Minor Loss Coefficient Calculator

Q: How do I convert a manufacturer's Cv or Kv into a K?

K = 890 d4/Cv² with the diameter d in inches, or K = 1.6×109 d4/Kv² with d in millimetres. The trap is which diameter: for a reduced-bore valve the maker usually refers Cv to the nominal line size, not the seat bore, and using the seat bore will give you a K several times too large. It is the same class of error as using the wrong velocity, and for the same reason — a coefficient is meaningless until the reference is named. Also check the definition of Cv your maker uses; it is US gallons per minute of water at one psi differential, and a few catalogues quote something else under the same name.

K factors turned into pressure drop, head loss and an equivalent length in diameters — with the sudden expansion and contraction derived rather than tabulated, the question of WHICH velocity the head belongs to answered explicitly, and the factor-of-two disagreement between published K tables shown as a range instead of hidden behind one number.

Last updated 30 Sep 2026

Fitting loss coefficient, pressure drop and equivalent length

fitting, velocity, bore and fluid → K, Δp, head loss and L_eq/D

FittingThe first three are computed, not looked up: a sudden expansion from the momentum theorem, a sudden contraction from the re-expansion out of its vena contracta, and an exit into a reservoir from the first of those in the limit. The rest are tabulated, and tabulated K values vary by up to a factor of two between sources for the same fitting — so each one here carries a low and a high value as well as a typical one, and all three pressure drops are printed. If you have a manufacturer’s own figure for the valve you are actually installing, use the last option and enter it.

Reference velocity V (m/s)The velocity the K value is defined against, and for an expansion or a contraction that is the question. K is a multiplier on a velocity head, so it is meaningless without naming the velocity. The convention for both an expansion and a contraction is the SMALLER pipe’s velocity — the faster one — and the selector below lets you say which one you have entered. Bulk mean velocity, not centreline.

Which pipe is that velocity inRead only for the expansion and the contraction, where the two velocities differ. Using the wrong one is the classic error in a minor-loss calculation, and it is not a small error: the two K values differ by a factor of β⁴, which at a diameter ratio of 0.5 is a factor of sixteen. The K printed below is always the one that goes with the velocity you entered, so the pressure drop is right either way — but if you copy a K from a table you must check which velocity the table meant.

Diameter ratio: smaller bore divided by largerβ = D_small/D_large, always at most 1 whichever way the flow is going. β → 0 with an expansion is a pipe discharging into a reservoir (K = 1 on the pipe velocity); β → 0 with a contraction is a pipe drawing from a reservoir (K → 0.42). β → 1 is no change of section and K → 0 as (1 − β²)².

Your own KRead only when the fitting above is set to your own K. A manufacturer’s figure for the specific valve or fitting you are installing beats any table, and if it is given as a flow coefficient C_v or K_v rather than a K, convert it first: K = 890 d⁴/C_v² with d in inches, or K = 1.6×10⁹ d⁴/K_v² with d in millimetres. Otherwise this field shows the K the selection above has produced.

How many identical fittingsFittings in series at the same velocity. Beware of adding two bends that are close together: two elbows within about ten diameters of each other do not cost twice one elbow — depending on whether they are in the same plane or at right angles the pair can be anywhere from 0.6 to 1.4 times the sum, because the second one meets a distorted profile rather than a developed one. That interaction is outside what any K table can express.

Pipe bore (mm) — the pipe your velocity is inUsed for the Reynolds number, the friction factor and therefore the equivalent length in diameters. Not used for the pressure drop itself, which needs only K, the density and the velocity.

FluidAir densities come from the ideal-gas law at 101.325 kPa and air viscosities from Sutherland’s formula as the US Standard Atmosphere 1976 states it; water comes from the NIST Chemistry WebBook at 0.101325 MPa. Every number here is identical to the one the y+ page uses, so the two pages cannot disagree on a Reynolds number. Choose Custom for sea water, a gas at pressure, oil or anything compressible.

Density ρ (kg/m³)Read only when the fluid above is Custom; otherwise this field shows the preset’s value. For a compressible run use the density at the edge of the boundary layer, not the stagnation density.

Dynamic viscosity μ (Pa·s)Dynamic viscosity, not kinematic. If your solver reports ν in m²/s, multiply by the density. In water the confusion is a factor of a thousand and obvious; in air it is a factor of only 1.2 — 1.81×10⁻⁵ Pa·s against 1.51×10⁻⁵ m²/s — so it passes unnoticed and biases everything by 20 per cent. Check the unit, not the magnitude.

