Question

Consider a composite slab consisting of two different materials having equal thickness and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is

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

Answer 1

Answer: (D)

$\frac{4}{3}K$

Answer 2

The equivalent thermal conductivity when the slabs are connected in series is

$K_{eq} = \frac{\Sigma x_{n}}{\frac{\Sigma x_{n}}{\Sigma K_{n}}} = \frac{x + x}{\frac{x}{K} + \frac{x}{2K}} = \frac{2x(2K^{2})}{(2Kx + Kx)} = \frac{4}{2}K$