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: Understanding the Concept:
Heat flows in a steady state through series slabs similarly to current through series resistors.
The total thermal resistance is the sum of the individual thermal resistances.
Step 2: Key Formula or Approach:
Thermal resistance $R = \frac{d}{KA}$.
Equivalent resistance $R_{\text{eq}} = R_1 + R_2$.
Rate of heat transfer $H = \frac{\Delta T}{R_{\text{eq}}}$.
Step 3: Detailed Explanation:
Let cross-sectional area be $A$.
Resistance of first slab: $R_1 = \frac{x}{K \cdot A}$.
Resistance of second slab: $R_2 = \frac{4x}{(2K) \cdot A} = \frac{2x}{KA}$.
Total equivalent thermal resistance: \[ R_{\text{eq}} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA} \] The heat transfer rate $H$ across the entire slab is: \[ H = \frac{\Delta T}{R_{\text{eq}}} = \frac{T_2 - T_1}{\frac{3x}{KA}} \] Rearranging the terms: \[ H = \frac{KA(T_2 - T_1)}{3x} \] We can express this in the format given in the question: \[ H = \left[ \frac{A(T_2 - T_1)K}{x} \right] \cdot \frac{1}{3} \] Comparing this to the provided expression $H = \left[ \frac{A(\text{T}_2-\text{T}_1)K}{x} \right] f$, we identify the factor: \[ f = \frac{1}{3} \] Step 4: Final Answer:
The factor $f$ is $1/3$.