Wall roughness, for the equivalent lengthOnly the equivalent length needs this, because L_eq/D = K/f and f depends on roughness. The values are the same ones the pipe friction page uses, and the friction factor is computed with the same Haaland correlation, so the equivalent length here and the straight-pipe loss there are consistent.

Custom roughness ε (mm)In millimetres. Read only when the roughness above is Custom.

1.123kPaExample

A sudden expansion from a 50 mm to a 100 mm bore, water at 20 °C at 2 m/s in the 50 mm pipe, commercial steel

One definition, two exact derivations, and the equivalent length

Δp = K ½ρV² · h_L = K V²/2g · L_eq/D = K/f
Sudden expansion (Borda–Carnot, exact): h_L = (V_s − V_l)²/2g → K_s = (1 − β²)², K_l = K_s/β⁴
Static pressure rise across it: p_l − p_s = ρV_s²β²(1 − β²), a fraction 2β²/(1 + β²) of the frictionless rise
Sudden contraction: K_s = (1/C_c − 1)² = min[0.42(1 − β²), (1 − β²)²], branches meeting at β = √0.58 = 0.761577
Exit into a reservoir: the β → 0 limit of the expansion, K = 1 exactly
Flow coefficients: K = 890 d⁴/C_v² (d in inches) = 1.6×10⁹ d⁴/K_v² (d in mm)

K: loss coefficient, dimensionless. The number of velocity heads the fitting destroys. Meaningless until you name the velocity, which is the whole point of this page
V: the reference velocity, bulk mean. For an expansion or a contraction the convention is the smaller pipe’s velocity, and a K quoted against the other one differs by β⁴ — a factor of 16 at β = 0.5, and 625 at β = 0.2
β: D_small/D_large, always at most 1. The area ratio is β², and every derived K on this page is a function of β² alone
C_c: contraction coefficient: the vena contracta area divided by the downstream pipe area. Empirical. Kirchhoff’s free-streamline solution gives exactly π/(π+2) = 0.61102 for a two-dimensional slot in an infinite wall, which is the nearest thing to a derived value
f: Darcy friction factor, needed only for the equivalent length. Haaland above Re = 2300 and 64/Re below, identically to the pipe friction page, so the two pages agree
L_eq/D: K/f: the length of straight pipe of the same bore that would lose the same pressure. A convenience, not a physical length, and it inherits the roughness dependence of f — the same fitting in a rough pipe is worth fewer equivalent diameters, because the pipe itself is worth more per diameter
Δp rise: for a sudden expansion only: the static pressure genuinely rises, because the flow decelerates. ρV_s²β²(1−β²), maximum at β² = 1/2 where it equals half the small-pipe velocity head

Worked example

A sudden expansion from a 50 mm to a 100 mm bore, water at 20 °C at 2 m/s in the 50 mm pipe, commercial steel
β = 50/100 = 0.5, so the area ratio is β² = 0.25 and the downstream velocity is 2 × 0.25 = 0.5 m/s. The 2 m/s is in the smaller pipe, which is the conventional reference.
Borda–Carnot: K on the small-pipe velocity is (1 − 0.25)² = 0.5625. Nothing empirical enters that. Referred to the downstream velocity instead it would be 0.5625/0.25² = 9.0 — a factor of sixteen, which is what β⁴ = 0.0625 means. Both describe the same fitting and the same loss; only one goes with a 2 m/s velocity head.
The velocity head: ½ρV² = 0.5 × 998.21 × 4 = 1996.4 Pa. So Δp = 0.5625 × 1996.4 = 1123.0 Pa = 1.123 kPa, and the head loss is 1123.0/(998.21 × 9.80665) = 0.1147 m of water.
The static pressure nevertheless rises. p_large − p_small = ρV_s²β²(1 − β²) = 998.21 × 4 × 0.25 × 0.75 = +748.7 Pa. The frictionless value would have been ½ρV_s²(1 − β⁴) = 1996.4 × 0.9375 = 1871.6 Pa, so the recovery achieved is 748.7/1871.6 = 40.0 per cent — exactly 2β²/(1 + β²) = 0.5/1.25. And 1871.6 − 748.7 = 1122.9 Pa, which is the loss, as it must be.
Equivalent length. Re = 2 × 0.05 / 1.003196×10⁻⁶ = 99,681 and ε/D = 0.04572/50 = 9.144×10⁻⁴, so Haaland gives f = 0.021672. L_eq/D = K/f = 0.5625/0.021672 = 25.96 diameters = 1.298 m of 50 mm pipe. Note that in a rougher pipe the same fitting is worth FEWER equivalent diameters, because the pipe is worth more per diameter.
For contrast, the sudden CONTRACTION of the same ratio — 100 mm down to 50 mm, measured on the same 2 m/s in the small pipe — is min[0.42 × 0.75, 0.75²] = min[0.315, 0.5625] = 0.315, so 629 Pa. Contracting costs 56 per cent of what expanding costs, because a contracting flow does not separate and an expanding one does.

