(
4
3
x
x
VkC
q (u)
PROBLEM 5.8 (continued)
(5) Checking. Dimensional check: (i) Equations (g), (p), (r) and (u) are dimensionless. (ii) Equations (e), (f), (k), (o), (s) and (t) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (e) satisfies the four boundary conditions.
Limiting check: (i) If surface temperature is the same as free stream temperature, the heat
flux will be zero. According to equation (i), TxTs )( when C = 0. Setting C = 0 in (u)
gives 0sq .
(ii) According to equation (i), surface temperature is the same as free stream temperature for .0 this case should give zero heat flux. Setting 0 in (t) and applying L’Hospital’s
rule, give 0sq .
(6) Comments. Solution (t) shows that surface heat flux increases with x. Recall that for uniform surface temperature Pohlhausen’s solution shows that the heat transfer coefficient and surface flux decrease with x. However, in this problem surface temperature and the heat transfer coefficient increase with x. This results in a flux that increases with x.
PROBLEM 5.9
A square array of chips of side L is mounted flush on a flat
plate. The chips dissipate non-uniform surface flux according to
x
Cqx
The plate is cooled by forced convection with uniform upstream
velocity V and temperature T . Assume laminar boundary
layer flow with .1/t Use third degree polynomials for the
axial velocity and temperature.
[a] Show that the local Nusselt number is given by
2/13/1331.0 eRPrNux
[b] Show that surface temperature is uniform.
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Surface heat flux is variable. It decreases with distance x. (iii) Surface temperature is unknown. (iv) Newton’s law of cooling gives surface temperature.
This requires knowing the local heat transfer coefficient. (v) .1/t
(2) Problem Definition. Determine the velocity and temperature distribution for boundary layer flow over a flat plate at variable surface temperature. This reduces to determining the viscous and thermal boundary layer thickness.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the momentum and energy equations using a third degree polynomial profiles to determine the velocity and temperature distribution. Apply Newton’s law of cooling to determine surface temperature.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow
(Rex < 5 105), (4) two-dimensional, (5) viscous boundary layer flow (Rex > 100), (6) thermal boundary layer (Pe > 100), (7) uniform upstrea m velocity and temperature , (8) flat plate, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11)
negligible dissipation and (12) no buoyancy ( = 0 or g = 0).
(ii) Analysis. [a] The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TxT
y
xTk
hs )(
)0,(
(1.10)
PROBLEM 5.9 (continued)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
The velocity solution, ),( yxu , for an assumed third degree polynomial is solved in Section 5.7.1
and given by equation (5.9) 3
2
1
2
3 yy
V
u (5.9)
where the integral solution to )(x is
xRex
64.4 (5.10)
The local Reynolds number is defined as
xVRex (b)
For the temperature profile we assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (c)
The boundary conditions on the temperature are
(1) x
C
y
xTk
0,
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (c) and the four boundary conditions give the coefficients )(xbn
,3
20 t
x
C
kTb
x
C
kb
11 , ,02b
x
C
kb
t23
1
3
1
Substituting the above into (c)
3
23
1
3
2),( yy
xk
CTyxT
t
t (d)
Surface temperature is obtained by setting 0y in (d)
PROBLEM 5.9 (continued)
tsxk
CTxTxT
3
2)0,()( (e)
Substituting (d) and (e) into (1.10)
t
kh
2
3 (f)
Combining (a) and (f)
tx
xNu
2
3 (g)
Thus to determine surface temperature and Nusselt number requires the determination of t .
Application of the energy equation gives .t Substitute (5.9) and (d) into (5.7)
)(
0
3
2
3
3
1
3
2
2
1
2
311
xt
t
t dyyyyy
xdx
d
xV
Expanding the integrand and evaluating the integral in the above
532
/140
1/
10
11tt
xdx
d
xV (h)
Since 1/t , it follows that
35/
10
1/
140
1tt
Dropping the last term in (h), gives 3
1110 t
xdx
d
xV (i)
Separate variables and integrate
tt
xdx
ddx
xV
x 311
10
Evaluate the integrals
ot C
xx
V
31
20 (j)
where oC is constant of integration. The boundary condition on t is
0)0(t (k)
Apply (k) to (j) gives 0oC . Substitute into (j) and solve for t
3/1
20 xV
t (l)
Use (5.10) to eliminate in (l)
PROBLEM 5.9 (continued)
3/12
)64.4)(20(x
tRe
x
V (m)
Express (m) in dimensionless form
3/1
1)64.4)(20(
x
t
RexVx
Note that pck / , the above is rewritten as
3/1
15274.4
xp
t
RexVc
k
x
Introduce the definition of the Prandtl number, the above gives
2/13/1
5274.4
x
t
RePrx (n)
Substitute (n) into (g) gives the local Nusselt number
2/13/1331.0 xx RePrNu (o)
[b] Surface temperature is obtained by apply Newton’ s law of cooling or substituting (e) into (e). Newton’s law gives
h
qTT s
s (p)
Substitute (a) into (o) gives h
2/13/1331.0 xRePrx
kh (q)
Surface flux is given by
x
Cqs (r)
(b), (q) and (r) into (p)
xxV
Prk
xCTTs
3/1331.0
Note that in the above variable x cancels out to give
VPrk
CTTs
3/1331.0
(s)
PROBLEM 5.9 (continued)
This result can be expressed in dimensionless form as
3/1331.0
1
Pr
Vk
C
TTs (t)
This result shows that surface temperature is uniform.
(iii) Checking. Dimensional check: (i) Equations (5.9), (5.10), (b) and (n) are dimensionless. (ii) Equations (d), (f), (l), (m) and (s) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary conditions.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream
temperature. Zero flux corresponds to C = 0. Setting C = 0 in (s) gives .)( TxTs
(5) Comments. (i)The solution is valid for .1/t This corresponds to .1Pr
(ii) Pohlhausen’s solution for a plate at uniform surface temperature gives h(x) as
d
d
x
Vkxh
)0()( (4.66)
The corresponding surface heat flux is
d
d
x
VTTkTTxhq sss
)0()())((
This result can be expressed as
x
Cqs (u)
In this problem a surface flux of the form (u) results in uniform surface temperature. This is in agreement with Pohlhausen’s solution. Thus this problem is identical to Pohlhausen’s problem of flow over a plate at uniform surface temperature.
(iii) Pohlhausen’s solution for the local Nusselt number is
2/13/1331.0 xx RePrNu 10Pr (4.72c)
This is in good agreement with the integral solution(o).
PROBLEM 5.10
A square array of chips of side L is mounted flush on a flat plate. The forward edge of the array
is at a distance ox from the leading edge of the plate. The heat dissipated in each row increases
with successive rows as the distance from the forward edge increases. The distribution of surface
heat flux for this arrangement may be approximated by
2Cxqs
where C is constant. The plate is cooled by forced
convection with uniform upstream velocity V and
temperature T . Assume laminar boundary layer
flow. Assume further that the axial velocity within
the thermal boundary layer is equal to the free
stream velocity, Vu . Use a third degree
polynomial temperature profile.
[a] Show that the local Nusselt number is given by
2/12/12/1
3)/(13.1 eRPrxxNu ox
[b] Determine the maximum surface temperature
[c] How should the rows be rearranged to reduce the maximum surface temperature?
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be uniform,
Vu . This represents a significant simplification. (iii) Surface heat flux is variable. It
increases with distance x. (iv) Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x.Thus maximum surface temperature is at the trailing end x = L. (v) Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a non-uniformly heated flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow
(Rex < 5 105), (4) two-dimensional, (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) negligible changes in kinetic and potential energy,
(9) negligible axial conduction, (10) negligible dissipation, (11) no buoyancy ( = 0 or g = 0) and (12) uniform velocity within the thermal boundary layer ( 1Pr ).
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
PROBLEM 5.10 (continued)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
However, For 1Pr the velocity boundary layer thickness is much smaller than the thermal boundary layer thickness. Thus we assume
Vu (b)
For the temperature profile we assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (c)
The boundary conditions on the temperature are
(1) 20,Cx
y
xTk
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (c) and the four boundary conditions give the coefficients )(xbn
,3
2 20 tx
k
CTb 2
1 xk
Cb , ,02b
2
2
33
t
x
k
Cb
Substituting the above into (c)
2
32
3
1
3
2),(
tt
yyx
k
CTyxT (d)
Surface temperature is obtained by setting 0y in (d)
ts xk
CTxTxT 2
3
2)0,()( (e)
Substituting (d) and (e) into (1.10)
t
kh
2
3 (f)
PROBLEM 5.10 (continued)
Combining (a) and (f)
tx
xNu
2
3 (g)
Thus to determine surface temperature and Nusselt number requires the determination of t .
Application of the energy equation gives .t Substituting (b) and (d) into (5.7)
)(
0
2
322
3
1
3
2
xt
tt dy
yyx
dx
dVx
Evaluating the integral in the above
2222222
412
1
2
1
3
2tttt x
dx
dVx
dx
dVx
Separating variables and rearranging
dxxV
xd t222 4
Integrating
dxxV
xd t222 4
Performing the integration
ot CxV
x 322
3
4 (h)
where oC is constant of integration. The boundary condition on t is
0)( ot x (i)
Applying (i) to (h) gives
3
3
4oo x
VC
Substituting into (h) and solving for t
xxxV
ot3)/(1
3
4 (j)
Substituting (j) into (g)
3)/(1
1
4
33
xx
xVNu
o
x (k)
Noting that
p
p
ckk
cPr
/
/(l)
[a] Using (l), equation (k) is expressed in terms of Prandtl and Reynolds numbers
2/12/12/1
3)/(13.1 eRPrxxNu ox (m)
PROBLEM 5.10 (continued)
[b] Surface temperature is obtained by substituting (j) into (e)
32/5 )/(133
4xxx
Vk
CTT os (n)
This result shows that surface temperature increases with x. Thus maximum temperature is at the trailing end :Lx
32/5max )/(1
33
4)( LxL
Vk
CTT os (o)
[c] Since heat transfer coeffici ent decreases with distance from the leading end, rows of high power density chips should be placed near the leading edge and low density rows towards the trailing end.
(iii) Checking. Dimensional check: Dimensional check: (1) Equations (g), (k) and (l) are dimensionless. (2) Equations (d), (f), (j) and (n) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary conditions on temperature.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream
temperature. Zero flux corresponds to C = 0. Setting C = 0 in (n) gives .)( TxTs
(5) Comments. (i) For the special case of no leading insulated section, equations (m) and (n) reduce to
xx PrReNu 3.1 (p)
2/5
33
4x
Vk
CTTs (q)
(ii) Application of (m) at oxx gives infinite Nusselt number. This anomaly is due to the fact
that boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.
PROBLEM 5.11
Repeat Problem 5.10 using a linear surface flux distribution .Cxqs
[a] Show that the local Nusselt number is given by
2/12/12/1
3)/(106.1 eRPrxxNu ox
[b] Determine the maximum surface temperature
[c] How should the rows be rearranged to reduce the
maximum surface temperature?
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be uniform,
Vu . This represents a significant simplification. (iii) Surface heat flux is variable. It
increases with distance x. (iv) Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x.Thus maximum surface temperature is at the trailing end x = L. (v) Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a non-uniformly heated flat plate with insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties, (3) laminar flow
(Rex < 5 105), (4) two-dimensional, (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) negligible changes in kinetic and potential energy,
(9) negligible axial conduction, (10) negligible dissipation, (11) no buoyancy ( = 0 or g = 0) and (12) uniform velocity within the thermal boundary layer ( 1Pr ).
(ii) Analysis. [a] The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution )(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
PROBLEM 5.11 (continued)
However, For 1Pr the velocity boundary layer thickness is much smaller than the thermal boundary layer thickness. Thus we assume
Vu (b)
For the temperature profile we assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (c)
The boundary conditions on the temperature are
(1) xCy
xTk
0,
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (c) and the four boundary conditions give the coefficients )(xbn
,3
20 tx
k
CTb x
k
Cb1 , ,02b
233
t
x
k
Cb
Substituting the above into (c)
3
2
1
3
1
3
2),( yy
k
xCTyxT
tt (d)
Surface temperature is obtained by setting 0y in (d)
ts xk
CTxTxT
3
2)0,()( (e)
Substituting (d) and (e) into (1.10)
t
kh
2
3 (f)
Combining (a) and (f)
tx
xNu
2
3 (g)
Thus to determine surface temperature and Nusselt number requires the determination of t .
Application of the energy equation gives .t Substituting (b) and (d) into (5.7)
)(
0
2
3
3
1
3
2
xt
tt dy
yyx
dx
dVx
Evaluating the integral in the above
PROBLEM 5.11 (continued)
2222
4
1
12
1
2
1
3
2tttt x
dx
dVxxx
dx
dVx
Separating variables and rearranging
xdxV
xd t
42
Integrating
xdxV
xd t
42
Evaluating the integral
ot CxV
x 22 2 (h)
where oC is constant of integration. The boundary condition on t is
0)( ot x (i)
Applying (i) to (h) gives
22oo x
VC
Substituting into (h) and solving for t
xxxV
ot2)/(1
2 (j)
Substituting (j) into (g)
2)/(1
1
22
3
xx
xVNu
o
x (k)
Noting that pck / , the above is expressed as
2
2/12/1
)/(106.1
xx
eRPrNu
o
x (l)
[b] Surface temperature is obtaine d by substituting (j) into (e)
32)/(13
22xxx
Vk
CTT os (m)
This result shows that surface temperature increases with x. Thus maximum temperature is at the trailing end :Lx
32)/(13
22max, LLx
Vk
CTT os (n)
[c] According to (l), the heat transfer coefficien t decreases with distance from the leading end. Thus rows of high power density chips should be placed near the leading edge and low density rows towards the trailing end.
PROBLEM 5.11 (continued)
(iii) Checking. Dimensional check: Dimensional check: (i) Equations (g), (k) and (l) are dimensionless. (ii) Equations (d), (f), (j) and (m) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the four boundary conditions.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as free stream
temperature. Zero flux corresponds to C = 0. Setting C = 0 in (m) gives .)( TxTs
(5) Comments. (i) For the special case of no leading insulated section, equations (l) and (m) reduce to
xx PrReNu 06.1 (o)
3
3
22x
Vk
CTTs (p)
(ii) Application of (l) at oxx gives infinite Nusselt number. This anomaly is due to the fact
that boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.
PROBLEM 5.12
A fluid at temperature oT and flow rate om is injected radially between parallel plates. The
spacing between the plates is H. The upper plate is insulated and the lower plate is maintained at
uniform temperature sT along oRr and is insulated along .0 oRr Consider laminar
boundary layer flow and assume that the radial velocity u does not vary in the direction normal
to the plates (slug flow).
[a] Show that for a cylindrical element rdrt 2 the external mass flow edm to the thermal
boundary layer is
to
e dH
mdrru
dr
ddm
t
0
2
[b] Show that the integral form of conservation of energy is
dyTTdr
d
H
cm
r
rTkr o
pot
)(2
)0,(
0
[c] Assume a linear temperature profile, show that the local Nusselt number is
2/12/12/1
2/1
2
1ror eRPrrRNu
where
H
murRe o
r2
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) Velocity variation with y is negligible. (iii) Conservation of mass requires that radial velocity decrease with radial distance r. (iv) Surface temperature is uniform.
(2) Problem Definition. Determine the integral formulation of conservation of mass and energy. Determine the temperature distribution within the thermal boundary layer. This requires the
determination of ).(xt
(3) Solution Plan. (i) Apply conservation of mass and energy to a cylindrical element
.2 rdrt (ii) Use the integral form of the energy equation to determine ).(xt
PROBLEM 5.12 (continued)
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional temperature field, (3) slug flow (velocity varies radially only), (4) constant properties, (5) laminar flow, (6) thermal boundary layer flow, (7) flat surface, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation and (12) no
buoyancy ( = 0).
(ii) Analysis. [a] External mass flow rate. Consider a
volumetric cylindrical element rdrt 2 . The diagram
shows the cross section of the element and the external and radial mass flow rates. Conservation of mass gives
drdr
dmmmdm r
rre
or
re dmdm (a)
where
edm external mass flow element into the boundary layer
rm radial mass flow rate entering element
Integral formulation of rm is
rdyum
t
r
0
2
where u is radial velocity and is density. For constant density and slug flow the above
simplifies to
dyurm
t
r
0
2 (b)
Evaluating the integral
tr urm 2 (c)
Application of conservation of mass between the inlet and channel section location r, gives
omrHu2
Solving for u
rH
mu o
2 (d)
(d) into (c)
to
rH
mm (e)
Differentiating (e)
to
r dH
mdm (f)
drdr
dmm r
rrm
edm
dr
t
PROBLEM 5.12 (continued)
(f) into (a)
to
e dH
mdm (g)
[b] Conservation of energy fo r the volumetric cylindrical
element rdrt 2 shown gives
drdr
dEEdEEdq r
rers
This simplifies to
drdr
dEdEdq r
es (h)
where
edE = energy supplied by the external mass
rE = energy convected radially through the boundary layer
sdq = energy conducted to the element through the surface
We now formulate the three energy components:
eope dmTcdE
(d) into the above
to
ope dH
mTcdE (i)
ruTdycE pr 2
(f) into the above
TdyH
mcE
top
r
0
Differentiating the above
Tdydr
d
H
mc
dr
dE topr
0
(j)
Fourier’s law gives
dry
rTrkdqs
)0,(2 (k)
Substituting (i)-(k) into (h)
drTdydr
d
H
mcd
H
mTcdr
y
rTrk
top
to
op
0
)0,(2
Dividing through by dr2 and rearrange the above
dr
t
drdr
dEE r
rrE
edE
sdq
PROBLEM 5.12 (continued)
dr
dTTdy
dr
d
H
mc
y
rTrk t
oop
t
0
)0,( (l)
However, the last term in (l) can be written as
t
dyTdr
d
dr
dT o
to
0
(m)
(m) into (l)
dyTTdr
d
H
cm
r
rTkr o
pot
)(2
)0,(
0
(n)
[c} Nusselt number. The local Nusselt number is defined as
k
hrNur (o)
where the heat transfer coefficient h is given by equation (1.10)
os TT
y
rTk
h
)0,(
(1.10)
Thus h depends on the temperature distribution ).,( yrT The integral form of the energy equation
is used to determine the temperature distribution. We assume a linear temperature profile
yrbrbyrT )()(),( 10 (p)
The boundary conditions on the temperature are
(1) sTrT )0,(
(2) oTrT t ),(
Equation (p) and the two boundary conditions give
tss
yTTTyrT o )(),( (q)
Substituting (q) into (1.10)
t
kh
When this is substituted into(o), we obtain
tr
rNu (s)
Thus the problem becomes one of determining the thermal boundary layer thickness .t The
integral form of conservation of energy is used to determine .t Substituting (q) into (n)
PROBLEM 5.12 (continued)
dyy
dr
dTT
H
cmTTkr
t
tso
po
t
so
0
1)(2
Evaluating the integral and simplifying
dr
d
H
cmkr tpo
t 4
Separating variables
rdrcm
kHd
pott
4
Integrating
opo
t Crcm
kH 22
2
2 (t)
where oC is constant of integration. The boundary condition on t is
0)( ot R (u)
(t) and (u) give
22o
poo R
cm
kHC
Substituting into (t) and solving for t
224o
pot Rr
cm
kH (v)
(v) into (s)
2/122
4o
por Rr
kH
cmrNu
Substituting (d) into the above and rearranging
2/122
2o
pr Rr
k
crurNu
This result can be expressed in terms of the Reynolds and Prandtl numbers as
2/12/12/1
2/1
2
1ror eRPrrRNu (w)
urRer
PROBLEM 5.12 (continued)
(iii) Checking. Dimensional check: (1) Units of (d), (e), (i), (k), (n) and (v) are correct. (2) Equations (s) and (w) are dimensionless.
Boundary conditions check: Assumed temperature profile (q) satisfies the two boundary conditions on temperature.
(5) Comments. (i) The assumption of bulk flow provided a significant simplification. (ii)
Application of (w) at oRr gives infinite Nusselt number. This anomaly is due to the fact that
boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.
PROBLEM 5.13
The lower plate in Problem 5.12 is heated with uniform flux sq along oRr and insulated
along .0 oRr
[a] Show that for a cylindrical element drrt 2 the external mass flow edm to the thermal
boundary layer is
to
e dH
mdrru
dr
ddm
t
0
2
[b] Show that the integral form of conservation of energy is
dyTTdr
d
H
cmq o
pos
t
)(2 0
[c] Assume a linear temperature profile show that the local Nusselt number is
2/12/12/1
2/1 ror eRPrrRNu
where
H
murRe o
r2
(1) Observations. (i) This problem is described by cylindrical coordinates. (ii) Velocity variation with y is negligible. (iii) Conservation of mass requires that radial velocity decrease with radial distance r. (iv) Surface heat flux is uniform. (v) Surface temperature is unknown.
(2) Problem Definition. Determine the integral formulation of conservation of mass and energy. Determine the temperature distribution within the thermal boundary layer. This requires the
determination of ).(xt
(3) Solution Plan. (i) Apply conservation of mass and energy to a cylindrical element
.2 rdrt (ii) Use the integral form of the energy equation to determine ).(xt
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional temperature field, (3) slug flow (velocity varies radially only), (4) constant properties, (5) laminar flow, (6) thermal boundary layer, (7) flat surface, (8) uniform surface heat flux, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation and (12) no
buoyancy ( = 0).
PROBLEM 5.13 (continued)
(ii) Analysis. [a] External mass flow rate. Consider a
volumetric cylindrical element rdrt 2 . The diagram
shows the cross section of the element and the external and radial mass flow rates. Conservation of mass gives
drdr
dmmmdm r
rre
or
re dmdm (a)
where
edm external mass flow element into the boundary layer
rm radial mass flow rate entering element
Integral formulation of rm is
rdyum
t
r
0
2
where u is radial velocity and is density. For constant density and slug flow the above
simplifies to
dyurm
t
r
0
2 (b)
Evaluating the integral
tr urm 2 (c)
Application of conservation of mass between the inlet and channel section location r, gives
omrHu2
Solving for u
rH
mu o
2 (d)
(d) into (c)
to
rH
mm (e)
differentiating (e)
to
r dH
mdm (f)
(f) into (a)
to
e dH
mdm (g)
[b] Conservation of energy for th e volumetric cylindrical element rdrt 2 shown gives
drdr
dmm r
rrm
edm
dr
t
PROBLEM 5.13 (continued)
drdr
dEEdEEdq r
rers
This simplifies to
drdr
dEdEdq r
es (h)
where
edE = energy supplied by the external mass
rE = energy convected radially through the boundary
layer
sdq = energy conducted to the element through the surface
We now formulate the three energy components:
eope dmTcdE
(d) into the above
to
ope dH
mTcdE (i)
ruTdycE pr 2
(f) into the above
TdyH
mcE
top
r
0
Differentiating the above
Tdydr
d
H
mc
dr
dE topr
0
(j)
Fourier’s law gives
rdrqdq ss 2 (k)
where
sq surface heat flux
Substituting (i)-(k) into (h)
drTdydr
d
H
mcd
H
mTcdrqr
t
op
to
ops
0
2
Dividing through by dr2 and rearrange the above
dr
dTTdy
dr
d
H
mcrq t
o
op
s
t
0
(l)
However, the last term in (l) can be written as
dr
t
drdr
dEE r
rrE
edE
sdq
PROBLEM 5.13 (continued)
t
dyTdr
d
dr
dT o
to
0
(m)
(m) into (l)
dyTTdr
d
H
cmrq o
po
s
t
)(2 0
(n)
[c} Nusselt number. The local Nusselt number is defined as
k
hrNur (o)
where the heat transfer coefficient h is given by equation (1.10)
oTxT
y
rTk
hs )(
)0,(
(1.10)
Thus h depends on the temperature distribution ).,( yrT The integral form of the energy equation
is used to determine the temperature distribution. We assume a linear temperature profile
yrbrbyrT )()(),( 10 (p)
The boundary conditions on the temperature are
(1) sqy
rTk
)0,(
(2) oTrT t ),(
Equation (p) and the two boundary conditions give
)(),( yk
qTyrT t
so (q)
Surface temperature is obtained by setting 0y in (q)
tk
qTrTxT s
os )0,()( (r)
Substituting (q) and (r) into (1.10)
t
kh
When this is substituted into(o), we obtain
tr
rNu (s)
Thus the problem becomes one of determining the thermal boundary layer thickness .t The
integral form of conservation of energy is used to determine .t Substituting (q) into (n)
PROBLEM 5.13 (continued)
dyydr
d
Hk
cmr
t
t
po
0
)(2
Evaluating the integral
dr
d
kH
cmr tpo
2
4
Separating variables
rdrcm
kHd
po
t
42
Integrating
o
po
t Crcm
kH 22 2 (t)
where oC is constant of integration. The boundary condition on t is
0)( ot R (u)
(t) and (u) give
22o
poo R
cm
kHC
Substituting into (t) and solving for t
222o
po
t Rrcm
kH (v)
(v) into (s)
)(
1
2 22o
po
rRrkH
cmrNu
Substituting (d) into the above and rearranging
2/122o
p
r Rrk
crurNu
This result can be expressed in terms of the Reynolds and Prandtl numbers as
2/12/12/1
2/1 ror eRPrrRNu (w)
urRer
Substituting (v) in (r) gives surface temperature distribution
tk
qTrTxT s
os )0,()(
This result can be written in dimensionless form as
PROBLEM 5.13 (continued)
22)0,()( o
p
sos Rr
ruc
k
k
qTrTxT
1)/(1)( 2
oos
os RrPrRe
k
Rq
TxT
r
(iii) Checking. Dimensional check: (1) Units of (d), (e), (i), (k), (n) and (v) are correct. (2) Equations (s) and (w) are dimensionless.
Boundary conditions check: Assumed temperature profile (q) satisfies the two boundary conditions on temperature.
Limiting check: If surface heat flux vanishes, surface temperature will be the same as inlet fluid
temperature .oT Setting 0sq in (r) gives .)( os TxT
(5) Comments. (i) The assumption of bulk flow provided a significant simplification. (ii) The
solution does not apply to the limiting case of 0oR since the flow is three-dimensional and
thus cannot be approximated by bulk conditions. (ii) Application of (w) at oRr gives infinite
Nusselt number. This anomaly is due to the fact that boundary layer approximations (neglecting axial conduction) is not valid near the leading edge of the thermal boundary layer.
PROBLEM 5.14
A porous plate with an impermeable and insulated leading section of length ox is maintained at
uniform temperature sT along .oxx The plate is cooled by forced convection with a free
stream velocity V and temperature .T
Fluid at temperature oT is injected through
the porous surface with uniform velocity .ov
The injected and free stream fluids are
identical. Assume laminar boundary layer
flow, introduce axial velocity simplification
based on 1Pr and use a linear
temperature profile to determine the local
Nusselt number.
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be uniform,
Vu . This represents a significant simplification. (iii) The plate is porous. (iv) Fluid is
injected through the plate with uniform velocity. (v) The plate is maintained at uniform surface temperature. (vi) A leading section of the plate is insulated.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a porous flat plate with surface injection and insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar
flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface temperature, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation, (12) no buoyancy
( = 0 or g = 0), (13) uniform axial velocity within the thermal boundary layer ( 1Pr ), (14) uniform porosity and (15) injected fluid is at uniform velocity and temperature and is the same as the external fluid.
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
PROBLEM 5.14 (continued)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution
TTPcudydx
dTcuTdyc
dx
d
y
xTPk oop
t
p
t
p
xx
v
)()(
00
0,1 (5.6)
This equation is simplified for constant properties
TTPdyTTudx
d
y
xTP oo
tx
v
)(
0
)(0,
1 (b)
However, For 1Pr the velocity boundary layer thickness is much smaller than the thermal boundary layer thickness. Thus we assume
Vu (b)
Assume a linear temperature profile
yxbxbyxT 10, (c)
The boundary conditions on the temperature are
(1) sTxT 0,
(2) TxT t,
Equation (d) and the two boundary conditions give the coefficients )(xbn
,0 sTbt
sTTb1
)(1
Substituting the above into (d)
tss
yTTTyxT )(),( (d)
Substituting (e) into (1.10)
t
kh (e)
Combining (a) and (e)
t
x
xNu (f)
The problem reduces to finding t which is obtained using the energy equation. Substituting (c)
and (e) into (b)
TTPdyy
dx
dTTV
TTP oo
t
ts
t
s
x
v
)(
0
1)()(
1
Evaluating the integral and rearranging
PROBLEM 5.14 (continued)
TT
TT
V
P
dx
d
V
P
s
oot
t
v
2
111 (h)
Rewriting (h)
dx
d
V
P t
t 2
111 (i)
where is constant, defined as
TT
TT
V
Pv
s
oo (j)
Equation (i) is solved for t by separating variables
t
tt
V
P
ddx
212
(k)
Integrating (k) and noting that 0)( ot x
t
t
tt
o
V
P
ddx
x
x 0 212
(l)
Evaluating the integrals
t
to
s
oo
P
VV
P
TT
TT
P
Vxx
)1(1
1ln
2
1
)(
)(
2 2v (m)
This result gives an implicitly solution for t . The procedure for determining the Nusselt number
at a given x is to select a value for x, use (m) to determine the corresponding t and substitute into
(g).
(iii) Checking. Dimensional check: Equations (h), (i) and (j) are dimensionally correct.
Boundary conditions check: Assumed temperature profile (d) satisfies the two boundary conditions on temperature.
(5) Comments. Limiting checks are on solution (m) do not yield useful results. For example, for a solid plate, 0P , the first term on the right side of (m) becomes infinite. For the limiting
case of TTs corresponds to . When this is substituted into (m) gives ./ These
difficulties arise because solution (m) is not valid for 0P or . The type of
differential equation (i) changes for these limiting values and consequently solutions different from (m) must be obtained.
PROBLEM 5.15
A porous plate with an impermeable and
insulated leading section of length ox is
heated with uniform surface flux sq along
.oxx The plate is cooled by forced
convection with a free stream velocity V
and temperature .T Fluid at temperature
oT is injected through the porous surface
with uniform velocity .ov The injected and free stream fluids are identical. Assume laminar
boundary layer flow and introduce axial velocity simplification based on 1Pr . Use a third
degree polynomial temperature profile to determine the local Nusselt number.
(1) Observations. (i) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (ii) Fluid velocity for 1Pr is assumed to be uniform,
Vu . This represents a significant simplification. (iii) The plate is porous. (iv) Fluid is
injected through the plate with uniform velocity. (v) The plate is heated with uniform surface flux. (vi) Surface temperature is unknown, (vii) A leading section of the plate is insulated.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a uniformly heated porous flat plate with surface injection and insulated leading section.
(3) Solution Plan. Start with equating Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar
flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform upstream velocity and temperature , (7) flat plate, (8) uniform surface heat flux, (9) negligible changes in kinetic and potential energy, (10) negligible axial conduction, (11) negligible dissipation, (12) no buoyancy
( = 0 or g = 0), (13) uniform axial velocity within the thermal boundary layer ( 1Pr ), (14) uniform porosity and (15) injected fluid is at uniform velocity and temperature and is the same as the external fluid.
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
PROBLEM 5.15 (continued)
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy equation
is used to determine the temperature distribution
TTPcudydx
dTcuTdyc
dx
d
y
xTPk oop
t
p
t
p
xx
v
)()(
00
0,1 (5.6)
This equation is simplified for constant properties
TTPdyTTudx
d
y
xTP oo
tx
v
)(
0
)(0,
1 (b)
However, For 1Pr the velocity boundary layer thickness is much smaller than the thermal boundary layer thickness. Thus we assume
Vu (c)
For the temperature profile we assume a third degree polynomial
33
2210, yxbyxbyxbxbyxT (d)
The boundary conditions on the temperature are
(1) sqy
xTk
0,
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (d) and the four boundary conditions give the coefficients )(xbn
,3
20 t
s
k
qTb
k
qb s
1 , ,02b2
1
33
tk
qb s
Substituting the above into (d)
2
3
3
1
3
2),(
t
ts y
yk
qTyxT (e)
Surface temperature is obtained by setting 0y in (d)
ts
sk
qTxTxT
3
2)0,()( (f)
Substituting (e) and (f) into (1.10)
t
kh
2
3 (g)
PROBLEM 5.15 (continued)
Combining (a) and (g)
tx
xNu
2
3 (h)
Thus to determine surface temperature and Nusselt number requires the determination of t .
Substituting (c) and (e) into (b)
TTqV
Pkdy
yy
dx
d
V
Po
s
o
t
t
t
x
v
)(
0
2
3
3
1
3
21
Evaluating the integral and rearranging
)(4
112
TTqV
P
dx
d
V
Po
s
ot vk (h)
Rewriting (h)
)(414
2
TTqV
Pk
V
P
dx
do
s
ot v (i)
Note that the right side of (i) is constant. The boundary condition on (i) is
0)( ot x (j)
Integrating (i) and using (j)
)()(4142
oos
ot xxTT
qV
Pk
V
P v
Solving for t
2/1
2/1
)()(414
oos
ot xxTT
qV
Pk
V
P v (k)
(k) into (h) gives the local Nusselt number
2/1
2/1
)()(
414
2
3
o
os
ox
xx
xTT
qV
Pk
V
PNu
v (l)
This result can be expressed in terms of the Prandtl and local Reynolds number as
2/1
2/1
2/12/1
)()()1(
Re
4
3
oos
op
xx
xxTTq
PcP
PrNu
v
(m)
Surface temperature is obtained by substituting (k) into (f)
2/1
2/1
)()(414
3
2)( oo
s
oss xxTT
qV
Pk
V
P
k
qTxT
v (n)
PROBLEM 5.15 (continued)
(iii) Checking. Dimensional check: (1) Equations (e), (h), (k) and (n) are dimensionally correct. (2) Equations (l) and (m) are dimensionless.
Boundary conditions check: Assumed temperature profile (e) satisfies the two boundary conditions on temperature.
Limiting check: For the special case of solid plate, P = 0, (m) and (n) reduce to
2/12/12/1 )(Re4
3oxx xxPrNu (o)
2/1)(3
2)( o
ss xx
k
qTxT (p)
Equation (o) is the correct result for this case (see Problem 5.3). For 0ox , equation (p) shows
that surface temperature varies with x . This result is correct (see Problem 5.6).
(5) Comments. The effect of wall injection on surface temperature is can be evaluated using
solution (n) for ).(xTs If the temperature of the injected fluid is greater than the free stream
temperature, TTo , injection increases surface temperature. On the other hand, if TTo ,
injection will lower surface temperature.
PROBLEM 5.16
Consider steady two-dimensional laminar flow in the inlet region of two parallel plates. The
plates are separated by a distance H. The lower plate is maintained at uniform temperature oT
while heat is removed from the upper plate at uniform flux .oq The inlet temperature is .iT
Determine the distance from the inlet where the lower and upper thermal boundary layers meet.
Use a linear temperature profile and assume that velocity is uniform equal to .iV Express your
result in terms of dimensionless quantities.
(1) Observations. (i) There are two thermal boundary layers in this problem. (ii) The upper and lower plates have different boundary conditions. Thus, temperature distribution is not symmetrical. (iii) The lower plate is at uniform temperature while heat is removed at uniform flux along the upper plate. (iv) Fluid velocity is assumed uniform throughout the channel.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over the lower and upper plates.
(3) Solution Plan. Apply the integral form of the energy equation using linear temperature profiles for both plates.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) two-dimensional, (4) laminar
flow (Rex < 5 105), (5) thermal boundary layer (Pe > 100), (6) uniform velocity throughout, (7) uniform upstream temperature , (7) flat plates, (8) uniform surface temperature at the lower plate, (9) uniform surface flux at the upper plate, (10) negligible changes in kinetic and potential
energy, (11) negligible axial conduction, (12) negligible dissipation and (13) no buoyancy ( = 0 or g = 0).
(ii) Analysis. At the location where the two thermal boundary layers meets, we have
HLL tt )()( 21 (a)
where
H spacing between the two plates L distance from inlet to location where the two thermal layers meet
1t thermal boundary layer for the lower plate
2t thermal boundary layer for the upper plate
The integral form of conservation of energy is given in equation (5.7)
)(
0
)(0,
xt
dyTTudx
d
y
xT (5.7)
However, we assume
iViT
0 x
L
H
oT
oq
1t
2t
PROBLEM 5.16 (continued)
iVu (b)
where
iV = fluid axial velocity
Substitute (b) into (a) )(
0
)(0,
xt
ii
dyTTdx
d
y
xT
V (c)
where
iT fluid temperature outside the thermal boundary layers
(1) Lower plate. Assume a linear temperature profile
yxbxbyxT 10, (d)
The boundary conditions on the temperature are
(1) oTxT )0,(
(2) it TxT ),( 1
Equation (d) and the two boundary conditions give the coefficients )(xbn
,0 oTb1
1)(1
t
oi TTb
Substitute the above into (d)
1)(),(
t
oio
yTTTyxT (e)
Substitute (e) into (c) 1
11
0
1)()(
t
t
oi
t
oi
i
dyy
dx
dTT
TT
V (f)
Evaluate the integral and simplify
)(2
11 1
1 t
ti dx
d
V
Separate variables
dx
ddx
V
tt
i
112
Integrate and use boundary condition 0)0(1t
2)(4 1t
i
xV
Solve for 1t
PROBLEM 5.16 (continued)
xVi
t 21 (g)
(1) Upper plate. Assume a linear temperature profile
yxbxbyxT 10, (d)
The boundary conditions on the temperature are
(1) oqy
xTk
)0,(
(2) it TxT ),( 2
Equation (d) and the two boundary conditions give the coefficients )(xbn
20 t
oi
k
qTb
k
qb o
1
Substitute the above into (d)
)(),( 2 yk
qTyxT t
oi (h)
Substitute (h) into (c) 2
2
0
)(
t
too
i
dyyk
q
dx
d
k
q
V
Evaluate the integral and simplify
)(2
1 2t
i dx
d
V
Separate variables
)(2 2t
i
ddxV
Integrate and use boundary condition 0)0(2t
2)(2 2t
i
xV
Solve for 2t
xVi
t 22 (i)
Let Lx in (g) and (i), substitute into (a)
LV
LV
Hii
22
Solve for L
PROBLEM 5.16 (continued)
VHL
2
2
)22( (j)
Rewrite (j) in dimensionless form and use the pck /
k
HVc
H
L p
2)22(
1
This can be written in terms of the Prandtl number and local Reynolds number
HPrReH
L2)22(
1 (k)
where
HVReH (l)
(iii) Checking. Dimensional check: (1) Equations (e), (g), (h), (i) and (j) are dimensionally correct. (2) Equations (k) and (l) are dimensionless.
Boundary conditions check: Assumed temperature profiles (e) and (h) satisfy their respective boundary conditions.
Qualitative check: Increasing the free stream velocity decreases the thermal boundary layer,
resulting in an increase in L. Solution (j) shows that L is directly proportional to V .
(5) Comments. (1) To increase L the Reynolds number should be increased. (2) Taking the ratio of (g) and (i)
22
1
t
t (m)
Thus the thermal boundary layer for constant wall temperature is thicker than that of uniform
surface flux by a factor of .2
PROBLEM 6.1
Use scaling to determine the ratio ./ ht LL Compare scaling estimates with exact solutions.
(1) Observations. (i) This is an internal forced convection problem. (ii) Scaling gives estimates
of hL and .tL (iii) Exact solutions for hL and tL are available for laminar flow through
channels. (iv) Exact solutions for tL depend on channel geometry and surface boundary
conditions.
(2) Problem Definition. Determine the ratio ht LL / using scaling and using exact solutions.
(3) Solution Plan. Apply scaling results (6.2) and (6.3) to estimate ./ ht LL Apply (6.5) and (6.6)
to obtain an exact solution for ./ ht LL
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant properties, (4) uniform surface heat flux or uniform surface temperature, (5) negligible axial conduction (6) negligible changes in kinetic and potential energy and (7) negligible dissipation
(ii) Analysis.
Scaling estimates of hL and tL are given by equations (6.2) and (6.3)
1~/
2/1
D
h
Re
DL (6.2)
1~/
2/1
rPRe
DL
D
t (6.3)
From (6.2) and (6.3) we obtain
1~/
Pr
LL ht (a)
Exact solutions for hL for laminar flow is given by equation (6.5)
eDhh ReC
D
L
e
(6.5)
where eD is the equivalent diameter, defined as
P
AD
fe
4
where fA is channel flow area and P is channel perimeter. The coefficient hC depends on
channel geometry and is given in Table 6.1. Similarly, exact solutions for tL for laminar flow is
given by equation (6.6)
PROBLEM 6.1 (continued)
eDtt PrReC
D
L
e
(6.6)
where tC is a constant which depends on channel geometry as well as boundary conditions and
is given in Table 6.1. Taking the ratio of (6.6) to (6.5) and rearranging
h
tht
C
C
Pr
LL / (b)
(iii) Computations. Using Table 6.1,
the ratio ht CC / is computed for the six
geometries listed in the table for both uniform surface flux and uniform surface temperature. Comparisons between scaling estimate and exact solutions are tabulated. Tabulation results show that scaling estimate is close to exact solutions for the six geometries examined.
(5) Comments. (i) Scaling estimate of hL
and tL does not take into consideration
channel geometry. In addition, scaling does not distinguish between laminar and turbulent flow. (ii) Examination of the tabulated results show that scaling
provides reasonable estimates of ht LL /
for all Prandtl numbers.
Scaling Estimate
1~/
Pr
LL ht
Exact Solution
h
tht
C
C
Pr
LL /
Geometry uniform surface flux
uniform surface temperature
0.77 0.60
ab
a
a/b =10.73 0.46
a
b a/b = 2 0.67 0.57
a
b a/b = 4 0.56 0.72
1.09 0.73
PROBLEM 6.2
Use scaling to estimate the hydrodynamic and thermal entrance lengths for the flow of air in a
cm3cm3 square duct . The mean velocity is 0.8 m/s. Compare scaling estimates with exact
solutions. Evaluate properties at C.50o
(1) Observations. (i) This is an internal forced convection problem. (ii) Scaling gives estimates
of hL and .tL (iii) Exact solutions for hL and tL are available for laminar flow through
channels. (iv) Exact solutions for tL depend on channel geometry and surface boundary
conditions.
(2) Problem Definition. Determine the ratio ht LL / using scaling and using exact solutions.
(3) Solution Plan. Apply scaling results (6.2) and (6.3) to estimate ./ ht LL Apply (6.5) and (6.6)
to obtain an exact solution for ./ ht LL
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant properties, (4) uniform surface heat flux or uniform surface temperature, (5) negligible axial conduction (6) negligible changes in kinetic and potential energy and (7) negligible dissipation
(ii) Analysis.
Scaling estimates of hL and tL are given by equations (6.2) and (6.3)
1~/
2/1
D
h
Re
DL (6.2)
1~/
2/1
rPRe
DL
D
t (6.3)
Exact solutions for hL for laminar flow is given by equation (6.5)
eDhh ReC
D
L
e
(6.5)
where the Reynolds number is defined as
eD
DuRe
e (a)
where eD is the equivalent diameter, defined as
P
AD
fe
4 (b)
fA = channel flow area, 2m
P = channel perimeter, m u mean flow velocity = 0.8 m/s
kinematic viscosity, 2m /s
PROBLEM 6.2 (continued)
The coefficient hC depends on channel geometry and is given in Table 6.1. Similarly, exact
solutions for tL for laminar flow is given by equation (6.6)
eDtt PrReC
D
L
e
(6.6)
where tC is a constant which depends on channel geometry as well as boundary conditions and
is given in Table 6.1.
(iii) Computations. For a square duct of side 0.03 m,
m03.0m))(03.0(4
)m()03.0(4
22
eD
Properties of air at C50o are
Pr 0.709
s
m1093.17
26
The Reynolds number is
3.1339s)/(m1093.17
)m)(03.0)(s/m(08.026eDRe
Scaling estimates: Substituting into (6.2)
1~3.1339
)m(03.0/2/1
hL
2.40~hL m
Equation (6.3) gives
1~)3.1339)(709.0(
)m(03.0/2/1
tL
28.5~tL m
Exact solution: For a square channel, Table 6.1 gives:
09.0hC
066.0tC , for uniform surface heat flux
041.0tC , for uniform surface temperature
Equation (6.5) gives hL
3.1339)m)(03.0(09.0hL = 3.61 m
PROBLEM 6.2 (continued)
Equation (6.5) gives tL
88.1)3.1339)(709.0)(m)(03.0(066.0tL m, for uniform surface heat flux.
17.1)3.1339)(709.0)(m)(03.0(041.0tL m, for uniform surface temperature.
(5) Comments. (i) Scaling estimate of hL and tL does not take into consideration channel
geometry or surface thermal condition.(ii) Scaling overestimates hL and tL by and order of
magnitude.
PROBLEM 6.3
Far away from the entrance of a channel the velocity and temperature become fully developed. It
can be shown that under such conditions the Nusselt number becomes constant. Consider air
flowing with a mean velocity of 2 m/s through a long tube of diameter 1.0 cm. The mean
temperature at a section in the fully developed
region is 35oC. The surface of the tube is
maintained at a uniform temperature of 130oC.
What is the length of the tube section needed for
the mean temperature to reach 105oC? The Nusselt
number for this case is given by
657.3DNu
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a long tube. (iii) The surface is maintained at a uniform temperature. (iv) Since the tube section is far away from the entrance, the velocity and temperature can be assumed fully developed. (v) Tube diameter, mean velocity and inlet, outlet and surface temperatures are known. The length is unknown. (vi) The fluid is air.
(2) Problem Definition. Determine the tube length needed to raise the mean temperature to a specified level.
(3) Solution Plan. Use the analysis of flow in tubes at uniform surface temperature to determine the required tube length.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed flow, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)
][)()( exp xmc
hPTTTxT
psmism (a)
cp = specific heat, J/kg-oC
h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate, kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 35oCTs = surface temperature = 130oCx = distance from inlet of heated section, m
Applying (a) at the outlet of the heated section (x = L) and solving for L
mos
misp
TT
TT
hP
cmL ln (b)
usT
L
PROBLEM 6.3 (continued)where
Tmo = mean outlet temperature = 105oC
To compute L using (b), it is necessary to determine cp, P, m , and h . Air properties are determined
at the mean temperature mT , defined as
mT = T Tmi mo
2 (c)
The perimeter P and flow rate m are given by
P = D (d) And
uD
m4
2
(e)
where
D = inside tube diameter = 1 cm = 0.01 m u = mean flow velocity = 2 m/s
= density, kg/m3
The heat transfer coefficient for this case is determined from the Nusselt number, given by
657.3k
DhNuD (f)
(iii) Computations. Properties are determined at the mean temperature mT . Using (c)
mT = C702
)C)(10535( oo
Properties of air at this temperature are:
cp = 1008.7 J/kg-oC
k = 0.02922 W/m-oCPr = 0.707
= 19.9 10-6 m2/s
= 1.0287 kg/m3
Substituting into (d), (e) and (f)
P = 0.01(m) = 0.03142 m
kg/s0.0001616)2(m/s)m1.0287(kg/4
)(m(0.01) 322
m
h = 3.657 )m(01.0
)Cm/W(02922.0 o
= 10.69 W/m2-oC
Substituting into (b)
)C)(105130(
)C)(35130(ln
)Cm/W(69.10)m(03142.0
)Ckg/J(7.1008)s/kg(0001616.0o
o
o2
o
L = 0.65 m
PROBLEM 6.3 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and (f) are dimensionally consistent.
Limiting checks: (1) For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo = Tmi in (b) gives L = 0.
(2) The required length for the outlet temperature to reach surface temperature is infinite. Setting
Tmo = Ts in (b) gives L = .
Quantitative checks: (1) An approximate check can be made using conservation of energy and Newton’s law of cooling. Conservation of energy is applied to the air between inlet and outlet
Energy added at the surface = Energy gained by air (g)
Assuming that air temperature in the tube is uniform equal to mT , Newton’s law of cooling gives
Energy added at surface = h D L (Ts mT ) (h)
Neglecting axial conduction and changes in kinetic and potential energy, energy gained by air is
Energy gained by air = m cp(Tmo miT ) (i)
Substituting (h) and (i) into (g) and solving for the resulting equation for L
)(
)(
ms
mimop
TTDh
TTmcL (j)
Equation (j) gives
)C)(70130)(m)(01.0()Cm/W(69.10
)C)(35105)(Ckg/J(7.1008)s/kg(0001616.0oo2
oo
L = 0.57 m
This is in reasonable agreement with the more exact answer obtained above.
(2) The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. However, it shoul d be kept in mind that values of h in Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. Equation (f) gives the Nusselt number and heat transfer coefficient for this case. This equation is valid under certain conditions. Key among the restrictions are: fully developed laminar flow in tubes at uniform surface temperature.
PROBLEM 6.4
A fluid is heated in a long tube with uniform surface flux. The resulting surface temperature
distribution is found to be higher than design specification. Two suggestions are made for
lowering surface temperature without changing surface flux or flow rate: (1) increasing the
diameter, (2) decreasing the diameter. You are asked to determine which suggestion to follow. The
flow is laminar and fully developed. Under such conditions the Nusselt number is given by
364.4DNu
(1) Observations. (i) This is an internal force convection in a tube. (ii) The surface is heated at uniform flux. (iii) Surface temperature increases along the tube and is unknown. (iv) The flow is assumed laminar and fully developed. (v) The heat transfer coefficient for fully developed flow through channels is constant. (vi) According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.
(2) Problem Definition. Derive an equation for surface temperature variation in terms of tube diameter.
(3) Solution Plan. Apply surface temperature solution for fully developed laminar flow through a tube with constant surface flux.
(4)Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar, (3) fully developed flow, (4) axisymmetric flow, (5) constant properties, (6) uniform surface heat flux, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) negligible dissipation and (10) no energy generation.
(ii) Analysis. Application of conservation of energy and Newton’s law of cooling give the variation of surface temperature with distance. The solution is given by equation (7.61)
hcm
PxqTxT
psmis
1)( (a)
where
cp = specific heat, J/kg-oC h = heat transfer coefficient, W/m2-oC
m = mass flow rate, kg/s x = distance from inlet, m
P = tube perimeter, m
sq = surface heat flux, W/m2
Ts (x) = local surface temperature, oC T
mi = mean inlet temperature, oC
The perimeter is given by
P = D (b)
PROBLEM 6.4 (continued)
For fully developed laminar flow through tubes at constant surface flux the Nusselt number is given by
k
hDNuD = 4.364 (c)
where
D = tube diameter, m k = thermal conductivity, W/m-oCNuD = Nusselt number
Solving (c) for h
h = 4.364k
D (d)
Substituting (b) and (d) into (a)
Ts (x) = Tmi + qD x
mc
D
ks
p .4 364 (e)
Examination of (e) shows that decreasing the diameter will decrease surface temperature.
(iii) Checking. Dimensional check: The right hand side of (e) should have units of oC.
qs (W/m2)D x
m c
D
kp
m m
kg / s J / kg C
m
W / m Co o
( )
( ) ( )
( )
. ( )4 364= W
o oC
J / s
C
W= oC
Qualitative check: Increasing surface flux, increases Ts. Decreasing the mass flow rate, increases Ts. This behavior is confirmed by equation (e).
Limiting check: If surface flux qs = 0, fluid outlet temperature remains constant equal to the inlet
temperature. Setting qs = 0 in (e) gives Ts = Tmi.
(5) Comments. The effect of diameter, surface flux, mass flow rate, distance along the tube and fluid properties on surface temperature can be evaluated using the result obtained in (e). However, attention should be given to the assumptions leading to this result.
PROBLEM 6.5
Two identical tubes are heated with the same uniform flux at their surfaces. Air flows through
one tube while water flows at the same rate through the other. The mean inlet temperature for
both tubes is the same. Which tube will have a higher surface temperature distribution? Assume
laminar flow and neglect entrance effects. For this case the Nusselt number is given by
364.4DNu
(1) Observations. (i) This is an internal force convection in a tube. (ii) The surface is heated at uniform flux. (iii) Surface temperature increases along the tube and is unknown. (iv) The flow is assumed laminar and fully developed. (v) The heat transfer coefficient for fully developed flow through channels is constant. (vi) According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.
(2) Problem Definition. Derive an equation for surface temperature variation in terms of tube diameter.
(3) Solution Plan. Apply surface temperature solution for fully developed laminar flow through a tube with constant surface flux.
(4)Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar, (3) fully developed flow, (4) axisymmetric flow, (5) constant properties, (6) uniform surface heat flux, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) negligible dissipation and (10) no energy generation.
(ii) Analysis. Application of conservation of energy and Newton’s law of cooling give the variation of surface temperature with distance. The solution is given by equation (6.10)
Ts (x) = Tmi + hcm
Pxq
ps
1 (6.10)
where
cp = specific heat, J/kg-oCh = heat transfer coefficient, W/m2-oCm = mass flow rate, kg/s
x = distance from inlet, m
moT
tL
hL
L
miT
m
u D
sq
sq
sq
x
PROBLEM 6.5 (continued)
P = tube perimeter, m qs = surface heat flux, W/m2
Ts (x) = local surface temperature, oCT
mi = mean inlet temperature, oC
The perimeter is given by
P = D (a)
For fully developed laminar flow through tubes at constant surface flux the Nusselt number is given by
NuD = hD
k = 4.364 (b)
where
D = tube diameter, m k = thermal conductivity, W/m-oCNuD = Nusselt number
Solving (b) for h
h = 4.364k
D (c)
Substituting (a) and (c) into (6.10)
Ts (x) = Tmi + qD x
mc
D
ks
p .4 364 (d)
Examination of this result shows that decreasing the diameter will decrease surface temperature.
(iii) Checking. Dimensional check: The right hand side of (d) should have units of oC.
qs (W/m2))CW/m(364.4
)m(
)CJ/kg()kg/s(m
)m(moo k
D
c
xD
p
= W
o oC
J / s
C
W= oC
Qualitative check: Increasing surface flux, increases Ts. Decreasing the mass flow rate, increases Ts. This behavior is confirmed by equation (d).
Limiting check: If surface flux qs = 0, fluid outlet temperature remains constant equal to the
inlet temperature. Setting qs = 0 in (d) gives Ts = Tmi.
(5) Comments. The effect of diameter, surface flux, mass flow rate, distance along the tube and fluid properties on surface temperature can be evaluated using the result obtained in (d). However, attention should be given to the assumptions leading to this result.
PROBLEM 6.6
Water flows through a tube with a mean velocity of 0.2 m/s. The mean inlet temperature is 20oC
and the inside diameter of the tube is 0.5 cm. The water is heated to 80oC with uniform surface
heat flux of 0.6 W/cm2. Determine surface temperature at the outlet. If entrance effects can be
neglected the Nusselt number for fully developed flow is constant given by
364.4DNu
Is it justifiable to neglect entrance effects?
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The surface is heated at uniform flux. (iii) Surface temperature changes along the tube and is unknown. (iv) The Reynolds number should be checked to determine if the flow is laminar or turbulent. (v) If hydrodynamic and thermal entrance lengths are small compared to tube length, the flow can be assumed fully developed throughout. (vi) For fully developed flow, the heat transfer coefficient is uniform. (vii) The length of the tube is unknown. (viii) The fluid is water.
(2) Problem Definition. (i) Find the required length to heat the water to a given temperature and (ii) determine the surface temperature at the outlet.
(3) Solution Plan. (i) Since surface flux, mean velocity, diameter, inlet and outlet temperatures are known, apply conservation of energy between the inlet and outlet to determine the required tube length. (ii) Check the Reynolds number to determine if the flow is laminar or turbulent. (iii) Calculate the hydrodynamic and thermal entrance lengths and compare with the tube length. (iv) Apply surface temperature solution for flow through a tube with constant surface flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) axisymmetric flow, (4) uniform surface heat flux, (5) negligible changes in kinetic and potential energy, (6) negligible axial conduction, (7) negligible dissipation and (8) no energy generation.
(ii) Analysis. Determination of tube length. Application of conservation of energy between the inlet and outlet of the tube gives
sqLD = mcp(Tmo - Tmi) (a)
where
pc = specific heat, J/kg-oC
D = tube diameter = 0.5 cm = 0.005 m L = tube length, m m = mass flow rate, kg/s
miT = mean temperature at the inlet = 20oC
moT = mean temperature at the outlet = 80oC
sq = surface heat flux = 0.6 W/cm2 = 6000 W/m2
Solving (a) for L gives
L =s
mimop
qD
TT mc )( (b)
PROBLEM 6.6 (continued)
The mass flow rate m is given by
uDm 2)4/( (c)
where
u = mean flow velocity = 0.2 m/s
= density, kg/m3
To determine surface temperature at the outlet, use the solution for surface temperature distribution for flow through a tube with uniform surface flux, given by equation (6.10)
Ts (x) = Tmi + )(
1
xhcm
Pxq
ps (d)
where
h = local heat transfer coefficient, W/m2-oCP = tube perimeter, m Ts (x) = local surface temperature, oCx = distance from inlet of heated section, m
The perimeter P is given by
P = D (e)
Surface temperature at the outlet Ts(L) is obtained by setting x = L in (d). Substituting (e) into (d) and letting x = L gives
Ts (L) = Tmi +)(
1
Lhcm
LDq
ps (f)
The determination of h(L) requires establishing if the flow is laminar or turbulent and if it is fully developed at the outlet. Thus, the Reynolds number should be determined. It is defined as
ReD
u D (g)
where
ReD = Reynolds number
= kinematic viscosity, m2/s
Properties of water are determined at the mean temperature T defined as
T = T Tmi mo
2 (h)
Substituting into (h)
T = C502
)C)(8020( oo
Properties of water at this temperature are given in Appendix D
cp = 4182 J/kg-oC k = 0.6405 W/m-oC Pr = 3.57
PROBLEM 6.6 (continued)
= 0.5537 10-6
m2/s
= 988 kg/m3
Substituting into (g)
ReD
0 2 0 005
05537 1018066 2
. ( / ) . ( )
. ( / )
m s m
m s
Since the Reynolds number is less than 2300, the flow is laminar. The next step is calculating the hydrodynamic and thermal entrance lengths Lh and Lt to see if the flow is fully developed at the outlet. For laminar flow in a tube the hydrodynamic and thermal lengths are given by (7.43)
Lh = Ch D ReD (i)
Lt = Ct D ReD Pr (j)
where
Ch = hydrodynamic entrance length constant (Table 6.1) = 0.056 Ct = thermal entrance length constant (Table 6.1) = 0.043 Lh = hydrodynamic entrance length, m Lt = thermal entrance length, m
Substituting numerical values into (i) and (j)
Lh = 0.056 0.005 (m) 1806 = 0.506 m
and
Lt = 0.043 0.005 (m) 1806 3.57 = 1.386 m
If tube length L is larger than Lh and Lt, the flow is fully developed. Thus, it is necessary to compute L using equation (b). The mass flow rate in equation (b) is given by (c)
m = 988(kg/m3) 0.2(m/s) (0.005)2(m2)/4 = 0.00388kg/s
Substituting into (b)
L =)m/cm(10)cm/W(6.0)m(005.0
)C)(2080)(Ckg/J(4182)s/kg(00388.02242
oo
= 10.33 m
Since L > Lt> Lh, the flow is fully developed at the outlet. The heat transfer coefficient for fully developed laminar flow through a tube with uniform surface flux is given by
NuD = hD
k = 4.364 (k)
where
k = thermal conductivity = 0.6405 W/m-oCNuD = Nusselt number
Solving (k) for hDkh /364.4 (l)
PROBLEM 6.6 (continued)
(iii) Computations. To determine surface temperature at the outlet we first use (l) to compute h(L)
h(L) = 4.364 0 6405
0 005
. ( / )
. ( )
W m C
m
o
= 559 W/m2-oC
With L, m and h(L) determined, equation (f) gives the surface temperature at the outlet
Ts (L) = 20oC + )Cm/W(559
1
)Ckg/J(4182)s/kg(00388.0
)m(43.10)m(005.0)m/W(6000
o2o
2 = 91.3oC
(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (e), (f), (g), (i), (j) and (l) are dimensionally correct.
Quantitative checks: (1) Alternate approach to determining Ts(L): Application of Newton’s law of cooling at the outlet gives
sq = h [Ts(L) - Tmo ] (m)
solving for Ts(L)
Ts(L) = Tmo + q
h
s= 80 (oC) +
0 6 10
559
2 4 2 2
2
. ( / ) ( / )
( / )
W cm cm m
W m Co = 90.7oC
(2) The value of h is within the range reported in Table 1.1 for forced convection of liquids.
Limiting check: If Tmi = Tmo, the required length should vanish. Setting Tmi = Tmo into (b) gives L= 0.
(5) Comments. (i) As long as the outlet is in the fully developed region, surface temperature at the outlet is determined entirely by the local heat transfer coefficient. Therefore, it is not necessary to justify neglecting entrance length to solve the problem. (ii) In solving internal forced convection problems, it is important to establish if the flow is laminar or turbulent and if it is developing or fully developed.
PROBLEM 6.7
Fluid flows with a mean axial velocity u in a tube of
diameter D. The mean inlet temperature is miT . The
surface is maintained at uniform temperature .sT
Show that the average Nusselt number for a tube of
length L is given by
sm
smiDL
TLT
TTPrReNu
)(ln
4
wherek
LhNu L
L ,Du
ReD and Lh is the average
heat transfer coefficient over the length L.
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a tube. (iii) The surface is maintained at a uniform temperature. (iv) Entrance effect is important in this problem. (v) The average Nusselt number for a tube of length L depends on the average heat transfer coefficient over the length.
(2) Problem Definition. Determine the average heat transfer coefficient for a tube of length L
which is maintained at uniform surface temperature.
(3) Solution Plan. Start with the definition of the average Nusselt number. Use the analysis of flow in tubes at uniform surface temperature to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) axisymmetric flow, (3) constant properties, (4) uniform surface temperature, (5) negligible changes in kinetic and potential energy, (6) negligible axial conduction, (7) negligible dissipation and (8) no energy generation.
(ii) Analysis. The average Nusselt number for a tube of length L is given by
k
LhNu L
L (a)
where Lh is the average heat transfer coefficient over the length L, defined as
L
L dxxhL
h
0
)(1
(6.12)
For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)
][)()( exp xmc
hPTTTxT
psmism (6.13)
pc = specific heat, J/kg-oC
Lhh = average heat transfer coefficient for a tube of length x, W/m2-oC
m = mass flow rate, kg/s
usT
L
miT
PROBLEM 6.7 (continued)
P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature, oCTs = surface temperature, oCx = distance from inlet of heated section, m
Applying (a) at the outlet of the heated section (x = L) and solving for Lh
)(ln
LTT
TT
LP
cmh
ms
mispL (b)
where
)(LTm = mean outlet temperature
Substitute (b) into (a)
)(ln
LTT
TT
kP
cmNu
ms
mispL (c)
However, the mass flow rate m is given by
uD
m4
2
(d)
where
D = inside tube diameter, m u = mean flow velocity, m/s
= density, kg/m3
The perimeter of a tube is DP (e)
Substitute (d) and (e) into (c)
)(ln
4 LTT
TT
k
cuDNu
ms
mispL (f)
The coefficient in (f) can be expressed in terms of Prandtl and Reynolds number as
Dpp
PrReuD
k
c
k
cuD (g)
(g) into(f)
)(ln
4 LTT
TTPrReNu
ms
misDL (h)
(iii) Checking. Dimensional check: (1) Equations (6.12), (6.13), (b) and (d) are dimensionally consistent. (2) Equations (c) and (f) are dimensionless.
(5) Comments. Equation (f) or (h) can be used to experimentally determine the average heat transfer coefficient and average Nusselt number.
PROBLEM 6.8
Water flows through a cm75.0cm75.0 square duct with
a mean velocity of 0.12 m/s. The duct is heated with a
uniform surface flux of 0.25 W/cm2. The mean inlet
temperature is 25oC. The maximum allowable surface
temperature is 95oC. Justify neglecting entrance effects.
And determine maximum outlet mean temperature.
(1) Observations. (i) This is an internal forced convection problem. (ii) The fluid is heated at uniform wall flux. (iii) Surface temperature changes with distance along the channel. It reaches a maximum value at the outlet. (iv) The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem. (v) The channel has a square cross-section. (vi) Application of Newton’s law of cooling at the outlet relates outlet temperature to surface temperature, surface flux and heat transfer coefficient. (vii) Application of conservation of energy gives a relationship between heat added, inlet temperature, outlet temperature, specific heat and mass flow rate.
(2) Problem Definition. [a] Determine the outlet temper ature corresponding to a specified surface temperature and flux. [b] Determine the require d channel length to heat the water to outlet temperature and compare with entrance lengths.
(3) Solution Plan. Apply Newton’s law of cooling at the outlet to determine the mean outlet temperature. This requires determining the heat transfer coefficient. Check the Reynolds and Peclet numbers to establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem. Obtain a solution to the heat transfer coefficient (Nusselt number).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface heat flux, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible dissipation and (7) no energy generation.
(ii) Analysis. [a] Determination of Tmo. Applying Newton’s law of cooling at the outlet
qs = h(L) [Ts (L) moT ] (a)
Solving (a) for Tmo
Tmo = Ts(L)Lh
qs (b)
moT
tL
hL
L
miT
m
u
sqsq
sq
)(LTs
S
S
moT
miT
u
L
sT
PROBLEM 6.8 (continued)
where
)(Lh = heat transfer coefficient at the outlet, W/m2-oC
L = channel length, m qs = surface heat flux = 0.25 W/cm2 = 2500 W/m2
moT = mean outlet temperature, oC
)(LTs = surface temperature at the outlet = 95oC
Equation (b) gives Tmo in terms of the heat transfer coefficient at the outlet, h(L). The value of h(L) depends on whether the flow is laminar or turbulent and if the flow is developing or fully developed at the outlet. To establish these conditions, the Reynolds and Prandtl numbers are determined. The Reynolds number for flow through a square channel is defined as
ReDeeuD
(c)
where
De = equivalent diameter, m ReDe = Reynolds numberu = mean velocity = 0.12 m/s
= kinematic viscosity, m2/s
The equivalent diameter for a square channel is defined as
De = 4A
P = 4
4
2S
S = S (d)
where
A = channel flow area = S2, m2
P = channel perimeter in contact with the fluid = 4S, m S = side dimension of the square channel = 0.75 cm = 0.0075 m
Water properties are evaluated at the mean temperature, Tm , defined as
mT = (Tmi + Tmo)/2 (e)
where
mT = mean fluid temperature in channel, oC
miT = mean inlet temperature = 25oC
However, since Tmo is unknown, a solution is obtained using a trial and error procedure. A value
for Tmo is assumed, (e) is used to calculate mT and (b) is used to calculate Tmo. The calculated Tmo
is compared with the assumed value. The procedure is repeated until a satisfactory agreement is obtained between assumed and calculated values of Tmo.
Let Tmo = 85oC
Equation (e) gives
Tm = (25 + 85)( oC)/2 = 55oC
Properties of water at this temperature are
cp = specific heat = 4184 J/kg-oCk = thermal conductivity = 0.6458 W/m-oC
PROBLEM 6.8 (continued)Pr = Prandtl number = 3.27
= kinematic viscosity = 0.5116 10-6 m2/s
= density =985.7 kg/m3
Using (d) to calculate eD
De = 0.0075 m
Substituting into (c)
DeRe = )s/m(105116.0
)m(0075.0)s/m(12.026
= 1759
Since the Reynolds number is smaller than 2300, the flow is laminar. To establish if the flow is developing or fully developed at the outlet, the hydrodynamic length, thermal entrance length and tube length must be determined. Equations (7.43a) and (7.43b) give
Lh = Ch Dee ReD (f)
and
Deett ReDCL Pr (g)
where
Ch = velocity entrance length constant (Table 7.2) = 0.09 Ct = temperature entrance length constant (Table 7.2) = 0.066 Lh = hydrodynamic entrance length, m Lt = thermal entrance length, m
Substituting numerical values into (f) and (g)
Lh = 0.09 (0.0075)(m)(1759) = 1.19 m and
Lt = 0.066(0.0075)(m)(1759) (3.27) = 2.85 m
These two lengths should be compared with the tube length L. To determine L, conservation of energy is applied to the fluid between inlet and outlet
Energy added at the surface = Energy gained by the fluid (h)
Neglecting axial conduction and changes in kinetic and potential energy, (h) gives
qs (4 S L) = mcp(Tmo miT )
or
L =m c T T
S q
p mo mi
s4 (i)
where the mass flow rate m is given by
m = u A = u S2 (j)
Since Tmo is unknown, L cannot be computed. To proceed, assume that the flow is fully developed at the outlet, determine h(L), use (b) to compute Tmo and (i) to compute L. If the computed length is larger than Lh and Lt, the assumption of fully developed flow is verified. The Nusselt number for fully developed laminar flow in a square channel with uniform surface heat flux is given by equation (7.58) and Table 7.3
PROBLEM 6.8 (continued)
k
hDNu e
De = 3.608 (k)
or
h = 3.608 eDk / (l)
(iii) Computations. [a] Determination of Tmo. Equation (l) gives h
h = 3.608(0.6458)(W/m-oC)/0.0075(m) = 310.7 W/m2-oC
Substituting into (b)
Tmo = 95(oC))Cm/W(7.310
)m/W(2500o2
2
= 87oC
This is close to the assumed value of 85oC used to obtain approximate water properties.
[b] Determination of channel length L. Equation (j) gives m
m = 985.7(kg/m3)0.12(m/s)(0.0075)2(m2) = 0.00665 kg/s
Substituting into (i)
L =)m/W(2500)m)(0075.0(4
)C(2587)Ckg/J(4184)s/kg(00665.02
oo
= 23 m
Since L > Lt > Lh, the flow is fully developed at the outlet.
(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (f), (g), (i), (j) and (l) are dimensionally consistent.
Quantitative check. The value of h is within the range listed in Table 1.1.
Qualitative check: As surface heat flux is decreased, channel length should increase. Equation (i) shows that L is inversely proportional to qs .
Limiting check. If Tmo = Tmi, channel length should be zero. Setting Tmo = Tmi in (i) gives L = 0.
(5) Comments. (i) This problem illustrates how analysis cannot always be completed without carrying out some computations. This can occur if it is necessary to establish if the flow is laminar or turbulent or if entrance effects can be neglected or not. (ii) A solution is obtained without the need to neglect entrance effects. As long as the outlet is in the fully developed region, water outlet temperature is determined entirely by the local heat transfer coefficient.
PROBLEM 6.9
Two experiments were conducted on fully developed laminar flow through a tube. In both
experiments surface temperature is C180o and the mean inlet temperature is C20o . The mean
outlet temperature for the first experiment is found to be C120o . In the second experiment the
flow rate is reduced by a factor of 2. All other conditions remained the same. Determine:
[a] The outlet temperature of the second experiment.
[b] The ratio of heat transfer rate for the two experiments.
(1) Observations. (i) This is an internal forced convection problem in tubes. (ii) The flow is laminar and fully developed. (iii) The surface is maintained at uniform temperature. (iv) All conditions are identical for two experiments except the flow rate through one is half that of the other. (v) The total heat transfer rate depends on the outlet temperature.
(2) Problem Definition. Determine the outlet temperature for fully developed laminar flow through a tube at uniform surface temperature.
(3) Solution Plan. Use the analysis of fully developed laminar flow in tubes at uniform surface temperature to determine the outlet temperature. Apply conservation of energy to obtain an equation for the heat transfer.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed laminar flow, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) no axial conduction, (8) no dissipation and (9) no energy generation.
(ii) Analysis. For flow through tubes with uniform surface temperature, conservation of energy and Newton’s law of cooling lead to equation (6.13)
][)()( exp xcm
hPTTTxT
psmism (a)
pc = specific heat, CJ/kg o
h = average heat transfer coefficient for a tube of length x, CW/m o2
m = mass flow rate, kg/s
P = tube perimeter, m
)(xTm = mean temperature at x, Co
miT = mean inlet temperature = 20 Co
sT = surface temperature = 180 Co
x = distance from inlet of heated section, m
Applying (a) at the outlet ( Lx )
][)()( exp Lcm
hPTTTxT
psmismo (b)
where
L = tube length, m
moT = mean outlet temperature, Co
PROBLEM 6.9 (continued)
The quantities L, P and pc are the same for both experiments. Furthermore, the flow remains
laminar and fully developed when the flow rate is reduced in the second experiment. Thus the heat transfer coefficient is the same for both experiments. Equation (b) is rewritten in dimensionless form as
)/exp( mTT
TTC
smi
som (c)
where
pc
hPLC (d)
Rewriting (c)
mTT
TT C
som
smiln (e)
Let the subscripts 1 and 2 refer to the first an d second experiments. Applying (e) to the two experiments gives
11
lnmTT
TT C
som
smi (f)
22
lnmTT
TT C
som
smi (g)
Taking the ratio of (g) to (f) and rearranging
som
smi
som
smi
TT
TT
mTT
TT m
12
1
2
lnln
Or
2
1
12
m
m
som
smi
som
smi
TT
TT
TT
TT
Solving for 2moT
2
1
12 )(
m
m
smi
somsmisom
TT
TTTTTT (h)
This result gives the outlet temperature when the flow rate is reduced. Application of conservation of energy to the fluid between the inlet and outlet gives the heat transfer rate q
mimop TTcmq (i)
Applying (i) to the flow in the two tubes and taking the ratio of the two results
mimo
mimo
TTm
TTm
q
q
11
22
1
2 (j)
(iii) Computations. Substituting numerical values into (h) and noting that 2/ 21 mm
PROBLEM 6.9 (continued)
2
o
ooo
2)C)(18020(
)C)(180120()C)(18020()C(180
omT = 157.5 Co
Equation (j) gives
6875.020120
205.157
2
1
1
2
q
q
(iv) Checking. Dimensional check: Each term in (h) has units of temperature.
