Water topic 12 of 18 — free theory

Pipe Networks

Hydraulics
FE Civil
PE WR&E

A distribution system is a maze of loops and branches, but the whole maze obeys two rules: what flows into a junction flows out, and the head lost around any loop sums to zero. Everything else — equivalent pipes, flow splits, Hardy Cross — is bookkeeping on those two rules.

FE + PE bridgeFE Civil · Hydraulics and Hydrologic Systems (8–12)PE WRE · Hydraulics—Closed Conduit (7–11)

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The two governing rules

Write every pipe's loss as h_L = r Q|Q|ⁿ⁻¹ (i.e. h_L = rQⁿ with the sign of Q). Then a network is solved when both of these hold everywhere:

Node continuity: ΣQ_in = ΣQ_out + q_demand

q_demand	water withdrawn at the junction (0 for a plain pipe junction)
Sign convention	inflows positive, outflows negative — the algebraic sum is zero

Loop energy: Σh_L = 0 around every closed loop

h_L	signed head loss: positive when the assumed flow follows the loop direction
n	head-loss exponent: ≈ 2.0 for Darcy-Weisbach (f treated as constant), 1.852 for Hazen-Williams

The exponent matters. Double the flow and a Darcy-Weisbach loss roughly quadruples (2² = 4); a Hazen-Williams loss grows by 2^1.852 ≈ 3.61. An exam option showing ×4 is the Darcy answer; ×3.6 is the Hazen-Williams one.

Series and parallel — equivalent pipes

Two pipes in series carry the same Q and their losses add, so their resistances add. Two pipes in parallel share the same head loss and their flows add, so it is the square-roots of the resistances that combine:

Series: r_eq = r₁ + r₂ + …

Parallel: 1√r_eq = 1√r₁ + 1√r₂ + … (for h_L = rQ²)

r	pipe resistance from h_L = rQⁿ (Darcy: r = 8fL/(π²gD⁵))
Flow split (parallel)	Q₁/Q₂ = (r₂/r₁)^1/n — the lower-resistance branch takes the larger share

Sanity check a split before trusting it: the branch with the bigger diameter (or shorter length) should carry the bigger flow. If your numbers say otherwise, you flipped the ratio.

The Hardy Cross loop method

For a looped network you guess the flows, compute the head-loss imbalance around each loop, then correct every pipe by ΔQ. One iteration usually gets close; exam problems ask for exactly one.

ΔQ = −Σh_Ln · Σ|h_L/Q| applied to every pipe in the loop

Σh_L	algebraic sum of signed head losses around the loop (drives the correction)
Σ\|h_L/Q\|	sum of absolute values — always positive, always in the denominator
n	the same exponent as the head-loss law you used (2 for Darcy, 1.852 for Hazen-Williams)
Corrected flow	Q_new = Q_old + ΔQ for pipes in the loop direction; a negative result just means the real flow opposes your arrow

Pipes shared between two loops get two corrections — one from each loop — with the sign depending on whether the loop directions agree on that pipe. Forgetting the second correction is the classic multi-loop error.

PE depth: reading a network

PE questions reward qualitative network sense as much as calculation. Three patterns cover most of them:

Closing a valve redistributes flow. Shut a branch and the remaining parallel paths carry more — their head losses rise roughly as Q², so the system head at the pump climbs and the total flow drops. Sketch the new system curve mentally before touching numbers.

Negative corrected flow is information, not an error. If Hardy Cross flips a pipe's sign, that branch genuinely flows opposite to your guess — common in loops fed from two sides. Keep the negative value and continue; do not re-guess from scratch.

Equivalent length for minor losses. Valves and fittings in a network are often rolled into an equivalent pipe length (L_e = KD/f) so the whole system collapses to series/parallel resistances. Watch whether the exam gives you K values or expects you to look them up.

PE trap: Hardy Cross balances loops; node continuity must already be satisfied by your initial guess. If inflows do not match outflows at a junction before you start, no number of loop corrections will fix the imbalance.

Worked example One Hardy Cross iteration on a loop

Given: a single loop with h_L = rQ|Q| (n = 2). Assumed flows (clockwise positive): pipe 1, r = 200, Q = +0.40 m³/s; pipe 2, r = 100, Q = +0.30 m³/s; pipe 3, r = 150, Q = −0.20 m³/s.

