A cylinder of radius R made of a material of thermal conductivity & is surrounded by a cylindrical shell of inner radius .E and outer radius 2 .R made of a material of thermal conductivity K2. 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 systems in steady state. The effective thermal conductivity of the system is

Heat flowing per second through cylinder of radius R Q1=K1πR2T1−T2lHeat flowing per second through outer shell of radius 2 RQ2=K2π(2R)2−π(R)2T1−T2lTotal Q=Q1+Q2 =K1+3K2πR2T1−T2l ....(1)Let K be the equivalent thermal conductivity of the system. Then Q=Kπ(2R)2T1−T2l ....(2)From eqs. (1) and (2), we have K1+3K2=4K K=K1+3K24

A cylinder of radius R made of a material of thermal conductivity & is surrounded by a cylindrical shell of inner radius .E and outer radius 2 .R made of a material of thermal conductivity K2. 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 systems in steady state. The effective thermal conductivity of the system is

Heat flowing per second through cylinder of radius R Q1=K1πR2T1−T2lHeat flowing per second through outer shell of radius 2 RQ2=K2π(2R)2−π(R)2T1−T2lTotal Q=Q1+Q2 =K1+3K2πR2T1−T2l ....(1)Let K be the equivalent thermal conductivity of the system. Then Q=Kπ(2R)2T1−T2l ....(2)From eqs. (1) and (2), we have K1+3K2=4K K=K1+3K24

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A cylinder of radius R made of a material of thermal conductivity & is surrounded by a cylindrical shell of inner radius .E and outer radius 2 .R made of a material of thermal conductivity K₂. 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 systems in steady state. The effective thermal conductivity of the system is

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$K_{1} + K_{2}$

$\frac{K_{1} K_{2}}{K_{1} + K_{2}}$

$\frac{K_{1} + 3 K_{2}}{4}$

$\frac{3 K_{1} + K_{2}}{4}$

answer is C.

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Detailed Solution

Heat flowing per second through cylinder of radius R
$Q_{1} = K_{1} ({πR}^{2}) (\frac{T_{1} - T_{2}}{l})$
Heat flowing per second through outer shell of radius 2 R
$Q_{2} = K_{2} [π (2 R)^{2} - π (R)^{2}] (\frac{T_{1} - T_{2}}{l})$
Total $Q = Q_{1} + Q_{2}$
$= (K_{1} + 3 K_{2}) {πR}^{2} (\frac{T_{1} - T_{2}}{l}) . . . . (1)$
Let K be the equivalent thermal conductivity of the system. Then
$Q = Kπ (2 R)^{2} (\frac{T_{1} - T_{2}}{l}) . . . . (2)$
From eqs. (1) and (2), we have
$(K_{1} + 3 K_{2}) = 4 K$
$K = \frac{K_{1} + 3 K_{2}}{4}$