Question

The temperature of the two outer surfaces of a composite slab consisting of two materials with thermal conductivity k and 2k and thickness x and 4x respectively are Tâ and Tâ (Tâ > Tâ).
The rate of heat transfer through the slab is [A(Tâ - Tâ)k / x] f. So f = ...... .
(A) 1
(B) 1/2
(C) 2/3
(D) 1/3

Answer 1

Dear Student,

◆ Answer - (D)

f = 1/3

● Explanation -

Rate 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.

Thanks dear...