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Hydraulics, open channels and fluid power
Equivalent straight length of a fitting
Length of straight pipe that would give the same head loss as a fitting with loss coefficient K, in a pipe of diameter D with Darcy friction factor f at the same mean velocity.
- Ref
- HYD-12
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- Le [m]
A fitting with loss coefficient K gives a head loss K V²/(2g), and a straight pipe of length L gives f (L/D) V²/(2g). Setting these equal gives the equivalent length Le = K D/f. It replaces a fitting by a length of pipe of the same diameter. It depends on the friction factor, so it changes with the Reynolds number and the pipe roughness.
Length of straight pipe that would give the same head loss as a fitting with loss coefficient K, in a pipe of diameter D with Darcy friction factor f at the same mean velocity.
Calculator
- Equivalent length in pipe diameters, K/f36
- Equivalent length of ten fittings36 m
- Head loss of the fitting at 2 m/s (g = 9.81)0.1835 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A fitting with K = 0.9 loses as much head as 3.6 m (36 diameters) of 100 mm pipe.
- Ten such fittings add the equivalent of 36 m of pipe, which can outweigh a short run of straight pipe.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| K Loss coefficient of the fitting | 0.01 to 100 | 0.04 | 400 | 400 | +10% | +10 % |
| D Pipe diameter | 0.005 to 3 m | 0.18 | 108 | 107.8 | +10% | +10 % |
| f Darcy friction factor | 0.008 to 0.1 | 0.9 | 11.25 | 10.35 | +10% | −9.09 % |
Sweep of Loss coefficient of the fitting
| Loss coefficient of the fitting | Equivalent length (m) |
|---|---|
| 0.01 | 0.04 |
| 0.03162 | 0.1265 |
| 0.1 | 0.4 |
| 0.3162 | 1.265 |
| 1 | 4 |
| 3.162 | 12.65 |
| 10 | 40 |
| 31.62 | 126.5 |
| 100 | 400 |
Sweep one variable
Variables and how each one contributes
Equivalent length
What the equation gives you. Everything else feeds it.
Loss coefficient of the fitting
The loss coefficient of the fitting, referred to the mean velocity in the pipe. Raises the equivalent length in direct proportion. Raising it by 10 % (to 0.99) moves the result by +10 %, from 3.6 to 3.96 m.
Pipe diameter
The pipe diameter, in m. Raises the equivalent length in direct proportion. Raising it by 10 % (to 0.11) moves the result by +10 %, from 3.6 to 3.96 m.
Darcy friction factor
The Darcy friction factor of the pipe at the operating flow. A rougher pipe (larger f) makes a fitting worth less pipe length, so Le falls in inverse proportion. Raising it by 10 % (to 0.0275) moves the result by −9.09 %, from 3.6 to 3.273 m.
Worked example with the values above
- Ratio: K/f = 0.9 / 0.025 = 36 diameters.
- Multiply by the diameter D = 0.1 m.
- Le = 3.6 m.
Assumptions and limits
- Steady incompressible flow, with the friction factor and loss coefficients held at their duty-point values.
- Head losses proportional to the square of the flow, with positive flow in each pipe.
Common mistakes
- Using a loss coefficient that is referred to a different velocity from the one in the formula.
- Mixing the Darcy friction factor with the Fanning factor, which is a quarter of it.
- Adding resistances in parallel as if they were in series.
Statistics and uncertainty
Loss coefficients and friction factors are empirical and are usually known to within 10 % to 20 %, so network results carry that uncertainty. The quadratic forms assume the friction factor stays fixed as the flow changes.
Where it is used
- Building system curves for pumps and checking the effect of fittings.
- Balancing flows in looped water networks, and simplifying branches into one resistance.
Maths you need
- Head loss: energy lost per unit weight, in metres
- Darcy friction factor: f, as in hL = f (L/D) V²/2g
- Continuity: flow into a node equals flow out