Wilson Braz
MANE 6630
September 30, 2009
Homework 2
One Dimensional Steady State Heat Conduction
1
Consider a steady state heat conduction through the homogeneous, isotropic wall of a
hollow cylindrical shell of thermal conductivity k with internal heat generation inside the
wall. Use a differential volume element of radial thickness r, unit length and
circumference 2 to obtain the steady state heat equation by performing a differential
thermal energy balance on the volume element.
For a homogeneous isotropic solid in the steady state condition, k is constant and the
general form of the heat equation is:
q
 2T   0 eq. 1
k
In cylindrical coordinates,  2T is given by:
 2T 1 T 1  2T  2T
 2T  2 


r r r 2  2 z 2
r
Thus the equation becomes:
 2T 1 T 1  2T  2T
q

 2
 2 
2
2
r
r r r 
z
k
The above equation can be derived from inspection of a differential control volume.
NOTE: q  Heat transfer rate W; q ' '  Heat flux in W/m2; q  Heat generation per
unit volume W/m3
Given a differential control volume in cylindrical coordinate system ( r ,  , z )
The volume of the differential element is:
2
V  r  dr   d  dz  r 2  d  dz


V  r  dr   r 2  d  dz

2

V  r  rdr  dr  r 2  d  dz
2
2
V  r  dr   dr  d  dz
Developing the basic equation for conservation of energy:
Ein  E generated  Eout  Estored
Where:
Ein  qr  q  q z
Eout  qr dr  q d  q z dz
E generated  q  r  dr   dr  d  dz
Estored  c
T
 r  dr   dr  d  dz
t
Substituting into the energy equation:
qr  q  q z  q  r  dr   dr  d  dz  qr dr  q d  q z  dz  c
T
 r  dr   dr  d  dz
t
--------qr  dr  qr 
qr
dr ;
r
q d  q 
q

d ;
q z  dz  q z 
q z
dz .
z
Substitute above equations to obtain:
qr  q  qz  q  r  dr   dr  d  dz  qr 
q
qr
q
T
dr  q   d  qz  z dz  c
 r  dr   dr  d  dz
r

z
t
Simplify the above equation:
q
q
q
T
q  r  dr   dr  d  dz  r dr 
d  z dz  c
 r  dr   dr  d  dz
r

z
t
Fourier’s Law:
qr  k  r  d  dz
q  k  dr  dz

T

T
EQ Fourier 1 ;
r
EQ Fourier 2;
 Tz
q z   k  r  dr  d  r 2 d 
2
 
 Tz
q z   k  r 2  rdr  dr 2  r 2 d
q z   k  r  dr dr  d
q z   k r  dr dr  d
T
z
T
z
EQ Fourier 3
Substitute Fourier’s equations into energy balance:
q  r  dr   dr  d  dz 
q  r  dr   dr  d  dz 
q 
q
qr
q
T
dr 
d  z dz  c
 r  dr   dr  d  dz
r

z
t
 
T 
 
T 
 
T 
T
  k  dr  dz
d    k r  dr dr  d
 r  dr   dr  d  dz
  k  r  d  dz
dr 
dz  c
r 
r 
 
 
z 
z 
t
  k  r T    k
T    T 
T

 

   k
  c
r  r  dr  r    r  dr    z  z 
t
Using the chain rule and taking lim dr0 of above equation:
1   T  1   T    T 
T
 k
   k
 kr
 2
  q  c
r r  r  r     z  z 
t
Since our cylinder is in steady state,
T
 0 , hence:
t
1   T  1   T    T 
k
  k
 kr

  q  0
r r  r  r 2     z  z 
2
Consider a steady state heat conduction through the homogeneous, isotropic wall of a
hollow spherical shell of thermal conductivity k with internal heat generation inside the
wall. Use a differential volume element of radial thickness r, and circumferences 2 in
both axial and azimuthal directions to obtain the steady state heat equation by performing
a differential thermal energy balance on the volume element.
The general form of the heat equation is given as:
  kT   q  c
T
t
In spherical coordinates:
1   2 T 
1
 
T 
1
  T 
T
 k
  q  c
 kr
 2
 k sin 
 2 2
2
r r 
r  r sin   
  r sin     
t
3
Consider the problem of properly insulating a metal pipe that carries a hot fluid inside.
The pipe is insulated by covering its outside surface with a hollow shell of an insulating
material. Assume convective heat transfer conditions apply both along the inner radius of
the metal pipe and the outer radius of the insulation. Use the concept of overall thermal
resistance to obtain the critical radius of insulation.
Te
he
Tf
hf
Tf
R1
T0
R2
T1
R3
T2
R4
Te
The overall heat flow per unit length of a pipe is given to be:
q T f  Te

