Hardy-Cross for Looped Pipe Networks, Without the Mystery
Continuity first, then loop head balance. How ΔQ is calculated, a four-pipe worked loop, equivalent pipes, and when to move on to Newton-Raphson.
Key takeaways
- Initial flows must satisfy node continuity; the iteration enforces loop energy.
- h = K Q|Q| with K = 8 f L /(π² g D⁵) keeps the sign when flow reverses.
- ΔQ = −Σh / Σ(2K|Q|) is a Newton step on the loop residual.
- Skeletonize series/parallel groups with Dupuit before you draw loops.
A looped water network has more pipes than you can solve by hand with “same Q in series, same h in parallel.” Hardy-Cross is the classic loop method: guess flows that already satisfy continuity, then correct them until the head around every loop closes.
Two laws
- Nodes: mass is conserved. Inflow equals outflow at every junction (including demands).
- Loops: energy is conserved. Signed head loss around a closed loop is zero (no pump) or equals the pump head.
Series pipes automatically satisfy (1) with one Q. Parallel pipes automatically satisfy (2) with one h. A rectangle of four pipes satisfies neither until you iterate.
Resistance and the correction
Add +ΔQ to pipes that run with the loop direction and −ΔQ to pipes that run against it. Repeat. When |Σh| is a few millimetres of head, you are done for hand work.
Four-pipe loop (the textbook case)
Inflow Qin at N1, outflow Qin at N3. Clockwise: P1 (N1→N2), P2 (N2→N3), P3 (N4→N3, against clockwise), P4 (N1→N4, against clockwise). A legal first guess is Q1 = Q2 = 0.6 Qin and Q3 = Q4 = 0.4 Qin. Continuity holds; energy does not, until Hardy-Cross moves flow onto the lower-resistance path.
Open solver: Hardy-Cross loop solverSkeletonizing with equivalent pipes
Before you draw loops, collapse obvious series chains and parallel twins into one equivalent pipe (Dupuit). Series: K_eq = Σ K_i. Parallel: 1/√K_eq = Σ 1/√K_i. Fewer elements, same K. That is how large city models stay tractable even today.
Open solver: Equivalent pipe (Dupuit)Open solver: Series networkOpen solver: Parallel networkHardy-Cross vs Newton-Raphson
Hardy-Cross corrects one loop at a time and is easy to see. Newton-Raphson (or the global gradient algorithm) solves all loops and nodes together and converges faster on large GIS models. For a two-loop exam problem, Hardy-Cross is still the right tool to learn.
Open solver: Three-reservoir problemFrequently asked questions
A tree has no loops, so there is nothing to correct. Solve from known heads or known outflows with series/parallel reductions, or use a junction solver such as the three-reservoir problem.