 2001, W. E. Haisler
22
Chapter 7: Heat Transfer Applications in a Solid
1-D Heat Conduction through a Single Flat Wall with
Convection Boundary Conditions
k
T ,1
L
h1
x
T ,2
T(x)=C C 2 x
h2
From previous work for a single flat wall with no heat source, we
know that the temperature is a linear function of x:
T ( x)  C1  C2 x
 2001, W. E. Haisler
23
Chapter 7: Heat Transfer Applications in a Solid
1-D Heat Conduction through a 2-layer Composite Flat Wall
with Convection Boundary Conditions
T ,1
h1
Layer 1
Layer 2
k1
k2
L1
L2
A
B
x
C
h2
x=0
T1(x)=a 1+b1x
x=x1
x=x2
T ,2
T2(x)=a2+b 2x
Again, for no heat sources, temperature field in each layer is a
linear function of x. At point A, x=0; at B, x=x1; at C, x=x2.
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
24
The heat flux through the boundary between layers 1 and 2
must be continuous. This provides two B.C. at the interface
between two layers that must be satisfied.
Consider two bodies that are touching so that they have an
interface between them as shown below. Assume that the
temperature distribution in bodies 1 and 2 are T1 ( x, y, z , t ) and
T2 ( x, y, z , t ) , respectively. Thermal conductivity for the two
bodies is k1 and k2.
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
q2
k1
1
interface boundary conditions:
T1  T2
n
n
2
q1
25
n  q 1 n  q 2
k2
At the interface of two solids, there are two “boundary”
conditions that must be met:
1. The temperature of each body at the interface must be
equal: T1  T2 .
2. The heat flux leaving body 1 at the interface must equal
the heat flux entering body 2, i.e., the heat flux is
constant across the interface: n  q1  n  q 2 .
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
26
For the planar wall considered here, the normal n  i .
Layer 1
k1
T1 ( x )
q2
ni
q1
x  x1
x
Layer 2
k2
T2 ( x )
interface
boundary
Temperatures at the interface located at x  x1 must be equal:
T1 ( x1 )  T2 ( x1 )
Conservation of heat flux requires that: n  q  n  q
1
2
 i  q1  i  q2

dT1
dT2
 k2
q1x  q2 x  k1
dx x  x1
dx
x  x1
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
27
For each layer, steady-state heat transfer (without any heat
 2T
 0. For layers 1 and 2, this gives
sources) requires
 x2
T ( x)  a  b x
T ( x)  a  b x .
and
1 1
2 2
1
2
Apply boundary conditions at surfaces A, B, and C.
A: Convection B.C. at x=0 (where n  i )
n  q x  0  n  (k T ) x0  h(T  T )
s  x 0
T
or
(  k 1 )
 h (T  T )
1x
1 1 ,1 x0
x0
or
k b  h (a  T )
1 1 1 1 ,1
(1)
T ( x)  a  b x , T ( x)  a  b x 28
1 1 2
2 2
1
B. Interface B.C. at x=x1:
T1(x1)=T2(x1)
or a1  b1x  a2  b2 x
(2)
1
1
and
T
 T1
2
k b k b
k1
k
or
(3)
1
1
2
2
2
 x x x
x
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
1
x  x1
C: Convection B.C. at x=x2 (where n  i )
n  q x x  n  (k T ) x x  h(T  T )
s  x x2
2
2
T
2)
or
(k
 h (T  T
)
2 x
2 2 ,2 x x
x x
2
2
k b  h (a  b x  T
)
or
2 2 2 2 2 2 ,2
(4)
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
29
We have the 4 equations above to solve for a1, b1, a2, b2 .
Writing these in matrix notation gives:
 h1