,
L
Rt
where Tf is the temperature of the fluid within the pipe, Te is the temperature of the
fluid surrounding the system and Rt is the overall thermal resistance of the system.
Rt can be written as:
Rt 
1
2
1
1
1
1 
 

  or,
 R1 R2 R3 R4 
Rt 
1
2
 1
1 r
1 r
1 

 ln 2  ln 3 
h r k
r1 k 2 r2 he r3 
1
 f 1
To find the critical radius, we differentiate Rt with respect to r then set the result
equal to zero to find the min/max:
dRt
1 1 1 

  
dr 2r  k hr 
0
1 1 1  1 r
   
2r  k hr  h k
rcr 
k
h
4
Consider the steady state heat conduction problem inside a homogeneous, isotropic
cylindrical shell of wall thickness L = b − a, constant thermal conductivity, with internal
heat generation and subjected to Dirichlet conditions:
T (r = a) = Ta
T (r = b) = Tb
Write down the governing equation of the problem. Then, use the finite difference
method to obtain a system of simultaneous linear algebraic equations that can produce an
approximation to the solution. Solve the problem corresponding to the following data:
a = 0.10, b = 0.11, Ta = Tb = 0, k = 20 W/mK, Qv = 103W.
The heat equation for a 1-D cylindrical steady-state problem with internal heat
generation reduces to:
d 2T 1 dT q v

 0
dr 2 r dr k
The finite difference approximation can be written as:
Tm1  2Tm  Tm1 1 Tm1  Tm1 q v

 0
rm
2r
k
r 2
This equation can be rearranged to solve for Tm:
Tm1  2Tm  Tm1 

r
Tm1  Tm1    qv
2rm
k
 q


r 
r 
  Tm1 1 
 2
Tm   v  Tm1 1 
 2rm 
 2rm 
k
We can modify the finite difference equation shown above slightly to allow for the
solution to be found by using the Gauss-Seidel iterative scheme.
 q


