Question

A composite slab consists of two materials having coefficients of thermal conductivity $K$ and $2K$, thickness $x$ and $4x$ respectively. The temperatures of two outer surfaces of a composite slab are $\text{T}_2$ and $\text{T}_1$ respectively ($\text{T}_2 > \text{T}_1$). The rate of heat transfer through the slab in a steady state is $\left[ \frac{A(\text{T}_2-\text{T}_1)K}{x} \right] f$, where $f$ is equal to}

• A. 1
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Hint:

Thermal conductivity is analogous to electrical resistance: $R = L/(KA)$. Series sum is $R_1 + R_2$.

Answer 1

The Correct Option is D

Step 1: Concept
For slabs in series, the equivalent thermal resistance $R_{eq}$ is the sum of individual resistances: $R = \frac{L}{KA}$.

Step 2: Meaning
$R_1 = \frac{x}{KA}$ and $R_2 = \frac{4x}{(2K)A} = \frac{2x}{KA}$.

Step 3: Analysis
$R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}$. Rate of heat flow $Q = \frac{\Delta T}{R_{eq}} = \frac{T_2 - T_1}{3x/KA} = \frac{KA(T_2 - T_1)}{3x}$.

Step 4: Conclusion
Comparing with the given formula, $f = \frac{1}{3}$. Final Answer: (D)