---
title: "Permissible roughness for minimum energy loss"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:one-metre-pipe-permissible-roughness"
source: https://www.bctengineer.com/q/semester_4/hydraulics/one-metre-pipe-permissible-roughness
site: BCT Engineering
---

# Permissible roughness for minimum energy loss

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2075 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/one-metre-pipe-permissible-roughness
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

One meter diameter pipe is to carry a water discharge of 1.0 m³/s at the minimum loss of energy. What will be the permissible height of surface roughness?

## Short answer

For minimum energy loss the pipe must remain hydrodynamically smooth, i.e., k ≤ 4ν/u_*. With V=1.273 m/s, Re≈1.27×10⁶, smooth pipe f=0.01118, u_*=0.0476 m/s, the permissible roughness height is about 0.084 mm.

## Answer

**Idea.** A pipe has the minimum energy loss when it is hydrodynamically smooth, since then $f$ depends only on $Re$ and the roughness adds nothing. The pipe stays smooth as long as the roughness is buried in the laminar sub-layer:
$$
\frac{u_*k}{\nu} \le 4 \quad\Rightarrow\quad k \le \frac{4\nu}{u_*}
$$

**Data.** $D = 1$ m, $Q = 1.0$ m³/s, $\nu = 1.0\times10^{-6}$ m²/s (water at about 20 °C).

**Step 1: velocity and Reynolds number**
$$
V = \frac{Q}{\pi D^2/4} = 1.2732\ \text{m/s},\qquad Re = \frac{VD}{\nu} = 1273240
$$

**Step 2: smooth-pipe friction factor** (Prandtl: $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$)
$$
f = 0.01118
$$

**Step 3: shear velocity**
$$
u_* = V\sqrt{\frac{f}{8}} = 1.2732\sqrt{\frac{0.01118}{8}} = 0.04759\ \text{m/s}
$$

**Step 4: permissible roughness**
$$
k \le \frac{4\nu}{u_*} = \frac{4(1.0\times10^{-6})}{0.04759} = 0.0840\ \text{mm}
$$

**Answer:** The surface roughness height must not exceed about $0.084$ mm (0.08 mm) for minimum loss. With the looser limit $u_*k/\nu \le 5$, the value is 0.105 mm.

## Check yourself

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

1. Why is the pipe taken as hydrodynamically smooth for minimum energy loss?
   A. Roughness then adds nothing to f, which depends only on Re
   B. Smooth pipes have zero friction
   C. Roughness always reduces loss
   D. Velocity is zero
   Answer: A. Within the laminar sub-layer roughness has no effect on f, so loss is a minimum. (Key point: Minimum loss idea)

2. Which condition gives the permissible roughness?
   A. k <= 100 nu / u*
   B. k <= 4 nu / u*
   C. k = D / 100
   D. k >= 4 nu / u*
   Answer: B. Smooth requires u*k/nu <= 4. (Key point: Limit on k)

3. How is u* obtained?
   A. u* = V sqrt(f/8), with smooth-pipe f from Prandtl's law
   B. u* = V
   C. u* = Q/D
   D. u* = V f
   Answer: A. f = 0.01118 gives u* = 0.04759 m/s. (Key point: Shear velocity)

4. What is the permissible roughness?
   A. About 0.84 mm
   B. About 0.084 mm
   C. About 8.4 mm
   D. About 0.0084 mm
   Answer: B. k = 4e-6/0.04759 = 0.084 mm. (Key point: Result)

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