# 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) 1b) 1/2c) 2/3d) 1/3

Swapnil Saxena
102 Points
12 years ago

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(T2-T1)/3x ---------(ans)

In this way the answer is 1/3

ankit singh
4 years ago
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
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.