---
title: "Discharge and head loss in horizontal pipe"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:pipe-discharge-head-loss"
source: https://www.bctengineer.com/q/semester_6/hydro_power/pipe-discharge-head-loss
site: BCT Engineering
---

# Discharge and head loss in horizontal pipe

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2081 Chaitra, 1+2 marks
- Video: [Fluid Mechanics: Topic 8.3 - Pressure drop and head loss in pipe flow](https://www.youtube.com/watch?v=pU_I5sc_-B8) — CPPMechEngTutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/pipe-discharge-head-loss
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Water flows through a 1 km long horizontal pipe of 150 mm diameter with a velocity of 2 m/s. Calculate the discharge and head loss in the pipe if the friction factor is 0.02.

## Short answer

Discharge Q = A·V = 0.0353 m³/s. Head loss from Darcy–Weisbach: h_f = f (L/D)(V²/2g) = 0.02×(1000/0.15)×(2²/19.62) = 27.18 m.

## Answer

**Given:** $L = 1000$ m, $D = 0.15$ m, $V = 2$ m/s, $f = 0.02$ (Darcy).

#### Discharge
$$
\begin{aligned}
Q &= \frac{\pi}{4}D^2 V = \frac{\pi}{4}(0.15)^2 \times 2 \\
&= 0.01767 \times 2 = 0.03534\ \text{m}^3/\text{s}
\end{aligned}
$$

#### Head loss (Darcy–Weisbach)
$$
\begin{aligned}
h_f &= f\frac{L}{D}\frac{V^2}{2g} = 0.02 \times \frac{1000}{0.15} \times \frac{2^2}{2 \times 9.81} \\
&= 133.33 \times 0.2039 = 27.18\ \text{m}
\end{aligned}
$$

**Answer:** $Q = 0.0353$ m³/s (35.3 L/s); head loss $h_f = 27.18$ m of water.

## Check yourself

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

1. How is the discharge found?
   A. Q = pi D V
   B. Q = (pi/4) D^2 V
   C. Q = (pi/4) D V^2
   D. Q = D V
   Answer: B. Q = area x velocity = 0.01767 x 2 = 0.03534 m^3/s. (Key point: Discharge formula)

2. Which equation gives the head loss?
   A. h_f = f D V^2/L
   B. h_f = V^2/2g
   C. h_f = f L V
   D. h_f = f (L/D) V^2/2g
   Answer: D. Darcy-Weisbach. (Key point: Head loss formula)

3. What is the head loss?
   A. 2.72 m
   B. 13.6 m
   C. 27.18 m
   D. 54.4 m
   Answer: C. 0.02 x (1000/0.15) x (4/19.62) = 133.33 x 0.2039 = 27.18 m. (Key point: Numerical result)

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