---
title: "Velocity, head loss, discharge in pipes BC, CD, BF, FD (0.1 m)"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:loop-c72"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/loop-c72
site: BCT Engineering
---

# Velocity, head loss, discharge in pipes BC, CD, BF, FD (0.1 m)

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2072 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-c72
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Determine the velocity, head loss and discharge in the pipes BC, CD, BF and FD. The length of pipe BC, CD, BF and FD are 100 m, 200 m, 300 m and 100 m respectively and dia of all pipes are 0.1 m and Darcy's coefficient of friction, f = 0.03.

[Figure: 5 m$^3$/s enters at A and reaches B; the loop B-C-D-F; 1 m$^3$/s leaves at C, 2 m$^3$/s leaves at F, and 2 m$^3$/s leaves from D towards E.]

## Short answer

Hardy-Cross solution gives: BC 2.809 m³/s (357.7 m/s, hf=195605 m); CD 1.809 m³/s (230.3 m/s, 162256 m); BF 2.191 m³/s (279.0 m/s, 356948 m); FD 0.192 m³/s (24.4 m/s, 913 m).

## Answer

The 5 m³/s entering at A reaches B and splits between the paths B–C–D and B–F–D. All pipes are only 0.1 m in diameter, so the numbers are not realistic (velocities of hundreds of m/s), but the method is the same.

#### 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
BC = $x$, CD = $x - 1$ (1 m³/s leaves at C), BF = $5 - x$ and FD = $3 - x$ (2 m³/s leaves at F). At D: CD + FD = 2 ✓. Trial: BC = 3, CD = 2, BF = 2, FD = 0 (a very small value 0.001 is used so that the correction can be calculated).

#### Constants k for each pipe ($f = 0.03$)
| Pipe | L (m) | D (m) | k = 8fL/(π²gD⁵) |
|---|---|---|---|
| BC | 100 | 0.1 | 24788.0572 |
| CD | 200 | 0.1 | 49576.1143 |
| BF | 300 | 0.1 | 74364.1715 |
| FD | 100 | 0.1 | 24788.0572 |

#### Hardy-Cross iterations (clockwise positive; flows in m³/s)
| Iteration | ΔQ | BC | CD | BF | FD |
|---|---|---|---|---|---|
| 1 | -0.1923 | 2.8077 | 1.8077 | 2.1923 | 0.1933 |
| 2 | +0.0014 | 2.8091 | 1.8091 | 2.1909 | 0.1919 |
| 3 | +0.0000 | 2.8091 | 1.8091 | 2.1909 | 0.1919 |
| 4 | +0.0000 | 2.8091 | 1.8091 | 2.1909 | 0.1919 |

#### Results
| Pipe | Discharge (m³/s) | Velocity (m/s) | Head loss (m) |
|---|---|---|---|
| BC | 2.8091 | 357.667 | 195604.8111 |
| CD | 1.8091 | 230.343 | 162256.3632 |
| BF | 2.1909 | 278.953 | 356948.4203 |
| FD | 0.1919 | 24.432 | 912.7540 |

Check: upper path = 357861.2 m; lower path = 357861.2 m ✓

**Answer: BC = 2.809 m³/s (v = 357.7 m/s, h_f = 195605 m); CD = 1.809 m³/s (230.3 m/s, 162256 m); BF = 2.191 m³/s (279.0 m/s, 356948 m); FD = 0.192 m³/s (24.4 m/s, 913 m).** The very large velocities and head losses show that the data (5 m³/s through 100 mm pipes) is not physically practical; the pipe sizes must be increased in a real design.

## Check yourself

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

1. Why are velocities in hundreds of m/s?
   A. The friction factor is too small
   B. Hardy-Cross fails
   C. 5 m^3/s is forced through only 0.1 m diameter pipes
   D. The lengths are too short
   Answer: C. A huge discharge through tiny pipes gives absurd velocity and head loss; sizes must be increased in a real design. (Key point: Why results are unrealistic)

2. If BC carries x, then BF and FD are:
   A. BF = 5 - x, FD = 3 - x
   B. BF = x, FD = x - 2
   C. BF = 5 + x, FD = 3 + x
   D. BF = 5 - x, FD = 2 - x
   Answer: A. 5 enters at B so BF = 5 - x; 2 leaves at F so FD = 3 - x. (Key point: Continuity)

3. k in h_f = kQ^2 depends on:
   A. Only the discharge
   B. Only the diameter
   C. f, L and D (k = 8fL/(pi^2 g D^5))
   D. Only the velocity
   Answer: C. k = 8fL/(pi^2 g D^5), so it depends on pipe friction, length and diameter. (Key point: Constant k)

4. How is the final answer checked?
   A. Velocities of both paths equal
   B. Flows in all pipes equal
   C. Total head loss is zero in each pipe
   D. Head loss along B-C-D equals head loss along B-F-D
   Answer: D. Both paths were found to lose 357861 m, confirming loop balance. (Key point: Convergence check)

## 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
- [Head loss and velocity in loop pipes AB, BC, AC, CD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-c73) — Water Supply Engineering, chapter 7