module 6: worked-out problems
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MODULE 6: Worked-out Problems
Problem 1:
For laminar free convection from a heated vertical surface, the localconvection coefficient may be expressed as h x=Cx
-1/4 , where h x is thecoefficient at a distance x from the leading edge of the surface and thequantity C, which depends on the fluid properties, is independent of x.Obtain an expression for the ratio x x hh /
, where xh
is the average
coefficient between the leading edge (x=0) and the x location. Sketch thevariation of h
xand xh
with x.
Known : Variation of local convection coefficient with x for freeconvection from a heated vertical plate.
Find: ratio of average to local convection coefficient.
Schematic:
Boundary layer,h x=C x
-1/4 where Cis a constant
T s
x
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Analysis: It follows that average coefficient from 0 to x is given by
x1/4
x
x
0
1/4x
0xx
h34
Cx34
x3/4xC
34
h
dxxxC
dxhx1
h
===
==
Hence34
hh
x
x =
The variation with distance of the local and average convection coefficient isshown in the sketch.
x x hh
x x hh 3
4=
4/1= Cxh x
Comments: note that
hh x / x =4/3, independent of x. hence the average
coefficients for an entire plate of length L is Lh
=4/3L, where h L is the localcoefficient at x=L. note also that the average exceeds the local. Why?
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Problem 2:Experiments to determine the local convection heat transfer coefficientfor uniform flow normal to heated circular disk have yielded a radial
Nusselt number distribution of the form
+==
n
oD r
ra1
k h(r)D
Nu
Where n and a are positive. The Nusselt number at the stagnation
point is correlated in terms of the Reynolds number (Re D=VD/ ) andPrandtl
0.361/2Do Pr0.814Rek
0)Dh(rNu ===
Obtain an expression for the average Nusselt number, k Dh Nu D /
= ,corresponding to heat transfer from an isothermal disk. Typically
boundary layer development from a stagnation point yields a decaying
convection coefficient with increasing distance from the stagnation point. Provide a plausible for why the opposite trend is observed for the disk.
Known: Radial distribution of local convection coefficient for flownormal to a circular disk.
Find: Expression for average Nusselt number.
Schematic:
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Assumptions: Constant properties.
Analysis: The average convection coefficient is
o
o
s
r
0no
2n2
3o
no0
r
02
o
As
s
2)r(nar
2r
rkNuo
h
]22 r)a(r/r[1NuDk
r1
h
hdAA1
h
++=
+=
=
+
Where Nu o is the Nusselt number at the stagnation point (r=0).hence,
( )
0.361/2D
o
r
o
2n2
o
Pr2)]0.841Re2a/(n[1NuD
2)]2a/(n[1NuNuD
ror
2)(na
2r/ro
2Nuk Dh
NuDo
++=
++=
++==
+
Comments: The increase in h(r) with r may be explained in terms of the sharp turn, which the boundary layer flow must take around theedge of the disk. The boundary layer accelerates and its thicknessdecreases as it makes the turn, causing the local convection coefficientto increase.
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Problem 3:
In a flow over a surface, velocity and temperature profiles are of the forms
u(y)=Ay+By 2-Cy 3 and T(y)=D+Ey+Fy 2-Gy 3
Where the coefficients A through G are constants. Obtain expressionsfor friction coefficients C f and the convection coefficient h in terms of u, T and appropriate profile coefficients and fluid properties.
Known: form of the velocity and temperature profiles for flow over asurface.
Find: expressions for the friction and convection coefficients.
Schematic:
u,TT(y)=D+Ey+Fy 2-Gy 3
Ts=T(0)=D
U(y)=Ay-By2
-Cy3
y y
Analysis: The shear stress at the wall is
.]32[ 02
0
ACy By A yu
y
y
s =+== =
=
Hence, the friction coefficient has the form,
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Problem 4:
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as
=
v
yuPrexp1
TTTT
s
s
Where y is the distance normal to the surface and the Prandtl number,Pr=c p/k=0.7, is a dimensionless fluid property. If T =400K, T s=300K,and u /v=5000m
-1, what is the surface heat flux?
Known: Boundary layer temperature distribution
Find: Surface heat flux.
Schematic:
Pr exp(1)(
yu
T T
T yT
s
s
=
Properties:Air ( k T S 300= ): k = 0.0263 W/m.k
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Analysis:Applying the Fouriers law at y=0, the heat flux is
0y
s
0y
"s yuPrexp uPr)Tk(TyTk q =
=
==
u)PrTk(Tq s
"s
=
1/m0.7x5000.K(100K)0.02063w/mq "s =
Comments: (1) Negligible flux implies convection heat transitedsurface (2) Note use of k at sT to evaluate from
"sq Fouriers law.
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Problem 5:
Consider a lightly loaded journal bearing using oil having the constant properties =10 -2 kg/s-m and k=0.15W/m. K. if the journal and the
bearing are each mentioned at a temperature of 400C, what is themaximum temperature in the oil when the journal is rotating at 10m/s?
Known: Oil properties, journal and bearing temperature, and journalspeed for lightly loaded journal bearing.
Find: Maximum oil temperature.
Schematic:
Assumptions: (1) steady-state conditions, (2) Incompressible fluidwith constant properties, (3) Clearances is much less than journalradius and flow is Couette.
Analysis: The temperature distribution corresponds to the resultobtained in the text example on Couette flow.
+=2
2o L
yLy
U2k
TT(y)
The position of maximum temperature is obtained from
== 22
L2y
L1
U2k
0dydT
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y=L/2.
Or,The temperature is a maximum at this point since hence 0.T/dyd 22 <
22
2 U8k
To41
21
U2k
ToT(l/2)axTm, +=
+==
C40.83T0.15W/m.K 8
/s)kg/s.m(10m10C40T
max
22
max
=
+=
Comments: Note that T max increases with increasing and U, decreaseswith increasing k, and is independent of L.
