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. If 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 the two ends be $\Delta \theta$ and length of the cylinder be $L$.
Let $d{q}_1$ and $d{q}_2$ be the heat transferred by inner cylinder and outer shell in time $dt$.
Heat transferred by inner cylinder in this time is
$d{q}_1=\frac{K_1\pi R^2\Delta \theta }{L}\times dt$
Heat transferred by outer shell in this time is
$d{q}_2=\frac{K_2\pi \left(\right(2R{)}^2-R^2\left)\right(\Delta \theta )}{L}\times dt$
$d{q}_2=\frac{K_2\left(3\pi R^2\right)\left(\Delta \theta \right)dt}{L}$

Total heat transferred is $dq=d{q}_1+d{q}_2$
$\therefore dq=\frac{(K_1+3K_2)\pi R^2\left(\Delta \theta \right)dt}{L}$
Now, total radius of the system $R^{\prime }=2R$
Hence, we can write
$\Rightarrow dq=\frac{(K_1+3K_2)\pi \left(2R{)}^2\right(\Delta \theta )dt}{4L}$
$\Rightarrow dq=\frac{(K_1+3K_2)}{4}\frac{\pi \left(R^{\prime }{)}^2\right(\Delta \theta )dt}{L}$
[since $R^{\prime }=2R$]
$\Rightarrow dq=K_{eq}\frac{\pi (R^{\prime }{)}^2\Delta \theta dt}{L}$
where equivalent conductivity $K_{eq}=\frac{K_1+3K_2}{4}$
Therefore, option (d) is correct.