1

0

 0
k
0
0
  a1   h T

1
1

,1
  

x
1
x
  b1   0 
1
1
   

k
0
k
a   0 


2
1
2
  

h
T
0 h (k  h x )   b2   2 ,2 
2
2 2 2
You can substitute numerical values for all given values like k1,
h1, x1, T , etc. Solve for a1, b1, a2, b2, and substitute back into
T ( x)  a  b x and T ( x)  a  b x . You can now calculate
2 2
T
temperature, T, and heat flux, q  k
, at any point x.
x
x
1
1
1
2
 2001, W. E. Haisler
30
Chapter 7: Heat Transfer Applications in a Solid
Example: Consider a two-layer composite with 1-D heat transfer
through the layers and free convection of air on either side. Assume
two layers made of copper [k1=398 J/(sec m oC)], and teflon
[k2=0.25 J/(sec m oC)], and thickness of L1=L2=10 cm. Assume the
convection coefficient is approximately h1=h2=5 BTU/(hr ft2 oF) and
T
 100C , T
 50C .
,1
,2
T,1  100C
h1
Layer 1
(copper)
k1
Layer 2
(teflon)
k2
L1
L2
B
A
T1 ( x)
x=0
C
T2 ( x)
x=x1
x
h2
x=x2
T,2  50C
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
31
Convert h to SI units:
1 BTU
1,055 J
=
 5.68 J / (m 2 sec C)
2
2
(hr ft F) (3,600 sec) (.3048 m) (5/9 C)
Thus h1=h2=28.39 J/(sec m2 oC).
Assuming solutions for layers 1 and 2 as T1 ( x)  a1  b1 x and
T2 ( x)  a2  b2 x , integrating the ODE and applying boundary
conditions, results in the system of equations:
0
 h1 k1 0
  a1   h T

1

,1

  

x
1
x
b
1


 1   0 
1
1

   

k
0
k
0
 a2   0 
1
2

  

h
T
0 h (k  h x )   b2   2 ,2 
 0
2
2 2 2
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
32
Substituting values of h, k, x, etc. gives the following set of
equations:
 a1 
0
0     2839. 
 28.39 398

b  

1
0.1

1

0.1
0


  1   

 0

398
0
0.25 a   0 

 2
0
28.39 5.928   b  1419.5
 0
 2 
Solving for the constants gives:
a1  96.258, b1  0.2669 , a2  138.72 and b2  424.90 .
Hence, the temperature equations for each layer are given by:
T1( x)  96.258  0.2669 x o C and T2 ( x)  138.72  424.9 x o C
(x in meters)
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
33
The temperatures at the boundary and interface are given by:
T1(0)  96.258 o C (left boundary)
T1(0.1)  T2 (0.1)  96.232 o C (interface)
T2 (0.2)  53.742 o C (right boundary)
The heat flux through each layer is given by
 T1
qx1  k1
  k1b1  398(0.2669)  106.225W / m 2
x
 T2
q x 2   k2
 k2b2  0.25(424.90)  106.225W / m 2
x
Note that the heat flux is positive (indicating heat flow to the
right)--as one would expect since higher temperature on left.
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
34
To simplify the solution for a composite wall, we seek to
develop a simplified relation between the overall heat flux
through the composite wall and the given temperature
gradient from one side of the composite wall to the other:
qx  U T
 