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Problem 6:
Consider two large (infinite) parallel plates, 5mm apart. One plate isstationary, while the other plate is moving at a speed of 200m/s. both plates
are maintained at 27 C. Consider two cases, one for which the plates areseparated by water and the other for which the plates are separated by air.
For each of the two fluids, which is the force per unit surface arearequired to maintain the above condition? What is the correspondingrequirement?What is the viscous dissipation associated with each of the two fluids?What is the maximum temperature in each of the two fluids?
Known: conditions associated with the Couette flow of air or water.
Find: (a) Force and power requirements per unit surface area, (2) viscousdissipation,(3) maximum fluid temperature.
Schematic:
Assumptions: (1) Fully developed Couette flow, (2) Incompressible fluidwith constant properties.
Properties: Air (300K); =184.6*10-7 N.s/m2, k=26.3*10-3W/m.K; water (300K): =855*106N.s/m2,k=0.613W/m.K
Analysis: (a) the force per unit area is associated with the shear stress.Hence, with the linear velocity profile for Couette flow = (du/dy) = (U/L).
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34.2N/m0.005m200m/s
N.s/m10855 :Water
0.738N/m0.005m200m/s
N.s/m10184.6 :Air
226-water
227-air
==
==
With the required power given by P/A= .U
6840W/m200m/s)(34.2N/m(P/A) :Water
147.6W/m200m/s)(0.738N/m(P/A) :Air22
water
22air
==
==
(b) The viscous dissipation is .hence(U/L)(du/dy) 22 ==
W/m101.370.005m200m/s
N.s/m10855( :Water
W/m102.950.005m200m/s
N.s/m10184.6 ( :Air
362
26-water
342
27-air
=
=
=
=
The location of the maximum temperature corresponds to y max=L/2. HenceTmax=To+U
2/8k and
C34.0K 80.613W/m.
(200m/s)2N.s/m10855C27(T :Water
C30.5K 0.0263W/m.8
(200m/s)2_ N.s/m10184.6C27)(T :Air
26-
watermax)
2-7
airmax
=+=
=
+=
Comments: (1) the viscous dissipation associated with the entire fluid layer,),( LA must equal the power, P.
(2) Although ater w)( >> air )( , k water >>k air . Hence, T max,water Tmax,air .
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Problem 7:
A flat plate that is 0.2m by 0.2 m on a side is orientated parallel to anatmospheric air stream having a velocity of 40m/s. the air is at atemperature of T =20 C, while the plate is maintained at T s=120 C. Thesir flows over the top and bottom surfaces of the plate, and measurementof the drag force reveals a value of 0.075N. What is the rate of heattransfer from both sides of the plate to the air?
Known: Variation of h x with x for flow over a flat plate.
Find: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x=L.
Schematic:
Analysis: The expressions for the local and average Nusselt number are
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2.NuLNuL
andk
2CLk
(L)2CLNuL
Hence
2CLLL
2Cdxx
LC
dxhL1
h
where
k Lh
Nu
k CL
k )L(CL
k Lh
Nu
-
1/21/2-
1/21/2L
0
1/2L
0xL
LL
1/21/2L
L
=
==
====
=
===
Comments: note the manner in which-
NuL is defined in terms of Lh
. Alsonote that
L
0xL dxNuL
1Nu
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Problem 8:
For flow over a flat plate of length L, the local heat transfer coefficient hx
is known to vary as x -1/2 , where x is the distance from the leading edge of the plate. What is the ratio of the average Nusslet number for the entire
plate to the local Nusslet number at x=L (Nu L)?
Known: Drag force and air flow conditions associated with a flat plate.
Find: Rate of heat transfer from the plate.
Schematic:
Assumptions: (1) Chilton-Colburn analogy is applicable.
Properties: Air(70 C,1atm): =1.018kg/m 3, c p=1009J/kg.K, pr=0.70, =20.22*10 -6m2/s.
Analysis: the rate of heat transfer from the plate is
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)T(T(L)h2q s2
=
Whereh may be obtained from the Chilton-Colburn analogy,
240Wq
C20)(120.K)(0.3m)2(30W/mq
is rateheatThe
.K 30W/mh
2/3.70)9J/kg.K)(0)40m/s(100(1.018kg/m105.76h
Prcu2
Ch
hence,
105.76/2(40m/s)1.018kg/m
(0.2m)(0.075N/2)21
/2u
21
2
C
Prcu
htPrS
2
C j
22
2-
34-
2/3p
f -
423
2
2sf
2/3
p
2/3f H
==
=
=
=
===
===
Comments: Although the flow is laminar over the entire surface( )100.4/1022.20/2.0/40/Re 526 === smmsm Lu L , the pressure gradient is zero
and the Chilton-Colburn analogy is applicable to average, as well as local,surface conditions. Note that the only contribution to the drag force is made
by the surface shear stress.
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Problem 9:
Consider atmospheric air at 25 C in parallel flow at 5m/s over both surfaces of 1-m-long flat plate maintained at 75 C. Determine the boundary layer thickness, thesurface shear stress, and the heat flux at the trailing edge. Determine the drag force onthe plate and the total heat transfer from the plate, each per unit width of the plate.
Known: Temperature, pressure, and velocity of atmospheric air in parallel flow over aPlate of prescribed length and temperature.
Find: (a) Boundary layer thickness, surface shear stress and heat flux at trailing edges,(b) drag force and total heat transfer flux per unit width of plate.
