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

K Values, and Why Minor Is a Misnomer

A K value is not a property of a fitting. It is a property of a fitting, a pipe, a Reynolds number, and whoever measured it.Module 4

By the end of this lesson

  • Compute a minor loss from a K value and a velocity
  • Total the K of a realistic station pipe run
  • Show the crossover length at which friction overtakes fittings

Module 3 priced straight pipe. A pump station is not straight pipe. Between the wet well and the property line the water is turned, squeezed, metered, stopped from coming back, joined to another pump's flow and thrown away into a manhole, and every one of those events costs head. The bookkeeping is easier than friction's — one dimensionless number per fitting times the velocity head — and the whole trap is in the word minor, an accurate description of a transmission main and a false one inside a station. This lesson gives you the length of pipe at which it flips. It is shorter than most people guess.

One form, one derivation, and a valve catalogue that became a standard

The form is older than the tables. Julius Weisbach (1806–1871) of the Bergakademie Freiberg wrote pipe losses as a dimensionless coefficient times a velocity head in his Lehrbuch der Ingenieur- und Maschinen-Mechanik of 1845 — the same book that gave §3.2 its friction equation. h = K·v²/2g is that bookkeeping with f·L/D replaced by a single number.

Exactly one coefficient in this lesson is derived rather than measured, and it is the oldest. Jean-Charles de Borda (1733–1799), the French naval officer and mathematician of the repeating circle, studied discharge from vessels through orifices in a memoir to the Académie royale des sciences in 1766. Lazare Carnot (1753–1823), father of Sadi Carnot, treated the loss of vis viva in an abrupt change of velocity in Principes fondamentaux de l'équilibre et du mouvement, 1803. Their result, h = (v1v2)²/2g for a sudden enlargement, is why the exit loss in the schedule below is exactly 1.0 and not a number someone fitted to data.

The tables everyone actually uses came from industry, not academia. Crane Co. of Chicago, a valve manufacturer, published Flow of Fluids Through Valves, Fittings, and Pipe as its Technical Paper No. 410; revised for decades and still in print, it is the de facto reference in the English-speaking world. I. E. Idelchik's Handbook of Hydraulic Resistance (first Russian edition 1960) is the exhaustive competitor, and D. S. Miller's Internal Flow Systems (BHRA, 2nd edition 1990) the most careful about fittings close enough to interfere. That the industry's loss data began as a manufacturer's pamphlet is worth knowing before §4.1.3.

4.1.1The currency is the velocity head

Put a bend in a pipe and the flow cannot follow the wall. It separates on the inside of the turn, reattaches downstream, and sets up a swirl that persists for tens of diameters. All of that costs energy, and the energy available to waste is the kinetic energy the stream already has. So the loss is a multiple of the velocity head:

h = K · v²/2g

and K — the loss coefficient, or K value — is dimensionless. It is a count of velocity heads, nothing more. Read it that way and the arithmetic is trivial and the errors are obvious.

At this station's 30 L/s the 150 mm main runs at 1.698 m/s (5.570 ft/s), so one velocity head is 0.1469 m (0.4821 ft). The 175 mm suction carries the same flow at 1.247 m/s, where one velocity head is only 0.0793 m. Those are the exchange rates for everything below, and there are two of them on purpose: a K value is not a head until a velocity is attached to it, and the velocity belongs to a pipe.

One entry in any schedule is not a measurement at all. Where a pipe discharges into water still enough to stop it, the whole velocity head is destroyed and K = 1.0 exactly. That follows from Borda and Carnot: for a sudden enlargement, a momentum balance gives h = (v1v2)²/2g, which referred to the upstream velocity head is K = (1 − β²)², β being the diameter ratio. Let the receiving vessel grow, β → 0, and K → 1. No table, no disagreement — the one number in §4.1.2 nobody can argue with.

4.1.2The station's budget, itemised

§1.2 counted eleven fittings between the wet well and the discharge manhole of the 30 L/s station, quoted their cost as ΣK = 7.10 for about 1.01 m, and promised this lesson would show the working. The K values are this course's table in src/core/hydraulics.js; the last column is what each fitting costs at the velocity head of the pipe it is installed in.

