---
title: "Hollow cylinder: heat transfer and radial temperature"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:hollow-cylinder-8-12cm-heat-transfer"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/hollow-cylinder-8-12cm-heat-transfer
site: BCT Engineering
---

# Hollow cylinder: heat transfer and radial temperature

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

- Asked in: 2073 Bhadra, 4 marks
- Video: [Radial Conduction Heat Transfer Through Hollow Cylinders HMT Tutorial- 16](https://www.youtube.com/watch?v=B3aq3vuvGxE) — Ujjwal Kumar Sen
- 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/hollow-cylinder-8-12cm-heat-transfer
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

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.

## 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.

## Answer

**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.)

## Check yourself

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

1. Heat transfer per unit length through a cylinder is
   A. 2π k ΔT (r2 - r1)
   B. 2π k ΔT / ln(r2/r1)
   C. 2π k ΔT ln(r2/r1)
   D. k A ΔT / L
   Answer: B. Q/L = 2π k (T1 - T2)/ln(r2/r1) = about 139.5 kW/m. (Key point: Formula)

2. T at radial distance r is
   A. (T1 + T2)/2
   B. T1 - (T1 - T2) ln(r/r1)/ln(r2/r1)
   C. T2 + (T1 - T2) ln(r2/r)
   D. T1 - (T1 - T2)(r - r1)/(r2 - r1)
   Answer: B. Temperature in radial conduction is logarithmic in r, giving 117.4 °C at 5 cm. (Key point: Temperature at r)

## Close variants set in other exams

- [Hollow cylinder: heat transfer; temperature at 10 cm](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Ahollow-cylinder-8-12cm-radial-10cm) — Fundamental of Thermodynamics & Heat Transfer, chapter 7, asked 2×

## Related in Fundamental of Thermodynamics & Heat Transfer

- [Overall heat transfer coefficient of hollow cylinder](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Ahollow-cylinder-overall-u-inside-outside) — Fundamental of Thermodynamics & Heat Transfer, chapter 7
- [Radial conduction through hollow cylinder; resistance](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aradial-conduction-hollow-cylinder-resistance) — Fundamental of Thermodynamics & Heat Transfer, chapter 7

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