Schematic:
Fluid
=5m/sT=25C
P=1 atm L=1m
x)
Ts=75C
Assumptions: (1) Critical Reynolds number is 5*10 5, (2) flow over top and bottomsurfaces
Properties : (T f =323K, 1atm) Air: =1.085kg/m3
, =18.2*10-6m 2/s,k=0.028W/m.K,pr=0.707
Analysis: (a) calculate the Reynolds number to know the nature of flow
526L 102.75/sm1018.2
m1m/s5
LuRe =
==
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Hence the flow is laminar, and at x=L
22Ls,
1/2523
1/2L
2Ls,
1/251/2L
0.0172N/m.s0.0172kg/m
)1050.664/(2.7(5m/s)mkg
21.085
/2)0.664Re(
9.5mm)101m/(2.7555LRe
==
==
===
Using the correct correlation,
22s
2L
1/31/21/31/2L
LL
C)217W/m25C.K(754.34W/m)ThL(Ts(L)q".K 4.34W/m8W/m.K)/1m155.1(0.02h
hence,
155.1(0.707)105)0.332(2.75Pr0.332Rek
hLNu
====
====
(b) The drag force per unit area plate width is Ls L D
= 2' where the factor of two isincluded to account for both sides of the plate. Hence with
868W/mC25).K(75/m2(1m)8.68W)T(T2Lhq
.K 8.68W/m2hLhwithalso
0.0686N/m3N/m2(1m)0.034D
0.0343N/m
isdrag The
)101.328(2.75/2)(5m/s)(1.085kg/m/2)1.328Re(
2s
-
L"
2-
L
2'
2Ls
1/25231/2L
2Ls
===
==
==
=
==
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Problem 10:Engine oil at 100 C and a velocity of 0.1m/s flows over both surfaces of a 1-m-longflat plate maintained at 20 C. Determine
a. The velocity and thermal boundary thickness at the trailing edge. b. The local heat flux and surface shear stress at the trailing edge.c. The total drag force and heat transfer per unit area width of the plate.
Known: Temperature and velocity of engine oil Temperature and length of flat plate.
Find: (a) velocity and thermal boundary thickness at the trailing edge, (b)Heat flux and surface shear stress at the trailing edge, (c) total drag force and heattransfer per unit plate width.
Schematic:
Assumptions: engine oil (T f =33K): =864kg/m 3, =86.1*10 6m2/s, k=0.140W/m /K,Pr=1081.
Analysis: (a) calculate the Reynolds number to know the nature of flow
1161/sm10*86.1
1m0.1m/s
LuReL 26- =
==
Hence the flow is laminar at x=L, and
0.0143m)0.147(1081Pr
0.147m)5(1m)(11615LRe1/31/3
t
1/21/2L
===
===
(b) The local convection coefficient and heat flux at x=L are
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22
s
21/31/31/2LL
1300W/mC100).K(2016.25W/m)ThL(Tq"
.K 16.25W/m)(1081)0.332(11611m
0.140W/m.K Pr0.3325LRe
Lk
h
===
===
Also the local shear stress is
22Ls
1/223
1/2L
2Ls
0.0842N/m.s0.0842kg/m
)0.664(1161(0.1m/s)2
864kg/m/2)0.664Re(
==
==
(c) With the drag force per unit width given by Ls L D
= 2' where the factor of 2 isincluded to account for both sides of the plate, is follows that
5200W/mC100).K(20/m2(1m)32.5W)T(Th2Lq
thatfollowsalsoit.K,32.5W/m2hhwith
0.673N/m)1.328(1161(0.1m/s)/m2(1m)864kg/2)1.328Re(
2sL
_
2LL
_
1/2231/2L
2Ls
===
==
===
Comments: Note effect of Pr on ( /t).
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Problem 11:
Consider water at 27 C in parallel flow over an isothermal, 1-m-long, flat plate with avelocity of 2m/s. Plot the variation of the local heat transfer coefficient with distancealong the plate. What is the value of the average coefficient?
Known: velocity and temperature of air in parallel flow over a flat plate of prescribedlength.
Find: (a) variation of local convection coefficient with distance along the plate, (b)Average convection coefficient.
Schematic:
Assumptions: (1) Critical Reynolds number is 5*10 5.
Properties: Water (300K): =997kg/m 3,=855*10 -6 N.s/m 2, =/=0.858*10 -6m2/s,k=0.613W/m. K, Pr=5.83
Analysis: (a) With
526L 102.33/sm100.858
1m2m/s
LuRe =
==
Boundary layer conditions are mixed and
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4055 4240 4491 4871 5514 .K)(W/mh
1.0 0.8 0.6 0.4 0.0215 x(m)
xPr
u0.0296k Pr0.0296Re
xk
hx0.215m,xfor
1206 1768 hx(W/m2.K)
0.215 0.1 x(m)
xPr
u0.332k Pr0.332Re
xk
hx 0.215m,xfor
0.215m)10/2.33101m(5)/ReL(Rex
x
0.21/34/5
1/34/5x
1/21/31/2
1/31/2x
65Lcx,c
2
==>
==
===
The Spatial variation of the local convection coefficient is shown above
(b) The average coefficient is
.K 4106W/mh
871(5.83))1033[0.0379(2.1m
0.613W/m.K 871)Pr(0.037Re
Lk
h
2L
_
1/34/561/34/5LL
_
=
==
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Problem 12:
A circular cylinder of 25-mm diameter is initially at 150C and is quenched byimmersion in a 80C oil bath, which moves at a velocity of 2m.s in cross flow over thecylinder. What is the initial rate of heat loss unit length of the cylinder?
Known: Diameter and initial temperature of a circular cylinder submerged in an oil bathf prescribed temperature and velocity.
Find: initial rate of heat loss unit per length.
Schematic:
Assumptions: (1) Steady-state conditions, (2) uniform surface temperature.
Properties: Engine oil (T =353K): =38.1*10 -6m2/s, k=0.138W/m. K, Pr =501;(T s=423K): Pr s=98.
