---
title: "Hardy-Cross network discharge AB, BC, AD, CD"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:hardy-cross-as69"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/hardy-cross-as69
site: BCT Engineering
---

# Hardy-Cross network discharge AB, BC, AD, CD

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2069 Asar, 10 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/hardy-cross-as69
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Calculate the discharge in pipes AB, BC, AD and CD for the water distribution network given below by using Hardy-cross method. The available data of network are as follows:

| Pipe | Length (in meter) | Diameter (in mm) |
|---|---|---|
| AB | 400 | 300 |
| BC | 300 | 200 |
| AD | 500 | 400 |
| CD | 500 | 300 |

Hazen Williams coefficient as 100 for all pipes. Assume other necessary data suitably.

[Figure: loop ABCD; 55 lps enters at A, 20 lps leaves at B, 10 lps leaves at C, 25 lps leaves at D.]

## Short answer

Use Hardy-Cross method with Hazen-Williams equation. After iterations, final discharges: AB = 20.97 lps, BC = 0.97 lps, AD = 34.03 lps, CD = 9.03 lps from D to C.

## Answer

Hardy-Cross method balances the flows in a loop by repeatedly applying a correction until the algebraic sum of head losses around the loop is zero.

#### Data
Hazen-Williams $h = r\,Q^{1.852}$ with $r = 10.67L/(C^{1.852}D^{4.87})$, C = 100. Inflow 55 lps at A; outflow 20 lps at B, 10 lps at C and 25 lps at D. Correction:
$$
\Delta Q = -\frac{\sum h}{1.852\sum |h/Q|}
$$
Take clockwise flow (A→B→C→D) as positive.

#### Trial flows (satisfying continuity at every node)
AB = 35 lps (A→B), BC = 15 lps (B→C), CD = 5 lps (C→D), AD = 20 lps (A→D, anticlockwise, so negative).
Check: A: 35 + 20 = 55 ✓; B: 35 − 20 = 15 ✓; C: 15 − 10 = 5 ✓; D: 5 + 20 = 25 ✓.

#### Iteration 1
| Pipe | Q (lps) | h = rQ^1.852 (m) | h/Q |
|---|---|---|---|
| AB | +35.00 | +0.597 | 0.0171 |
| BC | +15.00 | +0.672 | 0.0448 |
| CD | +5.00 | +0.020 | 0.0041 |
| AD | -20.00 | -0.065 | 0.0033 |

$\sum h = +1.224$ m, $\sum|h/Q| = 0.0692$, so $\Delta Q = -1.224/(1.852 \times 0.0692) = -9.555$ lps.
Corrected flows (the correction is added to clockwise pipes and subtracted from AD): AB = 25.45, BC = 5.45, CD = -4.55, AD = 29.55 lps.

#### Iteration 2
| Pipe | Q (lps) | h = rQ^1.852 (m) | h/Q |
|---|---|---|---|
| AB | +25.45 | +0.331 | 0.0130 |
| BC | +5.45 | +0.103 | 0.0189 |
| CD | -4.55 | -0.017 | 0.0038 |
| AD | -29.55 | -0.134 | 0.0046 |

$\sum h = +0.282$ m, $\Delta Q = -3.791$ lps.

#### Further iterations
Corrections: -0.662, -0.026, -0.000, -0.000 lps; the process stops when ΔQ is negligible (< 0.01 lps).

#### Final discharges
| Pipe | Discharge (lps) | Direction |
|---|---|---|
| AB | 20.97 | A → B |
| BC | 0.97 | B → C |
| AD | 34.03 | A → D |
| CD | 9.03 | D → C |

Checks: at B, 20.97 = 20 + 0.97 ✓; at C, 0.97 + 9.03 = 10 ✓; at D, 34.03 = 25 + 9.03 ✓.

**Answer: AB = 20.97 lps, BC = 0.97 lps, AD = 34.03 lps, CD = 9.03 lps (from D to C).**

## Check yourself

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

1. What does the Hardy-Cross method do?
   A. Fixes all pipe diameters
   B. Applies a correction repeatedly until the algebraic sum of head losses around a loop is zero
   C. Finds only the pressure at nodes
   D. Assumes equal flow in every pipe
   Answer: B. Flows are corrected iteratively until the sum of head losses around the loop is zero. (Key point: Principle of Hardy-Cross)

2. What condition must the trial flows satisfy?
   A. All pipes carry equal flow
   B. Continuity at every node
   C. Zero head loss
   D. Flows must be clockwise only
   Answer: B. Trial flows must satisfy continuity at each node (e.g. at A: 35 + 20 = 55 lps). (Key point: Initial trial flows)

3. With clockwise flow positive, how is pipe AD (flow A to D) treated?
   A. Positive
   B. Ignored
   C. Negative, as it flows anticlockwise
   D. Doubled
   Answer: C. A to D is against the A-B-C-D clockwise direction, so its flow and head loss are negative. (Key point: Sign convention)

4. Which statement about the final flow in CD is right?
   A. 9.03 lps from C to D
   B. 5 lps from C to D
   C. 20 lps from D to C
   D. 9.03 lps from D to C
   Answer: D. The flow in CD reverses to 9.03 lps from D to C; at C, 0.97 + 9.03 = 10 lps. (Key point: Final result)

## Related in Water Supply Engineering

- [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
- [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
- [Design pipes AB, BC, AD (min pressure 1.1 kg/cm2)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apipe-design-ab-bc-ad-110) — Water Supply Engineering, chapter 7, asked 2×
- [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
- [Design pipes AB, BC and AC, C = 110](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-j66) — Water Supply Engineering, chapter 7
- [Design pipes AB and BC, 20,000 population, f = 0.04](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-b66) — Water Supply Engineering, chapter 7

## Related in other subjects

- [Verify suggested discharge distribution (Hardy Cross)](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahardy-cross-verify-suggested-distribution) — Hydraulics, chapter 3
- [Hardy Cross correction derivation and network](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahardy-cross-correction-and-kathmandu-network) — Hydraulics, chapter 3
- [Hardy Cross network with valve on diagonal](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahardy-cross-network-six-links-resistance) — Hydraulics, chapter 3
- [Flow distribution in five-pipe network](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Anetwork-five-pipes-flow-distribution) — Hydraulics, chapter 3