Question

The temperature drop through a two layers furnace wall is ${900}^{\circ }C$. Each layer is of equal area of cross-section. Which of the following actions 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

H = rate of heat flow $=\frac{900}{\frac{l_i}{K_{i}A}+\frac{l_0}{K_{0}A}}...\left(1\right)$

Now, $100-\theta =\frac{H{l}_i}{K_{i}A}...\left(2\right)$

$\text{or}\theta =1000-\left[\frac{900}{\frac{l_i}{K_{i}A}+\frac{l_0}{K_{0}A}}\right]\frac{l_i}{K_{i}A}=1000-\frac{900}{1+\frac{l_0}{K_0}\frac{K_i}{l_i}}$

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