Limiting check: If 21 mm , the two outlet temperatures must be the same. Setting 21 mm in
(h) gives 12 momo TT .
(5) Comments. (i) Although tube size, fluid nature a nd flow rate are not known, it was possible to obtain a solution to the problem. Taking the ratio of two operating conditions results in the cancellation of the unknown factors. (ii) Although the outlet temperature increases as the flow rate is decreased, the rate of heat transfer decreases. (iii) Decreasing the flow rate without changing inlet and surface temperatures and heat transfer coefficient is expected to increase the
outlet temperature. In the limit as the flow rate approaches zero ),0( 2m the corresponding
outlet temperature becomes equal to surface temperature. This follows from (h).
PROBLEM 6.10
A long rectangular duct with a cm8cm4 cross section is
used to heat air from –19.6oC to 339.6
oC. The mean
velocity in the duct is 0.2 m/s and surface temperature is
340oC. Determine the required duct length. Is neglecting
entrance effects justified?
(1) Observations. (i) This is an internal forced convection problem. (ii) The channel has a rectangular cross section. (iii) Surface temperature is uniform. (iv) The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if entrance effects can be neglected. (v) Channel length is unknown. (vi) The fluid is air.
(2) Problem Definition. Determine the required channel length to heat air to a specified outlet temperature. Determine the hydrodynamic and thermal entrance lengths and compare with channel length.
(3) Solution Plan. Use the analysis of flow through tubes with uniform surface temperature to determine channel length. Check the Reynolds and Peclet numbers to establish if the flow is laminar or turbulent. Compute the hydrodynamic and thermal entrance lengths and compare with channel length to determine if entrance effects can be neglected. Obtain a solution to the heat transfer coefficient (Nusselt number).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface temperature, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible dissipation and (7) no energy generation.
(ii) Analysis. For flow through tubes with uniform surface temperature, conservation of energy and Newton’s law of cooling lead to equation (6.13)
][)()( exp xcm
hPTTTxT
psmism (a)
where
cp = specific heat, J/kg-oC
h = average heat transfer coefficient for a channel of length x, W/m2-oCm = mass flow rate, kg/s P = channel perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature, -19.6oC
Ts = surface temperature = 340oCx = distance from inlet of heated section, m
u
moT
miT
sTL
tL
hL
m
L
moT
miT
b
a
sT
uu
PROBLEM 6.10 (continued)
Applying (a) at the outlet, x = L, and solving for L
mos
misp
TT
TT
hP
cmL ln (b)
where
L = channel length, m
Tmo = mean outlet temperature = 339.6oC
Equation (b) gives L in terms of cp, P, m, and h . The average heat transfer coefficient depends on whether the flow is laminar or turbulent and if the flow is developing or fully developed. To establish these conditions, the Reynolds and Prandtl numbers are determined. The Reynolds number for flow through a rectangular channel is defined as
ReDeeuD
(c)
where
eD = equivalent diameter, m
ReDe = Reynolds number u = mean velocity = 0.2 m/s
= kinematic viscosity, m2/s
The equivalent diameter for a rectangular channel is defined as
eD = 4A
P =
)(2
ba
ab (d)
where
A = channel flow area = ab, m2
P = channel perimeter in contact with the fluid = 2(a + b), m a = width of rectangular channel = 8cm = 0.08 m b = height of rectangular channel = 4 cm = 0.04 m
The perimeter P and flow rate m are given by
P = 2(a + b) (e) and
uabuAm )( (f)
where
= density, kg/m3
Air properties are evaluated at the mean temperature, Tm , defined as
mT = (Tmi + Tmo)/2 (g)
where
mT = mean fluid temperature in channel, oC
The mean temperature is calculated in order that properties are determined. Substituting into (g)
mT = C1602
)C)(6.3396.19( oo
PROBLEM 6.10 (continued)
Properties of air at this temperature are given in Appendix C
cp = 1018.5 J/kg-oCk = 0.3525 W/m-oCPr = 0.701
= 29.75 10-6, m2/s
= 0.8342 kg/m3
Substituting into (d)
eD =)m)(08.004.0(
)m(08.0)m(04.02 = 0.0533 m
Substituting into (c)
ReDe 3.358)/sm(1075.29
)m(0533.0)/(2.026
sm
Since the Reynolds number is smaller than 2300, the flow is laminar. To determine if entrance effects can be neglected, the hydrodynamic and thermal entrance length must be compared with channel length. For laminar flow, equations (6.5) and (6.6) give
Lh = Ch Dee ReD (h)
and
Deett ReDCL Pr (i)
where
Ch = hydrodynamic entrance length constant (Table 6.1, for a/b = 0.08m/0.04m = 2) = 0.085 Ct = thermal entrance length constant (Table 6.1, for a/b = 0.08m/0.04m = 2) = 0.049 Lh = hydrodynamic entrance length, m Lt = thermal entrance length, m
Substituting numerical values into (h) and (i)
Lh = 0.085 (0.0533)(m)(358.3) = 1.623 m and
Lt = 0.049(0.0533)(m) (358.3) (0.706) = 0.656 m
These two lengths should be compared with channel length L. However, L can be determined
only after h is computed. To compute h , L must be known! To proceed, assume that entrance
length effects are negligible (fully developed flow throughout channel), determine h , use (b) to compute L and compare it with Lh and Lt. For fully developed laminar flow through a rectangular channel at constant surface temperature, the Nusselt number is given in Table 6.2
k
DhNu e
De = 3.391 (j)
or
h = 3.391 eDk / (k)
(iii) Computations. Equations (e), (f) and (k) give
P = 2(0.08 + 0.04)(m) = 0.24 m
PROBLEM 6.10 (continued)
m = 0.0.8342(kg/m3)0.08(m)0.04(m)0.2(m/s) = 0.0005339 kg/s
h = 3.391(0.03525)(W/m-oC)/0.0533(m) = 2.242 W/m2-oC
Substituting into (b)
C)339.6)((340
C)19.6)((340ln
)C42(W/m0.24(m)2.2
C)5(J/kgkg/s)1018.0.0005339(o
o
o2
o
L = 6.873 m
Comparing Lh and Lt with L
Lh/L = 1.623(m)/6.873(m) = 0.236
and
Lt/L = 0.656(m)/6.87(m) = 0.095
(iv) Checking. Dimensional check: Computations showed that equations (b) )f( , (h), (i)
and (l) are dimensionally consistent.
Limiting checks: (1) For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo = Tmi in (b) gives L = 0. (2) The required length for the outlet temperature to reach
surface temperature is infinite. Setting Tmo = Ts in (b) gives L = .
Quantitative check: The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. Howe ver, equation (l) shows that for laminar flow
through channels, h is inversely proportional to D. A large D can result in a small h . In
addition, values of h in Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. (i) This problem illustrates the importance of establishing if entrance effects can
be neglected or not. (ii) Neglecting thermal entrance length is justified since tL is less than 10%
of L. However, neglecting the viscous entrance length requires careful judgment. This
assumption underestimates h and consequently, according to (b), it overestimates L. Thus it is a conservative assumption.
PROBLEM 6.11
A rectangular duct with inside dimensions of
cm4cm2 is used to heat water from 25 Co to
115 Co . The mean water velocity is 0.018 m/s. The
surface of the duct is maintained at 145 Co . Determine
the required duct length. Assume fully developed flow
conditions throughout.
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a
rectangular duct. (iii) The surface is maintained at a uniform temperature. (iv) The velocity and
temperature are fully developed. (v) The Reynolds number should be checked to determine if the
flow is laminar or turbulent. (vi) Duct size, mean velocity and inlet, outlet and surface
temperatures are known. The length is unknown. (vii) Duct length depends on the heat transfer
coefficient. (vii) The fluid is water.
(2) Problem Definition. Determine the duct length needed to raise the mean temperature to a specified level. This requires determining the heat transfer coefficient.
(3) Solution Plan. Use the analysis of flow in channels at uniform surface temperature to determine the required duct length.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed flow, (3) constant properties, (4) uniform surface temperature, (5) negligible changes in kinetic and potential energy, (6) negligible axial conduction, (7) negligible dissipation and (8) no energy generation.
(ii) Analysis. For flow in a channel at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)
][)()( exp xcm
hPTTTxT
psmism (a)
where
pc = specific heat, CJ/kg o
h = average heat transfer coefficient for a channel of length L, CW/m o2
m = mass flow rate, kg/s
P = cross section perimeter, m
Tm(x) = mean temperature at x, Co
Tmi = mean inlet temperature = 25 Co
Ts = surface temperature = 145 Co
x = distance from inlet, m
Applying (a) at the outlet (x = L) and solving for L
mos
misp
TT
TT
hP
mcL ln (b)
u
moT
miT
sTL
PROBLEM 6.11 (continued)
where
Tmo = mean outlet temperature = 115 Co
To compute L using (b), it is necessary to determine cp, P, m , and h . Water properties are
determined at the mean temperature mT , defined as
mT = T Tmi mo
2 (c)
The perimeter P is given by P = 2(a + b) (d)
Where
a = duct width = 4 cm = 0.04 m b = duct height = 2 cm = 0.02 m
The mass flow rate m is given by uabm )( (e)
where
= density, 3kg/m
The heat transfer coefficient for fully developed flow is uniform along a channel. Its value depends on whether the flow is laminar or turbulent. To proceed, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a rectangular duct the Reynolds number is defined as
eDe
DuRe (f)
where
eD = equivalent diameter, m
DeRe = Reynolds number
u = mean velocity = 0.018 m/s
= kinematic viscosity, /sm2
The equivalent diameter for a rectangular channel is defined as
eD =P
A4 =
)(2
ba
ab (g)
where
A = duct flow area = ab, m2
The mean temperature is calculated to determine water properties. Substituting into (c)
mT = C702
)C)(11525( oo
Properties of water at this temperature are:
pc = 4191 CJ/kg o
PROBLEM 6.11 (continued)
k = 0.6594 CW/m o
Pr = 2.57
= 6104137.0 /sm2
= 977.7 3kg/m
Substituting into (h)
eD = m02667.0m))(02.004.0(
m)(02.0)m(04.02
Equation (h) gives
1160/s)m(104137.0
m)(02667.0)m/s(018.026DeRe
Since the Reynolds number is smaller than 2300, the flow is laminar. The Nusselt number for fully developed laminar flow through rectangular channels at uniform surface temperature is given by equation (7.57) and Table 7.3. Thus
k
DhNu e
De = 3.391 (h)
Solving for h
eD
kh 391.3 (i)
(iii) Computations. Substituting into (d), (e) and (i)
)m)(02.004.0(2P = 0.12 m
kg/s01408.0)m/s(018.00.02(m)m)(04.0)kg/m(7.977 3m
h = )m(02667.0
)CW/m(6594.0391.3
o
= 83.84 CW/m o2
Substituting into (b)
)C)(115145(
)C)(25145(ln
)CW/m(84.83m)(12.0
)s/kg(01408.0)CJ/kg(4191o
o
o2
o
L 8.13 m
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e), (f), (g) and (i) are dimensionally consistent.
Limiting checks: (1) For the special case of Tmo = Tmi , the required length should be zero. Setting Tmo = Tmi in (b) gives L = 0.
(2) The required length for the outlet temperature to reach surface temperature is infinite. Setting
Tmo = Ts in (b) gives L = .
Quantitative checks: (1) An approximate check can be made using conservation of energy and Newton’s law of cooling. Conservation of energy is applied to the water between inlet and outlet
Energy added at the surface = Energy gained by water (j)
PROBLEM 6.11 (continued)
Assuming that water temperature in the tube is uniform equal to mT , Newton’s law of cooling
gives
Energy added at surface = h P L (Ts mT ) (k)
Neglecting axial conduction and changes in kinetic and potential energy, energy gained by the water is
Energy gained by air = mcp(Tmo miT ) (l)
Substituting (k) and (l) into (j) and solving for the resulting equation for L
)(
)(
ms
mimop
TTPh
TTcmL (m)
Equation (m) gives
)C)(70145)(m)(12.0)CW/m(84.83
)C)(25115)(CJ/kg(4191)kg/s(01408.0oo2
oo
L = 7.04 m
This is in reasonable agreement with the more exact answer obtained above.
(2) The value of h within the range listed in Table 1.1 for forced convection of liquids.
(5) Comments. This problem is simplified by two conditions: fully developed and laminar flow.
PROBLEM 6.12
Air is heated in a cm4cm4 square duct at
uniform surface flux of 590 .W/m2 The mean air
velocity is 0.32 m/s. At a section far away from
the inlet the mean temperature is 40 Co . The
mean temperature is 120 Co . Determine the
maximum surface temperature.
(1) Observations. (i) This is an internal forced
convection problem in a channel. (ii) The surface is heated at uniform flux. (iii) Surface
temperature changes along the channel. It reaches a maximum value at the outlet. (iv) The
Reynolds number should be checked to determine if the flow is laminar or turbulent. (v) Velocity
and temperature profiles become fully developed far away from the inlet. (vi) The heat transfer
coefficient is uniform for fully developed flow. (vii) The channel has a square cross section.
(viii) tube length is unknown. (ix) The fluid is air.
(2) Problem Definition. (i) Find the required length to heat the air to a given temperature and (ii) determine surface temperature at the outlet.
(3) Solution Plan. (i) Since surface flux, mean velocity, duct size, inlet and outlet temperatures are known, application of conservation of energy between the inlet and outlet gives the required duct length. (ii) Check the Reynolds number to determine if the flow is laminar or turbulent. (iii) Apply surface temperature solution for flow through a channel with constant surface flux.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface heat flux, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction, (6) negligible dissipation and (7) no energy generation.
(ii) Analysis. Application of conservation of energy between the inlet and outlet gives the required channel length
)( mimops TTcmqLP (a)
where
pc = specific heat, CJ/kg o
L = channel length, m m = mass flow rate, kg/s P = perimeter, m
sq = surface heat flux = 590 2W/m
miT 40 Co
moT 120 Co
Solving (a) for L
qP
TTcmL
mimop )( (b)
moT
miT
L
u
sq
PROBLEM 6.12 (continued)
The mass flow rate and perimeter are given by
uSm 2 (c)
SP 4 (d)
where
S = duct side = 0.04 m
u = mean flow velocity = 0.32 m/s
= density, 3kg/m
Substituting (c) and (d) into (b)
q
TTcuSL
mimop
4
)( (e)
To determine surface temperature at the outlet, use the solution for surface temperature distribution for channel flow with uniform surface flux, given by equation (6.10)
Ts (x) = Tmi +)(
1
xhcm
Pxq
ps (f)
where
)(xh = local heat transfer coefficient, CW/m o2
)(xTs = local surface temperature, Co
x = distance from inlet of heated section, m
Surface temperature at the outlet, Ts(L), is obtained by setting x = L in (f). Substituting (c) and (d) into (f)
Ts (L) = Tmi + )(
14
LhcuS
Lq
p
s (g)
Finally, it remains to determine the heat transfer coefficient at the outlet, h(L). This requires establishing whether the flow is laminar or turbulent. Thus, the Reynolds number should be determined. The Reynolds number for flow through a square channel is defined as
eDe
DuRe (h)
where
eD = equivalent diameter, m
= kinematic viscosity, /sm2
The equivalent diameter for a square channel is defined as
De = P
A4 =
S
S
44
2
= S (i)
Substituting (i) into (h)
PROBLEM 6.12 (continued)
SuReDe (j)
Properties of air are determined at the mean temperature mT defined as
mT = T Tmi mo
2 (k)
Substituting into (k)
mT = C802
)C)(12040( oo
Properties of air at this temperature are:
pc = 1009.5 CJ/kg o
k = 0.02991 CW/m o
Pr = 0.706
= 20.92 610 m2/s
= 0.9996 kg/m3
Substituting into (j)
9.611/s)m(1092.20
m)(04.0)m/s(32.026DeRe
Since the Reynolds number is smaller than 2300, the flow is laminar. The heat transfer coefficient for fully developed laminar flow through a square channel with uniform surface flux is constant. It is given by equation (6.55) and Table 6.2
k
DhNu e
De = 3.608 (l)
where hh . Solving (k) for h
eD
kh 608.3 (m)
(iii) Computations. Substituting numerical values in (e) gives required channel length
4378.0)W/m(590)4(
C))(40120(C)J/kg-(5.1009m/s)(32.0)m(04.0)kg/m(9996.02
oo3
L m
To determine surface temperature at the outlet, the heat transfer coefficient is computed using (m)
h(L) = h =)m(04.0
)CW/m(02991.0608.3
o
= 2.7 CW/m o2
Equation (g) gives the surface temperature at the outlet
PROBLEM 6.12 (continued)
C)W/m(7.2
1
C)k/J(5.1009m/s)(32.0)m(04.0)kg/m(9996.0
)m)(4378.0(4)W/m(590)C(40
o2o3
2o)(g
LTs
)(LTs = 338.5 Co
(iv) Checking. Dimensional check: Computations showed that equations (e), (g), (j), and (m) are dimensionally correct.
Quantitative checks: (1) Alternate approach to determining Ts(L): Application of Newton’s law of cooling at the outlet gives
qs = h [Ts(L) - Tmo ] (n)
solving for Ts(L)
Ts(L) = Tmo + h
qs = C5.338)CW/m(7.2
)W/m(590)C(120 o
o2
2o
(2) The value of h is within the range reported in Table 1.1 for forced convection of liquids.
Limiting check: If Tmi = Tmo, the required length should be zero. Setting Tmi = Tmo into (e) gives L = 0.
(5) Comments. (i) As long as the outlet is in the fully developed region, surface temperature at the outlet is determined entirely by the local heat transfer coefficient. (ii) In solving internal forced convection problems, it is important to establish if the flow is laminar or turbulent and if it is developing or fully developed.
PROBLEM 6.13
Consider fully developed laminar flow in two tubes having the same length. The flow rate, fluid,
inlet temperature and surface temperature are the same for both tubes. However, the diameter
of one tube is double that of the other. Determine the ratio of the heat transfer rate from the two
tubes.
(1) Observations. (i) This is an internal forced convection problem in tubes. (ii) The flow is laminar and fully developed. (iii) The surface is maintained at uniform temperature. (iv) All conditions are identical for two tubes except the diameter of one is twice that of the other. (v) The total heat transfer in each tube depends on the outlet temperature.
(2) Problem Definition. Compare the outlet temperatures of the two tubes.
(3) Solution Plan. Apply conservation of energy to obtain an equation for the heat transfer in each tube. Use the analysis of fully developed laminar flow in tubes at uniform surface temperature to determine the outlet temperatures.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed laminar flow, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) no axial conduction, (8) no dissipation and (9) no energy generation.
(ii) Analysis. Application of conservation of energy to the fluid between the inlet and outlet of tube, gives
q = mimop TTcm (a)
where
cp = specific heat, J/kg-oCm = mass flow rate, kg/s q = rate of heat transfer, W Tmi = inlet mean temperature, oCTmo = outlet mean temperature, oC
Applying (a) to the flow in the two tubes, noting that cp, m and Tmi are the same for both tubes, and taking the ratio of the two results
mimo
mimo
TT
TT
q
q
1
2
1
2 (b)
where the subscripts 1 and 2 refer to the small tube and large tube, respectively. Thus, the problem becomes one of determining the outlet temperatures. For flow through tubes with uniform surface temperature, conservation of energy and Newton’s law of cooling lead to equation (7.13)
][)()( exp xcm
hPTTTxT
psmism (c)
h = average heat transfer coefficient for a tube of length x, W/m2-oCm = mass flow rate, kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature, 35oC
PROBLEM 6.13 (continued)
Ts = surface temperature, oCx = distance from inlet of heated section, m
Applying (c) at the outlet where x = L and Tm(x) = Tmo and rearranging
)]/1)[( exp( pmismiom cmLhPTTTT (d)
where
L = tube length, m Tmo = mean outlet temperature, oC
Examination of (d) shows that all quantities are identical for both tubes except P and h .Applying (d) to the two tubes and taking the ratio of the resulting equations
)/exp(1
)/exp(1
11
22
1
2
p
p
mimo
mimo
cmLhP
cmLhP
TT
TT (e)
Thus, the two outlet temperatures will differ according to how the product of P h differs for the two tubes. The perimeter P is given by
P = D (f)
where D is tube diameter. For fully developed laminar flow with uniform surface temperature, the heat transfer coefficient is uniform along the tube, given by (7.57)
NuD = hD
k= 3.66 (g)
where
NuD = Nusselt number k = thermal conductivity of fluid, W/m-oC
The product P h can now be constructed from (f ) and (g)
P h = 3.66 Dk
D= 3.66 k (h)
Thus, the product P h is independent of tube size. It follows from (e) that the two outlet temperatures are identical
1)/66.3exp(1
)/66.3exp(1
1
2
p
p
mimo
mimo
cmkL
cmkL
TT
TT (i)
Substituting (i) into (b) gives
1/ 12 qq (j)
(iii) Checking. Dimensional check: The exponent in equation (d) should be dimensionless.
pcm
LhP ( )( / )( )
( / )( / ) /
m W m C m
kg s J kg C
W
J s
W
W
o
o
2
= 1
Limiting check: In the limit as L , the outlet temperature becomes equal to the surface
temperature regardless of tube size. Setting L = in (d) gives Tmo = Ts.
(5) Comments. The result is somewhat surprising. One would expect that increasing the diameter, increases the heat transfer rate. However, according to (g), the heat transfer coefficient is inversely proportional to diameter. On the other hand, the perimeter is
PROBLEM 6.13 (continued)
(6) directly proportional to diameter. These two effects cancel each other resulting in identical outlet temperatures regardless of tube size. This is true for the assumptions listed above and as long as all conditions are the same for both tubes.
PROBLEM 6.14
To evaluate the accuracy of scaling prediction of the thermal entrance length and Nusselt number, compare scaling estimates with the exact results of Graetz solution for flow through
tubes.
(1) Observations. (i) This is an internal forced convection problem. (ii) Equation (6.3) gives scaling estimate of the thermal entrance length. (iii) Equation (6.20b) gives scaling estimate of the local Nusselt number. (iv) The Graetz problem d eals with laminar flow in the entrance of a tube at uniform surface temperature. (v) Grae tz solutions gives the thermal entrance length (distance to reach fully developed temperature) and local Nusselt number.
(2) Problem Definition. Determine the thermal entrance length and Nusselt number using scaling and compare with Graetz results.
(3) Solution Plan. (i) Use Graetz solution (Table 6.4 or Fig. 6.9) to determine the distance from the entrance to the section where the Nusselt number is constant (fully developed temperature). Compare with scaling estimate, equation (6.3) (ii) Use Graetz so lution (Table 6.4) to determine the Nusselt number at various distances from the entrance. Compare with scaling estimate, equation (6.20b).
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (3) uniform surface temperature, (4) negligible axial conduction (5) negligible changes in kinetic and potential energy and (6) negligible dissipation
(ii) Analysis. Equation (6.3) gives scaling estimate of the thermal entrance length tL
1~/
2/1
rPRe
DL
D
t (a)
where
D = diameter
Pr = Prandtl number
DRe = Reynolds number
Scaling estimate of the Nusselt number is given by equation (6.20b)
1~2/1
x/D
RerP
uN D (b)
Graetz solution for the variation of the local Nusse lt number with distance from the entrance is presented in Table 6.4. Since the Nusselt is constant in the fully developed region, Graetz solution can also be used to determine the entrance length. The fully developed Nusselt number is
66.3DNu
Table 6.4 gives the dimensionless distance corresponding to 66.3DNu as
PROBLEM 6.14 (continued)
1.0//
PrRe
DL
PrRe
Dx
D
t
D
(c)
To compare with scaling result (a), equation (c) is rewritten in the same form as (a)
316.01.0/
2/1
rPRe
DL
D
t (d)
Thus the Graetz solution constant 0.316 is replaced by unity in scaling.
Graetz solution, Table 6.4, shows that the Nusse lt number depends on the dimensionless axial distance , defined in (c). Rewriting scaling result (b) in terms of , gives
1~DuN (e)
To facilitate comparison of scaling estimate (e) with Graetz
solution, Table 6.4 is modified to include .DuN
Examination the result shown in Table 6.4a shows that
exact values of .DuN range from 0.286 to 1.157. Scaling
predicts these constants to be unity, as shown in (e).
(iv) Checking. Dimensional check: All equations are dimensionless.
(5) Comments. (i) To compare scaling estimate with exact solution for the Nusselt number, it is necessary to cast both results in the same form. (ii) Scaling estimate of the Nusselt number is surprisingly good.
Table 6.4a Local Nusselt number for tube at
uniform surface temperature
=PrRe
Dx
D
/ )(Nu )(Nu
0.0005 12.8 0.286
0.002 8.03 0.359
0.005 6.00 0.424
0.02 4.17 0.590
0.04 3.77 0.754
0.05 3.71 0.830
0.1 3.66 1.157
PROBLEM 6.15
Use scaling to estimate the heat transfer coefficient for plasma at a distance of 9 cm from the
entrance of a vessel. The mean plasma velocity is 0.042 m/s and the diameter is 2.2 mm.
Properties of plasma are:
CJ/kg3900 opc , CW/m5.0 ok , /sm1094.0 26 , 3kg/m1040
(1) Observations. (i) This is an internal forced convection problem. (ii) Equation (6.20b) gives scaling estimate of the local Nusselt number. (iii) The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface temperature.
(2) Problem Definition. Determine the Nusselt number using scaling.
(3) Solution Plan. Use (6.20b) to estimate the Nusselt number in the entrance region of a tube.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties and (3) uniform surface temperature.
(ii) Analysis. Scaling estimate of the Nusselt number is given by equation (6.20b)
2/1
~x/D
RerPuN D
D (a)
where
D = diameter = 2 mm = 0.002 m
Pr = Prandtl number
DRe = Reynolds number
The Reynolds number is defined as
DuReD (b)
where
u mean velocity = 0.042 m/s
The Prandtl number is given by
k
c
k
cPr
pp (c)
where
pc = specific heat = CJ/kg3900 o
k thermal conductivity = CW/m5.0 o
kinematic viscosity = /sm1094.0 26
= density = 3kg/m1040
The Nusselt number is defined as
PROBLEM 6.15 (continued)
k
hDNuD (d)
h heat transfer coefficient
Substitute (d) into (a) and solve for h
2/1
~x/D
RerP
D
kh D (e)
(iii) Computations. Use (b) and (c) to compute the Reynolds and Prandtl numbers
3.98/sm1094.0
)m(0022.0)m/s(024.026DRe
CW/m5.0
/s)m(1094.0)kg/m(1040C)J/kg(3900o
o 263
Pr = 7.63
Substitute into (e)
C-W/m9730022(m)0.09(m)/0.
98.37.63
0.0022(m)
C)0.5(W/m-~ o2
2/1o
h
The exact solution to this problem is given in Table 6.4 and Fig. 6.9. The local Nusselt is given in terms of the dimensionless distance , defined as
PrRe
Dx
D
/ (f)
Computing
98.37.63
0022(m)0.09(m)/0.= 0.05454
At this value of , Table 6.4 gives
7.3DNu
Substituting into (d)
C-W/m8410.0022(m)
C)0.5(W/m-3.7 o2
o
h
(iv) Checking. Dimensional check: (1) Each term in (e) has units of heat transfer coefficient. (2) Equations (a)-(d) are dimensionless.
(5) Comments. Scaling estimate of the heat transfer coefficient is surprisingly good.
PROBLEM 6.16
Air flows with fully developed velocity through a tube of inside diameter 2.0 cm. The flow is fully
developed with a mean velocity of 1.2 m/s. The surface is maintained at a uniform temperature
of 90oC. Inlet temperature is uniform equal 30
oC. Determine the length of tube needed to
increase the mean temperature to 70oC.
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is fully developed. (iii) The temperature is developing. (iv) Surface is maintained at uniform temperature. (v) The Reynolds number should be computed to establish if flow is laminar or turbulent. (vi) Tube length is unknown. (vii) The determination of tube length requires determining the heat transfer coefficient.
(2) Problem Definition. Find the required tube length to increase the air temperature to a specified level. This reduces to determining the heat transfer coefficient.
(3) Solution Plan. Compute the Reynolds number to establish that the flow is laminar. Use the results of Section 6.5 on flow through tubes at uniform surface temperature. Use Graetz solution for fully developed laminar flow and developing temperature in tubes to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed velocity, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.
(ii) Analysis.. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)
][)()( exp xcm
hPTTTxT
psmism (6.13)
cp = specific heat, J/kg-oC
h = average heat transfer coefficient for a tube of length L, W/m2-oCL = length of tube, m m = mass flow rate, kg/s
P = tube perimeter, m Tm(x) = mean temperature at x, oC
miT = mean inlet temperature = 30oC
sT = surface temperature = 90oC
x = distance from inlet of heated section, m
Applying (6.13) at the outlet (x = L) and solving for L
tL
hL
m
L
moT
miT
sT
uu
PROBLEM 6.16 (continued)
mos
misp
TT
TT
hP
cmL ln (a)
where
moT = mean outlet temperature = 70oC
To compute L using (b), it is necessary to determine cp, P, m , and h . All properties are
determined at the mean temperature mT defined as
mT = T Tmi mo
2 (b)
The perimeter P and flow rate m are given by
P = D (c) and
uD
m4
2
(d)
where
D = inside tube diameter = 2 cm = 0.02 m u = mean flow velocity = 1.2 m/s
= density, kg/m3
The average heat transfer coefficient, h , for fully developed velocity and developing temperature is given in Graetz solution, Section 6. 8 (Table 6.4 and Fig. 6.9). To proceed, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the Reynolds number is defined as
ReD
u D (e)
where
ReD = Reynolds number
= kinematic viscosity, m2/s
The mean temperature is calculated in order that properties are determined. Substituting into (b)
mT = C502
)C)(7030( oo
Properties of air at this temperature are
cp = 1007.4 J/kg-oCk = 0.02781 W/m-oCPr = 0.709
= 17.92 10-6, m2/s
= 1.0924 kg/m3
Substituting into (f)
ReD
12 0 02
17 92 101339 36 2
. ( / ) . ( )
. ( / ).
m s m
m s
PROBLEM 6.16 (continued)
Since the Reynolds number is smaller than 2300, the flow is laminar. Attention is focus on the
determination of h using either Table 6.4 or Fig. 6.9. However, h depends on the length L
which is unknown. Thus the problem is solved by trial and error. Assume L, use Table 6.4 or Fig.
6.9 to determine h and substitute into (a) to calculate L. If the calculated L is not the same as the
assumed value, the procedure is repeated until a satisfactory agreement between assume and calculated L is obtained.
Table 6.4 gives the average Nusselt number, )(Nu , as a function of dimensionless axial distance
. These are defied as
k
xhNu )( (f)
The variable is defined in (6.21) as
DPrRe
Dx / (g)
Solving (f) for h
D
kNuh )( (h)
(iii) Computations. Substituting into (d) and (e)
P = 0.02(m) = 0.06283 m
kg/s0.00041181.2(m/s))m1.0924(kg/4
)(m(0.02) 322
m
The result of the trial and error procedure described above is:
Assume: x = L = 1.15 m. Substitute into (g)
0606.03.1339709.0
)m)/0.02(m(14.1
At this value of Table 6.4 gives
536.4)(Nu
Substitute into (h)
C)W/m307.6536.40.02(m)
C)m0.02781(W/ o2o
h
Substitute into (a)
m15.1C)70)((90
C)30)((90ln
C)(W/m307.60.06283(m)
)C4(J/kgkg/s)1007.0.0004118(o
o
o2
o
L
Thus the calculated value of L is the same as the assumed value.
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e), (g) and (h) are dimensionally consistent.
PROBLEM 6.16 (continued)
Limiting check: For the special case of mimo TT , the required length should vanish. Setting
mimo TT in (a) gives x = L = 0.
Quantitative checks: The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. Howe ver, it should be remembered that values of h in Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. (i) The thermal entrance length tL is determined using equation (6.6)
Dtt PrReC
D
L (6.6)
Table 6.1 gives 033.0tC . Substituting into (6.6) gives 627.0tL m. Since this is not small
compared to L , entrance length must be taken into consideration in solving this problem.
(ii) If entrance effects are neglected and temperature is assumed full developed, the corresponding Nusselt number will be 3.66. Substituting this value in (a) gives L = 1.425 m. This is 24% larger than the more accu rate result of entrance length analysis.
PROBLEM 6.17
Air flows with a mean velocity of 2 m/s through a tube of diameter 1.0 cm and length 14 cm. The
velocity is fully developed throughout. The mean temperature at the inlet is 35oC. The surface of
the tube is maintained at a uniform temperature of 130oC. Determine the outlet temperature.
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is fully developed. (iii) The temperature is developing. (iv) Surface is maintained at uniform temperature. (v) The Reynolds number should be computed to establish if flow is laminar or turbulent. (vi) Outlet mean temperature is unknown. (vii) The determination of outlet temperature requires determining the heat transfer coefficient. (viii) Since outlet temperature is unknown, air properties can not be determined. Thus a trial and error procedure is needed to solve the problem.
(2) Problem Definition. Find the outlet temperature of air heated in tube at uniform surface temperature. This reduces to determining the heat transfer coefficient.
(3) Solution Plan. Compute the Reynolds number to establish that the flow is laminar. Use the results of Section 6.5 on flow through tubes at uniform surface temperature. Use Graetz solution for fully developed laminar flow and developing temperature in tubes to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) fully developed velocity, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation and (9) no energy generation.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)
][)()( exp xcm
hPTTTxT
psmism (6.13)
cp = specific heat, J/kg-oC
h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate, kg/s
P = tube perimeter, m Tm(x) = mean temperature at x, oC
miT = mean inlet temperature = 35oC
sT = surface temperature = 130oC
x = distance from inlet of heated section, m
Applying (6.13) at the outlet (x = L) and solving for moT
tL
hL
m
L
moT
miT
sT
uu
PROBLEM 6.17 (continued)
p
missmocm
hLPTTTT exp)( (a)
where
L = length of tube = 0.14 m
moT mean outlet temperature, oC
To compute moT using (b), it is necessary to determine cp, P, m , and h . All properties are
determined at the mean temperature mT defined as
mT = T Tmi mo
2 (b)
The perimeter P and flow rate m are given by
P = D (c) and
uD
m4
2
(d)
where
D = inside tube diameter = 1 cm = 0.01 m u = mean flow velocity = 2 m/s
= density, kg/m3
The average heat transfer coefficient, h , for fully developed velocity and developing
temperature is given in Graetz solution, Section 6.8 (Table 6.4 and Fig. 6.9). However, this solution is valid for laminar flow. Thus, to proceed with the analysis, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the Reynolds number is defined as
ReD
u D (e)
where
ReD = Reynolds number
= kinematic viscosity, m2/s
Since mean outlet temperature moT is unknown, properties can not be determined using (b). A
trial an error procedure is required in which a value for moT is assumed, properties determined
using the assumed value and (a) is used to calculate moT . If the calculated moT is equal to the
assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Assume C65omoT . (b) gives
mT = C502
)C)(6535( oo
Properties of air at this temperature are
cp = 1007.4 J/kg-oCk = 0.02781 W/m-oC
PROBLEM 6.17 (continued)
Pr = 0.709
= 17.92 10-6, m2/s
= 1.0924 kg/m3
Substituting into (f)
ReD 1.1116/s)(m1017.92
(m)2(m/s)0.0126
Since the Reynolds number is smaller than 2300, the flow is laminar. Attention is now focus on
the determination of h using either Table 6.4 or Fig. 6.9. Table 6.4 gives the average Nusselt
number, )(Nu , as a function of dimensionless axial distance . These are defied as
k
xhNu )( (f)
The variable is defined in (6.21) as
DPrRe
Dx / (g)
Solving (f) for h
D
kNuh )( (h)
(iii) Computations. Substituting into (d) and (e)
P = 0.01(m) = 0.03141 m
kg/s0001716.02(m/s))m1.0924(kg/4
)(m(0.01) 322
m
To determine h , use (g) and Table 6.4
01769.007.1116709.0
)m)/0.01(m(14.0
At this value of Table 6.4 gives
95.5)(Nu
Substitute into (h)
C)W/m55.1695.50.01(m)
C)m0.02781(W/ o2o
h
Substitute into (a)
C6.67)CJ/kg(4.1007)kg/s(0001716.0
)CW/m(55.16)m(03141.0)m(14.0exp)35130()C(130 o
o
o2o
moT
This is close to the assume value. Thus C6.67 omoT .
PROBLEM 6.17 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e), (g) and (h) are dimensionally consistent.
Limiting check: For the special case of mis TT , the outlet temperature should be the same as
the inlet. Setting mis TT in (a) gives mio TT .
Quantitative checks: The value of h appears to be low compared with typical values listed in Table 1.1 for forced convection of gases. Howe ver, it should be remembered that values of h in Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. (i) The thermal entrance length tL is determined using equation (6.6)
Dtt PrReC
D
L (6.6)
Table 6.1 gives 033.0tC . Substituting into (6.6) gives 261.0tL m. Since this is not small
compared to L , entrance length must be taken into consideration in solving this problem. In fact the thermal boundary layer is developing throughout the tube.
(ii) Because outlet temperature is unknown propertie s can not be determined a priori. Thus trial and error procedure is needed to solve the problem.
PROBLEM 6.18
A research apparatus for a pharmaceutical laboratory requires heating plasma in a tube 0.5 cm
in diameter. The tube is heated by uniformly wrapping an electric element over its surface. This
arrangement provides uniform surface heat flux. The plasma is monitored in a 15 cm long
section. The mean inlet temperature to this
section is C18o and the mean velocity is 0.025
m/s. The maximum allowable temperature is
C.42o You are asked to provide the designer of
the apparatus with the outlet temperature and
required power corresponding to the maximum
temperature. Properties of plasma are:
CJ/kg3900 opc , CW/m5.0 ok , /sm1094.0 26 , 3kg/m1040
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) The velocity is fully developed and the temperature is developing. (iii) The surface is heated with uniform flux. (iv) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (v) Compute thermal entrance length to determine if it can be neglected. (vi) Surface temperature varies with distance from entrance. It is maximum at the outlet. Thus surface temperature at the outlet is known. (vii) Analysis of uniformly heated channels gives a relationship between local surface temperature, heat flux and heat transfer coefficient. (viii) The local heat transfer coefficient varies with distance form the inlet. (ix) Knowing surface heat flux, the required power can be determined. (x) Newton’s law of cooling applied at the outlet gives outlet temperature.
(2) Problem Definition. Determine the local heat transfer coefficient at the outlet ).(Lh
(3) Solution Plan. Apply channel flow heat transfer analysis for uniform surface flux to determine surface heat flux. Compute the Reynolds number to establish if the flow is laminar or turbulent. Compute entrance length to determine if it can be neglected. Select an applicable equation for determining the Nusselt number at the outlet. Apply Newton’s law of cooling at the outlet to determine outlet temperature.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (e) uniform surface heat flux, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction and (6) negligible dissipation.
(ii) Analysis. Equation (6.10) gives the result of heat transfer analysis for channel flow with uniform surface heat
)(
1)(
xhcm
PxqTxT
psmis (6.10)
where
pc = specific heat, CJ/kg3900 o
)(xh = local heat transfer coefficient, CW/m o2
+
-L
miT
moT
uDsectiontest
PROBLEM 6.18 (continued)
m = mass flow rate, kg/s
P = tube perimeter, m
miT mean inlet temperature = C18o
sq = surface heat flux, 2W/m
)(xTs = local surface temperature, Co
x = distance from inlet of heated section, m
Apply (6.10) at the outlet, x = L, and solve for sq
1
)(
1)(
Lhcm
PLTLTq
pmiss (a)
where
L = 0.15 m
42)(LTs Co
The perimeter P is DP (b)
where
D = tube diameter = 0.005 cm
The flow rate is given by
uD
m4
2
(c)
where
u mean velocity = 0.025m/s
density = 3kg/m1040
The heat transfer coefficient at the outlet, h(L). This requires establishing whether the flow is laminar or turbulent. Thus, the Reynolds number should be computed
DuReD (d)
= kinematic viscosity /sm1094.0 26
The Peclet number, Pe, is computed to determine if axial conduction can be neglected
DPrRePe (e)
The Prandtl number, Pr, is given by
k
c
k
cPr
pp (f)
where is viscosity.
The thermal entrance length tL is determined using equation (6.6)
Dtt PrReC
D
L (6.6)
PROBLEM 6.18 (continued)
where tC is a constant given in Table 6.1. For tubes at uniform surface heat flux, 033.0tC .
To determine the required power, the surface flux is multiplied by the total surface are
sqDLPower (g)
To determine outlet temperature, apply Newton’s law of cooling at the outlet
moss TTLhq )(
Solve for moT
)(Lh
qTT s
smo (h)
Substitute into (c)
DRe 133/s)(m1094.0
0.005(m)0.025(m/s)26
Thus the flow is laminar. Substitute into (f)
rP 625.7C)-(W/m5.0
/s)(m1094.0)mC)1040(kg/3900(J/kg-o2
263o
Substitute into (e)
1014133625.7Pe
Thus axial conduction can be neglected.
Equation (6.6) is used to compute tL
m166.0132625.7)m(005.0033.0tL
Therefore, the thermal boundary layer is still developing at the outlet. It follows that entrance effects are important and that the heat transfer coefficient at the outlet should be obtained from Fig. 6.11. For laminar flow in the entrance region of a tube at fully developed velocity profile and uniform heat flux Fig. 6.11 gives the local Nusselt numbers Nu(x) as a function of dimensionless axial distance , defined as
rPRe
Dx
D
/ (i)
The local heat transfer coefficient is given by
)()( xNuD
kxh (j)
(iii) Computation. Compute at x = L
0296.0625.7133
)m(005.0/)m(15.0
At 0296.0 , Fig. 6.11 gives 5)(LNuD . Equation (j) gives
PROBLEM 6.18 (continued)
CW/m50050.005(m)
C)0.5(W/m)( o
o2Lh
Equation (b) is used to compute P
m01571.0)m)(005.0(P
Equation (c) gives the flow rate
kg/s0.0005105m/s)(025.0)kg/m(10404
)m()005.0( 32
m
Substitute into (a)
2
2W/m7540
CW/m500
1
)CJ/kg-(3900)kg/s(0005105.0
)m(15.0)m(0157.0)C(18)C(42
1
oooo
sq
Substitute into (g)
W77.17)W/m(7540)m(15.0)m(005.0 2Power
Substitute into (h)
C9.26)CW/m(500
)W/m(7540)C(42 o
o2
2o
moT
(iv) Checking. Dimensional check: (i) Computations showed that equations (a), (b), (c), (e), (g), (h) and (j) are dimensionally consistent. (ii) Equations (d), (f) and (i) are dimensionless.
Limiting checks: For the special case of mis TT , the required surface flux should vanish and
.mimo TT Setting mis TT in (a) gives .0sq When this result is substituted into (g) gives
.mimo TT
Quantitative checks: (i) The value of h within the range listed in Table 1.1 for forced convection of liquids.
Global energy balance: energy added at the surface (power) should be equal to energy change of mass flow rate
)( mimops TTmcqDL
or
1)( mimop
s
TTmc
qDL
The above gives
0026.1C))(189.26)(CJ/kg-(3900)kg/s(0005105.0
)W/m(7540)m(15.0)m(005.0oo
2
(5) Comments. (i) Using Fig. 6.1 to determine h introduces a small error. (ii) If entrance effects are neglected and the temperature is assumed fully developed at the outlet, the corresponding
PROBLEM 6.18 (continued)
Nusselt number is 4.364. Using this value gives C,W/m4.436 o2h ,W/m6978 2sq power =
16.45 W and C.28 omoT
(iii) In solving internal forced convection problems, it is important to establish if the flow is laminar or turbulent and if it is developing or fully developed.
PROBLEM 6.19
An experiment is designed to investigate heat transfer in rectangular ducts at uniform surface
temperature. One method for providing heating at uniform surface temperature is based on
wrapping a set of electric elements around the surface. Power supply to each element is
individually adjusted to provide uniform surface temperature. This experiment uses air flowing
in a cm8cm4 rectangular duct 32 cm
long. The air is to be heated from C22o
to C.98o The velocity is fully developed
with a mean value of 0.15 m/s. Your task
is to provide the designer of the
experiment with the heat flux distribution
along the surface. This data is needed to
determine the power supplied to the
individual elements.
(1) Observations. (i) This is an internal forced convection problem in a rectangular channel. (ii) The velocity is fully developed and the temperature is developing. (iii) The surface is maintained at uniform temperature. (iv) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (v) Compute entrance lengths to determine if they can be neglected. (vi) Surface flux varies with distance from entrance. It is minimum at outlet. (vii) Newton’s law gives surface flux in terms of the local heat transfer coefficient )(xh and the local mean
temperature )(xTm . (viii) The local and average heat transfer coefficient decrease with distance
form the inlet. (ix) The local mean temperature depends on the local average heat transfer
coefficient ).(xh (x) Surface temperature is unknown.
(2) Problem Definition. Determine the local and average heat transfer coefficient temperature to
a specified level. This requires determining the local heat transfer coefficients )(xh and )(xh and
surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling. Use the analysis of flow in channels at
uniform surface temperature to determine the local mean temperature ).(xTm Compute the
Reynolds number to establish if the flow is laminar or turbulent. Compute entrance lengths to determine if entrance or fully developed analysis is required. If the thermal entrance can be neglected, use fully developed Nusselt number results. If entrance region is significant, use Graetz solution to determine the local a nd average heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties, (e) uniform surface temperature, (4) negligible changes in kinetic and potential energy, (5) negligible axial conduction and (6) negligible dissipation.
(ii) Analysis. Newton’s law of cooling gives
)()()( xTTxhxq mss (a)
where
)(xh = local heat transfer coefficient, CW/m o2
+
L
miT mo
T
u
+ + + +
---- - -a
b
PROBLEM 6.19 (continued)
)(xqs = local surface heat flux, 2W/m
)(xTm = local mean temperature, Co
sT = surface temperature, Co
For flow through a channels at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (6.13)
][)()( exp xcm
hPTTTxT
psmism (6.13)
where
pc = specific heat, CJ/kg o
h = average heat transfer coefficient for a channel of length x, CW/m o2
m = mass flow rate, kg/s
P = duct perimeter, m
miT mean inlet temperature C22o
x = distance from inlet, m
Substituting (6.13) into (a)
])(
[))(()( exp xcm
xhPTTxhxq
pmiss (b)
Thus we need to determine: ,pc P, m, ,sT )(xh and ).(xh Properties are determined at the mean
temperature mT , defined as
mT = 2
momi TT (c)
Surface temperature is determined by applying (6.13) at the outlet (x = L) where mom TLT )( ,
and solving for sT
)/exp()/exp(1
1pmomi
ps mcLhPTT
mcLhPT (d)
where
L = channel length = 32 cm = 0.32 m
moT = outlet temperature = C98o
The perimeter P is )(2 baP (e)
where
a = channel height = 8 cm = 0.08 m b = channel width = 4 cm = 0.04 m
x
sT
0umiT
t
a
b
sq
PROBLEM 6.19 (continued)
Mass flow rate is given by uabm )( (f)
where
= density, 3kg/m
u = mean velocity = 0.15 m/s
The determination of )(xh and )(xh requires computing the Reynolds number to establish if the
flow is laminar or turbulent and computing the thermal entrance lengths to determine if it is important. The Reynolds number is
eeD
DuRe (g)
where
eD = equivalent diameter, m
= kinematic viscosity, m2/s
The equivalent diameter is defined as
)(
2
)(244
ba
ab
ba
ab
P
AD
fe (h)
where fA is flow area. To proceed with the analysis the Reynolds number must be computed
first. Properties are determined at mT
mT = C602
)C)(9822( oo
Properties of air at this temperature are
pc = 1008 J/kg- oC
k = 0.02852 W/m-oC
Pr = 0.708
= 18.9 10-6
m2/s
= 1.0596 kg/m3
Equation (h) gives
2m05333.0)m)(04.008.0(
)m)(04.0)m)(08.0(2eD
Substituting into (g)
2DRe 28.423/s)(m109.81
.053333(m)0.15(m/s)026
Since the Reynolds number is less than 2300, the flow is laminar. The next step is to compute
the thermal entrance length tL . For laminar flow through channels equation (6.6) gives
eDett PrReDCL (6.6)
PROBLEM 6.19 (continued)where
tC = thermal entrance length coefficient, uniform surface temperature (Table 6.1) = 0.049
Substituting numerical values into (6.6)
tL = 0.049 0.053333 (m) 423.28 0.708 = 0.783 m
Since tL is larger than channel length L, it follows that entrance effects must be taken into
consideration in determining )(xh and ).(xh For laminar flow in the entrance region of a tube at
fully developed velocity profile and uniform surface temperature, Graetz solution gives )(xh and
).(xh Fig. 6.9 and Table 6.4 give the average the local and average Nusselt numbers Nu(x) and
Nu as a function of dimensionless axial distance , defined as
rPRe
Dx
eD
e/ (i)
The average heat transfer coefficient )(xh is given by
)()( xNuD
kxh (j)
Similarly, the local heat transfer coefficient is given by
)()( xNuD
kxh (k)
(iii) Computation. Surface temperature is determined using (d). This requires computing
).(Lh Thus we compute at x = L
02002.0708.028.423
)m(053333.0/)m(32.0
At 02.0 , Table 6.4 gives 81.5Nu
Substituting into (j)
CW/m11.381.5)0.053333(m
C)m0.02852(W/)( o
o2Lh
Equation (c) is used to compute P
m24.0)m)(04.0()m)(08.0(2P
Equation (d) gives the flow rate
kg/s0.0005086m/s)(15.0m))(04.0(m))(08.0()kg/m(0596.1 3m
Before substituting into (d), the exponent of the exponential is calculated
4659.0C)J/kgkg/s)1008(0.0005086(
C)0.32(m)(W/m)11.3(m)(24.0o
o2
pmc
LhP
PROBLEM 6.19 (continued)
Substituting into (d)
C07.2264659.0exp()C(98)C(22)4659.0exp(1
1 ooosT
Equation (b) is used to determine the heat flux variation with x. The procedure is to select a value of x, use (i) to compute the corresponding , use Table 6.4 or Fig. 6.9 to determine )(xh and
).(xh Substituting in (b) gives the local heat flux. For example, to determine the heat flux at
,Lx we apply (b) at Lx
])(
[))(()( exp Lcm
LhPTTLhLq
pmiss (l)
Table 6.4 gives the local Nusselt number Nu(L) at x = L ( 02.0 ) as 17.4Nu . Thus
CW/m23.217.4)0.053333(m
C)m0.02852(W/)( o
o2Lh
Substituting into (l)
22 W/m6.285)4659.0)C)(2207.226)(CW/m(23.2)( (expooLqs
The flux at other locations x along the channel is given in the following table.
(iv) Checking. Dimensional check: (i) Computations showed that equations (a)-(k) are dimensionally consistent. (ii) The Reynolds number and the exponent of the exponential are dimensionless.
Limiting checks: For the special case of smmi TxTT )( , the required surface flux should
vanish. Setting sm TxT )( in (a) gives .0sq
Qualitative check: As anticipated, the local and average heat transfer coefficients and surface heat flux decrease with distance from the inlet.
Quantitative checks: (i) The value of h is outside the range listed in Table 1.1 for forced convection of gases. Examination of equation (j) of (k) shows that the heat transfer coefficient is inversely proportional to the diameter. Thus, as diameter increases the heat transfer coefficient decreases.
x(m)
Nu(x) h(x)
( CW/m o2 )
)(xNu )(xh
( CW/m o2 )
)(xqs
2W/m 0 0
0.008 0.0005 12.8 6.85 19.29 10.32 1344.9
0.016 0.001 10 5.35 16 8.56 1024
0.032 0.002 8.03 4.29 12.09 6.47 794.6
0.08 0.005 6.0 3.21 8.92 4.77 547.9
0.12 0.0075 5.5 2.94 8 4.28 471.9
0.16 0.01 5 2.67 7 3.74 411.8
0.32 0.02 4.17 2.23 5.81 3.11 285.6
PROBLEM 6.19 (continued)
(5) Comments. (i) Using Fig. 6.9 to determine h and h introduces a small error. (ii) If entrance effects are neglected and the flow is assumed fully developed throughout, the corresponding
Nusselt number is 3.66. Using this value gives C.W/m96.1 o2h
PROBLEM 7.1
Explain why
(a) t can not be larger than .
(b) can be larger than .t
Solution
[a] The driving force in free convection is buoyancy. Thus, wherever the local temperature is different from the ambient temperature, the
fluid will move. This means that t can not be
larger than .
[b] Fluid inside the thermal boundary layer moves due to buoyancy force. Because of viscous forces this moving fluid drags fluid layers outside the thermal boundary layer and cause it to
move. That is, fluid motion outside t is due to viscous force and not buoyancy. Consequently,
can be larger than t .
PROBLEM 7.2
A vertical plate 6.5 cm high and 30 cm wide is maintained at
82 Co . The plate is immersed in water at 18 Co . Determine:
(a) The viscous boundary layer thickness.(b) The thermal boundary layer thickness at the trailing end of
the plate.(c) The average heat transfer coefficient.
(d) Total heat added to water.
(1) Observations. (i) This is an external free convection problem over a vertical plate. (ii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (iii) The solution for laminar flow is given in Section 7.4. (iv) For laminar flow, Fig.7.2 gives the viscous
boundary layer thickness and Fig. 7.3 gives the thermal boundary layer thickness t . (v)
Newton’s law of cooling gives the heat transfer rate. (vi) Equation (7.23) gives the average heat
transfer coefficient h . (vii) The fluid is water.
(2) Problem Definition. Determine flow and heat transfer characteristics for free convection over a vertical plate at uniform surface temperature.
(3) Solution Plan. Compute the Rayleigh number to determine if the flow is laminar. For
laminar flow use the result of Section 7.4 to determine , t and .h Apply Newton's law of
cooling to determine the total heat transfer .Tq
(4) Plan Execution.
(i) Assumptions. (1) Continuum, (2) Newtonian, (3) steady state, (4) two-dimensional, (5) constant properties (except in buoyancy), (6) boundary layer flow, (7) laminar flow, (8) uniform surface temperature, (9) negligible radiation and (10) quiescent fluid.
(ii) Analysis. The Rayleigh number RaL is calculated first to determine the appropriate
correlation equation for h . The Rayleigh number is defined as
RaL= PrLTTg s
2
3
(a)
where
g = gravitational acceleration = 9.81 m/s2
L = plate dimension in the direction of gravity = 6.5 cm = 0.065 m Pr = Prandtl number RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
v = kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf=
T Ts
2 (b)
g
T
W
L
sT
PROBLEM 7.2 (continued)
Ts
= surface temperature = 82oC
T = ambient air temperature = 18oC
Viscous boundary layer thickness is determined using Fig. 7.2. This requires computing the
following:
x
yGrx4/1
4 (c)
and
2
3xTTgGr s
x (d)
Thermal boundary layer thickness t is determined using Fig. 7.3. The average heat transfer
coefficient is given by equation (7.27)
d
dGr
L
kh L )0(
43
44/1
(7.27)
h = average heat transfer coefficient, W/m2-oC
k = thermal conductivity, W/m-oC
Temperature gradient dd /)0( is given in Table 7.1. Newton’s law of cooling gives the total
heat transfer rate
Tq = h A (Ts - T ) (e)
where
A = surface area of the two vertical sides, m2
Tq = heat transfer from the surface to the ambient air, W
Surface area of the two vertical sides is given by
A = 2 LW (f)W = plate width = 30 cm = 0.3 m
(iii) Computations. Properties of air are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
2
)C)(1882( o
= 50oC
Water properties at this temperature
k = 0.64056 W/m-oC Pr = 3.57
310462.0 1/K6105537.0 m2/s
Substituting into (a)
RaL = 57.3)/s(m)10(0.5537
)(mC)(0.065)18)()(82m/s(1/K)9.81(100.4622426
33o23
= 0.927577 910
PROBLEM 7.2 (continued)
At Pr = 3.57, Fig. 7.2 gives
54
4/1
L
GrL (g)
Use (d) to compute LGr at the trailing end x = L
9
2426
33o23
10259826.0)/s(m)10(0.5537
)(mC)(0.065)18)()(82m/s(1/K)9.81(100.462LGr
Substituting into (g)
)m(065.04
10259826.05
4/19
Solving the above for
mm62.3m1062.3 3
Fig. 7.3 gives the thermal boundary layer thickness t . At Pr = 3.57, Fig. 7.3 gives
)m(065.04
10259826.05.2
4/19
tt
Solving for mm1.81m1081.1 3t
At Pr = 3.57, Table 7.1 gives the gradient dd /)0(
d
d )0(0.86
Substituting into (7.27)
Cm
W10140.86
4
100.25982
0.065(m)
C)0.6405(W/m
3
4o2
1/49o
h
Substituting into (e)
Tq = W1265C)18)(0.3(m)(82C)0.065(m)14(W/m10 oo2
(iv) Checking. Dimensional check: Computations showed that equations (a)-(g) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of gases.
Qualitative check: Increasing surface temperature Ts should increase the heat transfer rate.
According to equation (e), Tq is directly proportional to h and Ts. According to (7.27), an
increase in sT brings about an increase in .h
(5) Comments. The computed values of and t are approximate since the corresponding
values of cannot be read accurately from Fig. 7.2 and Fig. 7.3.
PROBLEM 7.3
Use Fig. 7.3 to determine dd /)0( for Pr = 0.01 and 100. Compare your result with the value
given in table 7.1.
(1) Observations. (i) This is an external free convection problem for flow over a vertical plate. (ii) Laminar flow solution for temperature distribution for a plate at uniform surface temperature is given in Fig. 7.3 . (iii) The dimensionless temperature gradient at the surface is given in Table 7.1. (iv) The solution depends on the Prandtl number.
(2) Problem Definition. Determine the normal temperature gradient at the surface, ,/)0( dd
for laminar boundary layer flow over a vertical plate.
(3) Solution Plan. Use the temperature solution presented graphically in Fig. 7.3 and compare with the exact value listed in Table 7.1.
(4) Plan Execution.
(i) Assumptions (a)
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) Boussinesq approximations, (4)
two-dimensional, (5) laminar flow ( 910xRa ),
(6) vertical flat plate (7) uniform surface temperature, (8) no dissipation and (9) no radiation.
(ii) Analysis. Fig. 7.3 is a plot of the dimensionless temperature vs. the similarity
variable . Temperature gradient dd /)0(
can be determined from this figure by graphically evaluating the slope is the slope at
.0 This slope is expressed as
0
)0(
d
d (a)
(iii) Computations. For Pr = 0.01, equation (a) and Fig. 7.3 give
076.06
545.01)0(
0d
d
Table 7.1 gives 0806.0)0(
d
d.
For Pr = 100, equation (a) and Fig. 7.3 give
PROBLEM 7.3 (continued)
04.249.0
1)0(
0d
d
Table 7.1 gives 191.2)0(
d
d.
(iv) Checking. Quantitative check: Numerical results obtained using Fig. 7.3 are in good agreement with the exact solutions of Table 7.1.
Comments. Using Fig. 7.3 to determine dd /)0( has an inherent error associated with reading
its scale. Nevertheless, for Prandtl numbers 0.01 and 100 the error is less than 7%.
PROBLEM 7.4
In designing an air conditioning system for a pizza restaurant an estimate of the heat added to
the kitchen from the door of the pizza oven is needed. The rectangular door is
cm201cm50 with its short side along the vertical direction. Door surface temperature is
110oC. Estimate the heat loss from the door if the ambient air temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) Heat is lost from the door to the surroundings by free convection and radiation. (iii) To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air. (iv) As a first approximation, radiation can be neglected. (v) Newton’s law of cooling gives the rate of heat transfer. (vi) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the solution of Section 7.4 is applicable.
(2) Problem Definition. Determine the average heat transfer coefficient for free convection from a vertical plate.
(3) Solution Plan. Apply Newton's law of cooling to the door. For laminar flow use results of Section 7.4.
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible radiation, (6) quiescent ambient fluid and (7) door is in the closed position at all times.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
q = h A ( sT - T ) (a)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC q = heat transfer rate, W
sT = surface temperature = 110oC
T = ambient air temperature = 20oC
Surface area is given by
A = LW (b) where
L = door height = 50 cm = 0.5 m W = door width = 120 cm = 1.2 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first to determine if the flow is laminar or turbulent.
L
W
T
gsT
PROBLEM 7.4 (continued)
RaL = PrLTTg s
2
3
(c)
where
g = gravitational acceleration = 9.81 m/s2
L = door side in the direction of gravity = 50 cm = 0.5 m Pr = Prandtl number of air RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
For ideal gases, the coefficient of thermal expansion is given by
= 003411.015.273)C(20
1
)K(
1oT
(1/K)
Properties are determined at the film temperature Tf defined as
Tf=
2
TTs = 2
)C)(20110( o
= 65oC
Properties of air at this temperature are
k = 0.02887 W/m-oCPr = 0.7075
= 19.4 10-6 m2/s
Substituting into (c) gives
RaL = 7075.0)s/m()104.19(
)m()5.0)(C)(20110)(s/m(81.9)C/1(003411.02426
33o2o
= 0.70766 109
Since RaL
< 109, the flow is laminar and h is given by (7.23)
d
dGr
L
kh L )0(
43
44/1
(7.23)
where LGr is the Grashof number given by
LGr =Pr
RaLTTg Ls
2
3
(d)
and dd /)0( is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
(iii) Computations. Substitution into (d) gives
99
1000226.17075.0
1070766.0LGr
PROBLEM 7.4 (continued)At Pr = 0.7075, Table 7.1 gives
501.0)0(
d
d
Substitute into (7.23)
CW/m853.4501.04
1000226.1
0.5(m)
C)m0.02887(W/
3
4 o2
4/19o
h
Substitute into (a) and use (b) give the heat transfer rate from door
q = 4.853 (W/m2-oC) (0.5)(m)(1.2)(m) (110 )20 (oC) = 261.9 W
(iii) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1.
(5) Comments. (i) The model used to solve this problem is not conservative due to neglecting radiation. If the door is treated as a small object enclosed by a much larger surface at the ambient temperature, an estimate can be made of the radiation heat loss. Assuming that the door is made of stainless steel with an emissivity of 0.25, radiation heat loss will be 120 W. This is significant when compared with free convection heat loss. (ii) Opening and closing the door results in transient effects not accounted for in the above model. In addition, when the door is open radiation from the interior of oven may be significant.
PROBLEM 7.5
To compare the rate of heat transfer by radiation with that by free convection, consider the following test case. A vertical plate measuring 12 cm 12 cm is maintained at a uniform
surface temperature of 125oC. The ambient air and the surroundings are at 25oC. Compare the
two modes of heat transfer for surface emissivities of 0.2 and 0.9. A simplified model for heat
loss by radiation rq is given by
)( 44sursr TTAq
where A is surface area, is emissivity and 428 KW/m1067.5 . Surface and
Surroundings temperatures are measured in degrees kelvin
(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a vertical plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate. (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vi) For laminar flow the solution of Section 7.4 is applicable. (vii) Since radiation heat transfer is considered in this problem, all temperatures should be expressed in degrees kelvin. (viii) The fluid is air.
(2) Problem Definition. Determine heat transfer rate by free convection and by radiation from a vertical plate in air. Convection requires the determination of the heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to determine the rate of heat loss by convection. Apply Stefan-Boltzmann radiation law to determine the rate of heat loss by radiation. Compute the Rayleigh number and select an appropriate correlation equations to obtain the average heat transfer coefficient. For laminar flow use results of Section 7.4.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) the plate is a small surface enclosed by a much larger surface at a uniform temperature and (6) quiescent ambient.
(ii) Analysis. Application of Newton's law of cooling to the vertical plate gives
)( TTAhq sc (a)
where
A = surface area of vertical side, m2
h = average heat transfer coefficient, W/m2-oC or W/m2-K
cq = convection heat transfer rate, W
sT = surface temperature = 125(oC) + 273.15 = 398.15 K
T = ambient temperature = 25(oC) + 273.13 = 298.15 K
Surface area is
L
T
g sT
gssurroundin
surT
L
PROBLEM 7.5 (continued)
A = L2 (b)where
L = side of square plate = 12 cm = 0.12 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first to determine if the flow is laminar or turbulent.
RaL = PrLTTg s
2
3
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number of air RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
( . . )( )29815 39815
2
K = 348.15K
Properties of air at this temperature are
k = 0.02957 W/m-oCPr = 0.7065
= 20.41 10-6 m2/s
For ideal gases the coefficient of thermal expansion is given by
= 002872.015.348
1
)K(
1
fT(1/K)
Substituting into (c) gives
RaL = 7065.0)s/m()1041.20(
)m()12.0)(C)(25125)(s/m(81.9)C/1(002872.02426
33o2o
= 8.257 610
Since RaL
< 109, the flow is laminar and h is given by (7.23)
d
dGr
L
kh L )0(
43
44/1
(7.23)
where LGr is the Grashof number given by
LGr =Pr
RaLTTg Ls
2
3
(d)
and dd /)0( is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
PROBLEM 7.5 (continued)
Radiation heat loss rq is given by the Stefan-Boltzmann law. Assuming that the plate is a small
surface which is surrounded by a much larger surface, rq is given by
rq = A ( 44surs TT ) (e)
where
rq = radiation heat loss, W
Tsur = surroundings temperature = 25(oC) + 273.13 = 298.15 K
= emissivity
= Stefan-Boltzmann constant = 5.67 10-8 W/m2-K4
(iii) Computations. Convection heat loss: substitution into (d) gives
66
10687.117065.0
10257.8LGr
At Pr = 0.7065, Table 7.1 gives
5009.0)0(
d
d
Substitute into (7.23)
CW/m804.65009.04
10687.11
0.12(m)
C)m0.02957(W/
3
4 o2
4/16o
h
Substitute into (a) and use (b) give the heat transfer rate from door
cq = 6.804 (W/m2-oC) (0.12)(m)(0.12)(m) (125 -25)(oC) = 9.798 W
Radiation heat loss: equation (e) for = 0.2, equation (e) gives
rq = 0.2 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 44 15.29815.398 )(K4) = 2.81 W
For = 0.9:
rq = 0.9 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 44 15.29815.398 )(K4) = 12.66 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (c), (7.23) and (e) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of gases.
Limiting check: For surs TTT , heat transfer by convection and radiation vanish. Setting
surs TTT in (a) and (e) gives .0rc qq
(5) Comments. (i) When compared with free convection, radiation heat loss can be significant
PROBLEM 7.5 (continued)
and in general should not be neglected. (ii) The magnitude of is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat. (iii) Because temperature in the Stefan-Boltzmann radiation law must be expressed in degrees kelvin, care should be exercised in using the correct units when carrying radiation computations.
PROBLEM 7.6
A sealed electronic package is designed to be cooled by free convection. The package consists of components which are mounted on the inside surfaces of
two cover plates measuring cm5.7cm5.7 cm each. Because the plates are
made of high conductivity material, surface temperature may be assumed uniform. The maximum allowable surface temperature is 70oC. Determine
the maximum power that can be dissipated in the package without violating design constraints. Ambient air temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) For laminar flow the solution of Section 7.4 is applicable. (ix) The fluid is air.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A ( TTs ) (a)
where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oC P = power dissipated in package, W
q = heat transfer from the surface to the ambient air, W
sT = surface temperature = 70oC
T = ambient air temperature = 20oC
Surface area of the two vertical sides is given by
A = 2 LW (b)
gT
components
air
g
T
W
L
sT
PROBLEM 7.6 (continued)
where
L = package height = 7.5 cm = 0.75 m W = package width = 7.5 cm = 0.75 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first to determine if the flow is laminar or turbulent.
RaL = PrLTTg s
2
3
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number of air RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
( )( )70 20
2
oC = 45oC
Air properties at this temperature are
k = 0.02746 W/m-oCPr = 0.7095
= 17.44 10-6 m2/s
For ideal gases the coefficient of thermal expansion is given by
= 003143.015.273)C(45
1
)K(
1o
fT(1/K)
Substituting into (c) gives
RaL = 7095.0)/s(m)10(17.44
)(mC)(0.075)20)()(70C)9.81(m/s/0.003143(12426
33o2o
= 1.5171 610
Since RaL
< 109, the flow is laminar and h is given by (7.23)
d
dGr
L
kh L )0(
43
44/1
(7.23)
where LGr is the Grashof number given by
LGr =Pr
RaLTTg Ls
2
3
(d)
and dd /)0( is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
(iii) Computations. Substitution into (d) gives
PROBLEM 7.6 (continued)
66
105171.17065.0
10257.8LGr
At Pr = 0.7095, Table 7.1 gives
5017.0)0(
d
d
Substitute into (7.23)
CW/m08.65017.04
105171.1
0.075(m)
C)m0.02746(W/
3
4 o2
4/16o
h
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 2 6.08 (W/m2-oC) (0.075)(m)(0.075)(m) (70-20)(oC) = 3.42 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23) dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of gases.
Qualitative check: Increasing the allowable surface temperature sT should increase the maximum
power P. According to equation (a), q is directly proportional to sT . Furthermore, h increases
when sT is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting radiation and heat loss from the side surfaces. (ii) The maximum power dissipated is relatively small, indicating the limitation of free convection in air as a cooling mode for such applications.
The maximum dissipated power in water is 494.7 W (Problem 7.7). (iii) The magnitude of is
the same whether it is expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 7.7
Assume that the electronic package of Problem 7.6 is to be used in an underwater application.
Determine the maximum power that can be dissipated if the ambient water temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) For laminar flow the solution of Section 7.4 is applicable. (ix) The fluid is water.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A ( TTs ) (a)
where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oC P = power dissipated in package, W
q = heat transfer from the surface to the ambient air, W
sT = surface temperature = 70oC
T = ambient air temperature = 20oC
Surface area of the two vertical sides is given by
A = 2 LW (b)where
L = package height = 7.5 cm = 0.75 m W = package width = 7.5 cm = 0.75 m
To determine the average heat transfer coefficient h , the Rayleigh number RaL is calculated first to determine if the flow is laminar or turbulent.
g
T
W
L
sT
PROBLEM 7.7 (continued)
RaL = PrLTTg s
2
3
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number of water RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
( )( )70 20
2
oC = 45oC
Water properties at this temperature are
k = 0.6286 W/m-oCPr = 4.34
= 0.000389 1/K
= 0.6582 10-6 m2/s
Substituting into (c) gives
RaL = 34.4)/s(m)10(0.6582
)(mC)(0.075)10)()(70C)9.81(m/s/0.000389(12426
33o2o
= 0.96778 910
Since RaL
< 109, the flow is laminar and h is given by (7.23)
d
dGr
L
kh L )0(
43
44/1
(7.23)
where LGr is the Grashof number given by
LGr =Pr
RaLTTg Ls
2
3
(d)
and dd /)0( is a dimensionless temperature gradient which depends on the Prandtl number. It
is listed in Table 7.1.
(iii) Computations. Substitution into (d) gives
99
1022299.034.4
1096778.0LGr
At Pr = 4.34, Table 7.1 gives
9108.0)0(
d
d
Substitute into (7.23)
PROBLEM 7.7 (continued)
CW/m5.8799108.04
1022299.0
0.075(m)
C)0.6286(W/m
3
4 o2
4/19o
h
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 2 879.5 (W/m2-oC) (0.075)(m)(0.075)(m) (70-20)(oC) = 494.7 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (c) and (7.23) dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of liquids
Qualitative check: Increasing the allowable surface temperature sT should increase the maximum
power P. According to equation (a), q is directly proportional to sT . Furthermore, h increases
when sT is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
(5) Comments. (i) The model used to solve this problem is conservative due to neglecting heat loss from the side surfaces. (ii) The maximum power dissipated is relatively large, indicating the effectiveness of water as a free convection medium. The maximum power in air (Problem 7.6) is
3.42 W. (iii) The magnitude of is the same whether it is expressed in units of degree Celsius or
kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 7.8
Consider laminar free convection from a vertical plate at uniform
surface temperature. Two 45 triangles are drawn on the plate as
shown. Determine the ratio of the heat transfer rates from two
triangles.