The two derived cases, exactly, against diameter ratio

The expansion columns are exact: K = (1 − β²)² on the smaller pipe’s velocity, and dividing by β⁴ re-refers it to the larger. **Two identities worth checking against this table.** At β = 0.7071 the two expansion K values are both 0.25, because β⁴ = 1/4 there; and at β = 0.761577 = √0.58 the contraction’s two branches give the same 0.1764, which is where the printed crossover of 0.76 comes from. The last column is the static pressure RISE across the expansion, which peaks at exactly half the small-pipe velocity head when the area ratio is a half.
Diameter ratio, small over large	Area ratio, that squared	Expansion K on V_small	Expansion K on V_large	Contraction K on V_small	Expansion pressure rise over the small-pipe velocity head
0 (reservoir)	0	1.0000 exactly	—	0.4200	0
0.2	0.04	0.9216	576.00	0.4032	0.0768
0.4	0.16	0.7056	27.563	0.3528	0.2688
0.5	0.25	0.5625	9.0000	0.3150	0.3750
0.6	0.36	0.4096	3.1605	0.2688	0.4608
0.7071	0.5	0.2500	1.0000	0.2100	0.5000, the maximum
0.761577	0.58	0.1764	0.5244	0.1764, branches meet	0.4872
0.9	0.81	0.0361	0.0550	0.0361	0.3078
0.99	0.9801	0.000396	0.000412	0.000396	0.039

Tabulated fittings, with the published spread rather than one number

**These are a band across the published engineering literature, not a reproduction of any one source’s table, and the band is the point.** The high-over-low column runs from 1.4 to 6, so for several of these fittings the K is genuinely uncertain by a factor of several. There are real reasons: K depends on nominal size (a 1 inch elbow and a 12 inch elbow of the same pattern differ by roughly a factor of two), on Reynolds number below about 10⁴, on the manufacturing pattern, on whether a valve is full bore or reduced, and on how much straight pipe preceded it. **For anything where the minor losses dominate, use the manufacturer’s own figure for the actual part.** The equivalent-length column is K/f at the friction factor of the worked example above and will change with roughness and bore.
Fitting	Low	Typical	High	High over low	L_eq/D at f = 0.0217
Sharp-edged pipe entrance from a reservoir	0.42	0.50	0.57	1.4	23
Slightly rounded or 45° bevelled entrance	0.12	0.20	0.28	2.3	9
Well-rounded entrance, r/D ≥ 0.15	0.03	0.04	0.05	1.7	2
Re-entrant (projecting) entrance	0.56	0.80	1.00	1.8	37
90° elbow, threaded, standard radius	0.90	1.50	2.00	2.2	69
90° elbow, flanged, long radius r/D ≈ 1.5	0.20	0.30	0.60	3.0	14
90° mitre bend, no vanes	0.90	1.10	1.30	1.4	51
45° elbow	0.20	0.40	0.50	2.5	18
180° return bend, flanged	0.20	0.30	0.40	2.0	14
Tee, flow through the run	0.10	0.20	0.40	4.0	9
Tee, flow through the branch	0.70	1.00	1.80	2.6	46
Gate valve, fully open	0.10	0.17	0.30	3.0	8
Globe valve, fully open	4.0	10.0	20.0	5.0	461
Ball valve, fully open, full bore	0.04	0.05	0.10	2.5	2
Butterfly valve, fully open	0.20	0.30	1.20	6.0	14
Swing check valve, fully open	2.0	2.5	5.0	2.5	115

