Question

A cuboidal slab has square faces of area $A$ and a width $d$. Across the square faces a temperature of $T$ is maintained. The conductivity of the material varies linearly with the distance from each face from $K$ to $2K$ at the middle. The rate of heat flow is

• A. $\frac{KAT}{d \ln 2}$
• B. $\frac{KAT \ln 2}{d}$
• C. $\frac{3KAT}{2d}$
• D. $\frac{3KAT}{d}$

Answer 1

Answer: (A)

$\frac{KAT}{d\ln2}$

Answer 2

1. Temperature Drop & Effective Resistance:
   For one-dimensional conduction, the effective thermal resistance is

   $$R_{\text{eff}}=\int_0^d \frac{dx}{A\,K(x)},$$

   where $K(x)$ is the thermal conductivity.

2. Conductivity Variation:
   The conductivity varies linearly from $K$ at the faces ($x=0$ and $x=d$) to $2K$ at the midplane ($x=\frac{d}{2}$). For $0\le x\le \frac{d}{2}$:

   $$K(x)=K\left(1+\frac{2x}{d}\right).$$

   By symmetry, the region $\frac{d}{2}\le x\le d$ gives an identical contribution.

3. Calculate Resistance for Half-slab:

   $$R_{1}=\int_0^{d/2} \frac{dx}{A\,K\left(1+\frac{2x}{d}\right)}.$$

   Use substitution: Let

   $$u=1+\frac{2x}{d}\quad\Rightarrow\quad dx=\frac{d}{2}du.$$

   When $x=0$, $u=1$; when $x=\frac{d}{2}$, $u=2$. Thus,

   $$R_{1}=\frac{d}{2A\,K}\int_1^2\frac{du}{u}=\frac{d}{2A\,K}\ln2.$$

4. Total Resistance:

   $$R_{\text{eff}}=2R_{1}=\frac{d\,\ln2}{A\,K}.$$

5. Heat Flow:
   With a temperature difference $T$, Fourier’s law gives:

   $$\dot{Q}=\frac{T}{R_{\text{eff}}}=\frac{A\,K\,T}{d\,\ln2}.$$

Thus, the rate of heat flow is $\displaystyle \frac{KAT}{d\ln2}$.