---
title: "Velocity, head loss, discharge in loop pipes BC, CD, BF, FD"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:loop-a76"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/loop-a76
site: BCT Engineering
---

# Velocity, head loss, discharge in loop pipes BC, CD, BF, FD

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2076 Asoj, 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-a76
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Determine velocity, head loss and discharge in the pipes BC, CD, BF and FD. If diameter of pipes BC, CD, BF and FD are 1200 mm, 800 mm, 1000 mm, and 1000 mm respectively. Assume length of pipes are 100 m and coefficient of friction, f = 0.03.

[Figure: 4 m$^3$/s enters at A and flows to B; the loop B-C-D-F has C and F as the upper and lower corners; outflows 1 m$^3$/s at C, 1 m$^3$/s at F and 2 m$^3$/s at D towards E.]

## Short answer

Using Hardy-Cross, the flows are: BC 2.025 m³/s (v=1.79 m/s, hf=0.408 m); CD 1.025 m³/s (2.04 m/s, 0.794 m); BF 1.975 m³/s (2.51 m/s, 0.967 m); FD 0.975 m³/s (1.24 m/s, 0.236 m).

## Answer

The 4 m³/s entering at A reaches B and splits between the upper path B–C–D and the lower path B–F–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
Let BC carry $x$ m³/s. Continuity gives CD = $x - 1$ (1 m³/s leaves at C), BF = $4 - x$ and FD = $3 - x$ (1 m³/s leaves at F). At D: CD + FD = 2 m³/s ✓. The loop condition is head loss B→C→D = head loss B→F→D. Trial: BC = 2, CD = 1, BF = 2, FD = 1 (the inflow at B is split equally between the two branches).

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

#### Hardy-Cross iterations (clockwise positive; flows in m³/s)
| Iteration | ΔQ | BC | CD | BF | FD |
|---|---|---|---|---|---|
| 1 | +0.0249 | 2.0249 | 1.0249 | 1.9751 | 0.9751 |
| 2 | -0.0001 | 2.0248 | 1.0248 | 1.9752 | 0.9752 |
| 3 | -0.0000 | 2.0248 | 1.0248 | 1.9752 | 0.9752 |

#### Results
| Pipe | Discharge (m³/s) | Velocity (m/s) | Head loss (m) |
|---|---|---|---|
| BC | 2.0248 | 1.790 | 0.4084 |
| CD | 1.0248 | 2.039 | 0.7944 |
| BF | 1.9752 | 2.515 | 0.9671 |
| FD | 0.9752 | 1.242 | 0.2357 |

Check: upper path = 0.4084 + 0.7944 = 1.2028 m; lower path = 0.9671 + 0.2357 = 1.2028 m ✓

**Answer: BC = 2.025 m³/s (v = 1.79 m/s, h_f = 0.408 m); CD = 1.025 m³/s (2.04 m/s, 0.794 m); BF = 1.975 m³/s (2.51 m/s, 0.967 m); FD = 0.975 m³/s (1.24 m/s, 0.236 m).**

## Check yourself

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

1. Which formula gives head loss for this problem (f given)?
   A. Hazen-Williams with C
   B. Manning
   C. Darcy-Weisbach h_f = 8fLQ^2/(pi^2 g D^5) = kQ^2
   D. Hardy-Cross h = Q/D
   Answer: C. Since a friction factor f is given, Darcy-Weisbach is used, written as h_f = kQ^2. (Key point: Head loss formula)

2. What condition fixes the split between BCD and BFD?
   A. Head loss B-C-D equals head loss B-F-D
   B. Both paths carry equal flow
   C. Both paths have equal diameter
   D. Velocity is equal in all pipes
   Answer: A. Both paths run between the same two nodes B and D, so their head losses are equal. (Key point: Loop condition)

3. If BC carries x m^3/s, what do CD and FD carry?
   A. CD = x and FD = 4 - x
   B. CD = x - 1 and FD = 3 - x
   C. CD = x + 1 and FD = 2 - x
   D. CD = 1 and FD = 1
   Answer: B. 1 m^3/s leaves at C so CD = x - 1; BF = 4 - x and 1 leaves at F so FD = 3 - x. (Key point: Continuity)

4. For h = kQ^2, the Hardy-Cross correction is:
   A. -sum kQ|Q| / (2 sum k|Q|)
   B. -sum kQ / sum k
   C. sum kQ^2 / sum Q
   D. -sum kQ|Q| / (1.852 sum k|Q|)
   Answer: A. With exponent 2, dQ = -sum kQ|Q| / (2 sum k|Q|). (Key point: Correction formula)

## Related in Water Supply Engineering

- [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
- [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