Question

The temperature drop through a two layer furnace wall is $900{}^{\circ }C$. Each layer is of equal area of cross-section. Which of the following action(s) will result in lowering the temperature $\theta$ of the interface?

• A. By increasing the thermal conductivity of outer layer
• B. By increasing the thermal conductivity of inner layer
• C. By increasing thickness of outer layer
• D. By increasing thickness of inner layer

Answer 1

Answer: (A), (D)

The correct options are
A By increasing the thermal conductivity of outer layer
D By increasing thickness of inner layer

Let the temperature of the interfeace be $\theta$ and thickness of inner and outer layer be $l_i$ and $l_0$ respectively.

Rate of heat flow for both layers $\left(H\right)=\frac{900}{\frac{l_i}{K_{i}A}+\frac{l_0}{K_{0}A}}$ ...(i)
Now rate of flow for inner layer
$H_i=\frac{1000-\theta }{\frac{l_i}{K_{i}A}}$
And we know heat flow is in accordance with the total heat flow. So $H_i=H$
$\Rightarrow 1000-\theta =\frac{H{l}_i}{K_{i}A}$
From eq(i)
$\theta =1000-[\frac{900}{\frac{l_i}{K_{i}A}+\frac{l_0}{K_{0}A}}\frac{l_i}{K_{i}A}]$
$\theta =1000-\frac{900}{1+\frac{l_0}{K_0}\frac{K_i}{l_i}}$

Now, we can see that $\theta$ decreases by increasing thermal conductivity of outer layer $\left(K_0\right)$ and thickness of inner layer $\left(l_i\right)$