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Standard VII

Chemistry

Prevention of Corrosion

The temperatu...

Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2 K$ , thickness $x$ and $4 x$ respectively, are $T_{2}$ and $T_{1} (T_{2} > T_{1})$ . Area of cross-section $A$ is the same. The rate of heat transfer through the slab, in steady state, is $F [\frac{A (T_{2} - T_{1}) K}{x}]$ . The value of $F$ is
(answer up to two decimal places)

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Solution

Using Fourier's law of conduction,
$\frac{d Q}{d t} = [\frac{A (T_{2} - T_{1}) K}{x}] = \frac{T_{2} - T_{1}}{R}$ ........(1)
where $R = \frac{x}{K A}$ is the thermal resistance.

Here, the two slabs are in series.
$\Rightarrow R_{e q} = R_{1} + R_{2} = \frac{x}{K A} + \frac{4 x}{2 K A} = \frac{3 x}{K A}$

$∴$ Rate of heat transfer through the composite slab
$i = \frac{d Q}{d t} = \frac{T_{2} - T_{1}}{R_{e q}}$
$= \frac{(T_{2} - T_{1})}{3 x / K A} = \frac{K A (T_{2} - T_{1})}{3 x}$
$= \frac{1}{3} [\frac{A (T_{2} - T_{1}) K}{x}]$ ..........(2)

Comparing (2) with the given rate of heat transfer,
$\Rightarrow F = \frac{1}{3} = 0.33$

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