1 chapter 2 one-dimensional steady state conduction 2.1 examples of one-dimensional conduction:...
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1
CHAPTER 2
ONE-DIMENSIONAL STEADY STATE
CONDUCTION
2.1 Examples of One-dimensional
Conduction:
2.1.1 Plate with Energy Generation and
Variable Conductivity
2
Example 2.1: Plate with internal energy generationq and a variable k
)1( Tkk o
Find temperature distribution.
L
q C0oC0o
Fig. 2.1
x0
(1) Observations • Variable k
• Symmetry
• Energy generation
• Rectangular system
• Specified temperature at boundaries
3
(2) Origin and Coordinates
Use a rectangular coordinate system
(3) Formulation
(i) Assumptions
• One-dimensional
• Steady
• Isotropic
• Stationary
• Uniform energy generation
4
(ii) Governing Equation
Eq. (1.7):
0
qdx
dTk
dx
d(2.1)
(a) )1( Tkk o
(a) into eq. (2.1)
(1 ) 0o
d dT qTdx dx k
(b)
(iii) Boundary Conditions.
Two BC are needed:
5
(c) 0)0( T
0)( LT (d)
(4) Solution
Integrate (b) twice
2122
22CxCx
k
qTT
o
(e)
BC (c) and (d)
ok
LqC
21
, 02 C (f)
6
(f) into (e)
(g) 0122
L
x
k
xLqTT
o
Take the negative sign
Solving for T
(h)
L
x
k
xLqT
o
111
2
(i)
L
x
k
xLqT
o
111
2
7
(5) Checking
• Dimensional check
• Boundary conditions check
• Limiting check: 0 0, Tq
• Symmetry Check:
(j) )12
)(()1(1
2
1 2
1
2
L
x
k
Lq
L
x
k
Lxq
dx
dT
oo
Setting x = L/2 in (j) gives dT/dx = 0
8
Conservation of energy and symmetry:
(k) 2
)0(ALq
q
2)(
ALqLq
(l)
Fourier’s law at x = 0 and x = L
2
)0()0(1)0(
ALq
dx
dTTkq o
(m)
• Quantitative Check
9
2
)()(1)(
ALq
dx
LdTLTkLq o
(n)
(6) Comments Solution to the special case:
k = constant: Set 0
2.1.2 Radial Conduction in a Composite Cylinder with Interface Friction
10
Example 2.2: Rotating shaft in sleeve, frictional
heat at interface, convection on
outside. Conduction in radial direction.
Determine the temperature
distribution in shaft and sleeve.
(1) Observations
• Composite cylindrical wall
• Cylindrical coordinates
• Radial conduction only
Fig. 2.2
0
sleeve
1T
2T
Th,
sRoR
r
iq
11
• Steady state:
• No heat is conducted through the shaft
Energy generated = heat conducted through the sleeve
• Specified flux at inner radius of sleeve, convection at outer radius
(2) Origin and Coordinates Shown in Fig. 2.2
12
(3) Formulation
(i) Assumptions
• One-dimensional radial conduction
• Steady
• Isotropic
• Constant conductivities
• No energy generation
• Perfect interface contact
• Uniform frictional energy flux
• Stationary
13
(ii) Governing Equation
Shaft temperature is uniform. For sleeve: Eq. (1.11)
(2.2) 01
dr
dTr
dr
d
(iii) Boundary Conditions
:sRSpecified flux at
(a) dr
RdTkq s
i)(1
1
14
Convection at :oR
])([)(
11
1 TRThdr
RdTk o
o (b)
(4) Solution
Integrate eq. (2.2) twice
(c) 211 ln CrCT
BC give C1 and C2
11 k
RqC si (d)
15
and
(d) and (e) into (c)
oo
si
hR
kR
k
RqTC 1
12 ln (e)
o
osi
hR
k
r
R
k
RqTrT 1
11 ln)( (f)
khRo / = Biot number
Shaft temperature T2: Use interface boundary condition
(g) )()()( 122 ss RTRTrT
16
Evaluate (f) at r = Rs and use (g)
os
osi
hR
k
R
R
k
RqTrT 1
12 ln)( (h)
(5) Checking
• Limiting check: 0iq
(6) Comments
• Conductivity of shaft does not play a role
Fig. 2.2
0
sleeve
1T
2T
Th,
sRoR
r
iq
• Dimensional check
• Boundary conditions check
17
• Problem can also be treated formally as a composite cylinder. Need 2 equations and 4 BC.
