---
title: "Verify suggested discharge distribution (Hardy Cross)"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:hardy-cross-verify-suggested-distribution"
source: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-verify-suggested-distribution
site: BCT Engineering
---

# Verify suggested discharge distribution (Hardy Cross)

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2072 Magh, 8+2 marks
- Video: [Problem on Hardy Cross method II step by step with example in 10 minutes only II](https://www.youtube.com/watch?v=-nYHbAGhd-8) — Sandhu's Corner
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-verify-suggested-distribution
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Verify whether the following suggested distribution of discharge in the pipelines of the network shown in figure below is satisfactory by using Hardy-Cross method. If not, determine the proper distribution. If the elevation at point B is 50 m and pressure head is 40 m and the elevation at D is 40 m, find the pressure at D.

| Line | AB | BC | CD | DA | AC |
|---|---|---|---|---|---|
| Suggested discharge (units) | 58 | 42 | 32 | 18 | 20 |

[Figure: network with nodes A, B, C, D and diagonal AC; resistances r: AB = 2, BC = 4, AC = 1, AD = 5, CD = 1; external flows 100 units at B, 20 units at A, 30 units at C and 50 units at D]

## Short answer

The suggested distribution is not exact; the corrected flows are BA=57.8, BC=42.2, AC=20.6, AD=17.3, CD=32.7. Pressure head at D is 50 – h_BD; the given data produce a negative pressure head (–8132.5), indicating physically inconsistent units.

## Answer

**Data.** Suggested discharges (units): B to A 58, B to C 42, A to C 20, A to D 18, C to D 32 (these satisfy continuity at every node with 100 in at B, 20 out at A, 30 out at C, 50 out at D). $r$: AB 2, BC 4, AC 1, AD 5, CD 1; $h_f = rQ^2$ ($n = 2$).

**Loop I** = A-C-B-A (AC +, CB $-$, BA +). **Loop II** = A-D-C-A (AD +, DC $-$, CA $-$).

#### Test of the suggested distribution (first iteration)
Loop I: $\sum rQ|Q| = 1(20)^2 - 4(42)^2 + 2(58)^2 = 400 - 7056 + 6728 = 72$, $2\sum r|Q| = 2(20 + 168 + 116) = 608$, so $\Delta Q_I = -\dfrac{72}{608} = -0.118$.

Loop II: $\sum rQ|Q| = 5(18)^2 - 1(32)^2 - 1(20)^2 = 1620 - 1024 - 400 = 196$, $2\sum r|Q| = 2(90 + 32 + 20) = 284$, so $\Delta Q_{II} = -\dfrac{196}{284} = -0.690$.

The loop head-loss sums are not zero (72 and 196), so strictly the suggested distribution is **not exact**, although the corrections are small (under 1.2% of the pipe discharges). Continuing the 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 | 72.00 | 608.00 | -0.118 | 196.00 | 284.00 | -0.690 |
| 2 | 27.90 | 609.62 | -0.046 | 6.32 | 279.62 | -0.023 |
| 3 | 0.93 | 609.75 | -0.002 | 1.88 | 279.40 | -0.007 |

#### Corrected distribution
| Pipe | Suggested | Corrected |
|---|---|---|
| B to A | 58 | 57.83 |
| B to C | 42 | 42.17 |
| A to C | 20 | 20.55 |
| A to D | 18 | 17.28 |
| C to D | 32 | 32.72 |

(Checks: $57.83 + 42.17 = 100$; at D, $17.28 + 32.72 = 50$.) The suggested values are therefore correct to within about 4%; the correction mainly shifts about 0.7 unit of flow from pipe AD to the route A-C-D.

#### Pressure at D
Piezometric head at B: $H_B = 50 + 40 = 90$ m. Head loss from B to D via A: $h_{BD} = h_{BA} + h_{AD} = 2(57.83)^2 + 5(17.28)^2 = 6689.5 + 1493.0 = 8182.5$ (the same through C: $7111.9 + 1070.6$). Then
$$
H_D = H_B - h_{BD}, \qquad \frac{p_D}{\gamma} = H_D - z_D = 90 - h_{BD} - 40 = 50 - h_{BD}
$$
With $r$ and $Q$ used exactly as given, $h_{BD} = 8182.5$, so $p_D/\gamma = 50 - 8182.5 = -8132.5$, which is physically impossible. This shows that, with the "units" of discharge in the data, $rQ^2$ does not come out in metres of head. The method is still the one above: the pressure head at D is $50 - h_{BD}$ (with $h_{BD}$ in metres), and the pressure is $p_D = \gamma\,(50 - h_{BD})$.

**Answer:** The suggested distribution is nearly, but not exactly, satisfactory; the proper values are BA = 57.8, BC = 42.2, AC = 20.6, AD = 17.3, CD = 32.7. Pressure head at D: $p_D/\gamma = 50 - h_{BD}$; with the data taken literally, $h_{BD} = 8182.5$ and the result is negative ($-8132.5$), so the given data do not give a physically meaningful pressure in metres.

## Check yourself

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

1. What shows that the suggested distribution is not exact?
   A. The corrections are negative
   B. The loop sums of rQ|Q| are not zero (72 and 196)
   C. Continuity fails at node A
   D. The pipe resistances differ
   Answer: B. A correct distribution has zero head-loss sum around every loop; non-zero sums mean corrections are needed. (Key point: Test of a suggested distribution)

2. For loop I, sum rQ|Q| = 72 and 2 sum r|Q| = 608. What is DeltaQ_I?
   A. -0.118
   B. +0.118
   C. -8.44
   D. -0.690
   Answer: A. DeltaQ = -72/608 = -0.118. (Key point: Loop I correction)

3. With elevation 50 m and pressure head 40 m at B, what is the piezometric head used to find the pressure at D?
   A. 10 m
   B. 40 m
   C. 50 m
   D. 90 m
   Answer: D. H_B = z + p/gamma = 50 + 40 = 90 m; subtracting the head loss B to D gives H_D, then p_D/gamma = H_D - z_D. (Key point: Piezometric head at B)

## Related in Hydraulics

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