Which velocity — the mistake, and what it costs

A sudden expansion or contraction has two velocities and therefore two K values, and they are related by exactly β⁴. Pairing a K quoted on one velocity with the other velocity head does not give a slightly wrong answer; at β = 0.5 it gives an answer wrong by a factor of sixteen. **Almost every published table quotes the smaller pipe’s velocity**, because that is where the loss is generated and because it makes the reservoir limits come out as the familiar 1.0 and 0.42. This page always prints the K that goes with the velocity you entered, and prints the other one beside it.
Diameter ratio	V_small / V_large	K on V_small	K on V_large	Factor between them, the ratio to the fourth power	Error if you use the wrong pair
0.2	25.0	0.9216	576.00	625	wrong by 625×, either way
0.4	6.25	0.7056	27.563	39.1	wrong by 39×
0.5	4.00	0.5625	9.0000	16.0	wrong by 16×
0.7071	2.00	0.2500	1.0000	4.0	wrong by 4×
0.9	1.235	0.0361	0.0550	1.524	wrong by 52 %

Which velocity the head belongs to, which cases can be derived, and why two published K values for one elbow differ by a factor of two

K is a number of velocity heads, so the first question is always: which velocity? For a bend, a valve or a tee in a constant-bore line the question does not arise, and that is why it gets forgotten. For a sudden expansion or contraction there are two velocities differing by 1/β², and the two K values that describe the same fitting differ by β⁴ — a factor of sixteen at β = 0.5 and 625 at β = 0.2. Pairing a K from a table with the wrong velocity head is therefore not a small error; it is an order-of-magnitude error, and it is the classic mistake in minor-loss work. The convention in almost every published table is the SMALLER pipe’s velocity — the faster one, where the loss is generated — and that convention is what makes the two reservoir limits come out as the familiar numbers: 1.0 for an exit and about 0.42 to 0.5 for an entrance. This page asks which velocity you have and always prints the K that matches it, with the other convention’s value beside it.

Two of these cases are derived and the rest are measured, and the page keeps the two kinds apart. A sudden expansion has an exact answer. Take a control volume spanning the step, apply the momentum theorem with the pressure on the annular face equal to the upstream pressure, and the head loss falls out as (V_s − V_l)²/2g: the Borda–Carnot result, with no empirical constant in it anywhere. In terms of K on the small-pipe velocity that is exactly (1 − β²)², and setting β = 0 gives K = 1, which is where the exit loss of one velocity head comes from. A sudden contraction does not have an exact answer, and it is worth being clear about why, because it is often presented as though it did. The loss is not in the contraction: the jet contracts to a vena contracta narrower than the downstream pipe and then re-expands, and it is that re-expansion which dissipates the energy. So K = (1/C_c − 1)² is exact GIVEN the contraction coefficient — and C_c is measured. The nearest thing to a derived value is Kirchhoff’s free-streamline solution for a two-dimensional slot, which gives C_c = π/(π+2) = 0.611015 exactly and hence K = (2/π)² = 4/π² = 0.405285, within 4 per cent of the 0.42 this page uses in the reservoir limit.

The static pressure rises across a sudden expansion, and a lot of people do not believe it. The flow is decelerating, so Bernoulli wants to convert dynamic pressure into static pressure; the loss only stops it converting all of it. The exact rise is ρV_s²β²(1 − β²), the frictionless rise would have been ½ρV_s²(1 − β⁴), and the ratio — the pressure recovery achieved — simplifies to 2β²/(1 + β²): 40 per cent at β = 0.5, 67 per cent at β = 0.707, 90 per cent at β = 0.95. The absolute rise is greatest at β² = 1/2, where it is exactly half the small-pipe velocity head. This is a useful CFD check: a horizontal sudden expansion should show a static pressure rise, and if your solution does not, something is wrong with the mesh, the outlet condition or the post-processing rather than with the physics.

Tabulated K values disagree by up to a factor of two and this is the dominant uncertainty in any minor-loss calculation. That is not sloppiness in the sources; it is that a single K for a named fitting is a simplification of something that genuinely depends on several things. Size: a 1 inch elbow and a 12 inch elbow of the same nominal pattern can differ by a factor of two, which is why the two-K and three-K correlations carry an explicit diameter term. Reynolds number: below about 10⁴ K rises as Re falls, and in laminar flow it rises roughly as 1/Re, so every tabulated value is a fully turbulent asymptote. Pattern: a globe valve’s seat and stem design change its K by a factor of five, and a butterfly valve’s disc thickness matters too. Installation: a fitting less than ten diameters downstream of another does not see a developed profile, and the pair can cost between 0.6 and 1.4 times the sum of two isolated ones. This page therefore gives a low, a typical and a high value for every tabulated fitting and computes the pressure drop for all three. Take the high value when you are sizing a pump and the low one when you are predicting a flow, so that neither answer is optimistic. And if the minor losses dominate your system, the manufacturer’s figure for the actual part beats every table including this one.

