HEAT TRANSFER 789
dharm
\M-therm\Th15-1.pm5
Since the quantity of heat transmitted per unit time through each slab/layer is same, we
have
Q = kAt t
L
kAt t
L
kAt t
L
A
A
B
B
C
C
.( 12 − )=. ( 23 − )=. ( 34 − )
(Assuming that there is a perfect contact between the layers and no temperature drop occurs
across the interface between the materials).
Rearranging the above expression, we get
t 1 – t 2 = QL
kA
A
A
.
.
...(i)
t 2 – t 3 =
QL
kA
B
B
.
.
...(ii)
t 3 – t 4 =
QL
kA
C
C
.
.
...(iii)
Adding (i), (ii) and (iii), we have
(t 1 – t 4 ) = Q
L
kA
L
kA
L
kA
A
A
B
B
C
...C
L ++
N
M
O
Q
P
or Q =
At t
L
k
L
k
L
k
A
A
B
B
C
C
() 14 −
++
L
N
M
O
Q
P
...(15.27)
or Q =
()
...
()
[]
tt
L
kA
L
kA
L
kA
tt
A RRR
A
B
B
C
C
th A th B th C
14 − 14
++
L
N
M
O
Q
P
= −
−−−++
...(15.28)
If the composite wall consists of n slabs/layers, then
Q =
[]tt()
L
kA
n
n
11
1
− +
∑
...(15.29)
A
C
B
Q
kA kC kG
kB D
kD
F
kF
E
kE
G
Q
Composite wall
Rth–A
Rth–C
Rth–B
Rth–D
Rth–E
Rth–G
Rth F–
Fig. 15.5. Series and parallel one-dimensional heat transfer through a composite wall and electrical analog.