qx   1 T
R
where U  effective heat transfer coefficient of composite
the wall, R=1/U  effective thermal resistance of the
composite wall and for this case the known temperature
gradient form left to right is given by T  T
T .
,2 ,1
We know from the solution of the ODE for heat transfer
that the temperature variation in any layer with no heat
source is a linear function of x: T(x) = a + bx where a and b
are constants. If the temperature on either side of a wall of
or
 2001, W. E. Haisler
35
Chapter 7: Heat Transfer Applications in a Solid
thickness L is TA and TB, then T(x)=TA+[(TB-TA)/L] x . For
the composite wall, use the following notation:
T ,1
h1
Layer 1
Layer 2
k1
k2
L1
L2
B
A
x
C
h
2
x=0
x=x
1
TB  TA
T1( x )  TA 
x
L1
x=x2
T ,2
TC  TB
T2 ( x )  TB 
( x  L1 )
L2
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
36
Note: T1(L1)=T2(L1)=TB. For a convection boundary
condition on the left boundary, T,1 is usually known
(specified) but TA on boundary in unknown. If TA is taken
as a known, then T,1 must be taken as an unknown.
Similarly, on the right boundary, T,1 is usually known but
TB is unknown.
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
37
From conservation of energy, the heat flow rate Qx through a
given area A (Qx = qx A) must be constant as it enters on the left
and leaves on the right boundary (since we assumed there is no
internal heat generation, ). Since heat flow is normal to wall,
each layer has same normal area (so area cancels out). Thus,
T T
T T
qx  h (T  T )  k B A  k C B  h (T  T
)
1 A ,1
2 C ,2
1 L
2 L
1
2
(q for air on left) = (q for layer 1) = (q for layer 2) = (q for air on right)
Separate into 4 equations:
L
T  TB  1 q x
(1);
A
k
1
L
TB  T  2 q x
C k
2
(2)
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
1
T
 T  qx
,1 A h
1
(3);
1
T T
 qx
C ,2 h
2
38
(4)
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
39
Add the four equations, (1) through (4), to obtain
L L
1 1
1
2
T
T
( 
  )q x
,1 ,2 k k
h h
1 2 1 2
or
where
qx  (T
T )/ R
,2 ,1
1 L1 L2 1
1  2 Li  1
R  

=
 

h k k
h
h i 1k  h
1 1 2
2
1 
i 2
R  equivalent thermal resistance of the composite wall.
Note that thermal resistance terms (like L/k or 1/h) are
additive similar to resistors in electrical theory.
 2001, W. E. Haisler
k
L
T ,1
h1
1
1
40
Chapter 7: Heat Transfer Applications in a Solid
k
2
L
2
q x  (T
T )/ R
,2 ,1
k3
L
3
T
A
x
TB
T
C
TD
h2
T,2
n L
1
1
i
R  

h i 1k h
1
i
2
Evaluate R and solve for qx.
Then evaluate temperatures
TA, TB, TC, etc. from
equations at right.
q x  h (T  T )
1 A ,1
k
1
qx   T  T
B A L
1
k
2
qx   T  T
C B L
2
k
3
qx   T  T
D C L
3
q x  h (T
T )
2 ,2 D






 2001, W. E. Haisler
Two other cases:
TA, T,2 specified
k
L
1
TA, TD specified
2
k3
k
L
2
L
L1
k
1
41
Chapter 7: Heat Transfer Applications in a Solid
3
TA
x
k
1
2
k3
L2
L3
TA
x
TB
TB
TC
TD
T
C
h2
T,2
TD
n L
1
i
R 

i  1 ki h2
n L
R  i
i  1 ki
qx  (T
T )/ R
,2 A
qx  (T  T ) / R
D A
 2001, W. E. Haisler
42
Chapter 7: Heat Transfer Applications in a Solid
Example 1. Consider a two-layer composite with 1-D heat
transfer through the layers and free convection of air on either
side. Assume two layers made of glass with k1= k2=1.7 J/(sec m
o
C) and thickness of L1=L2=10 cm. Assume the convection
coefficient is approximately h1=h2=5 BTU/(hr ft2 oF) and
T
 100C , T
 50C
,1
,2
Layer 1
T ,1
h1
A
Layer 2
k1
k2
L1
L2
B
x
C
h2
x=0
x=x1
x=x2
T ,2
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
43
Convert h to SI units:
BTU
1,055 J
=
 5.68 J / (m 2 sec C)
2
2
(hr ft F) (3,600 sec) (.3048 m) (5/9 C)
Thus h1=h2=28.39 J/(sec m2 oC). The effective thermal resistance
is given by
1
0.1m
0.1
1
R



