What condition on a loop leads to the Hardy Cross correction?

Algebraic sum of head losses around the loop is zero

All pipes carry equal discharge

Sum of discharges around the loop is zero

Why is the same DeltaQ added to every pipe of the loop?

It preserves continuity at the nodes

How are signs treated in the denominator, sum 2r|Q0|?

Only anticlockwise terms are used

2071 Magh · 2 marks

Derive Hardy Cross correction factor

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.

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.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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