Equivalent length is a convenience with a trap in it. L_eq/D = K/f converts a fitting into a length of straight pipe, which is handy for a hand calculation and for a pipe-network solver that only understands lengths. But it inherits the friction factor, so the same fitting is worth fewer equivalent diameters in a rough pipe than in a smooth one — not because the fitting has changed, but because the pipe is worth more per diameter. A fitting quoted as “30 diameters” therefore carries a hidden assumption about f, usually around 0.02. The K form has no such dependence and is the one to prefer. The friction factor used here is Haaland’s, the same as on the pipe friction page, so a system total computed from the two pages is consistent. For the profile question rather than the pressure question — how far downstream a fitting’s wake persists — see the entrance length page, and for a non-round section get D_h from the hydraulic diameter page first.

Frequently asked questions

Which velocity does K go with for an expansion or a contraction?

Conventionally the smaller pipe’s velocity, the faster one, for both. That is where the loss is generated and it is what makes the reservoir limits come out as the familiar 1.0 for an exit and 0.42 to 0.5 for an entrance. The two K values differ by exactly β⁴: at β = 0.5, K on the small pipe is 0.5625 and K on the large pipe is 9.0. Both describe the same fitting and the same loss. Pairing one with the other’s velocity head is wrong by a factor of sixteen at β = 0.5 and by 625 at β = 0.2, so whenever you copy a K from a source, find the sentence that says which velocity it is on. If the source does not say, the value is not usable for a change of section.

Why does the static pressure go UP across a sudden expansion if there is a loss?

Because the flow is decelerating and Bernoulli wants to turn dynamic pressure into static pressure; the loss only prevents the full conversion. At β = 0.5 with 2 m/s of water in the small pipe, the static pressure rises by 749 Pa where a frictionless diffuser would have given 1872 Pa, and the difference — 1123 Pa — is exactly the loss. The fraction recovered is 2β²/(1 + β²), which is 40 per cent here. This is a good check on a CFD solution: a horizontal sudden expansion that shows a static pressure drop is telling you about your mesh or your boundary conditions, not about the flow.

Why do two handbooks give different K for the same elbow?

Because a single K for a named fitting is a simplification. The real K depends on nominal size (roughly a factor of two between 1 inch and 12 inch for the same pattern), on Reynolds number below about 10⁴, on the manufacturing pattern and finish, on whether a valve is full bore or reduced, and on what is immediately upstream. That is why the two-K and three-K correlations take the form K = K₁/Re + K_∞(1 + K_d/D^0.3) — the Reynolds and diameter terms are there because the effects are real. The range shown for each tabulated fitting on this page is an honest summary of what the literature supports, and for several fittings it is a factor of two or more. That range, not the mesh or the friction factor, is the dominant uncertainty in most system pressure-drop calculations.

Can I just add up the K values of all my fittings?

Yes, if they are far enough apart to each see a redeveloped profile, and roughly ten diameters is the usual rule for that. Closer than that and the interaction matters: two elbows within a few diameters can cost between about 0.6 and 1.4 times the sum of two isolated ones, depending on whether they lie in the same plane (the second partly straightens the first’s secondary flow) or at right angles (the second amplifies it into a swirl). Swirl in particular decays very slowly — a single elbow can leave measurable swirl fifty diameters downstream — and no K table can express any of this. If a compact manifold is the thing you are predicting, that is what CFD is for.

How do I convert a manufacturer’s Cv or Kv into a K?

K = 890 d⁴/C_v² with the diameter d in inches, or K = 1.6×10⁹ d⁴/K_v² with d in millimetres. The trap is which diameter: for a reduced-bore valve the maker usually refers C_v to the nominal line size, not the seat bore, and using the seat bore will give you a K several times too large. It is the same class of error as using the wrong velocity, and for the same reason — a coefficient is meaningless until the reference is named. Also check the definition of C_v your maker uses; it is US gallons per minute of water at one psi differential, and a few catalogues quote something else under the same name.

Are these K values valid in laminar flow?

