---
title: "Derive Hardy Cross correction factor"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:hardy-cross-correction-factor-derivation"
source: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-correction-factor-derivation
site: BCT Engineering
---

# Derive Hardy Cross correction factor

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2071 Magh, 2 marks
- Video: [Hardy Cross Method](https://www.youtube.com/watch?v=pxCWxGHKo2M) — Kenneth Lamb
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-correction-factor-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Derive the expression of correction factor $\Delta Q = -\frac{\sum (rQ_0^2)}{\sum (2rQ_0)}$ for solution of pipe network using Hardy Cross method. Where r is resistance coefficient of pipe and $Q_0$ is initial assumed discharge.

## Short answer

For a closed loop, the sum of head losses must vanish: Σ rQ|Q| = 0. Setting Q = Q0 + ΔQ and neglecting ΔQ² gives Σ rQ0² + 2ΔQ Σ rQ0 ≈ 0, hence the correction ΔQ = –Σ rQ0² / (2 Σ r|Q0|). It is applied loop‑wise until ΔQ is negligible while maintaining flow continuity at junctions.

## Answer

**Setting.** For a loop of pipes, the head loss in each pipe is $h_f = rQ^2$ (clockwise flow positive). For the correct discharges the algebraic sum of head losses around a closed loop must be zero: $\sum rQ|Q| = 0$.

Let $Q_0$ be the assumed discharge in a pipe (satisfying continuity at the nodes) and $\Delta Q$ the correction, the same for all pipes of the loop so that continuity is maintained: $Q = Q_0 + \Delta Q$.
$$
\sum r(Q_0 + \Delta Q)^2 = \sum r\left(Q_0^2 + 2Q_0\Delta Q + \Delta Q^2\right) = 0
$$
Neglect $\Delta Q^2$ since $\Delta Q \ll Q_0$:
$$
\sum rQ_0^2 + 2\Delta Q\sum rQ_0 = 0
$$
$$
\boxed{\Delta Q = -\frac{\sum rQ_0^2}{\sum 2rQ_0}}
$$
In the sum, the signs follow the flow direction (clockwise positive); in the denominator all terms are taken as positive, $2r|Q_0|$. The correction is applied to every pipe of the loop and the process is repeated until $\Delta Q$ is negligible.

## Check yourself

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

1. What condition on a loop leads to the Hardy Cross correction?
   A. All pipes carry equal discharge
   B. Algebraic sum of head losses around the loop is zero
   C. Sum of discharges around the loop is zero
   D. Sum of heads at nodes is zero
   Answer: B. For true discharges, sum rQ|Q| = 0 around a closed loop (clockwise positive). (Key point: Starting condition)

2. Why is the same DeltaQ added to every pipe of the loop?
   A. It preserves continuity at the nodes
   B. It makes every head loss equal
   C. It removes the need for r
   D. It makes Q0 zero
   Answer: A. Adding the same amount around the loop leaves the net flow at each node unchanged. (Key point: Same correction for each pipe)

3. After expanding (Q0 + DeltaQ)^2, which term is dropped?
   A. 2 Q0 DeltaQ
   B. Q0^2
   C. r
   D. DeltaQ^2
   Answer: D. DeltaQ is much smaller than Q0, so DeltaQ^2 is negligible, leaving sum rQ0^2 + 2 DeltaQ sum rQ0 = 0. (Key point: Neglected term)

4. How are signs treated in the denominator, sum 2r|Q0|?
   A. Only clockwise terms are used
   B. Only anticlockwise terms are used
   C. All terms are positive
   D. Signs follow the flow direction
   Answer: C. The derivative of rQ|Q| is 2r|Q|, which is always positive, whereas the numerator keeps signs. (Key point: Denominator signs)

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