---
title: "Flow distribution in five-pipe network"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:network-five-pipes-flow-distribution"
source: https://www.bctengineer.com/q/semester_4/hydraulics/network-five-pipes-flow-distribution
site: BCT Engineering
---

# Flow distribution in five-pipe network

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2069 Poush, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/network-five-pipes-flow-distribution
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Determine the distribution of flow in the pipe network shown in figure below. The value of each pipe is as given below. use n = 2 ($h_f = kQ^n$). [Figure: network with nodes A, B, C, D and a diagonal AC; 100 units enter at A; 20 units leave at B, 50 units at C and 30 units at D]

| Pipe | Length (m) | Diameter (mm) | Friction factor |
|---|---|---|---|
| AB | 300 | 200 | 0.02 |
| BC | 250 | 150 | 0.03 |
| CD | 300 | 100 | 0.02 |
| AD | 250 | 150 | 0.03 |
| AC | 500 | 100 | 0.025 |

## Short answer

Using the Hardy‑Cross method with resistance factors k from the pipe data, the converged flow distribution is: AB = 52.4 A→B, BC = 32.4 B→C, AC = 11.2 A→C, AD = 36.4 A→D, and DC = 6.4 D→C (all in the given flow units), satisfying continuity at all nodes.

## Answer

**Data.** 100 units enter at A; 20 leave at B, 50 at C and 30 at D. Pipes: AB (300 m, 200 mm, f 0.02), BC (250 m, 150 mm, 0.03), CD (300 m, 100 mm, 0.02), AD (250 m, 150 mm, 0.03), AC (500 m, 100 mm, 0.025). $h_f = kQ^2$ with $k = \dfrac{8fL}{\pi^2gD^5}$ (flow in m³/s units; the final distribution depends only on the ratios of $k$, so it applies to any consistent flow unit).

| Pipe | $k$ (s²/m⁵) |
|---|---|
| AB | 1549.3 |
| BC | 8160.7 |
| CD | 49576.1 |
| AD | 8160.7 |
| AC | 103283.6 |

**Initial distribution (continuity):** A to B 30, A to C 30, A to D 40; B to C 10; D to C 10 (check: B: $30 = 20 + 10$; D: $40 = 30 + 10$; C: $30 + 10 + 10 = 50$).

**Loop I** = A-B-C-A (AB +, BC +, CA $-$). **Loop II** = A-C-D-A (AC +, CD $-$ as DC, DA $-$).

#### Hardy-Cross iterations
| Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ |
|---|---|---|---|---|---|---|
| 1 | -90744818.41 | 6453183.05 | 14.062 | 74940519.31 | 7841390.76 | -9.557 |
| 2 | 3527371.24 | 1847342.04 | -1.909 | -34798261.81 | 4066053.33 | 8.558 |
| 3 | -22562136.76 | 3972538.03 | 5.680 | 9604960.48 | 5240079.52 | -1.833 |

Converged after 68 iterations (corrections below $10^{-9}$).

#### Final distribution
| Pipe | Discharge | Direction |
|---|---|---|
| AB | 52.44 | A to B |
| BC | 32.44 | B to C |
| AC | 11.15 | A to C |
| AD | 36.40 | A to D |
| CD | 6.40 | D to C |

Checks: A: $52.44 + 11.15 + 36.40 = 100$; B: $52.44 = 20 + 32.44$; D: $36.40 = 30 + 6.40$; C: $32.44 + 11.15 + 6.40 = 50$.

**Answer:** AB = 52.4, BC = 32.4, AC = 11.2, AD = 36.4 and DC = 6.4 units (all in the directions shown).

## Check yourself

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

1. How is the resistance k of each pipe computed?
   A. k = 8fL/(pi^2 g D^2)
   B. k = fL/(2gD)
   C. k = pi^2 g D^5/(8fL)
   D. k = 8fL/(pi^2 g D^5)
   Answer: D. Darcy-Weisbach gives h_f = 8fLQ^2/(pi^2 g D^5), so k = 8fL/(pi^2 g D^5). (Key point: Resistance of each pipe)

2. Which check validates the initial distribution (A to B 30, A to C 30, A to D 40, B to C 10, D to C 10)?
   A. Equal flow in each loop
   B. k values of loops are equal
   C. Continuity at every node
   D. Equal head loss in all pipes
   Answer: C. At B: 30 = 20 + 10; at D: 40 = 30 + 10; at C: 30 + 10 + 10 = 50. (Key point: Initial distribution)

3. Which final distribution check at A is correct?
   A. 52.44 + 11.15 = 100
   B. 52.44 + 11.15 + 36.40 = 100
   C. 52.44 + 36.40 = 100
   D. 11.15 + 36.40 = 100
   Answer: B. Pipes AB, AC and AD leave A and together carry the 100 units entering. (Key point: Final distribution check)

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