A cylinder of radius $R$ made of a material of thermal conductivity $k_{1}$ is surrounded by a cylindrical sheet of inner radius $R$ and outer radius $2 R$ made of 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 a steady state. Calculate the effective thermal conductivity of the system. Assume that heat flow is always axial.

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Solution

Two cylinders are in parallel, therefore equivalent thermal resistance $R$ is given by
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
But $R = \frac{l}{k A}$
$∴ \frac{k A}{l} = \frac{k_{1} A_{1}}{l_{1}} + \frac{k_{2} A_{2}}{l_{2}}$
Here $l_{1} = l_{2} = l$ and $A_{1} = π R^{2}$
$A_{2} = π (2 R)^{2} - π R^{2} = 3 π R^{2}$ and $A = π (2 R)^{2} = 4 π R^{2}$

$∴ \frac{k 4 π R^{2}}{l} = \frac{k_{1} π R^{2}}{l} + \frac{k_{2} 3 π R^{2}}{l}$
i.e. $k = \frac{k_{1} + 3 k_{2}}{4}$ .

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Two cylinders are in parallel, therefore equivalent thermal resistance R is given by1R=1R1+1R2But R=lkA∴kAl=k1A1l1+k2A2l2Here l1=l2=l and A1=πR2A2=π(2R)2−πR2=3πR2 and A=π(2R)2=4πR2∴k4πR2l=k1πR2l+k23πR2li.e. k=k1+3k24.