2.1.1 Composite Wall with Energy Generation
Example 2.1: Plate 1 generates heat at .q is sandwiched between two plates.
Outer surfaces of two plates at .oT
Plate 1
Find the temperature distribution in
the three plates.
oT x
q
2L
2L
1L
2k
oT
0
Fig. 2.3
1k
2k
18
oT x
q
2L
2L
1L
2k
oT
0
Fig. 2.3
1k
2k
(1) Observations • Composite wall
• Heat flows normal to plates
• Symmetry and steady state:
Energy generated = Energy conducted out
(2) Origin and Coordinates Shown in Fig. 2.3
• Use rectangular coordinates
• Symmetry: Insulated center plane
19
(3) Formulation (i) Assumptions
• Steady
• One-dimensional
• Isotropic
• Constant conductivities
• Perfect interface contact
• Stationary
(ii) Governing Equations
Two equations:
20
(b) 022
2
dx
Td0
21
2
k
q
dx
Td(a)
(iii) Boundary Conditions
Four BC:
Symmetry:
0)0(1
dx
dT(c)
Interface:
dx
LdTk
dx
LdTk
)2/()2/( 122
111 (d)
21
)2/()2/( 1211 LTLT (e)
Outer surface:
oTLLT )2/( 212 (f)
(4) Solution
Integrate (a) twice
(g) BAxxk
qxT
2
11 2
)(
Integrate (b)
DCxxT )(2 (h)
22
Four BC give 4 constants: Solutions (g) and (h) become
21
2
12
21
1
21
1 4
1
2)(
L
x
Lk
Lk
k
LqTxT o (i)
11
2
2
21
2 2
1
2)(
L
x
L
L
k
LqTxT o (j)
(5) Checking
:2
k
Lq • Dimensional check: units of
23
• Boundary conditions check
• Quantitative check:
1/2 the energy generated in center plate = Heat conducted at 2/1Lx
(k) dx
LdTkq
L )2/(
211
11
CCW/m
mW/m oo
223
)(
)()(
k
Lq
24
(i) into (k)
qL
dx
LdTk
2
)2/( 1111
Similarly, 1/2 the energy generated in center plate= Heat conducted out
(l) dx
LLdTkq
L )2/(
2212
21
(j) into (l) shows that this condition is satisfied.
• Limiting check:
(ii) If 01 L then .)(1 oTxT
,0q then .)()( 21 oTxTxT (i) If
25
(6) Comments Alternate approach: Outer plate with a specified
flux at 2/1Lx and a specified temperature at
.2/ 21 LLx
2.2 Extended Surfaces - Fins
2.2.1 The Function of Fins Newton's law of cooling:
(2.3) )( TThAq sss
26
Options for increasing :sq
• Increase h
• Lower T
• Increase sA
Examples of Extended Surfaces (Fins):
• Thin rods on condenser in back of refrigerator
• Honeycomb surface of a car radiator
• Corrugated surface of a motorcycle engine
• Disks or plates used in baseboard radiators
27
2.2.2 Types of Fins
straight fin constant area (a)
2.5 Fig.
annular fin (d) pin fin (c)
straight fin variable area (b)
28
Terminology and types
• Fin base
• Fin tip
• Straight fin
• Variable cross-sectional area fin
• Spine or pin fin
• Annular or cylindrical fin
2.2.3 Heat Transfer and Temperature Distribution in Fins
• Heat flows axially and laterally (two-dimensional) • Temperature distribution is two-dimensional
29
2.2.4 The Fin Approximation
x
Th ,
T
Tr
6.2.Fig
Neglect lateral temperature variation
)(xTT
Criterion: Biot number = Bi
(2.4) Bi = h /k << 1
cetanresisexternal
cetanresisInternal
h/
k/Bi
1
30
2.2.5 The Fin Heat Equation: Convection at Surface
(1) Objective: Determine fin heat transfer rate.
