Question

The temperature drop through a two layer furnace wall is 900⁰C. Each layer is of equal area of cross-section. Which of the following actions will result in lowering the temperature q 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 decreasing thickness of inner layer

Answer 1

Answer: (A)

By increasing the thermal conductivity of outer layer

Answer 2

H = rate of heat flow

= $\frac{900}{\frac{\mathcal{l}_{i}}{K_{i}A} + \frac{\mathcal{l}_{0}}{K_{0}A}}$

Now 100 – q = $\frac{H\mathcal{l}_{i}}{K_{i}A}$

or q = 1000 – $\frac{\mathcal{l}_{i}}{K_{i}A}$

= 100 – $\frac{900}{1 + \frac{\mathcal{l}_{0}}{K_{0}}\frac{K_{i}}{\mathcal{l}_{i}}}$

Now, we can see that q can be decreased by increasing thermal conductivity of outer layer (K₀) and thickness of inner layer (l_i).