Page 321 - Bird R.B. Transport phenomena

§10.6 Heat Conduction Through Composite Walls 305

We now assume that k QU k , and k are constants. Then we integrate each equation over
23
l2
the entire thickness of the relevant slab of material to get
Region 01: T o - T, = (Ю.6-8)

Region 12: Т -Т 2 = (10.6-9)
г

Region 23: T - T = (10.6-10)
2 3 y
In addition we have the two statements regarding the heat transfer at the surfaces ac-
cording to Newton's law of cooling:

At surface 0: T a — (10.6-11)
К
At surface 3: T - T = (10.6-12)
3 b
Addition of these last five equations then gives
т ^ " т T = ~ I 1 l~X 0 X , 1 (10.6-13)
q \j-
b o + 7-
К 2Ъ П 3
or

i i

Sometimes this result is rewritten in a form reminiscent of Newton's law of cooling, ei-
2
ther in terms of the heat flux q (J/m • s) or the heat flow Q (J/s):
o
0
= U(T -T ) or Q 0 = U(WH)(T -T ) (10.6-15)
b
b
a
a
The quantity U, called the "overall heat transfer coefficient," is given then by the follow-
ing famous formula for the "additivity of resistances":
ao.6-16)
U
Here we have generalized the formula to a system with n slabs of material. Equations
10.6-15 and 16 are useful for calculating the heat transfer rate through a composite wall
separating two fluid streams, when the heat transfer coefficients and thermal conductivi-
ties are known. The estimation of heat transfer coefficients is discussed in Chapter 14.
In the above development it has been tacitly assumed that the solid slabs are con-
tiguous with no intervening "air spaces." If the solid surfaces touch each other only at
several points, the resistance to heat transfer will be appreciably increased.

EXAMPLE 10.6-1 Develop a formula for the overall heat transfer coefficient for the composite cylindrical pipe
wall shown in Fig. 10.6-2.
Composite Cylindrical
W a l l s SOLUTION
An energy balance on a shell of volume lirrL Ar for region 01 is

Region 01: q \ * lirrL - q \ M-' ?r(r + Ar)L = 0 (10.6-17)
2
r r
r r+