- 2069 Poush · 4 marks
Steam pipe with single insulation: heat loss rate
Short answer
With the insulation’s outer surface at atmospheric temperature, the total resistance is the sum of pipe and insulation resistances. The steady heat loss is Q = 2πL (T_h−T_atm)/[ln(r₂/r₁)/k_p + ln(r₃/r₂)/k_i].
Steam of temperature Th, is flowing through the pipe (with certain thickness) and to minimize the heat loss, the pipe is covered by a single insulation layer. Under steady state condition what is the rate of heat loss from steam, if the temperature of outer surface of insulation layer is equal to atmospheric temperature (Tatm).
Setup. Steam at T_h flows in a pipe of inner radius r₁ and outer radius r₂ (conductivity k_p). The pipe carries one insulation layer from r₂ to r₃ (conductivity k_i). The outer surface of the insulation is at the atmospheric temperature T_atm. Length L; steady radial heat flow. Resistance of the steam film is neglected because the steam temperature is taken at the pipe's inner surface.
T_h --[ pipe wall ]--[ insulation ]-- T_atm
ln(r2/r1) ln(r3/r2)
2 pi kp L 2 pi ki L
Derivation
Pipe wall: T_h − T₂ = Q ln(r₂/r₁)/(2π k_p L) Insulation: T₂ − T_atm = Q ln(r₃/r₂)/(2π k_i L)
Add the two equations:
T_h − T_atm = (Q/2πL) [ln(r₂/r₁)/k_p + ln(r₃/r₂)/k_i]
Q = 2πL (T_h − T_atm)/[ln(r₂/r₁)/k_p + ln(r₃/r₂)/k_i]
Per unit length: Q/L = 2π (T_h − T_atm)/[ln(r₂/r₁)/k_p + ln(r₃/r₂)/k_i]
In resistance form, Q = (T_h − T_atm)/(R_pipe + R_ins). If the pipe wall is thin or of high conductivity (k_p large), R_pipe is negligible and Q = 2πk_iL (T_h − T_atm)/ln(r₃/r₂).
✓ Answer checked by Claude Opus 5.5 (high effort)
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