---
title: "Equivalent roughness from head loss test"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:commercial-pipe-test-roughness"
source: https://www.bctengineer.com/q/semester_4/hydraulics/commercial-pipe-test-roughness
site: BCT Engineering
---

# Equivalent roughness from head loss test

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2076 Bhadra, 6 marks
- Video: [Piping Problem with Head Losses and Moody Diagram](https://www.youtube.com/watch?v=YLbp3ifkqd0) — Fluid Matters
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/commercial-pipe-test-roughness
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

Test on a 500 mm diameter commercial pipe indicated that head loss in 100 m length of pipe at different discharge of water is as given below.

| Q (l/s) | 40 | 200 | 400 | 800 |
|---|---|---|---|---|
| $h_f$ (m) | 0.01 | 0.210 | 0.820 | 3.27 |

(i) Determine the equivalent sand grain roughness of the pipe. (ii) What is the maximum water discharge at which this pipe will act as smooth pipe? (iii) What is the maximum discharge at which this pipe will act as rough pipe?

## Short answer

From the test data, f becomes constant around 0.0193 in fully rough region. Solving Colebrook-White gives equivalent sand-grain roughness k ≈ 0.45 mm. The pipe behaves smooth for Q ≤ 36 L/s and fully rough for Q ≥ 889 L/s, based on shear Reynolds number limits 4 and 100.

## Answer

**Data.** $D = 0.5$ m, $L = 100$ m, water with $\nu = 1.0\times10^{-6}$ m²/s. Area $A = 0.19635$ m².

#### Friction factor from the test
$f = \dfrac{h_f\,2gD}{LV^2}$ with $V = Q/A$:

| Q (l/s) | V (m/s) | Re | $h_f$ (m) | f |
|---|---|---|---|---|
| 40 | 0.204 | 1.019e+05 | 0.01 | 0.02364 |
| 200 | 1.019 | 5.093e+05 | 0.21 | 0.01986 |
| 400 | 2.037 | 1.019e+06 | 0.82 | 0.01938 |
| 800 | 4.074 | 2.037e+06 | 3.27 | 0.01932 |

At high flows $f$ levels off at about 0.0193, which means the pipe is fully rough there.

#### (i) Equivalent sand-grain roughness
Use the highest flow ($Q$ = 800 l/s, $Re = 2037183$, $f = 0.01932$) in the Colebrook-White equation:
$$
\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)
$$
Solving, $\dfrac{k}{D} = 0.000903$, so
$$
k = 0.000903(500) = 0.45\ \text{mm}
$$
Check: Colebrook with this $k$ gives $f$ = 0.0218, 0.0198, 0.0195 and 0.0193 for the four flows, which agree with the test values.

#### (ii) Maximum discharge for smooth behaviour
The pipe is smooth while $\dfrac{u_*k}{\nu} \le 4$:
$$
u_* = \frac{4\nu}{k} = \frac{4(10^{-6})}{0.45\times10^{-3}} = 0.0089\ \text{m/s}
$$
With smooth-pipe $f$ (Prandtl law) at the limiting $Re$ (found by iteration, $Re \approx 92619$, $f = 0.0183$):
$$
V = \frac{u_*}{\sqrt{f/8}} = 0.185\ \text{m/s},\qquad Q = AV = 36.4\ \text{l/s}
$$

#### (iii) Discharge for fully rough behaviour
The pipe is fully rough when $\dfrac{u_*k}{\nu} \ge 100$, i.e. $u_* = \dfrac{100\nu}{k} = 0.2214$ m/s. For the rough pipe, $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$ gives $f = 0.01914$:
$$
V = \frac{u_*}{\sqrt{f/8}} = 4.526\ \text{m/s}, \qquad Q = 889\ \text{l/s}
$$
This is the limit of the rough regime: for $Q$ above about 889 l/s (it is a minimum, not a maximum, flow for rough behaviour) the pipe is fully rough and $f$ no longer depends on $Re$.

**Answer:** (i) $k \approx 0.45$ mm; (ii) smooth for $Q \lesssim 36$ l/s; (iii) fully rough for $Q \gtrsim 889$ l/s (transition in between).

## Check yourself

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

1. Why is the highest-flow test point used to find k?
   A. f is constant there, so the pipe is fully rough and k/D follows from Colebrook-White
   B. It has the lowest Re
   C. It is laminar
   D. Head loss is smallest
   Answer: A. At high Q, f levels off near 0.0193, giving k/D = 0.000903 and k = 0.45 mm. (Key point: Finding k)

2. Condition for the smooth limit in part (ii)?
   A. u* k / nu <= 4
   B. u* k / nu >= 100
   C. Re <= 2000
   D. f = 0.02
   Answer: A. u* = 4 nu/k = 0.0089 m/s, giving Q about 36 l/s. (Key point: Smooth limit)

3. What is the discharge condition for fully rough flow?
   A. Q below 889 l/s
   B. Q at or above about 889 l/s, where u* k / nu >= 100
   C. Q below 36 l/s
   D. Q equals 400 l/s
   Answer: B. The rough limit is a minimum discharge: above it f no longer depends on Re. (Key point: Rough limit)

4. How is f computed from the test data?
   A. f = Q/(V L)
   B. f = h_f 2 g D / (L V^2)
   C. f = 64/Re
   D. f = h_f / L
   Answer: B. Rearranging Darcy-Weisbach h_f = f L V^2/(2 g D). (Key point: Friction factor from test)

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