Chapter 3 Steady-State Conduction

Forsberg Heat Transfer
Chapter 3
Steady-State Conduction




One-Dimensional Conduction
Heat Flow Through a Plane Wall




  T    T    T  T
 k  +  k  +  k  + qgen =  c
x  x  y  y  z  z  t
1-D, steady, no heat generation, constant k
d 2T
Equation reduces to: =0
dx 2




1

, Heat Flow Through a Plane Wall (Cont’d)




BC: At x = 0, T = T1
T1 = C1 (0) + C 2  C 2 = T1
At x = L. T = T2
2
dT T2 = C1L + C 2  C1 = (T2 − T1 ) / L
=0
dx 2
 T −T 
dT T (x) =  2 1  x + T1
= C1  L 
dx
dT kA
T (x) = C1 x + C 2 qx = −k Ax = (T1 − T2 )
dx L




Multilayered Walls




k1 A k2 A k3 A k4 A
q= (T1 − T2 ) q= (T2 − T3 ) q= (T3 − T4 ) q= (T4 − T5 )
L1 L2 L3 L4

 L   L   L   L 
q  1  = (T1 − T2 ) q  2  = (T2 − T3 ) q  3  = (T3 − T4 ) q  4  = (T4 − T5 )
 k1 A   k2 A   k3 A   k4 A 
Summing the left and right sides of these four equations, and rearranging
T1 − T5
q =
 L1 L L L 
 + 2 + 3 + 4 
 k1 A k2 A k3 A k4 A 




2

, Electric-Heat Analogy and Resistance Concept
E
Ohm's Law: I =
R
(I = current, E = potential causing current flow, R = resistance)
kA
Heat flow through a plane wall: q = (T )
L
T
Rearranging, we get q =
 L 
 
 kA 
Comparing this to Ohm's Law
 L 
q  I T  E    R
 kA 
 L 
So,   is the thermal resistance to heat flow
 kA 
for a plane wall




Back to Multilayered Walls



T1 − T5
q =
 L1 L L L 
 + 2 + 3 + 4 
 k1 A k2 A k3 A k4 A 

T1 − T5 is the overall temperature difference across the wall. This
is causing the heat flow.
The denominator is the sum of the resistances between the two
temperatures.
Toverall
This can be generalized to q=
R




3