---
title: "Hollow cylinder: heat transfer; temperature at 10 cm"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:hollow-cylinder-8-12cm-radial-10cm"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/hollow-cylinder-8-12cm-radial-10cm
site: BCT Engineering
---

# Hollow cylinder: heat transfer; temperature at 10 cm

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

- Asked in: 2068 Magh
- Video: [Heat Conduction Through Hollow Cylinder | Heat Transfer | Thermodynamics | Engineering |](https://www.youtube.com/watch?v=iWnUYw_CoLQ) — Educational Park
- 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-radial-10cm
- 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 diameters of 8 cm and 12 cm respectively has an inner surface temperature of 200°C and an 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 10 cm from the axis of the cylinder.

## Short answer

Taking 10 cm as diameter (radius 5 cm), heat transfer per unit length Q/L = 2π×60×(200−50)/ln(0.06/0.04) ≈ 139.5 kW/m. Temperature at r = 5 cm: T = 200 − 150×ln(1.25)/ln(1.5) ≈ 117.4 °C.

## Answer

**Reading of the data.** The inner and outer radii are 4 cm and 6 cm, so a radial distance of 10 cm lies outside the wall and cannot be a point in the material. The 10 cm is taken as a **diameter**, i.e. radius r = 5 cm (the same problem is asked with "5 cm"), which lies inside the wall.

**Given:** r₁ = 0.04 m, r₂ = 0.06 m, T₁ = 200 °C, T₂ = 50 °C, k = 60 W/m·K. Steady radial conduction, no heat generation.

#### 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
T(r) = T₁ − (T₁ − T₂) ln(r/r₁)/ln(r₂/r₁)

T(0.05) = 200 − 150 × ln(1.25)/ln(1.5) = 200 − 150 × 0.22314/0.40547

**T ≈ 117.4 °C**

(For r ≥ 6 cm the material ends; at r = 6 cm the temperature is 50 °C.)

## Check yourself

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

1. For inner and outer diameters of 8 cm and 12 cm, the radii used are:
   A. 8 cm and 12 cm
   B. 4 cm and 6 cm
   C. 4 cm and 12 cm
   D. 8 cm and 6 cm
   Answer: B. Radius is half the diameter: r1 = 0.04 m, r2 = 0.06 m. (Key point: Radii from diameters)

2. Which formula gives heat transfer per unit length?
   A. Q/L = 2πk(T1 - T2)/ln(r2/r1)
   B. Q/L = k(T1 - T2)/(r2 - r1)
   C. Q/L = 2πk ln(r2/r1)/(T1 - T2)
   D. Q/L = 2πk(T1 - T2)/(r2 - r1)
   Answer: A. Radial conduction through a cylinder uses the logarithmic form, giving about 139.5 kW/m here. (Key point: Heat transfer per unit length)

## Close variants set in other exams

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

## Related in Fundamental of Thermodynamics & Heat Transfer

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