Home / Heat & fluids

Valve and fitting pressure drop

Enter flow, pipe and the number of each valve and fitting. The page computes each fitting’s K, the equivalent length, the straight run friction and the total loss in pressure and in head.

Pre-sizing and estimating tool: results do not replace design, calculations or reports signed by a licensed professional, nor verification against the applicable codes and standards.

Pre-sizing
Fluid and flow
Pipe
Valves and fittings

Reducer and expander K from the Crane formulas for the β and angle entered, referred to the main pipe velocity. Sharp-edged entrance K = 0.5; exit K = 1.

—Total pressure drop (straight run + fittings)
—

Where the pressure drop goes

Head loss by fitting group, compared with the straight run.

Tap the chart to see the values.

Pressure drop versus flow

From 40% to 140% of the flow entered: loss grows with the square of flow.

Tap the chart to see the values.

Loss by fitting and equivalent length

Next step

Take this page’s results to another calculator with the fields already filled in.

The calculation report gathers this page's inputs, results, equations, assumptions and references, with fields for project, engineer and revision. The full report also includes the charts and interpretations, in color. In the print dialog, choose Save as PDF.

Method, assumptions and references

Equations used

Minor loss: hK = ΣK · v² ÷ (2g); straight run (Darcy–Weisbach): hf = f · L ÷ D · v² ÷ (2g); ΔP = ρ · g · (hf + hK)
Crane TP-410: K = fT · (L/D), with fT the fully turbulent friction factor of steel pipe; butterfly: L/D 45 (2" to 8"), 35 (10" to 14"), 25 (16" to 24")
Darby 3-K: K = K1 ÷ Re + Ki · (1 + Kd ÷ Dn^0.3), Dn = nominal size in inches
Reducer (θ ≤ 45°): K1 = 0.8 sin(θ/2)(1 − β²); expander (θ ≤ 45°): K1 = 2.6 sin(θ/2)(1 − β²)²; above 45°: 0.5(1 − β²)√sin(θ/2) and (1 − β²)²; in the main pipe K = K1 ÷ β⁴
Cv → K: K = 891 · d⁴ ÷ Cv² (d in inches) = 200 · (3600 · A ÷ Kv)² in SI; Cv = 1.156 · Kv; equivalent length: Le = K · D ÷ f
Colebrook–White: 1 ÷ √f = −2 log10(ε ÷ (3.7 D) + 2.51 ÷ (Re √f)); laminar: f = 64 ÷ Re

Assumptions and limits

  • New, fully open fittings with developed flow ahead of each one; interaction between closely spaced fittings is not included.
  • Inside diameter of Sch 40 or Sch 80 steel pipe (ASME B36.10M); valves and fittings of the same nominal size as the pipe.
  • Crane constant K is for turbulent flow; in laminar or transitional flow use 3-K. Check valves assume enough velocity to open the disc fully.
  • Incompressible liquid without cavitation; for gases and steam with a drop above about 10% of absolute pressure, compressibility must be considered.

Standards and technical documents

  • Crane Co., Technical Paper No. 410 (TP-410): Flow of Fluids Through Valves, Fittings and Pipe
  • Darby, R. Correlate pressure drops through fittings. Chemical Engineering, July 1999; Darby, R. Chemical Engineering Fluid Mechanics, 2nd ed., Marcel Dekker, 2001 (método 3-K)
  • Idelchik, I. E. Handbook of Hydraulic Resistance, 4th ed., Begell House
  • ISA-75.01.01 / IEC 60534-2-1: Flow equations for sizing control valves (Cv e Kv)
  • ASME B36.10M: Welded and Seamless Wrought Steel Pipe
  • Colebrook, C. F. (1939). Turbulent flow in pipes. J. Inst. Civil Eng.

U.S. regulations

  • ASME B31.1 (Power Piping) and B31.3 (Process Piping), adopted by states and owners
  • International Plumbing Code (IPC), Chapter 6, and Uniform Plumbing Code (UPC): water supply pipe sizing
  • OSHA 29 CFR 1910.119 (process safety management) for piping in covered processes

Frequently asked questions

What is minor (fitting) loss and when does it matter?

It is the energy a liquid loses through valves, elbows, tees, reducers, entrances and exits, on top of straight-pipe friction. In long straight lines it is small, but in pump rooms, skids and short lines full of fittings it can exceed half the total. In the default example, fittings account for 55% of the 12.7 psi drop.

What is the difference between Crane constant K and the 3-K method?

Crane TP-410 gives each fitting a fixed equivalent length (L/D), multiplied by the pipe’s fully turbulent friction factor, so K does not change with flow. Darby’s 3-K method adjusts K for Reynolds number and size and is more accurate for very small or very large pipe and viscous liquids. With water in 2" to 12" pipe the two are close: in the example, a sum of K of 26.0 versus 25.9.

How do I convert Cv or Kv to a K coefficient?

Cv is the water flow in gpm that passes with a 1 psi drop; Kv is the flow in m³/h with a 1 bar drop; Cv = 1.156 × Kv. With the inside diameter d in inches, K = 891 × d⁴ ÷ Cv² (Crane TP-410). In the example, a Cv 140 control valve in 4" pipe (d = 4.026 in) is worth K ≈ 12, or about 7.4 ft of head at 250 gpm.

What is equivalent length?

It is the length of straight pipe of the same size that would lose as much as the fitting: Le = K × D ÷ f. An open 4" globe valve equals about 105 ft of pipe; an open gate valve, less than 3 ft. In the example, all fittings together equal 481 ft, more than the 400 ft straight run.

Where do I get the input data?

Flow comes from the design, a flowmeter or the pump curve at the duty point; size and wall from the isometric or piping spec; fitting counts from the drawing or a field walkdown. For control valves, strainers and meters, use the Cv, Kv or pressure drop on the manufacturer’s data sheet.

Why does the loss grow so fast with flow?

Because it depends on velocity squared: 20% more flow gives about 44% more loss. In the example, total loss climbs from 29.4 ft to 42.1 ft. That is why an oversized design flow leads to bigger pumps that cost more to run.

What are the limits of this calculator?

It assumes an incompressible liquid, new fully open fittings and developed flow ahead of each one; closely spaced fittings lose a bit more. Steam and gas with a drop above about 10% of absolute pressure need a compressible calculation. The result excludes static lift and delivery pressure: add them in the pump sizing calculator.