§ 4.5  Module 4 — Head Loss: Everything That Is Not Straight Pipe

Headers, Manifolds and Unequal Paths

Three identical pumps on one header are not three identical systems. The equipment schedule cannot tell you which one loses; only the layout can.

4.5.1Your own branch, before anyone else is running

Start with one pump and count what stands between its discharge nozzle and the common header. A gradual increaser off the nozzle. A check valve, because the other units will push back down this pipe the moment this one stops. An isolation valve, because somebody has to be able to pull the pump. Two elbows, because the nozzle and the header are neither coaxial nor coplanar. And the tee into the header. Six fittings and five metres of pipe: that assembly is the branch resistance, and it belongs to this unit alone.

Put 40 L/s through 150 mm of it. The velocity is 2.264 m/s, the velocity head 0.2612 m, and ΣK·v²/2g 1.515 m. The five metres of pipe, at Re = 2.98 × 10⁵ and f = 0.0232 by Colebrook, contribute 0.202 m. The branch costs 1.717 m, and 88.2% of it is fittings — the module epigraph as a measurement rather than a slogan.

The lever on that number is diameter, and it is a violent one. Hold the flow at 40 L/s: the same six fittings cost 9.356 m on a 100 mm branch, 1.717 m on 150 mm, 0.525 m on 200 mm. The minor term goes as D⁻⁴ at fixed flow, because v goes as D⁻² and the velocity head squares it; friction goes as D⁻⁵. Shortening the branch barely registers — halving it to 2.5 m saves 0.101 m, six per cent, because length acts only on the 12% that is friction.

Run this unit on its own and the station is short: 18.0 m of static lift, 1.507 m in the force main, 2.727 m in the branch — the branch rises because the flow does — for 50.43 L/s at a total dynamic head of 22.235 m. The branch, which no drawing calls part of the pump and no schedule calls part of the main, is nearly twice the force main's contribution — and in a moment it will be the only thing every unit has in common.

4.5.2The header makes the units interact

Now start the other two. A manifold — a common discharge header — is not three parallel pipes. It is one pipe that collects, and the flow in any length of it is the sum of the flows from every unit beyond it. Here the branch tees land 2.0 m apart — the pump bay pitch, a structural decision taken before anyone computed a head loss. With the main leaving past the last unit, the run between tee 1 and tee 2 carries unit 1 alone at 39.87 L/s and 0.812 m/s; between tee 2 and tee 3, 79.85 L/s at 1.627 m/s; only the stub carries all 120.08 L/s, at 2.446 m/s. One header, three velocities, three loss rates.

Two charges accumulate along that path. The first is ordinary pipe friction from Module 3, at whatever velocity the segment carries. The second is the combining loss at every tee the flow passes straight through — K ≈ 0.4 on the velocity leaving the junction, against the K ≈ 1.3 the joining branch pays. At full flow that tee is 0.1219 m while the two metres of pipe it sits on are 0.0497 m: the fitting outweighs the pipe by 2.45 times, which is why a header cannot be costed by length either.

Put the takeoff in the middle of the header instead and the geometry becomes a mirror. The two outer units are the same distance from the main through the same number of junctions, so they deliver exactly the same flow: 40.01 L/s each, against the middle unit's 40.23 L/s, 120.26 L/s in total, a spread of 0.548% of the mean. Every unit is asked for about 28 m, of which 18.0 m is static lift, 8.227 m the force main and 1.7 m its own branch; the header contributes 0.128 m to the outer pair and nothing to the middle one. Notice how small that is. A well-proportioned header at 2.45 m/s barely biases anything, and the honest headline of this lesson is that the imbalance is a design outcome.

4.5.3Move the takeoff, and somebody starts losing

Take the force main off the end of the header instead — past the last unit, the way it gets drawn when the site plan puts the main at one end of the building. Nothing about the pumps changes; nothing about the branches changes. The flows become 39.87, 39.98 and 40.22 L/s, spread 0.875%, and the units are now strictly ordered by their distance from the takeoff. Unit 1 pays 0.234 m of header path, unit 3 pays 0.030 m, and the 0.204 m between them is the whole difference.

Where does that head come from? Only one place. Each unit sits where the head it makes equals the head its own path demands, so the disadvantaged unit slides back up its curve until it balances — less flow at a higher head. Unit 1 produces 28.142 m and unit 3 27.968 m: the opposite of what "works harder" suggests, and exactly what a rising demand against a falling supply must do. The pump enters this lesson only through that falling relation — a catalogue shutoff of 38.0 m and a quadratic through 28.08 m at 40 L/s, taken as a boundary condition. Where the shape comes from is §6.2; what parallel operation does to capacity is §7.3.

