Question

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities $K$ and $2\, K,$ respectively. The equivalent thermal conductivity of the slab is

• A. $\frac{2}{3} K$
• B. $\sqrt 2 K$
• C. 3K
• D. $\frac{ 4}{3} K$

Answer 1

The Correct Option is D

To calculate the equivalent thermal conductivity of a compound slab with two different materials of equal thicknesses, we need to consider the conductivities and how they contribute to the overall conductivity of the slab.

1. The slab is made up of two materials with equal thickness. Let's denote the thickness of each material by d. Therefore, the total thickness of the slab is 2d.
2. The thermal conductivities of the materials are given as K and 2K, respectively.
3. Since the slabs are in series, the heat current is the same through both materials. The equivalent thermal resistance is the sum of individual resistances because the slabs are placed sequentially.
4. For a slab of material with conductivity K, the thermal resistance R_1 is given by: R_1 = \frac{d}{K \cdot A} where A is the cross-sectional area.
5. Similarly, for the second material with conductivity 2K, the resistance R_2 is: R_2 = \frac{d}{2K \cdot A}
6. The total resistance R_\text{total} in series is: R_\text{total} = R_1 + R_2 = \frac{d}{K \cdot A} + \frac{d}{2K \cdot A}
7. Simplifying, we get: R_\text{total} = \frac{d}{K \cdot A} \left(1 + \frac{1}{2}\right) = \frac{3d}{2K \cdot A}
8. The equivalent thermal conductivity K_\text{eq} can be expressed by considering the entire slab as a single material: K_\text{eq} = \frac{2d}{R_\text{total} \cdot A}
9. Substituting R_\text{total} gives: K_\text{eq} = \frac{2d}{\frac{3d}{2K}} = \frac{4}{3} K

Therefore, the equivalent thermal conductivity of the slab is \frac{4}{3} K.

Thus, the correct answer is \frac{4}{3} K.