Need temperature distribution.
(2) Procedure: Formulate the fin heat equation.
Apply conservation of energy.
• Select an origin and coordinate axis x.
• Assume ,1.0Bi )(xTT • Stationary material, steady state
31
Th,dx
dy
sy
)(a )(b )(c
ds
CsdA
cdq
xq dxdx
dqq x
x dxx
y
0
7.2.Fig
Conservation of energy for the element dx:
(a) outgin EEE =
(b) xin qE
cx
xout dqdxdx
dqqE (c)
32
(b) and (c) into (a)
(d) cx
g dqdxdx
dqE
Fourier's law and Newton’s law
dx
dTkAq cx (e)
(f) sc dATThdq )( Energy generation
dy
)(b )(c
ds
CsdA
cdq
xq dxdx
dqq x
x dx
33
(e), (f) and (g) into (d)
(2.5a)
( ) ( ) ( ) 0s cc
dTdkA x dx h T T dA q A x dx
dxdx
Assume constant k
0)()()(
12
2
k
q
dx
dATT
xkA
h
dx
dT
dx
dA
xAdx
Td s
c
c
c(2.5b)
• (2.5b) is the heat equation for fins
• Assumptions: (1) Steady state (2) Stationary
dxxAqE cg )( (g)
34
(3) Isotropic
(4) Constant k
(5) No radiation
(6) Bi << 1
2.2.6 Determination of dAs /dxFrom Fig. 2.7b
dsxCdAs )( (a)
= circumference )(xCds = slanted length of the element
• ,/ dxdAc and dxdAs / are determined from
the geometry of fin.
,Ac
35
For a right triangle
(b) 2/122 ][ sdydxds
(b) into (a) 2/12
1)(
dx
dyxC
dx
dA ss (2.6a)
For dxdys / << 1
)(xCdx
dAs (2.6b)
2.2.7 Boundary ConditionsNeed two BC
36
2.2.8 Determination of Fin Heat Transfer Rate :fq
Fig. 2.8
sq
sq
sqx0
)0(q
Th,
Th,
Conservation of energy for :0q
sf qqq )0( (a)
Two methods to determine :fq
37
(2) Convection at the fin surface.
Newton's law applied at the fin surface
(2.8) s
dATxThqqA ssf ])([
• Fin attached at both ends: Modify eq. (2.7) accordingly
• Fin with convection at the tip: Integral in eq. (2.8) includes tip
(1) Conduction at base.
Fourier's law at x = 0
(2.7) dx
dTkAqq cf
)0()0()0(
38
• Convection and radiation at surface: Apply eq. (2.7). Modify eq. (2.8) to include heat exchange by radiation.
