The temperature of the two outer surfaces of a composite slab, consisting of two

Discussion Forum : Heat Transfer

Question -

The temperature of the two 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 $T_{2} a n d T_{1} (T_{2} > T_{1})$ . The rate of heat transfer through the slab, in a steady state is $(\frac{A (T_{2} - T_{1}) K}{x}) f$ , with f which equal to

Options:

A . 1

B .

\frac{1}{2}

C .

\frac{2}{3}

D .

\frac{1}{3}

Answer: Option D
:
D
Equation of thermal conductivity of the given combination

K_{e q} = \frac{l_{1} + l_{2}}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}} = \frac{x + 4 x}{\frac{x}{K} + \frac{4 x}{2 K}} = \frac{5}{3} K

. Hence rate of
flow of heat through the given combination is

\frac{Q}{t} = \frac{K_{e q} . A (T_{2} - T_{1}))}{(x + 4 x)} = \frac{\frac{5}{3} K A (T_{2} - T_{1})}{5 x} = \frac{\frac{1}{3} K A (T_{2} - T_{1})}{x}

On comparing it with given equation we get

f = \frac{1}{3}

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