---
title: "Head loss and velocity in loop pipes AB, BC, AC, CD"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:loop-c73"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/loop-c73
site: BCT Engineering
---

# Head loss and velocity in loop pipes AB, BC, AC, CD

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2073 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/loop-c73
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

A part of the water distribution network is shown in figure below. If the diameter of pipes AB, BC, AC and CD are 1200 mm, 800 mm, 1000 mm and 1000 mm respectively, calculate the head loss and velocity in pipes AB, BC, AC and CD. Assume length of all the pipes are 100 m and coefficient of friction, f = 0.03.

[Figure: loop A-B-C-D; 4 m$^3$/s enters at A, 1 m$^3$/s leaves at B, 1 m$^3$/s leaves at C, 2 m$^3$/s leaves at D.]

## Short answer

Hardy-Cross yields: AB 1.943 m³/s (1.72 m/s, 0.376 m); BC 0.943 m³/s (1.88 m/s, 0.673 m); AC 2.057 m³/s (2.62 m/s, 1.049 m); CD 2 m³/s (2.55 m/s, 0.992 m).

## Answer

The 4 m³/s entering at A splits into paths A–B–C and A–C; pipe CD takes 2 m³/s to D. The split is found by Hardy-Cross.

#### Formulas
Darcy-Weisbach: $h_f = \dfrac{8 f L Q^2}{\pi^2 g D^5} = kQ^2$, velocity $v = Q/(\pi D^2/4)$. Hardy-Cross correction for the loop:
$$
\Delta Q = -\frac{\sum kQ|Q|}{2\sum k|Q|}
$$

#### Flow distribution
Pipe CD carries 2 m³/s (outflow at D) and its flow is fixed. In the triangle A–B–C let AB carry $x$; then BC = $x - 1$ (1 m³/s leaves at B) and AC = $4 - x$. At C: BC + AC = 3 = 1 (outflow at C) + 2 (to D) ✓. Loop condition: $h_{AB} + h_{BC} = h_{AC}$. Trial: AB = 2, BC = 1, AC = 2.

#### Constants k for each pipe ($f = 0.03$)
| Pipe | L (m) | D (m) | k = 8fL/(π²gD⁵) |
|---|---|---|---|
| AB | 100 | 1.2 | 0.0996 |
| BC | 100 | 0.8 | 0.7565 |
| AC | 100 | 1.0 | 0.2479 |

#### Hardy-Cross iterations (clockwise positive; flows in m³/s)
| Iteration | ΔQ | AB | BC | AC |
|---|---|---|---|---|
| 1 | -0.0563 | 1.9437 | 0.9437 | 2.0563 |
| 2 | -0.0007 | 1.9430 | 0.9430 | 2.0570 |
| 3 | -0.0000 | 1.9430 | 0.9430 | 2.0570 |
| 4 | -0.0000 | 1.9430 | 0.9430 | 2.0570 |

#### Results
| Pipe | Discharge (m³/s) | Velocity (m/s) | Head loss (m) |
|---|---|---|---|
| AB | 1.9430 | 1.718 | 0.3761 |
| BC | 0.9430 | 1.876 | 0.6727 |
| AC | 2.0570 | 2.619 | 1.0488 |

Pipe CD: $k = 0.2479$, $Q = 2$ m³/s, $h_f = 0.2479 \times 4 = 0.9915$ m, $v = 2.546$ m/s.
Check: AB + BC = 0.3761 + 0.6727 = 1.0488 m = AC = 1.0488 m ✓

**Answer: AB: 1.943 m³/s, 1.72 m/s, 0.376 m; BC: 0.943 m³/s, 1.88 m/s, 0.673 m; AC: 2.057 m³/s, 2.62 m/s, 1.049 m; CD: 2 m³/s, 2.55 m/s, 0.992 m.**

## Check yourself

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

1. Why is the flow in pipe CD not part of the iteration?
   A. It is outside the loop and carries the fixed 2 m^3/s outflow at D
   B. It has zero friction
   C. It is the largest pipe
   D. It is the shortest pipe
   Answer: A. CD is only a branch to D, so its flow equals the 2 m^3/s demand at D. (Key point: Fixed flow in CD)

2. The loop condition for triangle A-B-C is:
   A. h_AB = h_BC
   B. h_AB + h_BC = h_AC
   C. h_AB = h_AC + h_BC
   D. h_AC = 0
   Answer: B. Path A-B-C and path A-C join the same nodes so their head losses are equal. (Key point: Loop condition)

3. If AB carries x, what do BC and AC carry?
   A. BC = x, AC = 4 - x
   B. BC = x - 1, AC = 4 - x
   C. BC = x + 1, AC = 4 + x
   D. BC = 1 - x, AC = x
   Answer: B. 1 m^3/s leaves at B so BC = x - 1; the rest of the 4 m^3/s goes via AC. (Key point: Continuity)

4. Velocity in a pipe is obtained from:
   A. v = Q x A
   B. v = Q/(pi D^2/4)
   C. v = h_f/L
   D. v = Q/D
   Answer: B. v = Q/A with A = pi D^2/4. (Key point: Velocity)

## Related in Water Supply Engineering

- [Velocity, head loss, discharge in loop pipes BC, CD, BF, FD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-a76) — Water Supply Engineering, chapter 7
- [Velocity, head loss, discharge in pipes BC, CD, BF, FD (0.1 m)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-c72) — Water Supply Engineering, chapter 7
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