Pipe friction loss
Compute friction head loss and pressure drop for a pipe run from flow rate, bore and roughness. The friction factor is solved from Colebrook rather than assumed, which matters once the flow is turbulent.
| Quantity | Value | Working |
|---|---|---|
| Flow rate | 18.00 m³/h | Entered |
| Velocity | 2.546 m/s | Q/A, A = 19.63 cm² |
| Reynolds number | 1.268e+5 | Turbulent |
| Relative roughness | 0.000900 | ε/D with ε = 0.045 mm |
| Friction factor f | 0.02135 | Colebrook |
| Friction head loss | 8.472 m | f·(L/D)·V²/2g |
| Minor losses (15%) | 1.271 m | Fittings allowance |
| Total head | 9.742 m | Friction plus minor |
| Pressure drop | 82.91 kPa | ρ·g·h_f |
| Hydraulic power | 0.477 kW | ρ·g·Q·H |
Minor losses are estimated at 15% of friction. For fitting-heavy runs, sum the K-factors explicitly instead.
V = Q/A · A = πD²/4
h_f = f·(L/D)·V²/2g · the exact friction head
ΔP = ρ·g·h_f · convert head to pressure
1/√f = −2log(ε/3.7D + 2.51/(Re√f)) · solved iteratively
A clear path from inputs to results.
Head loss is the energy the pump must supply just to move fluid through pipe. Darcy-Weisbach is the exact relation; the only difficulty is the friction factor, which this engine solves from Colebrook rather than assuming. The result is a head in metres and a pressure drop in kilopascals.
How to use
- Enter the volumetric flow rate.
- Set the internal diameter.
- Enter the total pipe length.
- Set the absolute roughness — 0.045 mm for drawn tubing, 0.15 for galvanised, 0.26 for cast iron.
- Enter the fluid density and viscosity.
- Read the head loss and check the velocity is in the 0.9–2.5 m/s design band.
The formula
Reference: Darcy (1857); Weisbach (1845); Colebrook-White (1939).
Worked example
18 m³/h through 50 mm bore over 60 m (default inputs)
- Flow 18 m³/h.
- Diameter 50 mm.
- Length 60 m.
- Roughness 0.045 mm.
- Water at 998 kg/m³ and 1.002 mPa·s.
- 1
Velocity: (18/3600) / 0.0019635 = 2.546 m/s.
- 2
Re = 998 × 2.546 × 0.05 / 1.002×10⁻³ = 126,816.
- 3
Relative roughness 0.0009; Colebrook gives f = 0.0214.
- 4
Head loss: 0.0214 × (60/0.05) × (2.546² / 19.61) = 8.47 m.
- 5
Pressure drop: 998 × 9.807 × 8.47 = 82.9 kPa.
2.55 m/s · f 0.0214 · 8.47 m head loss · 82.9 kPa pressure drop · 0.42 kW hydraulic power.
Interpreting the result
At 2.55 m/s this pipe is at the top of the recommended velocity band, and the loss gradient is 0.14 m per metre — a long run would need a large pump. Halving the flow would cut the head loss by roughly a factor of four, because h_f scales with V². That is why pipe sizing decisions matter more than pump selection.
Assumptions & limitations
- Steady, incompressible, isothermal flow in a circular pipe running full.
- Minor losses from fittings are estimated at 15% of friction; sum K-factors explicitly for fitting-heavy runs.
- Roughness is uniform and does not change with age; in practice it increases with scaling and corrosion.
- No elevation change is included — add static head separately.
- The fluid is Newtonian with viscosity at the operating temperature.
Frequently asked questions
Why does head loss rise with the square of flow?
Because velocity enters h_f as V². Doubling the flow quadruples the friction head, which is why a pump’s required power climbs so steeply with flow.
How do I include fittings?
Either sum K·V²/2g for each fitting or use the equivalent-length method. The 15% allowance here is a rough default for a normal industrial run.
What velocity should I design for?
0.9–2.5 m/s for water in process piping. Below 0.9 m/s air and solids settle; above 2.5 m/s noise, erosion and water hammer risk rise.
What roughness value should I use?
Drawn tubing 0.045 mm, commercial steel 0.045, galvanised iron 0.15, cast iron 0.26, concrete 0.3–3.0. Age and water quality move these substantially.
Does this work for gases?
Only at low pressure drop (under about 10% of absolute pressure). Above that, density changes along the pipe and a compressible-flow model is needed.
How do I convert head to pressure for other fluids?
Multiply by ρg. Ten metres of water is 98 kPa; ten metres of mercury would be 1,333 kPa, because the density is 13.6 times higher.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.