The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficient of thermal conductivity $K$ and $2 K$ and thickness $X$ and $4 X$ respectively are $T_{2}$ and $T_{1} (T_{2} > T_{1})$ . The rate of heat transfer through the slab in a steady state is $(\frac{A (T_{2} - T_{1}) K}{X}) f$ , then $f$ is equal to

1

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\frac{1}{2}

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\frac{2}{3}

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{3}$
At steady state the heat current entering and leaving the slab will be equal.

For two individual slabs in series combination, resultant thermal resistance can be written as,
$R_{e q} = R_{1} + R_{2}$
$\Rightarrow \frac{5 X}{K_{e q} A} = \frac{X}{K A} + \frac{4 X}{2 K A}$
$∴ K_{e q} = \frac{5 K}{3}$
At steady state, the rate of heat transfer is given by,
$˙ Q = \frac{A (T_{2} - T_{1}) K_{e q}}{5 X}$
$\Rightarrow ˙ Q = \frac{5 A (T_{2} - T_{1}) K}{15 X}$
$\Rightarrow ˙ Q = \frac{A (T_{2} - T_{1}) K}{3 X}$
Given
$˙ Q = [\frac{A (T_{2} - T_{1}) K}{X}] f = \frac{A (T_{2} - T_{1}) K}{3 X}$

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

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