(1) Observations. (i) This is a free convection problem. (ii) The surface is maintained at uniform temperature. (iii) Newton’s law of cooling determines the heat transfer rate. (iv) Heat transfer rate depends on the heat transfer coefficient. (v) The heat transfer coefficient decreases with distance from the leading edge of the plate. (vi) The width of each triangle changes with distance from the leading edge. (vii) For laminar flow the solution of Section 7.4 is applicable.
(2) Problem Definition. Examine the variation of local heat transfer coefficient with distance and determine the heat transfer rate from each triangle.
(3) Solution Plan.
Apply Newton’s law of cooling to an element of each triangle. Formulate an equation for element area and heat transfer coefficient )(xh for laminar free convection over a flat plate.
(4) Plan Execution.
(i) Assumptions. (1) Laminar flow, (2) steady state, (3) two-dimensional, (4) constant properties (except in buoyancy), (5) uniform surface temperature, (6) quiescent fluid and (7) no radiation.
(ii) Analysis. Consider triangle 1 first. Application of Newton’s law of cooling to the element
dxxb )(1
dxxbTTxhdq s )())(( 11
IntegrateH
s dxxbxhTTq0
11 )()()( (a)
where
)(1 xb width of element 1, m
)(xh local heat transfer coefficient, CW/m o2
1q = heat transfer rate, W
sT surface temperature, oC
T = ambient temperature, oC
x = distance along plate, m
Similarly, for the second triangle H
s dxxbxhTTq0
22 )()()( (b)
where
x
dx
dx
b1(x)
b2(x)
g
T
12
H
B
gT
1
2
PROBLEM 7.8 (continued)
)(2 xb width of element 2, m
To evaluate the integrals in (a) and (b) it is necessary to determine the variation with x of h(x),
)(1 xb and )(2 xb . The local heat transfer coefficient for free convection laminar flow over a
vertical plate is given by (7.21)
d
dGr
x
kh x )0(
4
4/1
(7.21)
where
k thermal conductivity, W/m-oC
dxd /)0( dimensionless temperature gradient at the surface
xGr = Grashof number, defined as
2
3)( xTTgGr s
x (c)
g = gravitational acceleration, m/s2
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
(c) into (7.21)
d
dxTTg
x
kh s )0(
4
)(4/1
2
3
(d)
According to (d), )(xh decreases varies with x as
4/1)( xCxh (e)
The geometric functions )(1 xb and )(2 xb are determined using similarity of triangles. Thus
b1(x) = B(1 - x
H) (f)
and
b2(x) = Bx
H (g)
where
B = base of triangle, m H = height of triangle, m
Substitute (e) and (f) into (a)
dxxHxTTBCqH
s0
4/11 /1
Evaluate the integral
4/31 ))(21/16( BCHTTq s (h)
Similarly, substitute (d) and (g) into (b)
PROBLEM 7.8 (continued)
dxxHxTTBCqH
s0
4/12 )/(
Evaluate the integral 4/3
2 ))(7/4( BCHTTq s (i)
Taking the ratio of (i) and (i) q
q
1
2
= 4
3 (j)
(iii) Checking. Dimensional check: Units of C are determined using (d):
C = W/m7/4-oC
Thus units of h(x) in (d) are
h(x) = C (W/m7/4-oC) x-1/4(m)-1/4 = W/m2-oC
Examining units of q1 in (h)
q1 = B(m) C(W/m7/4-oC) (Ts -T ) (oC) 4/3H (m3/4) = W
Limiting check: If = 0 or g = 0 or Ts = T , no free convection takes place and consequently q1
= q2 = 0. Any of these limiting cases give C = 0. Thus, according to (h) and (i), q1 = q2 = 0.
(5) Comments. (i) More heat is transferred from triangle 1 than triangle 2. This follows from the fact that h decrease with distance x. This favors triangle 1 since its large base is at x = 0 where his maximum. (ii) The result applies to any right angle triangles and is not limited to 45o
triangles. (iii) Heat transfer from a surface of fixed area depends on its orientation relative to the leading edge. (iv) This problem illustrates how integration is used to account for variations in element area and heat transfer coefficient. The same approach can be applied if surface temperature and/or ambient temperature vary over a surface area.
PROBLEM 7.9
A vertical plate measuring 21 cm 21 cm is at a uniform surface temperature of 80oC. The
ambient air temperature is 25oC. Determine the heat flux at 1 cm, 10 cm and 20 cm from the
lower edge.
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The surface is maintained at uniform temperature. (iii) Local heat flux is determined by Newton’s law of cooling. (iv) Heat flux depends on the local heat transfer coefficient. (v) Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases. (vi) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. For (vii) Laminar flow the solution of Section 7.4 is applicable. (viii) The fluid is air.
(2) Problem Definition. Determine the local heat transfer coefficient for free convection over a vertical plate at uniform surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling, compute the Rayleigh number and select an appropriate Nusselt number correlation equation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Application of Newton's law gives
q = h(x) (Ts - T ) (a)
where
h(x) = local heat transfer coefficient, W/m2-oCq = local heat flux, W
Ts
= surface temperature = 80oC
T = ambient temperature = 25oC
The Rayleigh is computed to determine if the flow is laminar or turbulent.
Rax = PrLTTg s
2
3
(b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
LRa = Rayleigh number at LL = plate height = 0.21 m
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature fT defined as
Tf= 2/)( TTs = 2/)C)(2580( o = 52.5oC
L
T
g sT
L
PROBLEM 7.9 (continued)
Properties of air at this temperature are
k = 0.02799 W/m-oCPr = 0.709
= 18.165 10-6 m2/s
For ideal gases the coefficient of thermal expansion is given by
= 003071.015.273)C(5.52
1
)(
1oKT
(1/K)
Substituting into (b)
RaL = 709.0)/s(m)10(18.165
)(mC)(0.2)25)()(80/sC)9.81(m/0.003071(12426
33o2o
= 2.8482 107
Since RaL
< 109, the flow is laminar over the region of interest. The local heat transfer
coefficient for free convection laminar flow over a vertical plate is given by (7.21)
d
dGr
x
kh x )0(
4
4/1
(7.21)
where
dxd /)0( dimensionless temperature gradient at the surface
xGr = Grashof number, defined as
2
3)( xTTgGr s
x (c)
Note that dxd /)0( depends on the Prandtl number and is listed in Table T.1.
(iii) Computations. At Pr = 0.709, Table 7.1 gives
5015.0)0(
d
d=
At x = 0.01 m, (c) gives
)/s(m)10(18.165
)(mC)(0.01)25)()(80/sC)9.81(m/0.003071(12426
33o2o
xGr = 0.5022 410
Substitute into (7.21)
)5015.0(4
5022
m)(01.0
)CW/m(02799.04/1o
h = 8.356 W/m2-oC
Substitute into (a)
2oo2 W/m459.6C))(2580)(CW/m(355.8q
PROBLEM 7.9 (continued)
The same procedure is followed to determine the flux at x = 10 and x = 20 cm. Results are tabulated below.
x (cm) Grx h(x)(W/m2-oC) q (W/m2)
1 4105022.0 8.355 459.6
10 7105022.0 4.699 258.4
20 8104018.0 3.951 217.3
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (7.21) are dimensionally consistent.
Quantitative check: The values of h are approximately within the range given in Table 1.1 for free convection of gases.
Limiting check: The flux should vanish for Ts = T . Setting Ts = T in (a) gives 0q .
(5) Comments. According to (7.21) and (a), surface heat flux decreases with distance from the leading edge as
4/1x
Cq
Thus, high heat flux components should be place close to the leading edge.
PROBLEM 7.10
200 square chips measuring cm1cm1 each are mounted on both
sides of a thin vertical board .cm01cm10 The chips dissipate 0.035
W each. Assume uniform surface heat flux. Determine the maximum surface temperature in air at 22oC.
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power dissipated in the chips is transferred to the air by free convection. (iii) This problem can be modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end). (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the analysis of Section 7.5 gives surface temperature
distribution. (vii) The fluid is air. (viii) Properties depend on the average surface temperature sT .
Since sT is unknown, the problem must be solved by trail and error.
(2) Problem Definition. Determine surface temperature distribution for a vertical plate with uniform surface heat flux under free convection conditions.
(3) Solution Plan. Apply the analysis of Section 7.5 for surface temperature distribution of a vertical plate with uniform surface heat flux in laminar free convection. Compute the Rayleigh number to confirm that the flow is laminar.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6) negligible radiation and (7) quiescent ambient fluid.
(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is given by equation (7.27)
)0()(
5)(
5/1
4
42
xkg
qTxT s
s (7.27)
where
g = gravitational acceleration = 9.81 m/s2
k = thermal conductivity, W/m-oCPr = Prandtl number
sq = surface flux, W/m2
RaL = Rayleigh number at x = L
sT = surface temperature, oC
T = ambient temperature = 22oCx = distance from leading edge, m
= coefficient of thermal expansion, 1/K )0( = dimensionless surface temperature
= kinematic viscosity, m2/s
The dimensionless surface temperature, )0( , depends the Prandtl number. Values corresponding
to four Prandtl numbers are listed in Table 7.2.
T g
PROBLEM 7.10 (continued)
For laminar flow
RaL = PrLTTg s
2
3
< 910 (a)
where
L = vertical side of plate = 10 cm = 0.1 m
If all dissipated power in the chip leaves the surface, conservation of energy gives
A
Pqs (b)
where
A = chip surface area = 1 cm2 = 0.0001 m2
P = power dissipated in chip = 0.035 W
Properties are evaluate at the film temperature defined as
2
)2/( TLTT s
f (c)
where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x) is unknown, an iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to calculate the film temperature at which properties are determined. Equation (a) is then used to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations. Equation (b) gives surface flux
sq = 0.035(W)/0.0001(m2) = 350 W/m2
Assume Ts (L/2) = 58oC. Equation (c) gives
Tf = (58 + 22)(oC)/2 = 40oC
Properties of air at this temperature are
cp = 1006.8 J/kg- oCk = 0.0271 W/m-oCPr = 0.71
= 16.96 10-6 m2/s
Coefficient of thermal expansion for an ideal gas is given by
= 003193.015.273)C(40
1
15.273
1o
fT1/K
At Pr = 0.71, Table 7.2 gives .806.1)0(
Substituting into (7.27) and letting x = L/2 = 0.1(m)/2 = 0.05 m
806.1)m(05.0C)0.0271(W/m
)350(W/m
)C)9.81(m/s/0.003193(1
(m/s))10(16.965)C(22/2
5/14
o
2
2o
226oLTs = 87.7 Co
PROBLEM 7.10 (continued)
Ts(L/2) = 87.7oC
Since this is higher than the assumed value of 58oC, the procedure is repeated with a new assumed temperature at mid-point. Assume Ts(L/2) = 78oC. The following results are obtained
Tf = 50oCcp = 1007.4 J/kg- oCk = 0.02781 W/m-oCPr = 0.709
= 25.27 10-6 m2/s
= 0.0030945 1/K
= 17.92 10-6 m2/s
= 1.0924 kg/m3
Substituting into (a) gives Ts(L/2) = 76.8oC. This is close to the assumed value of 78oC.
Surface temperature at the trailing end is now computed by evaluating (a) at x = L = 0.1 m
5/14
o
2
2o
26262/1o )m(1.0
)Cm/W(02781.0
)m/W(350
)s/m(81.9)C/1(0030945.0
)s/m(1092.17)s/m(1027.25
709.0
)709.0(10)709.0(94)C(22/2LTs
Ts(L) = 84.9oC
The condition on the Rayleigh number in equation (b) is verified next. Substituting into (c)
RaL = 709.0)s/m()1092.17(
)m()1.0)(C)(229.84)(s/m(81.9)C/1(0030945.02426
33o2o
= 4.22 106
This satisfies the condition on RaL given in equation (b).
(iv) Checking. Dimensional check: Equations (a), (c) and (d) are dimensionally consistent..
Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using Newton's law of cooling:
h(L/2) = ])2/(/[ TLTq ss = 350(W/m2)/(76.8 - 22)(oC) = 6.39 W/m2-oC
This is within the range given in Table 1.1 for free convection of gases.
Validity of correlation equation (a): The conditions listed in (b) are met.
(5) Comments. (i) Surface temperature is determined without calculating the heat transfer coefficient. This is possible because equation (a) combines the correlation equation for the heat transfer coefficient and Newton's law of cooling to eliminate h and obtain an equation for surface
temperature in terms of surface heat flux. (ii) The magnitude of is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 7.11
cm21cm12 power board dissipates 15 watts uniformly. Assume that all energy leaves the
board from one side. The maximum allowable surface temperature is .C82o The ambient fluid is
air at .C24o Would you recommend cooling the board by free convection?
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power dissipated in the chips is transferred to the air by free convection. (iii) This problem can be modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end). (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vii) For laminar flow the analysis of Section 7.5 gives surface temperature distribution. (vii) The fluid is air. (viii)
Properties depend on the average surface temperature sT . Since sT is unknown, the problem
must be solved by trail and error.
(2) Problem Definition. Determine surface temperature distribution for a vertical plate with uniform surface heat flux under free convection conditions.
(3) Solution Plan. Apply the analysis of Section 7.5 for surface temperature distribution of a vertical plate with uniform surface heat flux in laminar free convection. Compute the Rayleigh number to confirm that the flow is laminar.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6) negligible radiation and (7) quiescent ambient fluid.
(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is given by equation (7.27)
)0()(
5)(
5/1
4
42
xkg
qTxT s
s (7.27)
where
g = gravitational acceleration = 9.81 m/s2
k = thermal conductivity, W/m-oC
sq = surface flux, W/m2
Ts = surface temperature, oC
T = ambient temperature = 22oCx = distance from leading edge, m
= coefficient of thermal expansion, 1/K )0( = dimensionless surface temperature
= kinematic viscosity, m2/s
The dimensionless surface temperature, )0( , depends the Prandtl number. It can be determined
from Table 7.2 or using correlation equation (7.29):
L
T
g
L
0
x
sq
PROBLEM 7.11 (continued)
5/12/1 1094
)0(25Pr
PrPr, 1000001.0 Pr (7.29)
The heat flux is defined as
A
Pqs (a)
where
A = surface area, 2m
P = dissipated power = 15 W
Surface area is2LA (b)
where
L = side of power board = 0.12 m
The Rayleigh number is used to determine if the flow is laminar. The criterion is
RaL = PrLTTg s
2
3
< 910 (c)
Equation (7.27) is used to determine maximum surface temperature, )(LTs , corresponding to the
specified surface flux. Properties are evaluate at the film temperature defined as
2
)2/( TLTT s
f (d)
where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x) is unknown, an iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to calculate the film temperature at which properties are determined. Equation (7.27) is then used to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations. Assume Ts (L/2) = 76oC. Equation (d) gives
Tf = (76 + 24)(oC)/2 = 50oC
Properties of air at this temperature are
k = 0.02781 W/m-oCPr = 0.709
= 17.92 10-6 m2/s
Coefficient of thermal expansion for an ideal gas is given by
= 0030945.015.273)C(50
1
15.273
1o
fT1/K
At Pr = 0.709, equation (7.29) gives )0(
PROBLEM 7.11 (continued)
49337.1)709.0(5
)709.0(10709.094)0(
5/1
2
Equations (a) and (b) give surface heat flux
22 m
W67.1041
)0.12(m0.12
15(W)sq
Substituting into (7.27) and letting x = L/2 = 0.12(m)/2 = 0.06 m
C9.15949337.1)m(06.0C)m0.02781(W/
)m1041.67(W/
)C)9.81(m/s/0.00309451
(m/s))10(17.925)C(24/2 o
5/14
o
2
2o
226oLTs
Since this is higher than the assumed value of 58oC, the procedure is repeated with a new assumed temperature at mid-point. Assume Ts(L/2) = 156oC. The following results are obtained
Tf = 90oCk = 0.03059 W/m-oCPr = 0.705
= 0.0027537 1/K
= 21.35 10-6 m2/s4958.1)0(
Substituting into (7.27) gives Ts(L/2) = 162.4oC. Further iteration will bring )2/(LTs between
156 oC and 162 oC. Surface temperature at the trailing end will be even higher. Therefore, board
temperature will exceed the maximum allowable of .C82o It follows that cooling by free
convection is not recommended.
Surface temperature at the trailing end is now computed by evaluating (7.27) at x = L = 0.12 m
4958.10.12(m)C)m0.03059(W/
)m1041.67(W/
)C)9.81(m/s1/0.0027537(
/s)(m)10(21.355C)(24/2
5/14
o
2
2o
2226oLTs
)(LTs 183.1oC
The Rayleigh number is computed to confirm that the flow is laminar. Substitute into (a)
RaL = 7
2426
33o2o
101149.050.70)/s(m)10(21.35
)(mC)(0.12)24)()(183.1C)9.81(m/s1/0.0027537(
Since this is less than 910 , the flow is laminar.
(iv) Checking. Dimensional check: Equations (7.27) and (a)-(d) are dimensionally consistent..
Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using Newton's law of cooling:
PROBLEM 7.11 (continued)
h(L/2) = ])2/(/[ TLTq ss = 1041.67(W/m2)/(159.0 - 24)(oC) = 7.66 W/m2-oC
This is within the range given in Table 1.1 for free convection of gases.
(5) Comments. (i) High surface temperature is due to high surface heat flux. Forced convection cooling is required to meet design conditions on maximum temperature. (ii) A trial and error procedure was required to solve this problem because properties depend on surface temperature which is unknown a priori.
PROBLEM 7.12
Use the integral method to obtain a solution to the local Nusselt number for laminar flow
over a vertical plate at uniform surface temperature .sT Assume t and a velocity and
temperature profiles given by
33
2210 )()()()(, yxayxayxaxayxu
and3
32
210 )()()()(),( yxbyxbyxbxbyxT
Since there is a single unknown ),(x either the momentum or energy equation may be
used. Select the energy equation to determine ).(xt
(1) Observations. (i) This is a free convection problem over a vertical plate at uniform surface temperature. (ii) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (iii) The integral method can be used to determine the velocity and temperature distribution. (iv) Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate at uniform surface temperature.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a third degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties except for buoyancy, (3) Boussinesq approximations are valid, (4) two-dimensional, (5) laminar flow
( 910LRa ), (6) flat plate, (7) uniform surface temperature, (8) negligible changes in kinetic
and potential energy, (9) negligible axial conduction, (10) negligible dissipation, and (11)
t .
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where
h local heat transfer coefficient, CW/m o2
k thermal conductivity, CW/m o
xNu local Nusselt number
x distance along plate measured from the leading edge, m
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
where
PROBLEM 7.12 (continued)
sT surface temperature, Co
T ambient temperature, Co
y coordinate normal to plate, m
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy
equation, (7.36), is used to determine the temperature distribution
t
dyTTudx
d
y
xT
0
)(0,
(b)
where
u axial velocity, m/s
thermal diffusivity, /sm2
Assume a third degree polynomial for the axial velocity u(x,y)
33
2210 )()()()(, yxayxayxaxayxu (c)
The coefficients )(xan are determined using the following known exact and approximate
boundary conditions on the velocity
(1) 0)0,(xu
(2) 0),( txu
(3) 0),(
y
xu t
(4) )()0,(
2
2
TTg
y
xus
Condition (4) is obtained by setting 0y in the x-component of the equations of motion,
(7.5). Equation (c) and the four boundary conditions give the coefficients )(xan
,00a ,4
)(1 t
s TTga ,
2
)(2
TTga s
t
s TTga
1
4
)(3
Substituting the above into (c) and rearranging
2
2
214
)(
ttt
s yyy
TTgu (d)
Note that velocity profile (d) is based on the assumption that .t
For the temperature profile we assume a third degree polynomial
PROBLEM 7.12 (continued)
33
2210 )()()()(),( yxbyxbyxbxbyxT (e)
The boundary conditions on the temperature are
(1) sTxT 0,
(2) TxT t,
(3) 0,
y
xT t
(4) 00,
2
2
y
xT
Equation (c) and the four boundary conditions give the coefficients )(xbn
,0 sTbt
sTTb1
)(2
31 , ,02b
33
1)(
2
1
t
sTTb
Substituting the above into (e)
3
3
2
1
2
3)(),(
ttss
yyTTTyxT (f)
Substituting (f) into (1.10)
t
kh
2
3 (g)
Combining (a) and (g)
tx
xNu
2
3 (h)
The problem reduces to finding t which is obtained using the energy equation. Substituting
(d) and (f) into (b))(
0
3
3
2
2
2
1
2
31)(1
4
)(1
2
)(3
xt
tts
ttt
s
t
s dyyy
TTyy
ydx
dTTgTT
Expand the integrand and simplify
t
tttttt
s
t
dyyyyyy
ydx
dTTg
0
5
6
4
5
2
3
3
42
2
14
2
7
4
)(1
2
3
Evaluate the integral
333333
14
1
6
1
5
1
6
7
2
1
4
)(1
2
3tttttt
s
t dx
dTTg
Simplify and rearrange
PROBLEM 7.12 (continued)
)(2
3153
TTgdx
d
s
tt
Rewrite the above
)(2
1053
TTgdx
d
s
tt
Separating variables and integrating
t
stt
t
dxTTg
d
0
3
0)(2
105
Evaluate the integrals and rearrange
xTTg s
t
)(2
105
4
4
Solve for xt /
4/1
3)(
210
xTTgxs
t (i)
This result can be expressed in terms of the Rayleigh number as
4/1
806.3
x
t
Rax (j)
Substitute (j) into (h) 4/1394.0 xx RaNu (k)
An alternate form is 4/1
394.0 PrGrNu xx (l)
(iii) Checking. Dimensional check: Equations (a), (h), (i), (j) and (k) are dimensionless. Units of (b), (d) and (f) are correct.
Boundary conditions check: Velocity profile (d) and temperature profile (f) satisfy their respective boundary conditions.
(5) Comments. (i) The integral form of the momentum equation was not used in the solution. Therefore the result does not satisfy momentum and thus it is not expected to be accurate. (ii) To examine the accuracy of this model, equation (l) is rewritten as
4/14/1
(394.04
Pr)NuGr
xx (m)
This result is compared with similarity solution (7.49) and integral solution (7.50) which satisfies both momentum and energy. As expected, the accuracy of the integral method
PROBLEM 7.12 (continued)
deteriorates when the momentum equation is neglected. Only at Prandtl numbers of order unity good accuracy is obtained.
xx Nu
Gr4/1
4
Exact Integral
Pr Eq. (7.49) Momentum & energy, Eq (7.50)
EnergyEq. (m)
0.01 0.0806 0.176 0.0725
0.09 0.219 0.305 0.2166
0.5 0.442 0.469. 0.4627
0.72 0.5045 0.513 0.5361
1.0 0.5671 0.557 0.6078
2.0 0.7165 0.663 0.7751
10 1.1649 0.991 1.2488
100 2.191 1.762 2.2665
PROBLEM 7.13
Consider laminar free convection over a vertical plate at uniform surface flux sq . Assume
t and a third degree polynomial velocity profile given by
2
1)(,yy
xuyxu o
Show that:
[a] An assumed second degree polynomial for the temperature profile gives
2
22
1),(
yy
k
qTyxT s
[b] The local Nusselt number is given by
5/1
4
2
2
4536
)(4x
k
qg
Pr
PrNu s
x
(1) Observations. (i) This is a free convection problem over a vertical plate at uniform surface heat flux. (ii) In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution. (iii) The integral method can be used to determine the velocity and temperature distribution. (iv) Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.
(2) Problem Definition. Determine the temperature distribution for boundary layer flow over a flat plate which is heated with uniform surface flux.
(3) Solution Plan. Start with the definition of local Nusselt number. Equate Newton’s law with Fourier’s law to obtain an equation for the heat transfer coefficient h. Apply the integral form of the energy equation using a second degree polynomial temperature profile.
(4) Plan Execution. (i) Assumptions. (1) Steady state, (2) constant properties except for buoyancy, (3) Boussinesq approximations are valid, (4) two-dimensional, (5) laminar flow
( 910LRa ), (6) flat plate, (7) uniform surface temperature, (8) negligible changes in kinetic
and potential energy, (9) negligible axial conduction, (10) negligible dissipation, and (11)
t .
(ii) Analysis. The local Nusselt number is defined as
k
hxNux (a)
where
h local heat transfer coefficient, CW/m o2
PROBLEM 7.13 (continued)
k thermal conductivity, CW/m o
xNu local Nusselt number
x distance along plate measured from the leading edge, m
where the heat transfer coefficient h is given by equation (1.10)
TT
y
xTk
hs
)0,(
(1.10)
where
sT surface temperature, Co
T ambient temperature, Co
y coordinate normal to plate, m
Thus h depends on the temperature distribution ).,( yxT The integral form of the energy
equation, (7.36), is used to determine the temperature distribution
t
dyTTudx
d
y
xT
0
)(0,
(b)
where
u axial velocity, m/s
thermal diffusivity, /sm2
A third degree polynomial for the axial velocity u(x,y) gives
2
1)(,yy
xuyxu o (c)
Assume a second degree polynomial temperature profile
2210 )()()(),( yxbyxbxbyxT (d)
The boundary conditions on the temperature are
(1) sqy
xTk
)0,(
(2) TxT t,
(3) 0,
y
xT t
Equation (d) and the three boundary conditions give the coefficients )(xbn
,2
0k
qTb s
k
qb s
1 ,1
22
k
qb s
Substituting the above into (d)
PROBLEM 7.13 (continued)
2
22
),(y
yk
qTyxT s (e)
Surface temperature is determined by setting y = 0 in (e)
k
qTyxT s
2),( (f)
Substituting (f) into (1.10)
kh 2 (g)
Combining (a) and (g)
xNux 2 (h)
The problem reduces to finding . The two unknown functions )(xuo and )(x are
determined using momentum equation (7.35) and energy equation (7.36):
dyudx
ddyTTg
y
xu
0
2
0
)()0,(
(7.35)
)(
0
)(0,
x
dyTTudx
d
y
xT (7.36)
Substitute (c) and (e) into (7.35)
dyy
yu
dx
ddy
yy
k
qg
u oso
0
42
2
2
0
2
122
(i)
Evaluate the integrals in (i)
22
105
1
6o
so udx
d
k
qgu (j)
Similarly, substitute (c) and (e) into (7.36)
)(
0
22
212
x
dyy
yy
yu
dx
d
k
q
k
q oss (k)
Evaluate the integrals and rearrange
260 oudx
d (l)
Equations (j) and (l) are two simultaneous firs t order ordinary differential equations. The
two dependent variables are )(x and ).(xuo We assume a solution of the form
PROBLEM 7.13 (continued)
mo Axxu )( (m)
nBxx)( (n)
where A, B, m and n are constants. To determine these constants we substitute (m) and (n) into (j) and (l) to obtain
12222
105
2
6
nmnsnm BxAnm
xBk
qgvx
B
A (o)
and122)2(60 nmxABnm (p)
To satisfy (o) and (p) at all values of x, the exponents of x in each term must be identical. Thus, (o) requires that
122 nmnnm (q)
Similarly, (p) requires that
012nm (r)
Solving (q) and (r) for m and n gives
5
3m ,
5
1n (s)
Introducing (s) into (o) and (p) gives two simultaneous algebraic equations for A and B
BAnm
Bk
qg
B
A s 22
105
2
6 (t)
and2)2(60 ABnm (u)
Solving equations (t) and (u) for A and B, gives
5/2
5
436060
sqg
kA (v-1)
and5/1
2
)6/(
60)75/3600(
kqgB
s
(w-1)
Note that
Pr (x)
Substitute into (v-1) and (w-1) 5/2
2
5
436060 Pr
qg
kA
s
(v-2)
PROBLEM 7.13 (continued)
5/12 )5/4(360
sqg
PrkaB (w-2)
Substitute (s) and (w-2) into (n), rearranging and introducing the definition of Rayleigh number, gives the solution to xx /)(
1)5/1(
5/12 )5/4(360
xqg
Prka
x s
(y)
Introduce (y) into (h), use (x) and rearrange gives the local Nusselt number
5/1
4
24536
4x
k
qg
Pr
PrNu s
2
x (z)
(iii) Checking. Dimensional check: Equations (c), (e), (j), (l), (t), (u), (v-1), (v-2), (w-1) and are dimensionally correct. Equations (a), (h), (y) and (z) are dimensionless.
Boundary conditions check: Temperature profile (e) satisfy the three listed boundary conditions.
Limiting check: If 0sq , the Nusselt number should vanish. Setting 0sq in (z) gives
.0xNu
(5) Comments. (i) The same approach can be used to solve the corresponding problem of
variable surface flux, ).(xqs
(ii) The accuracy of the integral solution can be evaluated by comparing (z) with the exact solution. Equation (7.32) gives the exact solution to the local Nusselt number for free convection over a vertical plate at uniform surface flux
)0(
1
5
5/1
4
2x
k
qgNu s
x (7.32)
where the parameter )0( is determined using correlation equation (7.33)
5/1
2
2/1
5
1094)0(
Pr
PrPr, 1000001.0 Pr (7.33)
To facilitate the comparison, the two solutions are rearranged. Integral solution (z) is rewritten as
5/15/1
4
2 4536
4
Pr
Prx
k
qgNu
2s
x (A-1)
Similarly, exact solution (7.32) is rewritten using (7.33) to eliminate )0(
PROBLEM 7.13 (continued)
5/1
2/1
215
4
2 1094 PrPr
Prx
k
qgNu s
x (A-2)
Thus the accuracy of the integral solution can be evaluated by comparing (A-1) and (A-2). The following table compares the two solutions.
The agreement between the two solution is excellent. At Pr = 0.01 the error is 8.3%. At all other Prandtl numbers from 0.1 to 100 the error ranges from 1% to 3.3%.
15
4
2x
k
qgNu s
x
Pr Integral (A-1) Exact (A-2
0.01 0.1019 0.1111
0.1 0.2505 0.2637
0.5 0.4432 0.4388
1.0 0.5479 0.5340
5 0.8254 0.8046
10 0.9618 0.9453
100 1.5455 1.5567
PROBLEM 8.1
Water at 120oC boils inside a channel with a flat
surface measuring 45 cm 45 cm. Air at 62 m/s
and 20oC flows over the channel parallel to the
surface. Determine the heat transfer rate to the air.
Neglect wall resistance.
(1) Observations. (i) This is an external forced convection problem. (ii) The geometry can be modeled as a flat plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives heat transfer rate from the surface to the air. (v) The average heat transfer coefficient must be determined. (vi) The Reynolds number should be evaluated to establish if the flow is laminar, turbulent or mixed. (vii) Analytic or correlation equations give the heat transfer coefficient.
(2) Problem Definition. Determine the average heat transfer coefficients for forced convection over a flat plate.
(3) Solution Plan. Compute the Reynolds number at the trailing edge to establish if it is laminar or turbulent. Apply Newton’s law of cooling. Use analytic or correlation equations to determine the local heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation
(ii) Analysis. Newton’s law of cooling gives the total heat transfer rate
)(2 TTLhq sT (a)
h = average heat transfer coefficient, CW/m o2
L = length and width of the flat surface = 0.45 m
Tq = total heat transfer rate, W
Ts = surface temperature = 120 Co
T = free stream temperature = 20 Co
To determine the average heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as
xRe =xV
(b)
where
xRe = Reynolds number
V = upstream velocity = 62 m/s
x = distance from the leading edge of the plate, m
= kinematic viscosity, /sm2
Properties are evaluated at the film temperature fT defined as
T
V
air
water water
PROBLEM 8.1 (continued)
2
TTT s
f (c)
For the flow over a flat plate, transition Reynolds number txRe is
txRe = 5 105 (d)
The flow is laminar if txx ReRe . Substituting into (c)
fT = (120 + 20)( Co )/2 = 70oC
Properties of air at this temperature are given in Appendix D
k = 0.02922 CW/m o
Pr = 0.707
= 19.9 10-6 /sm2
Evaluating the Reynolds number in (b) at Lx
LRe = 6
2610402.1
)/sm(109.19
)m(45.0)m/s(62
Therefore, the flow is mixed over the plate. The average Nusselt, LNu , number for a plate with
laminar and turbulent flow is given by equation (8.7b)
3/15/45/42/1 )()(037.0)(664.0 PrReReRek
LhNu
tLtL xx (e)
Equation (e) is subject to the limitations on th e Pohlhausen’s solution and following conditions
flat plate, constant sT
5 510 < xRe < 710
0.6 < Pr < 60
properties at fT
All conditions are satisfied.
(iii) Computations. Solving (e) for h
h = 3/15/455/462/15o
)707.0(])105()104.1[(037.0)105(664.0)m(45.0
)CW/m(02922.0
h = 126.6 CW/m o2
Substituting into (a)
Tq = 126.6( CW/m o2 ) )m()45.0( 22 (120 - 20)( Co ) = 2,562.8 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (e) are dimensionally consistent.
Quantitative check: The magnitudes of the heat transfer coefficient is within the range given in Table 1.1 for force convection of gases.
PROBLEM 8.1 (continued)
Limiting check: If the flow is laminar over the entire plate, ReL =txRe , equation (e) reduces to
Pohlhausen’s solution.
(5) Comments. (i) It is important to compute the Reynolds number to determine if the flow is laminar or turbulent. (ii) For laminar forced convection over a flat plate, the heat transfer coefficient decreases as the distance from the leading edge is increased. However, if transition takes place, the heat transfer coefficient increases at the transition location and drops as the distance from the leading edge is increased.
PROBLEM 8.2
Steam at 105oC flows inside a specially designed narrow channel.
Water at 25oC flows over the channel with a velocity of 0.52 m/s.
Assume uniform outside surface temperature Ts = 105oC.
[a] Determine surface heat flux at 20 cm and 70 cm down- stream
from the leading edge of the channel.
[b] Determine the total heat removed by the water if the length is L
= 80 cm and the width is W = 100 cm.
(1) Observations. (i) This is an external forced convection problem. (ii) The geometry can be modeled as a flat plate. (iii) Surface temperature is uniform. (iv) To determine the heat flux at a given location, the local heat transfer coefficient must be determined. (v) The average heat transfer coefficient is needed to determine the total heat transfer rate. (vi) Newton’s law of cooling gives surface flux and total heat transfer rate. (vii) The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed. (viii) Analytic or correlation equations give the heat transfer coefficient.
(2) Problem Definition. Determine the local and average heat transfer coefficients for forced convection over a flat plate.
(3) Solution Plan.
[a] Apply Newton's law of cooling to determine the local heat flux. Check the Reynolds number at 0.2 m and 0.7 m from the leading edge to see if it is laminar or turbulent. Use analytic or correlation equations for the local heat transfer coefficient.
[b] Apply Newton's law of cooling to the entire plate to determine the total heat loss. Check the Reynolds number at 0.8 m from the leading edge. Select an appropriate equation to determine the average heat transfer coefficient for the plate.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation.
(ii) Analysis.
[a] Newton’s law of cooling gives the local heat flux
qs = h (Ts - T ) (a)
where
h = local heat transfer coefficient, W/m2-oCqs = local surface heat flux, W/m2
Ts = surface temperature = 105oC
T = free stream temperature = 25oC
To determine the local heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as
Rex =V x
(b)
Where
steam
water
x
sT
0
L
WVT
sq
steam
water
W
L
PROBLEM 8.2 (continued)
Rex = Reynolds number
V = upstream velocity = 0.52 m/s x = distance from the leading edge of the plate, m
= kinematic viscosity, m2 /s
Properties are evaluated at the film temperature Tf defined as
Tf = 2
TTs (c)
For the flow over a flat plate, transition Reynolds number tx
Re is
txRe = 5 105 (d)
The flow is laminar if Rex < txRe .Substituting into (c)
Tf = (105 + 25)(oC)/2 = 65oC
Properties of water at this temperature are given in Appendix D
k = 0.6553 W/m-oCPr = 2.77
= 0.4424 10-6 m2/s
The Reynolds number at x = 20 cm = 0.2 m is
Rex = 081,235)/sm(104424.0
)m(2.0)m/s(52.026
Comparing this with the transition Reynolds number in (d) shows that the flow is laminar at x = 20 cm. Similarly, the Reynolds number at x = 70 cm = 0.7 m is
Rex = 785,822)/(104424.0
)(7.0)/(52.026 sm
msm
Thus, the flow is turbulent at x = 70 cm. The local heat transfer coefficient for laminar flow is given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.24b) gives
h = 0.332 kV
xPr
1/3 (e)
The local Nusselt number for turbulent flow over a flat plate is given by equation (8.4a)
3/15/4 )()(0296.0 PrRek
hxNu xx (f)
This correlation equation applies toflat plate, constant Ts
5 105 < Rex < 107
0.6 < Pr < 60 properties at Tf (g)
Since all conditions in (g) are satisfied, equation (f) is applicable to this case.
[b] Newton’s law of cooling gives the total heat transfer rate
qT = )( TTLWh s (h)
PROBLEM 8.2 (continued)
where
h = average heat transfer coefficient, W/m2-oCL = length of plate = 80 cm = 0.8 m
Tq = total heat transfer rate, W
W = width of plate = 100 cm = 1.0 m
The Reynolds number at x = L = 0.8 m is
ReL = 325,940)s/m(104424.0
)m(8.0)s/m(52.026
Therefore, the flow is mixed over the plate. The average Nusselt number for a plate with laminar and turbulent flow is given by equation (8.7b)
3/15/45/42/1 )()(037.0)(664.0 PrReReRek
LhNu
tt xLxL (i)
where
h = average heat transfer coefficient, W/m2-oC
Equation (i) is subject to the condi tions listed in (g) and the limitations on Pohlhausen’s solution. Both are satisfied by this case.
(iii) Computations.
[a] Heat flux at x = 20 cm. The local heat transfer coefficient for laminar flow at x = 20 cm is determined using (e). Solving (e) for h
h = 3/12/1o
)77.2()081,235(332.0)m(2.0
)Cm/W(6553.0= 740.7 W/m2-oC
Substituting this result in (a)
sq = 740.7(W/m2-oC)(105 - 25)(oC) = 59,256 W/m2
Heat flux at x = 70 cm. The local heat transfer coefficient is given by (f). Solving (f) for h
h = 0.0296)m(7.0
)Cm/W(6553.0 o
(822,785)4/5 (2.77)1/3 = 2,101 W/m2-oC
Substituting into (a)
sq = 2101(W/m2-oC)(105 - 25)(oC) = 168,080 W/m2
[b] Total heat transfer from plate. Solving (i) for h
3/15/45/42/1 )()(037.0)(664.0 PrReReReL
kh
tt xLx
h = 3/15/45/42/1o
)77.2(])000,500()325,940[(037.0)000,500(664.0)m(8.0
)Cm/W(6553.0= 1554.3 W/m2-oC
Substituting into (h)
qT
= 1554.3(W/m2-oC) 0.8(m)1.0(m) (105 - 25)(oC) = 99,475 W
PROBLEM 8.2 (continued)
(iv) Checking. Dimensional check: Computations showed that units of equations (a), (b), (e), (f), (h) and (i) are dimensionally consistent.
Quantitative check: The magnitudes of heat transfer coefficients are within the range given in Table 1.1 for force convection of liquids.
Limiting check: If the flow is laminar over the entire plate, ReL =txRe , equation (i) reduces to
Pohlhausen’s solution.
(5) Comments. (i) It is important to check the Reynolds number to determine if the flow is laminar or turbulent. (ii) For laminar forced convection over a flat plate, the heat transfer coefficient decreases as the distance from the leading edge is increased. However, if transition takes place, the heat transfer coefficient increases at the transition location and drops as the distance from the leading edge is increased. (iii) The fact that the average heat transfer coefficient for the entire plate (L = 0.8 m) is larger than the local heat transfer coefficient at x = 0.2 m is due to transition from laminar to turbulent flow.
PROBLEM 8.3
Electronic components are mounted on one side of a circuit board. The board is cooled on the
other side by air at 23oC flowing with a velocity of 10 m/s. The length of the board is L = 20 cm
and its width is W = 25 cm. Assume uniform board temperature.
[a] Determine the maximum power that can be dissipated in the package if surface temperature
is not to exceed 77oC. Assume that all dissipated power is conducted through the plate to the air.
[b] To increase the maximum power without increasing surface temperature, it is recommended
that the boundary layer be tripped to turbulent flow very close to the leading edge. Is this a valid
recommendation? Substantiate your view.
(1) Observations. (i) This is an external forced convection problem of flow over a flat plate. (ii) Surface temperature is assumed uniform. (iii) The heat transfer coefficient in turbulent flow is greater than that in laminar flow. Thus higher heat transfer rates can be sustained in turbulent flow than laminar flow. (iv) The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed. (v) Heat loss from the surface is approximately equal to the power dissipated in the package. (vi) Newton’s law of cooling gives a relationship between heat transfer rate, surface area, heat transfer coefficient, surface temperature and ambient temperature. (vii) The fluid is air.
(2) Problem Definition. Determine the average heat transfer coefficient.
(3) Solution Plan. Apply Newton’s law of cooling to determine the maximum heat removed, check the Reynolds number to establish if the flow is laminar, turbulent or mixed and use appropriate average Nusselt number solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) energy leaves the top surface only, (9) no buoyancy ( = 0 or g = 0) and (9) negligible radiation.
(ii) Analysis.
[a] Applying Conservation of energy to the package gives
P = q (a) where
P = power dissipated in package, W q = heat removed from top surface, W
Application of Newton’s law of cooling to the top surface
q = h A (Ts - T ) (b) where
A = plate area, m2
L
W
components
V
T
PROBLEM 8.3 (continued)
h = average heat transfer coefficient, W/m2-oCTs = surface temperature = 77oC
T = ambient temperature = 23oC
Surface area is A = LW (c)
where
L = plate length = 20 cm = 0.2 m W = plate width = 25 cm = 0.25m
To determine h it is necessary to first establish if the flow is laminar or turbulent. If the Reynolds number at the trailing end is smaller than the transition number, the flow is laminar throughout. Define
ReL =LV
(d)
and
txRe = 5 105 (e)
where
LRe = Reynolds number at the trailing end of plate
txRe = transition Reynolds number for flow over a flat plate
V = free stream velocity = 10 m/s
= kinematic viscosity, m2/s
Properties are evaluated at the film temperature Tf
Tf = (Ts + T )/2 = (77 + 23)(oC)/2 = 50oC
Properties of air at this temperature are given in Appendix C
k = thermal conductivity = 0.02781 W/m-oCPr = Prandtl number = 0.709
= kinematic viscosity = 17.92 10-6 m2/s
Substituting into (d)
LRe =10 0 2
17 92 10 6 2
( / ) . ( )
. ( / )
m s m
m s = 111,607
Since this is smaller than txRe , it follows that the flow is laminar throughout. Therefore,
Pohlhausen's solution (7.26)for the average Nusselt number is applicable.
3/12/1664.0 PrRe
k
LhNu LL (f)
where LNu is the average Nusselt number. Solving (f) for h
h3/12/1
664.0 PrReL
kL (g)
[b] The average Nusselt number for mixed flow over a flat plate of length L is given by (8.7b)
PROBLEM 8.3 (continued)
3/15/45/42/1 )()(037.0)(664.0 PrReReRek
LhNu
tt xLxL (h)
If the boundary layer is tripped at the leading edge, the flow will be turbulent throughout. Since transition is assumed to take place at x = 0, it follows that
txRe = 0 (i)
Substituting (i) into (h)
037.0k
LhNu L ( LRe )4/5 (Pr)1/3 (j)
Solving (j) for h
h = 0.037 k
L ( LRe )4/5 (Pr)1/3 (k)
(iii) Computations.
[a] Laminar flow. Equation (c) gives surface area
A = 0.2(m) 0.25(m) = 0.05 m2
Use (g) to calculate h
h 0 6640 02781
0 2111 607 0 709
1 2 1 3.
. ( / )
. ( ), .
/ /W m C
m
o
= 27.5 W/m2-oC
Substituting into (b) and using (a) gives
P = q = 27.5 (W/m2-oC) 0.05(m2) (77 - 23)(oC) = 74.25 W
[b] Turbulent flow. Use (k) to obtain h
h = 0.037 0 02781
0 2
. ( / )
. ( )
W m C
m
o
(111,607)4/5 (0.709)1/3 = 50.09 W/m2-oC
Substituting into (b) and using (a)
P = q = 50.09 (W/m2-oC) 0.05(m2) (77 - 23)(oC) = 135.2 W
(iv) Checking. Dimensional check: Computations showed that equations (b), (c), (d), (g) and (k) are dimensionally consistent.
Quantitative check: The magnitude of the heat transfer coefficient for both laminar and turbulent flow is within the range of values listed in Table 1.1. Furthermore, as expected, the average heat transfer coefficient for turbulent flow is higher than laminar flow.
(5) Comments. (i) Tripping the boundary layer increases the maximum power by 82 % without increasing surface temperature. (ii) The disadvantage of tripping the boundary layer is the corresponding increase in pressure drop.
PROBLEM 8.4
Water at 15oC flows with a velocity of 0.18 m/s over a plate of length L = 20 cm and width W =
25 cm. Surface temperature is 95oC. Determine the heat transfer rate from the leading and
trailing halves of the plate.
(1) Observations. (i) This is an external forced convection problem. (ii) The geometry is a flat plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives the heat transfer rate. (v) The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed. (vi) Analytic or correlation equations give the heat transfer coefficient. (vii) If the flow is laminar throughout, heat transfer from the first half should be greater than that from the second half. (viii) Second half heat transfer can be obtained by subtracting first half heat rate from the heat transfer from the entire plate. (ix) The fluid is water.
(2) Problem Definition. Determine the average heat transfer coefficient for the first half and for the entire plate.
(3) Solution Plan. Apply Newton’s law of cooling to the first half and to the entire plate. Check the Reynolds number at end of the first half and second half to establish if the flow is laminar, turbulent or mixed. Use analytic or correlation equations for the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy ( = 0 or g = 0 ) and (9) no radiation.
(ii) Analysis. Application of Newton’s law of cooling to the first half gives
1q = 1h A1(Ts - T ) (a)
where
1A = surface area of half plate, m2
1h = average heat transfer coefficient for the first half, W/m2-oC
1q = heat transfer rate from the first half, W
sT = surface temperature = 95oC
T = free stream temperature = 15oC
Heat transfer from the second half is given by
12 qqq T (b)
where
2q = heat transfer rate from the second half, W
Tq = heat transfer rate from entire plate, W
Heat transfer from the entire plate is given by
Tq = Th AT(Ts - T ) (c)
where
TA = surface area of entire plate of length L, m2
Th = average heat transfer coefficient for entire plate of length L, W/m2-oC
2q1q
x
2/L 2/L
T
V
0
PROBLEM 8.4 (continued)
The areas A1 and AT areA1 = WL/2 (d)
andAT = WL (e)
where
L = plate length = 20 cm = 0.2 m W = plate width = 25 cm = 0.25 m
To determine the average heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as
Rex =xV
(f)
where
Rex = Reynolds number
V = upstream velocity = 0.18 m/s x = distance from the leading edge of the plate, m
= kinematic viscosity, m2 /s
Properties of water are evaluated at the film temperature Tf defined as
Tf = T Ts
2 (g)
For the flow over a flat plate, transition Reynolds number txRe is
txRe = 5 105 (h)
The flow is laminar if Rex < txRe .Substituting into (g)
Tf = (95 + 15)(oC)/2 = 55oC
Properties of water at this temperature are given in Appendix D
k = 0.6458 W/m-oCPr = 3.27
= 0.5116 10-6 m2/s
The Reynolds number for the first half is evaluated at x = L/2 = 20 cm/2 = 10 cm = 0.1 m
ReL/2 = 184,35
s/m105116.0
)m(1.0)s/m(18.026
Therefore, the flow is laminar over the first half. The average Nusselt number for laminar flow is given by Pohlhausen’s solution. For 0.5 < Pr < 50, equation (7.26) gives
2/12/
1/312/ )(664.0
)2/(LL RePr
k
LhNu (i)
where
2/LNu = average Nusselt number for the first half
The Reynolds number for the entire plate is evaluated at x = L = 20 cm = 0.2 m
ReL = 367,70
s/m105116.0
)m(2.0)s/m(18.026
PROBLEM 8.4 (continued)
Therefore, the flow is laminar over the entire plate. The average Nusselt number for laminar is
2/11/3664.0 LT
L RePrk
LhNu (j)
(iii) Computations. Substituting into equations (d) and (e)
A1 = 0.2(m)0.25(m)/2 = 0.025 m2
andAT = 0.2(m)0.25(m) = 0.05 m2
Solving (i) for 1h
2/12/
1/31 664.0
2LRePr
L
kh = 2/13/1
o
)184,35()27.3(664.0)m(2.0
)Cm/W)(6458.0(2= 1193.9 W/m2-oC
Similarly, equation (j) gives
2/11/3 )(664.0 LT RePrL
kh = 2/13/1
o
)367,70()27.3(664.0)m(2.0
)Cm/W)(6458.0(= 844 W/m2-oC
Substituting into (a)
1q = 0.025(m2)1193.9(W/m2-oC)(95 15 )(oC) = 2388 W
Equation (c) gives
Tq = 0.05(m2)844(W/m2-oC)(95 15 )(oC) = 3376 W
Substituting into (b) gives the heat transfer rate from the second half
2q = 3376 (W) 2388(W) = 988 W
(iv) Checking. Dimensional check: Computations showed that units of equations (a), (c)-(f), (i) and (j) are dimens ionally consistent.
Quantitative check: The magnitudes of heat transfer coefficients are within the range given in Table 1.1 for forced convection of liquids.
Qualitative check: As anticipated, heat transfer from the first half is greater than that from the second half.
(5) Comments (i) For laminar flow, two half plates oriented in parallel (side by side) transfer
more heat than two placed in series. (ii) An alternate method for determining 2q is to determine
the average heat transfer coefficient for the second half. This requires integration of the local heat transfer coefficient from x = 0.1 m to x = 0.2 m.
PROBLEM 8.5
A chip measuring 5 mm 5 mm is placed flush on a flat plate 18 cm from the leading edge. The
chip is cooled by air at 17oC flowing with a velocity of 56 m/s. Determine the maximum power
that can be dissipated in the chip if its surface temperature is not to exceed 63oC. Assume no
heat loss from the back side of the chip.
(1) Observations. (i) The chip is cooled by forced convection. (ii) This problem can be modeled as a flat plate with an unheated leading section. (iii) Newton's law of cooling can be applied to determine the rate of heat transfer between the chip and the air. (iv) Check the Reynolds number to establish if the flow is laminar or turbulent.
(2) Problem Definition. Find the average heat transfer coefficient over the chip.
(3) Solution Plan. Apply Newton’s law of cooling to determine the maximum heat removed from the chip, check the Reynolds number to establish if the flow is laminar or turbulent, model the chip as flat heated surface with an insulated leading section and use appropriate Nusselt number solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) chip surface temperature is uniform, (7) constant properties, (8) all power dissipated in the chip is transferred to the air from
its surface, (9) the chip is mounted on an insulated plate, (10) no buoyancy ( = 0 or g = 0 ) and (10) no radiation.
(ii) Analysis. Applying Conservation of energy to the chip
P = q (a) where
P = power dissipated in chip, W q = heat removed from top surface, W
Application of Newton’s law of cooling to the top surface
q = h A (Ts - T ) (b) where
A = surface area of chip, m2
h = average heat transfer coefficient, W/m2-oCTs = surface temperature = 63oC
T = ambient temperature = 17oC
The surface area is A = LW (c)
where
L = chip length = 5 mm = 0.005 m W = chip width = 5 mm = 0.005 m
To determine h it is necessary to first establish if the flow over the chip is laminar or turbulent.
0x
S
xo
T
V chip
PROBLEM 8.5 (continued)
If the Reynolds number at the leading end of the chip is larger than the transition number, the flow is turbulent over the chip. Define
Rex =V x
(d)
and
txRe = 5 105 (e)
where
xRe = local Reynolds number
txRe = transition Reynolds number for flow over a flat plate
V = free stream velocity = 56 m/s x = variable, measured from the leading edge of plate, m
= kinematic viscosity, m2/s
Properties are evaluated at the film temperature Tf
Tf = (Ts + T )/2 = (63 + 17)(oC)/2 = 40oC
Properties of air at this temperature are given in Appendix C
k = thermal conductivity = 0.0271 W/m-oCPr = Prandtl number = 0.71
= kinematic viscosity = 16.96 10-6 (m2/s)
Evaluating the Reynolds number in (d) at the leading edge of the chip x = xo = 0.18 m
oxRe =56 018
16 96 10 6 2
( / ) . ( )
. ( / )
m s m
m s = 594,340
Since this is larger than txRe , it follows that the flow is turbulent over the chip. Equation (8.8)
gives the local Nusselt number for flow over a plate with an insulated leading section
Nuhx
k x xx
x0 0296
1
4 5
9 10 1 9
. ( ) (
( / )
/
/ /
Re Pr)1/3
o
(f)
where
Nux = local Nusselt number xo = distance from the leading edge of the plate to the chip = 0.18 m
However, what is needed in equation (b) is the average heat transfer coefficient. Since the chip is small compared to the distance xo, it is reasonable to assume that the average Nusselt number is approximately equal to the local value at the center of the chip, s
9/110/9o
1/35/4
)/(1
)()(0296.0
sx
PrRe
k
shNu s
s (g)
where
Nus = average Nusselt number over the chip
s = distance from leading edge of plate to center of chip, m
PROBLEM 8.5 (continued)
Reynolds number Res and distance s are given by
Res = sV
(h)
ands = xo + L/2 (i)
(iii) Computations. Substituting into (i) and (h)
s = 0.18(m) + 0.005(m)/2 = 0.1825 m
Res =56 01825
16 96 10 6 2
( / ) . ( )
. ( / )
m s m
m s = 602,594
Solving (g) for h and substituting numerical values
9/110/9o
1/35/4
)/(1
)()(0296.0
sx
PrRe
s
kh s
9/110/9
3/15/4o
])m1825.0/m18.0(1[
)71.0()594,602(0296.0
)m(1825.0
)Cm/W(0271.0
h = 268.8 W/m2-oC
Substituting into (b) and using (a) and (c)
P = q = 268.8 (W/m2-oC)0.005(m)0.005(m) ( 1763 )(oC) = 0.309 W
(iv) Checking. Dimensional check: Computations showed that equations (b)-(d) and (g)-(i) are dimensionally consistent.
Quantitative check: The magnitude of the heat transfer coefficient is within the range of values listed in Table 1.1 for forced convection of gases.
(5) Comments. The assumption that the average heat transfer coefficient over the chip is approximately equal to the local value at the center of the chip was made to avoid the need to integrate the local value, equation (f), over the surface of the chip. This approximation becomes less reasonable as the dimension of the chip in the x direction becomes large.
PROBLEM 8.6
A 1.2 m 1.2 m solar collector is mounted flush on
the roof of a house. The leading edge of the collector
is located 5 m from the leading edge of the roof.
Estimate the heat loss to the ambient air on a typical
winter day when wind speed parallel to the roof is 12
m/s and air temperature is 5oC. Outside collector
surface temperature is estimated to be 35oC.
(1) Observations. (i) Heat transfer from the collector to the air is by forced convection. (ii) This problem can be modeled as a flat plate with an unheated leading section. (iii) Newton's law of cooling can be applied to determine the rate of heat transfer between the collector and air. (iv) The heat transfer coefficient varies along the collector. (v) The Reynolds number should be computed to establish if the flow is laminar or turbulent.
(2) Problem Definition. Determine the heat transfer coefficient over the collector.
(3) Solution Plan. Apply Newton’s law of cooling to the collector, compute the Reynolds number to establish if the flow is laminar or turbulent, model the collector surface as flat heated surface with an insulated leading section and use appropriate Nusselt number solutions or correlation equations to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat surface, (6) collector surface temperature is
uniform, (7) constant properties, (8) roof is insulated, (9) no buoyancy ( = 0 or g = 0 ) and (9) no radiation.
(ii) Analysis. Since the heat transfer coefficient varies with distance along the collector, integration of Newton’s law of cooling gives the total heat transfer to the air
L
sT dxxhwTTq0
)()( (a)
where
)(xh = local heat transfer coefficient, CW/m o2
L = length of collector = 1.2 m
Tq = total heat transfer rate, W
Ts = surface temperature = 35 Co
T = ambient temperature = 5 Co
w = width of collector = 1.2 m
To determine the local heat transfer coefficient, the Reynolds number is computed to establish if the flow is laminar or turbulent. If the Reynolds number at the leading end of the collector is larger than the transition number, the flow is turbulent over the collector. Define
Rex =xV
(b)
and
txRe = 5 105 (c)
V
T
m6.5
solarcollector
PROBLEM 8.6 (continued) where
xRe = local Reynolds number
txRe = transition Reynolds number for flow over a flat plate
V = wind speed = 12 m/s
x = distance from the leading edge of the roof, m
= kinematic viscosity, /sm2
Properties are evaluated at the film temperature fT , defined as
2
TTT s
f (d)
Substituting into (d)
fT = (35 + 5)( Co )/2 = 20oC
Properties of air at this temperature are
k = 0.02564 CW/m o
Pr = 0.713
= 15.09 10-6 /sm2
Evaluating the Reynolds number in (b) at 5oxx m
LRe = 6
26109761.3
)/sm(1009.15
)m(5)m/s(12
Since the Reynolds number is greater than the transition number it follows that the flow is turbulent over the collector. Equation (8.8) gives the local Nusselt number for flow over a plate with an insulated leading section
Nuhx
k x xx
x0 0296
1
4 5
9 10 1 9
. ( ) (
( / )
/
/ /
Re Pr)1/3
o
(e)
where
Nux = local Nusselt number xo = distance from the leading edge of the roof to the collector = 5 m
Solving (e) for h
9/110/9o
1/35/4
)/(1
)()(0296.0
xx
PrRe
x
kh x (f)
However, when (f) is substituted in (a), the resulting integral can not be evaluated analytically. Thus, numerical integration is required. An approximate approach is to assume that the heat transfer coefficient over the collector is uniform equal to the local value at the center of the collector. Thus
)()( cxhxh (g)
where
cx = distance from the leading edge of the roof to the center of collector = 5.6 m
Substituting (g) into (a) and evaluating the integral
PROBLEM 8.6 (continued)
wLhTTq csT )( (h)
Evaluating (f) at cxx
9/110/9o
1/35/4
)/(1
)()(0296.0
c
cx
c
c
xx
PrRe
x
kh (i)
(iii) Computations. The Reynolds number at collector center 6.5cxx m is
xcRe = 6
26104533.4
)/sm(1009.15
)m(6.5)m/s(12
Substituting into (i)
CW/m7.32])m6.5/m5(1[
)713.0()104533.4(0296.0
)m(6.5
)CW/m(02564.0 o2
9/110/9
3/15/46o
ch
Substituting (g)
W6.1412)m(2.1)m(2.1)CW/m(7.32)C)(535( o2oTq
(iv) Checking. Dimensional check: Computations showed that equations (b), (h) and (i) are dimensionally consistent.
Quantitative check: The magnitude of the heat transfer coefficient is within the range of values listed in Table 1.1 for forced convection of gases.
(5) Comments. The assumption that the average heat transfer coefficient over the collector is approximately equal to the local value at the center of the collector was made to avoid the need to numerically evaluate the integral in equation (a). This approximation becomes less reasonable as the dimension of the collector in the x direction becomes large.
PROBLEM 8.7
Water at 20oC flows over a rectangular plate of length L =
1.8 m and width W = 0.3 m. The upstream velocity is 0.8
m/s and surface temperature is 80oC. Two orientations are
considered. In the first orientation the width W faces the
flow and in the second the length L faces the flow. Which
orientation should be selected to minimize heat loss from the
plate? Determine the heat loss ratio of the two orientations.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is over a flat plate. (iii) Surface temperature is uniform. (iv) Plate orientation is important. (v) Variation of the heat transfer coefficient along the plate affects the total heat transfer. (vi) The heat transfer coefficient for laminar flow decreases as the distance from the leading edge is increased. However, at the transition point it increases and then decreases again. (vii) Higher rate of heat transfer may be obtained if the wide side of a plate faces the flow. On the other hand, higher rate may be obtained if the long side of the plate is in line with the flow direction when transition takes place. (viii) The fluid is water.
(2) Problem Definition. Determine the average heat transfer coefficient for the flow over a rectangular plate for two orientations: [a] wide side facing the flow and [b] short side facing the flow.
(3) Solution Plan. Apply Newton's law of cooling for flow over a flat plate. Check the Reynolds number for the two orientations and select appropriate equations for the average Nusselt number to obtain the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface temperature, (7)
constant properties, (8) no buoyancy ( = 0 or g = 0 ) and (9) negligible radiation.
(ii) Analysis. Newton's law of cooling gives the total heat transfer rate
q = h A(Ts - T ) (a) where
A = plate area, m2
h = average heat transfer coefficient, W/m2-oCq = heat transfer rate, W
Ts = surface temperature = 80oC
T = ambient temperature = 20oC
Surface area is A = LW (b)
where
L = plate length = 1.8 m W = plate width = 0.3 m
To determine h it is necessary to first establish if the flow is laminar or turbulent. If the Reynolds
0
V
Tx
x
2
1
2W
1W
2L
1L
0
PROBLEM 8.7 (continued)
number at the trailing end is smaller than the transition number, the flow is laminar throughout. Define
ReL =V L
(c)
and
txRe = 5 105 (d)
where
ReL = Reynolds number at the trailing end of plate
txRe = transition Reynolds number for flow over plate
V = free stream velocity = 0.8 m/s
= kinematic viscosity, m2/s
Properties are evaluated at the film temperature Tf
Tf = (Ts + T )/2 = (80 + 20)(oC)/2 = 50oC
Properties of water at this temperature are given in Appendix D
k = thermal conductivity = 0.6405 W/m-oCPr = Prandtl number = 3.57
= kinematic viscosity = 0.5537 10-6 m2/s
Consider two orientations, 1 and 2 as shown. In orientation 1, the short side faces the flow. In orientation 2, the long side faces the flow. The Reynolds numbers corresponding to the two orientations are computed using (c)
1LRe =)s/m(105537.0
)m(8.1)s/m(8.026
= 2,600,686, turbulent
Similarly, for orientation 2
2LRe =)s/m(105537.0
)m(3.0)s/m(8.026
= 433,448, laminar
where
L1 = L = 1.8 m L2 = W = 0.3 m
Comparing1LRe and
2LRe withtxRe shows that the flow is mixed (laminar and turbulent) for
orientation 1, and laminar for orientation 2. The average Nusselt number for mixed flow over a flat plate of length L1 is given by (8.7b)
3/15/45/4
1
2/111
1037.0664.0 PrReReRe
k
LhNu
tt xLxL (e)
Valid for flat plate, constant Ts
5 105 < Rex < 107
0.6 < Pr < 60 properties at Tf (f)
Since all conditions in (f) are satisfied, equation (e) is applicable to this case. For orientation 2, Pohlhausen's solution (7.26) gives the average Nusselt number
PROBLEM 8.7 (continued)
3/12/1222 )(664.0
2PrRe
k
LhNu LL (g)
(iii) Computations. Solving (E) for 1h and substituting numerical values
3/15/45/42/1
1
1 )()(037.0)(664.01
PrReReReL
kh
tt xLx
1h 3/15/45/42/1o
)57.3()000,500()686,600,2(037.0)000,500(664.0)m(8.1
)Cm/W(6405.0
1h = 2253.5 W/m2-oC
Similarly, solving (g) for 2h and substituting numerical values
2h = 3/12/1
2
)(664.02
PrReL
kL
2h = 0.664 3/12/1o
)57.3()448,433()m(3.0
)Cm/W(6405.0= 1426.5W/m2-oC
Substituting into equation (a) and using (b) gives the heat transfer rate for each orientation
1q = 2253.5(W/m2-oC)1.8(m)0.3(m)( 2080 )(oC) = 73,013 W
and
2q = 1426.5(W/m2-oC)1.8(m)0.3(m)( 2080 )(oC) = 46,219 W
The ratio of the two heat transfer rates is
)W(219,46
)W(013,73
2
1
q
q= 1.58
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c), (e) and (g) are dimensionally consistent.
Quantitative check: Values of the heat transfer coefficients are within the range listed in Table 1.1 for forced convection of liquids.
Limiting check: For the special case of a square plate L1 = L2, the ratio of the two heat transfer
rates should be unity. Either equation (e) or (g) gives 1h = 2h . Equation (a) gives 1q = 2q .
(5) Comments. If the flow is laminar for both orientations, orientation 1 will have a lower heat transfer rate than orientation 2. However, if transition takes place in one or both orientations, numerical calculations must be carried out to determine which orientation has the higher heat transfer rate.
PROBLEM 8.8
100 flat chips are placed on a 10 cm 10 cm circuit board and cooled by forced convection of
air at 27oC. Each chip measures 1 cm 1 cm and dissipates 0.13 W. The maximum allowable
chip temperature is 83oC. Free stream air velocity is 5 m/s. Tests
showed that several chips near the trailing end of the board
exceeded the allowable temperature. Would you recommend
tripping the boundary layer to turbulent flow at the leading edge to solve the overheating problem? Substantiate your recommendation.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is over a flat plate. (iii) The problem can be modeled as flow over a flat plate with uniform surface heat flux. (iv) Surface temperature varies with distance along plate. The highest surface temperature is at the trailing end. (v) Tripping the boundary layer at the leading edge changes the flow from laminar to turbulent. This increases the heat transfer coefficient and lowers surface temperature. (vi) Newton’s law of cooling gives surface temperature.
(2) Problem Definition. Determine the local heat transfer coefficient at the trailing end for turbulent flow.
(3) Solution Plan. Apply Newton's law of cooling at the trailing end and use a correlation equation for turbulent flow over a flat plate at constant surface heat flux to determine the local heat transfer coefficient. Solve Newton's law for the surface temperature.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) flat plate, (6) uniform surface heat flux, (7)
constant properties, (8) no buoyancy ( = 0 or g = 0 ), (9) negligible radiation and (10) turbulent flow.
(ii) Analysis. Apply Newton's law of cooling
qs = h(x) [Ts(x) - T ] (a)
where
h(x) = local heat transfer coefficient, W/m2-oCqs = surface heat flux = 0.13 W/cm2 = 1300 W/m2
Ts(x) = local surface temperature, oC
T = free stream temperature = 27oC
x = distance from the leading edge, m
Solving equation (a) for Ts(x)
)()(
xh
qTxT s
s (b)
The local Nusselt number for turbulent flow over a flat plate at constant surface heat flux is given by equation (8.9)
Nuhx
kx x0 030 4 5 1 3. / /Re Pr (c)
V
T
x0
L
V
T
sq
)(xTs
PROBLEM 8.8 (continued)
where
k = thermal conductivity, W/m-oCNux = local Nusselt number
Rex = local Reynolds number = /xV
Pr = Prandtl number
V = free stream velocity = 5 m/s
= kinematic viscosity, m2/s
Equation (c) gives the local heat transfer coefficient. Properties are evaluated at the film temperature defined as
2/)( TTT sf (d)
where
fT = film temperature, oC
sT = average surface temperature, oC
Since surface temperature varies along the plate, an average value, sT , is used to determine film
temperature. sT is approximated by
2/)]([ LTTT ss (e)
(iii) Computations. To determine air properties, surface temperature at the trailing end, Ts(L), is needed to compute Tf. However, Ts(L) is unknown. In fact, the objective of the problem is determining Ts(L). To proceed, a solution is obtained by trial and error procedure. A value for Ts(L)is assumed, (d) and (e) are used to determine Tf, (c) is used to calculate h(L) and (b) is used to calculated Ts(L). If the calculated Ts(L) is not equal to the assumed value, the process is repeated until a satisfactory agreement is obtained between assumed and calculated values.
Assume Ts(L)= 79oC. Equations (e) and (d) give
sT = (27 + 79)(oC)/2 = 53oC
andTf = (53 + 27)(oC)/2 = 40oC
Air properties at this temperature are
k = 0.0271 W/m-oCPr = 0.71
= 16.96 10-6 m2/s
Thus, the Reynolds number at x = L = 0.1 m is
ReL =LV
481,29)s/m(1096.16
)m(1.0)s/m(526
Solving (c) for h and evaluating the resulting equation at x = L = 0.1 m
3/15/4 )()(03.0)( PrReL
kLh L
3/15/4o
)71.0()481,29()m(1.0
)Cm/W(0271.003.0 = 27.3 W/m2-oC
Substituting into (b)
PROBLEM 8.8 (continued)
Ts(L) = 27oC + )Cm/W(3.27
)m/W(1300o2
2
= 74.6oC
This is reasonably close to the assumed value of 79oC. Repeating the calculation with a new assumed value of 75oC will result in a minor change in the result.
(iv) Checking. Dimensional check: Computations showed that units for equations (b) and (c) are consistent.
Quantitative check: the value of h(L) is within the range given in Table 1.1 for forced convection of gases.
(5) Comments. (i) By tripping the boundary layer to cause transition to turbulent flow, surface temperature at the trailing end will not exceed the maximum allowable level. (ii) Since the
Reynolds number at the trailing end is less than the transition value of 5 105, the flow will be laminar if it is not tripped. To determine surface temperature under laminar flow conditions, the corresponding heat transfer coefficient must be computed. The Nusselt number for laminar flow over a flat plate with uniform surface flux is given by equation (7.31)
2/13/1453.0 xx RePrk
hxNu (f)
Solving this equation for h(L)
2/13/1 )()(453.0)( LRePrL
kLh 2/13/1
o
)481,29()71.0()m(1.0
)Cm/W(0271.0453.0 = 18.8 W/m2-oC
Substituting into (b)
Ts(L) = 27oC + )Cm/W(8.18
)m/W(1300o2
2
= 96.1oC
Since this is considerably higher than the assumed value of 79oC, the procedure is repeated with a new assumed value of Ts(L) = 99oC. This yields a calculated value of Ts(L) = 96.2oC. Thus for laminar flow, surface temperature exceeds the allowable maximum level of 83oC.
PROBLEM 8.9
Water at 27oC flows normally over a tube with a velocity of 4.5 m/s. The outside diameter of the
tube is 2 cm. Condensation of steam inside the tube results in a uniform outside surface
temperature of 98oC. Determine the length of tube needed to transfer 250,000 W of energy to the
water.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is normal to a tube. (iii) Surface temperature is uniform. (iv) Tube length is unknown. (v) Newton’s law of cooling can be used to determine surface area. Tube length is related to surface area. (vi) The fluid is water.
(2) Problem Definition. The required tube length can be determined from Newton's law of cooling. Thus, the problem is finding the average heat transfer coefficient for flow normal to a cylinder.
(3) Solution Plan. Apply Newton's law of cooling to the flow over tube. Use forced convection correlation equation to determine the average Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) uniform surface heat flux, (6) constant
properties, (7) no buoyancy ( = 0 or g = 0 ) and (8) no radiation.
(ii) Analysis. Newton's law of cooling gives
q = h A(Ts T ) (a)
where
A = plate area, m2
h = average heat transfer coefficient, W/m2-oC
q = heat transfer rate = 250,000 W
Ts = surface temperature = 98oC
T = free stream temperature = 27oC
Surface area is A = DL (b)
where
D = outside tube diameter = 2 cm = 0.02 mL = tube length, m
Substituting (b) into (a) and solving the resulting equation for L
)( TThD
qL
s
(c)
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder
5/48/5
4/13/2
3/12/1
000,282/1
/41
62.03.0 D
DD Re
Pr
PrRe
k
DhuN (d)
sTD
L
V T TV
PROBLEM 8.9 (continued)
Valid for: flow normal to cylinder Pe = Re PrD > 0.2
properties at Tf (e)where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pe = Peclet number = DPrRe
Pr = Prandtl number
DRe = Reynolds number
fT = film temperature, oC
Reynolds number is defined as
DVReD (f )
where
V = free stream velocity = 4.5 m/s
= kinematic viscosity, m2/s
Properties are determined at the film temperature fT defined as
Tf = 2/)( TTs (g)
The Reynolds number is computed to establish if (e) is satisfied. Substituting into (g)
Tf = (98 + 27)(oC)/2 = 62.5oC
Properties of water at this temperature are
k = thermal conductivity = 0.653 W/m-oCPr = Prandtl number = 2.885
= kinematic viscosity = 0.4586 10-6 m2/s
Thus
DRe)s/m(104586.0
)m(02.0)s/m(5.426
= 196,249
and
Pe = PrReD = 196,249 (2.885) = 5.66 105
Therefore, (e) is satisfied and correlation equation (d) is applicable.