2.2.9 Applications: Constant Area Fins with Surface Convection
Fig. 2.9
0 x
CTh,
Th,oT
cA
39
Use eq. (2.5b). Set
(a) 0/ dxdAc
A. Governing Equation
Fig. 2.9
0 x
CTh,
Th,oT
cA = constant sy
0/ dxdys
(a) and (b) into eq. (2.5b)
Eq. (2.6a)
CdxdAs / (b)
0)(2
2
TTkA
hC
dx
Td
c
(2.9)
40
Rewrite eq. (2.9)
TT (c)
ckA
hCm 2 (d)
022
2
m
dx
d(2.10)
Valid for:
(1) Steady state
(2) constant k, cA and T
= constant, (c) and (d) into (2.9)TAssume
41
(3) No energy generation
(4) No radiation
(5) Bi 1
(6) Stationary fin
B. Solution
(2.11a) )exp()exp()( 21 mxAmxAx
Assume: h = constant
mxBmxBx coshsinh)( 21 (2.11b)
42
C. Special Case (i):
• Finite length
• Specified temperature at base, convection at tip
Boundary conditions:
(e) o)0( TT
])([)(
TLThdx
LdTk t
(f)
o)0( (h)
Fig. 2.10cA
C
th0
oT
Th,
xTh,
43
)()(
Lhdx
Ldk t
(i)
Two BC give B1 and B2
fqEq. (2.7) gives
mLmkhmL
mLmkhmLTThCAkq
t
tcf sinh )/( cosh
] cosh)/()[sinh(][ o2/1
(2.13)
TT
TxTx
o o
)()(
(2.12)
mLmkhmL
xLmmkhxLm
t
t
sinhcosh
sinhcosh
44
C. Special Case (ii):
• Finite length
• Specified temperature at base, insulated tip
BC at tip:
0)(
dx
Ld(j)
0thSet eq. (2.12)
mL
xLm
TT
TxTx
cosh
)-( cosh)()(
oo
(2.14)
0thSet eq. (2.13)
(2.15) mLTTChkAq ocf tanh21
45
2.2.10 Corrected Length Lc
• Insulated tip: simpler solution
• Simplified model: Assume insulated tip, compensate by increasing length by cL
• The corrected length is cL
cc LLL (2.16)
• The correction increment Lc depends on the
geometry of the fin:
Circular fin:
cLrr o2
o 2
cLIncrease in surface area due to = tip area
46
2/oc rL
4/tLc
= total surface areasA
Square bar of side t
f2.2.11 Fin EfficiencyDefinition
maxq
q ff (2.17)
TThAq osmax
47
Eq. (2.17) becomes
(2.18) )( o
TTAh
q
s
ff
2.1.12 Moving Fins
Examples:
• Extrusion of plastics
• Drawing of wires and sheets
• Flow of liquids
Udx
Th,
oT x
surT
(a)
dx
hm ˆ )(ˆ
ˆ dxdx
hdhm
dxdx
dqq x
x xq
cdqrdq
Fig. 2.11
(b)
48
Heat equation:
• Steady state• Constant area• Constant velocity U• Surface convection and radiation
Conservation of energy for element dx
rcx
xx dqdqdx
hdmhmdx
dx
dqqhmq
ˆˆˆ
(a)
cUAm (b)
dx
hm ˆ )(ˆ
ˆ dxdx
hdhm
dxdx
dqq x
x xq
cdqrdq
Fig. 2.11(b)
Assume:
49
dTchd pˆ (c)
Fourier’s and Newton’s laws
dx
dTkAq cx (d)
sc dATThdq )( (e)
dxCdAs (f)
ssurr dATTdq )( 44 (g)
(b)-(g) into (a) assume constant k
0)()( 442
2
surc
p TTk
TTkA
hC
dx
dT
k
Uc
dx
Td
(2.19)
50
Assumptions leading to eq. (2.19): (1) Steady state
(3) Isotropic
(4) Gray body
(5) Small surface enclosed by a much larger surface and
,(2) Constant U, k, P and
(6) Bi << 1
51
Example 2.4
insulated bottomfurnace
Fig. 2.12
t
oT
xTh,
WU
2.2.13 Application of Moving Fins
Determine the temperature distribution in the sheet.
(2) Bi < 0.1(3) No radiation(4) No heat transfer from bottom
Plastic sheet leaves furnace at .oT
Sheet is cooled at top by convection.
