A system consist of a cylinder surrounded by a cylindrical shell. A cylinder is a radius $R$ and is made of material of thermal conductivity $K$ , whereas a cylindrical shell has inner radius $R$ and outer radius $2 R$ and is made of material of thermal conductivity twice as that of cylinder. Assuming the system in steady state and negligible heat loss across the cylindrical surface, find the effective thermal conductivity of the system, if the two ends of the combined system are maintained at two different temperatures.

3 K

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\frac{2}{3} K

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\frac{7 K}{4}

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\frac{5 K}{4}

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Solution

The correct option is D $\frac{7 K}{4}$
$∴ R_{1} = \frac{L}{K_{1} \cdot π R^{2}}$
and $R_{2} = \frac{L}{K_{2} (4 π R^{2} - π R^{2})} = \frac{L}{3 K_{2} π R^{2}}$
$R_{e q} = \frac{L}{K_{e q} \cdot 4 π R^{2}}$
$R_{1}$ and $R_{2}$ are in parallel.
$∴ \frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\Rightarrow \frac{4 π R^{2} K_{e q}}{L} = \frac{π R^{2} K_{1}}{L} + \frac{3 π R^{2} K_{2}}{L}$
$\Rightarrow 4 K_{e q} = K_{1} + 3 K_{2}$
On putting values of $K_{1}$ and $K_{2}$ , we get
$4 K_{e q} = K + 3 (2 K)$ $[∵ K_{1} = K_{1} K_{2} = 2 K]$
$K_{e q} = \frac{K + 6 K}{4} = \frac{7 K}{4}$

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A cylinder of radius R made of a material of thermal conductivity k, 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. 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