(iii) Computations. Equation (d) gives the average heat transfer coefficient
5/48/5
4/13/2
1/32/1
000,282/249,1961
885.2/4.01
(2.885))249,196(62.03.0
k
DhuN D = 589
Solving for h
PROBLEM 8.9 (continued)
h =D
k589 =
)m(02.0
)Cm/W(653.0589
o
= 19,231 W/m2-oC
Substituting into (c) gives the required length
L =)C)(2798)(CW/m(231,19)m)(02.0(
)W(000,250oo2
= 2.91 m
(iv) Checking. Dimensional check: Computations showed that units for equations (c), (d) and (f) are dimensionally consistent.
Quantitative check: The value of h is within the range given in Table 1.1 for forced convection of liquids.
Limiting check: An infinitely long tube is needed if TTs . Setting TTs in equation (c)
gives L = .
(5) Comments. (i) The required length to transfer 250 kw of heat is only 2.43 m. This appears unreasonably short. A review of the analysis and calculations uncovered no errors. The relatively short length needed is due to the high heat transfer coefficient associated with forced convection of water. Note that the calculated heat transfer coefficient is at the high end of values given in Table 1.1. (ii) It was not necessary to consider the thermal interaction between the surface and the fluid inside the tube because outside surface temperature was specified.
PROBLEM 8.10
A proposed steam condenser design for marine applications is based on the concept of rejecting
heat to the surrounding water while a boat is in motion. The idea is to submerge a steam-
carrying tube in the water such that its axis is normal to boat velocity. Estimate the rate of
steam condensation for a 75 cm long tube with an outside diameter of 2.5 cm. Assume a
condensation temperature of 90oC and a uniform surface temperature of 88
oC. Ambient water
temperature is 15oC and boat speed is 8 m/s.
(1) Observations. (i) Heat is removed by the water from the steam causing it to condense. (ii) The rate at which steam condenses inside the tube depends on the rate at which heat is removed from the outside surface. (iii) Heat is removed from the outside surface by forced convection. (iv) This is an external forced convection problem of flow normal to a tube. (v) Newton’s law of cooling gives the rate of heat loss from the surface.
(2) Problem Definition. Determine the rate of heat transfer from the outside surface to the ambient water.
(3) Solution Plan. Apply conservation of energy to the condensing steam inside the tube. Use Newton’s law of cooling to determine the heat removed from the tube. Use correlation equation for forced convection flow normal to a tube to determine the average heat transfer coefficient (Nusselt number).
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) uniform surface temperature, (6) constant properties, (7) no radiation, (8) negligible changes in kinetic and potential energy of steam, (9)
negligible axial conduction in tube and steam, constant properties, (10) no buoyancy ( = 0 or g= 0 ) and (11) no radiation.
. (ii) Analysis. Applying conservation of energy to the steam between the inlet and outlet of the tube gives
q = m h hi o = fghm ˆ (a)
or, solving (a) for m
m =
fgh
q
ˆ (b)
where
hi = steam enthalpy at inlet of tube, J/kg
ho = steam enthalpy at outlet of tube, J/kg
fgh = latent heat of condensation = 2283.2 kJ/kg (at 90 oC)
m = rate of steam condensation, kg/s q = rate of heat removed from steam, W
The latent heat, fgh , is defined by the temperature of the condensing steam. Thus to determine
steam condensation rate from (b), the rate of heat removal from steam must be obtained. Applying conservation of energy to the tube gives
PROBLEM 8.10 (continued)
q = Energy removed from steam and added to tube
= Energy removed from tube and added to water (c)
Newton’s law of cooling gives the energy removed from tube surface by convection and added to water
q = h DL (Ts - T ) (d) where
D = diameter of tube = 2.5 cm = 0.025 m
h = average heat transfer coefficient at the outer surface of tube, W/m2-oCL = length of tube = 75 cm = 0.75 m Ts = surface temperature = 88oC
T = ambient temperature = 15oC
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder
5/48/5
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000,282/1
/4.01
62.03.0 D
DD Re
Pr
PrRe
k
DhuN (e)
Valid for: flow normal to cylinder
Pe = Re PrD > 0.2
properties at Tf (f)where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pe = Peclet number = DPrRe
Pr = Prandtl number
DRe = Reynolds number
fT = film temperature, oC
Reynolds number is defined as
DVReD (g)
where
V = free stream velocity = 8 m/s
= kinematic viscosity, m2/s
Properties are determined at the film temperature fT defined as
Tf = 2/)( TTs (h)
The Reynolds number is computed to establish if (f) is satisfied. Substituting into (h)
Tf = (88 + 15)(oC)/2 = 51.5oC
Properties of water at this temperature are
k = thermal conductivity = 0.6421 W/m-oC
T
VsT
PROBLEM 8.10 (continued)
Pr = Prandtl number = 3.48
= kinematic viscosity = 0.5411 10-6 m2/s
Thus
DRe)s/m(105411.0
)m(025.0)s/m(826
= 369,617
and
Pe = PrReD = 369,617(3.48) = 1.286 106
Therefore, (f) is satisfied and correlation equation (e) is applicable.
(iii) Computations. Equation (e) gives the average Nusselt number and average heat transfer coefficient
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000,282/617,3691
48.3/4.01
(3.48))617,369(62.03.0
k
DhuN D = 1012.3
Solving for h
h =D
k3.1012 =
)m(025.0
)Cm/W(6421.03.1012
o
= 26,000 W/m2-oC
Substituting into (d)
q = 26,000(W/m2-oC) 0.025(m) 0.75(m) (88 - 15)(oC) = 111,801 W = 111.8 kW
Latent heat of condensation at 90oC is
fgh = 2283.2 kJ/kg
Substituting into (b) gives the condensation rate m
m = )kg/kJ(2.2283
)kW(8.111 = 0.04897
)kg/kJ(
)s/kJ(= 0.04897 kg/s
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and (g) are dimensionally consistent.
Quantitative check: The value of h is outside the range given in Table 1.1 for forced convection of liquids. A review of the analysis and calculations uncovered no errors. It should be kept in mind that Table 1.1 gives rough estimates of h for typical applications. Exceptions are expected.
Qualitative check: Increasing the tube’s length and/or boat speed, increases condensation rate. Equation (b) shows that condensation rate is directly proportional to q. According to equation
(d), q increases as L is increased. Similarly, according to (e), an increase in V results in an
increase in h which in turn increases q.
(5) Comments. Although condensation rate may be adequate when the boat is in motion, it decreases when the boat is stationary.
PROBLEM 8.11
An inventive student wanted to verify the speed of a boat
using heat transfer analysis. She used a 10 cm long
electrically heated tube with inside and outside diameters
of 1.1 cm and 1.2 cm, respectively. She immersed the tube
in the water such that its axis is normal to boat velocity.
She recorded the following measurements:
Water temperature = 16.5oC
Outside surface temperature of tube = 23.5oC
Electric energy dissipated in tube = 480 W
Determine the speed of the boat.
(1) Observations. (i) Electric power is dissipated into heat and is removed by the water. (ii) This velocity measuring concept is based on the fact that forced convection heat transfer is affected by fluid velocity. (iii) velocity affects the heat transfer coefficient which in term affects surface temperature. (iv) Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature. (v) This problem can be modeled as external flow normal to a cylinder. (vi) The fluid is water.
(2) Problem Definition. Formulate a relationship between fluid velocity, heat transfer rate and surface temperature for flow normal to a cylinder.
(3) Solution Plan. Apply conservation of energy and Newton's law of cooling to the tube. Use a correlation equation to relate heat transfer coefficient to fluid velocity.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) no axial conduction through tube, (5) constant boat velocity, (6) uniform surface flux, (7) uniform surface
temperature (8) uniform water temperature, (9) no buoyancy ( = 0 or g = 0) and (10) negligible radiation.
(ii) Analysis. Conservation of energy applied to the tube gives
qP (a)
where
P = electric power supplied to tube = 480 W q = heat transfer rate from tube surface to water, W
Application of Newton’s law of cooling to the tube and gives
)( TTLDhq so (b)
where
oD = outside tube diameter = 1.2 cm = 0.012 m
h = average heat transfer coefficient, CW/m o2
L = tube length = 10 cm = 0.1 m
sT = surface temperature = 23.5 Co
T = water temperature = 16.5 Co
+
-T
sT
V
PROBLEM 8.11 (continued)
Substituting (a) into (b)
)( TTLDhP so
Since h is expected to depend on velocity, equation (a) is solved for h
)( TTLD
Ph
so
(c)
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder
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000,282/1/4.01
62.03.0 D
DoD Re
Pr
PrRe
k
DhuN (d)
Valid for: flow normal to cylinder Pe = Re PrD > 0.2
properties at Tf
where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pe = Peclet number = DPrRe
Pr = Prandtl number
DRe = Reynolds number
fT = film temperature, oC
Reynolds number is defined as
oD
DVRe (e)
where
V = boat speed, m/s
= kinematic viscosity, m2/s
Substituting (e) into (d) and solving for h
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000,282/)(1/4.01
)(62.03.0 /DV
Pr
Pr/DV
D
kh o
o
o
(f)
Properties of water are determined at the film temperature fT defined as
2
TTT s
f (g)
Equation (f) gives a relationship between h and the boat speed V . Substituting (f) into (c) gives
the desired relationship between P , V and Ts. However, the resulting equation cannot be solved
explicitly for V . The solution is obtained by trial and error. Equation (c) is used to calculate h ,
a value for V is selected and substituted into (f) to calculate h . If the calculated h using (f) is
PROBLEM 8.11 (continued)
not the same as that obtained from (c), the procedure is repeated until a satisfactory agreement is obtained between the two values.
(iii) Computations. Equation (c) is used to calculate h
CW/m189,18)C)(5.165.23()m(1.0)m(012.0
)W(480 o2
ch
Equation (f) is used to calculate fT
Tf= (23.5 + 16.5)( Co )/2 = 20 Co
Properties of water at this temperature are:
k = 0.5996 CW/m o
Pr = 6.99
= 1.004 10-6 /sm2
Assume V = 10 m/s. Substituting into (f)
h)m(012.0
)CW/m(5996.03.0
o
)m(012.0
)CW/m(5996.0
000,282)/sm(10004.1
)m(012.0)m/s(101
99.6/4.01
)99.6()/sm(10004.1
)m(012.0)m/s(1062.0
o5/4
8/5
264/13/2
3/1
1/2
26
h = 28,606 CW/m o2
Since this is larger than h = 18,189 CW/m o2 obtained from (c), the procedure is repeated
using a lower value for V . Assume 5V m/s and substituting into (f) gives h = 18,089
CW/m o2 . This agrees within 0.5% of the value obtained from (c). Thus the speed of the boat is
5 m/s.
(iv) Checking. Dimensional check: Computations showed that equations (c) and (f) are dimensionally consistent.
Quantitative check: The calculated value of h = 18,189 CW/m o2 is within the range given in
Table 1.1 for forced convection in liquids.
(5) Comments. (i) The velocity measuring method suggested by the student is indeed sound. It is based on the observation that fluid velocity affects heat transfer coefficient. Thus heat transfer coefficient may be used as a measure of velocity. (ii) Since correlation equations for heat transfer coefficients are not exact, a velocity measuring instrument which is based on this concept must be calibrated to obtain accurate velocity measurements.
PROBLEM 8.12
A thin electric heater is wrapped around a rod of diameter 3 cm. The heater dissipates energy
uniformly at a rate of 1300 W/m. Air at 20oC flows normal to the rod with a velocity of 15.6 m/s.
Determine the steady state surface temperature of the heater.
(1) Observations. (i) This is an external forced convection problem. (ii) The flow is normal to a rod. (iii) Surface heat transfer rate per unit length is known. However, surface temperature is unknown. (iv) In general, surface temperature varies along the circumference. However, the rod can be assumed to have a uniform surface temperature. (v) This problem can be modeled as forced convection normal to a rod with uniform surface flux or temperature. (vi) Newton’s law of cooling gives surface temperature. (vii) The fluid is air.
(2) Problem Definition. Surface temperature can be determined from Newton's law of cooling if the heat transfer coefficient is known. Thus, the problem is finding the average heat transfer coefficient for flow normal to a cylinder.
(3) Solution Plan. Apply Newton's law of cooling to the flow over a rod. Use forced convection correlation equation to determine the average Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) two-dimensional flow, (4) uniform upstream velocity and temperature, (5) uniform surface flux, (6) constant properties, (7)
all energy dissipated in electric heater leaves surface (no axial conduction), (8) no buoyancy ( = 0 or g = 0) and (9) negligible radiation.
(ii) Analysis. Newton's law of cooling gives
q = h DL ( sT T ) (a)
where
D = diameter of rod = 3 cm = 0.03 m
h = average heat transfer coefficient, W/m2-oC
L = tube length, m q = heat transfer rate, W
sT = average surface temperature, oC
T = free stream temperature = 20oC
Solving (a) for sT
hD
LqTTs
/ (b)
where
q/L = energy dissipated per unit length = 1300 W/m
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder at uniform surface temperature or surface flux
D
L
V T TV
sT_
PROBLEM 8.12 (continued)
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62.03.0 D
DD Re
Pr
PrRe
k
DhuN (c)
Valid for: flow normal to cylinder Pe = Re PrD > 0.2
properties at Tf (d)where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pe = Peclet number = DPrRe
Pr = Prandtl number
DRe = Reynolds number
fT = film temperature, oC
Reynolds number is defined as
DVReD (e)
where
V = free stream velocity = 15.6 m/s
= kinematic viscosity, m2/s
Properties are determined at the film temperature fT defined as
Tf = 2/)( TTs (f)
Since surface temperature is unknown, a trial and error procedure is needed to solve the problem.
A value for sT is assumed, (f) is used to determine Tf and (b) is used to calculate sT . If the
calculated sT is not equal to the assumed value, the process is repeated until a satisfactory
agreement is obtained between assumed and calculated sT .
Assume sT = 100oC. Equation (f) gives
Tf = (100 + 20)(oC)/2 = 60oC
At this temperature air properties are
k = thermal conductivity = 0.02852 W/m-oCPr = Prandtl number = 0.708
= kinematic viscosity = 18.9 10-6 m2/s
The Reynolds number is computed to establish if (d) is satisfied. Substituting into (e)
DRe)s/m(109.18
)m(03.0)s/m(6.1526
= 24,762
and
PROBLEM 8.12 (continued)
Pe = PrReD = 24,762(0.708) = 1.753 104
Therefore, (d) is satisfied and correlation equation (c) is applicable.
(iii) Computations. Equation (c) gives the average Nusselt number and average heat transfer coefficient
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000,282/762,241708.0/4.01
(0.708))762,24(62.03.0
k
DhuN D = 89.42
Solving for h
h =D
k42.89 =
)m(03.0
)Cm/W(02852.042.89
o
= 85 W/m2-oC
Substituting into (b) gives surface temperature
sT = 20oC +)Cm/W(85)m)(03.0(
)m/W(300,1o2
= 182.3oC
This value is considerably higher than the assumed one of 100oC. The procedure is repeated with
another assumed temperature. Assume sT = 180oC. This gives Tf = 100oC, h = 83.2 W/m2- oC
and a calculated surface temperature sT = 185.8oC. This is close to the assumed value. Thus the
resulting surface temperature is 185.8oC.
(iv) Checking. Dimensional check: Computations showed that units for equations (b), (c) and (e) are dimensionally consistent.
Quantitative check: The value of h is within the range given in Table 1.1 for forced convection of gases.
Limiting check: If the heater is turned off, surface temperature should be the same as free stream
temperature. Setting q = 0 in (b) gives sT T .
(5) Comments. (i) The trial and error procedure converges rapidly towards a satisfactory solution. This is because heat transfer coefficient is not very sensitive to the temperature at which properties are determined. Changing film temperature from 60oC to 100oC in the above example
resulted in a 2% change in h . (ii) The assumption that radiation loss is negligible should be examined in view of the high surface temperature. Since surface emissivity is unknown one can only make a rough approximation of radiation. Using Stefan-Boltzmann law, assuming that the rod is a small surface enclosed by a much larger surface and assuming an emissivity of 1.0, radiation loss is found to be 296 W/m. However, if the surface emissivity is 0.1, then the radiation loss is 2.2%, which is negligible.
PROBLEM 8.13
A fluid velocity measuring instrument consists of a wire which is heated electrically. By
positioning the axis of the wire normal to flow direction and measuring surface temperature and
dissipated electric power, fluid velocity can be estimated. Determine the velocity of air at 25oC
for a wire diameter of 0.5 mm, dissipated power 35 W/m and surface temperature 40oC.
(1) Observations. (i) Electric power is dissipated into heat and is removed by the fluid. (ii) This velocity measuring instrument is based on the fact that forced convection heat transfer is affected by fluid velocity. (iii) velocity affects the heat transfer coefficient which in term affects surface temperature and heat flux. (iv) Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature. (v) This problem can be modeled as external flow normal to a cylinder. (vi) The fluid is air.
(2) Problem Definition. Formulate a relationship between fluid velocity, heat transfer and surface temperature for flow normal to a cylinder.
(3) Solution Plan. Apply Newton's law of cooling to the wire. Use a correlation equation to relate heat transfer coefficient to the fluid velocity.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) no axial conduction through wire, (5) uniform upstream velocity and temperature, (6) constant properties,
(7) uniform surface flux and surface temperature, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation.
(ii) Analysis. Application of Newton’s law of cooling to the wire gives
q = h DL (Ts - T ) (a) where
D = wire diameter = 0.5 mm = 0.0005 m
h = average heat transfer coefficient, W/m2-oCL = wire length, m Ts = surface temperature = 40oC
T = free stream temperature = 25oC
Since h is expected to depend on velocity, equation (a) is
solved for h
h =)(
/
TTD
Lq
s
(b)
whereq
L = power or heat dissipated in wire per unit length = 35 W/m
Equation (8.10a) gives a correlation for the average Nusselt number for forced convection normal to a cylinder at uniform surface temperature or surface flux
_
V
TLq /
sT
PROBLEM 8.13 (continued)
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000,282/1
/4.01
62.03.0 D
DD Re
Pr
PrRe
k
DhuN (c)
Valid for: flow normal to cylinder Pe = Re PrD > 0.2
properties at Tf (d)where
k = thermal conductivity, W/m-oC
NuD = average Nusselt number
Pe = Peclet number = DPrRe
Pr = Prandtl number
DRe = Reynolds number
fT = film temperature, oC
The Reynolds number is defined as
ReD = V D
(e)
where
V = free stream velocity, m/s
= kinematic viscosity = m2/s
Properties of air are evaluated at the film temperature Tf defined as
Tf = (Ts + T )/2 (f)
The objective is to express h in terms of free stream velocity V , substitute into (b) and obtain
an equation relating q/L, V and Ts. Substituting (e) into (c) and solving the resulting equation
for h gives
D
kDV
Pr
PrDV
D
kh
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3/11/2
000,282
/1
/4.01
/62.03.0 (g)
Equation (g) gives a relationship between h and the free stream velocity V . Substituting this
result into (b) gives the desired relationship between q/L, V and Ts. However, the resulting
equation cannot be solved explicitly for V . The solution is obtained by trial and error. Equation
(b) is used to calculate h , a value for V is selected and substituted into (g) to calculate h . If the
calculated h using (g) is not the same as that obtained from (b), the procedure is repeated until a satisfactory agreement is obtained between the two values.
(iii) Computations. Equation (f) is used to calculate Tf
Tf= (40 + 25)(oC)/2 = 32.5oC
Properties of air at this temperature are given in Appendix C
k = 0.02656 W/m-oCPr = 0.7115
PROBLEM 8.13 (continued)
= 16.2475 10-6 m2/s
Substituting into (b)
h =)C)(2540)(m(0005.0
)m/W(35o
= 1485.4 W/m2-oC
Assume V = 50 m/s. Substituting into (g)
h)m(0005.0
)Cm/W(02656.03.0
o
)(0005.0
)/(02656.0
000,282)/(1024.16
)(0005.0)/(501
7115./4.01
)7115.0()/(1024.16
)(0005.0)/(5062.0 5/4
8/5
264/13/2
3/1
1/2
26
m
CmW
sm
msm
0
sm
msm
o
h = 1060.1 W/m2-oC
Since this is less than h = 1485.4 W/m2-oC, the procedure is repeated using a higher value for
V . The results of five trials are tabulated below.
Assumed V Calculated h m/s W/m2-oC
50 1059.9
70 1259.7
90 1434.9
100 1515.6
96 1484.2
The result shows that the free stream velocity is V = 96 m/s. With V determined, it remains to verify that condition (d) is satisfied by calculating the Peclet number.
Pe = ReD Pr = PrDV )/( = 7115.0)]s/m(102475.16/)m(0005.0)s/m(96[ 26 = 2102
Therefore, condition (d) is satisfied.
(iv) Checking. Dimensional check: Computations showed that equations (b), (e) and (g) are dimensionally consistent.
Qualitative check: If q/L is held constant and V is increased, surface temperature should
decrease. According to (g), increasing V increases h . An increase in h results in a decrease in Ts, as indicated by equation (a).
Quantitative check: The calculated value of h = 1484.2 W/m2-oC is outside the range suggested in Table 1.1. This is due to the fact that the diameter of the wire is very small (0.0005 m). This is not among the typical application considered in Table 1.1.
(5) Comments. This velocity measuring instrument is based on the observation that fluid velocity affects heat transfer coefficient. Since correlation equations are not exact, it is necessary to calibrate such an instrument to obtain accurate velocity measurements.
PROBLEM 8.14
Students were asked to devise unusual methods for determining the height of a building. One
student designed and tested the following system. A thin walled copper balloon was heated to
133oC and parachuted from the roof of the building. Based on aerodynamic consideration, the
student reasoned that the balloon dropped at approximately constant speed. The following
measurements were made:
D = balloon diameter = 13 cm
M = mass of balloon = 150 grams
oT = balloon temperature at landing = 47oC
T = ambient air temperature = 20oC
U = balloon velocity = 4.8 m/s
Determine the height of the building.
(1) Observations. (i) The sphere cools off as it drops. Heat loss from the sphere is by forced convection. (ii) The height of the building can be determined if the time it takes the sphere to land is known. (iii) Time to land is the same as cooling time. (iv) Transient conduction determines cooling time. (v) If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature. (vi) Cooling rate depends on the heat transfer coefficient.
(2) Problem Definition. Determine transient temperature of sphere. This requires determining the drop time and the heat transfer coefficient.
(3) Solution Plan. Apply Newton’s law of motion to the falling sphere and use lumped capacity method to determine drop time. Use correlation equation to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) Constant heat transfer coefficient, (3) Biot number < 0.1 (to be verified ), (4) constant properties, (5) constant sphere velocity, (6) uniform ambient conditions, (7) negligible wind speed, (8) no radiation and (9) no heat loss to the sphere interior.
(ii) Analysis. Application of Newton’s law of motion to the falling sphere gives
otUH (a)
where
H = building height, m
ot = drop time, s
U = balloon velocity, m/s
The dorp time is determined from transient temperature analysis. For Bi << 1, the lumped capacity model gives the transient temperature solution for the sphere. Equation (5.7) gives
)()( TTTtT i ])/(exp[ tVcAh ps (b)
where
sA = surface area, m2
PROBLEM 8.14 (continued)
pc = specific heat of copper = 385 CJ/kg o
h = average heat transfer coefficient, CW/m o2
t = time, s
)(tT = temperature variable, Co
T = ambient temperature = 20oC
iT = initial temperature of sphere = 133 Co
V = volume, m3
= density of copper = 8933 kg/m3
The product of V in equation (b) is equal to the mass of the sphere. That is
V = M (c) where
M = mass of sphere = 150 g = 0.15 kg
Surface area of sphere is 2DAs (d)
where
D = diameter = 13 cm = 0.13 m
Substituting (c) and (d) into (b) and solving the resulting equation for t
TtT
TT
hD
Mct ip
)(ln
2 (e)
Applying this result at the drop time, ott
TT
TT
hD
Mct
o
ip
o ln2
(f)
where
oT = sphere temperature at landing = 47 Co
Thus the only unknown in (f) is the average heat transfer coefficient. Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a sphere
Nuh D
kD D D
s2 0 4 0 061 2 2 3 0 4
1 4
. ./ / ./
Re Re Pr (g)
subject to the following limitations:
3.5 < ReD < 7.6 104
0.71 < Pr < 380
1.0 < ( / s) <3.2
properties at T , s at Ts where
k = thermal conductivity of air = 0.02564 W/m-oC
NuD = average Nusselt number
Pr = Prandtl number of air = 0.713
DRe = Reynolds number
PROBLEM 8.14 (continued)
= viscosity of air at the free stream temperature = 18.17 10-6 kg/m-s
s = viscosity of air at surface temperature, kg/m-s
The Reynolds number is defined as
UDReD (h)
where
= kinematic viscosity of air = 15.09 10-6 m2/s
Since surface temperature changes with time, s should be evaluated at the average surface
temperature Ts defined as
2
ois
TTT (i)
(iii) Computations. The average heat transfer coefficient h is computed using (g). Substituting numerical values into (i)
C902
)C)(47133( oo
sT
The viscosity at this temperature is
61035.21s skg/m
The Reynolds number is
352,41)/sm(1009.15
m)(13.0)m/s(8.426DRe
Substituting into (g)
4/1
6
64.03/22/1
)skg/m(1035.21
)skg/m(1017.18)713.0()352,41(06.0)352,41(4.02
k
DhNu D
k
DhNu D 130.4
Solving for h
m)(13.0
)CW/m(025644.1304.130
o
D
kh 25.7 CW/m o2
Substituting into (f)
6.60)2047(
)20133(ln
)C-m/W(7.25)(m)13.0(
kg)(15.0)CJ/kg(385o222
o
ot s
Substituting into (a) gives the building height
9.290)s(6.60)m/s(8.4H m
The Biot number can now be computed to establish the validity of the lumped capacity method. It is defined as
PROBLEM 8.14 (continued)
ck
hBi (j)
where
ck = thermal conductivity of copper = 397 CW/m o (at 90 Co )
= wall thickness of sphere, m
c = density of copper = 8933 3kg/m
Sphere thickness is determined from its mass, density and volume
m10162.3)m()13.0()kg/m(8933
)kg(15.0 4
2232D
M
c
Substituting into (j)
5
o
4o2
10047.2)CW/m(397
m)(10162.3)CW/m(7.25Bi
Since the Biot number is much smaller than unity it follows that the lumped capacity method is applicable.
(iv) Checking. Dimensional check: Computations showed that equations (a), (f), (g), (h) and (j) are dimensionally consistent.
Limiting check: If building height is infinite, the final temperature should be the same as ambient
temperature. Setting TTo in (f) gives ot . When this is substituted into (a) gives H .
Quantitative check: The value of h is within the range listed in Table 1.1 for forced convection of gases.
(5) Comments. Neglecting wind speed is a key assumption in the method used. Wind speed can introduce significant error in estimating the height of the building.
PROBLEM 8.15
A 6 cm diameter sphere is used to study skin friction characteristics at elevated temperatures.
The sphere is heated internally with an electric heater and placed in a wind tunnel. To obtain a nearly uniform surface temperature the sphere is made of copper. Specify the required heater capacity to maintain surface temperature at 140oC. Air velocity in the wind tunnel is 18 m/s and its temperature is 20oC.
(1) Observations. (i) The electric energy dissipated inside the sphere is removed from the surface as heat by forced convection. (ii) This problem can be modeled as external flow over a sphere. (iii) Newton’s law of cooling relates heat loss from the surface to heat transfer coefficient, surface area and surface temperature. (iv) The fluid is air.
(2) Problem Definition. Determine the rate of heat transfer from the surface of a sphere by forced convection.
(3) Solution Plan. Apply conservation of energy to the sphere. Apply Newton's law of cooling to obtain a relationship between heat removed from the sphere and its surface temperature. Use a correlation equation to determine the average heat transfer coefficient for the flow over a sphere.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) uniform surface flux, (5) uniform surface temperature, (6) uniform wind tunnel conditions, (7) constant
properties, (8) no buoyancy ( = 0 or g = 0) and (9) no radiation.
(ii) Analysis. Applying Conservation of energy to the sphere
P = q (a)where
P = electric power dissipated in sphere, W q = heat removed from surface, W
Application of Newton’s law of cooling to the surface of sphere
q = h A (Ts - T ) (b) where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oCTs = surface temperature = 140oC
T = ambient temperature = 20oC
Surface area of sphere is
A = D
2 (c) where
D = diameter = 0.06 m
The only remaining unknown in (b) is the heat transfer coefficient h . Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a sphere
4/14.03/22/1 )(06.04.02
sPrReRe
k
DhuN DDD (d)
T
V
sT
P
q
+ -
PROBLEM 8.15 (continued)
Valid for:
3.5 < ReD < 7.6 104
0.71 < Pr < 380
1.0 < ( / s) <3.2
properties at T , s at Ts (e) where
k = thermal conductivity of air = 0.02564 W/m-oC
NuD = average Nusselt number
Pr = Prandtl number of air = 0.713 ReD = Reynolds number
= viscosity of air at the free stream temperature = 18.17 10-6 kg/m-s
s = viscosity of air at surface temperature = 23.44 10-6 kg/m-s
The Reynolds number is given by
ReD = V D
(f)
where
V = free stream velocity = 18 m/s
= kinematic viscosity of air = 15.09 10-6 m2/s
(iii) Computations. Appendix C gives air properties at T = 20oC. Substituting into (f)
ReD = 18 0 06
1509 10 6 2
( / ) . ( )
. ( / )
m s m
m s = 71,571
Thus, the Reynolds and Prandtl numbers are within the limitations in (e). Next compute / s
/ s = 18.17 10-6 (kg/m-s)/ 23.44 10-6 (kg/m-s) = 0.775
Although this is outside the range given in (e), it represents a small deviation particularly since this ratio is raised to the 1/4 power. Substituting into (d)
NuhD
kD =
4/1
6
64.03/22/1
)sm/kg(1044.23
)sm/kg(1017.18)713.0()571,71(06.0)571,71(4.02 = 174.5
Solving the above for h
h = 174.5 k
D = 174.5 (0.02564) (W/m-oC)/0.06(m) = 74.57 W/m2-oC
Substituting into (b) and using (c) gives the required heater capacity P
P = q = 74.57(W/m2-oC) (0.06)2(m2) (140 -20)(oC) = 101.2 W
(iv) Checking. Dimensional check: Computations showed that equations (b), (d) and (f) are dimensionally consistent.
Quantitative check: The value of heat transfer coefficient is within the range shown in Table 1.1.
PROBLEM 8.15 (continued)
Limiting check: For the special case of Ts = T the required power should be zero. This is confirmed by equations (a) and (b).
(5) Comments. (i) Since not all conditions listed in (e) on correlation equation (d) have been met, the result should be viewed as an approximation. An alternate approach is to search the literature for another correlation equation that will meet the conditions of this problem. (ii) If radiation is included in the analysis, the required heater capacity will be greater than that determined by the above model.
PROBLEM 8.16
A hollow aluminum sphere weighing 0.2 kg is initially at 200oC. The sphere is parachuted from
a building window m100 above street level. You are challenged to catch the sphere with your
bare hands as it reaches the street. The sphere drops with an average velocity of 4.1 m/s. Its
diameter is 40 cm and the ambient air temperature is 20oC. Will you accept the challenge?
Support your decision.
(1) Observations. (i) The sphere cools off as it drops. Heat loss from the sphere is by forced convection. (ii) This is an external flow problem with a free stream velocity that changes with time. (iii) This is a transient conduction problem. The cooling time is equal to the time it takes the sphere to drop to street level. (iv) If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature. (v) Cooling rate depends on the heat transfer coefficient.
(2) Problem Definition. Determine transient temperature of sphere. This requires determining the drop time and the heat transfer coefficient.
(3) Solution Plan. Use the lumped capacity method to determine transient temperature. Use correlation equation to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) constant heat transfer coefficient, (3) Biot number < 0.1 (to be verified ), (4) constant properties, (5) constant sphere velocity, (6) uniform ambient conditions, (7) negligible wind speed, (8) no heat loss to the air inside the sphere, (9) no
buoyancy ( = 0 or g = 0) and. (10) no radiation and
(ii) Analysis. Equation (5.7) gives the transient temperature for the lumped capacity model
]exp[)()( tVc
AhTTTtT
p
si (a)
where
sA = surface area, m2
pc = specific heat of aluminum = 902 J/kg- oC
h = average heat transfer coefficient, W/m2-oCt = time, s
)(tT = temperature variable, oC
T = ambient temperature = 20oC
iT = initial temperature of sphere = 200oC
V = volume, m3
= density of aluminum = 2702 kg/m3
The product of V in equation (a) is equal to the mass of the sphere. That is
V = M (b) where
M = mass of sphere = 0.2 kg
Surface area of sphere is
sA = D2 (c)
PROBLEM 8.16 (continued)where
D = diameter = 0.4 m
There are two unknowns in (a): drop time to and the average heat transfer coefficient h . For constant sphere velocity, drop time is
V
sto (d)
where
s = drop distance = 100 m
V = sphere velocity = 4.1 m/s
Equation (8.12a) gives a correlation for the average Nusselt number for forced convection over a sphere
Nuh D
kD D D
s2 0 4 0 061 2 2 3 0 4
1 4
. ./ / ./
Re Re Pr (e)
Valid for:
3.5 < ReD < 7.6 104
0.71 < Pr < 380
1.0 < ( / s) <3.2
properties at T , s at Ts (f) where
k = thermal conductivity of air = 0.02564 W/m-oC
NuD = average Nusselt number
Pr = Prandtl number of air = 0.713 ReD = Reynolds number
= viscosity of air at the free stream temperature = 18.17 10-6 kg/m-s
s = viscosity of air at surface temperature, kg/m-s
The Reynolds number is defined as
ReD = V D
(g)
where
= kinematic viscosity of air = 15.09 10-6 m2/s
Since surface temperature changes with time, s should be evaluated at the average surface
temperature Ts defined as
sT = (Ti + To)/2 (h)
where To is the final sphere temperature obtained from (a). Since this temperature is unknown, the procedure becomes one of trial and error. A value for To is assumed and air viscosity
s determined at sT . Equation (a) is used to calculate To. The calculated To is compared with the
assumed value and the process is repeated until a satisfactory agreement is obtained between assumed and calculated values.
(iii) Computations. To calculate the final sphere temperature from (a), the drop time to and
the average heat transfer coefficient h need to be determined. Equation (d) gives to
to =100(m)/4.1(m/s) = 24.39 s
PROBLEM 8.16 (continued)
To determine s at the mean surface temperature sT , assume a final sphere temperature To =
80oC. Equation (h) gives
sT = (200 + 80)(oC)/2 = 140oC
Thus
s = 23.44 10-6 kg/m-s
The Reynolds number is
ReD = )s/m(1009.15
)m(4.0)s/m(1.426
= 108,681
This is somewhat outside the range of applicability of correlation equation (e). Therefore, results based on using this equation are approximate. Substituting into (e)
k
DhNu D =
4/1
6
64.03/22/1
)sm/kg(1044.23
)sm/kg(1017.18)713.0()108681(06.0)108681(4.02 = 222.1
h = 222.1(k/D) = 222.1 (0.02564) (W/m-oC)/0.4(m) = 14.2 W/m2-oC
Substituting into (a) and using (c)
To = 20(oC) + [(200 20 )(oC)])Ckg/J(902)kg(2.0
)s(39.24)m()4.0()Cm/W(2.14exp
o
22o2
= 88.6oC
Although this is slightly different from the assumed value of To = 80oC, repeating the procedure with a new value of To = 88.6oC will have a minor effect on the result.
The Biot number can now be checked to establish the validity of the lumped capacity method. The Biot number is defined as
Bi = h /kal (i)where
Bi = Biot number kal = thermal conductivity of aluminum = 236 W/m-oC
= wall thickness of sphere, m
The thickness is determined from its mass and volume
= M/ al D2 = 0.2(kg)/2702(kg/m3) (0.4) 2(m2) = 0.000147 m
When this is substituted into (i) gives Bi = 8.8 10-6. Thus, the use of the lumped capacity method is justified.
(iv) Checking. Dimensional check: Computations showed that equations (a), (d), (e) and (g) are dimensionally consistent.
Limiting check: If the sphere is dropped from an infinite height, its temperature at landing should
be equal to the ambient temperature. Setting s = in (d) gives to = . When this is substituted
into (a) gives To = T .
Quantitative check: The value of h is slightly outside the range listed in Table 1.1 for forced convection of gases.
PROBLEM 8.16 (continued)
(5) Comments. (i) The analysis shows that it is not safe to catch the sphere since its temperature at landing is 88.6oC. However, assumptions (8) and (10) are conservative since they result in an overestimate of surface temperature at landing. (ii) Both the Reynolds number and the viscosity correction factor in (e) are outside the range of applicability of correlation equation (e). However, the effect on the accuracy of the result should be minor.
PROBLEM 8.17
Steam condenses on the outside surface of a 1.6 cm diameter tube. Water enters the tube at
12.5oC and leaves at 27.5
oC. The mean water velocity is 0.405 m/s. Outside surface temperature
is 34 oC. Neglecting wall thickness, determine tube length.
(1) Observations. (i) This is an internal forced convection problem. (ii) The channel is a tube. (iii) The outside surface is maintained at a uniform temperature. (iv) Neglecting tube thickness resistance means that the inside and outside surface temperatures are identical. (v) Fluid temperature is developing. (vi) Inlet and outlet temperatures are known. (vii) The Reynolds number should be determined to establish if the flow is laminar or turbulent. (viii) The required tube length depends on the heat transfer coefficient. (ix) The fluid is water.
(2) Problem Definition. Determine the required tube length to heat water at a specified flow rate to a specified temperature. This requires the determination of the average heat transfer coefficient in a tube.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Compute the Reynolds number to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) axisymmetric flow, (4) constant properties, (5) uniform surface temperature, (6) negligible changes in kinetic and potential energy, (7) negligible axial conduction, (8) negligible dissipation, (9) negligible wall thickness, (10) no energy generation and (11) smooth tube.
(ii) Analysis. For flow through a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)
])()( [exp xTTTxT
p
smismcm
hP (a)
where
pc = specific heat, CJ/kg o
h = average heat transfer coefficient for a tube of length L, CW/m o2
m = mass flow rate, kg/s
P = tube perimeter, m
)(xTm = mean temperature at x, Co
miT = mean inlet temperature = 12.5 Co
sT = surface temperature = 34 Co
x = distance from inlet of heated section, m
The perimeter P is given by DP (b)
where
L
miTm
u
sT
D
moTx0
PROBLEM 8.17 (continued)
D = inside tube diameter = 1.6 cm = 0.016 m
Substituting (b) into (a) and applying the resulting equation at the outlet, x = L
])( [exp LTTTT
p
smismocm
hPD (c)
where
moT = outlet temperature = 27.5 Co
Solving (c) for the length L
smo
smip
TT
TT
hD
cmL ln (d)
Conservation of mass gives the flow rate m
4
2DuAum (e)
where
u = mean velocity = 0.405 m/s
= density, 3kg/m
To compute L using (d), it is necessary to determine h . The Reynolds number is computed to establish if the flow is laminar or turbulent. Reynolds number is defined as
DuReD (f)
where
DRe = Reynolds number
= kinematic viscosity, /sm2
Water properties are determined at the mean temperature mT , defined as
mT = T Tmi mo
2 (g)
Substituting numerical values into (g)
mT = C202
)C)(5.275.12( oo
Properties of water at this temperature are
pc = 4182 CJ/kg o
k = 0.5996 CW/m o
Pr = 6.99
= 6101004 /sm2
= 998.3 3kg/m
Substituting into (e)
PROBLEM 8.17 (continued)
4
)(m)016.0(m/s)(405.0)kg/m(34.998 223
m = 0.08129 kg/s
Finally, the heat transfer coefficient is needed to determine L from equation (d). The Reynolds number is computed to determine if the flow is laminar or turbulent. Equation (f) gives
6454)/sm(10004.1
m)(016.0)m/s(405.026DRe
Since this is greater than the transition number of 2300, it follows that the flow is turbulent. The appropriate correlation for the Nusselt number is given by the Gnielin ski equation (8.17a)
1)8/(7.121
1000)8/(
3/22/1 Prf
PrRef
k
DhuN D
D 1 2 3( / ) /D L (h)
Valid for valid for 0 < D/L <1 developing or fully developed turbulent flow through tubes
2300 < ReD < 5 106
0.5<Pr < 2000 properties at Tm
where
DuN = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by 2)64.1ln79.0( DRef (i)
(iii) Computations. The mass flow rate is calculated using (e)
kg/s08129.04
)m()016.0()m/s(405.0)kg/m(3.998 223
m
The Nusselt number is calculated using (h). The friction factor f is determined using (i)
0357.0)64.15.3746ln79.0( 2f
To determine h from equation (h) the length L must be known. Since L is unknown, a trial and
error procedure is used. A value for L is assumed and equation (h) is used to obtain a first
approximation for h . This approximate value is used in (d) to compute L. An improved value for
h is then obtained by substituting the calculated L into (h). Equation (d) is used again to obtain
an new value for L. This procedure is repeated until a satisfactory agreement is obtained between assumed and calculated L. As a first approximation, let L = 0. Equation (f) gives
k
DhuN D 3.52
1)99.6()8/0401.0(7.121
99.610006454)8/0357.0(
3/22/1
Solving for h
DNuD
kh = 52.3
)m(016.0
)CW/m(5996.0 o
= 1960 CW/m o2
PROBLEM 8.17 (continued)
Substituting into (d)
m13.4)C)(345.27(
)C)(345.12(ln
)C-W/m(1960)m()016.0(
)CJ/kg-(4182)kg/s(08129.0o
o
o2
o
L
Substituting this value into (h) gives 2.1984h CW/m o2 . With this value of h equation (d)
gives 077.4L m. Further iteration gives 1985h CW/m o2 and 075.4L m.
(iv) Checking. Dimensional check: Computations showed that equations (d)-(i) are dimensionally consistent.
Quantitative check: (1) The value of h is within the range given in Table 1.1 for forced
convection of liquids.
(2) An approximate value for L can be obtained based on the assumption that the fluid is at an
average temperature C20omT . Conservation of energy gives
)()( msmomip TThDLTTcm
Solving for L
)(
)(
ms
momip
TThD
TTcmL
Substituting numerical values into the above
m65.3)C)(2034(
)C)(5.125.27(
)C-W/m(1985)m()016.0(
)CJ/kg-(4182)kg/s(08129.0o
o
o2
o
L
This result differs from the exact solution by 10%.
(5) Comments. The fact that assuming L = 0 in calculating the heat transfer coefficient from
equation (h) introduces a small error in h implies that entrance effects are negligible.
PROBLEM 8.18
A 150 cm long tube with 8 mm inside diameter passes through a laboratory chamber. Air enters
the tube at 12oC with fully developed velocity and a flow rate 0.0005 kg/s. Assume uniform
surface temperature of 25oC, determine outlet air temperature.
(1) Observations. (i) This is an internal force convection problem. (ii) The channel is a tube. (iii) The surface is maintained at a uniform temperature. (iv) The velocity is fully developed. (v) The temperature is developing. (vi) The outlet temperature is unknown..(vii) The Reynolds number should be checked to establish if the flow is laminar or turbulent. (viii) The fluid is air.
(2) Problem Definition. Determine the outlet air temperature. This requires the determination of the average heat transfer coefficient.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Check the Reynolds number to determine if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) fully developed velocity, (4) axisymmetric flow, (5) constant properties, (6) uniform surface temperature, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) no dissipation, (10) no energy generation and (11) smooth tube.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)
])()( [exp xTTTxT
p
smismcm
hP (a)
where
cp = specific heat, J/kg- oC
h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate = 0.0005 kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 12oCTs = surface temperature = 25oCx = distance from inlet of heated section, m
The perimeter P is given by
P = D (b) where
D = inside tube diameter = 8 mm = 0.008 m
Substituting (b) into (a) and setting x = L gives the outlet temperature
])()( [exp LTTTxT
p
smismocm
hD (c)
L
miTm
u
sT
D
moTx0
PROBLEM 8.18 (continued)
where
L = tube length = 150 cm = 1.5 m Tmo = mean outlet temperature, oC
To compute Tmo using (c), it is necessary to determine h . The Reynolds number is calculated to establish if the flow is laminar or turbulent. Reynolds number is defined as
ReD /Du (d)
where
ReD = Reynolds number u = mean flow velocity, m/s
= kinematic viscosity, m2/s
The mean velocity is determined from mass flow rate
4/2DuAum
or2/4 Dmu (e)
where
A = cross section area, m2
= density, kg/m3
Air properties are determined at the mean temperature mT , defined as
mT = T Tmi mo
2 (f)
Since outlet temperature Tmo is unknown, the solution is obtained by a trial and error procedure.
A value for Tmo is assumed, (f) is used to calculate mT , a first approximation of properties is
determined at this temperature and (c) is used to calculate Tmo. If the calculated value is not the same as the assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Assume Tmo = 18oC. Equation (f) gives
mT = C152
)C)(1812( oo
Properties of air at this temperature are
cp = 1005.95 J/kg- oCk = 0.02526 W/m-oCPr = 0.7145
= 14.64 10-6 m2/s
= 1.22545 kg/m3
The mean velocity is obtained from (e)
u =)m()008.0)(m/kg)(22545.1(
)s/kg)(0005.0(4223
= 8.117 m/s
The Reynolds number is obtained from (d)
ReD 5.4435)s/m(1064.14
)m(008.0)s/m(117.826
PROBLEM 8.18 (continued)
Since this is greater than the transition number of 2300, it follows that the flow is turbulent. The
appropriate correlation for the Nusselt number is given by the Gnielin ski equation (8.17a)
1)8/(7.121
1000)8/(
3/22/1 Prf
PrRef
k
DhuN D
D 1 2 3( / ) /D L (g)
Valid for valid for 0< D/L <1 developing or fully developed turbulent flow through tubes
2300 < ReD < 5 106
0.5<Pr < 2000 properties at Tm (h)where
DuN = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
f = (0.79lnReD 1.64)–2 (i)
(iii) Computations. The Nusselt number is calculated using properties at the assumed outlet temperature Tmo = 18oC. The friction factor f is determined first using (i)
f = (0.79 ln 4435.5 - 1.64)-2 = 0.0401
Substituting into (g)
k
DhuN D
3/2
3/22/1)]m(5.1/)m(008.0[1
1)7145.0()8/0401.0(7.121
7145.010005.4435)8/0401.0(= 15.47
Solving for h
DNuD
kh = 15.47
)m(008.0
)Cm/W(02526.0 o
= 48.8 W/m2-oC
Substituting into (c) gives the outlet temperature
C7.24)m(5.1)CJ/kg(95.1005)kg/s(0005.0
)CW/m(8.48)m)(008.0([exp)C)(2512()C(25 o
o
o2oo
moT
Since this is greater than the assumed value of 18oC, the procedure is repeated with a new assumed value of 23oC. Based on this value, the calculated outlet temperature is found to be 24.7oC. Thus, the outlet temperature is 24.7oC.
(iv) Checking. Dimensional check: Computations showed that equations (c), (d), (e) and (g) are dimensionally consistent.
Quantitative check: The value of h is within the range given in Table 1.1 for forced convection
of gases.
(5) Comments. (i) The outlet temperature is within 0.3oC of the maximum value that it can reach. (ii) If one incorrectly assumes that the flow is laminar, the corresponding Nusselt number is 3.66. This is considerably smaller than the turbulent Nusselt number calculated above.
PROBLEM 8.19
Water enters a tube with a fully developed velocity and uniform temperature Tmi = 18oC. The
inside diameter of the tube is 1.5 cm and its surface temperature is uniform at Ts = 125oC.
Neglecting wall thickness, determine the length of the tube needed to heat the water to 82oC at a
flow rate of 0.002 kg/s.
(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is maintained at uniform temperature. (iii) The velocity is fully developed. (iv) The length of tube is unknown. (v) The temperature is developing. However, depending on tube length relative to the thermal entrance length, temperature may be considered fully developed throughout. (vi) The Reynolds number should be checked to determine if the flow is laminar or turbulent. (vii) The fluid is water.
(2) Problem Definition. Determine the length of tube needed to increase the temperature of water to a specified level at a given flow rate.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity, (7) no axial conduction, (8) negligible changes in potential and kinetic energy, (8) no dissipation, (9) no energy generation and (10) smooth tube.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)
])()( [exp xTTTxT
pcm
hPsmism (a)
where
cp = specific heat, J/kg- oC
h = average heat transfer coefficient, W/m2-oC
m= mass flow rate = 0.002 kg/s P = perimeter, m Ts = surface temperature = 125oCTmi = inlet temperature = 18oCTm (x) = mean temperature at any location x along the tube, oCx = distance along the tube measured from the inlet, m
To determine the required tube length, set x = L and Tm (L) = Tmo and solve (a) for L
mos
misp
TT
TT
Ph
cmL ln (b)
where
LTs
Ts
Du
Tmi
xTmo
PROBLEM 8.19 (continued)
Tmo = mean outlet temperature = 82oC
The perimeter P is given by
P = D (c)where
D = inside diameter = 1.5 cm = 0.015 m
Thus, all quantities in (b) are known except h . To determine h , the Reynolds number is calculated to establish if the flow is laminar or turbulent. The Reynolds number is defined as
ReD = u D
(d)
where
u = mean velocity, m/s
= kinematic viscosity, m2/s
The mean velocity is determined from mass flow rate equation
m uD2
4 (e)
where
= density, kg/m3
Solving (e) for u
u = 4
2
m
D (f)
Properties are evaluated at the mean temperature Tm , the average of inlet and outlet temperatures
C502
)C)(8218(
2
oo
momim
TTT
Properties of water at this temperature are given in Appendix D
cp = specific heat = 4182 J/kg- oCk = thermal conductivity = 0.6405 W/m-oCPr = Prandtl number = 3.57
= kinematic viscosity = 0.5537 10-6 m2/s
= density = 988 kg/m3
Equations (f) and ( d) give
u = 4 0 002
988 0 0153 2 2
( . )( / )
( / ) ( . ) ( )
kg s
kg m m= 0.01146 m/s
and
ReD =0 01146 0 015
05537 10 6 2
. ( / ) . ( )
. ( / )
m s m
m s= 310
Since ReD < 2300, the flow is laminar. The next step is establishing if thermal entrance length Lt
is small compared to tube length L. If this is the case, the flow can be considered thermally fully developed throughout. The thermal entrance length for laminar flow in a constant temperature tube is given by equation (7.43b)
Lt = 0.033 D ReD Pr (g)
PROBLEM 8.19 (continued)
Substituting into (h)
Lt = 0.033 (0.015)(m) (310) (3.57) = 0.548 m
Since L is unknown, comparison can not be made with Lt. Assuming that Lt is not small compared to L, the flow must be treated as thermally developing and correlation equation (8.14a) should be used
3/1))/(0.041
)/(0668.066.3
PrReLD
PrReLD
k
DhuN
D
DD (h)
Valid for entrance region of tube uniform surface temperature Ts
fully developed laminar flow (ReD < 2300) developing temperature
properties at 2/)( momim TTT (i)
To calculate h from this equation, the length L must be known. Thus, a trial and error procedure
is required to solve the problem. A value for L is assumed, equation (h) is used to calculate h
and the result substituted into (b) to determine L. The procedure is repeated until a satisfactory agreement is obtained between assumed and calculated values.
(iii) Computations. Assume L = 0.5 m. Substituting into (h)
k
DhuN D 229.5
310(3.57).5(m)0.015(m)/004.01
)57.3(310])m(5.0/)m(015.0[0668.066.3
3/2
Solving for h
DNuD
kh = Cm/W29.223229.5
)m(015.0
)Cm/W(6405.0 o2o
Substituting into (b)
m725.0)C)(82125(
)C)(18125(ln
)m)(015.0()Cm/W(29.223
)Ckg/J(4182)s/kg(002.0o
o
o2
o
L
Since calculated L is not equal to the assumed value of 0.5 m, the process is repeated with another assumed value of L = 0.73 m. Results of four trials are tabulated below.
Assumed L Calculated h Calculated L
m W/m2-oC m
0.5 223.29 0.725
0.73 205.38 0.788
0.79 202.23 0.8
0.80 201.74 0.802
Therefore, L = 0.802 m
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (f), (g) (h)
PROBLEM 8.19 (continued)
and (i) are dimensionally consistent.
Quantitative check: The value of the heat transfer coefficient is within the range shown in Table 1.1 for forced convection of liquids.
Qualitative check: The value of h for developing temperature should be larger than h for fully developed temperature. Equation (7.57) gives the Nusselt number for the fully developed case
66.3k
DhuN D (j)
Cm/W3.156)m(015.0
)Cm/W(6405.066.366.3 o2
o
D
kh
This is smaller than h = 201.74 W/m2-oC for the developing case.
Limiting check: If Tmo = Tmi, the required length should be zero. Setting Tmo = Tmi in (b) gives the L = 0.
(5) Comments. (i) Since Lh is not small compared to L, temperature entrance effects can not be neglected. (ii) Equation (h) converges to the limiting case of fully developed temperature when L
. Setting L = in (h) gives the fully developed solution (j).
PROBLEM 8.20
Cold air is supplied to a research apparatus at a rate of 0.14 g/s. The air enters a 20 cm long
tube with uniform velocity and uniform temperature of 20oC. The inside diameter of the tube is
5 mm. The inside surface is maintained at 30oC. Determine the outlet air temperature.
(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is maintained at a uniform temperature. (iii) The velocity and temperature are developing. Thus, entrance effects may be important. (iv) The outlet temperature is unknown. (v) The fluid is air.
(2) Problem Definition. Determine air outlet temperature. This requires determining the average heat transfer coefficient.
(3) Solution Plan. Apply the analysis of internal flow through a tube at uniform surface temperature. Check the Reynolds number to determine if the flow is laminar or turbulent. Compute the hydrodynamic and thermal entrance lengths to establish if entrance effects can be neglected. Use an appropriate correlation equation to compute the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) uniform inlet velocity and temperature, (7) no axial conduction, (8) negligible changes in potential and kinetic energy, (9) no dissipation and (10) no energy generation.
. (ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)
])()( exp[ xTTTxT
p
smismcm
hP (a)
where
cp = specific heat, J/kg- oC
h = average heat transfer coefficient for a tube of length L, W/m2-oCm = mass flow rate = 0.14 g/s = 0.00014 kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 20 oC
Ts = surface temperature = 30oCx = distance from inlet of heated section, m
The perimeter P is given by
P = D (b)where
D = inside tube diameter
= 5 mm = 0.005 m
Substituting (b) into (a) and setting x = L gives the outlet temperature
L
Ts
Du
Tmi
xTmo
PROBLEM 8.20 (continued)
])()( exp[ LTTTxT
p
smismocm
hD (c)
where
L = tube length = 20 cm = 0.2 m Tmo = mean outlet temperature, oC
To compute Tmo using (c), it is necessary to determine h . The Reynolds number is calculated to establish if the flow is laminar or turbulent. Reynolds number is defined as
ReD
u D (d)
where
ReD = Reynolds number u = mean flow velocity, m/s
= kinematic viscosity, m2/s
The mean velocity is determined from mass flow rate
4/2DuAum
or2/4 Dmu (e)
where
A = cross section area, m2
= density, kg/m3
Air properties are determined at the mean temperature mT , defined as
mT = T Tmi mo
2 (f)
Since outlet temperature Tmo is unknown, the solution is obtained by a trial and error procedure.
A value for Tmo is assumed, (f) is used to calculate mT , a first approximation of properties is
determined at this temperature and (c) is used to calculate Tmo. If the calculated value of Tmo is not the same as the assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Assume Tmo = 0oC. Equation (f) gives
mT = C102
)C)(020( oo
Properties of air at this temperature are
cp = 1005.6 J/kg- oC
k = 0.02329 W/m-oCPr = 0.721
= 12.46 0-6 m2/s
= 1.3414 kg/m3
The mean velocity is obtained from (e)
u =)m()005.0)(m/kg)(3414.1(
)s/kg)(00014.0(4223
= 5.315 m/s
PROBLEM 8.20 (continued)
The Reynolds number is obtained from (d)
ReD 2133)s/m(1046.12
)m(005.0)s/m(315.526
Since this is less than the transition number of 2300, it follows that the flow is laminar. The next step is to determine if entrance effects can be neglected. For laminar flow, the hydrodynamic entrance length Lh for a constant surface temperature tube is given by equation (7.43a)
Lh/D = 0.056 ReD (g)
The thermal entrance length Lt is given by equation (7.43b)
Lt/D = 0.033 ReD Pr (h)
Equations (g) and (h) give
Lh = 0.056(2,133)0.005(m) = 0.597m
andLt = 0.033(2,133)(0.721) 0.005(m) = 0.254 m
Thus, both velocity and temperature are developing. The appropriate correlation for the average Nusselt number is given by equation (8.15a)
14.03/1
)/(86.1s
DD PrReLDk
DhNu (i)
Valid for: entrance region of tube uniform surface temperature Ts
laminar flow (ReD < 2300) developing velocity and temperature 0.48 < Pr < 16700
0.0044 < s < 9.75
properties at Tm , s at Ts (j)
where
= viscosity at mean temperature = 16.71 10-6 kg/s-m
s = viscosity at surface temperature = 18.65 10-6 kg/s-m
Conditions (j) are satisfied.
(iii) Computations. Equation (i) is used to calculate h
k
DhNu D
14.0
6
63/1
)ms/kg(1065.18
)ms/kg(1071.16)721.0(1332
0.2(m)
)m(005.086.1 = 6.18
or
DNuD
kh 18.6
)m(005.0
)Cm/W(02329.0 o
= 28.8 W/m2-oC
Substituting into (c)
PROBLEM 8.20 (continued)
moT = 30(oC) )2030( (oC) 7.3)Ckg/J(6.1005)s/kg(00014.0
)m(2.0)Cm/W(8.28)m)(005.0(exp
o
o2oC
Since this is higher than the assumed temperature, the above procedure is repeated with assumed
Tmo = 4oC. The corresponding heat transfer coefficient and calculated outlet temperature are h =28.94 W/m2-oC and Tmo = 3.8oC.
(iv) Checking. Dimensional check: Computations showed that units of equations (b)-(i) are dimensionally consistent.
Qualitative check: Since entrance effects are important, it follows that the heat transfer coefficient is greater than that of fully developed flow. For laminar fully developed flow, the Nusselt number is given by Equation (7.57)
66.3k
DhuN D (k)
Using this equation to compute h
Cm/W17)m(005.0
)Cm/W(02329.066.366.3 o2
o
D
kh
This is smaller than h = 28.8 W/m2-oC for the developing case.
Limiting check: In the limit as L 0, the outlet temperature approaches inlet temperature. Setting L = 0 in equation (c) gives Tmo = Tmi.
(5) Comments. (i) If entrance effects are neglected, h will be underestimated ( h = 17 W/m2-oC) and the corresponding outlet temperature will be Tmo = 4 oC. (ii) The Reynolds number is very close to the transition Reynolds number of 2300. Since this value of transition Reynolds number is not exact, depending on surface roughness and other factors, it is uncertain if the flow is laminar or turbulent for this case.
PROBLEM 8.21
Water flows through a tube of inside diameter 2.5 cm. The inside surface temperature is 230oC
and the mean velocity is 3 cm/s. At a section far away from the inlet the mean temperature is
70oC.
[a] Calculate the heat flux at this section
[b] What will the flux be if the mean velocity is increased by a factor of ten?
(1) Observations. (i) This is an internal forced convection problem. (ii) Tube surface is maintained at uniform temperature. (iii) The section of interest is far away from the inlet. This means that flow and temperature can be assumed fully developed and the heat transfer coefficient uniform. (iv) It is desired to determine the surface flux at this section. Newton’s law of cooling gives a relationship between local flux, surface temperature and heat transfer coefficient. (v) The Reynolds number should be checked to determine if the flow is laminar or turbulent. (vi) The fluid is water.
(2) Problem Definition. Determine surface heat flux corresponding to two mean flow velocities. Since the flux can be obtained from Newton's law of cooling, the problem is one of finding the heat transfer coefficient or fully developed flow corresponding to the two velocities.
(3) Solution Plan. Apply Newton's law of cooling at the specified section of the tube. Use analytic solutions or correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (4) uniform surface temperature, (5) fully developed velocity and temperature, (6) no axial conduction, (7) negligible changes in potential and kinetic energy, (8) no dissipation, (9) no energy generation and (10) smooth tube.
(ii) Analysis. Apply Newton’s law of cooling
q = )( TTh s (a)
where
h = local heat transfer coefficient, W/m2-oC
q = surface flux, W/m2
Ts = surface temperature = 230oCTm = mean temperature of water = 70oC
Since flow and temperature are fully developed, the heat transfer coefficient is uniform (local and average coefficients are identical). To determine h, the Reynolds number is calculated to establish if the flow is laminar or turbulent. The Reynolds number is defined as
ReD = u D
(b)
where
D = inside tube diameter = 2.5 cm = 0.025 m ReD = Reynolds number u = mean velocity = 3 cm/s = 0.03 m/s
= kinematic viscosity = 0.4137 10-6 m2/s
Dx mT
sT
u
sT
fully developed
PROBLEM 8.21 (continued)
Properties are evaluated at the mean temperature Tm = 70oC. Appendix D gives
k = thermal conductivity = 0.6594 W/m-oCPr = Prandtl number =2.57
Equation (b) gives
ReD =0 03 0 025
0 4137 10 6 2
. ( / ) . ( )
. ( / )
m s m
m s= 1,813
Since ReD < 2300 it follows that the flow is laminar. The Nusselt number for fully developed laminar flow in tubes at uniform surface temperature is given by equation (7.57)
DNu = NuD =k
hD= 3.66 (c)
When the velocity is increased by a factor of 10 the Reynolds number increases to
ReD = 18,130
Thus, the flow becomes turbulent. For this case the average Nusselt number is given by the Gnielinski equation (8.17a)
Nuh D
kPr
D
D
f
f
8
8
1000
1 12 7 11 2
2 3
Re Pr
./
/
1 2 3( / ) /D L (d)
Valid for valid for 0< D/L <1 developing or fully developed turbulent flow through tubes
2300 < ReD < 5 106
0.5 <Pr < 2000 properties at Tm (e)where
NuD = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
f = (0.79lnReD 1.64)–2 (f)
For fully developed flow the factor D/L in equation (d) is set to zero.
(iii) Computations.
[a] For the case where u = 0.03 m/s and the flow is laminar. Equation (c) gives
h = 3.66k
D = 3.66
0 6594
0 025
. ( / )
. ( )
W m C
m
o
= 96.5 W/m2-oC
Substituting into (a)
q = 96.5(W/m2-oC)(230 - 70)(oC) = 15,440 W/m2
PROBLEM 8.21 (continued)
[b] For the case where the flow is turbulent, equations (d) and (f) give h . The Reynolds number for this case is 18,130. Equation (f) gives f
f = [0.79ln (18,130) 1.64]–2 = 0.02682
Substituting into (d) and setting D/L = 0
Nuh D
kD
0 02682 8 18 130 1000 2 57
1 12 7 0 02682 8 2 57 11 2 2 3
. / , .
. . / ./ /
= 89.76
Solving the above for h
h = 89.76k
D = 89.76
0 6594
0 025
. ( / )
. ( )
W m C
m
o
= 2367.5 W/m2-oC
Equation (a) gives the flux
q = 2367.5 (W/m2-oC)(230 - 70)(oC) = 378,800 W/m2
(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) are dimensionally consistent.
Qualitative check: As expected turbulent heat flux is greater than that of laminar flow.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. Although the velocity is increased by a factor of 10, the flux is increased by a factor of 25. This large increase is a result of transition from laminar to turbulent. If the flow remains laminar, then according to (c), there will be no change in h even though the Reynolds number is increased.
PROBLEM 8.22
Air flows through a tube of inside diameter 5 cm. At a section far away from the inlet the mean
temperature is 30oC. At another section further downstream the mean temperature is 70
oC.
Inside surface temperature is 90oC and the mean velocity is 4.2 m/s. Determine the length of this
section.
(1) Observations. (i) This is an internal forced convection problem in a tube. (ii) Both velocity and temperature are fully developed. (iii) Tube surface is maintained at uniform temperature. (iv) The Reynolds number should be computed to establish if flow is laminar or turbulent. (v) Mean velocity, mean inlet and outlet temperatures and tube diameter are known. (vi) The fluid is air.
(2) Problem Definition. Find the required tube length to increase the air temperature by a given amount.
(3) Solution Plan. Use the analysis of fully developed laminar flow in tubes at uniform surface temperature to determine the required tube length.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) fully developed flow, (4) axisymmetric flow, (5) constant properties, (6) uniform surface temperature, (7) negligible changes in kinetic and potential energy, (8) negligible axial conduction, (9) no dissipation, (10) no energy generation and (11) smooth tube.
(ii) Analysis. For flow in a tube at uniform surface temperature, conservation of energy and Newton's law of cooling lead to equation (7.64)
])()( exp[ xTTTxT
p
smismcm
hP (a)
cp = specific heat, J/kg- oC
h = average heat transfer coefficient for a tube of length L, W/m2-oCL = length of tube, m m = mass flow rate, kg/s P = tube perimeter, m Tm(x) = mean temperature at x, oCTmi = mean inlet temperature = 30oCTs = surface temperature = 90oCx = distance from inlet of heated section, m
Applying (a) at the outlet of the heated section (x = L) and solving for L
tL
hL
iT
iV
t
L
miT m
u
sT
D
moT
x0
PROBLEM 8.22 (continued)
Lmc
P h
T T
T T
p s mi
s mo
ln (b)
where
Tmo = mean outlet temperature = 70oC
To compute L using (b), it is necessary to determine cp, P, m , and h . All properties are
determined at the mean temperature mT defined as
mT = T Tmi mo
2 (c)
The perimeter P and flow rate m are given by
P = D (d) and
4/2 uDm (e)
where
D = inside tube diameter = 5 cm = 0.05 m u = mean flow velocity = 4.2 m/s
= density, kg/m3
The heat transfer coefficient for fully developed flow is uniform along a channel. Its value depends on whether the flow is laminar or turbulent. To proceed, it is necessary to calculate the Reynolds number to determine if the flow is laminar or turbulent. For flow in a tube the Reynolds number is defined as
ReD
u D (f)
where
ReD = Reynolds number
= kinematic viscosity, m2/s
The mean temperature is calculated in order that properties are determined. Substituting into (c)
mT = C502
)C)(7030( oo
Properties of air at this temperature are given in Appendix C
cp = 1007.4 J/kg- oCk = 0.02781 W/m-oCPr = 0.709
= 17.92 10-6, m2/s
= 1.0924 kg/m3
Substituting into (f)
ReD 719,11)s/m(1092.17
)m(05.0)s/m(2.426
Since the Reynolds number is greater than 2300, the flow is turbulent. The Nusselt number for turbulent flow through a tube is given by the Gnielinski equation (8.17a)
PROBLEM 8.22 (continued)
17.121
1000
3/22/1
8
8
Pr
PrRe
k
DhuN
f
fD
D 1 2 3( / ) /D L (g)
Valid for valid for 0< D/L <1 developing or fully developed turbulent flow through tubes
2300 < ReD < 5 106
0.5<Pr < 2000 properties at Tm (h)where
DuN = average Nusselt number in turbulent flow
f = friction factor
For a smooth pipe f is given by
f = (0.79lnReD 1.64)–2 (i)
For fully developed flow, set D/L = 0 in (g)
1)8/(7.121
1000)8/(3/22/1 Prf
PrRef
k
DhuN D
D (j)
where the average and local heat transfer coefficients, h and h are identical in the fully
developed region.
(iii) Computations. Substituting into (d) and (e)
P = 0.05(m) = 0.1571 m
s/kg009009.0)s/m(2.4)m/kg(0924.14
)m()05.0( 322
m
The Nusselt number is calculated using equation (j). The friction factor f is determined using (i)
f = (0.79 ln11,719 64.1 )-2 = 0.0301
Substituting into (j)
k
DhuN D
1)709.0()8/0301.0(7.121
709.01000719,11)8/0301.0(
3/22/1= 34.2
Solving for h
DNuD
kh = 34.2
)m(05.0
)Cm/W(02781.0 o
= 19.02 W/m2-oC
Substituting into (b) gives the required tube length
)C)(7090(
)C)(3090(ln
)Cm/W(3.17)m(1571.0
)Ckg/J(4.1007)s/kg(009009.0o
o
o2
o
L = 3.34 m
(iv) Checking. Dimensional check: Computations showed that equations (b), (d), (e), (f), and (j) are dimensionally consistent.
PROBLEM 8.22 (continued)
Limiting check: For the special case of Tmo = Tmi , the required length should vanish. Setting Tmo
= Tmi in (b) gives L = 0.
Quantitative check: The value of h is on the low end of values listed in Table 1.1 for forced convection of gases. It should be kept in mind that values of h in Table 1.1 are for typical applications. Exceptions should be expected.
(5) Comments. For fully developed laminar flow in tubes at uniform surface temperature, the Nusselt number is given by equation (7.57)
k
hDNuD = 3.66 (k)
If one incorrectly uses this equation, the heat transfer coefficient and required length become
h = 1.85 W/m2-oC
and
L = 34.3 m
Thus, the error in using (k) is significant.
PROBLEM 8.23
Two identical tubes have inside diameters of 6 mm. Air flows through one tube at a rate of 0.03
kg/hr and through the other at a rate of 0.4 kg/hr. Far away from the inlets of the tubes the
mean temperature is 120oC for both tubes. The air is heated at a uniform surface temperature
which is identical for both tubes. Determine the ratio of the heat flux of the two tubes at this
section.
(1) Observations. (i) This is an internal forced convection problem. (ii) The surface of each tube is maintained at uniform temperature which is the same for both. (iii) The velocity and temperature are fully developed. Thus, the heat transfer coefficient is uniform. (iv) Air flows through each tube at different rates. (v) The Reynolds number should be computed to establish if the flow is laminar or turbulent. (vi) Surface heat flux depends on the heat transfer coefficient.
(2) Problem Definition. Determine the ratio of surface heat flux for the two tubes at sections here the mean temperature is the same for both.
(3) Solution Plan. Apply Newton’s law of cooling to each tube. Compute the Reynolds number to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) no axial conduction, (7) fully developed velocity and temperature, (8) no changes in potential and kinetic energy and (0) no dissipation.
(ii) Analysis. Local heat flux is given by Newton’s law of cooling
q = h (Ts - Tm) (a)
where
h = heat transfer coefficient, W/m2-oCq = surface heat flux, W/m2
Tm = mean temperature = 120oCTs = surface temperature, oC
Applying (a) to the two tubes, noting that Ts and Tm are the same for both tubes, and taking the ratio of the two equations
q
q
h
h
1
2
1
2
(b)
where the subscripts 1 and 2 refer to tubes 1 and 2. To determine the ratio of the heat transfer coefficients, the Reynolds number for each tube should be computed to establish if the flow is laminar or turbulent. The Reynolds number is defined as
ReD = uD
(c)
where
D = tube diameter = 0.006 m
u = mean velocity, m/s ReD = Reynolds number
= kinematic viscosity = 25.19 10-6 m2/s
PROBLEM 8.23 (continued)
Properties of air are determined at the mean temperature Tm = 120oC. The mean velocity is determined from the flow rate
/m u D2 4 (d)
or
um
D
42 (e)
where
m = mass flow rate, kg/s
= density = 0.8979 kg/m3
The mass flow rates for the two tubes are
m1 = 0.03 kg/hr
m2 = 0.4 kg/hr
Substituting into (e)
u1)/(3600)()006.0)(/(8979.0
)/)(03.0(4223 hrsmmkg
hrkg = 0.328 m/s
and
u2)/(3600)()006.0)(/(8979.0
)/)(4.0(4223 hrsmmkg
hrkg = 4.38 m/s
Substituting into (c)
ReD1 = = 78.1 and
ReD2 = = 1,043.3
Thus, the flow is laminar in both tubes. For laminar fully developed flow through tubes with uniform surface temperature, the Nusselt number is given by
NuD = hD/k = 3.66 or
h = 3.66 k/D (f) where
NuD = Nusselt number k = thermal conductivity of air = 0.03261 W/m-oC
(iii) Computations. Applying (f) to the two tubes and noting that D and k are the same for both tubes
h1 = h2 = 3.66 (0.03261)(W/m-oC)/ 0.006(m) = 19.9 W/m2-oC
Substituting into equation (b)
q
q
1
2
19 9
19 91
2
2
. ( / )
. ( / )
W m C
W m C
o
o
PROBLEM 8.23 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (c), (e) and (f) are dimensionally correct.
Quantitative check: The value of h is within the range given in Table 1.1 for force convection of gases.
(5) Comments. It is surprising that although the velocity in tube 2 is over 13 times greater than that in tube 1, the heat flux is the same for both tubes. This is due to the fact that for laminar fully developed flow, the Nusselt number is independent of the Reynolds number.
