Fitting and Valve Loss Coefficients (K Factors)

Typical loss coefficients (K) and equivalent lengths for 17 common fittings and valves.

Every elbow, tee and valve in a pipe adds a pressure loss on top of the friction of the straight pipe. It is written as a loss coefficient K: the loss equals K times the dynamic pressure, Δp = K × ρ v² ÷ 2, or in head, h = K × v² ÷ 2g. This table lists typical K values for 17 fittings and valves, for screwed and for flanged or welded connections, and the equivalent length of straight pipe in pipe diameters.

The table

17 rows
Fitting or valveLoss coefficient K, screwedLoss coefficient K, flanged or weldedEquivalent length in diameters, screwed (L/D)Equivalent length in diameters, flanged (L/D)
Standard 90 degree elbow0.90.34515
Long radius 90 degree elbow0.60.23010
45 degree elbow0.40.22010
180 degree return bend1.50.27510
Tee, flow straight through0.90.24510
Tee, flow into or out of the branch2110050
Gate valve, fully open0.20.21010
Gate valve, one quarter closed0.30.31515
Gate valve, half closed2.12.1105105
Gate valve, three quarters closed1717850850
Globe valve, fully open1010500500
Angle valve, fully open55250250
Ball valve, fully open0.050.0533
Swing check valve22100100
Pipe entrance, sharp edge0.50.52525
Pipe entrance, well rounded0.030.0322
Pipe exit into a tank115050

Source and notes: Typical loss coefficients for turbulent flow (Cengel and Cimbala, Fluid Mechanics, Table 8-4; Crane TP-410). The pressure loss is K x (density x velocity squared / 2); the head loss is K x velocity squared / (2 g). The equivalent length is K divided by a typical friction factor of 0.02 (fully rough flow in medium pipe); makers' data for the exact valve is always better.

How to use this table

  1. Find each fitting and valve in the pipe run and read its K.
  2. Add the K values and multiply by ρ v² ÷ 2 for the pressure loss (or v² ÷ 2g for the head loss).
  3. Add this to the friction loss of the straight pipe.

Worked example: six screwed elbows and a globe valve with water at 2 m/s

  • Total K = 6 × 0.9 + 10 = 15.4.
  • Pressure loss = K × ρ v² ÷ 2 = 15.4 × 998 × 2² ÷ 2 = 30.7 kPa.
  • In head: K v² ÷ 2g = 15.4 × 4 ÷ 19.62 = 3.14 m of water.

Notes and cautions

  • K varies with pipe size and Reynolds number; the figures are typical for turbulent flow in medium-size pipe. Large valves and small valves differ.
  • A globe valve costs far more pressure than a ball valve or a fully open gate valve: choose by the loss when flow control is not needed.
  • Partly closed valves have much higher K, which is how throttling works.

FAQ

What is the K factor of a 90 degree elbow?

About 0.9 for a standard screwed elbow and 0.3 for a flanged one, or 0.2 for a long-radius flanged elbow.

What is a loss coefficient?

A number that multiplies the dynamic pressure (ρ v² ÷ 2) to give the pressure lost in a fitting.

What is equivalent length?

The length of straight pipe that would give the same pressure loss: K divided by the pipe friction factor, here taken as 0.02.

Sources

  • Cengel and Cimbala, Fluid Mechanics, Table 8-4; Crane Technical Paper 410, Flow of Fluids Through Valves, Fittings and Pipe; Hooper’s two-K method for more accuracy.
  • Typical loss coefficients for turbulent flow (Cengel and Cimbala, Fluid Mechanics, Table 8-4; Crane TP-410). The pressure loss is K x (density x velocity squared / 2); the head loss is K x velocity squared / (2 g). The equivalent length is K divided by a typical friction factor of 0.02 (fully rough flow in medium pipe); makers' data for the exact valve is always better.