2
28.39( J /( sm C ) 1.7 J /( smC ) 1.7 28.39
 0.188 (m 2 °C)/W
and the heat flux is
50°C-100°C
2
qx  (T
T
) / R= =
265.8
W/m
,2 ,2
0.188(m 2 °C)/W
 2001, W. E. Haisler
44
Chapter 7: Heat Transfer Applications in a Solid
2. Same as Case 1 above except copper and teflon layers:
1
0.1 0.1
1
R



 0.47 (m2 C)/W
28.39 398 0.25 28.39
50-100°C
2
qx  (T
 T ) / R= =
106.2
W/m
,2 ,1
0.47(m 2 °C)/W
k
3. Same as Case 2 except BC are
TA=200 C and TC=25 C:
L
k
1
L
1
2
2
TA
0.1 0.1
R

 0.40 (m2 C)/W
398 0.25
x
T
B
T
copper
teflon
C
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
25  200°C
2
qx  (T  T ) / R  
=
437.5
W/m
C
A
0.40(m 2 °C)/W
45
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
46
Example 4. Same as Case 3 except determine L2 such that
qx=200 W/m2.
25°C-200°C
2
R  (T  T ) / q x  

0.875
(°C
m
)/W
2
C
A
200W/m
L
L
R  0.875(sm 2 °C )/J  1  2
k k
1 2
L
0.1m
2


398( J /( sm°C)) 0.25( J /( sm°C))
 L  0.22 m
2
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
47
Refrigerator.
a) Inside air temperature must be maintained at 0 F, outside air
temperature is 75 F. Three layers: 1/16" aluminum, 2"
polyurethane foam insulation (see
www.gwsford.com/~fjlawson/matlprops.html#thermal_cond) ,
1/16" aluminum. If the refrigerator is a rectangular box with
outside dimensions of 6'x3'x2.5', how much energy in Watts must
the cooling device provide to the interior of the refrigerator?
b) For the refrigerator above, if you want to reduce the energy
requirement to 1/2 of that required above, what is the required
insulation thickness?
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
outside
k1
k2
k3
T ,1
L1 L 2
L3
k1  k3  247W /(mC )
inside
k2  0.026W /(mC )
L1  L3  1/16"  0.00159m
L2  2"  0.0508m
TA
TB
h1
h2
TC
TD
Al
48
Al
polyurethane
foam
T ,1  75 F  23.89C
T ,2  0 F  17.78C
T ,2 h1  h2  10W /(m 2C )
A  (3' x6 ')2  (2.5 x6)2  (3x 2.5)2
 78.5 ft 2  7.525m 2
1 N Li 1 1 .00159 .0508 .00159 1
R    =




=2.154 m 2C/W
h1 i 1 ki h2 10
247
.026
247 10
qx  
T ,2  T ,1
R
17.78C  (23.89C )
2


19.35
W
/
m
2.154m 2C / W
Q  qx A  19.35W / m2 (7.525m2 )  145.6W
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
49
b) Required R to reduce heat flux by half:
qx  (19.35W / m ) / 2  
2
T ,2  T ,1
R
 R  4.307(m 2C / W )

17.78C  (23.89C )
R
Determine thickness of insulation:
N
Li 1
1
2
R  4.307( m C / W )    
h1 i 1 ki h2
tins
1 .00159 tins .00159 1
=




=0.200013+
(m 2C/W)
10
247
.026
247 10
.026
 tins  0.107 m  4.2"
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
Heat Conduction through a Cylindrical Wall
rB
TB
r
rA
Tinf,1
h1
TA
h2
Tinf,2
k
For steady-state, 2T=0. Radial flow only, T=T(r). Heat
conduction equation in cylindrical coordinates reduces to:
d 2T 1 dT