PROBLEM 8.24
Two concentric tubes of diameters 2.5 cm and 6.0 cm are used as a heat exchanger. Air flows
through the inner tube with a mean velocity of 2 m/s and mean temperature of 190oC. Water
flows in the annular space between the two tubes with a mean velocity of 0.5 m/s and a mean
temperature of 30oC. Determine the inside and outside heat transfer coefficients.
(1) Observations. (i) This is an internal forced convection problem. (ii) The geometry consists of two concentric tubes. (iii) Air flows in the inner tube while water flows in the annular space between the two tubes. (iv) The Reynolds number should be computed for both fluids to establish if the flow is laminar or turbulent. (v) Convection resistance depends on the heat transfer coefficient.
(2) Problem Definition. Determine the air side and water side heat transfer coefficients.
(3) Solution Plan. Compute the Reynolds numbers for air in the tube and for water in the annular space to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity and temperature, (7) negligible wall conduction resistance, (8) negligible wall thickness and (9) smooth tubes.
(ii) Analysis. To determine hi and ho, the two Reynolds numbers are calculated first to establish if the flow is laminar or turbulent. For air flow inside the inner tube the Reynolds number is
aiaDi DuRe / (a)
where
iD = diameter of inner tube = 2.5 cm = 0.025 m
DiRe = Reynolds number for air flow through inner tube
au = mean air velocity in inner tube = 2 m/s
a = kinematic viscosity of air = 33.34 10-6 m2/s
Substituting into (b)
DiRe =)s/m(1034.33
)m(025.0)s/m(226
= 1499.7
Thus, the flow is laminar in the inner tube. The Nusselt number for fully developed laminar flow in tubes depends on surface boundary condition. Assuming uniform surface temperature, the Nusselt number is given by equation (7.57)
DiNu =a
ii
k
Dh = 3.66 (b)
where
ka = thermal conductivity of air = 0.03718 W/m-oC
DiNu = average Nusselt number
The Reynolds number for water flow in the annular space between the two tubes is defined as
air
hi
ho
water
air
DiDo
PR0BLEM 8.24 (continued)
w
ewDe
DuRe (c)
where
De = equivalent diameter of the annular space, m
DeRe = Reynolds number for water flow through the annular space
wu = mean water velocity in the annular space = 0.5 m/s
w = kinematic viscosity of water = 0.8012 10-6 m2/s
The equivalent diameter is defined as
P
ADe
4 (d)
where
A = flow area, m2
P = wet perimeter, m
The flow area and wet perimeter for the annular space are given by
A = 4/)( 22io DD (e)
and
)( oi DDP (f)
where
Do = diameter of outer tube = 6 cm = 0.06 m
Substituting (e) and (g) into (d)
ioe DDD (g)
Using (h), equation (d) becomes
eDRe =)s/m(108012.0
)m)(025.006.0)(s/m(5.026
= 21,842
Therefore, the flow in the annular space is turbulent. The Nusselt number for turbulent flow through channels is given by the Gnielinski equation (8.17a)
)1()8/(7.121
)1000)(8/(2/32/1
w
weD
w
eoDe
Prf
PrRef
k
DhNu 3/2)/(1 LDe (h)
Valid for valid for 0< De/L <1 developing or fully developed turbulent flow through channels
2300 < ReDe < 5 106
0.5 <Pr < 2000 properties at Tm (i)
where
f = friction factor kw = thermal conductivity of water = 0.615 W/m-oC
wPr = Prandtl number of water = 5.42
PR0BLEM 8.24 (continued)
DeNu = Nusselt number in turbulent flow
For a smooth tube f is given by
f = (0.79lnReDe 1.64)–2 (j)
For fully developed flow, set De/L = 0 in (h)
)1()8/(7.121
)1000()8/(
3/22/1 Prf
PrRef
k
hDNu eD
w
eeD (k)
where the average and local heat transfer coefficients, oh and ho are identical in the fully
developed region.
(iii) Computations. Equation (b) gives hi
hi = 3.66 (0.03718)(W/m-oC)/0.025(m) = 5.44 W/m2-oC
Equation (k) gives f
f = (0.79 ln 21,842 64.1 )-2 = 0.02557
Substituting into (k)
w
eoD
k
DhNu
e)142.5(8/02557.07.121
42.5)1000842,21)(8/02557.0(3/22/1
= 144.6
and
oh = 144.6 (0.615)(W/m-oC)/(0.06-0.025)(m) = 2541 W/m2-oC
(iv) Checking. Dimensional check: Computations showed that the units of equations (a)- (c) and (k) are dimensionally correct.
Quantitative check: The values hi for air and ho for water are within the approximate range shown in Table 1.1 for forced convection.
(5) Comments. Resistance to heat transfer is inversely proportional to the heat transfer coefficient. Thus, the total resistance in this example is dominated by air side convection resistance. The contribution to the total resistance of the much larger water side heat transfer coefficient is insignificant and therefore can be neglected.
PROBLEM 8.25
A heat exchanger consists of a tube and square duct. The tube is
placed co-axially inside the duct. Hot water flows through the tube
while cold water passes through the duct. The inside and outside
diameters are 5 cm and 5.2 cm, respectively. The side of the duct is
10 cm. At a section far away from the inlet the mean hot water
temperature is 90oC and the mean cold water temperature is 30
oC.
The mean hot water velocity is 1.32 m/s and the mean cold water
velocity is 0.077 m/s. Determine the inside and outside heat transfer
coefficients.
(1) Observations. (i) This is an internal forced convection problem. (ii) The geometry consists of a tube concentrically placed inside a square duct,. (iii) Water flows in the tube and the duct. (iv) The Reynolds number should be computed for the two fluids to establish if the flow is laminar or turbulent. (v) Far away from the inlet the velocity and temperature may be assumed fully developed.
(2) Problem Definition. Determine the heat transfer coefficient for the flow inside the tube and for the flow in the duct.
(3) Solution Plan. Compute the Reynolds number for the flow in the tube and in the duct to establish if the flow is laminar or turbulent. Use analytic solutions or correlation equations to determine the heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Newtonian fluid, (2) steady state, (3) constant properties, (4) axisymmetric flow, (5) uniform surface temperature, (6) fully developed velocity and temperature and (7) smooth channels.
(ii) Analysis. To determine ih and oh , the two Reynolds numbers are calculated first to establish
if the flow is laminar or turbulent. For water flow inside the tube the Reynolds number is
i
iiDi
DuRe (a)
where
iD = inside tube diameter = 5 cm = 0.05 m
DiRe = Reynolds number for flow in tube
iu = mean water velocity in tube = 1.32 m/s
i = kinematic viscosity of water in tube = 6103264.0 /sm2
Substituting into (a)
DiRe = 5
2610022.2
)/sm(103264.0
)m(05.0)m/s(32.1
Thus the flow is turbulent in the tube. The Nusselt number for fully developed turbulent flow is given by (set D/L = 0 in eq. 8.17a)
iDoD
S
cold water
waterhot
PROBLEM 8.25 (continued)
)1()8/(7.121
)1000)(8/(2/32/1
i
iiD
i
iiD
Prf
PrRef
k
DhNu i (b)
Valid for
fully developed turbulent flow through channels
1/0 LDi
2300 < < 5 106
51052300iDRe prop
0.5 < iPr < 2000
properties at miT
where
f = friction factor
ik = conductivity of water in tube = 0.6727 CW/m o
iPr = Prandtl number of water in tube = 1.97
miT = mean temperature of water in tube = 90 Co
For a smooth tube f is given by
f = (0.79ln iDRe 1.64)–2 (c)
The Reynolds number for fluid flow in the annular duct space is defined as
o
eoDe
DuRe (d)
where
eD = equivalent diameter of the duct annular space, m
DeRe = Reynolds number for water flow through the duct
ou = mean water velocity through the duct = 0.077 m/s
o = kinematic viscosity of water in duct = 6108012.0 /sm2
The equivalent diameter is defined as
P
ADe
4 (e)
where
A = flow area, m2
P = wet perimeter, m
The flow area and wet perimeter for the annular duct space are given by
A = 4/22oDS (f)
and
SDP o 4 (g)
where
S 10 cm = 0.1 m
Substituting (f) and (g) into (e)
PROBLEM 8.25 (continued)
SD
DSD
o
oe
4
4 22
(h)
Using (h)
05592.0)m)(1.0(4)m)(052.0(
)m()052.0()m()1.0(4 2222
eD m
eDRe =)/sm(108012.0
)m(05592.0)m/s(077.026
= 5,374
Therefore, the flow in the duct space is turbulent. The Nusselt number for fully developed
turbulent flow through channels is given by equation (b) with iD replaced by eD
)1()8/(7.121
)1000)(8/(2/32/1
o
oeD
o
eoDe
Prf
PrRef
k
DhNu (i)
properties at moT
where
ok = conductivity of water in duct = 0.615 CW/m o
oPr = Prandtl number of water in duct = 5.42
moT = mean temperature of water in duct = 30 Co
(iii) Computations. For the flow through the tube, equation (c) gives f
01558.064.110022.2ln79.025f
Substituting into (b)
i
iiD
k
DhNu i
)197.1(8/01558.07.121
97.1)100010022.2)(8/01558.0(3/22/1
5
= 584.6
Thus
CW/m7865m)(05.0
)CW/m(6727.06.5846.584 o2
o
i
ii
D
kh
For the flow in the annular duct space equation (c) gives f
03777.064.15374ln79.02
f
Applying (b) to the annular duct space and setting ei DD
o
ooD
k
DhNu e 74.39
)142.5(8/03777.07.121
42.5)10005374)(8/03777.0(3/22/1
Thus
CW/m1.437m)(05592.0
)CW/m(615.074.3974.39 o2
o
e
oo
D
kh
PROBLEM 8.25 (continued)
(iv) Checking. Dimensional check: Computations showed that the units of equations (a), (b), and (h) are dimensionally correct.
Quantitative check: The values hi for and ho for water are within the range shown in Table 1.1 for forced convection of liquids.
(5) Comments. Since convection resistance to heat transfer is inversely proportional to the heat transfer coefficient, The total resistance in this example is dominated by the hot water convection resistance in the tube.
PROBLEM 8.26
In designing an air conditioning system for a pizza restaurant an estimate of the heat added to
the kitchen from the door of the pizza oven is needed. The rectangular door is 50 cm 120 cm
with its short side along the vertical direction. Door surface temperature is 110oC. Ambient air
and surroundings temperatures are 20oC and 24
oC, respectively. Door surface emissivity is 0.08.
Estimate the heat loss from the door.
(1) Observations. (i) Heat is lost from the door to the surroundings by free convection and radiation. (ii) To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air and by radiation to a large surroundings. (iii) Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation.
(2) Problem Definition. Determine the average heat transfer coefficient for free convection from a vertical plate.
(3) Solution Plan. Apply Newton's law of cooling to the door. Use correlation equations to determine the average heat transfer coefficient. Apply Stefan-Boltzmann relation, equation (1.12), to determine the heat transfer by radiation
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent ambient fluid, (6) door is in the closed position at all times, (7) surroundings is at uniform temperature and (8) the door is small compared to the surroundings.
(ii) Analysis. Total heat transfer q is given by
rc qqq (a)
where
cq heat transfer by convection, W
rq heat transfer by radiation, W
Application of Newton's law of cooling to the surface gives
cq = h A ( sT - T ) (b)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
sT = surface temperature = 110oC = 283.15 K
T = ambient air temperature = 20oC
Surface area is given by A = LW (c)
where
L = door height = 50 cm = 0.5 m
gT
W
L
sT
PROBLEM 8.26 (continued)
W = door width = 120 cm = 1.2 m
The average heat transfer coefficient h is determined from correlation equations for free convection over a vertical plate. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL = PrLTTg s
2
3
(d)
where
g = gravitational acceleration = 9.81 m/s2
L = door side in the direction of gravity = 50 cm = 0.5 m Pr = Prandtl number RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf=
2
TTs = 2
)C)(20110( o
= 65oC
Air properties at this temperature are
k = 0.02887 W/m-oCPr = 0.7075
= 19.4 10-6 m2/s
For ideal gases, the coefficient of thermal expansion is given by
= 002957.015.273)C(65
1
)(
1
of KT
(1/K)
Substituting into (d) gives
RaL = 7075.0)/s(m)10(19.4
)(mC)(0.5)20)()(110/sC)9.81(m/0.002957(1
2426
33o2o
= 0.61349 109
Since RaL
< 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt
number is given by (8.25b)
4/14/1
2/1 953.4884.4435.2LL Ra
PrPr
Pr
k
LhuN (e)
Valid for vertical plate constant surface temperature Ts
laminar free convection 104 < RaL < 109
0 < Pr < properties at Tf (f)
where
PROBLEM 8.26 (continued)
LuN = average Nusselt number
Radiation heat transfer is given by equation (1.12)
)( 44sursr TTAq (g)
where
surT surroundings temperature = 24 Co = 24 + 273.15 = 297.15 K
surface emissivity = 0.08
Stefan-Boltzmann constant = 428 KW/m1067.5
(iii) Computations. Substitution into (e) gives
k
LhuN L
4/19
4/1
2/1)1061349.0(
)7075.0(953.4)7075.0(884.4435.2
7075.0 = 81.07
Solving the above for h
h = 84.02k
L = 81.07 (0.02887)(W/m-oC)/0.5(m) = 4.68 W/m2-oC
Equation (b) gives
cq = 4.68 (W/m2-oC) (0.5)(m)(1.2)(m) (110 )20 (oC) = 252.7 W
Equation (g) gives
)08.0(rq )KW/m(1067.5 428 (0.5)(m)(1.2)(m) )K()15.297()K()15.383( 4444 37.4 W
Substituting into (a)
q = 252.7 + 37.4 =290.1 W
(iii) Checking. Dimensional check: Computations showed that equations (b), (d), (e) and (g) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1.
Validity of correlation equation (d): The conditions listed in (f) are met.
(5) Comments. (i) Equation (8.26a) applies to both laminar and turbulent flow. However, it is
less accurate than (8.25b) or (e) above. Equation (8.26a) gives h = 6.37 W/m2-oC and q = 344.2 W. This is 31% higher than that found using the laminar free convection equation (e). (ii) Although radiation accounts for only 13% of the total heat added to the room, the contribution of radiation is minimized due to the low surface emissivity of the door. (iii) Opening and closing the door results in transient effects not accounted for in the above model. In addition, when the door is open radiation from the interior of oven may be significant.
PROBLEM 8.27
To compare the rate of heat transfer by radiation with that by free convection, consider the
following test case. A vertical plate measuring 12 cm 12 cm is maintained at a uniform
surface temperature of 125oC. The ambient air and the surroundings are at 25
oC. Compare the
two modes of heat transfer for surface emissivities of 0.2 and 0.9.
(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a vertical plate. (iii) Surface temperature is uniform. (iv) Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate. (v) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (vi) Since radiation heat transfer is considered in this problem, all temperatures should be expressed in degrees kelvin. (vii) The fluid is air.
(2) Problem Definition. Determine heat transfer rate by free convection and by radiation from a vertical plate in air.
(3) Solution Plan. Apply Newton's law of cooling to determine the rate of heat loss by convection. Apply Stefan-Boltzmann radiation law to determine the rate of heat loss by radiation. Compute the Rayleigh number and select an appropriate correlation equations to obtain the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) the plate is a small surface enclosed by a much larger surface at a uniform temperature and (6) quiescent ambient.
(ii) Analysis. Application of Newton's law of cooling to the vertical plate gives
qc = h A (Ts - T ) (a) where
A = surface area of vertical side, m2
h = average heat transfer coefficient, W/m2-oC or W/m2-K
qc = convection heat transfer rate, W Ts = surface temperature = 125(oC) + 273.15 = 398.15 K
T = ambient temperature = 25(oC) + 273.13 = 298.15 K
surface area is A = L2 (b)where
L = side of square plate = 12 cm = 0.12 m
The average heat transfer coefficient h is determined from correlation equations for free convection over vertical plates. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL =g T T Ls
3
2 Pr (c)
where
g
T
L
sTL
qrqc
surroundings at Tsur
PROBLEM 8.27 (continued)
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number Ra
L = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
( . . )( )29815 39815
2
K = 348.15K
Appendix C gives air properties at Tf = 348.15(K) - 273.15 = 75oC
k = 0.02957 W/m-oCPr = 0.7065
= 20.41 10-6 m2/s
For ideal gases the coefficient of thermal expansion is given by
= 002872.015.348
1
)K(
1
fT(1/K)
Substituting into (c) gives
RaL
= 7065.0)s/m()1041.20(
)m()12.0)(C)(25125)(s/m(81.9)C/1(002872.02426
33o2o
= 8.257 610
Since RaL
< 109, the flow is laminar. Thus the appropriate equation for the average Nusselt
number is given by equation (8.25b)
4/14/1
2/1 953.4884.4435.2LL Ra
PrPr
Pr
k
LhuN (d)
Valid for vertical plate constant surface temperature Ts
laminar free convection 104 < RaL < 109
0 < Pr < properties at Tf (e)
where
NuL = average Nusselt number
Radiation heat loss qr is given by the Stefan-Boltzmann law. Assuming that the plate is a small surface which is surrounded by a much larger surface, qr is given by equation (1.15)
qr = A ( 44surs TT ) (f)
where
qr = radiation heat loss, W Tsur = surroundings temperature = 25(oC) + 273.13 = 298.15 K
PROBLEM 8.27 (continued)
= emissivity
= Stefan-Boltzmann constant = 5.67 10-8 W/m2-K4
(iii) Computations. Convection heat loss. Substitution into (d) gives
k
LhuN L
4/16
4/1
2/1)10257.8(
)7065.0(953.4)7065.0(884.4435.2
7065.0 = 27.6
Solving the above for h
h = 27.6 k
L = 27.6(0.02957)(W/m-oC) / 0.12(m) = 6.8 W/m2-oC
Substituting into (a) and using (b) gives the rate of heat loss by convection
qc = 6.8 (W/m2-oC) (0.12)(m)(0.12)(m) (125 -25)(oC) = 9.79 W
Radiation heat loss. Equation (f) is used to determine qr. For = 0.2:
qr = 0.2 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 39815 298154 4. . )(K4) = 2.81 W
For = 0.9:
qr = 0.9 5.67 10-8 (W/m2-K4) 0.12(m) 0.12(m) ( 39815 298154 4. . )(K4) = 12.66 W
(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) and (f) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of gases.
Limiting check: For Ts = T = Tsur, both convection and radiation vanish. Setting Ts = T = Tsur
in (a) and (f) gives qc = qr = 0.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. (i) When compared with free convection, radiation heat loss can be significant
and in general should not be neglected. (ii) The magnitude of is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat. (iii) Because temperature in the Stefan-Boltzmann radiation law must be expressed in degrees kelvin, care should be exercised in using the correct units when carrying radiation computations.
PROBLEM 8.28
A sealed electronic package is designed to be cooled by free convection.
The package consists of components which are mounted on the inside
surfaces of two cover plates measuring 10 cm 10 cm each. Because the
plates are made of high conductivity material, surface temperature may
be assumed uniform. The maximum allowable surface temperature is
70oC. Determine the maximum power that can be dissipated in the
package without violating design constraints. Ambient air temperature is 20
oC. Neglect radiation.
(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) The fluid is air.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A (Ts - T ) (a) where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oCP = power dissipated in package, W q = heat transfer from the surface to the ambient air, W T
s= surface temperature = 70oC
T = ambient air temperature = 20oC
Surface area of the two vertical sides is given by
A = 2 LW (b)where
L = package height = 10 cm = 0.1 m
gT
components
air
g
T
W
L
sT
PROBLEM 8.28 (continued)
W = package width = 10 cm = 0.1 m
The average heat transfer coefficient h is determined from correlation equations for free convection over vertical plates. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL=g T T Ls
3
2 Pr (c)
where
g = gravitational acceleration = 9.81 m/s2
L = plate dimension in the direction of gravity = 10 cm = 0.1 m Pr = Prandtl number RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
( )( )70 20
2
oC = 45oC
Appendix C gives air properties at this temperature
k = 0.02746 W/m-oCPr = 0.7095
= 17.44 10-6 m2/s
For ideal gases the coefficient of thermal expansion is given by
= 003143.015.273)C(45
1
)K(
1o
fT(1/K)
Substituting into (c) gives
RaL = 7095.0)s/m()1044.17(
)m()1.0)(C)(2070)(s/m(81.9)C/1(003143.02426
33o2o
= 3.5962 610
Since RaL < 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt number is (8.25b)
4/14/1
2/1 953.4884.4435.2LL Ra
PrPr
Pr
k
LhuN (d)
Valid for vertical plate constant surface temperature Ts
laminar free convection 104 < RaL < 109
0 < Pr < properties at Tf (e)
where
PROBLEM 8.28 (continued)
LNu = average Nusselt number
(iii) Computations. Substitution into (d) gives
k
LhuN
L
4/16
4/1
2/1105962.3
)7095.0(953.4)7095.0(884.4435.2
7095.0 = 22.44
Solving the above for h
h = 22.44k
L = 22.44(0.02746)(W/m-oC)/0.1(m) = 6.16 W/m2-oC
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 6.16 (W/m2-oC)2(0.1)(m)(0.1)(m) (70 - 20)(oC) = 6.16W
(iv) Checking. Dimensional check: Computations showed that equations (a)-(d) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1 for free convection of gases.
Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum
power P. According to equation (a), q is directly proportional to Ts. Furthermore, h increases when Ts is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting radiation and heat loss from the top and bottom surfaces. (ii) The maximum power dissipated is relatively small, indicating the limitation of free convection in air as a cooling mode for such
applications. (iii) The magnitude of is the same whether it is expressed in units of degree
Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.29
Assume that the electronic package of Problem 8.28 is to be used in an undersea application.
Determine the maximum power that can be dissipated if the ambient water temperature is 10oC.
(1) Observations. (i) This is a free convection problem. (ii) The power dissipated in the electronic package is transferred to the ambient fluid by free convection. (iii) As the power is increased, surface temperature increases. (iv) The maximum power dissipated corresponds to the maximum allowable surface temperature. (v) Surface temperature is related to surface heat transfer by Newton’s law of cooling. (vi) The problem can be modeled as free convection over a vertical plate. (vii) The Rayleigh number should be computed to determine if the flow is laminar or turbulent. (viii) The fluid is water.
(2) Problem Definition. Since dissipated power is related to heat loss from the surface to the surroundings, Newton's law of cooling should be applied. Thus, the problem becomes finding the average heat transfer coefficient.
(3) Solution Plan. Apply Newton's law of cooling to the surface of the package. Check the Rayleigh number to determine if the flow is laminar or turbulent. Use appropriate correlation equations to determine the average heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible heat loss from the top and bottom surfaces, (6) the sides are vertical flat plates, (7) negligible radiation and (8) quiescent fluid.
(ii) Analysis. Application of Newton's law of cooling to the surface gives
P = q = h A (Ts - T ) (a) where
A = surface area of the two vertical sides, m2
h = average heat transfer coefficient, W/m2-oCP = power dissipated in package, W q = heat transfer from the surface to the water, W T
s= surface temperature = 70oC
T = ambient water temperature = 10oC
Surface area of the two vertical sides is given by
A = 2 LW (b)where
L = package height = 10 cm = 0.1 m W = package width = 10 cm = 0.1 m
gT
components
water
g
T
W
L
sT
PROBLEM 8.29 (continued)
The average heat transfer coefficient h is determined from correlation equations for free convection over vertical plates. The Rayleigh number RaL is calculated first to determine the
appropriate correlation equation for h . The Rayleigh number is defined as
RaL=g T T Ls
3
2 Pr (c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf=
T Ts
2 =
2
)C)(1070( o
= 40oC
Appendix C gives air properties at this temperature
k = 0.6286 W/m-oCPr = 4.34
= 0.000389 1/K
= 0.6582 10-6 m2/s
Substituting into (c) gives
RaL = 34.4)s/m()106582.0(
)m()1.0)(C)(1070)(s/m(81.9)C/1(000389.02426
33o2o
= 2.294 910
Since RaL > 109, the flow is turbulent. Thus, the appropriate equation for the average Nusselt number is (8.26a)
k
LhuN L
2
8/279/16
6/1
/0.4921
0.3870.825
Pr
RaL (d)
Valid for vertical plate uniform surface temperature Ts
laminar, transition, and turbulent
10 1 < RaL <1012
0 < Pr < properties at Tf (e)
where
LNu = average Nusselt number
PROBLEM 8.29 (continued)
(iii) Computations. Substitution into (d) gives
k
LhuN L
2
8/279/16
6/19
34.4/0.4921
]1040.387[2.290.825 = 191.7
Solving the above for h
h = 191.7L
k = 191.7(0.6286)(W/m-oC)/0.1(m) = 1205 W/m2-oC
Equations (a) and (b) give the maximum power dissipated from the two sides of the package
P = q = 1205(W/m2-oC)2(0.1)(m)(0.1)(m) (70 - 10)(oC) = 1446 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c) and (d) are dimensionally consistent.
Quantitative check: The magnitude of h is within the approximate range given in Table 1.1 for free convection of liquids.
Qualitative check: Increasing the allowable surface temperature Ts should increase the maximum
power P. According to equation (a), q is directly proportional to Ts. Furthermore, h increases when Ts is increased.
Limiting check: The dissipated power should vanish (P = 0) if heat cannot be removed by free
convection (q = 0). Setting h = 0 in (a) gives P = q = 0.
Limitations on correlation equation (d): The conditions listed in (e) are met.
(5) Comments. (i) The model used to solve this problem is conservative due to both neglecting heat loss from the top and bottom surfaces. (ii) The maximum power dissipated is relatively
large, indicating the effectiveness of water as a free convection medium. (iii) The magnitude of
is the same whether it is expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.30
A plate 20 cm high and 25 cm wide is placed vertically in water at 29.4oC. The plate is
maintained at 70.6oC. Determine the free convection heat transfer rate from each half.
(1) Observations. (i) This is a free convection problem. (ii) The surface is maintained at uniform temperature. (iii) The heat transfer coefficient decreases with distance from the leading edge of the plate. (iv) The heat transfer rate from the lower half 1 is greater than that from the upper half 2. (v) Total heat transfer from each half can be determined using the average heat transfer coefficient. (vi) Heat transfer from the upper half is equal to the heat transfer from the entire plate minus heat transfer from the lower half.
(2) Problem Definition. Determine the heat transfer rate from each vertical rectangle.
(3) Solution Plan. Since the heat transfer coefficient changes with distance, the average heat transfer coefficient should be used in Newton's law of cooling. The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible radiation and (6) quiescent fluid.
(ii) Analysis. Application of Newton's law to rectangle 1
2/)(11 HwTThq s (a)
where
1h = average heat transfer coefficient for rectangle 1, CW/m o2
H = plate height = 20 cm = 0.2 m
1q = heat transfer rate from rectangle 1, W
sT = surface temperature = 70.6 Co
T = ambient temperature = 29.4oC
w = plate width = 25 cm = 0.25 m
The heat transfer rate from rectangle 2 is given by
1212 qqq (b)
where
2q = heat transfer rate from rectangle 2, W
21q = total heat transfer rate from rectangles 1 and 2, W
Application of Newton’s law of cooling to the entire plate gives 21q
HwTThq s )(2121 (c)
where
21h = average heat transfer coefficient for rectangles 1 and 2, CW/m o2
g
w
H
0
x1
2T
PROBLEM 8.30 (continued)
To determine the average heat transfer coefficient for free convection over a vertical plate, the Rayleigh number is computed to establish if the flow is laminar or turbulent. The Rayleigh number for rectangle 1 is defined as
PrHTTg
Ra s2
3
1
)2/)(( (d)
where
g = gravitational acceleration = 9.81 2m/s
Pr = Prandtl number
1Ra = Rayleigh number for rectangle 1
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature fT defined as
2/)( TTT sf (e)
Substituting into (e)
502/)4.296.70(fT Co
Properties of water at this temperature are
k = 0.6405 CW/m o
Pr = 3.57 310462.0 1/K 6105537.0 /sm2
Substituting into (d)
9
2426
33o23
1 10174.257.3)/sm()105537.0(
)m()2/2.0)(C)(4.296.70()m/s(81.9)K/1(10462.0Ra
Since the Rayleigh number is greater than 910 , it follows that the flow turbulent. Similarly, the
Rayleigh number for the entire plate, 21Ra is
PrHTTg
Ra s2
3
21
)( (f)
Substituting into (f)
9
2426
33o23
21 103947.1757.3)/sm()105537.0(
)m()2.0)(C)(4.296.70()m/s(81.9)K/1(10462.0Ra
Thus the flow is turbulent over the entire upper half of the plate. The applicable correlation equation for the average Nusselt number for a vertical plate of height H/2 is
2
27/816/9
6/111
2/
)/492.0(1
387.0825.0
)2/(
Pr
Ra
k
HhNuH (g)
PROBLEM 8.30 (continued)
Valid for vertical plate
constant surface temperature Ts
laminar, transition, and turbulent 12
2/1 1010 HRa
0 < Pr < properties at Tf
Similarly, the average Nusselt number for the entire plate is
2
27/816/9
6/12121
)/492.0(1
387.0825.0
Pr
Ra
k
HhNu H (h)
(iii) Computations. Substituting into (g)
81.185)57.3/492.0(1
10174.2387.0825.0
)2/(2
27/816/9
6/191
2/k
HhNu H
Solving for 1h
CW/m1190m))(2/2.0(
)CW/m(6405.081.185
2/81.185 o2
o
1H
kh
Substituting into (a)
2/m)(0.2)(m)(25.0)C()4.296.70)(CW/m(1190 oo21q = 1225.7 W
Similarly, (h) gives the average heat transfer coefficient for the entire plate
58.358)57.3/492.0(1
103947.17387.0825.0
2
27/816/9
6/1921
k
HhNu H
Solving for 21h
CW/m3.1148m)(2.0
)CW/m(6405.058.35858.358 o2
o
21H
kh
Substituting into (c)
W5.3652.2(m)0m)(25.0C)()4.296.70)(CW/m(3.1148 oo221q
Substituting into (b) gives the heat transfer rate from rectangle 2
W8.1139)W(7.1225)W(5.23652q
(iii) Checking. Dimensional check: Computations showed that units of equations (a)-(d), (g)
and (h) are dimensionally consistent.
Limiting check: If TTs , no free convection takes place and consequently 0211 qq .
PROBLEM 8.30 (continued)
Setting TTs in (a) and (c) gives the anticipate result.
Qualitative check: The heat transfer rate from the leading rectangle 1 should be higher than that from the trailing rectangle 2. This is in agreement with the results obtained.
(5) Comments. (i) More heat is transferred from rectangle 1 than rectangle 2. This was predicted prior to solving the problem analytically. However, the difference between the two heat transfer rates is not very significant. (ii) The average heat transfer coefficients are slightly outside the range given in Table 1.1 for free convection in liquids. It should be remembered that values listed in Table 1.1 are for typical applications. Exceptions should be expected.
PROBLEM 8.31
Consider laminar free convection from a vertical plate at uniform surface temperature. Two 45
triangles are drawn on the plate as shown.
[a] Explain why free convection heat transfer from triangle 1 is greater than that from the
triangle 2. [b] Determine the ratio of the heat transfer from two triangles.
(1) Observations. (i) This is a free convection problem. (ii) The surface is maintained at uniform temperature. (iii) The heat transfer coefficient decreases with distance from the leading edge of the plate. (iv) The width of each triangle changes with distance from the leading edge.
(2) Problem Definition. Examine the variation of local heat transfer coefficient with distance and determine the heat transfer rate from each triangle.
(3) Solution Plan.
[a] Formulate an equation for )(xh for laminar free convection over a flat plate.
[b] Since the heat transfer coefficient and area change with distance, Newton's law of cooling should be applied to an infinitesimal area of each triangle. Integration over the area gives the total heat transfer rate.
(4) Plan Execution.
(i) Assumptions. (1) Laminar flow, (2) steady state, (3) two-dimensional, (4) constant properties (except in buoyancy), (5) uniform surface temperature, (6) quiescent fluid and (7) no radiation.
(ii) Analysis.
[a] The heat transfer coefficient for laminar free convection over a vertical plate is given by equation (8.25a)
4/14/1
2/1 953.4884.4435.24
3xx Ra
PrPr
Pr
k
hxNu (a)
and
Rax = Pr)(
2
3xTTg s (b)
where
g = gravitational acceleration, m/s2
h = local heat transfer coefficient, W/m2-oCk = thermal conductivity, W/m-oCPr = Prandtl number Rax = local Rayleigh number x = vertical distance from the leading edge of plate, m
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Valid for
g
T1
2
PROBLEM 8.31 (continued)
vertical plate constant surface temperature Ts
laminar free convection 104 < Rax < 109
0 < Pr < properties at Tf (c)
Substituting (b) into (a) and solving for h
h(x) = C x-1/4 (d)
where the constant C is defined as
C =
4/1
22/1
2
953.4884.4435.24
3
PrPr
TTgPrk s (e)
According to (d), )(xh decreases with distance along the plate x.
It is infinite at .0x Triangle 1 has its wide base starting at
0x while triangle 2 has its apex at 0x . Since heat transfer
rate is proportional to the product of heat transfer coefficient and area, it follows that triangle 1 transfers more heat than triangle 2.
[b] Application of Newton's law to an infinitesimal area gives
dq = (Ts - T )h(x) dA (f) where
A = surface area, m2
q = heat transfer rate, W T
s= surface temperature, oC
T = ambient temperature, oC
The area dA for each triangle is dA1 = b1(x) dx (g)
anddA2 = b2 (x) dx (h)
Similarity of triangles give b1(x) and b2(x)
b1(x) = B(1 - x
H) (i)
and
b2(x) = Bx
H (j)
where
B = base of triangle, m H = height of triangle, m
Substituting (i) into (g) and (j) into (h)
dA1 = B(1 - x
H)dx (k)
and
x
dx
dx
b1(x)
b2(x)
g
T
12
H
B
PROBLEM 8.31 (continued)
dA2 = Bx
H dx (m)
Applying (f) to triangle 1, substituting (d) and (k) into (f) and integrating from x = 0 to x = Hgives the total heat transfer from triangle 1
q1 = dxxHxTTBCdqH
s
H
0
4/1
01 /1 = (16/21) B C (Ts -T ) 4/3H (n)
Similarly, applying (f) to triangle 2, substituting (d) and (m) into (f) and integrating from x = 0 to x = H gives the total heat transfer, q2, from triangle 2
q2 = dxxHxTTBCdqH
s
H
0
4/1
02 / = (4/7) B C (Ts - T ) 4/3H (o)
Taking the ratio of (n) and (o) q
q
1
2
= 4
3
(iii) Checking. Dimensional check: To check the units of (d), units of C in equation (e) is determined first
C =
4/1
2422/1
os
2o
o )s/m(555.2(
)C)()(s/m(g)C/1(
Cm
W
Pr)Pr
TTPrk
2
= W/m7/4-oC
Thus units of h(x) in (d) are
h(x) = C (W/m7/4-oC) x-1/4(m)-1/4 = W/m2-oC
Examining units of q1 in (n)
q1 = B(m) C(W/m7/4-oC) (Ts -T ) (oC) 4/3H (m3/4) = W
Limiting check: If = 0 or g = 0 or Ts = T , no free convection takes place and consequently q1
= q2 = 0. Any of these limiting cases give C = 0. Thus, according to (n) and (o), q1 = q2 = 0.
(5) Comments. (i) More heat is transferred from triangle 1 than triangle 2. This was predicted prior to solving the problem analytically. (ii) The result applies to any right angle triangles and is not limited to 45o triangles. (iii) Heat transfer from a surface of fixed area depends on its orientation relative to the leading edge. (iv) This problem illustrates how integration is used to account for variations in element area and heat transfer coefficient. The same approach can be applied if surface temperature and/or ambient temperature vary over a surface area.
PROBLEM 8.32
A vertical plate measuring 21 cm 21 cm is at a uniform surface temperature of 80oC. The
ambient air temperature is 25oC. Determine the free convection heat flux at 1 cm, 10 cm and 20
cm from the lower edge.
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The surface is maintained at uniform temperature. (iii) Local heat flux is determined by Newton’s law of cooling. (iv) Heat flux depends on the local heat transfer coefficient. (v) Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases. (vi) The Rayleigh number should be computed to select an appropriate Nusselt number correlation equation. (vii) The fluid is air.
(2) Problem Definition. Determine the local heat transfer coefficient for free convection over a vertical plate at uniform surface temperature.
(3) Solution Plan. Apply Newton’s law of cooling, compute the Rayleigh number and select an appropriate Nusselt number correlation equation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Application of Newton's law gives
q = h(x) (Ts - T ) (a)
where
h(x) = local heat transfer coefficient, W/m2-oCq = local heat flux, W
Ts
= surface temperature = 80oC
T = ambient temperature = 25oC
The local heat transfer coefficient is determined from correlation equations for free convection over a vertical plate. The Rayleigh number Rax is calculated to select an appropriate correlation equation for h. The Rayleigh number is defined as
Rax = PrxTTg s
2
3
(b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number Rax = local Rayleigh number x = distance from leading edge, m
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf= 2/)( TTs = 2/)C)(2580( o = 52.5oC
Appendix C gives air properties at this temperature
g
T
W
L
sT
x
PROBLEM 8.32 (continued)
k = 0.02799 W/m-oCPr = 0.709
= 18.165 10-6 m2/s
For ideal gases the coefficient of thermal expansion is given by
= 003071.015.273)C(5.52
1
)(
1
of KT
(1/K)
Substituting into (b) to evaluate the Rayleigh number at the trailing end x = L = 20 cm = 0.2 m
RaL = 709.0)/()10165.18(
)()2.0)()(2580)(/(81.9)/1(003071.02426
332
sm
mCsmC oo
= 2.8482 107
Since RaL
< 109, the flow is laminar over the region of interest. Thus, the appropriate equation
for the local Nusselt number xNu is given by (8.25a)
4/14/1
2/1 953.4884.4435.24
3xx Ra
PrPr
Pr
k
hxNu (c)
Valid for vertical plate constant surface temperature Ts
laminar free convection 104 < Rax < 109
0 < Pr < properties at Tf (d)
(iii) Computations. To evaluate the flux at x = L = 20 cm = 0.2 m, the heat transfer coefficient at this location is computed using (c)
k
hLNuL
4/17
4/1
2/1108482.2
)709.0(953.4)709.0(884.4435.2
709.0
4
3 = 28.2
Solving the above for h(L)
h(L)= 28.2 k
L = 28.2(0.02799)(W/m-oC)/0.2(m) = 3.95 W/m2-oC
Substituting into (a)
q = 3.95(W/m2-oC)(80 25 )(oC) = 217.3 W/m2
The same procedure is followed to determine the flux at x = 1 cm and x = 10 cm. Results for the three locations are tabulated.
x (cm) Rax Nux h(x)(W/m2-oC) q (W/m2)
1 0.356 104 2.98 8.34 458.7
10 0.356 107 16.78 4.7 258.3
20 2.8482 107 28.2 3.95 217.3
PROBLEM 8.32 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (c) are dimensionally consistent.
Quantitative check: The magnitudes of h are approximately within the range given in Table 1.1 for free convection of gases.
Limiting check: The flux should vanish for Ts = T . Setting Ts = T in (a) gives 0q .
Limitations on correlation equation (c): The conditions listed in (d) are met except at x = 1 cm where the Rayleigh number is below the lower limit. The accuracy of the computed flux at the location is in doubt.
Validity of correlation equation (c): The conditions listed in (d) are met.
(5) Comments. (i) This problem illustrates the importance of verifying the applicability of correlation equations to specific cases. It should be noted that correlation equations usually do not suddenly break down outside the limits of their applicability. Instead, their accuracy begins to deteriorate. (ii) The heat transfer literature should be consulted for applicable correlation
equations whenever the need arises. (iii) The magnitude of is the same whether it is expressed
in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.33
200 square chips measuring 1 cm 1 cm each are mounted on both sides of
a thin vertical board measuring 10 cm 10 cm. The chips dissipate 0.035
W each. Assume uniform surface heat flux. Determine the maximum
surface temperature in air at 22oC. Neglect heat exchange by radiation.
(1) Observations. (i) This is a free convection problem over a vertical plate. (ii) The power dissipated in the chips is transferred to the air by free convection. (iii) This problem can be modeled as free convection over a vertical plate with constant surface heat flux. (iv) Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end). (v) Newton’s law of cooling relates surface temperature to heat flux and heat transfer coefficient. (vi) The fluid is air.
(2) Problem Definition. Determine surface temperature distribution for a vertical plate with uniform surface heat flux under free convection conditions.
(3) Solution Plan. Apply the analysis of surface temperature distribution of a vertical plate with uniform surface heat flux in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface heat flux, (5) all dissipated power leaves surface as heat, (6) negligible radiation and (7) quiescent ambient fluid.
(ii) Analysis. Surface temperature distribution for a vertical plate with uniform surface flux is given by equation (8.29a)
T x TPr Pr
Pr g
q
kxs
s4 9 101 2 4 1 5/
/
(a)
Valid for: vertical plate uniform surface flux qs
laminar, 104 < RaL < 109
0 < Pr < bwhere
g = gravitational acceleration = 9.81 m/s2
k = thermal conductivity, W/m-oCPr = Prandtl number
sq = surface flux, W/m2
RaL = Rayleigh number at x = LTs = surface temperature, oC
T = ambient temperature = 22oCx = distance from leading edge, m
= thermal diffusivity, W/m2-s
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
g
T
PROBLEM 8.33 (continued)
The Rayleigh number is defined as
RaL = PrLTTg s
2
3
(c)
where
L = vertical side of plate = 10 cm = 0.1 m
If all dissipated power in the chip leaves the surface, conservation of energy gives
sq = P/A (d)
where
A = chip surface area = 1 cm2 = 0.0001 m2
P = power dissipated in chip = 0.035 W
Properties are evaluate at the film temperature defined as
2
)2/( TLTT s
f (e)
where Ts(L/2) is surface temperature at the mid-point. However, since Ts(x)is unknown, an iterative procedure is required to obtain a solution. An assumed value for Ts(L/2) is used to calculate the film temperature at which properties are determined. Equation (a) is then used to calculate Ts(L/2). If the calculated value does not agree with the assumed temperature, the procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations. Equation (d) gives surface flux
sq = 0.035(W)/0.0001(m2) = 350 W/m2
Assume Ts (L/2) = 58oC. Equation (e) gives
Tf = (58 + 22)(oC)/2 = 40oC
Properties of air at this temperature are
cp = 1006.8 J/kg- oCk = 0.0271 W/m-oCPr = 0.71
= 16.96 10-6 m2/s
= 1.1273 kg/m3
Coefficient of thermal expansion for an ideal gas is given by
= 003193.015.273)C(40
1
15.273
1o
fT1/K
Thermal diffusivity is defined as
pc
ks/m10877.23
)Ckg/J(8.1006)m/kg(1273.1
)Cm/W(0271.0 26
o3
o
Substituting into (a) and letting x = L/2 = 0.1(m)/2 = 0.05 m
5/14
o
2
2o
26262/1o )m(05.0
)Cm/W(0271.0
)m/W(350
)s/m(81.9)C/1(003193.0
)s/m(1096.16)s/m(10877.23
71.0
)71.0(10)71.0(94)C(22/2LTs
PROBLEM 8.33 (continued)
Ts(L/2) = 76.3oC
Since this is higher than the assumed value of 58oC, the procedure is repeated with a new assumed temperature at mid-point. Assume Ts(L/2) = 78oC. The following results are obtained
Tf = 50oCcp = 1007.4 J/kg- oCk = 0.02781 W/m-oCPr = 0.709
= 25.27 10-6 m2/s
= 0.0030945 1/K
= 17.92 10-6 m2/s
= 1.0924 kg/m3
Substituting into (a) gives Ts(L/2) = 76.8oC. This is close to the assumed value of 78oC.
Surface temperature at the trailing end is now computed by evaluating (a) at x = L = 0.1 m
5/14
o
2
2o
26262/1o )m(1.0
)Cm/W(02781.0
)m/W(350
)s/m(81.9)C/1(0030945.0
)s/m(1092.17)s/m(1027.25
709.0
)709.0(10)709.0(94)C(22/2LTs
Ts(L) = 84.9oC
The condition on the Rayleigh number in equation (b) is verified next. Substituting into (c)
RaL = 709.0)s/m()1092.17(
)m()1.0)(C)(229.84)(s/m(81.9)C/1(0030945.02426
33o2o
= 4.22 106
This satisfies the condition on RaL given in equation (b).
(iv) Checking. Dimensional check: Equations (a), (c) and (d) are dimensionally consistent..
Quantitative check: The heat transfer coefficient at the mid-point, h(L/2), can be calculated using Newton's law of cooling:
h(L/2) = ])2/(/[ TLTq ss = 350(W/m2)/(76.8 - 22)(oC) = 6.39 W/m2-oC
This is within the range given in Table 1.1 for free convection of gases.
Validity of correlation equation (a): The conditions listed in (b) are met.
(5) Comments. (i) Surface temperature is determined without calculating the heat transfer coefficient. This is possible because equation (a) combines the correlation equation for the heat transfer coefficient and Newton's law of cooling to eliminate h and obtain an equation for surface
temperature in terms of surface heat flux. (ii) The magnitude of is the same whether it is
expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.34
An apparatus is designed to determine surface emissivity of
materials. The apparatus consists of an electrically heated
cylindrical sample (disk) of diameter D and thickness . the
disk is insulated along its heated side and rim. It is placed
horizontally with its heated surface facing down in a large
chamber whose surface is maintained at uniform temperature
.surT The sample is cooled by free convection and radiation
from its upper surface. To determine the emissivity of a
sample, measurements are made of the diameter D, electric
power input P, surface temperature sT , surroundings
temperature surT and ambient temperature .T Determine
the emissivity of a sample using the following data:
D = 12 cm, 0.5 cm, P 13.2 W, C98osT , C27o
surT , C22oT
(1) Observations. (i) Power supply to the disk is lost from the surface to the surroundings by free convection and radiation. (ii) To determine the rate of heat loss, the disk can by modeled as a horizontal plate losing heat by free convection to an ambient air and by radiation to a large surroundings. (iii) Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation. (iv) Free convection correlations give the heat transfer coefficient. (v) Conservation of energy at the surface gives the emissivity, if it is the only unknown.
(2) Problem Definition. Determine the average heat transfer coefficient for free convection from a vertical plate.
(3) Solution Plan. Apply conservation of energy at the exposed surface. Use Newton's law of cooling Stefan-Boltzmann relation, equation. Use correlation equations to determine the average heat transfer coefficient for free convection from a horizontal surface.
(4) Plan Execution.
(i) Assumptions. (1) steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent ambient fluid, (5) the heated surface and the rim are perfectly insulated, (6) surroundings is at uniform temperature and (7) the disk is small compared to the surroundings.
(ii) Analysis. Conservation of energy at the surface gives
rc qqP (a)
where
P = electric energy power supplied to the disk = 13.2 W
cq heat transfer by convection, W
rq heat transfer by radiation, W
Application of Newton's law of cooling to the surface gives
cq = h A ( sT - T ) (b)
+ -
D
T
surT
sTg
PROBLEM 8.34 (continued)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
sT = surface temperature = 98oC = 371.15 K
T = ambient air temperature = 22oC = 295.15 K
Surface area is given by
2
4DA (c)
where
D = disk diameter = 12 cm = 0.12 m
Radiation heat transfer is given by equation (1.12)
)( 44sursr TTAq (d)
where
surT surroundings temperature = 27 + 273.15 = 300.15 K
surface emissivity
Stefan-Boltzmann constant = 428 KW/m1067.5
(c) and (d) into (a)
P = h A ( sT - T ) + )( 44surs TTA (e)
Solving (e) for
)(
)()/(44
surs
s
TT
TThAP (f)
The average heat transfer coefficient h for a heated plate facing up is determined from (8.29)
4/154.0 LL Rak
LhuN for 75 10210 LRa (8.29a)
3/114.0 LL Rak
LhuN for 107 103102 LRa (8.29b)
(8.29c)
The Rayleigh number RaL is defined as
RaL = PrLTTg s
2
3
(g)
where
g = gravitational acceleration = 9.81 m/s2
L = characteristic length for the horizontal disk, m
horizontal plate hot surface up or cold surface down
properties, except , at fT
at fT for liquids, T for gases
PROBLEM 8.34 (continued)
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
The length L is defined as
4
4/2D
D
D
p
AL (i)
Properties of air are determined at the film temperature Tf defined as
Tf=
2
TTs = 2
)C)(2298( o
= 60oC
at this temperature air properties are
k = 0.02852 W/m-oCPr = 0.708
= 18.9 10-6 m2/s
For ideal gases, the coefficient of thermal expansion for a horizontal surface is given by
= 0033881.015.273)C(22
1
)(
1
oKT(1/K)
Equation (i) gives L
L = 0.12 (m)/4 =0.03 m
Substituting into (g)
RaL = 708.0)/s(m)10(18.9
)(mC)(0.03)22)()(98/sC)9.81(m1/0.0033881(2426
33o2o
= 1.3518 510
Thus equation (8.29a) is applicable. Radiation heat transfer is given by equation (1.12)
)( 44sursr TTAq (j)
where
surT surroundings temperature = 27 Co = 27 + 273.15 = 300.15 K
Stefan-Boltzmann constant = 428 KW/m1067.5
(iii) Computations. Substitution into (8.29a) gives
k
LhuN L
4/15 )103518.1(54.0 = 10.35
Solving the above for h
h = 10.35k
L = 10.35 (0.02852)(W/m-oC)/0.03(m) = 9.84 W/m2-oC
Surface area is given by (c)
PROBLEM 8.34 (continued)
)m()12.0(4
22A = 0.01131 2m
Equation (f) gives
681.0)](K(300.15)[(371.15)105.67
295.15)C)(371.159.84(W/m).01131(m[13.2(W)/04448
o22
(iii) Checking. Dimensional check: Computations showed that equations (c), (f) and (g) are dimensionally consistent.
Quantitative check: The magnitude of h is within the range given in Table 1.1.
Validity of correlation equation (8.29c): The conditions listed in (8.29c) are met.
(5) Comments. (i) Disk thickness is not needed for the determination of . (ii) It is important
to provide good insulation of the heated surface and the rim. (iii) According to equation (f), increasing the operating surface temperature minimizes the error in the measurement of the surroundings temperature.
PROBLEM 8.35
It is desired to increase the heat by free convection from a wide vertical plate without increasing
its surface temperature. Increasing the height of the plate is ruled out because of the limited
vertical space available. It is suggested that a taller plate can be accommodated in the same
vertical space by tilting it 45o. Explore this suggestion and make appropriate recommendations.
Assume laminar flow.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a flat plate. (iii) Heat transfer from two plates is to be compared. One plate is vertical and the other is inclined. Both plates fit in the same vertical space. Thus, the inclined plate is longer than the vertical plate. (iv) Both plates are maintained at uniform surface temperature. (v) Heat transfer depends on surface area and average heat transfer coefficient.
(2) Problem Definition. Determine the average heat transfer coefficient for a vertical plate and an inclined plate under free convection conditions.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficient for each orientation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) negligible radiation and (6) quiescent fluid.
(ii) Analysis. Newton’s law of cooling gives
q = h A (Ts - T ) = h LS(Ts - T ) (a) where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oCL = plate length, m q = heat transfer rate, W S = plate width, m Ts = surface temperature, oC
T = ambient temperature, oC
Applying (a) to the two plates and using the subscripts i and v to refer to the inclined and vertical
plates, respectively
qi = hi iL S (Ts - T ) (b)
qv= vh vL S (Ts - T ) (c)
Taking the ratio of (e) and (f)
vq
qi =vv Lh
Lh ii (d)
The average Nusselt number for laminar free convection flow over a vertical plate of height vL
is
PROBLEM 8.35 (continued)
4/14/1
2/1)(
953.4884.4435.2 v
vvv LL Ra
PrPr
Pr
k
LhNu (e)
where
PrLTTg
Ras
L 2
3)(v
v (f)
where
g = gravitational acceleration, m/s2
vh = local heat transfer coefficient for vertical plate, W/m2-oC
k = thermal conductivity, W/m-oCPr = Prandtl number
vLRa = Rayleigh number for vertical plate
vL = length of vertical plate, m
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Valid for vertical plate constant surface temperature Ts
laminar free convection 104 < Rax < 109
0 < Pr < properties at Tf (g)
Solving (a) and (b) for vh4/1
v
vL
gCh (h)
where C is defined as4/1
22/1
2
)953.4884.4435.2(
)(
PrPr
TTPrkC s (i)
Applying (h) to the inclined plate4/1
cos
i
iL
gCh (j)
Substitute (h) and (j) into (d)
vq
qi = 4/1
4/3
)(cosvL
Li (k)
However, the length of the inclined plate is given by
Li = cos
vL (m)
Combining (k) and (m)
PROBLEM 8.35 (continued)
vq
qi =2/1)(cos
1(n)
(iii) Computations. For a plate inclined at 45o equation (m) gives
vq
qi = 2/1o )45(cos
1 = 1.19
(iv) Checking. Dimensional check: Equation (n) is dimensionless.
Limiting check: For the special case of = 0, the heat transfer from the two plates should be
identical. Setting = 0 in (n) gives
vq
qi = 1
(5) Comments. (i) The inclined plate loses more heat than the vertical plate. The increase in heat transfer rate is due to an increase in area rather than heat transfer coefficient. In fact if the two plates have the same length, equation (k) shows that the inclined plate loses less heat than the vertical plate. (ii) According to equation (n), the greater the angle of inclination, the more heat will transfer from the plate. (iii) The above analysis assumes that conditions (g) are satisfied
by both plates. (iv) The magnitude of is the same whether it is expressed in units of degree
Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.36
Estimate the free convection heat transfer rate from five sides of a cubical
ceramic kiln. Surface temperature of each side is assumed uniform at
70oC and the ambient air temperature is 20
oC. Each side measures 48
cm.
(1) Observations. (i) This is a free convection problem. (ii) The kiln has four vertical sides and a horizontal top. (iii) All surfaces are at the same uniform temperature. (iv) Newton’s law of cooling gives the heat transfer rate. (v) The sides can be modeled as vertical plates and the top as a horizontal plate. (vi) The fluid is air.
(2) Problem Definition. Determine the average heat transfer coefficient for a vertical plate and for a horizontal plate.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficients.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid, (5) no heat loss from the bottom surface and (6) negligible radiation.
(ii) Analysis. Heat transfer from the five sides is given by
ts qqq 4 (a)
where
q = heat transfer rate from five surfaces, W qs = heat transfer rate from each of four sides, W qt = heat transfer rate from the top, W
Application of Newton's law of cooling gives
)( TTAhq sss (b)
and
)( TTAhq stt (c)
where
A = surface area of one side of cube, m2
sh = average heat transfer coefficient for a vertical side, W/m2-oC
th = average heat transfer coefficient for top surface, W/m2-oC
sT = surface temperature = 70oC
T = ambient air temperature = 20oC
Surface area of each side is given by A = L2 (d)
where
L = length of each square side = 48 cm = 0.48 m
T
g
PROBLEM 8.36 (continued)
The average heat transfer coefficient is determined from correlation equations for free convection over a vertical plate. The Rayleigh number RaL is calculated first to select an
appropriate correlation equation for h . The Rayleigh number is defined as
RaL = PrLTTg s
2
3)( (e)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties of air are determined at the film temperature Tf defined as
Tf=
2
TTs = 2
)C)(2070( o
= 45oC
At this temperature air properties are
k = 0.02746 W/m-oCPr = 0.7095
= 17.44 10-6 m2/s
For ideal gases the coefficient of thermal expansion for the sides is given by
s = 003143.015.273)C(45
1
)K(
1o
fT(1/K)
Substituting into (e) gives
RaL = 7095.0)s/m()1044.17(
)m()48.0)(C)(2070)(s/m(81.9)C/1(003143.02426
33o2o
= 0.3977 109
Since RaL
< 109, the flow is laminar. Thus, the appropriate equation for the average Nusselt
number for a vertical plate is given by (8.25b)
4/14/1
2/1 953.4884.4435.2LL Ra
PrPr
Pr
k
LhuN (f)
This equation is valid for vertical plate constant surface temperature Ts
laminar free convection 104 < RaL < 109
0 < Pr < properties at Tf (g)where
LuN = average Nusselt number
PROBLEM 8.36 (continued)
Correlation equations for a heated horizontal plate is given by equation (8.29)
4/154.0 LL RauN for 75 10210 LRa (8.29a)
3/114.0 LL RauN for 107 103102 LRa (8.29b)
(8.29c)
Thus the Rayleigh number for the top surface must be determined. It is based on the characteristic length L defined in equation (8.31) as
p
ALt (h)
where p is perimeter. Using (h)
12.04
)m(48.0
44
2 L
L
LLt m
For a horizontal plat in air t is evaluated at T
t = 003411.015.273)C(20
1
)(
1oKT
(1/K)
(e) gives
RaL = 6
2426
33o2o
10744.67095.0)/s(m)10(17.44
)(mC)(0.12)20)()(70/sC)9.81(m/0.003411(1
Therefore, (8.29a) is applicable.
(iii) Computations. Substituting into (f) gives sh
k
LhuN s
L
4/19
4/1
2/1103977.0
)7095.0(953.4)7095.0(884.4435.2
7095.0 = 72.77
Solving the above for sh
sh = 72.77k
L = 72.77 (0.02746)(W/m-oC)/0.48(m) = 4.16 W/m2-oC
Equations (b) and (d) give the heat transfer rate from each side
sq = 4.16 (W/m2-oC) (0.48)2(m2)(70 )20 (oC) = 47.9 W
Heat transfer coefficient for the top surface is computed from equation (8.29a)
horizontal plate hot surface up or cold surface down
properties, except , at fT
at fT for liquids, T for gases
PROBLEM 8.36 (continued)
4/16 )10744.6(54.0k
LhuN t
L = 27.52
Solving for th
th = 27.520.12(m)
C)m0.02746(W/52.27
o
cL
k = 6.3 W/m2-oC
Equations (c) and (d) give the heat transfer rate from the top surface
tq = 6.3(W/m2-oC) (0.48)2(m2)(70 )20 (oC) = 72.6 W
Equation (a) gives the total heat loss from the five surfaces
q = 4(47.9)(W) + 72.6(W) = 264.2 W
(iv) Checking. Dimensional check: Computations showed that equations (a)-(f) and (h) are dimensionally consistent.
Quantitative check: Heat transfer coefficients are within the approximate range of Table 1.1.
Validity of correlation equations (f) and (h): The conditions listed in (g) and (i) are met.
(5) Comments. (i) Modeling each side as a free standing vertical plate is an approximation if the kiln rests on a flat surface. (ii) Heat loss from the top surface is considerably more than form a side surface. (iii) Depending on surface emissivity, radiation loss may be appreciable. (iv) The
magnitude of is the same whether it is expressed in units of degree Celsius or kelvin. The
reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
POBLEM 8.37
Determine the surface temperature of a single burner electric stove when its power supply is 70
W. The diameter of the burner is 18 cm and its emissivity is 0.32. The ambient air temperature is
30 Co and the surroundings temperature is 25 Co .
(1) Observations. (i) Heat transfer from the surface is by free convection and radiation. (ii) The burner can be modeled as a horizontal disk with its heated side facing down. (iii) Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relations gives heat transfer by radiation. (iv) Both convection and radiation depend on surface temperature. (v) If the burner is well insulated at the bottom heated surface and its rim, then the electric power supply is equal to surface heat transfer.
(2) Problem Definition. Determine the average heat transfer coefficient for a for a horizontal plate.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficients. Use Stefan-Boltzmann relation to determine heat loss by radiation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid, (5) no heat loss from the bottom surface and rim, (6) uniform surroundings temperature and (7) surface of burner is small compared to surroundings area.
(ii) Analysis. Based on assumption (5), conservation of energy at the surface gives
rc qqP (a)
where
P = power supply to burner = 70 W
cq convection heat transfer rate, W
rq radiation heat transfer rate, W
Application of Newton's law of cooling gives
)( TTAhq sc (b)
where
A = surface area of one side of cube, m2
h = average heat transfer coefficient, W/m2-oC
sT = surface temperature, oC
T = ambient air temperature = 30oC
Surface area is given by
A = 2
4D (c)
where
D = diameter = 18 cm = 0.18 m
g
D
T
sT
+ -
cqrq
PROBLEM 8.37 (continued)
The average heat transfer coefficient is determined from correlation equations for free convection over a horizontal plate given by (8.29)
4/154.0 LL RauN for 75 10210 LRa (8.29a)
3/114.0 LL RauN for 107 103102 LRa (8.29b)
(8.29c)
The Rayleigh number RaL should be calculated to select an appropriate correlation equation from the above. The Rayleigh number is defined as
RaL = PrLTTg s
2
3)( (d)
where
g = gravitational acceleration = 9.81 m/s2
L = characteristic lengthPr = Prandtl number RaL = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
The characteristic length L is defined in equation (8.31) as
p
AL (e)
where p is perimeter. Examination of (d) shows that the Rayleigh number depends on surface
temperature .sT However, sT is unknown. This suggest that the an iterative procedure must be
used to solve the problem.
Radiation heat transfer is given by equation (1.12)
)( 44sursr TTAq (f)
where
surT surroundings temperature = 25 + 273.15 = 298.15 K
surface emissivity = 0.3
Stefan-Boltzmann constant = 428 KW/m1067.5
(b) and (f) into (a)
P = h A ( sT - T ) + )( 44surs TTA (g)
Equation (g) is used to determine sT using an iterative procedure. A value for sT is assumed,
properties of air are determined, (d) is used to compute the Rayleigh number and an equation is
horizontal plate hot surface up or cold surface down
properties, except , at fT
at fT for liquids, T for gases
PROBLEM 8.37 (continued)
selected for the free convection Nusselt number, and finally (g) is used to calculate the dissipated power P. If the calculated P is not equal to 70 W, the procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations. Assume C190osT . Using (e)
m045.04
(m)18.0
4
4/2 D
D
DL
Properties of air are determined at the film temperature Tf defined as
Tf=
2
TTs = 2
C)()30190( o
= 110oC
At this temperature air properties are
k = 0.03194 W/m-oCPr = 0.704
= 24.1 10-6 m2/s
For a horizontal plate in air is evaluated at T
= 0032987.015.273)C(30
1
)(
1oKT
(1/K)
(d) gives
RaL = 5
2426
33o2o
107188.5704.0)/s(m)10(24.1
)(mC)(0.045)30)()(190/sC)9.81(m1/0.0032987(
Therefore, (8.29a) is applicable.
4/15 )107188.5(54.0k
LhuN L = 14.85
Solving for h
h = 14.85k
L = 14.85(0.03194)(W/m-oC)/0.045(m) = 10.54 W/m2-oC
Surface area is given by
222 m025447.04/)m18.0(4/DA
Substituting into (g)
)](K273.15)(25273.15))[(190K)(W/m100.3(5.67
C)30)()(190(mC)0.0254470.54(W/m1
444428
o2o2P
P = 59.4 W
This calculated value is not close to the given power 70P W. The procedure is repeated for an
assume value C.210osT The following result is obtained at this assume surface temperature:
PROBLEM 8.37 (continued)
C120ofT
k = 0.03261 W/m-oCPr = 0.703
= 25.19 10-6 m2/s0032987.0
5108806.5LRa
95.14k
LhuN L
h =10.83 W/m2-oC
64.49cq W
17.20rq
P = (49.64 + 20.17) = 69.8 W
This is close to the given value of P = 70 W.
(iv) Checking. Dimensional check: Computations showed that equations (d,) (f) and (g) are dimensionally consistent.
Quantitative check: Heat transfer coefficient is within the approximate range of Table 1.1.
Validity of correlation equations (8.29a): The conditions listed in (8.29c) are met.
(5) Comments. (i) Radiation loss is appreciable and thus can not be neglected. (ii) In practice some of the electric power supplied to the burner is lost through the bottom side and rim. This has the effect of lowering surface temperature. Thus the model used to solve the problem overestimates surface temperature.
PROBLEM 8.38
A test apparatus is designed to determine surface emissivity of material. Samples are machined
into disks of diameter D. A sample disc is heated electrically on one side and allowed to cool off
on the opposite side. The heated side and rim are well insulated. The disk is first placed
horizontally with its exposed surface facing up in a large chamber. At steady state the exposed
surface temperature is measured. The procedure is repeated, without changing the power
supplied to the disk, with the exposed surface facing down. Ambient air temperature in the
chamber is recorded.
[a] Show that surface emissivity is given by
)(
)()(41
42
2211
ss
ss
TT
TThTTh
where subscripts 1 and 2 refer to the exposed surface facing up and down, respectively, and
h average heat transfer coefficient,
CW/m o2
sT surface temperature,= K
T ambient temperature, K
Stefan-Boltzmann constant, 42 KW/m
[b] Calculate the emissivity for the following case:
D 12 cm, KCT os 15.5332601 , 15.5733002 CT o
s K, 15.29320 CT o K
(1) Observations. (i) Heat transfer from the surface is by free convection and radiation. (ii) The sample can be modeled as a horizontal disk with its heated side facing down or up. (iii) Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relation gives heat transfer by radiation. (iv) Radiation depends on surface emissivity. (v) If the disk is well insulated at the heated surface and its rim, then the electric power supply is equal to surface heat transfer. (vi) Since the electric power is the same for both orientations, it follows that surface heat transfer rate is also the same. (vii) Each orientation has its own Nusselt number correlation equation.
(2) Problem Definition. Determine surface heat transfer by convection and radiation for two horizontal orientations of a plate: heated side facing up and heated side facing down.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficients. Use Stefan-Boltzmann relation to determine heat loss by radiation. Equate total surface heat transfer rate for both orientations.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid, (5) no heat loss from the heated power supply surface and rim, (6) uniform surroundings temperature and (7) surface of disk is small compared to surroundings area.
surT
-
D
T sT
+
g
1 2
+
-
TsTD
PROBLEM 8.38 (continued)
(ii) Analysis. Based on assumption (5), conservation of energy at the surface gives
rc qqP (a)
where
P power supply to disk, W
cq convection heat transfer rate, W
rq radiation heat transfer rate, W
Application of Newton's law of cooling gives
)( TTAhq sc (b)
where
A = surface area, m2
h = average heat transfer coefficient, W/m2-oC
sT = surface temperature, oC
T = ambient air temperature = 20oC
Radiation heat transfer is given by equation (1.12)
)( 44sursr TTAq (c)
where
surT surroundings temperature
surface emissivity
Stefan-Boltzmann constant = 428 KW/m1067.5
(b) and (c) into (a)
P = h A ( sT - T ) + )( 44surs TTA (d)
This result is applied to orientation 1 (heated surface facing up) and orientation 2 (heated surface facing down), noting that P is the same for both
P = 1h A ( 1sT - T ) + )( 441 surs TTA (e)
P = 2h A ( 2sT - T ) + )( 442 surs TTA (f)
Equating (e) and (f) and solving for gives
)(
)()(41
42
2211
ss
ss
TT
TThTTh(g)
[b] The average heat transfer coefficient is determined from correlation equations for free convection over a horizontal plate. For a plate with the heated side facing up (orientation 1), use (8.29)
4/111 54.0 LL RauN for 7
15 10210 LRa (8.29a)
3/111 14.0 LL RauN for 10
17 103102 LRa (8.29b)
PROBLEM 8.38(continued)
(8.29c)
The Rayleigh 1LRa number is defined as
12
1
31 )(
1 PrLTTg
Ra sL (h)
where
g = gravitational acceleration = 9.81 m/s2
L = characteristic length
1Pr = Prandtl number
RaL1 = Rayleigh number
= coefficient of thermal expansion, 1/K
1v = kinematic viscosity, m2/s
For the average heat transfer coefficient for horizontal plate with the heated side facing down (orientation 2), use (8.30)
4/122 27.0 LL RauN for 10
25 103103 LRa (8.30a)
(8.30b)
The Rayleigh number 2LRa is given by
222
32 )(
2 PrLTTg
Ra sL
v
(i)
The characteristic length L is defined as
44perimeter
areasurface 2 D
D
DL (j)
where
D = diameter = 14 cm = 0.14 m
(iii) Computations. Consider orientation 1 first. Equation (j) gives
m035.04
)m(14.0L
Properties of air are determined at the film temperature 1fT defined as
horizontal plate hot surface up or cold surface down
properties, except , at fT
at fT for liquids, T for gases
horizontal plate hot surface down or cold surface up
properties, except , at fT
at fT for liquids, T for gases
PROBLEM 8.38(continued)
2
11
TTT s
f = 2
C)()20260( o
= 140oC
At this temperature air properties are
1k = 0.03394 W/m-oC
1Pr = 0.702
1v = 27.44 10-6 m2/s
For a horizontal plate in air is evaluated at T
= 0034112.015.273)C(20
1
)(
1
oKT(1/K)
Equation (h) gives
RaL1 = 5
2426
33o2o
1021.3702.0)/s(m)10(27.44
)(mC)(0.035)20)()(260/sC)9.81(m1/0.0034112(
Substituting into (8.29a)
85.12)1021.3(54.0 4/15
1
11
k
LhuN L
Solving for 1h
1h = 14.85 46.12m)(035.0
C)(W/m-03394.0 o1
L
k W/m2-oC
Consider orientation 2. Properties of air are determined at the film temperature 2fT defined as
2
22
TTT s
f = 2
C)()20300( o
= 160oC
At this temperature air properties are
2k = 0.03525 W/m-oC
2Pr = 0.701
2v = 29.75 10-6 m2/s
Equation (I) gives
RaL2 = 5
2426
33o2o
10182.3701.0)/s(m)10(29.75
)(mC)(0.035)20)()(300/sC)9.81(m1/0.0034112(
Substituting into (8.29a)
41.6)10182.3(27.0 4/15
2
22
k
LhuN L
Solving for 2h
PROBLEM 8.38(continued)
2h = 6.41 46.6m)(035.0
C)(W/m-03525.0 o2
L
k W/m2-oC
Substituting into (g)
769.0]K)()15.273260(K)()15.273300[()K(W/m1067.5
C))(20300(C)(W/m46.6C))(20260(C)(W/m46.124444428
oo2oo2
(iv) Checking. Dimensional check: Computations showed that equations (g)-(j) are dimensionally consistent.
Quantitative check: Heat transfer coefficients are within the approximate range of Table 1.1.
Validity of correlation equations (8.29a) and (8.30a): The conditions listed in (8.29c) and (8.30b) are met.
(5) Comments. (i) The determination of emissivity using this method does not require measuring the surroundings temperature. However, surroundings temperature must be uniform and the same for both orientations. (ii) Emissivity can be determined without knowing the power supply as long as it is the same for both orientations. (iii) The disk need not be perfectly insulated as long as heat loss from the insulated surfaces is the same for both orientations. (iv). Based on the calculated emissivity, heat transfer from the surface q is given by (b), (c) and (d). However, to compute heat transfer by radiation the surroundings temperature must be known.
Assume C20osurT . Thus, for orientation 1:
46.03C)(20))(260(m0.14/2)C)12.46(W/m o22o2cq W
)K]()15.27320()15.273260)[(m()2/14.0()K(W/m)1067.5(769.0 44422428rq = 49.27W
03.46q W + 49.27 W = 95.3 W
Similarly, for orientation 2
84.27cq W
46.67rq W
q 95.3 W
This confirms that the electric power supply to both positions is the same.
POBLEM 8.39
A hot water tank of diameter 65 cm and height 160 cm loses heat by free convection. Estimate
the free convection heat loss from its cylindrical and top surfaces. Assume a surface temperature
of 50oC and an ambient air temperature of 20
oC.
(1) Observations. (i) This is a free convection problem. (ii) Heat is transferred from the cylindrical surface and top surface of tank to the ambient air. (iii) Under certain conditions a vertical cylindrical surface can be modeled as a vertical plate. (iv) Newton’s law of cooling gives the heat transfer rate from tank. (v) The fluid is air.
(2) Problem Definition. Determine the average heat transfer coefficient for a vertical cylindrical surface and a heated horizontal surface facing up.
(3) Solution Plan. Apply Newton’s law of cooling and use appropriate correlation equations for the average Nusselt number to determine the average heat transfer coefficient for a vertical cylinder and a heated horizontal plate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) quiescent fluid and (6) negligible radiation.