Analysis: The initial heat loss per unit length is
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( ) 8.8W/mC80)0.025m(150.K 1600W/mq'
.K 1600W.mh
98501
(501)0.26(1312)0.025m
0.138W/m.K )1/4/Pr(PrPrCRe
Dk
h
hence.7.4tablefrom0.6mand 0.26CFind
1312/sm1038.1
m)2m/s(0.025
VDRe
whenrelation.Zhukauskasthefromcomputedbemay where
)TD(Thq
2
2 _
1/40.370.6
snm
D
_
26D
s
_
==
=
==
==
=
==
=
=
_
h
Comments: Evaluating properties at the film temperature, T f =388K( =14.0*10 -6m2/s,k=0.135W/m. K, Pr=196), find Re D=3517.
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Problem 13:
An uninsulated steam pipe is used to transport high-temperature steam from one building to one another. The pipe is 0.5-m diameter, has a surface temperature of 150C, and is exposed to ambient air at -10C.the air moves in cross flow over the pipe
with a velocity of 5m.s.What is the heat loss per unit length of pipe?
Known: Diameter and surface temperature of uninsulated steam pipe. Velocity andtemperature of air in cross flow.
Find: Heat loss per unit length.
Schematic:
V=5m/sT=263K
Air D=0.5mm
Ts=150 C
Assumptions: (1) steady-state conditions, (2) uniform surface temperature
Properties: Air (T=263K, 1atm): =12.6*10 -6m2/s, k=0.0233W/m. K, Pr=0.72;(Ts=423K,1atm); Prs=0.649.
Analysis: the heat loss per unit length is
)TD(Thq s _
=
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.K 1600W.mh
98501
(501)0.26(1312)0.025m
0.138W/m.K )1/4/Pr(PrPrCRe
Dk
h
hence.7.4tablefrom0.6mand 0.26CFind
1312/sm1038.1
m)2m/s(0.025
VDRe
whenrelation.Zhukauskasthefromcomputedbemay where
2 _
1/40.370.6
snm
D
_
26D
=
==
==
=
==
=
_
h
Hence the heat rate is
( ) 4100W/mC10)(-0.5m(150.K 16.3W/mq' 2 ==
Comments: Note that q Dm, in which case the heat loss increases significantly withincreasing D.
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Problem 14:
Atmospheric air at 25 C and velocity of 0.5m/s flows over a 50-W incandescent bulbwhose surface temperature is at 140 C. The bulb may be approximated as a sphere of 50-mm diameter. What is the rate of heat loss by convection to the air?
Known: Conditions associated with airflow over a spherical light bulb of prescribeddiameter and surface temperature.
Find: Heat loss by convection.
Schematic:
Assumptions: (1) steady-state conditions, (2) uniform surface temperature.
Properties: Air (T f =25 C, 1atm): =15.71*10 -6m2/s, k=0.0261W/m. /K.Pr=.0.71, =183.6*10 -7 N.s/m 2; Air (T s=140 C, 1atm): =235.5*10 -7 N.s/m 2
Analysis:)TD(Thq s
_
=
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10.3WC25)(140 (0.05m).K m
W11.4q
israteheattheand
.K 11.4W/mh
235.5183.6
](0.71)0.06(1591)[0.4(1591)20.05m
K 0.0261W/m.h
hence
1591/sm1015.71
0.05m0.5m/s
VDRe
Where
])(/)Pr0.06Re(0.4Re[2Dk
h
whenrelation.Whitakerthefromcomputedbemay where
22
2 _
1/40.42/31/2
_
26D
1/4s
0.42/3D
1/2D
_
==
=
++=
=
==
++=
_
h
Comments: (1) The low value of _
h suggests that heat transfer by free convection may
be significant and hence that the total loss by convection exceeds 10.3W(2) The surface of the bulb also dissipates heat to the surroundings by radiation. Further,in an actual light bulb, there is also heat loss by conduction through the socket.
(3) The Correlation has been used its range of application ( /s )
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L2D
uf p
2m=
e=2.6*10 -4 m for clean cast iron; hence e/D=1.04*10 -3. With
5326
mD 102.92/1000kg/mN.s/m10855
0.25m1m/s
DuRe =
==
0.42barN/m104.2P
.mkg/s104.21000m2(0.25m)
(1m/s)(1000kg/m0.021P
,hence0.021.f that8.3figfromfind
24
2423
==
==
)
The pump power requirement is
2.06kW1m/s/4)m0.25(N/m104.2P
/4)up( DVp.P
2224
m2
.
==
==
(b) Increasing by 25% it follows that =3.25*10-4m and e/D=0.0013. WithReD unchanged, from fig 8.3, it follows that f 0.0225. Hence
2.21kW)P/f (f P bar0.45)/f (f P 1122122 ==== 1 p
Comments: (1) Note that L/D=4000>>(x fd,h/D) 10 for turbulent flow andthe assumption of fully developed conditions is justified.(2) Surface fouling results in increased surface and increases operating costs
through increasing pump power requirements.
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Problem 16:
Consider flow in a circular tube. Within the test section length (between1 and 2) a constant heat flux q s is maintained.
(a) For the two cases identified, sketch, qualitatively, the surfacetemperature T s(x) and the fluid mean temperature Tm(x) as a functionof distance along the test section x. in case A flow is hydrodynamically and thermally fully developed. In case B flow is notdeveloped.
(b) Assuming that the surface flux q s and the inlet mean temperatureTm,1 are identical for both cases, will the exit mean temperature T m,2for case A be greater than, equal to , or less than T m,2 for case B?Briefly explain why?
Known: internal flow with constant surface heat flux, "sq .
Find: (a) Qualitative temperature distributions (x), under developing andfully developed flow, (b) exit mean temperature for both situations.
Schematic:
Flow
1 2
qs=constant
Assumptions: (a) Steady-state conditions, (b) constant properties, (c)incompressible flow.
