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

Equivalent Length, and Double Counting

A fitting can be charged as a coefficient or as a length of pipe. Charge it as both and the pump is bought for a head the station does not have.

Two traditions wrote two ledgers for the same energy. One charges a fitting as a coefficient on the velocity head; the other charges it as a length of straight pipe that would lose the same amount. Both are exact and they convert in one line. Each is invited by a different document — a drawing carries lengths, a schedule carries coefficients — and where one design carries both, the two ledgers are not two contributions to the loss. They are the same contribution, written twice.

4.2.1Two ledgers for the same energy

How much head does an open gate valve cost? §4.1 answered with a coefficient: h = K·v²/2g. The other answer names a length — this valve behaves like so many metres of the pipe it is bolted into. Write the two expressions for the same loss side by side, K·v²/2g = f·(Le/D)·v²/2g, and the velocity head cancels off both sides. Nothing survives but the conversion:

Le = K·D / f   and   K = Le·f / D

Forwards, on round numbers: a standard 90° elbow at K = 0.6 in a 250 mm main whose friction factor is 0.020 is 0.6 × 0.250 / 0.020 = 7.50 m of pipe, or 30 diameters. Backwards: a supplier quoting 28.76 m of equivalent pipe for the 250 mm swing check valve in a main at f = 0.017384 is asserting K = 28.76 × 0.017384 / 0.250 = 2.00. Do it both ways on any number you are handed; the answer says which document it came from.

The cancellation deserves a stare, because it makes the rest of this lesson inevitable. L_e was defined to make the two losses equal, so pipe friction plus ΣK, and one Darcy-Weisbach over a developed length of L + ΣL_e, are not two estimates that ought to agree: they are one number by two arrangements of the same algebra, pinned to twelve significant figures by the instrument's verifier. An exact identity, added to itself, is exactly wrong.

fitting        one swing check valve, K = 2.0
pipe           250 mm inside diameter, cement-lined ductile iron, e = 0.10 mm
flow           80 L/s of water at 15 C

area           pi x 0.250^2 / 4                        =  0.049087 m^2
velocity       0.080 / 0.049087                        =  1.6297 m/s
velocity head  1.6297^2 / (2 x 9.80665)                =  0.13542 m
Reynolds       1.6297 x 0.250 / 1.139497e-6            =  3.5756e5
e/D            0.10e-3 / 0.250                         =  4.000e-4
f  (Colebrook, solved iteratively)                     =  0.017384

LEDGER 1, coefficient:
  h            2.0 x 0.13542                           =  0.27084 m of head

LEDGER 2, length:
  Le           K D / f = 2.0 x 0.250 / 0.017384        =  28.762 m of pipe
  h            f x (Le/D) x v^2/2g
               0.017384 x (28.762/0.250) x 0.13542     =  0.27084 m   <- identical
  in diameters 28.762 / 0.250                          =  115.0 D

A PRINTED TABLE, at its own f_T = 0.020:
  L/D          K / f_T = 2.0 / 0.020                   =  100 D
  Le           100 x 0.250                             =  25.000 m   <- 13.1% short

One valve, both ledgers, and the check that they are the same number (src/core/hydraulics.js at 15 °C; ν = 1.139497 × 10⁻⁶ m²/s). Every intermediate is shown to the digits the next line uses.

4.2.2What that length is actually a property of

Read Le = K·D/f as a statement about ownership. K belongs to the fitting; D and f belong to the pipe, and f belongs to the pipe at a flow, through the Reynolds number. An equivalent length is a property of all three together and of nothing less.

The contrast that matters at a desk is with the loss itself, a property of the fitting and the velocity and nothing else: K·v²/2g contains no diameter, no friction factor and no wall. One standard 90° elbow at 1.63 m/s costs 0.0813 m of head in 100 mm of aged cast iron and in 600 mm of new HDPE alike. The length of pipe matching that identical 0.0813 m runs from 1.36 m to 29.98 m — a factor of 22 — and even in diameters, which sound as though they ought to be scale-free, from 13.6 to 50.0.

The direction of the roughness dependence is where intuition fails, and it fails in a way that feels like physics. A smoother pipe loses less head per metre — therefore you need more of it before it has lost as much as the valve does. The 250 mm swing check valve above is 15.45 m of aged cast iron and 35.55 m of new HDPE at the same 80 L/s through the same bore: a factor of 2.300, with nothing changed but the wall. An equivalent length carried over from another drawing was never about your fitting.

Nor is it constant for one fitting in one pipe. f falls as Reynolds rises, so L_e grows with flow: the same elbow in the same 250 mm lined main is 26.4 diameters at a night minimum of 10 L/s (0.204 m/s) and 35.9 diameters at 160 L/s (3.26 m/s), 36% across one day. An equivalent length is a snapshot at a flow, exactly as §3.4's field-measured Hazen-Williams C is a snapshot at a velocity.

And there is the circularity. The method is sold as the one that lets you skip the friction factor and add up lengths, and it cannot be evaluated without the friction factor. Not fatal — the pipe needed f anyway — but it disposes of any claim that the method is simpler. What it buys is one Darcy-Weisbach instead of two terms, which matters to a solver and not to you.