(ii) Analysis. The total heat transfer from tank is
q = qc + qt (a) where
q = total rate of heat transfer from tank, W qc = rate of heat transfer from cylindrical surface, W qt = rate of heat transfer from top surface, W
Applying Newton’s law of cooling to the two surfaces
qc = hc Ac (Ts - T ) (b)
and
qt = ht At (Ts - T ) (c)
where
Ac = area of cylindrical surface, m2
At = area of top surface, m2
hc = average heat transfer coefficient for the cylindrical surface, W/m2-oC
ht = average heat transfer coefficient for the top surface, W/m2-oC
Ts = surface temperature = 50oC
T = ambient temperature = 20oC
The two areas Ac and At are
Ac = D L (d) and
At = D2/4 (e)
where
D
L
qc
qt
g
T
qc
PROBLEM 8.39 (continued)
D = tank diameter = 0.65 m L = tank height = 1.6 m
The heat transfer coefficients are determined form correlation equations for the Nusselt number. To determine if the cylindrical surface can be modeled as vertical plate the following criterion is checked
4/1)(
35
LGrL
D (f)
where the Grashof number GrL is defined as
GrL =2
3LTTg s (g)
where
g = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf defined as
Tf = 2/)( TTs = (50 + 20)(oC)/2 = 35oC
Properties of air at this temperature are given in Appendix C
k = 0.02674 W/m-oCPr = 0.711
= 16.485 10-6 m2/s
For an ideal gas is given by
=)K(
1
fT =
15.273C35
1o
= 0.003245 1/K
Substituting into (g)
GrL =2226
33o2o
)s/m()10485.16(
)m()6.1)(C(2050)s/m(81.9)C/1(003245.0 = 14.394 109
The right hand side of (f) is
(35/GrL)1/4 = (35/14.394 109)1/4 = 0.00702
The left hand side of (f) is
D/L = 0.65(m)/1.6(m) = 0.406
Equation (f) is satisfied and thus, the cylindrical surface can be treated as a vertical plate.
The Rayleigh number RaL is computed next to select an appropriate correlation equation for the average Nusselt number for the cylindrical surface
RaL = GrL Pr = 14.394 109 0.711 = 10.234 109
For this Rayleigh number the correlation equation is given by (8.26a)
PROBLEM 8.39 (continued)
2
27/816/9
6/1
)/492.0(1
387.0825.0
Pr
Ra
k
LhuN Lc
L (h)
valid for vertical plate uniform surface temperature Ts
laminar, transition, and turbulent
10 1 < RaL <1012
0 < Pr < properties at Tf (i) where
LNu = average Nusselt number
Heat loss from the top surface is modeled as a horizontal heated plate facing up. The Rayleigh umber RaD for the top is computed to select an appropriate correlation equation for the Nusselt number. The Rayleigh number is defined as
RaD = 2
3)( DTTg s Pr
RaD = 711.0)s/m()10485.16(
)m()65.0)(C(2050)s/m(81.9)C/1(003245.02226
33o2o
= 0.6862 109
Thus, the appropriate equation for this Raleigh number is given by (8.31b)
DNu =k
Dht = 0.15 (RaD)1/3 (j)
valid for96 106.1108 LRa (k)
(iii) Computations. For the cylindrical surface, equation (h) gives
2
27/816/9
6/19
711.0/492.01
)10234.10(387.0825.0
k
LhuN c
L = 254.2
Solving the above for ch
ch = 254.2 k/L = 254.2(0.02674)(W/m-oC)/1.6(m) = 4.25 W/m2-oC
Equation (i) gives the heat transfer coefficient for the top surface
DNu =k
Dht = 0.15 (0.6862 109)1/3 = 132.3
Solving for ht
ht = 132.3 k/D = 132.3 (0.02674)(W/m-oC)/0.65(m) = 5.44 W/m2-oC
The surface areas are calculated from (d) and (e)
PROBLEM 8.39 (continued)
Ac = 0.65(m) 1.6(m) = 3.267 m2
and
At = (0.65)2(m2)/4 = 0.3318 m2
Equations (b) and (c) give the rate of heat transfer from the two surfaces
qc = 4.25(W/m2-oC) 3.267(m2)(50 - 20)(oC) = 416.5 W
and
qt = 5.44 (W/m2-oC) 0.3318(m2) (50 - 20)(oC) = 54.1 W
The total heat loss from the tank is
q = 416.5 (W) + 54.1 (W) = 470.6 W
(iv) Checking. Dimensional check: Computations showed that equations (b)-(h) and (j) are dimensionally consistent.
Quantitative check. The computed values of the heat transfer coefficient are representative of values listed in Table 1.1 for free convection of gases.
Validity of correlation equations (h) and (j): The conditions lis ted in (i) and (k) are met.
(5) Comments. (i) It is important to check the Rayleigh and Grashof number s to determine the applicable correlation equations for the Nusselt number. (ii) The heat loss from the tank is
significant. To conserve energy the tank should be insulated. (iii) The magnitude of is the
same whether it is expressed in units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.40
Hot gases from a furnace are discharged through a round horizontal duct 30 cm in diameter.
The average surface temperature of a 3 m duct section is 180oC. Estimate the free convection
heat loss from the duct to air at 25oC.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal round duct. (iii) Heat is transferred from duct surface to the ambient air. (iv) According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal round duct (cylinder).
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the duct
q = h A (Ts - T ) = h DL(Ts - T ) (a) where
A = surface area, m2
D = duct diameter = 3 cm = 0.3 m
h = average heat transfer coefficient,W/m2-oC
L = duct length = 3 m q = heat transfer rate, W Ts = surface temperature = 180oC
T = ambient temperature = 25oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection given by equation (8.35a)
2
27/816/9
6/1
/559.01
)(387.060.0
Pr
Ra
k
DhNu D
D (b)
valid for: horizontal cylinder uniform surface temperature or flux
10 5 < RaD < 1012
properties at Tf (c)where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pr = Prandtl number RaD = Rayleigh number
The Rayleigh number is defined as
D
L
sT
T
g
T
PROBLEM 8.40 (continued)
RaD =2
3)( DTTg s Pr (d)
where
g = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
For an ideal gas is given by
= K15.273
1
fT (e)
Properties are determined at the film temperature Tf, defined as
Tf = (Ts + T )/2 (f)
(iii) Computations. The film temperature is calculated using (f)
Tf = (180 +25)(oC)/2 = 102.5oC
Properties of air at this temperature are
k = 0.03144W/m-oCPr = 0.704
= 1/(102.5 + 273.15)(K) = 0.0026621(1/K)
= 23.29 10-6 m2/s
Substituting into (d)
RaD = 2226
33o2o
)s/m()1029.23(
)m()3.0)(C)(25180)(s/m(81.9)C/1(0026621.0 0.704 = 1.418 108
Equation (b) is applicable for this Rayleigh number. Substituting into (b)
2
27/816/9
6/18
D
)704.0/559.0(1
)10418.1(387.060.0
k
DhuN =66.86
Solving for h
h = 66..86 k/D = 66.86(0.03144)(W/m-oC)/0.3(m) = 7.01 W/m2-oC
Equation (a) gives
q = 7.01(W/m2-oC) 0.3(m)2.5(m) ( 25180 )(oC) = 2560 W
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b), (d) and (e) are dimensionally consistent.
Quantitative check: The value of the heat transfer coefficient is within the approximate range given in Table 1.1 for free convection of gases.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. Since surface temperature is relatively high, radiation heat loss may be significant.
PROBLEM 8.41
A 6 m long horizontal steam pipe has a surface temperature of 120oC. The diameter of the pipe
is 8 cm. It is estimated that if the pipe is covered with a 2.5 cm thick insulation material its
surface temperature will drop to 40oC. Determine the free convection heat loss from the pipe
with and without insulation. The ambient air temperature is 20oC.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal pipe. (iii) Heat is transferred from pipe surface to the ambient air. (iv) Adding insulation material reduces heat loss from pipe. (v) According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures. (vi) Heat transfer coefficient and surface area change when insulation is added. (vii) The fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal pipe (cylinder) with and without insulation.
(3) Solution Plan. Apply Newton’s law of cooling. Use correlation equations for the Nusselt number to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the bare and insulated pipe
qo = oh Ao (Tso - T ) = oh DoL(Tso - T ) (a)
and
qi = ih Ai(Tsi - ) = ih DiL(Tsi - T ) (b)
where the subscripts i and o refer to insulated and bare pipes, respectively, and
A = surface area, m2
Di = insulation diameter = 8 cm + 2(2.5)(cm) = 13 cm = 0.13 m Do = pipe diameter = 8 cm = 0.08 m
ih = average heat transfer coefficient for insulated pipe, W/m2-oC
oh = average heat transfer coefficient for bare pipe, W/m2-oC
L = pipe length = 6 m Tsi = surface temperature of insulated pipe = 40oCTso = surface temperature of bare pipe = 120oC
T = ambient temperature = 20oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection given by equation (8.35a)
2
27/816/9
6/1
/559.01
387.060.0
Pr
Ra
k
DhuN D
D (c)
D
L
sT
T
g
PROBLEM 8.41 (continued)Valid for
horizontal cylinder uniform surface temperature or flux
10 5 < RaD < 1012
properties at Tf (d)where
D = diameter, m k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pr = Prandtl number RaD = Rayleigh number
The Rayleigh number is defined as
RaD =2
3DTTg s Pr (e)
whereg = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K (or 1/oC)
= kinematic viscosity, m2/s
For an ideal gas is given by
= )(K)15.273(
1
fT (f)
Properties are determined at the film temperature Tf
Tf = (Ts + T )/2 (g)
(iii) Computations. Consider the bare pipe first. Film temperature is given by (g)
Tf = (120 +20)(oC)/2 = 70oC
Properties at this temperature are
k = 0.02922 W/m-oCPr = 0.707
= 1/(70 + 273.15) K = 0.002914 1/K
= 19.9 10-6 m2/s
Substituting into (e)
oDRa =2226
33o2o
)s/m()109.19(
)m()08.0)(C)(20120)(s/m(81.9)C/1(002914.0 0.707 = 0.261 107
Equation (c) is applicable for this Rayleigh number. Substituting into (c)
2
27/816/9
6/17
)707.0/559.0(1
)10261.0(387.060.0
k
DhuN oo
oD = 19.08
Solving for oh
PROBLEM 8.41(continued)
oh = 19.08k/Do = 19.08(0.02922)(W/m-oC)/0.08(m) = 6.97 W/m2-oC
Equation (a) gives
qo = 6.97(W/m2-oC) 0.08(m)6(m) (120-20)(oC) = 1051.1 W
Consider next the insulated pipe. Film temperature is calculated using (g)
Tf = (40 +20)(oC)/2 = 30oC
Properties at this temperature are
k = 0.02638 W/m-oCPr = 0.712
= 1/(30 + 273.15) K = 0.003299 1/K (ideal gas)
= 16.01 10-6 m2/s
Substituting into (e)
iDRa =2226
33ooo
)s/m()1001.16(
)m(13.0)C(2040)C/1(81.9)C/1(003299.0 0.712 = 0.395 107
Equation (c) is applicable for this Rayleigh number. Substituting into (c)
2
27/816/9
6/17
)712.0/559.0(1
)10395.0(387.060.0
k
DhuN ii
Di= 21.53
Solving for ih
ih = 21.53k/Di = 21.53(0.02638)(W/m-oC)/0.13(m) = 4.37 W/m2-oC
Equation (b) gives
qi = 4.37(W/m2-oC) 0.13(m)6(m) (40-20)(oC) = 214.2 W
(iv) Checking. Dimensionless check: Computations showed that equations (a)-(c) and (e) are dimensionally consistent.
Quantitative check: The value of the heat transfer coefficient is within the approximate range given in Table 1.1 for free convection of gases.
Validity of correlation equation (c): The conditions listed in (d) are met.
(5) Comments. (i) Heat loss from the pipe is significantly reduced when insulation is added. The reduction is due to a decrease in surface temperature and heat transfer coefficient. (ii) The
magnitude of is the same whether it is expressed in units of degree Celsius or kelvin. The
reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.42
An electric wire dissipates 0.6 W/m while suspended horizontally in air at 20oC. Determine its
surface temperature if the diameter is 0.1 mm. Neglect radiation.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal wire (cylinder). (iii) Under steady state conditions the power dissipated in the wire is transferred to the surrounding air. (iv) According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature. (v) The fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal wire.
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface flux and temperature, (5) no axial conduction, (6) quiescent fluid and (7) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the wire
q = h A (Ts - T ) = h DL(Ts - T )Solving the above for Ts
Ts = T + hD
Lq / (a)
where
A = surface area, m2
D = wire diameter = 0.1 mm = 0.0001 m
h = average heat transfer coefficient, W/m2-oCL = wire length, m q/L = power dissipated per unit length = 0.6 W/m Ts = surface temperature, oC
T = ambient temperature = 20oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection, given by equation (8.35a)
2
27/816/9
6/1
)/559.0(1
387.060.0
Pr
Ra
k
DhuN D
D (b)
Valid for horizontal cylinder uniform surface temperature or flux
10 5 < RaD < 1012
properties at Tf (c)where
k = thermal conductivity, W/m-oC
Lq/ sT Tg
PROBLEM 8.42 (continued)
DNu = average Nusselt number
Pr = Prandtl number RaD = Rayleigh number
The Rayleigh number is defined as
RaD =g T T Ds
3
2 Pr (d)
where
g = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + T )/2 (e)
(iii) Computations. Examining (a), (b) and (c) shows that h depends on Ts through the Rayleigh number. Thus, equation (a) must be solved for Ts by trial and error. A value for Ts is assumed, properties are determined based on the assumed temperature, equation (d) is used to
calculate RaD, (b) is used to determine h and (a) is used to calculated Ts. If the calculated temperature is not close to the assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Assume Ts = 200oC. Using (e)
Tf = (200 + 20)(oC)/2 = 110oC
Properties at this temperature are
k = 0.03194 W/m-oCPr = 0.704
= 1/(Tf + 273.15) K = 1/(110 + 273.15) K = 0.00261 (1/K) (ideal gas)
= 24.1 10-6 (m2/s)
Substituting into (d)
RaD = 2226
33o2o
)s/m()101.24(
)m(0001.0)C(20200)s/m(81.9)C/1(00261.0 0.704 = 5.5863 10-3
Equation (b) can be used for this Rayleigh number. Substituting into (b)
2
27/816/9
6/1
)704.0/559.0(1
0055863.0387.060.0
k
DhuN D = 0.5406
Solving for h
h = 0.5406 k/D = 0.5406(0.03194)(W/m-oC)/0.0001(m) = 172.7 W/m2-oC
Equation (a) gives
Ts = 20(oC) +Cm/W7.172m0001.0
m/W6.0o2
= 31.1oC
PROBLEM 8.42 (continued)
This calculated value of Ts is significantly different from the assumed value. Repeating the
procedure with a new assumed value of Ts = 30oC gives h = 127.84 W/m2-oC and a calculated Ts
= 34.9oC. Since this is close to the assumed value, it follows that the estimated surface temperature is 34.9oC.
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d) are dimensionally consistent.
Limiting check: If no energy is dissipated in the wire, its temperature should be the same as the
ambient temperature. Setting q = 0 in (a) gives Ts = T .
Quantitative check: The value of the heat transfer coefficient departs significantly from typical free convection values listed in Table 1.1. A review of the computations revealed no errors in the analysis or calculations. The departure from the range of h in Table 1.1 is due to the unusual nature of this application in which the diameter is very small.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. (i) Solution by trial and error was necessary because properties depend on the unknown surface temperature and equations (a), (b) and (d) can not be solved explicitly for Ts.(ii) Table 1.1 should be used as a guide only, keeping in mind that exceptions to values listed should be expected. (iii) Taking radiation into consideration has the effect of reducing surface
temperature. (iv) The magnitude of is the same whether it is expressed in units of degree
Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.43
The diameter of a 120 cm long horizontal section of a neon sign is 1.5 cm. Estimate the surface
temperature in air at 25oC if 12 watts are dissipated in the section. Neglect radiation heat loss.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a horizontal tube. (iii) Under steady state conditions the power dissipated in the neon tube is transferred to the surrounding air. (iv) According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature. (v) The fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient for a horizontal tube.
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) no axial conduction, (6) quiescent fluid and (7) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the wire
q = h A (Ts - T ) = h DL(Ts - T )Solving the above for Ts
Ts = T + hDL
q (a)
where
A = surface area, m2
D = tube diameter = 1.5 cm = 0.015 m
h = average heat transfer coefficient, W/m2-oCL = wire length = 120 cm = 1.2 m q = power dissipated in wire per unit length = heat transfer rate from wire =12 W Ts = surface temperature, oC
T = ambient temperature = 25oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection, given by equation (8.35a)
2
27/816/9
6/1
)/559.0(1
387.060.0
Pr
Ra
k
DhuN D
D (b)
valid for horizontal cylinder uniform surface temperature or flux
10 5 < RaD < 1012
properties at Tf (c)where
k = thermal conductivity, W/m-oC
PROBLEM 8.43 (continued)
DNu = average Nusselt number
Pr = Prandtl number RaD = Rayleigh number
The Rayleigh number is defined as
RaD =g T T Ds
3
2 Pr (d)
where
g = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + T )/2 (e)
(iii) Computations. Examining (a), (b) and (c) shows that h depends on Ts through the Rayleigh number. Thus, equation (a) must be solved for T- by trial and error. A value for Ts is assumed, properties are determined based on the assumed temperature, equation (d) is used to
calculate RaD, (b) is used to determine h and (a) is used to calculated Ts. If the calculated temperature is not close to the assumed value, the procedure is repeated until a satisfactory agreement is obtained.
Assume Ts = 75oC. Using (e)
Tf = (25 + 75)(oC)/2 = 50oC
Air properties at this temperature are given in Appendix C
k = 0.02781 W/m-oCPr = 0.709
= 1/(50oC + 273.15) = 0.0030945 1/K (ideal gas)
= 17.92 10-6 m2/s
Substituting into (d)
RaD =0 0030945 1 9 81 75 25 0 015
17 92 100 709
2 3 3
6 2 4 2
. ( / ) . ( / )( )( )( . ) ( )
( . ) ( / ).
o oC m s C m
m s= 1.131 104
and
Nuh D
kD
2
27/816/9
6/14
709.0/559.01
)10131.1(387.060.0 = 4.503
Solving the above for h
h = )m(015.0
)CW/m(02781.0503.4
o
= 8.35 W/m2-oC
Substituting into (a) gives the surface temperature
PROBLEM 8.43 (continued)
Ts = 25oC + )Cm/W(35.8)m(2.1)m(015.0
)W(12o2
= 50.4oC
This is not close to the assumed value of 75oC. The procedure is repeated for Ts = 50oC and 55oC. The tabulated results below shows that Ts 53.4oC.
Assumed Ts (oC) h (W/m2-oC) Calculated Ts (
oC)
75 8.348 50.4
50 7.16 54.6
55 7.46 53.4
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d) are dimensionally consistent.
Limiting check: If no power is dissipated (neon sign is off), surface temperature should be the
same as the ambient temperature. Setting q = 0 in (a) gives Ts = T .
Quantitative check: The value of the heat transfer coefficient is within the range of values listed in Table 1.1 for free convection of gases.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. (i) Solution by trial and error was necessary because properties depend on the unknown surface temperature and equations (a), (b) and (d) can not be solved explicitly for Ts.(ii) Taking radiation into consideration has the effect of reducing surface temperature. (iii) The
magnitude of is the same whether it is expressed in units of degree Celsius or kelvin. The
reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.44
An air conditioning duct passes horizontally a distance of 2.5 m through the attic of a house. The
diameter is 30 cm and the average surface temperature is 10oC. The average ambient air
temperature in the attic during the summer is 42oC. Duct surface emissivity is 0.1. Estimate the
rate of heat transfer to the cold air in the duct.
(1) Observations. (i) This is a free convection problem. (ii) The geometry is a round horizontal round duct. (iii) Heat is transferred from the ambient air to the duct. (iv) According to Newton’s law of cooling, the rate of heat transfer to the surface depends on the heat transfer coefficient, surface area and surface and ambient temperatures. (v) The fluid is air.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a horizontal round duct.
(3) Solution Plan. Apply Newton’s law of cooling. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a horizontal cylinder in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) two-dimensional (neglect end effects), (3) constant properties (except in buoyancy), (4) uniform surface temperature, (5) quiescent fluid and (6) negligible radiation.
(ii) Analysis. Applying Newton’s law of cooling to the round duct
q = h A(T - Ts) = h DL(T - Ts) (a) where
A = surface area, m2
D = duct diameter = 0.3 m
h = average heat transfer coefficient, W/m2-oC
L = duct length = 2.5 m q = rate of heat transfer, oCTs = surface temperature = 10oC
T = ambient temperature = 42oC
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a horizontal cylinder in free convection, given by equation (8.35a)
2
27/816/9
6/1
/559.01
387.060.0
Pr
Ra
k
DhuN D
D (b)
valid for: horizontal cylinder uniform surface temperature or flux
10 5 < RaD < 1012
properties at Tf (c)where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pr = Prandtl number
D
L
sTTgq
PROBLEM 8.44 (continued)
RaD = Rayleigh number
The Rayleigh number is defined as
RaD =2
3)( DTTg s Pr (d)
where
g = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K (or 1/oC)
= kinematic viscosity, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + T )/2 (e)
For an ideal gas, is given by
= K))(15.273(
1
fT (f)
(iii) Computations. The film temperature is calculated first using (e)
Tf = (10 +42)(oC)/2 = 26oC
Properties of air at this temperature are
k = 0.02608W/m-oCPr = 0.7124
= 1/(26 + 273.15) K = 0.003343 (1/K) = 0.003343(1/oC)
= 15.64 10-6 (m2/s)
Substituting into (d)
RaD = 2226
33o2o
)s/m()1064.15(
)m(3.0)C(1042)s/m(81.9)C/1(003343.0 0.7124 = 8.2522 107
Equation (b) is applicable for this Rayleigh number. Substituting into (b)
2
27/816/9
6/17
7124.0/559.01
102522.8387.060.0
k
DhuN D = 53.37
Solving for h
h = 53.37 k/D = 53.37(0.02608)(W/m-oC)/0.3(m) = 4.64 W/m2-oC
Equation (a) gives
q = 4.64 (W/m2-oC) 0.3(m) 2.5(m) (42 - 10)(oC) = 349.8 W
(iv) Checking. Dimensionless check: Computations showed that equations (a), (b) and (d) are dimensionally consistent.
Limiting check: No heat will transfer to the duct if the ambient temperature is the same as
surface temperature. Setting T = Ts in (a) gives q = 0.
PROBLEM 8.44 (continued)
Quantitative check: The value of the heat transfer coefficient within the approximate range given in Table 1.1 for free convection of gases.
Validity of correlation equation (b): The conditions listed in (c) are met.
(5) Comments. (i) Heat added to the cold air in the duct is significant. At $0.15/Kw-Hr it represents a cost of $38/month. (ii) In practice, air conditioning and heating duct passing through
unoccupied areas are insulated. (iii) The magnitude of is the same whether it is expressed in
units of degree Celsius or kelvin. The reason is that is measured in terms of degree change. One degree change on the Celsius scale is equal to one degree change on the kelvin scale. This is also true of units of heat transfer coefficient and specific heat.
PROBLEM 8.45
Estimate the surface temperature of a light bulb if its capacity is 150 W and the ambient air is at
23oC. Model the bulb as a sphere of diameter 9 cm. Neglect radiation.
(1) Observations. (i) This is a free convection and radiation problem. (ii) The geometry is a sphere. (iii) Under steady state conditions the power dissipated in the bulb is transferred to the surroundings by free convection and radiation and through the base by conduction. (iv) According to Newton’s law of cooling and Stefan-Boltzmann radiation law, heat loss from the surface depends on surface temperature. (v) The ambient fluid is air.
(2) Problem Definition. Determine the free convection heat transfer coefficient and radiation heat loss for the sphere.
(3) Solution Plan. Apply Newton’s law of cooling and Stefan-Boltzmann law. Use Nusselt number correlation equations to determine the average heat transfer coefficient for a sphere in free convection.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) no conduction through the bulb base, (5) all bulb power is transmitted to the surroundings by surface convection and radiation, (6) filament does not radiate to the surroundings directly. It radiates to the bulb surface, (7) bulb is a small surface surrounded by a much larger surface, (8) bulb is spherical, (9) surroundings is at the ambient air temperature and (10) quiescent fluid.
(ii) Analysis. Application of conservation of energy to the bulb gives
P = qc + qr (a) where
P = bulb power capacity = 150 W qc = heat transfer by convection, Wqr = heat transfer by radiation, W
Applying Newton’s law of cooling to the sphere
qc = h A (Ts - T ) = h D2 )( TTs (b)
Application of Stefan-Boltzmann law
)( 442sursr TTDq (c)
where
A = surface area, m2
D = bulb diameter = 9 cm = 0.09 m
h = average heat transfer coefficient, W/m2-oCP = bulb power = 150 W qc = convection heat transfer rate, W qr = radiation heat transfer rate, W Ts = surface temperature, K Tsur = surroundings temperature = 23oC + 273.15 = 296.15 K
T = ambient air temperature = 23oC + 273.15 = 296.15 K = emissivity of glass = 0.94
= Stefan-Boltzmann constant = 5.67 10-8 W/m2-K4
sTTg
rqcq
D
PROBLEM 8.45 (continued)
Substituting (b) and (c) into (a)
P = h D2 )( TTs + )( 442
surs TTD (d)
The heat transfer coefficient is obtained from correlation equations for the Nusselt number for a sphere in free convection, given by equation (8.36a)
9/416/9
4/1
469.01
)(589.02
Pr
Ra
k
DhuN D
D (e)
Valid for: sphere uniform surface temperature or flux RaD < 1011
Pr > 0.7 properties at Tf (f)
where
k = thermal conductivity, W/m-oC
DNu = average Nusselt number
Pr = Prandtl number RaD = Rayleigh number
The Rayleigh number is defined as
RaD =g T T Ds
3
2 Pr (g)
where
g = gravitational acceleration = 9.81 m/s2
= coefficient of thermal expansion, 1/K
= kinematic viscosity of air, m2/s
Properties are determined at the film temperature Tf
Tf = (Ts + T )/2 (h)
Equation (d) can not be solved explicitly for Ts because air properties depend Ts, the heat transfer
coefficient depends on Ts through the Rayleigh number, and the non-linear 4sT term in radiation.
Thus, equation (d) must be solved for Ts by trial and error. A value for Ts is assumed, properties are determined based on the assumed temperature, equation (g) is used to calculate RaD, (e) is
used to determine h and (d) is used to calculate P. If the calculated power is not close to the given value of 150 W, the procedure is repeated until a satisfactory agreement is obtained.
(iii) Computations.
Assume Ts = 177oC + 273.15 = 450.15 K. Using (h)
Tf = (296.15 + 450.15)(K)/2 = 373.15 K = 100oC
Properties of air at this temperature are given in Appendix C
k = 0.03127 W/m-oCPr = 0.704
= 1/(100oC + 273.15) = 0.00268 1/K (ideal gas)
PROBLEM 8.45 (continued)
= 23.02 10-6 m2/s
Substituting into (g) and (e)
RaD = 704.0)s/m()1002.23(
)m()09.0)(K)(15.29615.450)(s/m(81.9)K/1(00268.02426
332
= 3.921 106
and
Nuh D
kD 9/416/9
4/16
704.0/469.01
)10921.3(589.02 = 22.21
Solving the above for h
h = )m(09.0
)Cm/W(03127.021.22
o
= 7.72 W/m2-oC
Substituting into (d)
P = 7.72(W/m2-oC) (0.09)2(m2)(177 )23 (oC) +
0.94(5.67 10-8)(W/m2-K4) (0.09)2(m2)[(450.15)4 4)15.296( ](K4)
P = 30.25 (W) + 45.26 (W) = 75.51 W
This is not close to the given power of 150 W. The procedure is repeated for other assumed values of Ts and the results are tabulated below. At Ts = 535.15 K the calculated power is 151.46 W. This is close enough to the given power.
Assume Ts
KTf
KRaD h
W/m2-oC
qc
Wqr
Wqc +qr
W
450.15 373.13 3.921 106 7.72 30.25 45.26 75.51
530.15 413.15 3.776 106 8.3 49.42 96.7 146.12
535.15 415.65 3.757 106 8.33 50.66 100.8 151.46
(iv) Checking. Dimensionless check: Computations showed that equations (b)-(e) and (g) are dimensionally consistent.
Limiting check: If no power is dissipated (bulb is off), surface temperature should be the same as ambient temperature. Examination of (d) shows that the sum of the two terms will vanish only if
each term vanishes. Setting each term equal to zero gives Ts = Tsur = T .
Quantitative check: The value of the heat transfer coefficient is within the range of values listed in Table 1.1 for free convection of gases.
Validity of correlation equation (e): The conditions listed in (f) are met.
(5) Comments. (i) Solution by trial and error was necessary because equation (d) can not be solved explicitly for Ts. (ii) Radiation accounts for 66% of the total heat loss from the bulb. (iii) The estimated surface temperature of 535 K or 262oC appears to be high. Direct radiation between the filament and surroundings acts to lower surface temperature.
PROBLEM 8.46
A sphere of radius 2.0 cm is suspended in a very large water bath at
25oC. The sphere is heated internally using an electric coil. Determine
the rate of electric power that must be supplied to the sphere so that its
average surface temperature is 85oC. Neglect radiation.
(1) Observations. (I) At steady state, power supply to the sphere must be equal to the heat loss from the surface. (ii) Heat loss from the surface is by free convection. (iii) The surface is maintained at uniform temperature.
(2) Problem Definition. Determine the free convection heat transfer rate from the surface of a sphere.
(3) Solution Plan. Apply Newton’s law of cooling to the sphere. Use an appropriate correlation equation to determine the heat transfer coefficient.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) constant properties (except in buoyancy), (3) uniform surface temperature, (4) negligible radiation and (5) quiescent fluid.
(ii) Analysis. Conservation of energy for the sphere
qP (a)
where
P = power supply to the sphere, W
q = surface heat transfer rate from the sphere, W
Application of Newton's law of cooling
)( TTAhq s (b)
where
A = surface area of sphere
h = average heat transfer coefficient, CW/m o2
sT = surface temperature = 85 Co
T = ambient fluid temperature = 25 Co
Surface area of a sphere is 2DA (c)
where
D = Diameter of sphere = 4 cm = 0.04 m
Substituting (c) into (b) and (a)
)( TTDhP s (d)
The average heat transfer coefficient is determined from correlation equations. For free convection over a sphere the appropriate equation is given in equation (8.36)
water
PROBLEM 8.46 (continued)
9/416/9
4/1
)/469.0(1
589.02
Pr
Ra
k
DhNu D
D (e)
valid for uniform surface temperature
1110DRe
7.0Pr
properties at fT
The Rayleigh number is defined as
PrDTTg
Ra sD 2
3)( (f)
where
g = gravitational acceleration = 9.81 2m/s
k = thermal conductivity, CW/m o
Pr = Prandtl number
DRa = Rayleigh number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Properties are determined at the film temperature fT defined as
2/)( TTT sf (g)
(iii) Computations. Substituting into (g)
552/)2585(fT Co
Properties of water at this temperature are
k = 0.6458 CW/m o
Pr = 3.27 310496.0 1/K 6105116.0 /sm2
Substituting into (f)
8
2426
33o23
1 1033436.227.3)/sm()105116.0(
)m()04.0)(C)(2585()m/s(81.9)K/1(10496.0Ra
Substituting into (e)
02.66)27.3/469.0(1
1033436.2589.02
9/416/9
4/18
k
DhNuD
Solving for h
CW/m9.1065m)(04.0
)CW/m(6458.002.6602.66 o2
o
D
kh
PROBLEM 8.46 (continued)
Substituting into (d)
)C(2585)(m()04.0)(CW/m()9.1065( o22o2P = 321.5 W
(iii) Checking. Dimensional check: Computations showed that units of equations (d)-(f) are dimensionally consistent.
Limiting check: If TTs , no free convection takes place and consequently 0P . Setting
TTs in (d) gives the anticipate result.
(5) Comments. The average heat transfer coefficient is slightly outside the range given in Table 1.1 for free convection in liquids. It should be remembered that values listed in Table 1.1 are for typical applications. Exceptions should be expected.
PROBLEM 8.47
A fish tank at a zoo is designed to maintain water
temperature at 4 Co . Fish are viewed from outdoors
through a glass window L = 1.8 m high and w = 3 m wide.
The average ambient temperature during summer months
is 26 Co . To reduce water cooling load it is proposed to
create an air enclosure over the entire window using a
pexiglass plate. Estimate the reduction in the rate of heat
transfer to the water if the air gap thickness is 6 cm.
Neglect radiation. Assume that the cold side of the
enclosure is at the same temperature as the water and the
warm side is at ambient temperature.
(1) Observations. (i) Heat is transferred from the ambient air to the water in the fish tank. (ii) Adding an air enclosure reduces the rate of heat transfer. (iii) To estimate the reduction in cooling load, heat transfer from the ambient air to the water with and without the enclosure must be determined. (iv) Neglecting the thermal resistance of glass, the resistance to heat transfer form the air to the water is primarily due to the air side free convection heat transfer coefficient. (v) Installing an air cavity introduces an added thermal resistance. (vi) The problem can be modeled as a vertical plate and as a vertical rectangular enclosure. (vii) The outside surface temperature of the enclosure is unknown. (viii) Newton’s law of cooling gives the heat transfer rate. (ix) The Rayleigh number should be determined for both vertical plate and rectangular enclosure so that appropriate correlation equations for the Nusselt number are selected. However, since the outside surface temperature of the enclosure is unknown, the Rayleigh number can not be determined. The problem must be solved using an iterative procedure.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for (1) vertical plate at uniform surface temperature, and (2) for a rectangular cavity at uniform surface temperatures.
(3) Solution Plan. (i) Apply Newton's law of cooling to the window with and without the enclosure. (ii) Assume an outside enclosure surface temperature and compute the Rayleigh number. (iii) Select appropriate Nusselt number correlations equations for the two cases. (iv) Check the assumed temperature using conservation of energy from the tank to the ambient air.
(4) Plan Execution.
(i) Assumptions. (1) Outside surface temperature of the window is the same as water temperature, (2) insulated top and bottom surfaces of the enclosure, (3) negligible radiation, (4) uniform surface temperatures and (5) quiescent ambient air.
(ii) Analysis and Computations. Consider first the window without the added cavity. Newton's law of cooling gives
)(11 TTAhq w (a) where
A = surface area of window and enclosure = 1.8(m) 3.0(m) = 5.4 m2
1h = average heat transfer coefficient, W/m2-oC
1q heat transfer rate to water, W
T = ambient air temperature = 26oC
wT = cold surface temperature = outside surface temperature of window = 4oC
g
T
L
PROBLEM 8.47 (continued)
The problem reduces to determining the heat transfer coefficients 1h for the
window without the enclosure. The window is modeled as a vertical plate at uniform temperature. The Rayleigh number is calculated to determine the
appropriate correlation equation for the average heat transfer coefficient .1h
The Rayleigh number is defined as
PrLTTg
Ra wL 2
3
(b)
where
g = gravitational acceleration = 9.81 m/s2
L = window height = 1.8 m
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Air properties are evaluated at the film temperature fT defined as
C152
)C)(264(
2
ooTT
T wf
At this temperature air properties are
k = thermal conductivity = 0.02526 W/m-oCPr = 0.7145
= 14.64 10 6 m2/s
For an ideal gas is given by
fT
1 (c)
where Tf in this equation is in degrees kelvin. Thus
= 1/(15 + 273.15)K = 0.0034704 1/K
Substituting into (b)
10
2426
332
104562.17145.0)/sm()1064.14(
)m(8.1C))(426()m/s(81.91/K0034704.0LRa
Thus the flow is turbulent and the appropriate correlation equation is (8.22a)
2
27/816/9
6/1
/492.01
387.0825.0
Pr
Ra
k
LhuN L
L (d)
Valid for: vertical plate uniform surface temperature
laminar, transition, and turbulent 110 < RaL < 1210
0 < Pr <
properties at fT (e)
g
L
wT
T
PROBLEM 8.47 (continued)
Applying (d) to the window gives 1h
4.284
7145.0/492.01
104562.1387.0825.0
2
27/816/9
6/1101
k
LhuN L
C-W/m99.3.8(m)1
)C-W/m(02526.04.284 21h
Substituting into (a)
1q = 3.99(W/m2-oC) 1.8(m)(3)(m) (26–4)(oC) = 474.1 W
Consider next the window with the added enclosure. To determine the appropriate correlation equation for a vertical rectangular cavity the aspect ratio and Rayleigh number are calculated. The aspect ration is defined as
aspect ratio = L
(f)
where
L length of rectangle = 1.8 m
width of rectangle = 0.06 m
Equation (b) gives
30(m)06.0
.8(m)1L
The Rayleigh number for the enclosure is defined as
PrTTg
Ra ws
2
3)( (g)
where sT is outside surface temperature of enclosure. This temperature is needed to determine
both the enclosure heat transfer coefficient h and the outside surface coefficient .2h An iterative
procedure is required to determine sT , h and .2h A value for sT is assumed and h and 2h are
determined from applicable correlation equations. To check the assumed sT , conservation of
energy is applied to heat transfer from the water to the air. This gives
)()(2 wss TThTTh (h)
Assume C14osT . First determine the outside heat transfer coefficient 2h using (b) and (d). Air
properties are evaluated at the film temperature at T which is the average temperature of the two vertical surfaces of the enclosure given by
C202
)C)(2614(
2
ooTT
T s
At this temperature air properties are
g
LT
wT sT
PROBLEM 8.47 (continued)
k = thermal conductivity = 0.02564 W/m-oCPr = 0.713
= 15.09 10 6 m2/s
= 1/(20 + 273.15)K = 0.003411 1/K
Substituting into (b)
9
2426
332
103326.7713.0)/sm()1009.15(
)m(8.1C))(1426()m/s(81.91/K003411.0LRa
Thus the flow is turbulent and the appropriate correlation equation is (d)
89.228713.0/492.01
103326.7387.0825.0
2
27/816/9
6/192
k
LhuN L
C-W/m26.3.8(m)1
)C-W/m(02564.089.228 22h
To determine cavity heat transfer coefficient 2h Rayleigh number Ra is computed. Air
properties are determined at T , given by
C92
)C)(414(
2
oo
ws TTT
At this temperature air properties are
k = 0.02479 W/m-oCPr = 0.716
= 14.102 10 6 m2/s
= 1/(9 + 273.15)K = 0.0035442 1/K
5
2426
332
107039.2716.0)/sm()10102.14(
)m(06.0C))(414()m/s(81.91/K0035442.0Ra
Thus correlation equation (8.39a) is applicable
3.025.0012.0
42.0L
RaPrk
hNu (8.39a)
Valid for
(8.39b)
Substituting into (8.39a)
vertical rectangular enclosure
4010L
41021 Pr74 1010 Ra
properties at 2/)( hc TTT
PROBLEM 8.47 (continued)
438.330107039.20.71642.03.05012.0 25.0
k
hNu
C-W/m42.1.06(m)0
)C-W/m(02479.0438.3 2h
Use (h) to check the assume temperature
)C)(414)(CW/m(42.1?)1426)(CW/m(26.3 oo2o2
)W/m(2.14)W/m(1.39 22
Since (h) is not satisfied the procedure is repeated until a satisfactory agreement is obtained.
Assume C18osT . At this temperature the following result is obtained:
9107391.4LRa
9953.1LNu
86.22h )CW/m( o2
5107173.3Ra
723.3Nu
548.1h
Substituting into (h)
)C)(418)(CW/m(548.1?)C)(1826)(CW/m(86.2 oo2oo2
)W/m(67.21)W/m(88.22 22
The heat transfer rate is given by applying Newton’s law of cooling between the ambient air and the outside surface of the enclosure
2q = 6.123)C)(1826(m)3(m)(8.1)CW/m(86.2 oo2 W
Thus the reduction in the cooling load due to the addition of the cavity is
Load reduction = W5.350W123.6-W1.474
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (d) and (8.39a) are dimensionally consistent.
Quantitative check: The magnitude of ,h 1h and 2h are in line with typical free convection
values for air given in Table 1.1.
Validity of correlation equations (d) and (8.39a): Conditions listed in equations (e) (8.39b) are satisfied.
(5) Comments. The addition of a rectangular cavity reduces the heat transfer to the water by 74%. This is a significant saving in energy.
PROBLEM 8.48
It is proposed to replace a single pane observation window with double pane. On a typical
winter day the inside and outside air temperatures are C20oiT and C10o
oT . The inside
and outside heat transfer coefficients are ih CW/m4.9 o2 and CW/m37 o2oh . The height
of the window is 28.0L m and its width is w = 3 m. The thickness of glass is t = 0.3 cm and its
conductivity is .CW/m7.0 ogk Estimate the savings in energy if the single pane window is
replaced. Note that for the single pane window there are three resistances in series and the heat
transfer rate 1q is given by
ogi
oi
hk
t
h
TTAq
11
)(1
For the double pane window, two additional resistances are added. The width of the air space in
the double pane is 3 cm. In determining the heat transfer coefficient in the cavity, assume
that enclosure surface temperatures are the same as the inside and outside air temperatures.
(1) Observations. (i) Heat is transferred from the inside to the outside. (ii) Adding an air enclosure reduces the rate of heat transfer. (iii) To estimate the savings in energy, heat transfer through the single and double pane windows must be determined. (iv) The double pane window introduces an added glass conduction resistance and a cavity convection resistance. (v) the problem can be modeled as a vertical rectangular enclosure. (vi) Newton’s law of cooling gives the heat transfer rate. (vii) The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a rectangular cavity.
(3) Solution Plan. (i) Apply Newton's law of cooling to the single pane and double pane windows taking into consideration the multiple resistances in series. (ii) Select an appropriate Nusselt number correlation equation for the rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Insulated top and bottom surfaces of the enclosure, (2) negligible radiation and (3) uniform surface temperatures.
(ii) Analysis. Consider first the single pane window
ogi
oi
hk
t
h
TTAq
11
)(1 (a)
where
A = surface area of window and enclosure = 0.28(m) 3.0(m) = 0.84 m2
ih = inside heat transfer coefficient = 9.4 W/m2-oC
oh = outside heat transfer coefficient = 37 W/m2-oC
gk glass conductivity = 0.7 W/m-oC
1q heat transfer rate form single pane window, W
PROBLEM 8.48 (continued)
t glass thickness = 0.3 cm = 0.003 m
iT inside air temperature = 20oC
oT outside air temperature = -10oC
For the double pane window, two additional resistances are added in series: a glass conduction resistance and a rectangular cavity convection resistance. Equation (a) is modified to
hhk
t
h
TTAq
ogi
oi
1121
)(2 (b)
where
h rectangular cavity convection heat transfer coefficient, W/m2-oC
2q heat transfer rate from the double pane window, W
To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number Ra are
computed. The aspect ratio is given by
Aspect ratio = 33.9)m(03.0
)m(28.0L
The Rayleigh number for the enclosure is defined as
PrTTg
Ra ch
2
3)( (c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
cT enclosure cold surface temperature, Co
hT enclosure hot surface temperature, Co
= coefficient of thermal expansion, 1/K
enclosure air thickness = 3 cm = 0.03 m
= kinematic viscosity, m2/s
Air properties are determined at T which is the average temperature of the two vertical surfaces
of the enclosure, cT and .hT Both temperatures are unknown . They can be determined by an
iterative procedure. A simpler approach is to assume that the two surfaces are at the inside and
outside temperatures. That is, ih TT and oc TT . This assumption is conservative in that it
will overestimate the heat loss from the double pane window. Using this approximation gives
C52
)C)(1020(
2
oo
oi TTT
At this temperature air properties are
k = 0.02448 W/m-oCPr = 0.717
= 13.75 10 6 m2/s
PROBLEM 8.48 (continued)
= 1/(5 + 273.15)K = 0.003595 1/K
5
2426
332
100834.1717.0)/sm()1075.13(
)m(03.0C))(1020()m/s(81.91/K003595.0Ra
Thus correlation equation (8.38a) is applicable
25.028.0
2.022.0
LRa
Pr
Pr
k
hNu (8.38a)
Valid for
(8.38b)
(iii) Computations. Equation (a) gives
W02703.000429.010638.0
2.25
C)W/m(37
1
C)W/m(7.0
)m(003.0
C)W/m(4.9
1
)C)(1020)(m(84.0
o2o2o2
o
1q
1831q W
Substituting into (8.38a)
018.3333.90834.1717.2.0
22.025.0
28.0
0
0.717
k
hNu
CW/m463.2)m(03.0
)CW/m(02448.0018.3 o2
o
h
Equation (b) gives
W46.3W406.002703.000429.010638.0
2.25
C)W/m(463.2
1
C)W/m(37
1
C)W/m(7.0
)m(003.02
C)W/m(4.9
1
)C)(1020)(m(84.0
o2o2o2o2
o
2q
Thus the percent savings in energy are
%Savings = %75W186
W)3.46183(
vertical rectangular enclosure
102L
510Pr103 1010 Ra
properties at 2/)( hc TTT
PROBLEM 8.48 (continued)
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (c) and (8.38a) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.38a): Conditions listed in equation (8.38b) are satisfied.
(5) Comments. (i) Using double pane window reduces the heat loss by 75%. This is a
significant saving in energy. (ii) The assumption that ih TT and oc TT overestimates h . Thus
in fact heat loss form the double pane window is less than 46.3 W and the savings are more than 75%..
PROBLEM 8.49
To reduce heat loss form an oven, a glass door with a rectangular air
cavity is used. The cavity has a baffle at its center. The height of the
door is 65L cm and its width cm70w . The air space thickness is
cm5.1 . Estimate the heat transfer rate through the door if the
inside and outside surface temperatures of the cavity are C198o and
C42o .
(1) Observations. (i) Heat is transferred through the door from the inside to the outside. (ii) Newton’s law of cooling gives the heat transfer rate. (iii) The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected. (iv) The baffle divides the vertical cavity into two equal parts. This has the effect of decreasing the aspect ratio by a factor of two.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a rectangular cavity.
(3) Solution Plan. (i) Apply Newton's law of cooling (ii) Select an appropriate Nusselt number correlation equation for the rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Insulated top and bottom surfaces of the enclosure, (2) negligible radiation and (3) uniform surface temperatures.
(ii) Analysis. Newton’s law of cooling gives
)( ch TTAhq (a)
where
A = surface area of door = wL 0.065(m) 0.7(m) = 0.455 m2
h = enclosure heat transfer coefficient, W/m2-oC
q heat transfer rate, W
cT cavity cold side surface temperature = 42oC
hT cavity hot side temperature = 198oC
To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number Ra are
computed. The aspect ratio is given by
Aspect ratio = 667.21)m(015.0
)m)(2/65.0(L
The Rayleigh number for the enclosure is defined as
PrTTg
Ra ch
2
3)( (b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
enclosure air thickness = 1.5 cm = 0.015 m
= kinematic viscosity, m2/s
L
L/2
PROBLEM 8.49 (continued)
Air properties are determined at T
C1202
)C)(19842(
2
oo
hc TTT
At this temperature air properties are
k = 0.03261 W/m-oCPr = 0.703
= 25.19 10 6 m2/s
= 1/(120 + 273.15)K = 0.002544 1/K
4
2426
332
104557.1703.0)/sm()1019.25(
)m(015.0C))(42198()m/s(81.91/K002544.0Ra
Thus correlation equation (8.39a) is applicable
3.025.0012.0
42.0L
RaPrk
hNu (8.39a)
Valid for
(8.39b)
(iii) Computations. Equation (8.39a)) gives
826.1667.21104557.10.70342.03.04012.0 25.0
k
hNu
CW/m969.3)m(015.0
)CW/m(03261.0826.1 o2
o
h
Substituting into (b)
W7.281)C)(42198)(m(455.0)CW/m(969.3 o2o2q
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (8.39a) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.39a): Conditions listed in equation (8.39b) are satisfied.
(5) Comments. (i) Even with a double pane door the heat loss is significant. This added energy to the surroundings raises the temperature in the cooking area. (ii) Without the baffle the aspect
vertical rectangular enclosure
4010L
41021 Pr74 1010 Ra
properties at 2/)( hc TTT
PROBLEM 8.49 (continued)
ratio is 43.334. Although this is slightly over the limit of (8.39a), this equation can still be used without introducing significant error. With no baffle (8.39a)
CW/m224.3 o2h
8.228q W
Therefore the cavity acts to increase the heat loss and thus should be eliminated.
.
PROBLEM 8.50
The ceiling of an exhibit room is designed to provide
natural light by using an array of horizontal skylights.
Each unit is rectangular with an air gap cm5.6 thick.
The length and width of each unit are cm54L and
cm120w . On a typical day the inside and outside glass
surface temperatures are C15o and C15o . Estimate the rate of heat loss from each unit.
(1) Observations. (i) Heat is transferred through the skylight from the inside to the outside. (ii) Newton’s law of cooling gives the heat transfer rate. (iii) The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.
(2) Problem Definition. Determine the average free convection heat transfer coefficient for a rectangular cavity.
(3) Solution Plan. (i) Apply Newton's law of cooling (ii) Select an appropriate Nusselt number correlation equation for the rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Insulated end surfaces of the enclosure, (2) negligible radiation, (3) uniform surface temperatures and (4) negligible glass resistance no temperature drop across the glass).
(ii) Analysis. Newton’s law of cooling gives
)( ch TTAhq (a)
where
A = surface area of skylight = wL 0.54(m) 1.2(m) = 0.648 m2
h = enclosure heat transfer coefficient, W/m2-oC
q heat transfer rate, W
cT cavity cold side surface temperature = -15oC
hT cavity hot side temperature = 15oC
To determine cavity heat transfer coefficient h the aspect ratio and Rayleigh number Ra are
computed. The aspect ratio is given by
Aspect ratio = 3077.8)m(065.0
)m)(54.0(L
The Rayleigh number for the enclosure is defined as
PrTTg
Ra ch
2
3)( (b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
enclosure air thickness = 6.5 cm = 0.065 m
= kinematic viscosity, m2/s
L
w g
PROBLEM 8.50 (continued)
Air properties are determined at T
C02
)C)(1515(
2
oo
hc TTT
At this temperature air properties are
k = 0.02408 W/m-oCPr = 0.718
= 25.19 10 6 m2/s
= 1/(0 + 273.15)K = 0.003661 1/K
6
2426
332
101922.1718.0)/sm()1031.13(
)m(065.0C))(1515()m/s(81.91/K003661.0Ra
Thus correlation equation (8.39a) is applicable
074.03/1069.0 PrRa
k
hNu (8.41a)
Valid for
(8.41b)
(iii) Computations. Equation (8.39a)) gives
1389.7718.0101922.1069.0074.03/16
k
hNu
CW/m645.2)m(065.0
)CW/m(02408.01389.7 o2
o
h
Substituting into (b)
W4.51)C)(1515)(m(648.0.0)CW/m(645.2 o2o2q
(iv) Checking. Dimensional check: Computations showed that equations (a), (b) and (8.41a) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air
given in Table 1.1.
Validity of correlation equations (8.41a): Conditions listed in equation (8.41b) are satisfied.
(5) Comments. Using double pane skylight reduces heat loss to the surroundings.
horizontal rectangular enclosure heated from below
95 107103 Ra
properties at 2/)( hc TTT
PROBLEM 8.51
Repeat Example 8.4 using inclination angles of ,0o ,60o ,90o ,120o o150 and .175o Plot heat
transfer rate q vs. inclination angle .
(1) Observations. (i) Power requirement is equal to the heat transfer rate through the enclosure. (ii) The problem can be modeled as a rectangular cavity at specified hot and cold surface
temperatures. (iii) The inclination angle varies from o0 to .175o (iv)
Newton’s law of cooling gives the heat transfer rate. (v) The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.
(2) Problem Definition. Determine the average free convection
heat transfer coefficient h for a rectangular enclosure at various inclination angles.
(3) Solution Plan. (i) Apply Newton's law of cooling. (ii) Compute the aspect ratio and critical inclination angle. (iii) Select an appropriate Nusselt number correlation equation for convection for a horizontal, vertical and inclined rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Uniform hot and cold surface temperatures and (2) insulated end surfaces, (3) negligible radiation.
(ii) Analysis. Newton's law of cooling gives
)( ch TTAhqP (a)
where
A = surface area of rectangle = 0.7(m) 0.7(m) = 0.49 m2
h = average heat transfer coefficient, W/m2-oC
P power requirement, W q heat transfer rate through cavity, W
hT = hot surface temperature = 27oC
cT = cold surface temperature = 23oC
The aspect ratio is defined as
aspect ratio = L
(b)
where
L length of rectangle = 0.7 m
width of rectangle = 0.05 m
Equation (b) gives
14(m)05.0
(m)7.0L
According to Table 8.1, the critical angle is o70c . For 12/L and o700 , the
applicable correlation equation for the Nusselt number is (8.42a)
g
hT
cT
abso
rber
plat
e
L
PROBLEM 8.51 (continued)
*3/16.1*
118
)cos(
cos
)sin8.1(17081
cos
1708144.11
Ra
RaRak
hNu (8.42a)
Valid for
(8.42b)
The Rayleigh number is defined as
PrTTg
Ra ch
2
3
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Water properties are evaluated at the film temperatureT defined as
2
)( ch TTT (d)
To determine the applicable correlation equations for the horizontal, vertical and inclination
angles 90175 , the Rayleigh number is computed. Equation (d) gives
C252
)C)(2327( oo
T
Properties of water at this temperature are:
k = thermal conductivity = 0.6076 CW/m o
Pr = 6.13 310259.0 1/K
6108933.0 m2/s
Substituting into (c)
6
2426
3323
1075898.913.6)/sm()108933.0(
)m(05.0C))(2327()m/s(81.91/K10259.0Ra
Thus the applicable correlation equation for the horizontal position is (8.41a)
074.03/1069.0 PrRa
k
hNu (8.41a)
Valid for
inclined rectangular enclosure
12/L
c0
set 0* when negative
properties at 2/)( hc TTT
PROBLEM 8.51 (continued)
(8.41b)
For the vertical orientation the applicable correlation equation is (8.39a)
3/1046.0 Ra
k
hNu (8.39a)
Valid for
(8.39b)
For 90175 the applicable correlation equation is
sin1)90(1 oNuk
hNu (8.45a)
Valid for
(8.45b)
(iii) Computations.
(1) Horizontal position: .0 Substituting into (8.41a)
074.03/16o
o 6.131075898.9069.0)0(
)0(k
hNu = 16.862
CW/m9.204m)(05.0
)CW/m(6076.0862.16862.16)0( o2
oo k
h
The corresponding power is given by (a)
C))(2327)(C)(0.49)(mW/m(9.204)0( o2o2oP 401.6 W
horizontal rectangular enclosure heated from below
95 107103 Ra
properties at 2/)( hc TTT
vertical rectangular enclosure
401L
201 Pr96 1010 Ra
properties at 2/)( hc TTT
inclined rectangular enclosure
all /Loo 18090
properties at 2/)( hc TTT
PROBLEM 8.51 (continued)
(2) Inclination angle .60 Use (8.42a)
86.10118
)60cos1075898.9(
60cos1075898.9
)60sin8.1(17081
60cos1075898.9
1708144.11
)60()60(
*3/16
o6
6.1o*
o6
oo
k
hNu
CW/m132m)(05.0
)CW/m(6076.086.1086.10)60( o2
oo k
h
Equation (a) gives the required power
C))(2327)(C)(0.49)(mW/m(132)60( o2o2oP 258.7 W
(3) Vertical orientation, .90 Use (8.39a)
825.91075898.9046.0)90(
)90(3/16
oo
k
hNu
CW/m4.119m)(05.0
)CW/m(6076.0825.9825.9)90( o2
oo k
h
C))(2327)(C)(0.49)(mW/m(4.119)90( o2o2oP 234 W
(4) Inclination angle .120 Use (8.45a)
oo
o 120sin1825.91)120(
)120(k
hNu 8.643
CW/m03.105m)(05.0
)CW/m(6076.0643.8643.8)120( o2
oo k
h
C))(2327)(C)(0.49)(mW/m(03.105)120( o2o2oP 206 W
(5) Inclination angle .150 Use (8.45a)
oo
150sin1825.91)150(
)150(k
hNu 5.413
CW/m8.65m)(05.0
)CW/m(6076.0413.5413.5)150( o2
oo k
h
C))(2327)(C)(0.49)(mW/m(8.65)150( o2o2oP 129 W
(6) Inclination angle .175 Use (8.45a)
oo
o 175sin1825.91)175(
)175(k
hNu 1.77
CW/m5.21m)(05.0
)CW/m(6076.077.1647.8)175( o2
oo k
h
PROBLEM 8.51 (continued)
C))(2327)(C)(0.49)(mW/m(5.21)175( o2o2oP 42.1 W
Using the result of Example 8.4 and the above data, the required power at various angles is tabulated and plotted.
(iv) Checking. Dimensionalcheck: Computations showed that equations (a), (c), (8.39a), (8.41a), (8.42a) and (8.45a) are dimensionally correct.
Quantitative check: The magnitude of h is in line with typical free convection values for liquids given in Table 1.1.
Validity of correlation equations (8.39a), (8.41a), (8.42a) and (8.45a): Conditions listed in equations (8.39b), (8.41b), (8.42b) and (8.45b)are satisfied.
(5) Comments. (i) If the device is to be used continuously, the estimate power requirement is relatively high. Decreasing the temperature difference between the hot and cold surfaces will reduce the power requirement. (ii) The ambient temperature plays a role in the operation of the proposed device. The design must take into consideration changing ambient temperature. (iii) Specification of the driving motor should be based on highest power which corresponds to the horizontal orientation.
)(o P(W)
0 401.6
30 303.8
60 258.7
90 234
120 206
150 129
180 42.1
o
P(W)
030 60 90 120 150 1800
100
200
300
400
PROBLEM 8.52
A rectangular solar collector has an absorber plate of length
m5.2L and width m.0.4w A protection cover is used to
form a rectangular air enclosure of thickness cm4 to
provide insulation. Estimate the heat loss by convection from
the plate when the enclosure inclination angle is o45 and its
surfaces are at C28o and C.72o
(1) Observations. (i) The absorber plate is at a higher temperature than the ambient air. Thus heat is lost through the rectangular cavity to the atmosphere. (ii) The problem can be modeled as an inclined rectangular cavity at specified hot and cold surface temperatures. (iii) Newton’s law of cooling gives the heat transfer rate. (iv) The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.
(2) Problem Definition. Determine the average free convection heat transfer coefficient h for an inclined rectangular enclosure.
(3) Solution Plan. (i) Apply Newton's law of cooling. (ii) Compute the aspect ratio and critical inclination angle. Select an appropriate Nusselt number correlation equation for convection in an inclined rectangular cavity.
(4) Plan Execution.
(i) Assumptions. (1) Uniform hot and cold surface temperatures, (2) insulated end surfaces and (3) negligible radiation.
(ii) Analysis. Newton's law of cooling gives
)( ch TTAhq (a)
where
A = surface area of rectangle = 2.5(m) 4(m) = 10 m2
h = average heat transfer coefficient, W/m2-oC
q heat transfer rate through cavity, W
hT = hot surface temperature = 72oC
cT = cold surface temperature = 28oC
The aspect ratio is defined as
aspect ratio = L
(b)
where
L length of rectangle = 2.5 m
width of rectangle = 4 cm = 0.04 m
Equation (b) gives
5.62(m)04.0
.5(m)2L
g
hT
cT
abso
rber
plat
e
L
PROBLEM 8.52 (continued)
According to Table 8.1, the critical angle is o70c . Since 12/L and c0 , it
follows that the applicable correlation equation for the Nusselt number is
*3/16.1*
118
)cos(
cos
)sin8.1(17081
cos
1708144.11
Ra
RaRak
hNu (8.42a)
Valid for
(8.42b)
The Rayleigh number is defined as
PrTTg
Ra ch
2
3
(c)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
Water properties are evaluated at the film temperatureT defined as
2
)( ch TTT (d)
(iii) Computations. Equation (d) gives
C502
)C)(2872( oo
T
Properties of water at this temperature are:
k = thermal conductivity = 0.0.02781 CW/m o
Pr = 0709 0.0030946(K))15.27350/(1 1/K
61092.17 m2/s
Substituting into (c)
5
2426
332
1088745.1709.0)/sm()1092.17(
)m(04.0C))(2872()m/s(81.91/K0030946.0Ra
Substituting into (8.42a)
inclined rectangular enclosure
12/L
c0
set 0* when negative
properties at 2/)( hc TTT
PROBLEM 8.52 (continued)
2338.4118
)45cos1088742.1(
45cos1088742.1
)45sin8.1(17081
45cos1088742.1
1708144.11
*3/15
o5
6.1o*
o5k
hNu
CW/m9435.2m)(04.0
)CW/m(02781.02338.42338.4 o2
okh
Equation (a) gives the heat transfer rate
C))(2872)(C)(10)(mW/m(9435.2 o2o2q 1295 W
(iv) Checking. Dimensional check: Computations showed that equations (a), (8.42a) and (c) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air given in Table 1.1.
Validity of correlation equation (8.42a): Conditions listed in equation (8.42b) are satisfied.
(5) Comments. The rate of energy loss from the collector is significant. Increasing the thickness of the air gap will reduce the heat lost to the atmosphere.
PROBLEM 8.53
A liquid-vapor mixture at C20oiT flows inside a tube of diameter
cm4iD and length m.3L The tube is placed concentrically
inside another tube of diameter cm.6oD Surface temperature of the
outer tube is at C.10ooT Air fills the annular space. Determine the
heat transfer rate from the mixture.
(1) Observations. (i) Heat is transferred through the annular space from the outer cylinder to the inner. (ii) Newton’s law of cooling gives the heat transfer rate. (iii) The Rayleigh number should be determined for the enclosure formed by the concentric cylinders so that an appropriate correlation equation can be selected. (iv) The cylinders are horizontally oriented.
(2) Problem Definition. Determine the average free convection heat transfer coefficient annular cavity between two concentric horizontal cylinders.
(3) Solution Plan. (i) Use equation (8.46) describing heat transfer between two concentric cylinders. (ii) Select an appropriate correlation equation for this geometry.
(4) Plan Execution.
(i) Assumptions. (1) negligible radiation and (2) uniform surface temperatures.
(ii) Analysis. Heat transfer between two concentric cylinders is given by equation (8.46)
)()/ln(
2iTT
DD
kq o
io
eff (8.46)
where
iD 4 cm = 0.04 m
oD 6 cm =0.06 m
effk effective conductivity, CW/m o
q heat transfer rate per unit length of tube, W/m
iT = inner surface temperature (cold) = -20oC
oT = outer surface temperature (hot) = 10oC
Using (8.46), the total heat transfer rate from a tube of length L is
LTTDD
kq io
io
eff)(
)/ln(
2 (a)
where
L tube length = 3 m q total heat transfer rate, W
Correlation equation for the effective conductivity effk is
4/1*
861.0386.0 Ra
rP
rP
k
keff (8.47a)
where
iD
oD
iT
oT
PROBLEM 8.53 (continued)
Ra
DD
DDRa
oi
io5
5/35/33
4*
)()(
)/ln( (8.47b)
2
io DD (8.47c)
Valid for
(8.47d)
The Rayleigh number Ra in (8.47b) is defined as
PrTTg
Ra ch
2
3)( (b)
where
g = gravitational acceleration = 9.81 m/s2
Pr = Prandtl number
= coefficient of thermal expansion, 1/K
= kinematic viscosity, m2/s
(iii) Computations. Equation (8.47) gives
01.02
)(m)04.006.0(m
Air properties are determined at T
C52
)C)(1020(
2
oo
oi TTT
At this temperature air properties are
k = 0.023698 W/m-oCPr = 0.7195
= 12.885 10 6 m2/s
= 1/(-5 + 273.15)K = 0.0037293 1/K
Substituting into (b)
4.47567195.0)/sm()10885.12(
)m(01.0C))(1515()m/s(81.91/K0037293.02426
332
Ra
Use (8.47b)
2.4554.4756
)m60.0()m04.0()01.0(
)04.0/06.0ln(5
5/35/33
4*Ra
concentric cylinders 7*2 1010 Ra
properties at 2/)( oi TTT
PROBLEM 8.53 (continued)
Thus condition (8.47d) is satisfied. Substituting into (8.47a)
465.12.4557195.0861.0
7195.0386.0
4/1
k
keff
C)(W/m)02369.0(465.1465.1 okk ffe
Equation (a) gives q
m4.48)m(3C))(2010()m04.0/m06.0ln(
C)W/m)(0347.0(2 oo
q
(iv) Checking. Dimensional check: Computations showed that equations (a), (b), (8.47a) and (8.47b) are dimensionally consistent.
Quantitative check: The magnitude of h is in line with typical free convection values for air given in Table 1.1.
Validity of correlation equations (8.47a): Conditions listed in equation (8.47d) are satisfied.
(5) Comments. The concentric annular space provides good insulation. This is indicated by the
low value of kkeff / = 1.465. A ratio of unity corresponds to pure conduction with no fluid
circulation.
PROBLEM 9.1
The speed of sound, c, in an ideal gas is given by
TRc
where is the specific heat ratio R is gas constant and T is temperature. Show that
Re
MKn
2
where M is Mach number defined as
c
VM
(1) Observations. (i) Definitions of Knudsen number, Reynolds number, and Mach number are needed. (ii) Fluid velocity appears in the definition of Reynolds number and Mach number.
(2) Problem Definition. Show that the Knudsen number can be expressed in terms of Reynolds and Mach numbers.
(3) Solution Plan. Star with the definition of Knudsen number and multiply and divide by variables to form the Reynolds and Mach numbers
(4) Plan Execution.
(i) Analysis. The Knudsen number is defined as
eDKn (1.2)
The mean free path for an ideal gas is given by
RTp 2
(9.2)
Since the Reynolds number is expressed in terms of density , use the ideal gas law to
eliminate p in (9. 2) RTp (9.31)
Substitute (9.31) into (9.2)
22 RTRT
RT(a)
(a) into (1.2)
2RTDKn
e
(b)
Multiply and divide (b) by V
PROBLEM 9.1 (continued)
RT
V
VDKn
e2 (c)
Introduce the definition of the Reynolds and Mach number
eDVRe (d)
RT
V
c
VM (e)
Substitute (d) and (e) into (c)
Re
MKn
2 (f)
(ii) Checking: Dimensional check: Both sides of (f) are dimensionless.
(5) Comments. In determining the number of governing parameters in flow through microchannels, it should be noted that the three parameters, Kn, Re and M are not independent.
PROBLEM 9.2
Reported discrepancies in experimental data on the fiction factor f are partially attributed to
errors in measurements. One of the key quantities needed to calculate f is channel diameter D.
Show that5Df
(1) Observations. (i) The definition of friction factor shows that it depends on pressure drop, diameter, length and mean velocity. (ii) Mean velocity is determined from flow rate measurements and channel flow area.
(2) Problem Definition. Determine the dependency of friction factor on diameter.
(3) Solution Plan. Starting with the definition of friction factor f, express it in terms of diameter.
(4) Plan Execution.
(i) Assumptions. Continuum.
(ii) Analysis. Friction factor f is defined as
22
1
mu
p
L
Df (9.6b)
Pressure drop is determined by measuring the pressure at the inlet and outlet chambers. If pressure drop at the inlet and outlet can be neglected, then p is independent of diameter. Mean
velocity is determine from flow rate measurements:
4
2DuAum mm (a)
where
A = flow area m = mass flow rate
Solve (a) for mu
2
14
D
mum (b)
Substitute (b) into (9.6b)
5
2
24
2
2
32162
1D
mL
pD
mp
L
Df (f)
This result shows that f is proportional to the fifth power of diameter.
(iii) Checking.
Dimensional check: (f) should be dimensionless:
PROBLEM 9.2 (continued)
unityDmL
pf )m(
)kg/s(m)(
)mkg/s()kg/m(
32
55
22
232
(5) Comments. Accurate measurements of diameter or channel spacing in microchannels is critical in obtaining accurate data on friction factor.
PROBLEM 9.3
Consider shear driven Couette flow between parallel plates separated by a distance H. The
lower plate is stationary while the upper plate moves with a velocity .su Assume that no heat is
conducted through the lower plate and that the upper plate is maintained at uniform temperature
.sT Taking into consideration dissipation, velocity slip and temperature jump, determine the
Nusselt number. Assume steady state ideal gas flow.
(1) Observations. (i) The determination of the Nusselt number requires the determination of the temperature distribution. (ii) Temperature field depends on the velocity field. (iii) The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2. (iv) The solution to the energy equation gives the temperature distribution.
(2) Problem Definition. Determine the temperature distribution for Couette flow with insulated stationary plate and uniform temperature moving plate.
(3) Solution Plan. Start with the definition of the Nusselt number, use the velocity solution for Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve for the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Constant viscosity, conductivity and specific heat, (2) infinite plates, (3) uniform boundary conditions, (4) uniform spacing between plates, (5) no variation of density along y, (6) no gravity, (7)
,0.1Tu (8) velocity field is
independent of temperature, (9) ideal gas, and (10) continuum, slip flow regime conditions apply.
(ii) Analysis. The Nusselt number for flow between parallel plates is defined as
k
HhNu
2 (a)
The heat transfer coefficient h for channel flow is defined as
sm TT
y
HTk
h
)(
(b)
(b) into (a)
sm TT
y
HT
HNu
)(
2 (9.19)
where
k thermal conductivity of fluid
suy
xHu
sT
PROBLEM 9.3 (continued)
T fluid temperature function (variable)
mT fluid mean temperature
sT plate temperature
The mean temperature mT , as defined in Section 6.6.2, is
H
pmp dyTucWTmc
0
(9.21)
where
pc = specific heat
m = mass flow rate
T = temperature distribution u = velocity distribution
W = plate width
= density
The velocity distribution is given in Section 9
)(21
1Kn
H
y
Knu
u
s
(9.14)
2
suWHm (9.16)
(9.16) into (9.21) H
sm dyTu
HuT
0
2 (9.22)
Temperature distribution is governed by the energy equation. Based pm the above assumptions, energy equation (2.15) simplifies to
02
2
y
Tk (9.23)
where2
y
u (9.24)
(9.24) into (9.23)2
2
2
dy
du
kdy
Td(9.25)
Note that T is independent of x. Substitute (9.14) into (9.25)
2
2
2
)21( KnH
u
kdy
Td s (c)
Defining the constant as
PROBLEM 9.3 (continued)
2
)21( KnH
u
k
s (d)
Substituting (d) into (c)
2
2
dy
Td (e)
This energy equation requires two boundary conditions. They are:
0)0(
dy
dT (f)
The second boundary condition is at y = H. Plate temperature is specified at this boundary. However, the boundary condition must be associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature T(H). For
yHn and 1T , (9.11) gives
dy
HdT
PrHTTs
)(
1
2)( (9.20)
Solve for )(HT
dy
HdT
PrTHT s
)(
1
2)( (g)
Integration of (e) gives
DCyyT 2
2 (h)
where C and D are constants of integration. Application of boundary conditions (f) and (g) gives the two constants:
0C (i)
and
22
1
2
2H
Pr
KnHTD s (j)
Substituting into (h)
sTHPr
KnHyT 2
22
1
2
22(k)
To determine the Nusselt number using (9.19), equation (k) is used to formulate dy
HdT )( and
.mT Differentiating (k)
Hdy
HdT )( (l)
mT is determined by substituting (9.14) and (k) into (9.22)
PROBLEM 9.3 (continued)
dyDyKnH
y
KnHT
H
m )2
)(()21(
2 2
0
(m)
where D is defined in (j). Evaluating the integral, gives
DKnHHKn
Tm22
6
1
8
1
21
2
Substituting (j) into the above
sm THPr
KnHKnHH
KnT 2
222
1
2
26
1
8
1
21
2
or
sm THPr
KnKnHH
KnT 222
1
2
3
2
4
1
21
1 (n)
Using (l) and (n) into (9.19) gives the Nusselt number
222
2
1
2
3
2
4
1
21
1
2
HPr
KnKnHH
Kn
HNu
This simplifies to
Pr
KnKnKn
KnNu
)21(
1
8
3
81
)21(8 (o)
(iii) Checking.
Dimensional check: (i) Noting that units of are 2o C/m , each term in (n) has units of
temperature. (ii) The Nusselt number in (o) is dimensionless.
Limiting check: Setting Kn = 0 in (o) gives
8oNu (p)
This is the correct value of Nusselt number for macrochannel flow
(5) Comments.
(i) The Nusselt number is independent of the Reynolds number. This is also the case with macrochannel flows.
(ii) Unlike macrochannels, the Nusselt number depends on the fluid.
(iii) The Knudsen number in (o) represents the effect of rarefaction while the third term in the denominator represents the effect of temperature jump. Both act to reduce the Nusselt number.
PROBLEM 9.3 (continued)
(iv) If dissipation is neglected ( )0 , equation (k) gives the corresponding temperature solution
as
sTT
Thus, the temperature is uniform and no heat transfer takes place.
PROBLEM 9.4
A large plate moves with constant velocity su
parallel to a stationary plate separated by a
distance H. An ideal gas fills the channel
formed by the plates. The stationary plate is
at temperature oT and the moving plate is at
temperature sT . Assume laminar flow and
take into consideration dissipation and
velocity slip and temperature jump:
(a) Show that temperature distribution is given by
H
y
Pr
Kn
Pr
Kn
TT
H
y
H
y
Pr
Kn
Knk
uTT o
oss
1
2
1
221
1
2
)21(2 2
2
2
2
(b) Determine the heat flux at the plates.
