Question

A solid cylinder of radius R made of a material of thermal conductivity $K_1$ is surrounded by a cylindrical shell of inner radius R and outer radius $2R$ made of a material of thermal conductivity $K_2$. The two ends of the combined system are maintained at two different temperatures. Then there is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is:

• A. $\frac{4K_1+3K_2}{4}$
• B. $\frac{3K_1+4K_2}{4}$
• C. $\frac{3K_1+K_2}{4}$
• D. $\frac{K_1+3K_2}{4}$

Answer 1

The correct option is D $\frac{K_1+3K_2}{4}$

Let temperature difference between two ends be $\delta \theta$ and length of cylinder be L .
Let $d{q}_1$ and $d{q}_2$ be heat transferred by inner and outer shell in time dt.
Heat transferred by inner cylinder in this time is
$d{q}_1=\frac{K_1\pi R^2\delta \theta dt}{L}$
Heat transferred by outer cylinder in this time is $d{q}_2=\frac{K_2\pi \left(\right(2R{)}^2-R^2)\delta \theta dt}{L}$
$d{q}_2=\frac{K_{2}3\pi R^2\delta \theta dt}{L}$
Total heat transferred is dq$=$$d{q}_1$+$d{q}_2$
$\therefore$ $dq=\frac{(K_1+3K_2)\pi R^2\delta \theta dt}{L}$
Now total radius of the system R'=2R$dq=\frac{(K_1+3K_2)\pi (2R{)}^2\delta \theta dt}{4L}$
$dq=\frac{(K_1+3K_2)}{4}\frac{\pi (R^{\prime }{)}^2\delta \theta dt}{L}$
$dq=K\frac{\pi (R^{\prime }{)}^2\delta \theta dt}{L}$
where equivalent conductivity $K=\frac{K_1+3K_2}{4}$
Option D is correct.