---
title: "Cast iron pipe size for given head loss"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:cast-iron-pipe-size-for-head-loss"
source: https://www.bctengineer.com/q/semester_4/hydraulics/cast-iron-pipe-size-for-head-loss
site: BCT Engineering
---

# Cast iron pipe size for given head loss

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2068 Bhadra, 8 marks
- Video: [Solving Head Loss with Moody Diagram and Darcy-Weisbach: A Step-by-step Example](https://www.youtube.com/watch?v=f6aOzMx4ONc) — Tutor Jack Ph
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/cast-iron-pipe-size-for-head-loss
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

What size of new cast iron pipe is needed to transport 400 lps of water for 1 km long pipe with 2 m head loss? Take roughness height of the pipe is 0.26 mm and the viscosity of water 0.0014 Pa·s.

## Short answer

Iterating Darcy–Weisbach with Colebrook-White, required diameter converges to about 0.645 m (≈645 mm) for 2 m head loss over 1 km.

## Answer

**Data.** $Q = 0.4$ m³/s, $L = 1000$ m, $h_f = 2$ m, $e = 0.26$ mm, $\mu = 0.0014$ Pa s, so $\nu = \mu/\rho = 1.40\times10^{-6}$ m²/s. This is a "find the diameter" problem, so iterate.

#### Working equation
From Darcy-Weisbach with $V = \dfrac{4Q}{\pi D^2}$:
$$
h_f = \frac{8fLQ^2}{\pi^2gD^5} \;\Rightarrow\; D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5} = \left(\frac{8f(1000)(0.4)^2}{\pi^2(9.81)(2)}\right)^{1/5}
$$

#### Trials (f from Colebrook-White at each new D)
| Trial | f assumed | D (m) | Re | f (Colebrook) |
|---|---|---|---|---|
| 1 | 0.0200 | 0.6672 | 5.452e+05 | 0.0169 |
| 2 | 0.0169 | 0.6447 | 5.642e+05 | 0.0169 |
| 3 | 0.0169 | 0.6452 | 5.638e+05 | 0.0169 |
| 4 | 0.0169 | 0.6452 | 5.638e+05 | 0.0169 |

The diameter converges to $D = 0.6452$ m. Check: $V = 1.223$ m/s, $Re = 563815$, $e/D = 0.00040$, $f = 0.01692$, and
$$
h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.01692\left(\frac{1000}{0.6452}\right)\frac{(1.223)^2}{19.62} = 2.000\ \text{m}
$$

**Answer:** $D \approx 645$ mm; a 650 mm cast iron pipe would be used.

## Check yourself

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

1. Why must the diameter be found by trial?
   A. D appears in f through Re and e/D, while f appears in the D formula
   B. The head loss is unknown
   C. The discharge is unknown
   D. Because the pipe is not circular
   Answer: A. f depends on Re and e/D which depend on D, so D and f are iterated. (Key point: Why iteration)

2. Which relation gives D from the given data?
   A. D = (8 f L Q^2/(pi^2 g h_f))^(1/5)
   B. D = (8 f L Q^2/(pi^2 g h_f))^(1/2)
   C. D = 4Q/(pi V)
   D. D = (f L Q/h_f)^(1/3)
   Answer: A. Rearranging h_f = 8fLQ^2/(pi^2 g D^5). (Key point: Working equation)

3. How is f updated in each trial?
   A. Fixed at 0.02
   B. Colebrook-White using the new D (Re and e/D)
   C. f = 64/Re
   D. Taken from the head loss
   Answer: B. Re = 4Q/(pi D nu) and e/D are recomputed for the new D and Colebrook gives f. (Key point: Friction factor update)

4. What diameter results and what is selected?
   A. About 645 mm; a 650 mm pipe
   B. About 400 mm; a 400 mm pipe
   C. About 1000 mm
   D. About 300 mm
   Answer: A. The iteration converges at D = 0.6452 m, rounded up to a standard size. (Key point: Final size)

## Related in Hydraulics

- [Equivalent roughness from head loss test](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acommercial-pipe-test-roughness) — Hydraulics, chapter 1
- [Permissible roughness for minimum energy loss](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aone-metre-pipe-permissible-roughness) — Hydraulics, chapter 1
- [Roughness height from velocity ratio](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Arough-pipe-velocity-ratio-roughness) — Hydraulics, chapter 1
- [Siphon with 90 mm pipe, number and length](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-90mm-pipe-total-length) — Hydraulics, chapter 2
- [Viscosity of oil from laminar pipe flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahorizontal-oil-pipe-laminar-viscosity) — Hydraulics, chapter 1
- [Siphon with 75 mm pipes and summit height](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-75mm-pipes-summit-height) — Hydraulics, chapter 2

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