Solution:

Signed losses: h₁ = 200(0.40)(0.40) = 32.0 m; h₂ = 100(0.30)(0.30) = 9.0 m; h₃ = 150(−0.20)(0.20) = −6.0 m. Σh_L = 32.0 + 9.0 − 6.0 = 35.0 m.
Denominator: Σ|h_L/Q| = 200(0.40) + 100(0.30) + 150(0.20) = 80 + 30 + 30 = 140.
ΔQ = −35.0/(2 × 140) = −0.125 m³/s.
Corrected flows: Q₁ = 0.275, Q₂ = 0.175, Q₃ = −0.325 m³/s. The new loop imbalance is only 2.34 m — down from 35 m in one step.

Answer: ΔQ = −0.125 m³/s; corrected flows 0.275, 0.175 and −0.325 m³/s.

Pipe Networks

The two governing rules

Series and parallel — equivalent pipes

The Hardy Cross loop method

PE depth: reading a network

Worked example One Hardy Cross iteration on a loop

Pipe Networks

1. Two pipes with resistances r₁ = 40 and r₂ = 60 s²/m⁵ (h_L = rQ²) are connected in series. What is the equivalent resistance?

2. In the Hardy Cross loop-balancing method, the flow correction applied to each pipe in a loop is:

3. A total flow of 3.0 m³/s splits between two parallel branches with r₁ = 100 and r₂ = 400 s²/m⁵ (h_L = rQ²). What are the branch flows Q₁ / Q₂?

4. Two pipes deliver 0.50 and 0.30 m³/s into a junction, and one pipe carries 0.65 m³/s away. What demand is withdrawn at the junction?

5. A pipe’s head loss is computed with the Hazen-Williams equation. If the flow doubles, by what factor does the head loss increase?

6. A loop uses h_L = rQ|Q|. Assumed flows (clockwise positive): pipe 1, r = 200 s²/m⁵, Q = +0.40 m³/s; pipe 2, r = 100, Q = +0.30 m³/s; pipe 3, r = 150, Q = −0.20 m³/s. Perform one Hardy Cross iteration: what is ΔQ and the corrected flows?

7. Two parallel branches have r₁ = 120 and r₂ = 480 s²/m⁵ (h_L = rQ²). What single pipe is equivalent to the pair, and what total flow passes at 25 m of head loss?

Preparing for the PE Civil: Water Resources & Environmental exam?

Pipe Networks

The two governing rules

Series and parallel — equivalent pipes

The Hardy Cross loop method

PE depth: reading a network

Worked example One Hardy Cross iteration on a loop

Pipe Networks

1. Two pipes with resistances r₁ = 40 and r₂ = 60 s²/m⁵ (hL = rQ²) are connected in series. What is the equivalent resistance?

2. In the Hardy Cross loop-balancing method, the flow correction applied to each pipe in a loop is:

3. A total flow of 3.0 m³/s splits between two parallel branches with r₁ = 100 and r₂ = 400 s²/m⁵ (hL = rQ²). What are the branch flows Q₁ / Q₂?

4. Two pipes deliver 0.50 and 0.30 m³/s into a junction, and one pipe carries 0.65 m³/s away. What demand is withdrawn at the junction?

5. A pipe’s head loss is computed with the Hazen-Williams equation. If the flow doubles, by what factor does the head loss increase?

6. A loop uses hL = rQ|Q|. Assumed flows (clockwise positive): pipe 1, r = 200 s²/m⁵, Q = +0.40 m³/s; pipe 2, r = 100, Q = +0.30 m³/s; pipe 3, r = 150, Q = −0.20 m³/s. Perform one Hardy Cross iteration: what is ΔQ and the corrected flows?

7. Two parallel branches have r₁ = 120 and r₂ = 480 s²/m⁵ (hL = rQ²). What single pipe is equivalent to the pair, and what total flow passes at 25 m of head loss?

Preparing for the PE Civil: Water Resources & Environmental exam?

1. Two pipes with resistances r₁ = 40 and r₂ = 60 s²/m⁵ (h_L = rQ²) are connected in series. What is the equivalent resistance?

3. A total flow of 3.0 m³/s splits between two parallel branches with r₁ = 100 and r₂ = 400 s²/m⁵ (h_L = rQ²). What are the branch flows Q₁ / Q₂?

6. A loop uses h_L = rQ|Q|. Assumed flows (clockwise positive): pipe 1, r = 200 s²/m⁵, Q = +0.40 m³/s; pipe 2, r = 100, Q = +0.30 m³/s; pipe 3, r = 150, Q = −0.20 m³/s. Perform one Hardy Cross iteration: what is ΔQ and the corrected flows?

7. Two parallel branches have r₁ = 120 and r₂ = 480 s²/m⁵ (h_L = rQ²). What single pipe is equivalent to the pair, and what total flow passes at 25 m of head loss?