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 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)

By increasing thickness of inner layer

H = rate of heat flow
$=\frac{900}{\frac{l_i}{K_{i}A}+\frac{l_0}{K_{0}A}}$
Now, $1000-\theta =\frac{H{l}_i}{K_{i}A}$
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 $\left(l_i\right)$

Why this question? Key Concept: Rate of heat flow is same for both- the effective thermal resistance and the individual thermal resistances in series combination.