 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
1
Chapter 7 - Heat Transfer Applications in a Solid
In Chapter 6, the conservation of energy was derived for a
general 3-D body and included both mechanical and thermal
energy. For a body which has only thermal energy flow that is
one-dimensional, the derivation is considerably simpler.
Consider a slab of solid material shown below such that the heat
flow is only in the x direction through the
y
area, Ax (assume insulated along its
x
lateral surfaces). Assume the material has
density,  ( gm / cm3 ) , and specific heat,
C ( J / gm / oC ) . Assume that the material heat
contains a heat source  ( J / gm / s). A
flow
heat energy flux of qx ( J / m2 / s) flows
into and out of the body through the area Area, A x
x
z
Ax . Since the heat flow is only in the x
 2001, W. E. Haisler
2
Chapter 7: Heat Transfer Applications in a Solid
direction, the temperature T is a function of x only.
Conservation of thermal energy requires:
accumulation of
 thermal energy (J)


contained in
 the system for

 the time period


  thermal energy (J) 
entering the system 
 



 

during
the
time
 

 

period



 thermal energy (J)   net generation of
leaving the system   thermal energy (J) in

 


during
the
time

  the system during


period
 
 time time period
 C
 ˆ T 
y
x







heat
flow
Area, A x
ˆ T   Ax x
  C
t t
t

 qx x  Ax t   qx x x  Ax t   Ax xt
x
z
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
3
Note that each term in the above has units of J. Dividing by
Ax xt and taking the limit as x, t  0 gives
T
qx
 Ĉ  
 
t
x
1-D Conservation of thermal energy,
Heat Transfer in a Solid
Note that T=T(x,t).
We have 1 governing PDE (Conservation of Energy) and 2
unknowns (T and q x ). We need another equation!
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
4
For the static case (no mass velocity, v  0 ), Conservation
of Energy in 3-D becomes:

C
T
   q   Heat Conduction in Solid
t
Unknowns:
T  T ( x, y, z, t ) =temperature
q ( x, y, z, t )  qxi  q y j  qz k is the heat flux vector (W/m2)
Known:
C is the specific heat of the material,  is mass density.
  ( x, y, z ) =heat source (W/kg).
Notice that we have 1 governing PDE but there are a total of
4 unknowns (T, q x , q y , q y )!! And they are functions of x, y,
z, and t.
 2001, W. E. Haisler
5
Chapter 7: Heat Transfer Applications in a Solid
We need 3 more equations. Fourier made an experimental
observation which relates temperature (T) and heat flux (q). For
the case of heat flow in one direction only (say x direction), he
observed that the heat flux was proportional to the negative of
the temperature gradient:
qx
wall
qx
T1
T2
dT
qx   k
dx
k
dT T2  T1

dx
L
dT
L

x
dx
The proportionality constant k is called the coefficient of thermal
conductivity and is an experimental material "constant." Units: k
T
has units of J/(m sec K) = W/(m K). The equation q  k
x
x
is called a constitutive relation because it relates certain
o
o
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
6
variables by a material property. For an isotropic material
(same properties in all directions), we find experimentally
T
T
T
that q  k
, q  k
, and q  k
 q   k T
x
z
x y
z
y
For the general 3-D case, Fourier’s law of heat conduction
states that the heat flux due to conduction is proportional to
the negative of the temperature gradient. Negative in above
indicates that positive temperature gradient produces negative
heat flux (heat flow from hot to cold).