Analysis: Based upon the analysis, the constant surface heat flux conditions,
constantdx
dT m =
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Hence, regardless of whether the hydrodynamic or thermal boundary layer isfully developed, is follows that
Tm(x) is linear
T m,2 will be the same for all flow conditions.
The surface heat flux can be written as
(x)]Th[Tq ms"s =
Under fully developed flow and thermal conditions, h=h fd is a constant.When flow is developing h> h fd . Hence, the temperature distributions appear as below.
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Problem 17:
A thick-walled, stainless steel (AISI 316) pipe of inside and outsidediameter D i=20mm and D o=40m is heated electrically to provide a
uniform heat generation rate of 36.
/10 mW q = . The outer surface of the pipe is insulated while water flows through the pipe at a rate of
skgm /1.0.
= (a) If the water inlet temperature is T m,1=20 C and the desired outlet
temperature is T m,o=40 C, what is the required pipe length
(b) What are the location and value of the maximum pipe temperature?
Known: Inner and outer diameter of a steel pipe insulated on the outside and experiencinguniform heat generation. Flow rate and inlet temperature of water flowing through the
pipe.
Find: (a) pipe length required to achieve desired outlet temperature, (b)location and value of maximum pipe temperature.
Schematic:
Assumptions: (1) steady-state conditions, (2) constant properties, (3)negligible kinetic energy, potential energy and flow work changes, (4) one-dimensional radial conduction in pipe wall, (5) outer surface is adiabatic.
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Properties: Stainless steel 316 (T 400K): k=15W/m.k; water ( 303K T.
m = );c p=4178J/kg. K, k=0.617W/m. K, =803*10 -6 N.s/m 2, Pr=5.45
Analysis: (a) performing an energy balance for a control volumeabout the inner tube, it follows that
8.87mL
](0.02)4m( /4)[(0.W/m10C)20178(J/kg.K (0.1kg/s)4
)LD( /4)(q
)T(TmcL
)LD( /4)(qq)T(Tcm
22362
i
2
0
.
.
im,om,p
2i
20
.
im,om,p
.
=
=
=
==
(b) The maximum wall temperature exists at the pipe exit (x=L) and theinsulated surface (r=r o). The radial temperature distribution in the wall is of the form
Tnr2k qr
-r4k q
C Cnr2k
qrr
4k q-
T)T(r:rr
2k
qrC
rC
r2k
q0
drdT
;rr
;conditionsboundarythegconsiderin
CnrCr4k q
T(r)
si
.2
o2i
.
22i
.2
o2i
.
si1
.2
o1
o
1o
.
rro
212
.
o
+=++===
=+== =
++=
=
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The temperature distribution and the maximum wall temperature (r=r o) are
)Th(T4D
)D(DqLD
)LDq( ( /4)q
fromfollowsitexit,atwalltheof temperturesurafaceinnertheTs,where
Trr
n2k
qr )r-(r
4k q
-)T(rT
Trr
n2k
qr )r-(r
4k q
T(r)
om,s
i
2i
2o
.
i
.2i
2o"
s
si
o
.2
o2i
2o
.
omaxw,
si
.2
o2i
2
.
=
=
=
++==
++=
Where h is the local convection coefficient at the exit. With
7928N.s/m103 (0.02m)80
0.1kg/s4D
m4Re 26
i
.
D ===
The flow is turbulent and, with (L/D i)=(8.87m/0.02m)=444>>(x fd/D)10,it isalso fully developed. Hence, from the Dittus-Boelter correlation,
.K 1840W/m5.45)0.023(79280.02m
0.617W/m.K )pr(0.023Re
Dk
h 20.44/50.44/5Di
===
Hence the inner surface temperature of the wall at the exit is
C52.4C48.20.010.02
n215W/m.K
(0.02)W/m10](0.01)[(0.02)
415W/m.K W/m10
T
and
C48.2C40.K(0.02m)180W/m4
](0.02m)[(0.04m)W/m10T
4D)D(Dq
23622
36
maxw,
2
2236
om,i
2i
2o
.
s
=++=
=+
=+
=
T
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Comments: The physical situation corresponds to a uniform surface heatflux, and T m increases linearly with x. in the fully developed region, Ts alsoincreases linearly with x.
T
xfd
Ts(x)
Tm(x)
x
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Problem 18:
The surface of a 50-mm diameter, thin walled tube is maintained thinwalled tube is maintained at 100 C. In one case air is cross flow very the
tube with a temperature of 25 C and a velocity of 30m/s. In another caseair is in fully developed flow through the tube with a temperature of 25 Cand a mean velocity of 30m/s. compare the heat flux from the tube to theair for the two cases.
Known: surface temperature and diameter of a tube. Velocity andtemperature of air in cross flow. Velocity and temperature of air in fullydeveloped internal flow.
Find: convection heat flux associated with the external and internal flows.
Schematic:
Air
Um=30m/sTm=25C
V=30m/sT =25 C
Ts=100 C
D=0.05m
Air
Assumptions: (1) steady-state conditions, (2) uniform cylinder surfacetemperature, (3) fully developed internal flow
Properties: Air (298K): =15.71*10-6m 2/s, k=0.0261W/m.K, Pr=0.71
Analysis: for the external and internal flow
426 1055.9/1071.15
05./30Re =
=== smmsm DuVD m D
From the Zhukauskas relation for the external flow, with C=0.26 and m=0.6
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223)1()71.0()1055.9(26.0)Pr Pr(Pr/Re 4/137.06.044/1 _
=== sm D D C u N
Hence, the convection coefficient and heat rate are
232"
2 _ _
/1073.8)25100(./4.116)(
./4.11622305.0
./0261.0
mW C k mW T T hq
K mW m
K mW NuD
Dk
h
s ===
===
Using the Dittus-Boelter correlation, for the internal flow, which isTurbulent,
232
_ _
4.05/4244.05/4 _
/1058.7)25100(./101)(
.2/10119305.0
./0261.0
193)71.0()1055.9(023.0Pr Re023.0
mW C K mW T T h
K mW K mW
Nu Dk
h
NU
ms
D
D D
===
===
===
"q
isfluxheattheand
Comments: Convection effects associated with the two flow conditions arecomparable.