(1) Observations. (i) Temperature distribution depends on the velocity field. (ii) The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2. (iii) The solution to the energy equation gives the temperature distribution. (iv) Two temperature boundary conditions must be specified. (v) Temperature distribution and Fourier’s law give surface heat flux.
(2) Problem Definition. Determine the temperature distribution for Couette flow with specified surface temperature on both plates.
(3) Solution Plan. Use the velocity solution for Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve for the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Constant viscosity, conductivity and specific heat, (2) infinite plates, (3) uniform boundary conditions, (4) uniform spacing between plates, (5) no variation of density
along y, (6) no gravity, (7) ,0.1Tu (8) the velocity field is independent of temperature,
and (9), ideal gas, and (10) continuum, slip flow regime conditions apply.
(ii) Analysis.
(a) Temperature distribution is governed by the energy equation. Based pm the above assumptions, energy equation (2.15) simplifies to
02
2
y
Tk (9.23)
2
y
u (9.24)
where
k = thermal conductivity of fluid
u = axial velocity
= fluid viscosity
suy
xHu
sT
oT
PROBLEM 9.4 (continued)
(9.24) into (9.23)2
2
2
dy
du
kdy
Td(9.25)
The velocity distribution is given in Section 9
)(21
1Kn
H
y
Knu
u
s
(9.14)
2
2
2
)21( KnH
u
kdy
Td s (a)
Defining the constant as
2
)21( KnH
u
k
s (b)
Substituting (b) into (a)
2
2
dy
Td (c)
This energy equation requires two boundary conditions. The temperature of each plate is specified. However, the boundary conditions must be associated with the fluid at y = H and not the plates. Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature. At ,0y fluid temperature T(0) is obtained from (9.11). At n = y = 0 and for
1T , equation (9.11) gives
dy
dT
PrTT o
)0(
1
2)0( (d)
Similarly, at n = H – y and 1T , (9.11) gives
dy
HdT
PrTHT s
)(
1
2)( (e)
Integration of (c) gives
DCyyT 2
2 (h)
where C and D are constants of integration. Application of boundary conditions (d) and (e) gives the two constants:
Pr
KnHH
TTHPr
KnH
Cos
1
22
1
2
2
22
(i)
and
PROBLEM 9.4 (continued)
HCPr
KnTD o
1
2(j)
Substitute (i) and (j) into (h)
oos T
Pr
Kn
H
y
Pr
Kn
TT
Pr
Kn
H
y
H
yHT
1
2
1
41
1
2
2 2
22
(k)
Using the definition of in (b) into (k) and rearranging, gives
H
y
Pr
Kn
Pr
Kn
TT
H
y
H
y
Pr
Kn
Knk
uTT o
oss
1
2
1
41
1
2
)21(2 2
2
2
2
(l)
This result can be written in dimensionless form as
H
y
Pr
Kn
Pr
KnH
y
H
y
Pr
Kn
TTKnk
u
TT
TT
oo
o
s
s
s 1
2
1
41
1
1
2
)()21(2 2
2
2
2
(m)
(b) Heat flux .q Application of Fourier’s law at y = 0
dy
dTkq
)0()0( (n)
(l) into (m)
HPr
KnH
TTk
KnH
uq oss
1
4
)(
)21(2)0(
2
2
(o)
(o) is written in dimensionless form as
Pr
KnTTKn
u
H
TTk
q
oo s
s
s
1
41
1
)()21(2)(
)0(2
2
(p)
Similarly, at y = H
dy
HdTkHq
)()( (q)
(l) into (q)
HPr
KnH
TTk
KnH
uHq oss
1
4
)(
)21(2)(
2
2
(r)
Written in dimensionless form, (r) becomes
PROBLEM 9.4 (continued)
Pr
KnTTKn
u
H
TTk
Hq
oo s
s
s
1
41
1
)()21(2)(
)(2
2
(s)
(iii) Checking.
Dimensional check: (i) Noting that units of are 2o C/m , each term in (k) has units of
temperature. (ii) Each term in (o) and (r) has units of heat flux. (iii) each term in (m), (p) and (s) is dimensionless.
Limiting check: (i) Since the velocity profile is linear, dissipation is uniform. Thus, if oTTs ,
heat flux at each plate should be equal in magnitude and opposite in direction. Setting oTTs in
(o) and (r) gives
2)21(2)0(
2
KnH
uq s (t)
2)21(2)(
2
KnH
uHq s (u)
(ii) If dissipation is neglected, temperature distribution should be linear. Setting 0 in (l) gives
H
y
Pr
Kn
Pr
Kn
TTTT o
os
1
2
1
41
(s)
(iii) If dissipation and rarefaction are neglected, temperature distribution should be the same as one dimensional conduction. Setting Kn = 0 in (l) gives
H
yTTTT oo s )( (v)
(5) Comments.
(i) The Knudsen number in (l), (o) and (r) represents the effect of rarefaction while the Prandtl number terms represents the effect of temperature jump.
(ii) The solution is governed by two parameters:
Dissipation parameter = )()21( 2
2
oTTKnk
u
s
s (w)
Temperature jump parameter =Pr
Kn
1 (x)
PROBLEM 9.5
Consider Couette flow between two parallel plates separated by a distance H. The lower
plate moves with velocity 1su and the upper plate moves in the opposite direction with
velocity .2su The channel is filled with ideal gas. Assume velocity slip conditions,
determine the mass flow rate. Under what condition will the net flow rate be zero?
(1) Observations. (i) To determine mass flow rate it is necessary to determine the velocity distribution. (ii) Velocity slip takes place at both boundaries of the flow channel. (iii) Because plates move in opposite directions, the fluid moves in both directions. This makes it possible for the net flow rate to be zero.
(2) Problem Definition. Determine the velocity distribution in the channel.
(3) Solution Plan. Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) one-dimensional (no variation with axial distance x and normal distance z), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7) negligible lateral variation of density, (8) the velocity
accommodation coefficient is equal to unity, ,0.1u (9) continuum, slip flow regime
conditions apply, (10) ideal gas, and (11) no gravity.
(ii) Analysis. Mass flow rate is given The flow rate, m , for a channel of width W is
given by H
dyuWm
0
(9.15)
whereu axial velocity
density
To determine u we follow the analysis of Section 9.6.2. The axial component of the Navier-Stokes equations for Couette flow between parallel plates is given by (9.12)
02
2
dy
ud(9.12)
Boundary conditions for (9.12) are given by (9.10)
n
xuuxu
u
us
)0,(2)0,( (9.10)
Applying (9.10) to the lower surface, 0yn , and setting 1u
dy
duuu s
)0()0( 1 (a)
y
x
H
1su
2su
u
PROBLEM 9.5 (continued)
For the upper surface, n = H – y, (9.10) gives
dy
HduuHu s
)()( 2 (b)
The solution to (9.12) is ByAu (c)
Boundary conditions (a) and (b) give the two constants of integration A and B
)21(
21
KnH
uuA ss ,
)21(
)( 211
Kn
KnuuuB ss
s (d)
where Kn is the Knudsen number, defined as
HKn (9.13)
Substituting (d) into (c)
)()21(
)( 211
H
yKn
Kn
uuuu ss
s (e)
Substituting (e) into (9.15) and noting that is assumed constant along y, gives
H
dyKnH
y
Kn
uuuWm ss
s
0
211 )(
21 (f)
Evaluating the integral
212
ss uuWH
m (g)
Examination of this result shows that the net mass flow rate is zero when the two velocities are the same. That is m = 0 for
21 ss uu (h)
(iii) Checking.
Dimensional check: Equation (g) has the correct units for mass flow rate.
Limiting Check: For the special case of stationary lower plate, ,02su equation (g)
reduces to
12
suWH
m (i)
This agrees with (9.17) of Section 9.6.2.
Boundary conditions check: Solution (d) satisfies boundary conditions (a) and (b).
(5) Comments. (i) The effect of slip is to decrease fluid velocity at the upper and lower surfaces. (ii) Because the velocity distribution is linear, slip velocity is the same for both plates. (iii) The mass flow rate is independent of Knudsen number.
PROBLEM 9.6
Determine the frictional heat generated by the fluid in Example 9.1.
(1) Observations. (i) In Example 9.1, Couette flow between parallel plates is used to model the flow in the channel between the shaft and housing. (ii) At steady state, the heat generated due to friction (dissipation) is equal to the net heat conducted from the channel. (iii) Since no heat is transferred to the shaft, the net heat leaving the channel is equal to the heat conducted through the housing surface. (iv) Velocity slip takes place at both boundaries of the flow channel. (v) To determine heat transfer rate it is necessary to determine fluid temperature distribution. This requires the determination of the velocity field.
(2) Problem Definition. Determine the velocity and temperature distribution in the channel.
(3) Solution Plan. (i) Model channel flow as Couette flow between parallel plates. (ii) Apply Fourier’s law at the housing surface to determine heat leaving the channel. (iii) Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1. (iv) Use the energy equation to determine the temperature distribution
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) one-dimensional (no variation with axial distance x and normal distance z), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7) negligible lateral variation of density, (8) the velocity accommodation
coefficients are equal to unity, ,0.1Tu (9) continuum, slip flow regime conditions apply,
(10) ideal gas, and (11) no gravity.
(ii) Analysis. Apply Fourier’s law at the housing surface (upper plate)
dy
HdTkAHq
)()( (a)
where
A surface area k fluid thermal conductivity
q heat transfer rate
T fluid temperature distribution
Surface area A for a shaft of length L is
LHRA )(2 (b)
(b) into (a)
dy
HdTLHRkHq
)()(2)( (c)
y
x
H
1su
2su
u
2T
PROBLEM 9.6 (continued)
Flow Field. To determine u we follow the analysis of Section 9.6.2. The axial component of the Navier-Stokes equations for Couette flow between parallel plates is given by (9.12)
02
2
dy
ud(9.12)
Boundary conditions for (9.12) are given by (9.10)
n
xuuxu
u
us
)0,(2)0,( (9.10)
Applying (9.10) to the lower surface, 0yn , and setting 1u
dy
duuu s
)0()0( 1 (d)
For the upper surface, n = H – y, (9.10) gives
dy
HduuHu s
)()( 2 (e)
The solution to (9.12) is ByAu (f)
Boundary conditions (a) and (b) give the two constants of integration A and B
)21(
21
KnH
uuA ss ,
)21(
)( 211
Kn
KnuuuB ss
s (g)
where Kn is the Knudsen number, defined as
HKn (9.13)
Substituting (g) into (f)
)()21(
)( 211
H
yKn
Kn
uuuu ss
s (h)
Temperature Field. Temperature distribution is governed by the energy equation. Based pm the above assumptions, energy equation (2.15) simplifies to
02
2
y
Tk (9.23)
where2
y
u (9.24)
(9.24) into (9.23)2
2
2
dy
du
kdy
Td(9.25)
PROBLEM 9.6 (continued)
Note that T is independent of x. Substitute (h) into (9.25)
2
21
2
2
)21( KnH
uu
kdy
Td ss (i)
Defining the constant as
2
2
)21( KnH
uu
k
ss (j)
Substituting (d) into (c)
2
2
dy
Td (k)
This energy equation requires two boundary conditions. They are:
0)0(
dy
dT (l)
The second boundary condition is at y = H. Plate temperature is specified as 2T at this boundary.
However, the boundary condition must be associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature jump condition (9.11) gives fluid temperature T(H). For
yHn and 1T , (9.11) gives
dy
HdT
PrHTT
)(
1
2)(2 (9.20)
Solve for )(HT
dy
HdT
PrTHT
)(
1
2)( 2 (m)
Integration of (k) gives
DCyyT 2
2 (n)
where C and D are constants of integration. Application of boundary conditions (l) and (m) gives the two constants:
0C (o)
and
22
21
2
2H
Pr
KnHTD (p)
Substituting into (n)
22
22
1
2
22TH
Pr
KnHyT (q)
Rewriting (q) in dimensionless form
Pr
Kn
H
y
H
TT
1
21
2
12
2
2
2
PROBLEM 9.6 (continued)
Using the definition of in the above
Pr
Kn
H
y
Kn
uu
k
TT
ss1
21
2
1
)21(
2
2
2
2
2 (r)
Frictional Heat. Using (r) to form the temperature gradient at Hy and substituting into (c)
gives the heat generated by fluid friction
2
2
)21()(2)(
Kn
uu
H
LHRHq ss (s)
(iii) Checking.
Dimensional check: Each term in (h) has units of velocity. Each term in (q) has units of temperature. Each term in (r) is dimensionless. q(H) in (s) is expressed in watts.
Governing equations check: Velocity solution (h) satisfies (9.12). Temperature solution (r) satisfies (9.25).
Boundary conditions check: Velocity solution (h) satisfies boundary conditions (d) and (e). Temperature solution (r) satisfies boundary conditions (l) and (m).
Limiting checks:
(i) If the two plates are stationary )0( 21 ss uu , there is no fluid motion, gas temperature is
uniform, and there is no frictional energy. Setting 021 ss uu in (h) gives u = 0. Setting 0
in (q) gives .2TT Setting 021 ss uu in (s) gives q(H) = 0.
(5) Comments. (i) The velocity field is governed by a single parameter, Kn. The temperature
field is governed by the parameter .1
2
Pr
Kn
(ii) Equation (s) shows that the narrower the gap H between the rotor and housing, the greater is the frictional energy. In addition, frictional energy decreases as the Knudsen number is increased.
PROBLEM 9.7
Consider shear driven Couette flow between
parallel plates. The upper plate moves with velocity
su and is maintained at uniform temperature .sT
The lower plate is heated with uniform flux .oq The
fluid between the two plates is an ideal gas. Taking
into consideration velocity slip, temperature jump,
and dissipation, determine the temperature of the
lower plate.
(1) Observations. (i) To determine the temperature of the lower plate, fluid temperature distribution must be known. (ii) Temperature distribution depends on the velocity field. (iii) The velocity field for Couette flow with a moving upper plate is given in Section 9.6.2. (iv) The solution to the energy equation gives the temperature distribution. (v) Two temperature boundary conditions must be specified.
(2) Problem Definition. Determine the temperature distribution for Couette flow with uniform flux at the lower plate and specified temperature at the upper plate.
(3) Solution Plan. Use the velocity solution for Couette flow of Section 9.6.2, formulate the energy equation and boundary conditions, and solve for the temperature distribution.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) Constant viscosity, conductivity and specific heat, (4) infinite plates, (5) uniform boundary conditions, (6) uniform spacing between
plates, (7) no variation of density and pressure along y, (8) no gravity, (7) ,0.1Tu (9) the
velocity field is independent of temperature, (10), ideal gas, and (11) continuum, slip flow regime conditions apply.
(ii) Analysis. Surface temperature is related to fluid temperature through temperature jump condition (9.11)
n
xT
PrTxT
T
Ts
)0,(
1
22)0,( (9.11)
Solving the above for plate temperature at y = 0, oT , and assuming 1T , gives
n
xT
PrxTTo
)0,(
1
2)0,( (a)
where )0,(xT fluid temperature at the lower plate n = y . Thus fluid temperature distribution is
needed to determine oT . Temperature distribution is governed by the energy equation. Based on
the above assumptions, energy equation (2.15) simplifies to (see Section 9.6.2)
02
2
y
Tk (9.23)
where
suy
xHu
sT
oq
PROBLEM 9.7 (continued)
2
y
u (9.24)
where
k = thermal conductivity of fluid
u = axial velocity
= fluid viscosity
Noting that velocity and temperature are independent of axial distance, (9.24) into (9.23)2
2
2
dy
du
kdy
Td(9.25)
The velocity distribution is given in Section 9.6.2
)(21
1Kn
H
y
Knu
u
s
(9.14)
2
2
2
)21( KnH
u
kdy
Td s (b)
Defining the constant as
2
)21( KnH
u
k
s (c)
Substituting (c) into (b)
2
2
dy
Td (d)
This energy equation requires two boundary conditions. Heat flux is specified at the lower plate. Fourier’s law gives
oqdy
dTk
)0( (e)
Surface temperature is specified at the upper plate. However, this boundary condition must be associated with the fluid at y = H and not the plate. Knowing plate temperature, temperature
jump condition (9.11) gives fluid temperature. Thus at n = H – y and 1T , (9.11) gives
dy
HdT
PrTHT s
)(
1
2)( (f)
Integration of (d) gives
DCyyT 2
2 (g)
where C and D are constants of integration. Application of boundary conditions (e) and (f) gives the two constants:
PROBLEM 9.7 (continued)
k
qC o , s
o TPr
KnH
Pr
Kn
k
HqD
1
41
21
21
2
(h)
Substitute (h) into (g)
22
21
41
21
21 yy
k
q
Pr
KnH
Pr
Kn
k
HqTT oo
s (i)
Using the definition of in (c), the above becomes
2
222
)21(2
1
1
41
)21(21
21
H
y
Kn
u
ky
k
q
Pr
Kn
Kn
u
kPr
Kn
k
HqTT soso
s (j)
Surface temperature of the lower plate is determined by substituting (j) into (a) and setting y = 0
Pr
Kn
Kn
u
kk
HqTT s
so
o1
41
)21(2 2
2
(k)
(iii) Checking.
Dimensional check: (i) Noting that units of are 2o C/m , each term in (j) has units of
temperature.
Limiting check: (i) If dissipation is neglected, temperature distribution should be linear. Setting 0 in (j) gives
yk
q
Pr
Kn
k
HqTT oo
s1
21 (l)
(ii) If dissipation and rarefaction are neglected, the process reduces to pure conduction. Setting 0Kn in (j) gives
H
y
k
qTT o
s 1 (m)
This is the one dimensional conduction solution to the problem.
(5) Comments. (i) The Knudsen number in (j) and (l) represents the effect of rarefaction while the Prandtl number term represents the effect of temperature jump.
(ii) Fluid temperature adjacent to the lower surface is obtained by setting y = 0 in (j)
Pr
Kn
Kn
u
kPr
Kn
k
HqTT so
s1
41
)21(21
21)0(
2
(n)
PROBLEM 9.7 (continued)
To examine the difference between plate and fluid temperature at y = 0, (n) is subtracted from (k)
k
Hq
Pr
KnTT o
o1
2)0( (o)
This result shows that departure of plate temperature from fluid temperature at y = 0 increases with increasing heat flux and rarefaction.
(iii) To identify the governing parameters, solution (j) is expressed in dimensionless form
2
222
)21(21
41
)21(21
21
H
y
Kn
u
Hqh
y
Pr
Kn
Kn
u
HqPr
Kn
k
Hq
TT s
o
s
oo
s (p)
This result shows that temperature distribution is governed by two parameters:
Pr
Kn
1
2 and
2
)21(2 Kn
u
Hq
s
o
(q)
PROBLEM 9.8
Pressure distribution in Poiseuille flow between parallel plates is given by
L
x
p
pKn
p
p
p
pKnKn
p
xp
o
io
o
i
o
ioo
o
)1(12)1(66)(
2
22
(9.35)
This equation was derived in Section 9.6.3 using the continuity equation to determine the y
velocity component v. An alternate approach to derive (9.35) is based on the condition that for
steady state the flow rate is invariant with axial distance x. That is
022/
0
H
udyWdx
d
dx
dm
where W is channel width. Derive (9.35) using this approach.
(1) Observations. (i) To use the proposed approach, the solution to the axial velocity distribution must be know. (ii) The velocity distribution for Poiseuille flow between parallel plates is given by equation (9.30) of Section 9.6.3.
(2) Problem Definition. Determine the mass flow rate.
(3) Solution Plan. Use the solution to the axial velocity for Poiseuille flow, equation (9.30), to determine the mass flow rate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7) no
gravity, (8) ,0.1u (9) the velocity field is independent of temperature, (10) ideal gas, and
(11) continuum, slip flow regime conditions apply.
(ii) Analysis. To derive pressure solution (9.35), it is proposed to use the following conservation of mass equation
02
2/
0
H
udyWdx
d
dx
dm(a)
where W = plate width. The solution to the axial velocity u is given by (9.30)
2
22
4)(418 H
ypKn
dx
dpHu (9.30)
Substituting (9.30) into (a) and noting that varies along x and is assumed constant along y, we
obtain
04)(418
2
2/
02
22H
dyH
ypKn
dx
dpHW
dx
d
dx
dm
This result simplifies to
PROBLEM 9.8 (continued)
04)(41
2/
02
2H
dyH
ypKn
dx
dp
dx
d
dx
dm (b)
Evaluating the integral in (b)
0)(23
pKnHH
dx
dp
dx
d(c)
To proceed, the density and Knudsen number in (c) must be expressed in terms of pressure. Ideal gas law (9.31) gives
RT
p (d)
The Knudsen number is expressed in terms of pressure in (9.33)
pRT
HHKn
1
2 (9.33)
(d) and (9.33) into (c)
01
22
3 pRT
H
dx
dp
RT
p
dx
d
Assuming isothermal flow, the above simplifies to
01
2
2
3
1
pRT
Hdx
dpp
dx
d(e)
Integrating (e) once
Cp
RTHdx
dpp )
1
2
2
3
1
Rewriting the above
Cdx
dpRT
Hdx
pdp
2
2
3
1(f)
Integrating again
DCxpRTH
p 26
1 2 (g)
The boundary conditions on p are
ipp )0( , opLp )( (h)
Here L is channel length. Equation (g) and (h) are identical to (o) and (q) of Section 9.6.3. Thus the solution to p is the same for both, given in (9.35)
L
x
p
pKn
p
p
p
pKnKn
p
xp
o
io
o
i
o
ioo
o
)1(12)1(66)(
2
22
(9.35)
(4) Comments. This approach for determining p(x) is simpler than that used in Section 9.6.3 where it was necessary to first determine the normal velocity component v.
PROBLEM 9.9
One of the factors affecting mass flow rate through a microchannels is channel height H. To
examine this effect, consider air flow through two microchannels. Both channels have the same
length, inlet pressure and temperature and outlet pressure. The height of one channel is double that of the other. Compute the mass flow ratio for the following case:
m,51H m,102H kPa,420ip kPa,105op C30 oiT
(1) Observations. (1) This is a pressure driven microchannel Poiseuille flow between parallel plates. (ii) The solution to mass flow rate through microchannels is given in Section 9.6.3. (iii) Channel height affects the Knudsen number.
(2) Problem Definition. Determine the mass flow rate for microchannel Poiseuille flow between parallel plates.
(3) Solution Plan. Apply the mass flow solution, equation (9.39), to two channels having different heights and take their ratio.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7) no
gravity, (8) ,0.1u (9) the velocity field is independent of temperature, (10) ideal gas, (11)
continuum, slip flow regime conditions apply, and (12) fully developed flow.
(ii) Analysis. The mass flow rate through Poiseuille flow microchannels is given by equation (9.39):
)( 112124
12
223
o
io
o
io
p
pKn
p
p
LRT
pHWm (9.39)
Apply (9.39) to two channels having heights 1H and 2H , and outlets Knudsen numbers 1oKn
and 2oKn
)( 112124
112
2231
1
o
io
o
io
p
pKn
p
p
LRT
pHWm (a)
)( 112124
122
2232
2
o
io
o
io
p
pKn
p
p
LRT
pHWm (b)
Take the ratio of (b) and (a)
)(
)(
1121
1121
1
2
2
2
3
1
2
1
2
o
io
o
i
o
io
o
i
p
pKn
p
p
p
pKn
p
p
H
H
m
m (c)
Equation (9.34) gives the outlet Knudsen number
PROBLEM 9.9 (continued)
RTpH
Kno
o2
(9.34)
Apply (9.34) to the two channels
RTpH
Kno
o21
1 (d)
RTpH
Kno
o22
2 (e)
(iii) Computations. The following date is given
m105m5 -61H
m1001m10 -62H
m-s
kg000,105kPa105
2op
m-s
kg000,420kPa420
2ip
C30 ooi TT
Properties of air at this temperature are
712.0Pr
K-s
m287
K-kg
J287
2
2
R
m-s
kg1065.18
6
Substituting into (d) and (e)
01313.0)K)(303)(K-s
m)(287(
2)
m-s
kg)(000,105((m)105
)m-s
kg(1065.18
2
2
2
6-
6
1oKn
006566.0)K)(303)(K-s
m)(287(
2)
m-s
kg)(000,105((m)1001
)m-s
kg(1065.18
2
2
2
6-
6
2oKn
Substituting into (c)
PROBLEM 9.9 (continued)
878.7
1105
420)01313.0(121
105
420
1105
420)006566.0(121105
420
5
10
)(
)(
2
2
3
1
2
m
m
(4) Checking.
Dimensional check: Computations showed that units of Kn in (d) and (e) are dimensionless.
Limiting check: If 21 HH , the mass ratio should be unity. Setting 21 HH in (d) and (e) gives
21 oo KnKn
Setting 21 HH and 21 oo KnKn in (c) gives
21 mm
(5) Comments. (i) The effect of channel size on mass flow rate is significant. (ii) Setting
021 oo KnKn in (c) gives the mass ratio for macrochannels
81
2
m
m
This indicates that the effect of rarefaction on the mass ratio is 1.5%.
PROBLEMT 9.10
A micro heat exchanger consists of rectangular channels of height m,25H width
m,600W and length mm.10L Air enters the channels at temperature C20 oiT and
pressure kPa.420ip The outlet pressure is kPa.105op The air is heated with uniform
surface heat flux .W/m1100 2sq Taking into
consideration velocity slip and temperature jump,
assume fully developed conditions, compute the
following:
(a) Mass flow rate, m.
(b) Mean outlet temperature, .moT
(c) Heat transfer coefficient at the outlet, ).(Lh
(d) Surface temperature at the outlet, ).(LTs
(1) Observations. (i) This is a pressure driven microchannel Poiseuille flow. (ii) Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates. (iii) Channel surface is heated with uniform flux. (iv) The solution to mass flow rate, temperature distribution, and Nusselt number for fully developed Poiseuille channel flow with uniform surface flux is presented in Section 9.6.3.
(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed Poiseuille channel flow with uniform surface heat flux.
(3) Solution Plan. Apply the analysis and results of Section 9.6.3.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7)
,0.1Tu (8) no gravity, (9) the velocity field is independent of temperature, (10) ideal
gas, (11) continuum, slip flow regime conditions apply, and (12) fully developed flow.
(ii) Analysis.
(a) Mass flow rate m. Equation (9.39) gives the mass flow rate through the channel:
)( 112124
12
223
o
io
o
io
p
pKn
p
p
LRT
pHWm (9.39)
where
m-61052m25heightchannelH
oKn outlet Knudsen number = )( opKn
L channel length = 0.01 m
q
q
H
W
L
PROBLEMT 9.10 (continued)
ip inlet pressure m-kg/s000,420kPa420 2
m-kg/s000,105kPa105pressureoutlet 2op
2W/m1100sq
m10600 widthchannel 6W
K-s/m287K-J/kg287constantgas 22R
m-kg/s1017.18 6
Equation (9.34) gives the outlet Knudsen number
oooo
pRT
HHpKnKn
1
2)( (9.34)
where moo TT is the mean outlet temperature.
(b) Mean outlet temperature, .moT The local mean temperature )(xTm is given by equation
(9.60):
mimp
sm Tx
Huc
qxT
2)( (9.60)
where
CJ/kg3.998heatspecific opc
2W/m1100fluxheatsurfacesq
C20eratureinlet tempmean oimi TT
s
mity,mean velocmu
3kg/mdensity,
The mean outlet temperature is obtained by setting x = L in (9.60)
mimp
smmo TL
Huc
qLTT
2)( (a)
The product mu is determined from the mass flow rate using the continuity equation:
HW
mum (b)
(b) into (a)
mip
smmo TLW
mc
qLTT
2)( (c)
CW/m02564.0 ok
C20 ooi TT
PROBLEMT 9.10 (continued)
(c) Heat transfer coefficient at the outlet, ).(Lh The Nusselt number is used to determine the
heat transfer coefficient. Nusselt number for channel flow is defined as
k
hHNu
2 (d)
where
Cm
W02564.0tyconcuctivithermal
ok
Applying (d) at the outlet, x = L and solving for h(L)
)(2
)( LNuH
kLh (e)
The Nusselt number is given by (9.64)
KnPr
1KnKn
KnKn
Kn
Nu
1
2
560
13
40
13)(
)61(
1
48
5
2
1
)61(
3
2
2
(9.64)
where
Kn local Knudsen number 0.713numberPrandtlPr
= specific heat ratio = 1.4
Evaluation (9.64) at x = L where oKnKn
zzzKnPr
1KnKn
KnKn
Kn
Nu
oooo
oo 1
2
560
13
40
13)(
)61(
1
48
5
2
1
)61(
3
2
2
(f)
(d) Surface temperature at the outlet, ).(LTs Surface temperature distribution is given by
(9.63):
)(1
2
48
5
2
1
)61(
3)( xgKn
Prk
HqKn
Knk
HqxT ss
s (9.63)
where g(x) is given by (9.62):
560
13
40
13)(
)61(
32)( 2
2KnKn
Knk
Hqx
Huc
qTxg s
mp
smi (9.62)
Substituting (b) into (9.62), setting x = L and oKnKn , gives
560
13
40
13)(
)61(
32)( 2
2 oos
p
smi KnKn
Knk
HqL
mc
WqTLg (g)
PROBLEMT 9.10 (continued)
To determine surface temperature at the outlet, ),(LTs the Knudsen number in (9.62) and (9.63) is
evaluated at outlet pressure and g(x) is evaluated at x = L.
(iii) Computations.
(a) Mass flow rate m. Equation (9.39) for m is based on the assumption that the flow is
isothermal. Since the outlet temperature oT is not yet determined, as a first approximation we
assume K293io TT in (9.34)
002516.0K))(293)(K-s/m)(287(2)m-kg/s)(000,105(m)(1052
)m-kg/s(1017.18 22
26-
6
oKn
Substituting into (9.39)
)( 1105000
420000002516.0(121
105000
420000
K))(293)(K-s/m(287m))(01.0()m-kg/s(1017.18
m-kg/s)000,105((m))1025((m)10600
24
12
226
222366 3
m
kg/s1025126.4 6m
(b) Mean outlet temperature, .moT Equation (c) gives
mip
smmo TLW
mc
qLTT
2)( (d)
)C(20m))(10m)(600)(01.0()kg/s)(1025126.4)(CJ/kg(3.998
)W/m()1100(2)( o6-
6-o
2
LTT mmo
C11.23 omoT
(c) Heat transfer coefficient at the outlet, ).(Lh Setting oKnKn , and substituting into (9.64),
gives the Nusselt number at the outlet
(0.713)
)002516.0(
14.1
)4.1(2
560
13)002516.0(
40
13)002516.0(
)002516.0(61
1
48
5)002516.0(
2
1
)002516.0(61
3
2)(
2
LNu
14.8)(LNu
Equation (e) gives h(L)
CW/m417414.8)(m)6-1052(2
)CW/m(02564.0)( o2
o
Lh
(d) Surface temperature at the outlet, ).(LTs Use (f) to compute g(L)
PROBLEMT 9.10 (continued)
560
13)002516.0(
40
13)002516.0(
)002516.061)(CW/m(02564.0
)(m)-61052)(W/m()1100(3
(m)01.0)(kg/s)-61025126.4)(CJ/kg(3.998
)(m)10600)(W/m()1100(2C)(20)(
2
2o
2
o
62oLg
04.23)(Lg Co
Substitute into (9.63)
)002516.0()713.0)(CW/m(02564.0
)(m)-61052)(W/m()1100(
14.1
)4.1(2
48
5)002516.0(
2
1
)002516.061)(CW/m(02564.0
)(m)-61052)(W/m()1100(3)(
o
2
o
2
LTs
C373.23)( oLTs
(iv) Checking.
Dimensional check: computations showed that equations (9.34), (9.39), (9.60), (9.62) and (9.63) are dimensionally correct.
Surface temperature check: Application of Newton’s law at the outlet gives
moss TLTLhq )()(
Solving for )(LTs
mos
s TLh
qLT
)()(
Using this equation to compute )(LTs , gives
C23.374C11.23)CW/m(4174
)W/m()1100()( oo
o2
2
LTs
This is close to the value determined above.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on the assumption that the flow is isothermal. Computation showed that the outlet temperature is
C11.23 omoT . Since the outlet is C11.3 o above the inlet temperature, it follows that the
assumption of isothermal flow is reasonable. (ii) The heat transfer coefficient at the outlet is high compared to values for air encountered in typical macrochannels applications. (iii) The Nusselt
number for slip theory for fully developed macrochannel flow is obtained by setting 0oKn in
(f). This gives
235.8oNu
Thus macrochannel theory overestimates the Nusselt number if applied to microchannels.
PROBLEM 9.11
Rectangular microchannels are used to remove heat
from a device at uniform surface heat flux. The height,
width, and length of each channel are m,29.6H
m,90W and mm,10L respectively. Using air
at C20oiT as the coolant fluid, determine the mass
flow rate and the variation of Nusselt number along
the channel. Inlet and outlet pressure are 410ip
kPa, 105op kPa. Assume steady state fully
developed slip flow and temperature jump conditions.
(1) Observations. (i) The problem can be modeled as pressure driven Poiseuille flow between two parallel plates with uniform surface flux. (ii) Assuming fully developed velocity and temperature, the analysis of Section 9.6.3 gives the mass flow rate and Nusselt number. (iii) The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the channel due to pressure variation, it follows that pressure distribution along the channel must be determined.
(2) Problem Definition. Determine the flow and temperature fields for fully developed Poiseuille flow with uniform surface flux.
(3) Solution Plan. Apply the results of Section 9.6.3 for the mass flow rate, pressure distribution, and Nusselt number.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no variation along the width W), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity and specific heats, (7) negligible lateral variation of density and pressure, (8) the
accommodation coefficients are equal to unity, ,0.1Tu (9) negligible dissipation, (10)
uniform surface flux, (11) negligible axial conduction, and (12) no gravity.
(ii) Analysis. Assuming isothermal flow, the results of Section 9.6.3 give the mass flow rate as
)( 112124
12
223
o
io
o
i
o
o
p
pKn
p
p
LRT
pHWm (9.39)
The Knudsen number at the exit, oKn is
oo
oo RT
pHH
pKn
2
)( (9.34)
where the temperature oT at the outlet is assumed to be the same as inlet temperature and the
viscosity is based on inlet temperature.
The Nusselt number, ,Nu is given by
x
yHL
Wsq
sqm
PROBLEM 9.11 (continued)
KnPr
1KnKn
KnKn
Kn
Nu
1
2
560
13
40
13)(
)61(
1
48
5
2
1
)61(
3
2
2
(9.64)
The local Knudsen number, ,Kn depends on the local pressure p(x) according to
RTpHH
Kn2
(9.33)
Equation (9.35) gives )(xp
L
x
p
pKn
p
p
p
pKnKn
p
xp
o
io
o
i
o
ioo
o
)1(12)1(66)(
2
22
(9.35)
Thus, (9.35) is used to determine p(x), (9.33) to determine ),(xKn and (9.64) to determine the
variation of the Nusselt number along the channel.
(iii) Computations. Air properties are determined at .C20o To compute p(x), ),(xKn
and ,Nu the following data is used
m29.6H
ip m/skg10420 23
op m/skg10105 23
713.0Pr
Ks/m287Kkg/J287 22R
C20 ooi TTT
m90W
4.1
m/skg1017.18 6
Substituting into (9.34)
(K)K)(293.15)-s/m(2872m)/skg(10105m)(1029.6
m)/skg(1017.18 22
236
6
oKn
01.0oKn
Using (9.39) and noting that 4/ oi pp
)12(01.0121)4(93.15(K)2K)s/m)287(m(01.0m)/skg(1017.18
)m/skg()10105)(m()1090(m)(1090
24
1 2
226
242233366
m
kg/s1003776.1 12m
PROBLEM 9.11 (continued)
Axial pressure variation is obtain from (9.35)
L
x
p
xp
o
)41(01.012)4(1)401.06(01.06)( 22
L
x
p
xp
o
36.154836.1606.0)(
(a)
Equation (a) is used to tabulate pressure variation with x/L. Equations (9.33) and (9.64) are used to compute the corresponding Knudsen and Nusselt numbers.
(iii) Checking. Dimensional check: Units for equations (9.33), (9.35), (9.39) and (9.64) are consistent.
Limiting check: No-slip macrochannel Nusselt number is obtained by setting 0Kn in (9.64). This gives Nu =
8.235. This agrees with the value given in Table 6.2.
(5) Comments. (i) To examine the effect of rarefaction and compressibility on the mass flow
rate, equation (9.41) is used to calculate omm / :
56.2)01.01214(2
1121
2
1o
o
i
o
Knp
p
m
m
This shows that incompressible no-slip theory will significantly underestimate the mass flow
rate. If rarefaction is neglected )0( oKn , the above gives
5.2)14(2
11
2
1
o
i
o p
p
m
m
Thus, compressibility plays a dominant role in the mass flow rate.
(ii) No-slip Nusselt number for fully developed Poiseuille flow between parallel plates with uniform surface heat flux is Nu = 8.235. Thus, no-slip theory overestimates the Nusselt number if applied to microchannels.
(iii) It should be noted that the equations used to compute ,m ),(xp and Nu are based on the
assumptions of isothermal conditions in the determination of the flow field. This is a reasonable approximation for typical applications.
Knopp/ Nux/L
1.0
0.20.40.60.8
0
1.000
0.02504.03.6023.1552.6361.988
0.002780.003170.003790.005290.0100
8.1418.1308.1158.0928.0367.862
PROBLEM 9.12
A micro heat exchanger consists of rectangular
channels of height m,7.6H width m,400W
and length mm.8L Air enters the channels at
temperature C30 oiT and pressure kPa.510ip
The outlet pressure is kPa.102op Channel surface
is at uniform temperature C.50osT Assume fully
developed flow and temperature, compute:
(a) Mass flow rate, m.(b) Heat transfer coefficient at the inlet, ),0(h and outlet, ).(Lh
(c) Mean outlet temperature, .moT
(d) Surface heat flux at the outlet, ).(Lqs
(1) Observations. (i) This is a pressure driven microchannel Poiseuille flow. (ii) Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates. (iii) Channel surface is maintained at uniform temperature. (iv) The solution to velocity, pressure, and mass flow rate is presented in Section 9.63. (v) The solution to the temperature distribution and Nusselt number for fully developed Poiseuille channel flow with uniform surface temperature is presented in Section 9.6.4. (vi) Surface heat flux is determined using Newton’s law.
(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed Poiseuille channel flow with uniform surface temperature.
(3) Solution Plan. To determine velocity, mass flow rate and pressure, apply the analysis and results of Section 9.6.3. To determine temperature and Nusselt number, apply the analysis and results of Section 9.6.4.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) infinite plates, (5) uniform spacing between plates, (6) no variation of density and pressure along y, (7)
,0.1Tu (8) no gravity, (9) the velocity field is independent of temperature, (10) ideal
gas, (11) continuum, slip flow regime conditions apply, and (12) fully developed flow.
(ii) Analysis.
(a) Mass flow rate m. Equation (9.39) gives the mass flow rate through the channel:
)( 112124
12
223
o
io
o
io
p
pKn
p
p
LRT
pHWm (9.39)
where
m
H
W
L
sT
sT
PROBLEMT 9.12 (continued)
m-6107.6m7.6heightchannelH
oKn outlet Knudsen number = )( opKn
L channel length = 0.008 m
ip inlet pressure m-kg/s000,510kPa510 2
m-kg/s000,102kPa102pressureoutlet 2op
m10400 widthchannel 6W
K-s/m287K-J/kg287constantgas 22R
m-s
kg1065.18
6
Equation (9.34) gives the outlet Knudsen number
oooo
pRT
HHpKnKn
1
2)( (9.34)
where moo TT is the mean outlet temperature.
(b) Heat transfer coefficient at inlet, )0(h and outlet, ).(Lh The Nusselt number is used to
determine the heat transfer coefficient. Nusselt number for channel flow is defined as
k
hHNu
2 (a)
where
Cm
W02638.0tyconductivithermal
ok
Applying (a) at the inlet, x = 0 and solving for h(0)
)0(2
)0( NuH
kh (b)
Similarly, at the outlet, (a) gives
)(2
)( LNuH
kLh (c)
The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig. 9.11. The Peclet number is defined as
RePrPe (d)
where the Reynolds number is defined as
HuRe m 2
(e)
Here is density and mu is the mean velocity obtained from continuity
PROBLEMT 9.12 (continued)
HW
mum (f)
(f) into (e)
W
mRe
2 (g)
The Knudsen number at the inlet is given by
iiii
pRT
HHpKnKn
1
2)( (h)
(c) Mean outlet temperature, .moT The local mean temperature )(xTm for channel flow at
uniform surface temperature is given by equation (6.13):
][exp)()( xcm
hPTTTxT
psmism (6.13)
where
CJ/kg006.41heatspecific opc
P = channel perimeter = m10134.810)4007.6(2)(2 66WH
C30eratureinlet tempmean oimi TT
C50 osT
m/sity,mean velocmu
3kg/mdensity,
h is the average heat transfer along the channel between inlet and section x, defined in (6.12)
x
dxxhx
h
0
)(1
(6.12)
The mean outlet temperature is obtained by setting x = L in (6.13)
][exp)()( Lcm
hPTTTLTT
psmismmo (i)
(d) Surface heat flux at the outlet, ).(Lqs Application of Newton’s law at the outlet gives
surface heat flux ))(( moss TTLhq (j)
(iii) Computations.
(a) Mass flow rate m. Equation (9.39) for m is based on the assumption that the flow is
isothermal. Since the outlet temperature oT is not yet determined, as a first approximation we
assume K303io TT in (9.34)
PROBLEMT 9.12 (continued)
01009.0K))(303)(K-s/m)(287(2)m-kg/s)(000,102(m)(107.6
)m-kg/s(1065.18 22
26-
6
oKn
Substituting into (9.39)
)( 1102000
510000)01009.0(121
102000
510000
K))(303)(K-s/m(287m))(008.0()m-kg/s(10657.18
m-kg/s)000,102((m))107.6((m)10400
24
12
226
222366 3
m
kg/s10098368.0 6m
(b) Heat transfer coefficient at the outlet, )0(h and ).(Lh To use Fig. 9.11 for the determination
of the Nusselt number, the Prandtl and Reynolds number, Peclet number, and Knudsen number are needed. Air properties give
712.0Pr
(g) gives the Reynolds number
372.26(m)10m)400-(kg/s1065.18
)(kg/s10098368.0)2(
66
6
Re
Thus the Peclet number is
777.18712.0372.26Pe
At this Peclet number the curve corresponding to Pe gives the approximate Nusselt number
for this case. The Knudsen number at the inlet is computed using (h)
00218.0K))(303)(K-s/m)(287(2)m-kg/s)(000,105(m)(107.6
)m-kg/s(1065.18 22
26-
6
iKn
At this value of Knudsen number Fig. 9.11 gives
5.7)0(Nu
Substitute into (b)
Cm
W765,14)5.7(
7.6(2
)CW/m(02638.0)0(
o2
o
)(m)-610h
At the outlet where 01009.0oKn , Fig. 9.11 gives
25.7)(LNu
(b) gives
Cm
W273,14)25.7(
7.6(2
)CW/m(02638.0)(
o2
o
)(m)-610Lh
PROBLEMT 9.12 (continued)
(c) Mean outlet temperature, .moT The average heat transfer coefficient, h , is needed to
determine .moT Using (6.12) to determine h requires the numerical integration of the local heat
transfer coefficient. However, since the change in h between inlet and outlet is very small, the
arithmetical mean can be used to approximate h . Thus
Cm
W519,14
2
765,14273,14
2
)()0(
o2
Lhhh
][ m)(008.0)kg/s)(10.098368)0()CJ/kg(006.41
)CW/m(519,14)m(10134.8)C)(5030()C(50
6-o
o26oo expmoT
C00215.50 omoT
(d) Surface heat flux at the outlet, ).(Lqs Use (j) to compute ).(Lqs
2
o2
m
W7.305000215.50)CW/m(273,14)(Lqs
(iv) Checking.
Dimensional check: computations showed that equations (9.34), (9.39), (6.13), and (j) are dimensionally correct.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on the assumption that the flow is isothermal. Computation showed that the outlet temperature is
C50omoT . To improve the solution, an iterative procedure can be followed by repeating the
computation assuming an arithmetical average of mean temperature in the channel equal to
C.40C)]/250(C)(30[ ooo (ii) The heat transfer coefficient at the outlet is high compared to
values for air encountered in typical macrochannels applications. (iii) The Nusselt number for no-slip theory and negligible axial conduction is obtained from Fig. 9.11 at Kn = 0 and Pe .This gives
5407.7oNu
This is close to the inlet Nusselt number when rarefaction is included, indicating a small rarefaction effect. (iv) Unlike the Nusselt number for fully developed flow in macrochannels, the Nusselt number is not constant along microchannels.
PROBLEM 9.13
Consider isothermal Poiseuille flow of gas in a microtube of radius .or Taking into
consideration velocity slip, show that the axial velocity is given by
2
22
414
o
oz
r
rKn
dz
dprv
(1) Observations. (i) Cylindrical coordinates should be used to solve this problem. (ii) The axial component of the Navier-Stokes equations must be solved to determine the
axial velocity .zv (iii) The procedure
and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this case.
(2) Problem Definition. Solve the axial component of the Navier-Stokes equations of motion.
(3) Solution Plan. Start axial component of the governing Navier-Stokes equations of motion, introduce simplifying assumptions, write down the slip velocity boundary conditions and solve the governing equation.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (axial and radial), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity (7) negligible radial variation of density and pressure, (8) negligible gravity, (9) the velocity
accommodation coefficients is equal to unity, ,0.1u (10) isothermal flow, (11) the
dominant viscous force is r
vr
rr
z1, and (12) negligible inertia forces;
z
vv
v
r
v
r
vv zzz
zr
(ii) Analysis. Following the analysis of Section 9.6.2, we begin with the axial component of the Navier-Stokes equations
2
2
2
2
2
11
z
vv
rr
vr
rrz
pg
t
v
z
vv
v
r
v
r
vv
zzzz
zzzzzr
(2.11z)
Introducing the above assumptions, this equation simplifies to
z
p
r
vr
rr
z 11 (a)
z
orr r
PROBLEM 9.13 (continued)
The boundary conditions are
0),0(
r
zvz (b)
r
vzru z
o ),( (c)
Since pressure is assumed independent of r, this equation can be integrated directly to
give the axial velocity zv . Thus
21
2
ln4
1CrC
r
z
pvz
(d)
Application of boundary conditions (b) and (c) give
01C ,o
o
rdz
dprC
24
12
2 (e)
Susbtituting (e) into (d)
2
2
0
2
241
4o
oz
r
r
rz
prv (f)
Introducing the definition of Knudsen number for tube flow
orKn
2 (g)
Substituting (g) into (f)
2
22
414
o
oz
r
rKn
z
prv (h)
(iii) Checking. Dimensional check: (h) is dimensionally correct.
Governing equation check: (h) satisfies (a).
Boundary conditions check: (h) satisfies conditions (b) and (c).
(5) Comments. It is important to note the assumptions leading to solution (h).
PROBLEM 9.14
Consider fully developed isothermal Poiseuille flow through a microtube. Follow the analysis of Section 9.6.3 and use the continuity equation in cylindrical coordinates to derive the following:
(a) The radial velocity component rv
)(24
1
2
11
4 3
33
pKnr
r
r
r
r
r
dz
dpp
zp
rv
ooo
or
where Kn(p) is the local Knudsen number.
(b) The local pressure p(z)
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(9.78)
where ip is inlet pressure, op outlet pressure, and oKn is the outlet Knudsen number.
(1) Observations. (i) This a pressure driven Poiseuille flow through a microtube. (ii) The procedure for determining the radial velocity component and axial pressure distribution is identical to that for slip Poiseuille flow between parallel plates. (iii) The solution to the axial velocity is given by equation (9.74). (iv) Continuity equation gives the radial velocity component. (v) Axial pressure is determined by setting the radial velocity component equal to zero at the surface. (vi) Cylindrical coordinates should be used to solve this problem.
(2) Problem Definition. Determine the radial velocity component of slip flow through a tube.
(3) Solution Plant. Use the continuity equation in radial coordinates and the solution to the axial velocity component to determine the radial component. Set the radial component equal to zero at the surface to determine the axial pressure distribution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional, (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity, and specific heats, (7) negligible radial variation of density and pressure, (8) negligible dissipation (9) negligible gravity, and (10) The velocity accommodation coefficient is equal to unity, .0.1u
(ii) Analysis.
(a) To determine the radial velocity component, we follow the derivation of Section 9.6.3 for the
analogous problem of Poiseuille flow between parallel plates. The axial velocity zv for tube flow
is derived in Section 9.6.5
2
22
414
o
oz
r
rKn
dz
dprv (9.74)
The Knudsen number, Kn, for the flow through tubes is defined as
RTprr
Knoo 2
1
22(a)
The radial component is determined using the continuity equation for compressible flow in cylindrical coordinates
PROBLEM 9.14 (continued)
011
zrzr
rrrt
vvv (2.4)
This simplifies to
01
zr vz
vrrr
(b)
Using the ideal gas law, (9.31), to express in terms of pressure and rearranging, the above is
written as
01
zr vpz
vrprr
Substituting (9.74) into the above
)(2
22
414
1
o
or
r
rKn
dz
dpp
z
rvrp
rr (c)
Integration of (c) gives the radial velocity component .rv Boundary conditions on rv are
0),0( zvr (d)
0),( zrv or (e)
Multiplying (c) by ,rdr integrating with respect to r and using boundary condition (d)
r
rdrr
rKn
dz
dpp
z
rv
vrpr
o
or
r
00
)(2
22
414
Evaluating the integrals and noting that the integrand on the right hand side is a function of zonly, yields
)(2
2
42
24
42
2
o
or
r
rrKn
r
dz
dpp
zd
drvrp
Solving for rv
)(2
2
42
2
1
4
42
2
o
or
r
rrKn
r
dz
dpp
zd
d
pr
rv (f)
(b) To determine axial pressure distribution, (f) is applied to boundary condition (e)
)( 24
110 Kn
dz
dpp
zd
d
p
Using (a) to eliminate Kn
024
1)( RT
prdz
dpp
dz
d
o
(g)
Integration with respect to z twice gives
DCzpRTr
po 28
1 2
PROBLEM 9.14 (continued)
where C and D are constants of integration. Noting that pressure is positive, the solution to this quadratic equation is
)(82
162
42
2
DCzRTr
RTr
poo
(h)
The boundary conditions on pressure are
ipp )0( and opLp )( (i)
Using (i),the constants C and D are determined
)(2
)(8
1 22io
oio ppRT
Lrpp
LC (j)
io
i pRTr
pD2
)8
1 2 (k)
Substituting (j) and (k) into (h)
o
i
ooo
i
o
i
ooo
i
oo
ooo
p
pRT
prp
p
L
z
p
pRT
prp
p
p
RT
r
RTprp
zp
22
8)1(
2
81
2
16
24
)(
2
2
2
2
22
2 (l)
Using (a), this result is expressed in terms of the Knudsen number at the outlet
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(m)
(iii) Checking. Dimensional check: Each term in (h) has units of velocity. Each term in (l) is dimensionless.
(5) Comments. Unlike fully developed Poiseuille flow in macrochannels, the radial velocity component does not vanish (streamlines are not parallel) and axial pressure distribution is not linear.
PROBLEM 9.15
Taking into consideration velocity slip, show that the mass flow rate for laminar, fully developed
isothermal Poiseuille flow in a microtube is give by
)( 116116 2
224
o
io
o
ioo
p
pKn
p
p
LRT
prm (9.798a)
(1) Observations. (i) Cylindrical coordinates should be used to solve this problem. (i) Axial velocity component is needed to determine mass flow rate. (iii) Equation (9.74) gives the axial velocity for this case. (iv) Since axial velocity vary with radial distance, mass flow rate requires integration of the axial velocity over the flow cross section area. (v) The procedure and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this problem.
(2) Problem Definition. Integration of the axial velocity over flow cross section.
(3) Solution Plan. Formulate the flow rate integral, use the axial velocity for the Poiseuille flow through tubes and carry out the integration of over tube radius.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (axial and radial), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, (7) negligible radial variation of density, (8) negligible gravity, (9) the velocity accommodation coefficients is
equal to unity, ,0.1u (9) isothermal flow, (10) the dominant viscous force is r
vr
rr
z1,
and (11) negligible inertia forces; z
vv
v
r
v
r
vv zzz
zr
(ii) Analysis. The mass flow rate is given by o
z
r
rdrvm
0
2 (a)
Based on the above assumptions, the axial velocity component is given in Section 9.6.2 as
2
22
414
o
oz
r
rKn
dz
dprv (9.74)
Substituting (9.74) into (a) and recalling that density is assumed constant along the radial
distance r, gives o
o
o
r
rdrr
rKn
dz
dprm
0
2
22
412
(b)
Evaluating the integral in (b)
Kndz
dprm o 81
8
4
(c)
PROBLEM 9.15 (continued)
The density of an ideal gas is given by (9.31)
RT
p (9.31)
The Knudsen number for tube flow is
RTpr
Kno 2
1
2 (d)
Substituting (9.31) and (d) into (c)
RTr
pdz
dp
RT
rm
o
o
2
4
8
4
(e)
This result gives the mass flow rate in terms of pressure. The solution to the pressure distribution )(zp is
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(9.78)
where oKn is the Knudsen number at the discharge. Evaluating (d) at the discharges where
opp
RTpr
Knoo
o2
1
2 (f)
Rewriting (9.78) as
CzBAzp )( (g)
where
oo pKnA 8 (h)
28 ioo ppKnB (i)
)1(1612
22
o
io
o
io
p
pKn
p
p
L
pC (j)
Differentiating (g)
2/1)(2
CzBC
dz
dp (k)
Substituting (g) and (k) into (e)
RTr
CzBACzBC
RT
rm
o
o
2
4)(
28
2/14
(l)
This simplifies to
CRT
rm o
4
16 (m)
Substituting (j) into (m)
PROBLEM 9.15 (continued)
)1(16116 2
224
o
io
o
ioo
p
pKn
p
p
LRT
prm (9.79a)
(iii) Checking. Dimensional check: Equation (9.79a) has the correct mass flow units of kg/s.
Limiting check: If ,oi pp no flow takes place and thus the mass flow rate should be zero.
Setting oi pp in (9.78a) gives m = 0.
(5) Comments. The Knudsen number in (9.79a) represents the effect of rarefaction. Neglecting
rarefaction ( )0oKn , (9.79a) reduces to
116 2
224
o
ioo
p
p
LRT
prm (n)
If both rarefaction and compressibility are neglected, the flow rate is given by (9.79b)
18
24
o
ioo
p
p
LRT
prm (9.70b)
PROBLEM 9.16
Pressure distribution for fully developed Poiseuille flow through tubes is given by
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(9.78)
Derive this equation using the condition that, for steady state, the mass flow rate is invariant
with axial distance z. That is
02
0
o
z
r
drrvdz
d
dz
dm
(1) Observations. (i) To use the proposed approach, the solution to the axial velocity distribution must be known. (ii) The velocity distribution for Poiseuille flow through tubes is given by equation (9.74) of Section 9.6.5. (iii) Cylindrical coordinates should be used to solve this problem.
(2) Problem Definition. Determine the mass flow rate.
(3) Solution Plan. Use the solution to the axial velocity for Poiseuille flow, equation (9.74), to determine the mass flow rate.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant viscosity (4) no radial
variation of density and pressure, (5) no gravity, (6) ,0.1u (7) the velocity field is
independent of temperature, (8) ideal gas, and (9) continuum, slip flow regime conditions apply.
(ii) Analysis. To derive pressure solution (9.78), it is proposed to use the following conservation of mass equation
02
0
o
z
r
drrvdz
d
dz
dm (a)
The solution to the axial velocity zv is given by (9.743)
2
22
414
o
oz
r
rKn
dz
dprv (9.74)
Substituting (9.743) into (a)
0)(414
02
22o
o
or
rrdr
rpKn
zd
dpr
zd
d
zd
dm(b)
Noting that density and pressure p vary along z and are assumed constant along r, and that the
viscosity is constant, (b) is written as
PROBLEM 9.16 (continued)
0)(410
2
2o
o
r
rrdr
rpKn
zd
dp
zd
d
zd
dm(c)
Evaluating the integral in (c)
0)(242
222
ooo rpKn
rr
zd
dp
zd
d
zd
dm
This simplifies to
0)(24
1pKn
zd
dp
zd
d(d)
To proceed, the density and Knudsen number in (c) must be expressed in terms of pressure. Ideal gas law (9.31) gives
RT
p (9.31)
The Knudsen number for tube flow is
RTpr
Kno 2
1
2 (e)
(9.31) and (e) into (d)
01
24
1
pRT
rdz
dp
RT
p
zd
d
o
Since the flow is assumed isothermal, the above simplifies to
01
24
1
pRT
rdz
dpp
zd
d
o
(f)
Integrating (f) once
1
1
24
1C
pRT
rdz
dpp
o
Rewriting the above as
124
1C
zd
dpRT
rzd
pdp
o
Integrating again
212
28
1CzCpRT
rp
o
(g)
The boundary conditions on p are
ipp )0( , opLp )( (h)
Here L is tube length. Boundary conditions (h) give 1C and 2C
)(2
)(8
1 221 io
oio ppRT
Lrpp
LC (i)
PROBLEM 9.16 (continued)
io
i pRTr
pC28
1 22 (j)
The solution to quadratic equation (g) is
)(8162
4212
2
CzCRTr
RTr
p
oo
(k)
Substituting (i) and (j) into (k), normalizing the pressure by ,op and introducing the Knudsen
number (e), we obtain
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(9.78)
(iii) Checking. Dimensional check: Each term in (k) has units of pressure.
Limiting check: If oi pp , axial velocity will vanish and pressure should be uniform throughout
the tube. Setting oi pp in (9.78) gives
1188)(
ooo
KnKnp
zp
(4) Comments. This approach for determining p(z) is simpler than that proposed in Section 9.6.4 where it is necessary to first determine the radial velocity component .rv
PROBLEMT 9.17
Air is heated in a microtube of radius m5or and length mm.2L Inlet temperature and
pressure are C20 oiT and Outlet pressure is kPa.100op Uniform surface flux,
,W/m1500 2sq is used to heat the air. Taking into consideration velocity slip and
temperature jump and assuming fully developed flow and temperature, compute:
(a) Mass flow rate, m.
(b) Mean outlet temperature, .moT
(c) Heat transfer coefficient at outlet, ).(Lh
(d) Surface temperature at the outlet, ).(LTs
(1) Observations. (i) This is a pressure driven Poiseuille flow through a microtube. (ii) Tube surface is heated with uniform flux. (iii) The solution to mass flow rate, temperature distribution and Nusselt number for fully developed Poiseuille flow through a tube with uniform surface flux is presented in Section 9.6.5.
(2) Problem Definition. Determine flow and heat transfer characteristics of fully developed Poiseuille flow through a tube with uniform surface heat flux.
(3) Solution Plan. Apply the analysis and results of Section 9.6.5.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) constant conductivity, specific heat
and viscosity, (4) no radial variation of density and pressure, (5) ,0.1Tu (6) no gravity,
(7) the velocity field is independent of temperature, (8) ideal gas, (9) continuum, slip flow regime conditions apply, and (10) fully developed flow.
(ii) Analysis.
(a) Mass flow rate m. Equation (9.79a) gives the mass flow rate through the tube:
)( 116116 2
224
o
io
o
i
o
oo
p
pKn
p
p
LRT
prm (9.79a)
where
oKn outlet Knudsen number = )( opKn
L channel length = 0.002 m
ip inlet pressure m-kg/s000,600kPa600 2
m-kg/s000,100kPa100pressureoutlet 2op
or tube radius = m-6105m5
K-s/m287K-J/kg287constantgas 22R
sq
sq
z
orr r
PROBLEMT 9.17 (continued)
C20oio TT
m-kg/s1017.18 6
The outlet Knudsen number is given by
o
o
oo
oop
RTrr
pKnKn1
222)( (a)
(b) Mean outlet temperature, .moT Conservation of energy between inlet and outlet, gives
mi
p
soom T
mc
qLrT
2 (b)
where
CJ/kg3.998heatspecific opc
2W/m1500fluxheatsurfacesq
C20eratureinlet tempmean oimi TT
(c) Heat transfer coefficient at the outlet, ).(Lh The Nusselt number is used to determine the
heat transfer coefficient. Nusselt number for channel flow is defined as
k
hrNu o2
(c)
where
Cm
W02564.0tyconcuctivithermal
ok
Applying (c) at the outlet, z = L, and solving for h(L)
)(2
)( LNur
kLh
o
(d)
The Nusselt number is given by (9.98)
KnPr
KnKnKn
KnKn
Nu1
1
4
24
7
3
1416
)81(
1
16
3
)81(
4
2
2
2)(
(9.98)
where
Kn local Knudsen number 0.713numberPrandtlPr
= specific heat ratio = 1.4
Evaluation (9.98) at z = L where oKnKn
ooo
o
o KnPr
KnKnKn
KnKn
Nu1
1
4
24
7
3
1416
)81(
1
16
3
)81(
4
2
2
2)(
(e)
PROBLEMT 9.17 (continued)
(d) Surface temperature at the outlet, ).(LTs Surface temperature distribution is given by
(9.97):
)(1
4
16
3
)81(
4)( zgKn
Prk
rqKn
Knk
rqzT oo ss
s (9.97)
where g(z) is given by (9.96):
24
7
3
1416
)81(
2)( 2
2KnKn
Knk
rqz
vrc
qTzg os
mzop
smi (9.96)
where zmv is the mean velocity. Continuity equation gives zmv in terms of mass flow rate m
2o
zmr
mv (f)
To determine surface temperature at the outlet, ),(LTs the Knudsen number in (9.96) and (9.97) is
evaluated at outlet pressure and g(z) is evaluated at z = L. Equation (9.97) becomes
)(1
4
16
3
)81(
4)( LgKn
Prk
rqKn
Knk
rqLT o
oo
o
o sss (g)
Using (f), and setting z = L and oKnKn in (9.96) gives g(L)
24
7
3
1416
)81(
2)( 2
2 oo
o
os
p
somi KnKn
Knk
rqL
mc
qrTLg (h)
(iii) Computations.
(a) Mass flow rate m. Equation (9.79a) for m is based on the assumption that the flow is
isothermal. Since the outlet temperature oT is not yet determined, as a first approximation we
assume K293io TT in (9.79a). The outlet Knudsen number is computed using (a)
0066054.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100(m)(1010
)m-kg/s(1017.18 22
26-
6
oKn
Substituting into (9.79a)
)( 1100000
6000000066054.0(161
100000
600000
K))(293)(K-s/m(287m))(002.0()m-kg/s(1017.18
m-kg/s)000,100((m))105((m)10600
16
2
226
2223366
m
kg/s10426.1 8m
(b) Mean outlet temperature, .moT Equation (b) gives
)C(20)kg/s)(10426.1)(CJ/kg(3.998
m))(002.0(m)(105)W/m()1500(2 o
-8o
62
moT
C6204.26 omoT
PROBLEMT 9.17 (continued)
(c) Heat transfer coefficient at the outlet, ).(Lh Substituting into (e), gives the Nusselt number
at the outlet
(0.713)
)0044054.0(
14.1
)4.1(2
24
7)0066054.0(
3
14)0066054.0(16
)0066054.0(81
1
16
30066054.0
)0066054.0(81
4
2)(
2
LNu
278.4)(LNu
Equation (d) gives h(L)
CW/m969,10278.4)(m)-6105(2
)CW/m(02564.0)( o2
o
Lh
(d) Surface temperature at the outlet, ).(LTs Use (h) to compute g(L)
24
7)0066054.0(
3
14)0066054.0(16
)0066054.081)(CW/m(02564.0
)(m)6-105)(W/m(1500
)(kg/s)8-10426.1)(CJ/kg(3.998
)(m)002.0(m))(105()W/m()1500(2C)(20)(
2
2o
2
o
62oLg
53507.26)(Lg Co
Substitute into (g)
C)(04.23)0066054.0()713.0)(CW/m(02564.0
)(W/m()1500(
14.1
)4.1(4
16
3)0066054.0(
)0066054.081)(CW/m(02564.0
)(W/m()1500(4)( o
o
2
o
2 )(m)-6105)(m)-6105LTs
C7571.26)( oLTs
(iv) Checking.
Dimensional check: computations showed that equations (9.79a), (d), (g), and (h) are dimensionally correct.
Surface temperature check: Application of Newton’s law at the outlet gives
moss TLTLhq )()(
Solving for )(LTs
mos
s TLh
qLT
)()(
Using this equation to compute )(LTs , we obtain
PROBLEMT 9.17 (continued)
C26.7572C6204.26)CW/m(10969
)W/m()1500()( oo
o2
2
LTs
This is close to the value determined above.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on the assumption that the flow is isothermal. Computation showed that the outlet temperature is
C62.26 omoT . Since the outlet is C62.6 o above the inlet temperature, it follows that the
assumption of isothermal flow is reasonable. (ii) The heat transfer coefficient at the outlet is very high compared to values for air encountered in typical macrochannels applications. (iii) In computing moT and sT , results are presented showing four decimal points. This is done to avoid
errors where temperature differences are small. (iv) The Nusselt number for slip theory for fully
developed macrotube flow is obtained by setting 0oKn in (e). This gives
364.4oNu
Thus macrochannel theory overestimates the Nusselt number if applied to microchannels.
PROBLEM 9.18
Determine the axial variation of the Nusselt number and heat transfer coefficient of the microtube in Problem 9.17.
(1) Observations. (i) The problem is a pressure driven Poiseuille flow through microtube with uniform surface heat flux. (ii) The Nusselt number depends on the Knudsen number, Kn. Since Kn
varies along the tube due to pressure variation, it follows that pressure distribution along the tube must be determined. (iii) Assuming fully developed velocity and temperature, the analysis of Section 9.6.5 gives axial pressure and Nusselt number variation along tube. (iv) The definition of Nusselt number gives the heat transfer coefficient.
(2) Problem Definition. Determine the flow and temperature fields for fully developed Poiseuille tube flow with uniform surface flux.
(3) Solution Plan. Apply the results of Section 9.6.5 for pressure and Nusselt number distribution.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular variation), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity and conductivity, (7) negligible radial variation of density and pressure, (8) the accommodation
coefficients are assumed equal to unity, ,0.1Tu (9) negligible dissipation, (10) uniform
surface flux, (11) negligible axial conduction, and (12) no gravity.
(ii) Analysis. The Nusselt number for tube flow heat is defined as
k
hrNu o2
Solving the above for the heat transfer coefficient h
or
kNuh
2 (a)
The Nusselt number, ,Nu is given by (9.99)
KnPr
KnKnKn
KnKn
Nu1
1
4
24
7
3
1416
)81(
1
16
3
)81(
4
2
2
2)(
(9.98)
The local Knudsen number, ,Kn depends on the local pressure p(x) according to
pRT
rrpKnKn
oo
1
222)( (b)
sq
sq
z
orr r
PROBLEM 9.18 (continued)
Evaluating (b) at the outlet
ooooo
pRT
rrpKnKn
1
222)( (c)
Equation (9.78) gives )(zp
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(9.78)
Thus, (9.78) is used to determine p(z), (b) to determine ),( pKn (c) gives oKn , (9.98) to determine
the variation of the Nusselt number along the tube, and (a) the heat transfer coefficient.
(iii) Computations. Air properties are determined at .C20o To compute p(x), ),(xKn
and ,Nu the following data is used
L channel length = 0.002 m
ip inlet pressure m-kg/s000,600kPa600 2
m-kg/s000,100kPa100pressureoutlet 2op
or tube radius = m-6105m5
K-s/m287K-J/kg287constantgas 22R
C20oio TTT
m-kg/s1017.18 6
713.0Pr
Ks/m287Kkg/J287 22R
C20 ooi TTT
4.1
m/skg1017.18 6
Substituting into(c)
0066054.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100(m)(1010
)m-kg/s(1017.18 22
26-
6
oKn
Using (9.78) and noting that 6/ oi pp gives axial pressure variation
L
z
p
xp
o
)61(0066054.016)6(1)60066054.08(0066054.08)( 22
L
z
p
xp
o
528432.3563691.360528432.0)(
(d)
PROBLEM 9.18 (continued)
The local Knudsen number Kn(p) is computed by taking the ratio of (b) and (c)
oo
o
pppp
KnpKn
/
0066054.0
/)( (e)
Using (d), (e) and Nusselt number, and (9.98), results are tabulated below.
(iii) Checking. Dimensional check: Computations showed that units of equations (a), (b), and (9.98) are consistent.
Limiting check: No-slip macrochannel Nusselt number is obtained by setting 0Kn in (9.98).
This gives Nu = 4.364. This agrees with the value given in Table 6.2.
(5) Comments. (i) No-slip Nusselt number for fully developed Poiseuille flow between parallel plates with uniform surface heat flux is Nu = 4.364 (Table 6.2). Thus, no-slip theory overestimates the Nusselt number if applied to microtubes.
(ii) The Nusselt number for fully developed flow is constant along channels. This example shows that in microtubes the Nusselt number vary slightly along the tube.
(iii) It should be noted that the equations used to compute ),(xp and Nu are based on the
assumptions of isothermal conditions in the determination of the flow field. Determining the outlet temperature will give an indication of the validity of this assumption.
6.0
5.3814
z/Lopp/ Kn Nu
1.0
0.2
0.40.6
0.8
0
4.68273.8612
2.8132
1.0
0.001109
0.0012275
0.0014106
0.0017107
0.002348
0.0066054
4.3486
4.3462
4.3429
4.3344
4.2780
)CW/m( o2h
11,150
11,144
11,134
11,113
10,969
4.3501 11,153
PROBLEM 9.19
A micro heat exchanger uses micro tubes of
radius m3or and length mm.6L Inlet air
temperature and pressure are C20 oiT and
pressure kPa.600ip Outlet pressure is
kPa.100op Each tube is maintained at
uniform surface temperature C.60 osT Taking
into consideration velocity slip and temperature
jump and assuming fully developed flow and
temperature, determine the following:
(a) Heat transfer coefficient at the inlet, ),0(h and outlet, ).(Lh
(b) Mean outlet temperature .moT
(1) Observations. (i) This is a pressure driven Poiseuille flow through a tube at uniform surface temperature. (ii) Since the flow field is assumed independent of temperature, it follows that the velocity, mass flow rate and pressure distribution for tubes at uniform surface flux, presented in Section 9.6.6, are applicable to tubes at uniform surface temperature.. (iii) The heat transfer coefficient can be determined if the Nusselt number is known. (iv) The variation of the Nusselt number with Knudsen number for air is shown in Fig. 9.16. (v) The determination of Knudsen number at the inlet and outlet and Fig. 9.16 establish the Nusselt number at these locations. (vi) The use of Fig. 9.16 requires the determination of the Peclet number. (vii) Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.
(2) Problem Definition. Determine the Nusselt number at the inlet and outlet and the average hat transfer coefficient.
(3) Solution Plan. Compute the Knudsen number at the inlet and outlet, compute the Peclet number, and use Fig. 9.16 to determine the Nusselt number. Use (6.13) to compute the outlet temperature.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular variation, (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity and specific heats, (7) negligible radial variation of density and pressure, (8) the
accommodation coefficients are assumed to be equal to unity, ,0.1Tu (9) negligible
dissipation, (10) uniform surface temperature, and (11) negligible gravity.
(ii) Analysis.
(a) Heat transfer coefficient at inlet, )0(h and outlet, ).(Lh The Nusselt number is used to
determine the heat transfer coefficient. Nusselt number for tube flow is defined as
k
hrNu o2
(a)
where
z
orr r
sTsT
PROBLEM 9.19 (continued)
h heat transfer coefficient, CW/m o2
CW/m02564.0tyconductivithermal ok
Nu Nusselt number
or = tube radius = m103 6
Applying (a) at the inlet, z = 0 and solving for h(0)
)0(2
)0( Nur
kh
o
(b)
Similarly, at the outlet, (a) gives
)(2
)( LNur
kLh
o
(c)
L tube length = 0.006 m
The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig. 9.16. The Peclet number is defined as
RePrPe (d)
where
Pr = 0.713
The Reynolds number is defined as
omz rvRe
2 (e)
where
zmv mean velocity, m/s
viscosity = 18.17 CW/m o
density, 3kg/m
Continuity gives
2o
zmr
mu (f)
(f) into (e)
or
mRe
2 (g)
The mass flow rate m is given by equation (9.79a)
)( 116116 2
224
o
io
o
ioo
p
pKn
p
p
LRT
prm (9.798a)
where
PROBLEM 9.19 (continued)
ip inlet pressure m-kg/s000,600kPa600 2
m-kg/s000,100kPa100pressureoutlet 2op
K-s/m287K-J/kg287 22R
C20ooi TTT
The Knudsen number at the inlet and outlet is given by
ii
ooii
pRT
rrpKnKn
1
222)( (h)
oo
ooo
pRT
rrpKnKn
1
222)( 0 (i)
(b) Mean outlet temperature, .moT The local mean temperature )(xTm for channel flow at
uniform surface temperature is given by equation (6.13):
][exp)()( zpcm
hPTTTzT smism (6.13)
where
CJ/kg006.41heatspecific opc
P = channel perimeter = or2 , m
C20eratureinlet tempmean oimi TT
C60 osT
h is the average heat transfer along the tube between inlet and outlet, defined in (6.12)
L
zdzhL
h
0
)(1
(6.12)
An accurate method for computing h requires the determination of the variation of local heat
transfer coefficient, h(z), and evaluating the integral in (6.12) numerically. An approximate approach is to use the arithmetical average. This is justified if h(z) is linear or its change is small.