No. Every tabulated K on this page is a fully turbulent asymptote. In laminar flow the loss stops being dominated by turbulent dissipation in a separated region and starts being dominated by viscous shear, and K rises roughly as 1/Re: at Re = 1000 a tabulated K can understate the loss several-fold. The two-K and three-K correlations exist to handle this and their K₁/Re term is exactly that. The two derived cases are better behaved but not immune — Borda–Carnot assumes the pressure on the annular face equals the upstream pressure, which is a separated-flow argument. The page warns when Re is below 2300.

Should I model fittings in CFD or use K values?

Use K values for a system calculation and CFD for the fitting itself, and be clear which you are doing. A K value is a lumped statement about a fitting in isolation with developed flow entering it; it cannot tell you where the separation is, how far the wake persists, whether cavitation is likely, or what happens when two fittings are close together. CFD can tell you those things and is the right tool when the fitting IS the question. What CFD is bad at is the whole system: meshing a hundred metres of pipe with forty fittings to get a pump duty is a waste when the K method gives the same answer to within the uncertainty of the K values themselves. And the K values are worth checking against a CFD run of a single fitting when the loss matters and the table’s range is a factor of two.

Why is contracting cheaper than expanding?

Because a contracting flow accelerates and an accelerating flow does not separate. The boundary layer is thinned and stabilised, the streamlines follow the wall, and almost nothing is dissipated in the contraction itself — the loss comes entirely from the jet re-expanding out of its vena contracta downstream. An expanding flow decelerates, the boundary layer cannot follow the step, it separates at the corner, and the whole kinetic-energy difference is dumped into a recirculating region. At β = 0.5 the contraction costs 0.315 velocity heads against the expansion’s 0.5625, so 56 per cent as much; in the reservoir limit it is 0.42 against 1.0, so 42 per cent. It is the same asymmetry that makes a diffuser hard and a nozzle easy.

Related calculators

References

J.-C. de Borda (1766) and L. Carnot (1803) for the sudden-expansion loss (V₁ − V₂)²/2g. Derived rather than quoted on this page: the momentum theorem on a control volume spanning the step, with the annular-face pressure taken as the upstream pressure, gives K = (1 − β²)² on the smaller pipe’s velocity, and the static pressure rise ρV_s²β²(1 − β²) follows from the same balance. The two were checked against each other by the requirement that the frictionless rise minus the actual rise equals the loss exactly, which it does identically.
G. Kirchhoff (1869) free-streamline solution for efflux through a two-dimensional slot in an infinite wall, giving the contraction coefficient C_c = π/(π+2) = 0.6110155 exactly and hence K = (1/C_c − 1)² = (2/π)² = 4/π² = 0.4052847. This is the only analytic contraction coefficient there is, and it is within 4 per cent of the 0.42 used here in the reservoir limit — which is the check that the empirical constant is the right size.
F. M. White, Fluid Mechanics, for the sudden-contraction form K ≈ 0.42(1 − d²/D²) below d/D = 0.76 and (1 − d²/D²)² above it. The pair was verified rather than taken on trust: the two branches are equal when 1 − β² = 0.42, that is at β = √0.58 = 0.761577, so the printed 0.76 is the exact crossover and the pair is self-consistent. Because the second branch is the lower of the two above that point, the whole thing collapses to a single expression, min[0.42(1 − β²), (1 − β²)²], with no conditional needed.
The tabulated K values in the second table are a band across the published engineering literature rather than a reproduction of any one source’s table, which is a deliberate choice: no single source is authoritative for a quantity that depends on size, Reynolds number, pattern and installation, and presenting one number would misrepresent how well it is known. The band is offered as a factual statement about the spread of published values, and the page says in three places that a manufacturer’s figure for the actual part is better than any of them.
W. B. Hooper (1981), The two-K method predicts head losses in pipe fittings, Chemical Engineering, and R. Darby’s three-K extension, cited for the FORM of the Reynolds-number and diameter corrections, K = K₁/Re + K_∞(1 + K_d/D^0.3). The coefficients are not reproduced; what is used here is the qualitative conclusion, which the page states plainly: a single K is a fully turbulent, size-independent asymptote, and both of those assumptions fail in identifiable circumstances.
S. E. Haaland (1983), Simple and explicit formulas for the friction factor in turbulent pipe flow, J. Fluids Eng. 105, 89–90, for the friction factor used in the equivalent length. It is the same correlation and the same roughness values as the pipe friction page, so a fitting loss and a straight-pipe loss taken from this site add up consistently.
NIST Chemistry WebBook (US Government) for water density and viscosity at 0.101325 MPa, and the US Standard Atmosphere 1976 for air. Identical to the values used across this section.

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