r 
r 
  Tmi 11m1 1 
 2
Tmi 1   v  Tmi 1 1 
 2rm 
 2rm 
k
Where i denotes iteration number.
Using a 6 node model;
1
2
3
4
5
6
using the B.C.’s given in the problem and setting initial temperature values of zero,
Excel was used to obtain the temperature distribution:
1.200
1.000
iteration
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
T1
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
T2
0.000
25.000
43.873
59.683
73.577
83.010
89.218
93.284
95.946
97.688
98.828
99.575
100.064
100.383
100.593
100.730
100.820
100.878
100.917
100.942
100.958
100.969
100.976
100.981
100.984
100.986
100.987
100.988
100.989
100.989
100.989
100.989
100.989
100.989
100.990
100.990
T3
0.000
37.380
68.692
96.211
114.894
127.188
135.241
140.513
143.964
146.223
147.702
148.670
149.303
149.718
149.989
150.167
150.283
150.359
150.409
150.442
150.463
150.477
150.486
150.492
150.496
150.499
150.500
150.502
150.502
150.503
150.503
150.503
150.503
150.504
150.504
150.504
T4
0.000
43.514
82.519
105.901
121.001
130.865
137.320
141.545
144.310
146.121
147.305
148.081
148.589
148.921
149.139
149.281
149.374
149.435
149.475
149.501
149.518
149.530
149.537
149.542
149.545
149.547
149.548
149.549
149.550
149.550
149.550
149.550
149.551
149.551
149.551
149.551
T5
0.000
46.555
65.878
77.460
84.940
89.827
93.024
95.117
96.487
97.384
97.971
98.355
98.606
98.771
98.879
98.949
98.996
99.026
99.046
99.058
99.067
99.073
99.076
99.079
99.080
99.081
99.082
99.082
99.082
99.083
99.083
99.083
99.083
99.083
99.083
99.083
T6
Max Delta Delta T1
0.000
0.000
46.5553
0.000
39.0055
0.000
27.5190
0.000
18.6838
0.000
12.2938
0.000
8.0533
0.000
5.2720
0.000
3.4509
0.000
2.2589
0.000
1.4786
0.000
0.9678
0.000
0.6335
0.000
0.4147
0.000
0.2714
0.000
0.1777
0.000
0.1163
0.000
0.0761
0.000
0.0498
0.000
0.0326
0.000
0.0213
0.000
0.0140
0.000
0.0091
0.000
0.0060
0.000
0.0039
0.000
0.0026
0.000
0.0017
0.000
0.0011
0.000
0.0007
0.000
0.0005
0.000
0.0003
0.000
0.0002
0.000
0.0001
0.000
0.0001
0.000
0.0001
0.000
0.0000
Delta T2
25.0000
18.8731
15.8094
13.8944
9.4335
6.2072
4.0661
2.6618
1.7424
1.1405
0.7465
0.4887
0.3199
0.2094
0.1370
0.0897
0.0587
0.0384
0.0252
0.0165
0.0108
0.0071
0.0046
0.0030
0.0020
0.0013
0.0008
0.0006
0.0004
0.0002
0.0002
0.0001
0.0001
0.0000
0.0000
Delta T3
37.3798
31.3118
27.5190
18.6838
12.2938
8.0533
5.2720
3.4509
2.2589
1.4786
0.9678
0.6335
0.4147
0.2714
0.1777
0.1163
0.0761
0.0498
0.0326
0.0213
0.0140
0.0091
0.0060
0.0039
0.0026
0.0017
0.0011
0.0007
0.0005
0.0003
0.0002
0.0001
0.0001
0.0001
0.0000
Delta T4
43.5136
39.0055
23.3819
15.0997
9.8642
6.4549
4.2250
2.7655
1.8102
1.1849
0.7756
0.5077
0.3323
0.2175
0.1424
0.0932
0.0610
0.0399
0.0261
0.0171
0.0112
0.0073
0.0048
0.0031
0.0021
0.0013
0.0009
0.0006
0.0004
0.0002
0.0002
0.0001
0.0001
0.0000
0.0000
Delta T5
Delta T6
46.5553
19.3222
11.5827
7.4800
4.8864
3.1976
2.0929
1.3700
0.8967
0.5870
0.3842
0.2515
0.1646
0.1078
0.0705
0.0462
0.0302
0.0198
0.0129
0.0085
0.0055
0.0036
0.0024
0.0016
0.0010
0.0007
0.0004
0.0003
0.0002
0.0001
0.0001
0.0001
0.0000
0.0000
0.0000
The above table shows that the Gauss-Seidel method converges to 1e-4 after 35
iterations.
5
Consider the steady state heat conduction problem inside a homogeneous, isotropic flat
wall of thickness L = b − a, constant thermal conductivity, with internal heat generation
and subjected to Dirichlet conditions
T (r = a) = 0
T (r = b) = 0
Use the Galerkin finite element method to obtain a system of simultaneous linear
algebraic equations that can produce an approximation to the solution. Solve the problem
corresponding to the following data: a = 0.0, b = 1.0, Ta = Tb = 0, k = 1 W/mK, Qv = 20x3
W.
The heat equation for a 1-D steady-state problem with internal heat generation
reduces to:
d 2T q v
 0
dx 2 k
Integrate the above equation twice to yield:
q x 2
T x    v  C1 x  C2
k
0.800
0.600
0.400
0.200
0.000
0.098
0.1
0.1
v

b2  a
q v a C 2
C1 

; C2   k
;
k
a
b 
q
  1
a 
q v 
q v b 2  a

b2  a 
;
; C1   a  2
C2  
k 
a b  a  
k ab  a 

T x   





q v x 2 q v 
b 2  a  q v b 2  a
x 
  a  2
k
k 
k ab  a 
a b  a  




q v  2 
b2  a 
b2  a 
x 
T  x     x   a  2

k 
ab  a  
a b  a  

6
Obtain an expression for the temperature distribution of a fin constant cross section A and
perimeter P and of finite length L with base temperature Tb and heat transfer coefficient h
all over its surface into an environment at T1.
Performing an energy balance on the system shown in the above diagram yields:
qx   qx  x   qconv EQ. 1.
As x  0 ,
qx  x   qx  
dq
x and
dx
qconv  hP x xT  T 
EQ. 1 becomes:
qx   qx  
dq
x  hPx xT  T 
dx
0
dq
 hP x T  T  EQ. 2.
dx
From Fourier’s Law:
dT
q  x    kA x 
where Ax is the cross sectional area.
dx
Substitute Fourier’s Law into EQ 2.
d 
dT  hPx 
T  T   0 .
 Ax 

dx 
dx 
k
If Ax  const. then Px  const. :
A
d 2T hP
T  T   0 or

dx 2
k
d 2T hP
T  T   0

dx 2 kA