Assume:
(1) Steady state
52
Solution
(1) Observations
• Constant area fin
• Temperature is one-dimensional
• Convection at surface
• Specified furnace temperature
• Fin is semi-infinite
• Constant velocity
insulated bottomfurnace
Fig. 2.12
t
oT
xTh,
WU
(2) Origin and Coordinates
53
(3) Formulation
(i) Assumptions (1) One-dimensional
(2) Steady state
(3) Isotropic
(4) Constant pressure
0)(2
2
TTkA
hC
dx
dT
k
Uc
dx
Td
c
p(2.20)
, (5) Constant U, k, P and (6) Negligible radiation
(ii) Governing Equation
Eq. (2.19)
54
WtAc (a)
tWC 2 (b)
(a) and (b) into eq. (2.20)
cTmdx
dTb
dx
Td 2
2
2
2 (c)
where
,2k
Ucb p
,)2(2
kWt
tWhm
T
tWk
tWhc
)2(
(d)
(iii) Boundary Conditions
oTT )0( (e)
)(T finite (f)
55
Eq. (c) is:
• Linear• Second order• Constant coefficients
(4) Solution:
Eq. (A-6b), Appendix A
01 C (g)
222
2
221
)exp(
)exp(
m
cxmbbxC
xmbbxCT
(2.21)
B.C. (f)
56
B.C. (e)
202m
cTC (h)
(d), (g) and (h) into (2.21)
xtWk
tWh
k
Uc
k
Uc
TT
TxT pp
o
)2(2
22exp
)( )(
(2.22)
(5) Checking
Dimensional check: Each term in the exponential in
eq. (2.22) is dimensionless
57
Boundary conditions check: Eq.(2.22) satisfies (e) and (f).
Limiting checks:
:0h(i) If
:U(ii) If
(6) Comments
(i) Temperature decays exponentially
(ii) Motion slows decay
oTxT )(
oTxT )(
58
2.2.14 Variable Area Fins
)(xAA cc
Example: Cylindrical or annular fin
Governing equation:
Usually has variable coefficients
Case (i) : The Annular Fin
0)()()(
12
2
dr
dATT
rkA
h
dr
dT
dr
dA
rAdr
Td s
c
c
c (2.23)
2.13 Fig.
Th,
drr
Th,
59
trrAc 2)( (a)
tdrdAc 2/ (b)
Eq.(2.6a) gives drdAs /
2/12
1)(
dr
dyrC
dr
dA ss
0/ drdys
)2(2)( rrC (c)
For
sy = constant:
2.13 Fig.
Th,
drr
Th,
60
Eq. (2.6a):
)2(2 rdr
dAs (d)
(a), (b) and (d) into eq. (2.23)
0))(/2(1
2
2
TTkthdr
dT
rdr
Td (2.24)
Case (ii): Triangular straight fin
Fin equation: Constant k,eq. (2.5b)
2.14 Fig.
Th,
L
dx
xsy
0
y
tTh,
)(2 xyWA sc (e)
61
(f) )2/)(/( tLxys
xLtWAc )/( (g)
LtWdx
dAc / (h)
Eq. (2.6a):
2/122/12 ])2/(1[2])/(1[2 LtWdxdyWdx
dAs
s (i)
2.14 Fig.
Th,
L
dx
xsy
0
y
tTh,
62
(g), (h) and (i) into eq. (2.5b)
0)(/12/1/21 2/12
2
2
TTxLtkthLdxdT
xdx
Td
(2.25)
Equations (2.24) and (2.25) are:
•Linear
•Second order
•Variable coefficients
63
(2.26) 0)21(
2)21(
22222222
22
22
ynCAxABxBxDC
dxdy
xBxAdx
ydx
C
2.3 Bessel Differential Equations and Bessel Functions
2.3.1 General form of Bessel Equations
Note the following:
(1) Eq. (2.26) is linear, second order with variable coefficients
(2) A, B, C, D, and n are constants
64
(3) n is called the order of the differential equation
(4) D can be real or imaginary
2.3.2 Solution: Bessel Functions
• Form: Infinite power series solutions
• General solution: depends on the constants n and D
(1) n is zero or integer, D is real
)()()exp()( 21C
nC
nA xDYCxDJCxBxxy