With Fourier’s law:  C
T
   (kT )   .
t
If k does not vary spatially, then
  (kT )  k  (T )  k 2T
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
7
2
2 2


where 2     


 x2  y 2  z 2
2T  2T  2T

and
2T 


 x2  y 2  z 2
Heat transfer equation becomes

C
T
 k  2T  
for constant k
t
For steady state-heat conduction (no variation with time)
k2T    0
Steady-state, constant k
d 2T

d 2T

1-D, steady-state, constant k
k 2    0 
2
dx
k
dx
 2001, W. E. Haisler
8
Chapter 7: Heat Transfer Applications in a Solid
The previous differential equations for heat transfer in a
solid require knowledge of:
Boundary Conditions (values of temperature, T; or the heat
flux on the boundary):
n
1) T  TS ( x, y, z , t )
Types of B.C.
1) specified temperature
2) specified heat flux
3) convection
2) n  q  constant
3) n  q  h(T  T )
s 
TS
T

Note:T is the
temperature of
the environment
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
1. Specified surface temperature, T(x,y,z,t)=TS(x,y,z,t)
2. Specified heat flux on surface with unit outward
normal n :
n  q  n  (kT )  known / given function( x, y, z, t )
nq  0
(special case of insulated boundary)
3. Specified heat flux due to convective heat transfer:
n  q  h(T  T )
s 
where T = surface temperature (unknown)
s
T = far-field temperature (specified/known)

h = heat transfer coefficient (from experiment)
Note: h has units of J/(m2 sec oK) = W/(m2 oK).
9
 2001, W. E. Haisler
10
Chapter 7: Heat Transfer Applications in a Solid
Difference between Conduction and Convection
Conduction
 Occurs in a Solid
 Relates heat flux q x to
temperature gradient in solid
T1
solid
qx
dT T2  T1

dx
L
Convection
 Occurs in a Fluid (say, air)
 Relates heat flux q x to
temperature difference in fluid
air
solid
qx
TS
T2
T
L
x
x T  TS  T
qx
qx  k
k

dT
dx
dT
dx
qx
q x  h (TS  T )
h
TS  T
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
11
 Both are assumptions verified by experiment
 Both "laws" contain experimentally obtained material "constants"
Note: AT the boundary between a solid and a fluid, heat flux must
be the same. Therefore, on a boundary whose unit outward normal
is i (as shown above):
qx ( solid )
 qx ( fluid )
dT
k
dx
 h(TS  T )  h(T  TS )
surface
solid
dT
dx
fluid
TS
T
 2001, W. E. Haisler
12
Chapter 7: Heat Transfer Applications in a Solid
Example: 1-D, Steady-state heat conduction in an insulated bar
with specified temperatures on ends of bar (TL and TR, as
shown):
GEOMETRY AND BOUNDARY CONDITIONS
y
insulated, q y =0
a
x
TL
b
TR
L
z
insulated, q =0
z
The heat conduction equation reduces to  2T  0 or
 2T  2T  2T


0
 x2  y 2  z 2
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
Insulated B.C. on lateral surfaces
T
y=a/2, q  k
 0  T  T ( y) ,
y
y
T
 0  T  T ( z)
z=a/2, q  k
z
z
This implies heat flow only in x direction so that the heat
 2T d 2T

 0.
conduction equation reduces to
 x 2 dx 2
Solving the heat conduction equation gives
T ( x)  C x  C
1
2
Apply temperature B.C. on end surfaces
13
 2001, W. E. Haisler
at x 0:
T ( x)  C x  C
Chapter 7: Heat Transfer Applications in a Solid
T (0)  T  C (0)  C
2
L 1
1
2
14
 C T
2 L
T (L)  T  C L  C
2
R 1
 T  C L T
 C  (T T )/ L
R 1
L
1 R L
Substituting constants of integration back into T(x) gives the
solution for T(x) as
 TR  TL 
T ( x)  TL  
 x
L
at x  L :


If the temperature at the left end is TL=45C and the right
 35C  45C 
end, TR=35C, and L=1 m; then T ( x)  45C  
x
1m


Or, T ( x)  45C  (10C / m) x
(a linear function in x)
The heat flux in the bar is given by
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
15
dT
dT
dT
TR  TL
q  k    k ( i 
j
k  k
i
dx
dy
dz
L
If the temperature at the left end is TL=45C and the right
end, TR=35C, L=1m; and the bar is made of aluminum with
k=247 W/(m-C), then the heat flux is given by
TR  TL
J 35C  45C
q  k
i  247
i  2, 470(W / m 2 )i
L
msC
1m
 2.47( kW / m 2 )i
Note that the heat flux is positive, i.e., to the right (since the
bar is cooler on the right).
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
16
Another Example: Suppose the bar above contained a
constant heat source of   5W / cm3, a=b=2 cm, L=1 m and
the bar is made of aluminum with a thermal conductivity of
k=247 W/m/C. The temperature at the left end is TL=45C
and the right end, TR=35C. Determine T(x) in the bar.
Assuming steady state, the heat transfer equation is given by
d 2T
d 2T
k
  
