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Conduction Law

Steam at 120∘...

Question

Steam at $120^{\circ} C$ is continuously passed through a 50 -cm long rubber tube of inner and outer radii 1.0 cm and 1.2 cm. The room temperature is $30^{\circ} C$ . Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = $0.15 J s^{- 1} m^{- 1 \circ} C^{- 1}$ .

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Solution

Given

$K_{r u b b e r} = 0.15 J / m - s -^{\circ} C$

$T_{1} = 120^{\circ} C$ $T_{2} = 30^{\circ} C$

We know for radial conduction in a cylinder

$\frac{Q}{t} = \frac{2 π K l (T_{2} - T_{1})}{l n (R_{2} / R_{1})}$

$= \frac{2 \times 3.14 \times 15 \times 10^{- 2} \times 50 \times 10^{- 1} \times 30}{l n (1.2 / 1)}$

= 232.5 = 233 J/s

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