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- 2073 Bhadra · 4 marks
Hollow cylinder: heat transfer and radial temperature
Short answer
Heat transfer per unit length: Q/L = 2πk (T1−T2)/ln(r2/r1) = 2π×60×150/ln(1.5) ≈ 139.5 kW/m. Temperature at 5 cm: T = 200 − 150×ln(1.25)/0.4055 = 117.4°C.
A hollow cylinder with inner and outer diameter of 8 cm and 12 cm respectively has an inner surface temperature of 200°C and outer surface temperature of 50°C. If the thermal conductivity of the cylinder material is 60 W/mK, determine the heat transfer from the unit length of the pipe. Also determine the temperature at the surface at a radial distance of 5 cm from the axis of the cylinder.
Given: r₁ = 4 cm = 0.04 m, r₂ = 6 cm = 0.06 m, T₁ = 200 °C, T₂ = 50 °C, k = 60 W/m·K. Radial, steady conduction.
Heat transfer per unit length
Q/L = 2πk (T₁ − T₂)/ln(r₂/r₁) = 2π × 60 × 150/ln(1.5) = 56 549/0.4055
Q/L ≈ 139466 W/m (≈ 139.5 kW/m)
Temperature at r = 5 cm
For radial conduction, T(r) = T₁ − (T₁ − T₂) ln(r/r₁)/ln(r₂/r₁):
T(0.05) = 200 − 150 × ln(0.05/0.04)/ln(1.5) = 200 − 150 × 0.2231/0.4055
T = 117.4 °C
(Check by resistance: Q/L × ln(1.25)/(2πk) = 82.6 K drop, so T = 200 − 82.6 = 117.4 °C.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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