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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

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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​ = 120C, T2 = 30C → ΔT = 90C

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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पाठ 28: Heat Transfer - Exercises [पृष्ठ ९९]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 28 Heat Transfer
Exercises | Q 19 | पृष्ठ ९९

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