The temperature of the outer surfaces of a composite slab consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x respectively are T2 and T1(T2>T1). The rate of heat transfer through the slab in steady state is [{AK(T2-T1)}/x]f, where f is equal to

a) 1

b) 1/2

c) 2/3

d) 1/3

Question

The temperature of the outer surfaces of a composite slab consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x respectively are T2 and T1(T2>T1). The rate of heat transfer through the slab in steady state is [{AK(T2-T1)}/x]f, where f is equal to

a) 1

b) 1/2

c) 2/3

d) 1/3

Swapnil Saxena · Answer

Solution: When two materials are joined in series, then the reciprocal of their conductanes are added. Here the individual conductances of these two slabs are = KA/x and 2KA/4x the equivalent conducatnce =1/ (x/KA + 4x/2KA) = 1/( x/KA + 2x/KA )= 1/ (3x/KA) = KA/3x The the rate of transfer of heta through the slab = KA(T 2 -T 1 )/3x ---------(ans) In this way the answer is 1/3

ankit singh · Answer

ate of heat flow in 1st outer surface is - dQ1/dt = kA(T-T1)/x Rate of heat flow in 2nd outer surface is - dQ2/dt = 2kA(T2-T)/4x In steady state, dQ1/dt = dQ2/dt kA(T-T1)/x = 2kA(T2-T)/4x (T-T1) = (T2-T)/2 T = (2T1+T2)/3 Rate of heat transfer in steady state is given by - dQ/dt = kA(T-T1)/x dQ/dt = kA/x × [(2T1+T2)/3 - T1] dQ/dt = kA/x (T2-T1)/3 dQ/dt = [A(T2-T1)k/x] 1/3 Therefore, value of f must be 1/3.

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