Question

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 are T₂ and T₁ (T₂> T₁). The rate of heat transfer through the slab, in steady state is $\left( \frac{A(T_{2}–T_{1})K}{x} \right).f$with f equal to –

• A. 1
• B. 1/2
• C. 2/3
• D. 1/3

Answer 1

Answer: (D)

1/3

Answer 2

Series Req = R₁ + R₂ = $\frac{x}{KA} + \frac{4x}{2KA} = \frac{3x}{KA}$Rate of heat $\frac{dQ}{dt}$ = $\frac{T_{2}–T_{1}}{R_{eq}}$ = $\frac{T_{2}–T_{1}}{3x}$ KA

given $\frac{dQ}{dt}$ = $\frac{A\lbrack T_{2}–T_{1}\rbrack K}{x}$f = $\frac{T_{2}–T_{1}}{3x}$ KA

f = $\frac{1}{3}$