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Problem 19:
Cooling water flows through he 25.4 mm diameter thin walled tubes of astream condenser at 1m/s, and a surface temperature of 350K is maintained
by the condensing steam. If the water inlet temperature is 290 K and thetubes are 5 m long, what is the water outlet temperature? Water propertiesmay be evaluated at an assumed average temperature of 300K
Known: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature.
Find: Water outlet temperature.
Schematic:
L=5m
D=0.0254m
Tm,i=290KUm=1m/s
Ts=350K
Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligiblekinetic energy, potential energy and flow work changes.
Properties: Water (300K): =997kg/m 3, c p=4179J/kg.K, =855*10-
6kg/s.m,k=0.613W/m.K, Pr=5.83
Analysis:
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29,618kg/s.m10855
54m(1m/s)0.02997kg/m
DuRe
]h)cm( DL)exp[T(TTT
6
3m
D
_
p
.
im,ssom,
=
==
=
The flow is turbulent. Since L/D=197, it is reasonable to assume fullydeveloped flow throughout the tube. Hence
323K K)(4179J/kg.(0.505kg/s
.K)5m(4248W/m (0.0254m)(60K)exp350K T
0.505kg/s4m)m/s)(0.025(1/m( /4)997k /4)(um
With
.K 4248W/m4m)/m.K/0.025176(0.613WNuD(k/D)h
2
om,
232m
.
2 _
=
===
===
Comments: The accuracy of the calculations may be improved slightly by
reevaluating properties at 306.5K T _
m = .
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Problem 20:
The air passage for cooling a gas turbine vane can be approximated as a tubeof 3-mm diameter and 75-mm length. If the operating temperature of the
vane is 650 C, calculate the outlet temperature of the air if it enters the tubeat 427 C and 0.18kg/h.
Known: gas turbine vane approximation as a tube of prescribed diameter andlength maintained at an known surface temperature. Air inlet temperatureand flow rate.
Find: outlet temperature of the air coolant.
Schematic:
hkgm /18.0.
=
Assumptions: (1) Steady-state conditions, (2) negligible Kinetic and
potential energy changes.
Properties: Air (assume )1,780 _
atmK T m = : cp=1094J/kg. K, k=0.0563 W/m.K, =363.7*10 -7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650 C=923K, 1atm):=404.2*10 -7 N.s/m 2.
Analysis: For constant wall temperatures heating,
=
.
_
,
, exp
pims
oms
mc
hPLT T
T T
Where P= D. for flow in circular passage,
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N.s/m103.7 (0.03m)36/3600s/h)0.18kg/h(14
Dm4
Re 7
.
D ==
The flow is laminar, and since L/D=75mm/3mm=25, the Sieder-Tate
correlation including combined entry length fields.
.K 87.5W/m10404.210363.7
250.706584
1.860.003m
K 0.0563W/m.h
L/DPrRe
1.86k Dh
Nu
2
0.14
7
71/3 _
0.14
s
1/3D
_ _
D
=
=
==
Hence, the air outlet temperature is
=
1094J/kg.K )kg/s(0.18/3600
.K 87.5W/m0.075m (0.003m)exp
C427)(650
T650 2om,
Tm,o=578 C
Comments: (1) based upon the calculations for T m,o=578 C, _
mT =775K whichis in good agreement with our assumption to evaluate the thermo physical
properties.
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Problem 21
A household oven door of 0.5-m height and 0.7-m width reaches an averagesurface temperature of 32 C during operation. Estimate the heat loss to the
room with ambient air at 22 C. If the door has an emissivity of 1.0 and thesurroundings are also at 22 C, comment on the heat loss by free convectionrelative to that by radiation.
Known: Oven door with average surface temperature of 32 C in a roomwith ambient temperature at 22 C.
Find: Heat loss to the room. Also, find effect on heat loss if emissivity of door is unity and the surroundings are at 22 C.
Schematic:
0 .7 m
L =
0 . 5
m
Assumptions: (1) Ambient air in quiescent, (2) surface radiation effectsare negligible.
Properties: Air (T f =300K, 1atm): =15.89*10 -6 m2/s, k=0.0263W/m. K,=22.5*10 -6 m2/s, Pr=0.707, =1/T f =3.33*10-3K -1
Analysis: the heat rate from the oven door surface by convection to the
ambient air is
)T(TAhq ss _
=
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Where _
h can be estimated from the free convection correlation for avertical plate,
[ ]
2
8/279/16
1/6L
_ _`
(0.492/Pr)1
0.387Ra0.825k
LhLNu
++==
The Rayleigh number,
42626
322s
L 101.142/sm1022.5/sm1015.89
m322)K0.52(1/300k))39.8m/s
)LT(TgRa =
=
=
Substituting numerical values into equation, find
[ ]
.K 3.34W/m63.50.5m
K 0.0263W/m.LNu
Lh
63.5(0.492/Pr)1
)1020.387(1.140.825
k Lh
LNu
2 _` _
2
8/279/16
1/88 _
_`
===
=
+
+==
k L
The heat rate using equation is
11.7W22)K (320.7)m.k(0.53.34W/mq22
==
Heat loss by radiation, assuming =1 is
21.4W]k 22)(27332)[(273.K W/m5.67100.7)m1(0.5q
)T(TAq4244282
rad
4sur
4ssrad
=++=
=
Note that heat loss by radiation is nearly double that by free convection.
