---
title: "Heat loss from a cylinder in cross flow"
subject: "Heat Transfer"
chapter: 3
key: "semester_5/heat_transfer:cross-flow-air-cylinder-heat-loss"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/cross-flow-air-cylinder-heat-loss
site: BCT Engineering
---

# Heat loss from a cylinder in cross flow

Heat Transfer · Chapter 3: Convection Heat Transfer

- Asked in: year not given, 6 marks
- Video: [Using the Hilpert Correlation for Cylinder in Cross Flow](https://www.youtube.com/watch?v=9E2B85NGNTI) — LearnChemE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/cross-flow-air-cylinder-heat-loss
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-3

## Question

Air at 25 °C flows at 10 m/s across a long horizontal cylinder of outer diameter 50 mm whose surface is at 125 °C. Properties of air at the film temperature of 75 °C: nu = 20.92 x 10^-6 m^2/s, k = 0.02953 W/m K, Pr = 0.7202. Using the Churchill-Bernstein correlation, find the average heat transfer coefficient and the heat loss per metre length.
Nu = 0.3 + [0.62 Re^0.5 Pr^(1/3)] / [1 + (0.4/Pr)^(2/3)]^0.25 x [1 + (Re/282000)^(5/8)]^(4/5)

## Short answer

Using Churchill-Bernstein correlation, Re = 2.39×10^4, Nu = 88.5, average heat transfer coefficient h = 52.3 W/m^2·K, heat loss per metre length = 821 W/m.

## Answer

#### Reynolds number

$$
Re_D = \frac{VD}{\nu} = \frac{10\times0.05}{20.92\times10^{-6}} = 2.39\times10^4
$$

$Re_D\,Pr = 1.72\times10^4 > 0.2$, so the Churchill-Bernstein equation is valid.

#### Nusselt number
Compute the parts:

$$
\begin{aligned}
Re^{0.5} &= 154.6, \qquad Pr^{1/3} = 0.896\\
\text{numerator} &= 0.62\times154.6\times0.896 = 85.9\\
\left[1+(0.4/0.7202)^{2/3}\right]^{1/4} &= (1.676)^{0.25} = 1.138\\
\left[1+(Re/282000)^{5/8}\right]^{4/5} &= (1.2138)^{0.8} = 1.168
\end{aligned}
$$

$$
Nu = 0.3 + \frac{85.9}{1.138}\times1.168 = 0.3 + 88.2 \approx 88.5
$$

#### Heat transfer coefficient

$$
h = \frac{Nu\,k}{D} = \frac{88.5\times0.02953}{0.05} = 52.3\ \text{W/m}^2\text{K}
$$

#### Heat loss per metre

$$
\frac{Q}{L} = h\,(\pi D)(T_s-T_\infty) = 52.3\times\pi\times0.05\times100 = 821\ \text{W/m}
$$

**Answer:** $h \approx 52.3\ \text{W/m}^2$K; $Q/L \approx 821\ \text{W/m}$.

## Check yourself

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

1. The characteristic length in Re_D is:
   A. Cylinder length
   B. Radius
   C. Film thickness
   D. Cylinder diameter
   Answer: D. Re_D = V D/nu = 2.39 x 10^4. (Key point: Reynolds number)

2. Why can Churchill-Bernstein be used?
   A. Re Pr > 0.2
   B. Re < 2300
   C. Pr > 100
   D. Flow is natural
   Answer: A. Re_D Pr = 1.72 x 10^4 > 0.2. (Key point: Validity)

3. Heat loss per metre uses:
   A. h D (Ts - Tinf)
   B. h pi D (Ts - Tinf)
   C. h 2 pi D
   D. h pi D^2 (Ts - Tinf)
   Answer: B. Area per metre is pi D; Q/L = 52.3 x pi x 0.05 x 100 = 821 W/m. (Key point: Heat loss per metre)

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