The temperature of the two outer surface of a composite slab consisting of two materials having coefficient of thermal conductivity k and 2k and thickness x and 4x respectively. T2 and t1 (t2 > t1). The rate of heat transfer through the slab in steady state is y, then y is equal to
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Answer:
Let the temperature of common interface be TK rate of heat flow H=
t
Q
=
l
KAΔT
∴H
1
=(
t
Q
)
1
=
4x
2KA(T−T
1
)
and H
2
=(
t
Q
)
2
=
x
KA(T
2
−T)
In steady-state, the rate of heat flow should be the same in the whole system i.e.,
H
1
=H
2
or
4x
2KA(T−T
1
=
x
KA(T
2
−T)
2
T−T
1
=T
2
−T or T−T
1
=2T
2
−2T
T=
3
2T
2
+T
1
........(i)
Hence heat flow from composite slab is
H=[
x
KA(T
2
−T)
]=
x
KA
(T
2
−
3
2T
2
+T
1
) [from eqn.(i)]
=
3x
KA
(T
2
−T
1
) ..........(ii)
According to question,H=[
x
A(T
2
−T
1
)K
]f ...........(iii)
by comparing eqs.(ii) and (iii) we get f=
3
1
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