The temperature drop through a two–layer furnace wall is $900 ° C$ . Each layer is of equal area of cross–section. Which of the following actions will result in lowering the temperature $^{'} θ^{'}$ of the interface?

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By increasing the thermal conductivity of outer layer

By increasing thickness of outer layer

By increasing thickness of inner layer

By increasing the thermal conductivity of inner layer

answer is A, D.

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Detailed Solution

$H =$ Rate of Heat flow $= \frac{900}{(\frac{μ}{K_{i} A}) + (\frac{l_{e}}{K_{e} A})}$
Now, $1000 - θ = \frac{H l_{i}}{K_{i} A} (o r)$
$θ = 1000 - [\frac{900}{(\frac{l_{1}}{k_{1} A} + \frac{l_{0}}{K_{0} A})}] \frac{l_{i}}{K_{i} A} = 1000 - \frac{900}{1 + \frac{l_{0} K_{i}}{K_{0} l_{i}}}$
So we can see that $' θ'$ can be decreased by increasing thermal conductivity of outer layer $(K_{0})$ and thickness of inner layer $(l_{i})$