FittingPipeKNo.ΣKv²/2gHead
Suction bellmouth175 mm0.0510.050.0793 m0.004 m
90° long-radius elbow175 mm0.3010.300.0793 m0.024 m
Gradual reducer, 175 → 150 mm175 mm0.1510.150.0793 m0.012 m
Swing check valve150 mm2.0012.000.1469 m0.294 m
Plug valve, open — isolation150 mm1.0011.000.1469 m0.147 m
Magnetic flow meter150 mm0.2010.200.1469 m0.029 m
90° long-radius elbow, vault150 mm0.3010.300.1469 m0.044 m
Tee into the header, through the branch150 mm1.3011.300.1469 m0.191 m
45° bends along the main150 mm0.4020.800.1469 m0.118 m
Exit into the discharge manhole150 mm1.0011.000.1469 m0.147 m
Total7.101.0095 m

1.0095 m (3.312 ft), and the halves are worth separating: 0.50 of the ΣK sits in the suction and costs 0.040 m, while 6.60 sits in the discharge and costs 0.970 m. Now do it the quick way — charge the whole 7.10 at the main's velocity head — and you get 7.10 × 0.1469 = 1.0433 m, 3.35% high. High is the dangerous direction: it oversizes a pump. The sign is predictable, too, because a suction pipe is normally larger than the discharge. Widen the suction and the overstatement grows; make the bores equal and it vanishes exactly.

What makes a fitting budget intuitive is its equivalent length: how much of the same pipe would cost as much. Set f·(L/Dv²/2g = ΣK·v²/2g and the velocity head cancels, leaving L = ΣK · D / f. With f = 0.02344 from §3.2 at Re = 2.24 × 10⁵, this station's 7.10 is worth 45.4 m of 150 mm pipe; the itemised 1.0095 m is worth 44.0 m, slightly less because three fittings sit in the slower suction. Carry the general form: one unit of K buys 1/f diameters of pipe — 42.7 here (6.40 m), 65.0 in new PVC, 26.2 in aged cast iron at ε = 1.5 mm. Rough pipe makes fittings look cheap and smooth pipe makes them look expensive, with nothing about the fittings changed.

Interactive 3D instrument

The K budget — a staircase of velocity heads

A 3D instrument you drive yourself, one variable at a time. It needs JavaScript and WebGL, so it is not shown in this static copy of the page.

K values disagree between sources by more than the losses you are arguing aboutcontested

Look up the K for a standard 90° elbow and you will find 0.3, 0.5, 0.75 and 0.9 in four reputable places. The spread is real and has causes: some values are for threaded fittings and some for long-radius welded bends, some are single-K and some come from the two-K or three-K methods that make K a function of Reynolds number and nominal diameter, and some tabulate an equivalent length in pipe diameters instead.

Treating a single-K table as precise is the error. In a station where fittings contribute 4 m of a 35 m TDH, a ±40% spread on K is ±1.6 m — small. In a low-lift station where fittings are 4 m of an 8 m TDH, the same spread moves the operating point enough to change the pump you select.

What to do

Compute the fitting total as a range, not a value, and check whether the pump selection survives both ends. When it does not, the fix is to reduce the fitting count and straighten the pipework — not to find a table with friendlier numbers.

4.1.3Why two tables disagree about the same elbow

Look up "90° elbow" in four reputable places and you will find 0.3, 0.5, 0.75 and 0.9 — this course's table says 0.6. It is not sloppiness. The name of a fitting does not determine its geometry, and a K value is not a material property like density — it is a fitted result from one test on one part. Six reasons, in rough order of the damage they do:

  • The reference velocity is not stated. Any fitting that changes area has two candidate velocity heads, so the same K means two different losses — for a 250 → 150 mm reducer, a factor of 7.72. The largest error available in the subject, and the cure is one line of notation.
  • The geometry differs under one name. A "90° elbow" may be threaded, flanged, welded, mitred, or a long-radius bend at r/D = 1.5. Bend radius alone moves K by a factor of two or three, and threaded fittings are internally more abrupt than flanged ones.
  • K depends on size. Crane expresses fitting resistance as K = n·fT, with fT a fully-turbulent friction factor tabulated by pipe size and falling as size rises. Geometrically similar fittings are therefore not hydraulically similar, and one number for all diameters is a compromise.
  • K depends on Reynolds number. Published values are the high-Re asymptote; below about 10⁴ they climb. Hence W. B. Hooper's two-K method (Chemical Engineering, August 1981), which splits the coefficient into a Reynolds-dependent term and an asymptote, and Ron Darby's later three-constant version, which adds a diameter term. Station pipework at design flow is comfortably turbulent; a sludge line is not.
  • Neighbouring fittings interact. Two bends a few diameters apart are not two bends: the second sits in the swirl of the first, and the pair can lose more or less than the sum. Every table assumes fittings are isolated, and a valve vault is exactly where they are not.
  • The endpoint is not agreed. Some schedules include the exit loss; some stop at the outlet flange and book a residual velocity head instead. Both are right, doing both is a double count, and §4.3 settles it.