   / k
or
dx 2
dx 2
For this case,   5W / cm3  5kW / m3 (a constant). The
ODE becomes
d 2T
5kW / m3
o
2




/
k




20.243(
C
/
m
)
2
dx
247W /(mC )
Integrating the ODE twice gives:
T ( x)  10.121( oC / m2 ) x 2  C1x  C2
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
Applying the boundary conditions:
At x=0m:
17
T ( x)  10.121( oC / m2 ) x 2  C1x  C2
T (0)  45 oC  10.121( oC / m 2 )(0) 2  C1 (0)  C2  C2  45 oC
At x=1m:
T (1m)  35 oC  10.121( oC / m 2 )(1m) 2  C1 (1m)  C2
 35 oC  10.121( oC / m 2 )(1m) 2  C1 (1m)  45 oC
 C1  0.121( oC / m)
Substituting the constants of integration into T(x) gives:
T ( x)  10.121( oC / m 2 ) x 2  0.121( oC / m) x  45 oC
Note that for a constant heat flux  , the temperature distribution
is quadratic in x.
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
The heat flux becomes
45 oC

18
35 oC
x
dT
qx  k
 247(W /(mC ))  20.242( oC / m 2 ) x  0.121( oC / m) 
dx
at x=0 m: qx  29.9(W / m2 ) (to left)
x=0.5 m: qx  2, 470(W / m2 ) (to right)
x=1 m: qx  4,970(W / m2 ) (to right)
Note: W=J/s
Question: Why is heat flowing to the left at x=0 even though the
bar is hotter on the left end compared to the right end (and
therefore you might expect heat flow from left to right)?
 2001, W. E. Haisler
19
Chapter 7: Heat Transfer Applications in a Solid
If we increase to   10kW / m3 , the solution becomes:
T ( x)  20.242( oC / m 2 ) x 2  10.242( oC / m) x  45 oC . Plotting
these three cases gives:
  10kW / m3
  5kW / m3
T(x)
  0

45 oC
x
35 oC
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
20
Some values of k and h:
Material
k, J/(m sec oC)
Silver
428
Copper
398
Aluminum
247
Nickel
89.9
Iron
80.4
Glass
Polyethylene
Teflon
Nylon
Polystyrene
Polypropylene
Material
h
Air (free convec) 5-25 W/(m2 oK)
Air (forced conv) 25-250 “
1.7
0.38
0.25
0.24
0.13
0.12
W = J/sec = 3.41 BTU/hr, 1 HP = 550 ft-lb/sec = 746 W, 1 m = 3.281 ft,
1 W/(m2 oK) = 0.176 Btu/(hr ft2 oR), 1 W/(m oK) = 0.578 Btu/(hr ft oR),
o
K = 5/9 oR = oC + 273 = 5/9 (460 + oF), oC = 5/9 (oF - 32)
 2001, W. E. Haisler
Chapter 7: Heat Transfer Applications in a Solid
Note on conversion of terms like k (
21
W
W
) and h( 2 )
mC
m C
Since K and C are different only by a constant (i.e., K=C+273), then a
variable that has units of 1/K is same (equal to) as 1/C. This is
because a one-degree change in K is equal to a one-degree change in
C.
Warning: this is not the case with F and C since F=(9/5) C + 32.
Hence a one degree change in F is equal a 5/9 degree change in C.
So for a variable with 1/F units, there is a 9/5 multiplier to convert to
1/C units.
Suppose we have h=1 BTU/(hr ft2 oF). Convert to metric:
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)