That gives the one formula worth carrying out of this lesson. Write each unit's balance as H₀ − K·q² = h_node + path, subtract any two, and the squares factor: (q_a − q_b)(q_a + q_b) = Δh/K, so

    Δq = Δh / (2·K·q̄) = Δh / (pump-curve steepness + branch steepness)

Here those steepnesses are 496.6 and 86.0 m per m³/s, so every extra metre of header path costs the unit paying it 1.72 L/s. A difference of squares rather than a Taylor series, so it is very nearly exact — within 0.025% of the full network solve even at a 12% imbalance. It also names the two things that decide the split: the header's asymmetry, and the pump curve's steepness. A flat curve shares worst.

4.5.4What it costs: wear, and the duty rotation that does not work

Under a per cent, none of this would matter. So take the case where it does, and it is not hypothetical: a station uprated to three larger pumps on its original header. That header is 150 mm, the bays are 5.0 m apart, the main leaves past the last unit. The flows become 36.90, 37.89 and 40.64 L/s — a spread of 9.71%, a path difference of 2.106 m, and 115.44 L/s in total against the 120.29 L/s the same pumps would deliver on a 300 mm header. Widen the bays to 6.0 m and the spread passes 10%. The tell is the header velocity: 6.53 m/s. Nobody designs that; it arrives when the pumps are replaced and the header is not.

Now count the consequences, because the flow figures are the least of them.

1. Equal run hours do not equal equal duty. Give each unit 100 hours and they deliver 13 286, 13 642 and 14 630 m³. The near unit is 10.1% ahead on throughput, exactly the ratio of the flows — the arithmetic is unavoidable, not approximate.
2. The units are not at the same place on their curves. They are making 29.556, 29.097 and 27.760 m. Best efficiency happens at one flow, so three identical pumps at three different flows cannot all be near it, and the far unit is furthest left. The 70–120% of best-efficiency flow commonly cited as the preferred operating region in the Hydraulic Institute's guideline on operating regions (ANSI/HI 9.6.3) is a wide band and a 10% spread sits inside it — but the spread is permanent, and it eats margin that the wet well level and thirty years of pipe aging also want.
3. A lead pump and a lag pump are different systems. Running alone, the far unit has no shared header segment to pay for: on this station it delivers 50.33 L/s with 0.076 m of path. Starting third, it pays 2.503 m and delivers 36.90 L/s. Each unit therefore has as many system curves as there are combinations of its neighbours, and "lead" is the cheapest of them.
4. Rotation intended to equalise wear does the opposite of what it says. A scheme that hands out equal hours is silently handing out unequal volume, unequal energy and unequal distance from best efficiency — and it does it consistently, in the same direction, for the life of the station.

To equalise throughput the run hours have to be inverse to the flows: q·h equal for every unit means h ∝ 1/q. Over 300 pump-hours that is 104.09 : 101.38 : 94.53 hours, each unit then delivering 13 830 m³. Ten hours across a hundred is a small adjustment and one a controller can make; the reason it is rarely made is that nobody measured the split.

4.5.5The fixes, in order of how much they buy

Three things move the imbalance, and they are not equally worth doing. Shortening the bays from 5.0 m to 2.0 m takes the spread from 9.71% to 7.365% and buys 0.93 L/s — and the bay pitch is structural, so it is usually not on offer. Moving the takeoff to the header centre halves the longest path: 4.844% and 118.25 L/s, a gain of 2.81 L/s. Going up one header size to 300 mm gives 0.482% and 120.29 L/s, a gain of 4.85 L/s, and drops the header velocity from 6.53 to 1.70 m/s. Both together, 0.277%. The D⁻⁵ lever wins, as it does everywhere in this subject.

What does not work is equalising by adding loss. Throttling the near unit's isolation valve until the flows match, or trimming its impeller, does produce three equal numbers — by making the best unit as bad as the worst and lowering the station total. It is cheap, reversible, and what actually gets done. Call it what it is: buying fairness with capacity. Sometimes that is the right trade, but it belongs on a drawing, not on a valve somebody nudged.

So the design test is a comparison, not a limit. Compute each unit's path to the takeoff, take the largest difference, divide by the sum of the pump-curve and branch steepnesses. A fraction of a per cent of duty flow means the header is proportioned; several per cent means size the header or write the run-hour targets to match. And build system curves per unit and per combination of running units, never one curve for the station: a single station curve is the assumption that the header is free.

What is deliberately missing. The combining tee is a single K = 0.4, whereas the real loss at a junction depends on the ratio of joining to through flow and on the branch angle — the two-K and three-K methods this module already flagged as disagreeing by more than the losses being argued about. The junction where the main leaves is charged to the main and not resolved, which understates the outer units slightly in a centre takeoff. Every velocity here is steady: valves are resistances only, and what a check valve does while it closes is §8.4. What a unit's position on its curve costs in efficiency and bearing life is §7.4. And none of it is drawn yet: on the hydraulic profile the header is three closely spaced lines that must be plotted separately, which is §5.3.