---
title: "Minimum hydrodynamically smooth pipe size"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:smooth-pipe-minimum-size"
source: https://www.bctengineer.com/q/semester_4/hydraulics/smooth-pipe-minimum-size
site: BCT Engineering
---

# Minimum hydrodynamically smooth pipe size

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2082 Kartik, 4 marks
- Video: [Hydrodynamically Smooth & Rough Boundaries | Turbulent Flow | Smooth & Rough Boundaries in Turbul...](https://www.youtube.com/watch?v=MR0FSvWMHBM) — Professor Nidhi Trivedi
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/smooth-pipe-minimum-size
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

Galvanized iron pipes can be assumed to have equivalent roughness of 0.15 mm. What minimum size of pipe will be hydrodynamically smooth at Reynolds number $2 \times 10^5$?

## Short answer

At Re = 2e5, smooth-pipe friction factor f = 0.01564. Using u* k / nu <= 4, D >= (Re k sqrt(f/8))/4 gives minimum diameter D = 0.332 m (about 330 mm).

## Answer

**Criterion.** A pipe is hydrodynamically smooth when the roughness is buried in the laminar sub-layer, i.e. when the roughness Reynolds number (Nikuradse) satisfies
$$
\frac{u_* k}{\nu} \le 4
$$
with $u_* = V\sqrt{f/8}$ and $k = 0.15$ mm.

**Step 1: friction factor of a smooth pipe at $Re = 2\times10^5$.** Prandtl's smooth-pipe law $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$ gives (by iteration)
$$
f = 0.01564, \qquad \sqrt{f/8} = 0.04421
$$

**Step 2: express the criterion in terms of $D$.** Since $Re = VD/\nu$,
$$
\frac{u_* k}{\nu} = \frac{V\sqrt{f/8}\,k}{\nu} = Re\,\sqrt{\frac{f}{8}}\,\frac{k}{D} \le 4
$$
$$
D \ge \frac{Re\,k\sqrt{f/8}}{4} = \frac{(2\times10^5)(0.00015)(0.04421)}{4} = 0.332\ \text{m}
$$

**Answer:** Minimum diameter $\approx 0.332$ m (about 330 mm). Any smaller pipe at this Reynolds number is not smooth. (Using the limit $u_*k/\nu = 5$ instead of 4 gives 0.265 m.)

## Check yourself

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

1. What condition makes a pipe hydrodynamically smooth?
   A. u* k / nu <= 4
   B. u* k / nu >= 100
   C. Re < 2000
   D. k / D = 0
   Answer: A. A pipe is smooth when roughness is buried in the laminar sub-layer: roughness Reynolds number u*k/nu <= 4. (Key point: Smooth criterion)

2. How is f found at Re = 2x10^5 for the smooth pipe?
   A. f = 64/Re
   B. Rough-pipe law 1/sqrt f = 2 log(D/k) + 1.14
   C. Prandtl's smooth-pipe law 1/sqrt f = 2 log(Re sqrt f) - 0.8 by iteration
   D. Fixed f = 0.02
   Answer: C. The smooth-pipe law gives f = 0.01564 at this Re. (Key point: Smooth-pipe f)

3. Which expression gives the minimum diameter?
   A. D = Re k sqrt(f/8) / 4
   B. D = 4 Re k / sqrt(f/8)
   C. D = 4 k / Re
   D. D = Re k / (4 f)
   Answer: A. Since u*k/nu = Re sqrt(f/8) k/D <= 4, D >= Re k sqrt(f/8)/4 = 0.332 m. (Key point: Minimum diameter)

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