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Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivities K and 2K, and thickness x and 4x respectively are T2 and T1(T2>T1) as shown in figure. The cross-sectional area of the slab is A. The rate of heat transfer through the slab in a steady state is [A(T2T1)Kfx], where f is equal to  



Solution

Let T be the temperature of the interface.
We know that , Fourier conduction equation is dQdt=KAdTdx

Since the rate of flow of heat across the interface is same,
KA(T2T)x=2KATT14x
T=2T2+T13
Substituting the value of T in dQdt for the slab of thickness 4x we get,
dQdt=2K×A(TT1dx)
=2KA×(2T2+T13T1)4x
dQdt=KA3x×(T2T1)
Comparing this with the given expression KA(T2T1)fx,
we get, f=3

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