AME 60634
Int. Heat Trans.
1-D Steady Conduction: Plane Wall
Governing Equation:
d 2T
=0
2
dx
Dirichlet Boundary Conditions:
T(0) = Ts,1 ;
Solution:
T(x)  Ts,1  Ts,2  Ts,1
x
L
dT k
 Ts,1  Ts,2 
Heat Flux:
dx L
dT kA

q

kA

Ts,1  Ts,2 
Heat Flow:
x
dx L

T(L) = Ts,2
temperature is not a function of k
q
x k
heat flux/flow are a function of k
Notes:
• A is the cross-sectional area of the wall perpendicular to the heat flow

• both heat flux and heat flow are uniform  independent of position (x)
• temperature distribution is governed by boundary conditions and length
of domain  independent of thermal conductivity (k)
D. B. Go
1
AME 60634
Int. Heat Trans.
1-D Steady Conduction: Cylinder Wall
Governing Equation:
1 d æ dT ö
d æ dT ö
ç kr ÷ = 0 Þ ç r ÷ = 0
r dr è dr ø
dr è dr ø
Dirichlet Boundary Conditions:
T(r1)  Ts,1 ;
T(r2 )  Ts,2
Ts,1  Ts,2  r 
Solution: T(r) 
ln   Ts,2
ln r1 r2  r2 

dT k Ts,1  Ts,2 


q

k

heat flux is non-uniform
Heat Flux: r
dr
r ln r2 r1

2Lk Ts,1  Ts,2 
dT


q

kA

2

rL
q

heat flow is uniform
Heat Flow: r
r
dr
ln r2 r1 

qr 2k Ts,1  Ts,2 

qr  
heat flow per unit length
L
ln r2 r1 
Notes:
 • heat flux is not uniform  function of position (r)
• both heat flow and heat flow per unit length are uniform  independent
of position (r)
D. B. Go

2
AME 60634
Int. Heat Trans.
1-D Steady Conduction: Spherical Shell
Governing Equation:
1 d æ 2 dT ö
d æ 2 dT ö
ç kr
÷ = 0 Þ çr
÷=0
r 2 dr è
dr ø
dr è dr ø
Dirichlet Boundary Conditions:
T(r1)  Ts,1 ;
Solution:
1 r1 r
T(r)  Ts,1  Ts,1  Ts,2 
1 r1 r2 

Heat Flux: q
r k

T(r2 )  Ts,2
k Ts,1  Ts,2 
dT
 2
dr r 1 r1  1 r2 
4k Ts,1  Ts,2 
dT
2


q

kA

4

r
q

Heat Flow: r
r
dr
1 r1  1 r2 

heat flux is non-uniform
heat flow is uniform
Notes:
• heat flux is not uniform  function of position (r)
• heat flow is uniform  independent of position (r)
D. B. Go
3
AME 60634
Int. Heat Trans.
D. B. Go
Thermal Resistance
4
AME 60634
Int. Heat Trans.
Thermal Circuits: Composite Plane Wall
Circuits based on assumption of
(a) isothermal surfaces normal to x direction
or
(b) adiabatic surfaces parallel to x direction
1
LE kF A kG A
LH
Rtot 
 



kE A 2LF 2LG  k H A

Rtot 
1
1
1

 2LE 2LG 2L H 
2L
2L
2L
E
F
H





  
 

kE A kF A k H A  kE A kG A k H A  

Actual solution for the heat rate q is bracketed by these two approximations
D. B. Go

5
AME 60634
Int. Heat Trans.
Thermal Circuits: Contact Resistance
In the real world, two surfaces in contact do not transfer heat perfectly
 
Rt,c
R
TA  TB
 Rt,c  t,c
qx
Ac

Contact Resistance: values depend on materials (A and B), surface roughness,
interstitial conditions, and contact pressure  typically calculated or looked up
Equivalent total thermal resistance: Rtot 
D. B. Go
R
LA
L
 t,c  B
kA Ac Ac kB Ac
6
AME 60634
Int. Heat Trans.
D. B. Go
7
AME 60634
Int. Heat Trans.
D. B. Go
8
AME 60634
Int. Heat Trans.
Fins: The Fin Equation
• Solutions
D. B. Go
9
AME 60634
Int. Heat Trans.
Fins: Fin Performance Parameters
• Fin Efficiency
– the ratio of actual amount of heat removed by a fin to the ideal
amount of heat removed if the fin was an isothermal body at the base
temperature
• that is, the ratio the actual heat transfer from the fin to ideal heat transfer
from the fin if the fin had no conduction resistance
hf º
qf
q f ,max
=
qf
hA f q b
• Fin Effectiveness
– ratio of the fin heat transfer rate to the heat transfer rate that would exist
without the fin
ef º
qf
q f ,max
=
qf
hAc,bq b
=
Rt,b
Rt, f
• Fin Resistance
– defined using the temperature difference between the base and fluid as
the driving potential
Rt, f º
D. B. Go
qb
qf
=
1
hA f h f
10
AME 60634
Int. Heat Trans.
D. B. Go
Fins: Efficiency
11
AME 60634
Int. Heat Trans.
D. B. Go
Fins: Efficiency
12
AME 60634
Int. Heat Trans.
Fins: Arrays
• Arrays
– total surface area
At = NA f + Ab
N º number of fins
Ab º exposed base surface (prime surface)
– total heat rate
qt = Nh f hA f qb + hAbq b = ho hAtq b =
qb
Rt,o
– overall surface efficiency
NA f
ho = 11- h f )
(
At
– overall surface resistance
Rt,o =
D. B. Go
qb
qt
=
1
hAtho
13
AME 60634
Int. Heat Trans.
Fins: Thermal Circuit
• Equivalent Thermal Circuit
• Effect of Surface Contact Resistance
q
qt = ho(c )hA f q b = b
Rt,o(c )
NA f æ h f ö
ho(c ) = 1ç1- ÷
At è C1 ø
æ R¢¢ ö
C1 = 1- h f hA f ç t,c ÷
è Ac,b ø
Rt,o =
D. B. Go
1
hAtho(c )
14
AME 60634
Int. Heat Trans.
Fins: Overview
• Fins
– extended surfaces that enhance fluid heat transfer
to/from a surface in large part by increasing the
effective surface area of the body
– combine conduction through the fin and
convection to/from the fin
• the conduction is assumed to be one-dimensional
• Applications
– fins are often used to enhance convection when h is
small (a gas as the working fluid)
– fins can also be used to increase the surface area
for radiation
– radiators (cars), heat sinks (PCs), heat exchangers
(power plants), nature (stegosaurus)
Straight fins of (a) uniform
and (b) non-uniform cross
sections; (c) annular
fin, and (d) pin fin of nonuniform cross section.
D. B. Go
15
AME 60634
Int. Heat Trans.
Fins: The Fin Equation
• Solutions
D. B. Go
16