(2.27) where
21 ,CC = constant of integration
65
)( Cn DxJ = Bessel function of order n of the first kind
)( Cn DxY = Bessel function of order n of the second kind
Note the following:
(i) The term )( CxD is the argument of the Bessel
function
(ii) Values of Bessel functions are tabulated
(2) n is neither zero nor a positive integer, D is real
(2.28)
)()()exp()( 21C
nC
nA xDJCxDJCxBxxy
66
(3) n is zero or integer, D is imaginary
(2.29)
)()()exp()( 21C
nC
nA xpKCxpICxBxxy
where
,i
Dp i is imaginary = 1
= modified Bessel function of order n of the first kind nI
nK = modified Bessel function of order n of the second
kind
67
(4) n is neither zero nor a positive integer, D is imaginary
)()()exp()( 21C
nC
nA xpICxpICxBxxy
(2.30)
2.3.3 Form of Bessel Functions
nnnnn IIJYJ ,,,, :nKand
• Symbols for infinite power series
• The form of each series depends on n
(2.31)
0 !!k 2
)2/()1()(
22
2 kk
xxJ
kk
Example: ,2n )(2 xJBessel function
68
2.3.4 Special Closed-form Bessel Functions:
2
integer oddn
For 2/1n
(2.32) xx
xJ sin2
)(2/1
and
xx
xJ cos2
)(2/1 (2.33)
69
For n = 3/2, 5/2, 7/2, … use Eq. (2.32) or eq. (2.33) and the recurrence equation:
)()(12
)( 3/22/12/1 xJxJx
kxJ kkk
k = 1, 2, 3, … (2.34)
For the modified Bessel functions, n = 1/2
xx
xI sinh2
)(2/1 (2.35)
and
xx
xI cosh2
)(2/1 (2.36)
70
For n = 3/2, 5/2, 7/2, …. use eq. (2.35) or eq. (2.36) and the recurrence equation
(2.37)
)()(12
)( 3/22/12/1 xIxIx
kxI kkk
k = 1, 2, 3, …
2.3.5 Special Relations for n = 0, 1, 2, …
)()1()( xJxJ nn
n (2.38a)
)()1()( xYxY nn
n (2.38b)
)()( xIxI nn (2.38c)
)()( xKxK nn (2.38d)
71
2.3.6 Derivatives and Integrals of Bessel Functions
(2.39)
mxZmx
mxZmxmxZx
dx
d
nn
nn
nn
1
1 IYJZ ,,
KZ (2.40)
(2.41)
mxZmx
mxZmxmxZx
dx
d
nn
nn
nn
1
1 KYJZ ,,
IZ (2.42)
72
(2.45)
mxZx
nmxmZ
mxZx
nmxmZ
mxZdx
d
nn
nn
n
1
1 KYJZ ,,
IZ (2.46)
(2.43)
mxZx
nmxmZ
mxZx
nmxmZ
mxZdx
d
nn
nn
n
1
1 IYJZ ,,
KZ (2.44)
IYJZmxZxmdxmxZx nn
nn ,,)()/1()(1
(2.47)
73
2.3.7 Tabulation and Graphical Representation of Selected Bessel Functions
(2.48)
mxZxmdxmxZx nn
nn
11
KYJZ ,,
Table 2.1
x J0(x) Jn(x) I0(x) In(x) Yn(x) Kn(x)
0 1 0 1 0 - 0 0 0 0
74
Fig. 2.15 Graphs of selected Bessel functions
75
2.4 Equidimensional (Euler) Equation
(2.49) 0012
22 ya
xdyd
xaxd
ydx
Solution:Depends on roots r1 and r2
2
4)1()1( 02
112,1
aaar
(2.50)
Three possibilities:
(1) Roots are distinct
2121)( rr xCxCxy (2.51)
76
(2) Roots are imaginary
)logsin()logcos()( 21 xbCxbCxxy a (2.52)
(3) One root only
)log()(21
xCCxxy r (2.53)
2.5 Graphically Presented Fin Solutions to Fin
Heat Transfer Rate fq
Fin efficiency :f
)( o
TTAh
q
s
ff (2.18)
77
Fig. 2.16 Fin efficiency of three types of straight fins [5]
78
Fig. 2.17 Fin efficiency of annular fins of constant thickness [5]