0

T (r )  C ln r  C
1
2
dr 2 r dr
50
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
51
Assume convection B.C. at interior and exterior surfaces:
Convection B.C. at r=rA (where n  e ):
r
dT
dT
n  q  h (T  T )  (e )  (k
e )k
 h (T  T )
1 A ,1
r
dr r
dr 1 A ,1
Convection B.C. at r=rB (where n  e ):
r
dT
dT
n  q  h (T  T
)  (e )  (  k
e )  k
 h (T  T
)
2 B ,2
r
2 B ,2
dr r
dr
Substitute above B.C. into solution for T(r):
At r=rA:
1
kC
 h (C ln r  C  T )
1 rA 1 1 A
2 ,1
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
1
 h (C ln rB  C  T
)
At r=rB: kC
1 rB
2 1
2 ,2
We can now solve the following two equations for C1 and
C2:
k
(  h ln r )C  (h )C  h T
1 A 1
1 2
1 ,1
r
A
k
(  h ln r )C  (h )C   h T
2 B 1
2 2
2 ,2
r
B
52
 2001, W. E. Haisler
53
Chapter 7: Heat Transfer Applications in a Solid
Take the case of air at 50 oF (10 oC) flowing through a
cylindrical duct, with an internal radius of 5 inches and a wall
thickness of 1 inch. The duct is placed in an environment where
the temperature is 140 oF (60 oC). Assume that the duct is made
of a dense foam material with k=3 BTU/(hr ft oF). Assume that
h1=10 BTU/(Hr ft2 oF) [value depends on flow velocity of air,
interior surface, etc] and h2= 5 BTU/(hr ft2 oF). Solve for T(r)
and the heat flow through the exterior boundary per unit length
of duct.
Substituting these values into the two equations above and
solving for C1 and C2 gives C1= 42.84 and C2=118.29 so that
T (r )  42.84ln r  118.29
(F)
 2001, W. E. Haisler
54
Chapter 7: Heat Transfer Applications in a Solid
Thus, the temperature at the interior and exterior surfaces is
T(rA)=80.8 oF and T(rB)=88.6 oF.
The heat flux per unit length of duct on interior surface is
q(r )  h (T  T ) = 308.19 BTU/hr/ft2
A
1 A ,1
and on the exterior surface
q(r )  h (T  T
) = -257.03 BTU/hr/ft2
B
2 B ,2
Multiplying each heat flux term above by circumference gives
the heat flux per unit length of pipe (Note: energy flow rate at A
= energy flow rate at B):
Qr = radial energy flow rate
= 308.19 BTU/hr/ft2 x (2 x 0.417 ft)
(inside
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
55
boundary)
= -257.03 BTU/hr/ft2 x (2 x 0.5 ft)
(outside boundary)
= 807.5 BTU/hr/ft
(same on both boundaries)
 2001, W. E. Haisler
56
Chapter 7: Heat Transfer Applications in a Solid
Solution Summary
for Steady-State 1-D Heat Transfer in a Solid
What equations apply where for 1-D steady-state heat transfer in
a plane wall?
boundary
AIR
(T1 , h1 )
boundary
conditions:
T  TS
qx  given
interface
boundary
SOLID
AIR
SOLID
k1
k2
d 2T1
d 2T2
k1 2  11  0 k2
  2 2  0
2
...
dx
dx
qx   h(TS  T )
dT1
qx1   k1
dx
convection
conduction
dT2
qx2   k2
dx
conduction
(T 2 , h2 )
boundary
conditions:
T  TS
qx  given
qx  h(TS  T )
convection
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
57
There are two basic approaches that have been discussed in this
class for the solving 1-D steady-state heat transfer in a solid:
I. Solution of the governing differential equations.
1. ODE (for constant k) is given by
d 2T
   0
uniaxial (x) heat flow: k
dx 2
 d 2T 1 dT 
    0

radial heat flow: k 
 dr 2 r dr 


2. Integrate ODE and obtain two constants of integration,
C1 and C2 for each layer: T ( x)  C  C x   (  dx)dx
1
2
(plane) or T (r )  C ln r  C   (  dr )dr (cylindrical).
1
2
3. Evaluate constants by applying specified boundary
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
58
conditions. For a planar problem:
exterior surface temperature, T=given
exterior surface heat flux, n  (k dT i )  given or
dx
exterior surface convection, n  (k dT i )  h(T  T )
S 
dx
If there is more than one layer, apply interface boundary
conditions by equating temperatures (T) and heat flux
k dT at each interface between two layers.
dx
4. From solution for T(x) or T(r): evaluate T at any points
desired, evaluate heat flux q, evaluate total energy flow from
Q=q (area) (time) .
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
59
II. Application of Effective Resistance Method
(only for planar wall or uniaxial flow with no heat source)
1. Find effective resistance for the n-layered composite wall
from
n L
1
1
i
R
 

h
k
h
1 i 1 i
2
where < > means to
include the h term only if there is convection on left (h1) or
right (h2).
 
