---
title: "Steam pipe with single insulation: heat loss rate"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-pipe-single-insulation-heat-loss"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-pipe-single-insulation-heat-loss
site: BCT Engineering
---

# Steam pipe with single insulation: heat loss rate

Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer

- Asked in: 2069 Poush, 4 marks
- Video: [Calculating Rate of Heat Transfer Through Lagged Pipes](https://www.youtube.com/watch?v=X99yHFbTxaM) — Engineers Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-pipe-single-insulation-heat-loss
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

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).

## 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].

## Answer

**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.

```text
 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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How are the pipe wall and insulation resistances combined?
   A. In parallel, since both carry heat
   B. In series, since the same Q passes through both
   C. Only the insulation counts
   D. They are multiplied
   Answer: B. The same radial heat flow passes through the wall and then the insulation, so their temperature drops add and resistances are in series. (Key point: Series resistances)

2. The rate of heat loss from the steam is:
   A. Q = 2πL(T_h - T_atm)/[ln(r3/r1)/k_i]
   B. Q = 2πL(T_h - T_atm)[ln(r2/r1)/k_p + ln(r3/r2)/k_i]
   C. Q = 2πL(T_h - T_atm)/[ln(r2/r1)/k_p + ln(r3/r2)/k_i]
   D. Q = 2π(T_h - T_atm)/[ln(r2/r1)/k_p + ln(r3/r2)/k_i]
   Answer: C. Adding the two drops gives T_h - T_atm = (Q/2πL)[...], so Q = 2πL(T_h - T_atm)/[...]. (Key point: Heat loss expression)

3. What does the result reduce to if the pipe wall resistance is negligible?
   A. Q = 2πk_i L (T_h - T_atm)/ln(r3/r2)
   B. Q = 0
   C. Q = 2πk_p L (T_h - T_atm)/ln(r2/r1)
   D. Q = k_i A (T_h - T_atm)/(r3 - r2)
   Answer: A. With R_pipe dropped, only the insulation term ln(r3/r2)/k_i remains. (Key point: Thin or high-k pipe wall)

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