---
title: "Water heating in a tube, Dittus-Boelter"
subject: "Heat Transfer"
chapter: 3
key: "semester_5/heat_transfer:dittus-boelter-water-heating-tube-length"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/dittus-boelter-water-heating-tube-length
site: BCT Engineering
---

# Water heating in a tube, Dittus-Boelter

Heat Transfer · Chapter 3: Convection Heat Transfer

- Asked in: year not given, 8 marks
- Video: [Heat transfer Tutorials | 6-1 | Heating of Water in a Tube by Steam](https://www.youtube.com/watch?v=ZjfroQPSXSs) — المقررات المفتوحة - Open Courses
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/dittus-boelter-water-heating-tube-length
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-3

## Question

Water flows at a mean velocity of 1 m/s through a tube of inner diameter 25 mm and is heated from 30 °C to 50 °C. The tube wall is maintained at a uniform temperature of 90 °C. Evaluate water properties at the mean bulk temperature of 40 °C (rho = 992.2 kg/m^3, k = 0.631 W/m K, mu = 653 x 10^-6 Pa s, cp = 4179 J/kg K, Pr = 4.32). Using the Dittus-Boelter equation, calculate (a) the Reynolds number, (b) the heat transfer coefficient, (c) the heat gained by water per second, and (d) the length of the tube required.

## Short answer

Using Dittus-Boelter correlation Nu = 0.023 Re^0.8 Pr^0.4 and log mean temperature difference: Re = 3.8e4, h = 4805 W/m^2K, heat gained Q = 40.7 kW, required tube length L = 2.2 m.

## Answer

#### (a) Reynolds number

$$
Re = \frac{\rho V D}{\mu} = \frac{992.2\times1\times0.025}{653\times10^{-6}} = 3.80\times10^4
$$

The flow is turbulent ($Re > 10\,000$) and the tube is long, so the Dittus-Boelter equation applies.

#### (b) Heat transfer coefficient
For heating of the fluid the exponent is $n = 0.4$:

$$
\begin{aligned}
Nu &= 0.023\,Re^{0.8}Pr^{0.4}\\
&= 0.023\,(37\,986)^{0.8}(4.32)^{0.4} = 190.4
\end{aligned}
$$

$$
h = \frac{Nu\,k}{D} = \frac{190.4\times0.631}{0.025} = 4805\ \text{W/m}^2\text{K}
$$

#### (c) Heat gained
Mass flow rate:

$$
\dot m = \rho V\frac{\pi D^2}{4} = 992.2\times1\times\frac{\pi(0.025)^2}{4} = 0.487\ \text{kg/s}
$$

$$
Q = \dot m c_p (T_{out}-T_{in}) = 0.487\times4179\times20 = 40\,700\ \text{W}
$$

#### (d) Tube length
For constant wall temperature the mean temperature difference is the logarithmic mean:

$$
\Delta T_{lm} = \frac{(90-30)-(90-50)}{\ln\left(\dfrac{90-30}{90-50}\right)} = \frac{60-40}{\ln1.5} = 49.3\ ^\circ\text{C}
$$

$$
Q = h\,(\pi D L)\,\Delta T_{lm}\;\Rightarrow\; L = \frac{40\,707}{4805\times\pi\times0.025\times49.3} = 2.19\ \text{m}
$$

Check: $L/D = 87.5 > 60$, so the entrance effect is small and the correlation is acceptable.

**Answer:** $Re = 3.8\times10^4$; $h = 4805\ \text{W/m}^2$K; $Q = 40.7$ kW; $L \approx 2.2$ m.

## Check yourself

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

1. Re = rho V D / mu gives about:
   A. 3.8 x 10^2, laminar
   B. 3.8 x 10^6, laminar
   C. 3.8 x 10^4, turbulent
   D. 1.5 x 10^3, laminar
   Answer: C. 992.2 x 0.025/653e-6 = 38000 > 10000. (Key point: Reynolds number)

2. Why is Pr exponent 0.4 used?
   A. The fluid is being cooled
   B. The flow is laminar
   C. The wall is insulated
   D. The fluid is being heated
   Answer: D. Dittus-Boelter uses n = 0.4 for heating, 0.3 for cooling. (Key point: Exponent)

3. Heat gained by water is:
   A. m cp (Tout - Tin) = 40.7 kW
   B. h A delta T = 4805 W
   C. m cp Tw = 190 kW
   D. k A dT/dx
   Answer: A. 0.487 x 4179 x 20 = 40.7 kW. (Key point: Heat gained)

4. For constant wall temperature the delta T used is:
   A. Arithmetic of wall and inlet, 60 C
   B. Log mean, 49.3 C
   C. Wall minus outlet, 40 C
   D. Outlet minus inlet, 20 C
   Answer: B. LMTD = (60-40)/ln(1.5) = 49.3 C. (Key point: Mean temperature difference)

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