(iii Computations.
The inlet and outlet Knudson numbers are computed first
001835.0K))(293)(K-s/m)(287(2)m-kg/s)(000,600)(m)(103(2
)m-kg/s(1017.18 22
26-
6
iKn
01101.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100)(m)(103(2
)m-kg/s(1017.18 22
26-
6
oKn
PROBLEM 9.19 (continued)
Substitute into (9.79a)
)( 1100000
600000)01101.0(121
100000
600000
K))(293)(K-s/m(287m))(006.0()m-kg/s(1017.18
m-kg/s)000,10((m))103((m)
16
2
226
22236 3
m
910622154.0m kg/s
(g) gives the Reynolds number
7266.7(m)10m)6-(kg/s1017.18
)(kg/s10622154.0)4(66
6
Re
Thus the Peclet number is
181.5713.07266.7Pe
At the above values of iKn and Pe, Fig. 9.16 gives
75.3)0(Nu
Substitute into (b)
Cm
W910,15)75.3(
)(m)6-103(2
)CW/m(02546.0)0(
o2
o
h
At the outlet Fig. 9.16 gives
68.3)(LNu
(b) gives
Cm
W620,15)68.3(
)(m)6-103(2
)CW/m(02546.0)(
o2
o
Lh
Since the variation of h between inlet and outlet is small, using the arithmetical average heat transfer coefficient in (6.13) is justified
Cm
W765,15)620,15910,15(
2
1)()0(
2
1o2
Lhhh
substituting into (6.13) and setting z = L
][ m)(006.0)kg/s(10622154.0)CJ/kg(998.3
)CW/m(765,15)m)(103(2)C)(6020()C(60
9-o
o26oo expTmo
C60omoT
(iv) Checking.
PROBLEM 9.19 (continued)
Dimensional check: computations showed that equations (b), (c), (9.79a), and (6.13) are dimensionally correct.
(5) Comments. (i) The flow field solution (velocity, mass flow rate and pressure) is based on the assumption that the flow is isothermal. Computation showed that the outlet temperature is
C60omoT . To improve the solution, an iterative procedure can be followed by repeating the
computation assuming an arithmetical average of mean temperature in the channel equal to
C.40C)]/260(C)(20[ ooo (ii) The heat transfer coefficients are very high compared to values
for air encountered in typical macrotubes applications.
PROBLEM 9.20
Air enters a microtube at temperature C20 oiT and pressure kPa.600ip Outlet pressure is
kPa.100op Tube radius is m1or and its length is mm.2.1L The surface is maintained
at uniform temperature C.40 osT Taking into
consideration velocity slip and temperature jump
and assuming fully developed conditions,
determine the variation along the tube of the following:
(a) Nusselt number, ).(zNu
(b) Heat transfer coefficient, ).(zh
(c) Mean temperature, ).(zTm
(1) Observations. (i) The problem is a pressure driven Poiseuille flow through microtube at uniform surface temperature. (ii) The Nusselt number depends on the Knudsen number, Kn.Since Kn varies along the tube due to pressure variation, it follows that pressure distribution must be determined. (iii) Assuming fully developed velocity and temperature, the analysis of Section 9.6.6 gives axial pressure and Nusselt number variation along the tube. (iv) The definition of Nusselt number gives the heat transfer coefficient. (v) The variation of the Nusselt number with Knudsen number and Peclet number for air is shown in Fig. 9.16. (vii) Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.
(2) Problem Definition. Determine the flow and temperature fields for fully developed Poiseuille tube flow with uniform surface temperature.
(3) Solution Plan. Apply the results of Section 9.6.6 for pressure and Nusselt number distribution. Use Fig. 9.16 to determine Nusselt number variation along the tube.
(4) Plan Execution.
(i) Assumptions. (1) Steady state, (2) laminar flow, (3) two-dimensional (no angular variation), (4) slip flow regime (0.001 < Kn < 0.1), (5) ideal gas, (6) constant viscosity, conductivity, and specific heat, (7) negligible radial variation of density and pressure, (8) the
accommodation coefficients are assumed equal to unity, ,0.1Tu (9) negligible
dissipation, (10) uniform surface temperature, and (11) no gravity.
(ii) Analysis.
(a) Nusselt number ),(zNu and (b) Heat transfer coefficient, ).(zh The Nusselt number for
tube flow is defined as
k
hrNu o2
Solving the above for the heat transfer coefficient h
or
kNuh
2 (a)
z
orr r
sTsT
PROBLEM 9.20 (continued)
The Nusselt number for air at various Peclet numbers and Knudsen numbers is given in Fig. 9.16. The local Knudsen number, ,Kn depends on the local pressure p(z) according to
pRT
rrpKnKn
oo
1
222)( (b)
Equation (9.78) gives )(zp
L
z
p
pKn
p
p
p
pKnKn
p
zp
o
io
o
i
o
ioo
o
)1(16)1(88)(
2
22
(9.78)
The Peclet number is defined as RePrPe (c)
where
Pr = 0.713
The Reynolds number is defined as
omz rvRe
2 (d)
where
zmv mean velocity, m/s
viscosity = skg/m1018.17 -6
density, 3kg/m
Continuity gives
2o
zmr
mu (e)
(e) into (d)
or
mRe
2 (f)
The mass flow rate m is given by equation (9.79a)
)( 116116 2
224
o
io
o
ioo
p
pKn
p
p
LRT
prm (9.79a)
where
ip inlet pressure m-kg/s000,600kPa600 2
m-kg/s000,100kPa100pressureoutlet 2op
K-s/m287K-J/kg287 22R
C20ooi TTT
PROBLEM 9.20 (continued)
(b) Mean outlet temperature, .mT The local mean temperature )(zTm for channel flow at
uniform surface temperature is given by equation (6.13)
][exp)()( Lpcm
hPTTTzT smism (6.13)
where
pc specific heat = CJ/kg98.39 o
P = channel perimeter = or2 , m
C20eratureinlet tempmean oimi TT
C40 osT
h is the average heat transfer along the tube between inlet and location z, defined in (6.12)
z
zdzhL
h
0
)(1
(6.12)
An accurate method for computing h requires the determination of the variation of local heat
transfer coefficient, h(z), and evaluating the integral in (6.12) numerically. An approximate approach is to use the arithmetical average. This is justified if h(z) is linear or its change is small.
Thus, (9.78) is used to determine p(z), (b) to determine ),( pKn (c) gives the Peclet number, Fig.
9.16 gives the variation of the Nusselt number along the tube, (a) the heat transfer coefficient, and (6.13) the mean temperature.
(iii) Computations. Air properties are determined at .C20o To compute p(z), ),(zKn
and ,Nu the following data is used
L channel length = 0.0012 m
ip inlet pressure m-kg/s000,600kPa600 2
op outlet pressure m-kg/s000,100kPa100 2
or tube radius = m-6101m1
K-s/m287K-J/kg287constantgas 22R
C20oio TTT
m-kg/s1017.18 6
713.0Pr
Ks/m287Kkg/J287 22R
C20 ooi TTT
4.1
m/skg1017.18 6
Evaluating (b) at the outlet
PROBLEM 9.20 (continued)
033027.0K))(293)(K-s/m)(287(2)m-kg/s)(000,100(m)(102
)m-kg/s(1017.18 22
26-
6
oKn
Noting that 6/ oi pp , (9.79a) gives m
)16)(033027.0(161)6(K))(293)(K-s/m(287m))(0012.0()m-kg/s(1017.18
m-kg/s)000,100((m))101(
16
2
226
22246
m
kg/s1010403.0m
Substituting into (f) gives the Reynolds number
412.1(m)10m)2-(kg/s1017.18
)(kg/s10403.0)4(66
10
Re
Thus the Peclet number is
007.1713.0412.1Pe
Using (9.78) and noting that 6/ oi pp gives axial pressure variation
L
z
p
xp
o
)61(033027.016)6(1)6033027.08(033027.08)( 22
L
z
p
xp
o
64216.37240402.39264216.0)(
(g)
The local Knudsen number Kn(p) is given by
oo
o
pppp
KnpKn
/
033027.0
/)( (h)
With 1Pe , equations (g), (h), Fig. 9.16, and (a) are used to compute and tabulate pressure,
Knudsen number, Nusselt number, and heat transfer coefficient, respectively, as functions of z/L.
6.0
5.3814
z/L opp/ Kn Nu
1.0
0.2
0.40.6
0.8
0
4.68273.8612
2.8132
1.0
)CW/m( o2h
0.0055505
0.0061536
0.0070972
0.0086528
0.0119812
0.033027
4.01
4.0
3.99
3.97
3.95
3.75
51,410
51,280
51,150
51,900
50,640
48,080
C)( ozT
40
40
40
40
40
40
PROBLEM 9.20 (continued)
Since the variation of h between inlet and outlet is small, using the arithmetical average heat transfer coefficient in (6.13) is justified. For determining the outlet temperature, we set
Cm
W745,49)080,48410,51(
2
1)()0(
2
1o2
Lhhh
To determine the highest mean temperature, (6.13) is applied at z = L
][ m)(0012.0)kg/s(10403.0)CJ/kg(998.3
)CW/m(745,49)m)(101(2)C)(4020()C(40
10-o
o26oo expmoT
C40omoT
Application (6.13) at other values of z gives C40)( ozTm . At z = 0, (6.13) gives C20omoT .
(iii) Checking. Dimensional check: Computations showed that units of equations (a), (f), (g), (h), and (9.79a) are dimensionally consistent.
(5) Comments. (i) No-slip Nusselt number for fully developed Poiseuille flow through tubes at uniform surface temperature is Nu = 3.656 (Table 6.2). Thus, no-slip theory overestimates the Nusselt number if applied to microtubes.
(ii) The Nusselt number for fully developed flow in macrotubes is constant. This example shows that for microtubes the Nusselt number varies along the tube.
(iii) It should be noted that the equations used to compute ),(xp and Nu are based on the
assumptions of isothermal conditions in the determination of the flow field. Determining the outlet temperature will give an indication of the validity of this assumption.
(v) The fluid equilibrates with surface temperature very close to the inlet and remains essentially at constant temperature, equal to throughout the tube. At a distance mz 1 , fluid temperature
increases to 39.98 C.o
Problem 1.1
Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable.
Ambient temperature and heat transfer coefficient are uniform
Surface temperature varies along the rectangle.
Problem 1.2
Heat is removed from the surface by convection. Therefore, Newton's law of cooling may be helpful.
Ambient temperature and surface temperature are uniform.
Surface area and heat transfer coefficient vary along the triangle.
Problem 1.3
Heat flux leaving the surface is specified (fixed).
Heat loss from the surface is by convection and radiation.
Convection is described by Newton's law of cooling.
Changing the heat transfer coefficient affects temperature distribution.
Surface temperature decreases as the heat transfer coefficient is increased.
Surface temperature gradient is described by Fourier’s law
Ambient temperature is constant.
Problem 1.3
Metabolic heat leaves body at the skin by convection and radiation.
Convection is described by Newton's law of cooling.
Fanning increases the heat transfer coefficient and affects temperature distribution, including surface temperature.
Surface temperature decreases as the heat transfer coefficient is increased.
Surface temperature is described by Newton’s law of cooling.
Ambient temperature is constant.
Problem 1.4
Metabolic heat leaves body at the skin by convection and radiation.
Convection is described by Newton's law of cooling.
Fanning increases the heat transfer coefficient and affects temperature distribution, including surface temperature.
Surface temperature decreases as the heat transfer coefficient is increased.
Surface temperature is described by Newton’s law of cooling.
) Ambient temperature is constant.
Problem 1.5
Melting rate of ice depends on the rate of heat added at the surface.
Heat is added to the ice from the water by convection.
Newton's law of cooling is applicable.
Stirring increases surface temperature gradient and the heat transfer coefficient. An increase in gradient or h increases the rate of heat transfer.
Surface temperature remains constant equal to the melting temperature of ice.
Water temperature is constant.
Problem 1.6
This problem is described by cylindrical coordinates.
For parallel streamlines 0vvr .
Axial velocity is independent of axial and angular distance.
Problem 1.7
This problem is described by cylindrical coordinates.
Streamlines are concentric circles. Thus the velocity component in the radial direction
vanishes ( ).0rv
For one-dimensional flow there is no motion in the z-direction ( 0zv ).
The -velocity component, , depends on distance r and time t.v
Problem 1.8
This problem is described by Cartesian coordinates.
For parallel streamlines the y-velocity component 0v .
For one-dimensional flow there is no motion in the z-direction (w = 0).
The x-velocity component depends on distance y and time t.
Problem 1.9
This problem is described by Cartesian coordinates.
For parallel streamlines the y-velocity component 0v .
For one-dimensional flow there is no motion in the z-direction (w = 0).
The x-velocity component depends on distance y only.
Problem 1.10
Heat flux leaving the surface is specified (fixed).
Heat loss from the surface is by convection and radiation
Convection is described by Newton's law of cooling.
Changing the heat transfer coefficient affects temperature distribution.
Surface temperature decreases as the heat transfer coefficient is increased.
Surface temperature gradient is described by Fourier’s law
Ambient temperature is constant.
Problem 1.11
Heat is removed from the surface by convection. Therefore, Newton's law of cooling is applicable.
Ambient temperature and heat transfer coefficient are uniform.
Surface temperature varies along the area.
The area varies with distance x.
Problem 2.2
The fluid is incompressible.
Radial and tangential velocity components are zero.
Streamlines are parallel.
Cylindrical geometry.
Problem 2.3
The fluid is incompressible.
axial velocity is invariant with axial distance.
Plates are parallel.
Cartesian geometry.
Problem 2.4
The fluid is incompressible.
Radial and tangential velocity components are zero.
Streamlines are parallel.
Cylindrical geometry.
Problem 2.5
Shearing stresses are tangential surface forces.
xy and yx are shearing stresses in a Cartesian coordinate system.
Tangential forces on an element result in angular rotation of the element.
If the net external torque on an element is zero its angular acceleration will vanish.
Problem 2.6
Properties are constant.
Cartesian coordinates.
Parallel streamlines: no velocity component in the y-direction.
Axial flow: no velocity component in the z-direction.
The Navier-Stokes equations give the three momentum equations.
Problem 2.7
Properties are constant.
Cylindrical coordinates.
Parallel streamlines: no velocity component in the r-direction.
Axial flow: no velocity component in the -direction.
No variation in the -direction. The Navier-Stokes equations give the three momentum
equations.
Problem 2.8
Properties are constant.
Cartesian coordinates.
Two dimensional flow (no velocity component in the z-direction
The Navier-Stokes equations give two momentum equations.
Problem 2.9
Properties are constant.
Cylindrical coordinates.
Two dimensional flow (no velocity component in the -direction.
The Navier-Stokes equations give two momentum equations.
Problem 2.10
Motion in energy consideration is represented by velocity components.
Fluid nature is represented by fluid properties.
Problem 2.11
Properties are constant.
Cartesian coordinates.
Parallel streamlines: no velocity component in the y-direction.
Axial flow: no velocity component in the z-direction.
Problem 2.12
Properties are constant.
Cartesian coordinates.
Parallel streamlines: no velocity component in the y-direction.
Axial flow: no velocity component in the z-direction.
The fluid is an ideal gas.
Problem 2.13
This is a two-dimensional free convection problem.
The flow is due to gravity.
The flow is governed by the momentum and energy equations. Thus the governing equations are the Navier-Stokes equations of motion and the energy equation.
The geometry is Cartesian.
Problem 2.15
The flow is due to gravity.
For parallel streamlines the velocity component v = 0 in the y-direction.
Pressure at the free surface is uniform (atmospheric).
Properties are constant.
The geometry is Cartesian.
Problem 2.16
This is a forced convection problem.
Flow properties (density and viscosity) are constant.
Upstream conditions are uniform (symmetrical)
The velocity vanishes at both wedge surfaces (symmetrical).
Surface temperature is asymmetric.
Flow field for constant property fluids is governed by the Navier-Stokes and continuity equations.
If the governing equations are independent of temperature, the velocity distribution over the wedge should be symmetrical with respect to x.
The geometry is Cartesian.
Problem 2.18
The geometry is Cartesian.
Properties are constant.
Axial flow (no motion in the z-direction).
Parallel streamlines means that the normal velocity component is zero.
Specified flux at the lower plate and specified temperature at the upper plate.
Problem 2.19
The geometry is cylindrical.
No variation in the axial and angular directions.
Properties are constant.
Problem 2.20
The geometry is cylindrical.
No variation in the angular direction.
Properties are constant.
Parallel streamlines means that the radial velocity component is zero.
Problem 2.21
The geometry is cylindrical. (ii)
No variation in the axial and angular directions.
Properties are constant.
Problem 2.22
This is a forced convection problem.
The same fluid flows over both spheres.
Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient.
Problem 2.23
This is a free convection problem.
The average heat transfer coefficient h depends on the vertical length L of the plate.
L appears in the Nusselt number as well as the Grashof number.
Problem 2.24
This is a forced convection problem.
The same fluid flows over both spheres.
Sphere diameter and free stream velocity affect the Reynolds number which in turn affect the heat transfer coefficient. (iv) Newton’s law of cooling gives the heat transfer
Problem 2.25
Dissipation is important when the Eckert number is high compared to unity.
If the ratio of dissipation to conduction is small compared to unity, it can be neglected.
Problem 2.26
The plate is infinite.
No changes take place in the axial direction (infinite plate).
This is a transient problem.
Constant properties.
Cartesian coordinates.
Problem 2.27
The plate is infinite.
No changes take place in the axial direction (infinite plate).
This is a transient problem.
Constant properties.
Cartesian coordinates.
Gravity is neglected. Thus there is no free convection.
The fluid is stationary.
Problem 3.1
Moving plate sets fluid in motion in the x-direction.
Since plates are infinite the flow field does not vary in the axial direction x.
The effect of pressure gradient is negligible.
The fluid is incompressible (constant density).
Use Cartesian coordinates.
Problem 3.2
Moving plate sets fluid in motion in the x-direction.
Since plates are infinite the flow field does not vary in the axial direction x.
The effect of pressure gradient must be included.
The fluid is incompressible.
Using Fourier’s law, Temperature distribution gives surface heat flux of the moving plate.
Use Cartesian coordinates.
Problem 3.3
Moving plate sets fluid in motion in the x-direction.
Since plates are infinite the flow field does not vary in the axial direction x.
The fluid is incompressible (constant density).
Use Cartesian coordinates.
Problem 3.4
Moving plates set fluid in motion in the positive and negative x-direction.
Since plates are infinite the flow field does not vary in the axial direction x.
The fluid is incompressible (constant density).
The fluid is stationary at the center plane y = 0.
Symmetry dictates that no heat is conducted through the center plane.
Use Cartesian coordinates.
Problem 3.5
Fluid motion is driven by axial pressure drop.
For a very long tube the flow field does not vary in the axial direction z.
The fluid is incompressible (constant density).
Heat is generated due to viscous dissipation. It is removed from the fluid by convection at the surface.
The Nusselt number is a dimensionless heat transfer coefficient.
To determine surface heat flux and heat transfer coefficient requires the determination of temperature distribution.
Temperature distribution depends on the velocity distribution.
Use cylindrical coordinates.
Problem 3.6
Fluid motion is driven by axial pressure drop.
For a very long tube the flow field does not vary in the axial direction z.
The fluid is incompressible (constant density).
Use cylindrical coordinates.
Problem 3.7
Fluid motion is driven by axial motion of the rod. Thus motion is not due to pressure gradient.
For a very long tube the flow field does not vary in the axial direction z.
The fluid is incompressible (constant density).
Heat is generated due to viscous dissipation. It is removed from the fluid by conduction at the surface.
The Nusselt number is a dimensionless heat transfer coefficient.
To determine the heat transfer coefficient require the determination of temperature distribution.
Temperature distribution depends on the velocity distribution.
Use cylindrical coordinates.
Problem 3.8
Fluid motion is driven by gravity.
No velocity and temperature variation in the axial direction.
The fluid is incompressible (constant density).
Heat is generated due to viscous dissipation.
Temperature distribution depends on the velocity distribution.
Use Cartesian coordinates.
Problem 3.9
Fluid motion is driven by gravity.
No velocity and temperature variation in the axial direction.
The fluid is incompressible (constant density).
Heat is generated due to viscous dissipation.
Temperature distribution depends on the velocity distribution.
the inclined surface is at specified temperature and the free surface exchanges heat by convection with the ambient.
Use Cartesian coordinates.
Problem 3.11
Fluid motion is driven by shaft rotation
The housing is stationary.
Axial variation in velocity and temperature are negligible for a very long shaft.
Velocity and temperature do not vary with angular position.
The fluid is incompressible (constant density).
Heat generated by viscous dissipation is removed from the oil at the housing.
No heat is conducted through the shaft.
The maximum temperature occurs at the shaft.
Heat flux at the housing is determined from temperature distribution and Fourier’s law of conduction.
Use cylindrical coordinates.
Problem 3.12
Fluid motion is driven by sleeve rotation
The shaft is stationary.
Axial variation in velocity and temperature are negligible for a very long shaft.
Velocity and temperature do not vary with angular position.
The fluid is incompressible (constant density).
Heat generated by viscous dissipation is removed from the oil at the housing.
No heat is conducted through the shaft.
The maximum temperature occurs at the shaft. (ix) Use cylindrical coordinates.
Problem 3.13
Fluid motion is driven by shaft rotation
Axial variation in velocity and temperature are negligible for a very long shaft.
Velocity, pressure and temperature do not vary with angular position.
The fluid is incompressible (constant density).
Heat generated by viscous dissipation is conducted radially.
The determination of surface temperature and heat flux requires the determination of temperature distribution in the rotating fluid.
Use cylindrical coordinates.
Problem 3.14
Axial pressure gradient sets fluid in motion.
The fluid is incompressible.
The flow field is determined by solving the continuity and Navier-Stokes equations.
Energy equation gives the temperature distribution.
Fourier’s law and temperature distribution give surface heat flux.
Axial variation of temperature is neglected.
Problem 4.3
This is forced convection flow over a streamlined body.
Viscous (velocity) boundary layer approximations can be made if the Reynolds number Rex > 100.
Thermal (temperature) boundary layer approximations can be made if the Peclet number Pex = Rex Pr > 100.
The Reynolds number decreases as the distance along the plate is decreased.
Problem 4.4
The surface is streamlined.
The fluid is water.
Inertia and viscous effects can be estimated using scaling.
If a viscous term is small compared to inertia, it can be neglected.
Properties should be evaluated at the film temperature .2/)( TTT sf
Problem 4.5
The surface is streamlined.
The fluid is water.
Convection and conduction effects can be estimated using scaling.
If a conduction term is small compared to convection, it can be neglected.
The scale for Lt / depends on whether t or .t
Properties should be evaluated at the film temperature .2/)( TTT sf
Problem 4.6
The fluid is air.
Dissipation and conduction can be estimated using scaling.
Dissipation is negligible if the Eckert number is small compared to unity.
Problem 4.7
The surface is streamlined.
The fluid is air.
Problem 4.9
This is a forced convection problem over a flat plate.
At the edge of the thermal boundary layer, the axial velocity is Vu .
Blasius solution gives the distribution of the velocity components u(x,y) and v(x,y).
Scaling gives an estimate of v(x,y).
Problem 4.11
This is a laminar boundary layer flow problem.
Blasius solution gives the velocity distribution for the flow over a semi-infinite flat plate. (iii) A solution for the boundary layer thickness depends on how the thickness is defined.
Problem 4.12
Since the flow within the boundary layer is two-dimensional the vertical velocity component does not vanish. Thus stream lines are not parallel.
Blasius solution is valid for laminar boundary layer flow over a semi-infinite plate.
The transition Reynolds number from laminar to turbulent flow is .5105
Boundary layer approximations are valid if the Reynolds number is greater than 100.
Problem 4.13
This is an external flow problem over a flat plate.
Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable.
Of interest is the value of the local stress at the leading edge of the plate.
Problem 4.14
This is an external flow problem over a flat plate.
The force needed to hold the plate in place is equal to the total shearing force by the fluid on the plate.
Integration of wall shear over the surface gives the total shearing force.
Blasius’s solution for the velocity distribution and wall shearing stress is assumed to be applicable.
Problem 4.16
This is an external forced convection problem for flow over a flat plate.
Of interest is the region where the upstream fluid reaches the leading edge of the plate.
The fluid is heated by the plate.
Heat from the plate is conducted through the fluid in all directions.
Pohlhausen’s solution assumes that heat is not conducted upstream from the plate and therefore fluid temperature at the leading edge is the same as upstream temperature.
Problem 4.18
This is a forced convection problem over a flat plate.
At the edge of the thermal boundary layer, fluid temperature is TT .
Pohlhausen’s solution gives the temperature distribution in the boundary layer.
The thermal boundary layer thickness t increases with distance from the leading edge.
t depends on the Prandtl number.
Problem 4.19
This is an external forced convection problem for flow over a flat plate.
Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable.
Of interest is the value of the local heat flux at the leading edge of the plate.
Knowing the local transfer coefficient and using Newton’s law, gives the heat flux
Problem 4.20
This is an external forced convection problem for flow over a flat plate.
Pohlhausen’s solution for the temperature distribution is assumed to be applicable.
Of interest is the value of the normal temperature gradient at the surface.
Problem 4.22`
This is an external forced convection problem over two flat plates.
Both plates have the same surface area.
For flow over a flat plate, the heat transfer coefficient h decreases with distance from the leading edge.
Since the length in the flow direction is not the same for the two plates, the average heat transfer coefficient is not the same. It follows that the total heat transfer rate is not the same.
The flow over a flat plate is laminar if the Reynolds number is less than 5 105.
Problem 4.23
This is an external forced convection problem for flow over a flat plate.
Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable.
Of interest is the value of the heat transfer rate from a section of the plate at a specified location and of a given width.
Newton’s law of cooling gives the heat transfer rate.
Problem 4.24
This is an external forced convection problem for flow over a flat plate.
Of interest is the variation of the local heat transfer coefficient with free stream velocity and distance from the leading edge.
Pohlhausen's solution applies to this problem.
Problem 4.25
This is an external flow problem.
At the edge of the thermal boundary layer, ty , fluid temperature approaches free
stream temperature. That is, TT and .1)/()(*ss TTTTT
According to Pohlhausen's solution, Fig. 4.6, the thermal boundary layer thickness
depends on the Prandtl number, free stream velocity V , kinematic viscosity and location x.
Problem 4.26
This is an external forced convection problem for flow over a flat plate.
The Reynolds number and Peclet number should be checked to determine if the flow is laminar and if boundary layer approximations are valid.
Pohlhausen's solution is applicable if 100 < Rex < 100 105 and Pex = Rex Pr > 100.
Thermal boundary layer thickness and heat transfer coefficient vary along the plate.
Newton’s law of cooling gives local heat flux. (vi) The fluid is water.
Problem 4.27
This is an external forced convection problem over a flat plate.
Increasing the free stream velocity, increases the average heat transfer coefficient. This in turn causes surface temperature to drop.
Based on this observation, it is possible that the proposed plan will meet design specification.
Since the Reynolds number at the downstream end of the package is less than 500,000, it follows that the flow is laminar throughout.
Increasing the free stream velocity by a factor of 3, increases the Reynolds number by a factor of 3 to 330,000. At this Reynolds number the flow is still laminar.
The power supplied to the package is dissipated into heat and transferred to the surroundings from the surface.
Pohlhausen's solution can be applied to this problem.
The ambient fluid is unknown.
Problem 4.28
This is an external forced convection problem of flow over a flat plate.
Convection heat transfer from a surface can be determined using Newton’s law of cooling.
The local heat transfer coefficient changes along the plate. The total heat transfer rate can be determined using the average heat transfer coefficient.
For laminar flow, Pohlhausen's solution gives the heat transfer coefficient.
For two in-line fins heat transfer from the down stream fin is influenced by the upstream fin. The further the two fins are apart the less the interference will be.
Problem 4.29
This is an external forced convection problem for flow over a flat plate.
Pohlhausen’s solution for the temperature distribution and heat transfer coefficient is assumed to be applicable.
Knowing the heat transfer coefficient, the local Nusselt number can be determined.
the Newton’s law of cooling gives the heat transfer rate.
Pohlhausen’s solution gives the thermal boundary layer thickness.
Problem 4.30
This is an external forced convection problem of flow over a flat plate.
Convection heat transfer from a surface can be determined using Newton’s law of cooling.
The local heat transfer coefficient changes along the plate.
For each triangle the area changes with distance along the plate.
The total heat transfer rate can be determined by integration along the length of each triangle.
Pohlhausen's solution may be applicable to this problem.
Problem 4.31
This is an external forced convection problem of flow over a flat plate.
Heat transfer rate can be determined using Newton’s law of cooling.
The local heat transfer coefficient changes along the plate.
The area changes with distance along the plate.
The total heat transfer rate can be determined by integration along the length of the triangle.
Pohlhausen's solution may be applicable to this problem.
Problem 4.32
This is an external forced convection problem of flow over a flat plate.
Heat transfer rate can be determined using Newton’s law of cooling.
The local heat transfer coefficient changes along the plate.
The area changes with distance along the plate.
The total heat transfer rate can be determined by integration along over the area of the semi-circle.
Pohlhausen's solution gives the heat transfer coefficient.
Problem 4.33
This is an external forced convection problem of flow over a flat plate.
Heat transfer rate can be determined using Newton’s law of cooling.
The local heat transfer coefficient changes along the plate.
The area changes with distance along the plate.
The total heat transfer rate can be determined by integration along the length of the triangle.
Pohlhausen's solution may be applicable to this problem.
Problem 4.34
This is an external forced convection problem of flow over a flat plate.
This problem involves determining the heat transfer rate from a circle tangent to the leading edge of a plate
Heat transfer rate can be determined using Newton’s law of cooling.
The local heat transfer coefficient changes along the plate.
The area changes with distance along the plate.
The total heat transfer rate can be determined by integration along the length of the triangle.
Pohlhausen's solution may be applicable to this problem.
Problem 4.36
The flow field for this boundary layer problem is simplified by assuming that the axial velocity is uniform throughout the thermal boundary layer.
Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution.
The Nusselt number depends on the temperature gradient at the surface.
Problem 4.37
The flow field for this boundary layer problem is simplified by assuming that the axial velocity varies linearly in the y-direction.
Since velocity distribution affects temperature distribution, the solution for the local Nusselt number can be expected to differ from Pohlhausen’s solution.
The Nusselt number depends on the temperature gradient at the surface.
Problem 4.38
The flow and temperature fields for this boundary layer problem are simplified by assuming that the axial velocity and temperature do not vary in the x-direction.
The heat transfer coefficient depends on the temperature gradient at the surface.
Temperature distribution depends on the flow field.
The effect of wall suction must be taken into consideration.
Problem 4.39
This is a forced convection flow over a plate with variable surface temperature.
The local heat flux is determined by Newton’s law of cooling.
The local heat transfer coefficient and surface temperature vary with distance along the plate. The variation of surface temperature and heat transfer coefficient must be such that Newton’s law gives uniform heat flux.
The local heat transfer coefficient is obtained from the local Nusselt number.
Problem 4.40
This is a forced convection flow over a plate with variable surface temperature.
The Reynolds number should be computed to determine if the flow is laminar or turbulent.
The local heat transfer coefficient and surface temperature vary with distance along the plate.
The local heat transfer coefficient is obtained from the solution to the local Nusselt number.
The determination of the Nusselt number requires determining the temperature gradient at the surface.
Problem 4.41
This is a forced convection flow over a plate with variable surface temperature.
The Reynolds number should be computed to determine if the flow is laminar or turbulent.
Newton’s law of cooling gives the heat transfer rate from the plate.
The local heat transfer coefficient and surface temperature vary with distance along the plate. Thus determining the total heat transfer rate requires integration of Newton’s law along the plate.
The local heat transfer coefficient is obtained from the local Nusselt number.
Problem 4.42
This is an external forced convection problem of flow over a flat plate
Convection heat transfer from a surface can be determined using Newton’s law of cooling.
The local heat transfer coefficient and surface temperature vary along the plate.
For each triangle the area varies with distance along the plate.
The total heat transfer rate can be determined by integration along the length of each triangle.
Problem 4.43
This is a forced convection boundary layer flow over a wedge.
Wedge surface is maintained at uniform temperature.
The flow is laminar.
The fluid is air.
Similarity solution for the local Nusselt number is presented in Section 4.4.3.
The Nusselt number depends on the Reynolds number and the dimensionless temperature gradient at the surface ./)0( dd (vii) Surface temperature gradient depends on wedge
angle.
Problem 4.44
This is a forced convection boundary layer flow over a wedge.
Wedge surface is maintained at uniform temperature.
The flow is laminar.
The average Nusselt number depends on the average heat transfer coefficient..
Similarity solution for the local heat transfer coefficient is presented in Section 4.4.3.
Problem 4.45
Newton’s law of cooling gives the heat transfer rate from a surface.
Total heat transfer from a surface depends on the average heat transfer coefficient h .
Both flat plate and wedge are maintained at uniform surface temperature.
Pohlhausen’s solution gives h for a flat plate.
Similarity solution for the local heat transfer coefficient for a wedge is presented in Section 4.4.3.
Problem 4.46
The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer is the same as that of the external flow.
Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3.
The local Nusselt number depends the local heat transfer coefficient which depends on the temperature gradient at the surface.
Problem 4.47
The flow field for this boundary layer problem is simplified by assuming that axial velocity within the thermal boundary layer varies linearly with the normal distance.
Since velocity distribution affects temperature distribution, the solution for the local Nusselt number differs from the exact case of Section 4.4.3.
The local Nusselt number depends on the local heat transfer coefficient which depends on the temperature gradient at the surface.
Problem 5.1
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution
Fluid velocity for 1Pr is assumed to be uniform, Vu . This represents a
significant simplification.
Problem 5.2
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
Fluid velocity for 1Pr is assumed to be linear, )/(yVu .
Problem 5.3
The velocity is assumed to be uniform, ,Vu throughout the thermal boundary layer.
A leading section of length is unheated.ox
at , surface heat flux is uniform. oxx
The determination of the Nusselt number requires the determination of the temperature distribution.
Surface temperature is unknown.
The maximum surface temperature for a uniformly heated plate occurs at the trailing end.
Problem 5.4
The velocity distribution is known.
Surface temperature is uniform.
The determination of the Nusselt number requires the determination of the temperature distribution.
Newton’s law of cooling gives the heat transfer rate. This requires knowing the local heat transfer coefficient.
Problem 5.5
The velocity distribution is known
Total heat transfer is equal to heat flux times surface area.
Heat flux is given. However, the distance x = L at which 2/Ht is unknown.
Problem 5.6
The determination of the Nusselt number requires the determination of the velocity and temperature distributions.
Velocity is assumed uniform.
Surface temperature is variable.
Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.
Problem 5.7
The determination of the Nusselt number requires the determination of the velocity and temperature distributions.
Velocity is assumed linear.
Surface temperature is variable.
Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.
Problem 5.8
The determination of the Nusselt number requires the determination of the velocity and temperature distributions.
Velocity is assumed uniform.
Surface temperature is variable.
Newton’s law of cooling gives surface heat flux. This requires knowing the local heat transfer coefficient.
Problem 5.9
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
Surface heat flux is variable. It decreases with distance x.
Surface temperature is unknown.
Newton’s law of cooling gives surface temperature. This requires knowing the local heat
transfer coefficient. (v) .1/t
Problem 5.10
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
Fluid velocity for 1Pr is assumed to be uniform, Vu . This represents a
significant simplification.
Surface heat flux is variable. It increases with distance x.
Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x. Thus maximum surface temperature is at the trailing end x = L.
Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.
Problem 5.11
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
Fluid velocity for 1Pr is assumed to be uniform, Vu . This represents a
significant simplification.
Surface heat flux is variable. It increases with distance x.
Surface temperature is unknown. Since flux increases with x and heat transfer coefficient decreases with x, surface temperature is expected to increase with x. Thus maximum surface temperature is at the trailing end x = L.
Newton’s law of cooling gives surface temperature. This requires knowing the local heat transfer coefficient.
Problem 5.12
This problem is described by cylindrical coordinates.
Velocity variation with y is negligible.
Conservation of mass requires that radial velocity decrease with radial distance r.
Surface temperature is uniform.
Problem 5.13
This problem is described by cylindrical coordinates.
Velocity variation with y is negligible.
Conservation of mass requires that radial velocity decrease with radial distance r.
Surface heat flux is uniform
Surface temperature is unknown.
Problem 5.14
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
Fluid velocity for 1Pr is assumed to be uniform, Vu . This represents a
significant simplification.
The plate is porous.
Fluid is injected through the plate with uniform velocity.
The plate is maintained at uniform surface temperature.
A leading section of the plate is insulated.
Problem 5.15
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
Fluid velocity for 1Pr is assumed to be uniform, Vu . This represents a
significant simplification.
The plate is porous.
Fluid is injected through the plate with uniform velocity.
The plate is heated with uniform surface flux
Surface temperature is unknown, (vii) A leading section of the plate is insulated.
Problem 5.16
There are two thermal boundary layers in this problem.
The upper and lower plates have different boundary conditions. Thus, temperature distribution is not symmetrical.
The lower plate is at uniform temperature while heat is removed at uniform flux along the upper plate.
Fluid velocity is assumed uniform throughout the channel.
Problem 6.1
This is an internal forced convection problem.
Scaling gives estimates of and hL .tL
Exact solutions for and are available for laminar flow through channels. hL tL
Exact solutions for depend on channel geometry and surface boundary conditions. tL
Problem 6.2
This is an internal forced convection problem.
Scaling gives estimates of andhL
Exact solutions for and are available for laminar flow through channels. hL tL
Exact solutions for depend on channel geometry and surface boundary conditions. tL
Problem 6.3
This is an internal force convection problem.
The channel is a long tube.
The surface is maintained at a uniform temperature.
Since the tube section is far away from the entrance, the velocity and temperature can be assumed fully developed.
Tube diameter, mean velocity and inlet, outlet and surface temperatures are known. The length is unknown.
The fluid is air.
Problem 6.4
This is an internal force convection in a tube.
The surface is heated at uniform flux.
Surface temperature increases along the tube and is unknown.
The flow is assumed laminar and fully developed.
The heat transfer coefficient for fully developed flow through channels is constant.
According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.
Problem 6.5
This is an internal force convection in a tube.
The surface is heated at uniform flux.
Surface temperature increases along the tube and is unknown.
The flow is assumed laminar and fully developed.
The heat transfer coefficient for fully developed flow through channels is constant.
According to Newton’s law of cooling, surface temperature is related to mean fluid temperature, surface heat flux and heat transfer coefficient.
Problem 6.6
This is an internal forced convection problem in a tube.
The surface is heated at uniform flux.
Surface temperature changes along the tube and is unknown.
The Reynolds number should be checked to determine if the flow is laminar or turbulent.
If hydrodynamic and thermal entrance lengths are small compared to tube length, the flow can be assumed fully developed throughout.
For fully developed flow, the heat transfer coefficient is uniform.
The length of the tube is unknown.
The fluid is water.
Problem 6.7
This is an internal force convection problem.
The channel is a tube.
The surface is maintained at a uniform temperature.
Entrance effect is important in this problem.
The average Nusselt number for a tube of length L depends on the average heat transfer coefficient over the length.
Problem 6.8
This is an internal forced convection problem.
The fluid is heated at uniform wall flux.
Surface temperature changes with distance along the channel. It reaches a maximum value at the outlet.
The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if this is an entrance or fully developed problem.
The channel has a square cross-section.
Application of Newton’s law of cooling at the outlet relates outlet temperature to surface temperature, surface flux and heat transfer coefficient.
Application of conservation of energy gives a relationship between heat added, inlet temperature, outlet temperature, specific heat and mass flow rate.
Problem 6.9
This is an internal forced convection problem in tubes.
The flow is laminar and fully developed.
The surface is maintained at uniform temperature.
All conditions are identical for two experiments except the flow rate through one is half that of the other.
The total heat transfer rate depends on the outlet temperature.
Problem 6.10
This is an internal forced convection problem.
The channel has a rectangular cross section.
Surface temperature is uniform.
The Reynolds and Peclet numbers should be checked to establish if the flow is laminar or turbulent and if entrance effects can be neglected.
Channel length is unknown.
The fluid is air.
Problem 6.11
This is an internal force convection problem.
The channel is a rectangular duct.
The surface is maintained at a uniform temperature.
The velocity and temperature are fully developed.
The Reynolds number should be checked to determine if the flow is laminar or turbulent.
Duct size, mean velocity and inlet, outlet and surface temperatures are known. The
length is unknown. (vii) Duct length depends on the heat transfer coefficient.
The fluid is water.
Problem 6.12
This is an internal forced convection problem in a channel.
The surface is heated at uniform flux.
Surface temperature changes along the channel. It reaches a maximum value at the outlet.
The Reynolds number should be checked to determine if the flow is laminar or turbulent.
Velocity and temperature profiles become fully developed far away from the inlet.
The heat transfer coefficient is uniform for fully developed flow.
The channel has a square cross section.
tube length is unknown. (ix) The fluid is air.
Problem 6.13
This is an internal forced convection problem in tubes.
The flow is laminar and fully developed.
The surface is maintained at uniform temperature.
All conditions are identical for two tubes except the diameter of one is twice that of the other.
The total heat transfer in each tube depends on the outlet temperature.
Problem 6.14
This is an internal forced convection problem.
Equation (6.3) gives scaling estimate of the thermal entrance length.
Equation (6.20b) gives scaling estimate of the local Nusselt number.
The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface temperature.
Graetz solutions gives the thermal entrance length (distance to reach fully developed temperature) and local Nusselt number.
Problem 6.15
This is an internal forced convection problem.
Equation (6.20b) gives scaling estimate of the local Nusselt number.
The Graetz problem deals with laminar flow in the entrance of a tube at uniform surface temperature.
Problem 6.16
This is an internal forced convection problem in a tube.
The velocity is fully developed.
The temperature is developing.
Surface is maintained at uniform temperature.
The Reynolds number should be computed to establish if flow is laminar or turbulent.
Tube length is unknown.
The determination of tube length requires determining the heat transfer coefficient.
Problem 6.17
This is an internal forced convection problem in a tube.
The velocity is fully developed.
The temperature is developing.
Surface is maintained at uniform temperature.
The Reynolds number should be computed to establish if flow is laminar or turbulent.
Outlet mean temperature is unknown.
The determination of outlet temperature requires determining the heat transfer coefficient.
Since outlet temperature is unknown, air properties can not be determined. Thus a trial and error procedure is needed to solve the problem.
Problem 6.18
This is an internal forced convection problem in a tube.
The velocity is fully developed and the temperature is developing.
The surface is heated with uniform flux.
The Reynolds number should be computed to establish if the flow is laminar or turbulent.
Compute thermal entrance length to determine if it can be neglected.
Surface temperature varies with distance from entrance. It is maximum at the outlet. Thus surface temperature at the outlet is known.
Analysis of uniformly heated channels gives a relationship between local surface temperature, heat flux and heat transfer coefficient.
The local heat transfer coefficient varies with distance form the inlet.
Knowing surface heat flux, the required power can be determined.
Newton’s law of cooling applied at the outlet gives outlet temperature.
Problem 6.19
This is an internal forced convection problem in a rectangular channel.
The velocity is fully developed and the temperature is developing.
The surface is maintained at uniform temperature.
The Reynolds number should be computed to establish if the flow is laminar or turbulent.
Compute entrance lengths to determine if they can be neglected
Surface flux varies with distance from entrance. It is minimum at outlet.
Newton’s law gives surface flux in terms of the local heat transfer coefficient and
the local mean temperature
)(xh
)(xTm .
The local and average heat transfer coefficient decrease with distance form the inlet.
The local mean temperature depends on the local average heat transfer coefficient
).(xh (x) Surface temperature is unknown.
Problem 7.2
This is an external free convection problem over a vertical plate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
The solution for laminar flow is given in Section 7.4
For laminar flow, Fig.7.2 gives the viscous boundary layer thickness and Fig. 7.3 gives
the thermal boundary layer thickness t .
Newton’s law of cooling gives the heat transfer rate.
Equation (7.23) gives the average heat transfer coefficient h . (vii) The fluid is water.
Problem 7.3
This is an external free convection problem for flow over a vertical plate.
Laminar flow solution for temperature distribution for a plate at uniform surface temperature is given in Fig. 7.3 .
The dimensionless temperature gradient at the surface is given in Table 7.1.
The solution depends on the Prandtl number.
Problem 7.4
This is a free convection problem.
Heat is lost from the door to the surroundings by free convection and radiation.
To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air.
As a first approximation, radiation can be neglected.
Newton’s law of cooling gives the rate of heat transfer.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
For laminar flow the solution of Section 7.4 is applicable.
Problem 7.5
This is a free convection and radiation problem.
The geometry is a vertical plate.
Surface temperature is uniform.
Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
For laminar flow the solution of Section 7.4 is applicable.
Since radiation heat transfer is considered in this problem, all temperatures should be expressed
Problem 7.6
This is a free convection problem.
The power dissipated in the electronic package is transferred to the ambient fluid by free convection.
As the power is increased, surface temperature increases.
The maximum power dissipated corresponds to the maximum allowable surface temperature.
Surface temperature is related to surface heat transfer by Newton’s law of cooling.
The problem can be modeled as free convection over a vertical plate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
For laminar flow the solution of Section 7.4 is applicable.
The fluid is air.
Problem 7.7
This is a free convection problem.
The power dissipated in the electronic package is transferred to the ambient fluid by free convection.
As the power is increased, surface temperature increases.
The maximum power dissipated corresponds to the maximum allowable surface temperature.
Surface temperature is related to surface heat transfer by Newton’s law of cooling.
The problem can be modeled as free convection over a vertical plate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
For laminar flow the solution of Section 7.4 is applicable.
The fluid is water.
Problem 7.8
This is a free convection problem.
The surface is maintained at uniform temperature.
Newton’s law of cooling determines the heat transfer rate.
Heat transfer rate depends on the heat transfer coefficient.
The heat transfer coefficient decreases with distance from the leading edge of the plate.
The width of each triangle changes with distance from the leading edge.
For laminar flow the solution of Section 7.4 is applicable.
Problem 7.9
This is a free convection problem over a vertical plate.
The surface is maintained at uniform temperature.
Local heat flux is determined by Newton’s law of cooling.
Heat flux depends on the local heat transfer coefficient
Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent. For
Laminar flow the solution of Section 7.4 is applicable.
The fluid is air.
Problem 7.10
This is a free convection problem over a vertical plate.
The power dissipated in the chips is transferred to the air by free convection.
This problem can be modeled as free convection over a vertical plate with constant surface heat flux.
Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end).
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
For laminar flow the analysis of Section 7.5 gives surface temperature distribution.
The fluid is air.
Properties depend on the average surface temperature . Since is unknown, the sT sT
problem must be solved by trail and error.
Problem 7.11
This is a free convection problem over a vertical plate.
The power dissipated in the chips is transferred to the air by free convection
This problem can be modeled as free convection over a vertical plate with constant surface heat flux.
Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end).
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
For laminar flow the analysis of Section 7.5 gives surface temperature distribution.
The fluid is air.
Properties depend on the average surface temperature . Since is unknown, the
problem must be solved by trail and error.
sT sT
Problem 7.12
This is a free convection problem over a vertical plate at uniform surface temperature.
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
The integral method can be used to determine the velocity and temperature distribution.
Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.
Problem 7.13
This is a free convection problem over a vertical plate at uniform surface heat flux.
In general, to determine the Nusselt number it is necessary to determine the velocity and temperature distribution.
The integral method can be used to determine the velocity and temperature distribution.
Application of the integral method reduces to determining the velocity and temperature boundary layer thickness.
Problem 8.1
This is an external forced convection problem.
The geometry can be modeled as a flat plate.
Surface temperature is uniform.
Newton’s law of cooling gives heat transfer rate from the surface to the air.
The average heat transfer coefficient must be determined.
The Reynolds number should be evaluated to establish if the flow is laminar, turbulent or mixed.
Analytic or correlation equations give the heat transfer coefficient.
Problem 8.2
This is an external forced convection problem.
The geometry can be modeled as a flat plate.
Surface temperature is uniform.
To determine the heat flux at a given location, the local heat transfer coefficient must be determined.
The average heat transfer coefficient is needed to determine the total heat transfer rate.
Newton’s law of cooling gives surface flux and total heat transfer rate.
The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed.
Analytic or correlation equations give the heat transfer coefficient.
Problem 8.3
This is an external forced convection problem of flow over a flat plate.
Surface temperature is assumed uniform.
The heat transfer coefficient in turbulent flow is greater than that in laminar flow. Thus higher heat transfer rates can be sustained in turbulent flow than laminar flow.
The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed.
Heat loss from the surface is approximately equal to the power dissipated in the package.
Newton’s law of cooling gives a relationship between heat transfer rate, surface area, heat transfer coefficient, surface temperature and ambient temperature.
The fluid is air.
Problem 8.4
This is an external forced convection problem.
The geometry is a flat plate.
Surface temperature is uniform.
Newton’s law of cooling gives the heat transfer rate.
The Reynolds number should be checked to establish if the flow is laminar, turbulent or mixed.
Analytic or correlation equations give the heat transfer coefficient.
If the flow is laminar throughout, heat transfer from the first half should be greater than that from the second half.
Second half heat transfer can be obtained by subtracting first half heat rate from the heat transfer from the entire plate.
The fluid is water.
Problem 8.5
The chip is cooled by forced convection.
This problem can be modeled as a flat plate with an unheated leading section.
Newton's law of cooling can be applied to determine the rate of heat transfer between the chip and the air.
Check the Reynolds number to establish if the flow is laminar or turbulent.
Problem 8.6
Heat transfer from the collector to the air is by forced convection.
This problem can be modeled as a flat plate with an unheated leading section.
Newton's law of cooling can be applied to determine the rate of heat transfer between the collector and air.
The heat transfer coefficient varies along the collector.
The Reynolds number should be computed to establish if the flow is laminar or turbulent.
Problem 8.7
This is an external forced convection problem.
The flow is over a flat plate.
Surface temperature is uniform.
Plate orientation is important.
Variation of the heat transfer coefficient along the plate affects the total heat transfer.
The heat transfer coefficient for laminar flow decreases as the distance from the leading edge is increased. However, at the transition point it increases and then decreases again.
Higher rate of heat transfer may be obtained if the wide side of a plate faces the flow. On the other hand, higher rate may be obtained if the long side of the plate is in line with the flow direction when transition takes place
The fluid is water.
Problem 8.8
This is an external forced convection problem.
The flow is over a flat plate.
The problem can be modeled as flow over a flat plate with uniform surface heat flux.
Surface temperature varies with distance along plate. The highest surface temperature is at the trailing end.
Tripping the boundary layer at the leading edge changes the flow from laminar to turbulent. This increases the heat transfer coefficient and lowers surface temperature.
Newton’s law of cooling gives surface temperature.
Problem 8.9
This is an external forced convection problem.
The flow is normal to a tube.
Surface temperature is uniform.
Tube length is unknown.
Newton’s law of cooling can be used to determine surface area. Tube length is related to surface area.
The fluid is water.
Problem 8.10
Heat is removed by the water from the steam causing it to condense.
The rate at which steam condenses inside the tube depends on the rate at which heat is removed from the outside surface.
Heat is removed from the outside surface by forced convection.
This is an external forced convection problem of flow normal to a tube. (v) Newton’s law of cooling gives the rate of heat loss from the surface.
Problem 8.11
Electric power is dissipated into heat and is removed by the water.
This velocity measuring concept is based on the fact that forced convection heat transfer is affected by fluid velocity.
Velocity affects the heat transfer coefficient which in term affects surface temperature.
Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature.
This problem can be modeled as external flow normal to a cylinder.
The fluid is water.
Problem 8.12
This is an external forced convection problem.
The flow is normal to a rod.
Surface heat transfer rate per unit length is known. However, surface temperature is unknown.
In general, surface temperature varies along the circumference. However, the rod can be assumed to have a uniform surface temperature.
This problem can be modeled as forced convection normal to a rod with uniform surface flux or temperature.
Newton’s law of cooling gives surface temperature.
The fluid is air.
Problem 8.13
Electric power is dissipated into heat and is removed by the fluid.
This velocity measuring instrument is based on the fact that forced convection heat transfer is affected by fluid velocity.
velocity affects the heat transfer coefficient which in term affects surface temperature and heat flux.
Newton’s law of cooling relates surface heat loss to the heat transfer coefficient, surface area and surface temperature.
This problem can be modeled as external flow normal to a cylinder.
The fluid is air.
Problem 8.14
The sphere cools off as it drops. Heat loss from the sphere is by forced convection.
The height of the building can be determined if the time it takes the sphere to land is known.
Time to land is the same as cooling time.
Transient conduction determines cooling time.
If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature.
Cooling rate depends on the heat transfer coefficient.
Problem 8.15
The electric energy dissipated inside the sphere is removed from the surface as heat by forced convection.
This problem can be modeled as external flow over a sphere.
Newton’s law of cooling relates heat loss from the surface to heat transfer coefficient, surface area and surface temperature. (iv) The fluid is air.
Problem 8.16
The sphere cools off as it drops. Heat loss from the sphere is by forced convection.
This is an external flow problem with a free stream velocity that changes with time.
This is a transient conduction problem. The cooling time is equal to the time it takes the sphere to drop to street level.
If the Biot number is less than 0.1, lumped capacity method can be used to determine transient temperature.
Cooling rate depends on the heat transfer coefficient.
Problem 8.17
This is an internal forced convection problem.
The channel is a tube.
The outside surface is maintained at a uniform temperature.
Neglecting tube thickness resistance means that the inside and outside surface temperatures are identical.
Fluid temperature is developing.
Inlet and outlet temperatures are known.
The Reynolds number should be determined to establish if the flow is laminar or turbulent.
The required tube length depends on the heat transfer coefficient.
The fluid is water.
Problem 8.18
This is an internal force convection problem.
The channel is a tube.
The surface is maintained at a uniform temperature.
The velocity is fully developed.
The temperature is developing.
The outlet temperature is unknown.
The Reynolds number should be checked to establish if the flow is laminar or turbulent.
The fluid is air.
Problem 8.19
This is an internal forced convection problem
Tube surface is maintained at uniform temperature.
The velocity is fully developed.
The length of tube is unknown.
The temperature is developing. However, depending on tube length relative to the thermal entrance length, temperature may be considered fully developed throughout.
The Reynolds number should be checked to determine if the flow is laminar or turbulent.
The fluid is water.
Problem 8.20
This is an internal forced convection problem.
Tube surface is maintained at a uniform temperature.
The velocity and temperature are developing. Thus, entrance effects may be important.
The outlet temperature is unknown.
The fluid is air.
Problem 8.21
This is an internal forced convection problem.
Tube surface is maintained at uniform temperature.
The section of interest is far away from the inlet. This means that flow and temperature
can be assumed fully developed and the heat transfer coefficient uniform.
It is desired to determine the surface flux at this section. Newton’s law of cooling gives a relationship between local flux, surface temperature and heat transfer coefficient.
The Reynolds number should be checked to determine if the flow is laminar or turbulent.
The fluid is water.
Problem 8.22
This is an internal forced convection problem in a tube.
Both velocity and temperature are fully developed.
Tube surface is maintained at uniform temperature.
The Reynolds number should be computed to establish if flow is laminar or turbulent.
Mean velocity, mean inlet and outlet temperatures and tube diameter are known.
The fluid is air.
Problem 8.23
This is an internal forced convection problem.
The surface of each tube is maintained at uniform temperature which is the same for both.
The velocity and temperature are fully developed. Thus, the heat transfer coefficient is uniform.
Air flows through each tube at different rates.
The Reynolds number should be computed to establish if the flow is laminar or turbulent.
Surface heat flux depends on the heat transfer coefficient.
Problem 8.24
This is an internal forced convection problem.
The geometry consists of two concentric tubes.
Air flows in the inner tube while water flows in the annular space between the two tubes.
The Reynolds number should be computed for both fluids to establish if the flow is laminar or turbulent.
Convection resistance depends on the heat transfer coefficient.
Problem 8.25
This is an internal forced convection problem.
The geometry consists of a tube concentrically placed inside a square duct,.
Water flows in the tube and the duct.
The Reynolds number should be computed for the two fluids to establish if the flow is laminar or turbulent.
Far away from the inlet the velocity and temperature may be assumed fully developed.
Problem 8.26
Heat is lost from the door to the surroundings by free convection and radiation.
To determine the rate of heat loss, the door can by modeled as a vertical plate losing heat by free convection to an ambient air and by radiation to a large surroundings.
Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation.
Problem 8.27
This is a free convection and radiation problem.
The geometry is a vertical plate.
Surface temperature is uniform.
Newton’s law of cooling gives convection heat transfer rate while Stefan-Boltzmann law gives radiation heat transfer rate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
Since radiation heat transfer is considered in this problem, all temperatures should be expressed in degrees kelvin.
The fluid is air.
Problem 8.28
This is a free convection problem.
The power dissipated in the electronic package is transferred to the ambient fluid by free convection.
As the power is increased, surface temperature increases.
The maximum power dissipated corresponds to the maximum allowable surface temperature.
Surface temperature is related to surface heat transfer by Newton’s law of cooling.
The problem can be modeled as free convection over a vertical plate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
The fluid is air.
Problem 8.29
This is a free convection problem.
The power dissipated in the electronic package is transferred to the ambient fluid by free convection.
As the power is increased, surface temperature increases.
The maximum power dissipated corresponds to the maximum allowable surface temperature.
Surface temperature is related to surface heat transfer by Newton’s law of cooling.
The problem can be modeled as free convection over a vertical plate.
The Rayleigh number should be computed to determine if the flow is laminar or turbulent.
The fluid is water.
Problem 8.30
This is a free convection problem.
The surface is maintained at uniform temperature.
The heat transfer coefficient decreases with distance from the leading edge of the plate.
The heat transfer rate from the lower half 1 is greater than that from the upper half 2.
Total heat transfer from each half can be determined using the average heat transfer coefficient.
Heat transfer from the upper half is equal to the heat transfer from the entire plate minus heat transfer from the lower half.
Problem 8.31
This is a free convection problem.
The surface is maintained at uniform temperature.
The heat transfer coefficient decreases with distance from the leading edge of the plate.
The width of each triangle changes with distance from the leading edge.
Problem 8.32
This is a free convection problem over a vertical plate.
The surface is maintained at uniform temperature.
Local heat flux is determined by Newton’s law of cooling.
Heat flux depends on the local heat transfer coefficient.
Free convection heat transfer coefficient for a vertical plate decreases with distance from the leading edge. Thus, the flux also decreases.
The Rayleigh number should be computed to select an appropriate Nusselt number correlation equation.
The fluid is air.
Problem 8.33
This is a free convection problem over a vertical plate.
The power dissipated in the chips is transferred to the air by free convection.
This problem can be modeled as free convection over a vertical plate with constant surface heat flux.
Surface temperature increases as the distance from the leading edge is increased. Thus, the maximum surface temperature occurs at the top end of the plate (trailing end).
Newton’s law of cooling relates surface temperature to heat flux and heat transfer coefficient.
The fluid is air.
Problem 8.34
Power supply to the disk is lost from the surface to the surroundings by free convection and radiation.
To determine the rate of heat loss, the disk can by modeled as a horizontal plate losing heat by free convection to an ambient air and by radiation to a large surroundings.
Newton’s law of cooling gives the rate of heat transfer by convection and Stefan-Boltzmann relation gives the heat loss by radiation.
Free convection correlations give the heat transfer coefficient.
Conservation of energy at the surface gives the emissivity, if it is the only unknown.
Problem 8.35
This is a free convection problem.
The geometry is a flat plate.
Heat transfer from two plates is to be compared. One plate is vertical and the other is inclined. Both plates fit in the same vertical space. Thus, the inclined plate is longer than the vertical plate.
Both plates are maintained at uniform surface temperature.
Heat transfer depends on surface area and average heat transfer coefficient.
Problem 8.36
This is a free convection problem.
The kiln has four vertical sides and a horizontal top.
All surfaces are at the same uniform temperature.
Newton’s law of cooling gives the heat transfer rate.
The sides can be modeled as vertical plates and the top as a horizontal plate.
The fluid is air.
Problem 8.37
Heat transfer from the surface is by free convection and radiation.
The burner can be modeled as a horizontal disk with its heated side facing down.
Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relations gives heat transfer by radiation.
Both convection and radiation depend on surface temperature.
If the burner is well insulated at the bottom heated surface and its rim, then the electric power supply is equal to surface heat transfer.
Problem 8.38
Heat transfer from the surface is by free convection and radiation.
The sample can be modeled as a horizontal disk with its heated side facing down or up.
Newton’s law of cooling gives heat transfer by convection and Stefan-Boltzmann relation gives heat transfer by radiation.
Radiation depends on surface emissivity.
If the disk is well insulated at the heated surface and its rim, then the electric power supply is equal to surface heat transfer.
Since the electric power is the same for both orientations, it follows that surface heat transfer rate is also the same.
Each orientation has its own Nusselt number correlation equation.
Problem 8.39
This is a free convection problem.
Heat is transferred from the cylindrical surface and top surface of tank to the ambient air.
Under certain conditions a vertical cylindrical surface can be modeled as a vertical plate.
Newton’s law of cooling gives the heat transfer rate from tank.
The fluid is air.
Problem 8.40
This is a free convection problem.
The geometry is a horizontal round duct.
Heat is transferred from duct surface to the ambient air.
According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures.
Problem 8.41
This is a free convection problem.
The geometry is a horizontal pipe.
Heat is transferred from pipe surface to the ambient air.
Adding insulation material reduces heat loss from pipe.
According to Newton’s law of cooling, the rate of heat transfer depends on the heat transfer coefficient, surface area and surface and ambient temperatures
Heat transfer coefficient and surface area change when insulation is added.
The fluid is air.
Problem 8.42
This is a free convection problem.
The geometry is a horizontal wire (cylinder).
Under steady state conditions the power dissipated in the wire is transferred to the surrounding air.
According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature.
The fluid is air.
Problem 8.43
This is a free convection problem.
The geometry is a horizontal tube.
Under steady state conditions the power dissipated in the neon tube is transferred to the surrounding air.
According to Newton’s law of cooling, surface temperature is determined by the heat transfer rate, heat transfer coefficient, surface area and ambient temperature.
The fluid is air.
Problem 8.44
This is a free convection problem.
The geometry is a round horizontal round duct.
Heat is transferred from the ambient air to the duct.
According to Newton’s law of cooling, the rate of heat transfer to the surface depends on the heat transfer coefficient, surface area and surface and ambient temperatures.
The fluid is air.
Problem 8.45
This is a free convection and radiation problem
The geometry is a sphere.
Under steady state conditions the power dissipated in the bulb is transferred to the surroundings by free convection and radiation and through the base by conduction.
According to Newton’s law of cooling and Stefan-Boltzmann radiation law, heat loss from the surface depends on surface temperature.
The ambient fluid is air.
Problem 8.46
At steady state, power supply to the sphere must be equal to the heat loss from the surface
Heat loss from the surface is by free convection.
The surface is maintained at uniform temperature.
Problem 8.47
Heat is transferred from the ambient air to the water in the fish tank.
Adding an air enclosure reduces the rate of heat transfer.
To estimate the reduction in cooling load, heat transfer from the ambient air to the water with and without the enclosure must be determined.
) Neglecting the thermal resistance of glass, the resistance to heat transfer form the air to the water is primarily due to the air side free convection heat transfer coefficient.
Installing an air cavity introduces an added thermal resistance.
The problem can be modeled as a vertical plate and as a vertical rectangular enclosure.
The outside surface temperature of the enclosure is unknown.
Newton’s law of cooling gives the heat transfer rate.
The Rayleigh number should be determined for both vertical plate and rectangular enclosure so that appropriate correlation equations for the Nusselt number are selected. However, since the outside surface temperature of the enclosure is unknown, the Rayleigh number can not be determined. The problem must be solved using an iterative procedure.
Problem 8.48
Heat is transferred from the inside to the outside.
Adding i an air enclosure reduces the rate of heat transfer.
To estimate the savings in energy, heat transfer through the single and double pane windows must be determined.
The double pane window introduces an added glass conduction resistance and a cavity convection resistance.
the problem can be modeled as a vertical rectangular enclosure.
Newton’s law of cooling gives the heat transfer rate
The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.
Problem 8.49
Heat is transferred through the door from the inside to the outside.
Newton’s law of cooling gives the heat transfer rate.
The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.
The baffle divides the vertical cavity
Problem 8.50
Heat is transferred through the skylight from the inside to the outside.
Newton’s law of cooling gives the heat transfer rate.
The aspect ratio and Rayleigh number should be determined for the rectangular enclosure so that an appropriate correlation equation for the Nusselt number can be selected.
Problem 8.51
Power requirement is equal to the heat transfer rate through the enclosure.
The problem can be modeled as a rectangular cavity at specified hot and cold surface temperatures.
The inclination angle varies from too0 .175o
Newton’s law of cooling gives the heat transfer rate.
The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.
Problem 8.52
The absorber plate is at a higher temperature than the ambient air. Thus heat is lost through the rectangular cavity to the atmosphere
The problem can be modeled as an inclined rectangular cavity at specified hot and cold surface temperatures.
Newton’s law of cooling gives the heat transfer rate.
The aspect ratio and critical inclination angle should be computed to determine the applicable correlation equation for the Nusselt number.
Problem 8.53
Heat is transferred through the annular space from the outer cylinder to the inner.
Newton’s law of cooling gives the heat transfer rate.
The Rayleigh number should be determined for the enclosure formed by the concentric cylinders so that an appropriate correlation equation can be selected.
The cylinders are horizontally oriented.
Problem 9.1
Definitions of Knudsen number, Reynolds number, and Mach number are needed.
Fluid velocity appears in the definition of Reynolds number and Mach number.
Problem 9.2
The definition of friction factor shows that it depends on pressure drop, diameter, length and mean velocity.
Mean velocity is determined from flow rate measurements and channel flow area.
Problem 9.3
The determination of the Nusselt number requires the determination of the temperature distribution.
Temperature field depends on the velocity field.
The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2.
The solution to the energy equation gives the temperature distribution.
Problem 9.4
Temperature distribution depends on the velocity field.
The velocity field for Couette flow with a moving upper plate is give in Section 9.6.2.
The solution to the energy equation gives the temperature distribution.
Two temperature boundary conditions must be specified.
Temperature distribution and Fourier’s law give surface heat flux.
Problem 9.5
To determine mass flow rate it is necessary to determine the velocity distribution.
Velocity slip takes place at both boundaries of the flow channel.
Because plates move in opposite directions, the fluid moves in both directions. This makes it possible for the net flow rate to be zero.
Problem 9.6
Model channel flow as Couette flow between parallel plates.
Apply Fourier’s law at the housing surface to determine heat leaving the channel.
Apply the Navier-Stokes equations and formulate the velocity slip boundary conditions. Follow the analysis of Section 9.6.2 and Example 9.1.
Use the energy equation to determine the temperature distribution
Problem 9.7
To determine the temperature of the lower plate, fluid temperature distribution must be known.
Temperature distribution depends on the velocity field.
The velocity field for Couette flow with a moving upper plate is given in Section 9.6.2.
The solution to the energy equation gives the temperature distribution.
Two temperature boundary conditions must be specified.
Problem 9.8
To use the proposed approach, the solution to the axial velocity distribution must be know.
The velocity distribution for Poiseuille flow between parallel plates is given by equation (9.30) of Section 9.6.3.
Problem 9.9
This is a pressure driven microchannel Poiseuille flow between parallel plates.
The solution to mass flow rate through microchannels is given in Section 9.6.3.
Channel height affects the Knudsen number.
Problem 9.10
This is a pressure driven microchannel Poiseuille flow.
Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates.
Channel surface is heated with uniform flux.
The solution to mass flow rate, temperature distribution, and Nusselt number for fully developed Poiseuille channel flow with uniform surface flux is presented in Section 9.6.3.
Problem 9.11
The problem can be modeled as pressure driven Poiseuille flow between two parallel plates with uniform surface flux.
Assuming fully developed velocity and temperature, the analysis of Section 9.6.3 gives the mass flow rate and Nusselt number.
The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the channel due to pressure variation, it follows that pressure distribution along the channel must be determined.
Problem 9.12
This is a pressure driven microchannel Poiseuille flow.
Since channel height is much smaller than channel width, the rectangular channel can be modeled as Poiseuille flow between parallel plates.
Channel surface is maintained at uniform temperature.
The solution to velocity, pressure, and mass flow rate is presented in Section 9.63.
The solution to the temperature distribution and Nusselt number for fully developed Poiseuille channel flow with uniform surface temperature is presented in Section 9.6.4.
Surface heat flux is determined using Newton’s law.
Problem 9.13
Cylindrical coordinates should be used to solve this problem.
The axial component of the Navier-Stokes equations must be solved to determine the
axial velocity .zv
The procedure and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this case.
Problem 9.14
This a pressure driven Poiseuille flow through a microtube.
The procedure for determining the radial velocity component and axial pressure distribution is identical to that for slip Poiseuille flow between parallel plates.
The solution to the axial velocity is given by equation (9.74).
Continuity equation gives the radial velocity component.
Axial pressure is determined by setting the radial velocity component equal to zero at the surface.
Cylindrical coordinates should be used to solve this problem.
Problem 9.15
Cylindrical coordinates should be used to solve this problem.
Axial velocity component is needed to determine mass flow rate.
Equation (9.74) gives the axial velocity for this case.
Since axial velocity vary with radial distance, mass flow rate requires integration of the axial velocity over the flow cross section area.
The procedure and simplifying assumptions used in the solution of the corresponding Couette flow between parallel plates, detailed in Section 9.6.2, can be applied to this problem.
Problem 9.16
To use the proposed approach, the solution to the axial velocity distribution must be known.
The velocity distribution for Poiseuille flow through tubes is given by equation (9.74) of Section 9.6.5.
Cylindrical coordinates should be used to solve this problem.
Problem 9.17
This is a pressure driven Poiseuille flow through a microtube.
Tube surface is heated with uniform flux.
The solution to mass flow rate, temperature distribution and Nusselt number for fully developed Poiseuille flow through a tube with uniform surface flux is presented in Section 9.6.5.
Problem 9.18
The problem is a pressure driven Poiseuille flow through microtube with uniform surface heat flux.
The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the tube due to pressure variation, it follows that pressure distribution along the tube must be determined.
Assuming fully developed velocity and temperature, the analysis of Section 9.6.5 gives axial pressure and Nusselt number variation along tube
The definition of Nusselt number gives the heat transfer coefficient.
Problem 9.19
This is a pressure driven Poiseuille flow through a tube at uniform surface temperature.
Since the flow field is assumed independent of temperature, it follows that the velocity, mass flow rate and pressure distribution for tubes at uniform surface flux, presented in Section 9.6.6, are applicable to tubes at uniform surface temperature..
The heat transfer coefficient can be determined if the Nusselt number is known.
The variation of the Nusselt number with Knudsen number for air is shown in Fig. 9.16.
The determination of Knudsen number at the inlet and outlet and Fig. 9.16 establish the Nusselt number at these locations.
The use of Fig. 9.16 requires the determination of the Peclet number.
Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.
Problem 9.20
The problem is a pressure driven Poiseuille flow through microtube at uniform surface temperature.
The Nusselt number depends on the Knudsen number, Kn. Since Kn varies along the tube due to pressure variation, it follows that pressure distribution must be determined.
Assuming fully developed velocity and temperature, the analysis of Section 9.6.6 gives axial pressure and Nusselt number variation along the tube.
The definition of Nusselt number gives the heat transfer coefficient.
The variation of the Nusselt number with Knudsen number and Peclet number for air is shown in Fig. 9.16.
Mean temperature variation along the tube is given by equation (6.13). Application of this equation requires the determination of the average heat transfer coefficient.