So how much does it matter? Swap this course's table for the low end of the published range, then the high end. ΣK moves from 3.68 to 9.25 and the budget from 0.522 m to 1.308 m — a factor of 2.51 from the choice of reference alone. That sounds ruinous and usually is not. Across the whole band this station's TDH runs from 39.18 m to 39.97 m: a spread of 2.0%, less than the uncertainty in the pipe roughness thirty years out (§3.5). Put the same eleven fittings in a booster with a 6 m lift and 40 m of main and the identical band swings TDH by 10.6% — the difference between two impellers.

Check your understanding

Check your understanding

3 auto-graded questions with an explanation for every wrong answer. Requires JavaScript. (m4-l1-q1)

4.1.4When "minor" is fair, and when it is a lie

The word is a comparison, and the crossover length is what it compares against. Below L* = ΣK·D/f ≈ 44 m of this pipe the fittings are the majority of the losses; above it the pipe is. Hold the fittings and the 30 L/s fixed and move only the length of the main:

Force main lengthFriction, hfFittings, hmFittings' share of the lossesWhat the station is
20 m0.46 m1.01 m68.7%an in-station booster
40 m0.92 m1.01 m52.4%a lift into an adjacent trunk
120 m2.75 m1.01 m26.8%a short force main
300 m6.89 m1.01 m12.8%still worth counting properly
900 m20.66 m1.01 m4.7%the §1.2 station
2 000 m45.92 m1.01 m2.2%genuinely minor

Two thresholds fall out, both multiples of L*. The fittings pass a quarter of the losses below 132 m (three L*) and drop under a twentieth above 879 m (twenty L*). So the defensible version of "minor losses can be neglected" is: when the main is more than about twenty times ΣK·D/f long, and you have said so. A calculation, not a habit, and one line long. What is never defensible is estimating a station's own pipework from its length: those twenty-odd metres contain the entire ΣK, so there the fittings are not most of the loss but essentially all of it.

Now the result that catches people out. Enlarge the whole station from 150 mm to 300 mm at the same 30 L/s and every term falls, but not equally. The fitting budget falls by exactly 16, because a velocity head goes as 1/D⁴ — 1.043 m to 0.065 m. Friction falls by 34.9, more than the 32 that 1/D⁵ predicts, because f also drops from 0.02344 to 0.02148 as the relative roughness improves. So the fittings' share rises, 4.8% to 9.9%. Every size increase makes the "minor" losses relatively more important — the opposite of the intuition the name creates.

Name the failure this prevents. A station is designed with the force main's friction from a hydraulic calculation and the station pipework allowed for as "say 20% of the friction", and it turns out to be a booster with 40 m of main. The fittings were 1.01 m of a 1.93 m loss — 52% of it; the allowance booked 0.18 m; the pump was bought 0.83 m short of its duty and now runs off the left of its efficiency curve — §7.2, and expensive for thirty years.

4.1.5What this lesson leaves open

Four debts inside this module. The same losses can be booked as added pipe length instead of as K values, and mixing the conventions double-counts a fitting — §4.2. Entrances, exits and the residual velocity head get their own lesson, because that is where the double counting actually happens — §4.3. Every valve coefficient here is for a fully open valve at design velocity, and a swing check whose disc is not fully lifted has a far larger K§4.4. And on a shared header each pump sees a different ΣK, so a different system curve — §4.5.

Two debts outside it. Nothing here is dynamic: the check valve is 2.0 velocity heads and nothing else, and what it does when the pump stops — closing on a reversing column, with a pressure rise that can burst the main — is Module 8. And the pump is still a black box assumed to deliver 30 L/s against whatever this staircase adds up to; what it does when handed 0.83 m more head than expected is Module 6. What you can do now is build a K budget, defend every line, and say in one calculation whether it matters.

Lab 4.1

A K budget you can take to work

Three functions. Together they are a fitting schedule you can paste into a spreadsheet's script editor and defend in a review. minorLoss(kTotal, qM3s, dM) returns the head loss in metres for a total loss coefficient in one bore. Compute the area, then the velocity, then the velocity head, then multiply. G = 9.80665 m/s² is provided. budget(schedule, qM3s) takes an array of { K, count, dM } rows and returns the total head loss in metres, charging each row at the velocity head of its own dM . This is the whole lesson in one function, and the reason it takes a diameter per row rather than one for the arrangement. crossoverLength(kTotal, dM, f) returns the length of straight pipe, in metres, that loses the same head as kTotal — the equivalent length L = Σ K · D / f . No flow appears in it, because the velocity head cancels; if your answer needs qM3s , the algebra has gone wrong somewhere. Graded in the browser against 7 assertions; the editor and harness require JavaScript.

Head, Loss and Lift · Module 4, Lesson 1