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प्रश्न
Steam at 120°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°C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15 J s−1 m−1°C−1.
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उत्तर
For steady conduction through a hollow cylinder:
`Q/t = (2pikL(T_1 - T_2))/ln(r^2/r^1)`
Substitute given values
k = 0.15J s−1m−1∘C−1
L = 50cm = 0.50m
T1 = 120∘C, T2 = 30∘C → ΔT = 90∘C
r1=1.0 cm = 0.010 m
r2 = 1.2cm = 0.012m
`Q/t = (2pi(0.15)(0.50)(90))/ln(0.012/0.010)`
2π ⋅ 0.15 ⋅ 0.50 ⋅ 90 = 2π ⋅ 6.75 = 42.41
ln`(0.012/0.010) = ln(1.2) = 0.1823`
`Q/t = 42.41/0.1823 =` 232.6 J/s
= 233 J/s
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