Comments: (1) Note the characteristics length in the Rayleigh number is theheight of the vertical plate (door).
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Problem 22
An Aluminum alloy (2024) plate, heated to a uniform temperature of 227 C, is allowed to cool while vertically suspended in a room where the
ambient air and surroundings are at 27 C. The late is 0.3 m square with athickness of 15 mm and an emissivity of 0.25.a. Develop an expression for the time rate of change of the plate
temperature assuming the temperature to be uniform at anytime.
b. Determine the initial rate of cooling of the plate temperature is227 C.
c. Justify the uniform plate temperature assumption.
Known: Aluminum plate alloy (2024) at uniform temperature of 227 Csuspended in a room where the ambient air and the surroundings are at 27 C
Find: (1) expression for the time rate of change of the plate, (2) Initial
rate of cooling (K/s) when the plate temperature is 227 C. (3) justify the
uniform plate temperature assumption.
Schematic:
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Properties: Aluminium alloy 2024 (T=500K): =2270 kg/m 3,
k=186W/m.K, c=983 J/kg.K; Air (T f =400K, 1atm): =26.41*10-
6m2
/s,k=0.0338 W/m.K, =38.3*10-6
m2
/s, Pr=0.690.
Analysis :( a) from an energy balance on the plate considering free
convection and radiation exchange..
st out E E = .
)T (T)T(Th[c2
dtdT
ordtdT
A)T(T2A)T(T2Ah 4sur4ssL
.
cs4
sur4
ssss
_
L + ==
Where T s is the plate temperature assumed to be uniform at any time.
(b) To evaluate (dt/dx), estimate _
Lh . Find first the Rayleigh number
82626
322s
L 101.308/sm10.33/sm1026.41
0.3m27)K 27(1/400k)(29.8m/s
)LT(TgRa =
=
=
8
Substituting numerical values, find
[ ] [ ]5.55
) =
++=
++= 4/99/16
82
4/99/16
1/4L
_`
90)(0.492/0.61
1080.670(1.300.68
(0.492/Pr)1
0.670Ra0.68LNu
0.099K/s
)K 300.K)(500W/m100.25(5.6727)K .K(2276.25W/m983J/kg.K 0.015m2770Kg/m2
dtdT
.K 6.25W/mK/0.3m0.0338W/m.55.5k/L Nuh
442823
2 _`
L
_
=
+
=
===
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(c) The uniform temperature assumption is justified if the Biot number
criterion is satisfied. With == )/A.(A)(V/AL sssc
and _ _ _ _
1.0/, =+= k h Bihhh tot rad convtot . Using the linearized radiation coefficient
relation find,
3.86W/m3002)K 300)(5002)(500.K 8W/m100.25(5.67)T)(TT (Th 3422sur2
ssursrad
_
=++=++=
Hence Bi=(6.25+3.86) W/m 2.k (0.015m)/186W/m. K=8.15*10 -4. Since
Bi
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Problem 23
The ABC Evening News Report in news segment on hypothermia researchstudies at the University of Minnesota claimed that heat loss from the body
is 30 times faster in 1 0 C water than in air at the same temperature. Isthat a realistic statement?Known: Person, approximated as a cylinder, experiencing heat loss inwater or air at 10 C.
Find: Whether heat loss from body in water is 30 times that in air.
Assumptions: (1) Person can be approximated as a vertical cylinder of diameter D=0.3 m and length L=1.8m, at 25 C, (2) Loss is only from the
lateral surface.
Properties: Air ( _
T =(25+10) C/2=290K, 1atm); K=0.0293 W/m. K, ==19.91*10 -6m2/s, =28.4*10 -6 m2/s; Water (290K); k=0.598 W/m.K; =vf =1.081*10 -6m2/s, =k v f /cp =1.431*10 -7m2/s, f =174*10 -6K -1.
Analysis: in both water (wa) an air(a), the heat loss from the lateralsurface of the cylinder approximating the body is
)TDL(Thq s _
=
Where T s and T are the same for both situations. Hence,
_
a
wa
_
a
wa
h
hqq
=
Vertical cylinder in air:
92626
323
L 105.228/sm28.410/sm10*19.91
10)k(1.8m)5(1/290K)(29.8m/s
g TRa =
==
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.K 2.82W/mh 173.4)100.1(5.228CRa
k
LhNu
1/3,n and 0.1Cwith
2L
_ 1/39n
L
_ _
LL =====
==
Vertical cylinder in water.
112726
3162
L 109.643/sm101.431/sm101.08110)K(1.8m)(25K 101749.8m/s
Ra ==
.K 328W/mh 978.9)100.1(9.643CRak
hLNu
1/3,n and 0.1Cwith
2L
_ 1/311n
L
_ _
L =====
==
Hence, from this analysis we find
30of claimthewithpoorlycompareswhich 117.K 2.8W/m.K 328W/m
qq
2
2
a
wa ==
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Problem 24
In a study of heat losses from buildings, free convection heat transfer from
room air at 305 K to the inner surface of a 2.5-m-high wall at 295 K issimulated by performing laboratory experiments using water in a smaller testcell. In the experiments the water and the inner surface of the test cell aremaintained at 300 and 290K, respectively. To achieve similarity betweenconditions in the room and the test cell, what is the required test cell height?If the average Nusselt number for the wall may be correlated exclusively interms of the Rayleigh number, what is the ratio f the average convectioncoefficient for the room wall to the average coefficient for the test cell wall?
Known: Air temperature and wall temperature and height for a room. Water temperature and wall temperature for a simulation experiment.
Find: Required test cell height for similarity. Ratio and height convectioncoefficient for the two cases.
Schematic:
L a
= 2
. 5 m
L w
Assumptions: (1) Air and water are quiescent; (2) Flow conditionscorrespond to free convection boundary layer development on an isothermal
vertical plate, (3) constant properties.
