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(C) For a system where heat flows parallel to the interface (as in this composite cylinder),the equivalent thermal conductivity $K$ is given by the we

Vedclass · Accepted Answer

$\frac{K_1 + 3K_2}{4}$

Vedclass · Answer

(C) For a system where heat flows parallel to the interface (as in this composite cylinder),the equivalent thermal conductivity $K$ is given by the weighted average of the thermal conductivities based on their cross-sectional areas:$K = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$Here,the inner cylinder has radius $R$,so its cross-sectional area is:$A_1 = \pi R^2$The outer hollow cylinder has an outer radius of $2R$ and an inner radius of $R$,so its cross-sectional area is:$A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$The total area is:$A_{total} = A_1 + A_2 = \pi R^2 + 3\pi R^2 = 4\pi R^2$Substituting these values into the formula for $K$:$K = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{4\pi R^2}$$K = \frac{\pi R^2(K_1 + 3K_2)}{4\pi R^2}$$K = \frac{K_1 + 3K_2}{4}$

Find the equivalent thermal conductivity of the system.

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