4.2.3The diameter count, and the pipe it was measured in

Fitting tables rarely print L_e. They print L/D — a diameter count, one number per fitting type, no pipe attached — which looks like the scale-free form of the same information. By Crane's definition above, L/D = K/f_T: a coefficient divided by a friction factor belonging to the table's pipe. Across the sizes a station uses f_T is near 0.02, which is what reproduces the counts that get quoted: K = 0.6 for a standard 90° elbow gives exactly 30 diameters, K = 2.0 for a swing check exactly 100.

Apply a fixed count in a pipe whose f is not the table's and the fitting term scales by f/f_T, exactly and at every flow. In the 250 mm main at 80 L/s, against the honest K·v²/2g:

• PVC / HDPE, ε = 0.0015 mm, f = 0.01406 — fitting losses 29.7% low
• Ductile iron, cement lined, ε = 0.10 mm, f = 0.01738 — 13.1% low
• Ductile iron, unlined, ε = 0.26 mm, f = 0.02063 — 3.1% high
• Cast iron, aged, ε = 1.50 mm, f = 0.03235 — 61.8% high
• Concrete, rough, ε = 3.00 mm, f = 0.04050 — 102.5% high

The error is neither small nor centred. It nearly vanishes at unlined ductile iron — roughly the pipe the tables were assembled around — and grows both ways from there. Its worst direction is the one municipal practice keeps moving toward: in new plastic a fixed count under-states the fitting loss by nearly a third, the total dynamic head comes out short, and, as with §3.4's leading-coefficient trap, the pump is chosen against a head the station does not have. All of it lands on the fittings, so on §4.2.4's 12 m run the table route is 11.1% low rather than 13.1%.

4.2.4The double count

Nobody has to be careless for this to happen. The drawing gives a developed length. Someone — or a spreadsheet inherited with the template — converts the fitting schedule into equivalent lengths and adds them into the length column, because that is what the column is for. The minor-loss row stays, because deleting a row nobody asked about is how you get blamed. By §4.2.1's identity the fitting term is now charged precisely twice, and the surplus is not a tolerance: it is ΣK·v²/2g, the entire fitting loss, added again.

What it costs depends on where you look, and it is worst exactly where fittings live. Take the station's own discharge run: 12 m of 250 mm cement-lined ductile iron at 80 L/s, carrying a swing check, a gate valve, three standard elbows, a tee through the run and a magnetic meter, ΣK = 4.6. The honest total is 0.736 m: the pipe supplies 0.113 m and the fittings 0.623 m, whose equivalent length is 66.2 m — so the run is 12 m as built and 78.2 m hydraulically. Double count it and the answer is 1.359 m, an inflation of 84.6%.

The share has a closed form: the inflation is 1 / (1 + f·L / (D·ΣK)), the velocity head having cancelled. It does not depend on flow, only on how much pipe there is per unit of K — and station pipework is short and crowded, so the inflation runs toward 100%. Add the 900 m force main (four 45° bends and the exit, ΣK = 2.6) and the same double count moves 9.563 m of loss to 10.538 m, 10.2%; under an 18 m static lift the TDH goes from 27.56 m to 28.54 m, 3.5%.

Three and a half per cent sounds survivable, which is what lets the error get built. It also points the wrong way: an inflated resistance buys a pump with head to spare, which then runs further out along its curve than anybody intended — Module 6 draws the curve and Module 7 the crossing. It is also one of very few errors in this subject that is exactly quantifiable and completely avoidable.

Auditing a loss schedule — including your own from two years ago — is six questions.

1. Is there a ΣK row and a length that exceeds the drawing's developed length? That combination is the error, and nothing innocent looks like it.
2. Take the length column's excess over the drawing and compute ΣL_e·f/D. If it comes back equal to the ΣK row, the two rows are the same fittings.
3. Ask which friction factor the equivalent lengths were computed at. If nobody can say, they came from about 0.020 and §4.2.3 applies on top.
4. Check the sanity ratio. In station pipework the fittings' equivalent length is several times the physical length — 66 m against 12 m here. A schedule where the fittings are a rounding error has lost them, which is the opposite failure and just as wrong.
5. Check the supplier's allowance. A packaged station's quoted "station losses" may already include its own pipework and valves; adding your own schedule on top double counts by another route.
6. Check the pipe length itself. A round 950 m where the drawing says 900 m is usually 50 m of somebody's equivalent length, absorbed into the pipe and now invisible.

4.2.5One convention, named on the sheet

The rule that survives review: use K for anything you will have to defend, and equivalent length only where a tool demands a length. Those cases are real — a solver whose only fitting input is added length, a nomograph, an inherited sheet with a length column and no coefficient column, and Hazen-Williams practice, where there is no f to hang a K on.

When you supply them, compute them from your own K, diameter and f at the duty you care about, never from a printed diameter count: two minutes with Le = K·D/f, and §4.2.3's error is gone. Then record the basis. "Fitting losses 0.623 m" is not a result. "Fitting losses 0.623 m: ΣK = 4.6 from single-K values, v = 1.630 m/s, D = 250 mm, f = 0.01738 by Colebrook at 15 °C; equivalent length 66.2 m if a length is required, not to be added to the K total" is a result.

Two honest limits. None of this improves the K values themselves, which run from about 0.3 for a long-radius welded bend to 0.9 for a threaded fitting of the same angle, depending on which reputable source you open; §4.1's advice to carry the total as a range stands. The double count is exact and avoidable and the spread on K is neither — confusing the two is how a schedule ends up precise and wrong. And valves are resistances here and nothing else: check-valve closure is Module 8, and drawing these losses on a hydraulic profile is Module 5.