Properties: Air (T f =300K, 1 atm); K=0.0293 W/m.K, ==15.9*10 -6m2/s,=22.5*10 -6 m2/s; =1/T f =3.33*10 -3K -1, k=0.0263W/m.K; water (T f =295K): =998kg/m
3, =959*10 -6 N.s/m 2, c p=4181 J/kg.K, =227.5*10 -6K -1, k=0.606 W/m.K; Hence =/ =9.61/s, =k / cp =1.45*10 -7m2/s.
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Analysis: Similarity requires that Ra L,a=Ra L,w where
mm L
o H L
L
hence
w
a
air w
a
w
45.0)179.0(5.2
10*228.010*33.3
10*5.22*9.1510*45.1*61.9
)()(
,
)L-T(TgRa
3
3
12
143/1
2
3s
L
==
=
=
=
3
w
a
a
w
w
_
_
a
_
wL,
_
aL,wL,aL,
10*7.810.606
0.02632.5
0.45k k
LL
h
h
.hence Nu Nuthatfollowsit,RaRaif
===
==
Comments: Similitude allows us to obtain valuable information for one
system by performing experiments for a smaller system and a differentfluid.
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Problem 25
A square plate of pure aluminum, 0.5 m on a side and 16 mm hick, isinitially at 300C and is suspended in a large chamber. The walls of thechamber are maintained at 27C, as is the enclosed air. If the surfaceemissivity of the plate temperature during the cooling process? Is itreasonable to assume a uniform plate temperature during the cooling
process?
Known: Initial temperature and dimensions of an aluminum plate.Conditions of the plate surroundings.
Find: (a) initial cooling rate, (2) validity of assuming negligible temperaturegradients in the plate during the cooling process.
Schematic:
L
Assumptions: (1) plate temperature is uniform; (2) chamber air is quiescent,(3) Plate surface is diffuse-gray, (4) Chamber surface is much larger than
that of plate, (5) Negligible heat transfer from edges.Properties: Aluminum (573k); k=232W/m.k, c p=1022J/kg.K, =2702 kg/m
3:Air (T f =436K, 1 atm): =30.72*10
-6m2/s,=44.7*10 -6m2/s,k=0.0363W/m.K,Pr=0.687, =0.00229K -1.
Analysis: (a) performing an energy balance on the plate,
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psur ss
pst sur ss
wcT T T T h Adt dT
dt dT Vc E T T T T h Aq
/)]4()([2/
]/[)]4()([2
4 _
.4
_
+=
==+=
82626
323s
L 105.58/sm017.44/sm10*30.72)K(0.5m)721(300-0.00229K 9.8m/s)L-T(Tg
Ra ===
K mW h
Ra
Lk
h L
./8.5
])687.0/492.0(1[)1058.5(670.0
68.05.0
0363.0]Pr)/492.0(1[
670.068.0
2 _
9/416/9
4/18
9/416/9
4/1 _
=
++=
++=
Hence the initial cooling rate is
{ }
sK dt dT
k kg J mmkgK K K mW C K mW
dt dT
/136.0
./1022)016.0(/2702])300()573[(./1067.525.0)27300(./8.52
3
444282
=
+=
(b) To check the Validity of neglecting temperature gradients across the plate thickness,
4-
22"
103.832W/m.K (0.008m)/2(11W/m2.K)Bi
./0.11273//)14131583()/( /)2/(
===+===
hence
k mW K mW T T qhwherek wh Bi itot eff eff
And the assumption is excellent.
Comments: (1) Longitudinal (x) temperature gradients are likely to me moresevere than those associated with the plate thickness due to the variation of hwith x. (2) Initially q conv q rad.
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Problem 26
Beer in cans 150 mm long and 60 mm in diameter is initially at 27 C and isto be cooled by placement in a refrigerator compartment at 4 C. In the
interest of maximizing the cooling rate, should the cans be laid horizontallyor vertically in the compartment? As a first approximation, neglect heattransfer from the ends.
Known: Dimensions and temperature of beer can in refrigerator compartment.
Find: orientation which maximum cooling rate.
Schematic:
Air,T=4CP=1 atmD=0.06m
Horizontal, (h) Vertical, (V)Ts=27 C
L =1 5 0 mm
Assumptions: (1) End effects are negligible, (2) Compartment air isquiescent, (3) constant properties.
Properties: Air (T f =288.5K, 1 atm): =14.87*10-6 m2/s, k=0.0254W/m.K,
=21.0*10 -6m2/s, Pr=0.71, =1/T f =3.47*10-3K -1.
Analysis: The ratio of cooling rates may be expressed as
h
v
s
s
h
v
qh
v
hh
T T T T
DL DL
hhq _
_
_
_
)()( =
==
For the vertical surface, find
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8/14/2019 MODULE 6: Worked-Out Problems
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639L
39326-26-
-1-323s
L
1044.8)15.0(105.2Ra
L105.2L/s)m10/s)(21m10(14.87
C)(23K 103.479.8m/s)L-T(TgRa
==
=
==
Using the correct correlation
K mW m
K mW Lk
Nuhh
NuL
Lv L ./03.515.0./0254.0
7.29 hence,
7.29])71.0/492.0(1[
)1044.8(387.0825.0
2 _ _ _
2
27/816/9
6/16
====
=
++=
Using the correct correlation,
97.018.503.5
,
./18.506.0
./0254.024.12
24.12])71.0/559.0(1[
)104.5(387.060.0
2 _ _ _
2
27/861/9
6/15 _
==
====
=
+
+=
h
v
Dh D
D
q
qhence
K mW m
K mW Dk
Nuhh
Nu
Comments: in view of the uncertainties associated with equations and theneglect of end effects, the above result is inconclusive. The cooling rates areare approximately the same.