2. Determine heat flux from q x   1 T where
R
T=temperature difference between known right and left
boundary conditions.
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
60
3. Knowing heat flux from step 2, calculate temperature at
boundary of each wall starting at left (or right) where a
boundary temperature (or heat flux condition) is given, and
continuing for each layer in a composite wall.
For example, if the first layer has convection on left with
fluid/air temperature T,1 and the unknown wall temperature
on left is TA (see composite wall derivation) then solve for TA
from qx  h (T  T ) . Once you know TA, then solve
1 A
,1
for temperature TB on right side of the first layer from
T T
qx  k B A . Continue in same manner until you
1 L
1
have temperatures at boundary of each layer.
 2001, W. E. Haisler
61
Chapter 7: Heat Transfer Applications in a Solid
Summary of boundary conditions for heat transfer. Use the
following figure showing a plane wall with two layers.
T ,1
h1
Layer 1
Layer 2
k1
k2
L1
L2
A
B
x
C
h2
x=0
T1(x)=a 1+b1x
x=x1
x=x2
T ,2
T2(x) =a2+b 2x
For no heat sources, the temperature in each layer is given
by linear functions of x as shown in the figure: T1(x)=a1+a2x
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
62
and T2(x)=b1+b2x.
At left boundary, point A (x=0, outward normal points left,
nx=-1), choice of 3 B.C.
1. Temperature B.C., T(x=0) = TA(given) = T1(x=0) = a1
2. Heat flux B.C., qx(x=0)=qA(given) = -k1(dT1/dx) = -k1b1
3. Convection B.C.,
qx(x=0) in the fluid = qx(x=0) in the solid
(nx) [-k1 (dT1/dx)] = h1 [T1(x=0)-T,1]
(-1) [-k1 b1] = h1 (a1-T,1)
At interface between layers (x=L1), temperature and heat
flux must be continuous (must apply 2 B.C.)
1. Temperature B.C., T1(x=L1) = T2(x=L1)
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
63
a1+b1L1 = a2+b2L1
2. Flux B.C., qx(x=L1) in layer 1= qx(x=L1) in layer 2
-k1 b1 = -k2 b2
At right boundary, point C (x=L1+L2, nx=+1), choice of 3
B.C.
1. Temperature B.C., T(x=0) = TC(given) = T2(x=L1+L2)
= a2 + b2
(L1+L2)
2. Heat flux B.C., qx(x= L1+L2) = qC(given) = -k2(dT2/dx)
= -k2b2
3. Convection B.C.,
qx(x= L1+L2) in the solid = qx(x= L1+L2) in the fluid
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
64
(nx) [-k2 (dT2/dx)] = h2 [T2(x=L1+L2)-T,2]
(+1) [-k2 b2] = h2 [a2+b2(L1+L2)-T,2]
 2001, W. E. Haisler
65
Chapter 7: Heat Transfer Applications in a Solid
All of the above assumes you have defined the temperature in
layer 1 and 2 to be T1(x)=a1+a2x and T2(x)=b1+b2x. If you
have a heat source, then you have to solve the following
differential equation:
k (d2T/dx2) +  = 0
which will give you


T(x)=a +a x -  (  / k )dx dx
1 2
For example, if the heat source  is a constant in the layer so
that (/k)=C, then the integral above is Cx2/2 and the
solution is given by
T(x)=a +a x-Cx2 / 2
1 2