Pipe Pressure Drop Calculator (Darcy-Weisbach, Colebrook)

Understanding the pipe pressure drop calculator — concept and principle

This calculator estimates the friction pressure loss when a liquid flows through a straight, full, round pipe. From flow, inner diameter, length, wall roughness, density and viscosity it returns the pressure drop, the equivalent head loss, the velocity, the Reynolds number and the Darcy friction factor. It is used to size pipes and pumps.

How it is calculated

• Velocity: v = Q ÷ (πD²/4); Reynolds number: Re = ρvD/μ.
• Friction factor: if Re < 2300, f = 64/Re (laminar). Otherwise f is found by iterating the Colebrook-White equation: 1/√f = −2 log₁₀(ε/(3.7D) + 2.51/(Re√f)).
• Darcy-Weisbach: Δp = f · (L/D) · (ρv²/2), in Pa, shown in kPa.
• Head loss: h = Δp ÷ (ρg), in metres of the flowing liquid.

Typical roughness ε: commercial steel 0.045 mm, PVC and drawn tubing about 0.0015 mm.

The principle

Friction at the wall turns part of the flow's energy into heat. Dimensional analysis shows the loss must be proportional to the dynamic pressure ρv²/2 and to the number of diameters L/D, with a dimensionless factor f that depends only on Re and relative roughness ε/D. In laminar flow f = 64/Re follows exactly from the Hagen-Poiseuille solution. In turbulent flow C. F. Colebrook (1939) combined smooth-pipe and fully rough laws into one equation, which L. F. Moody (1944) plotted as the Moody chart. Since f appears on both sides, it is solved iteratively.

Worked example

1. Inputs: Q = 3 L/s, D = 50 mm, L = 100 m, ε = 0.045 mm, ρ = 998 kg/m³, μ = 1.0 mPa·s.
2. v = 0.003 ÷ (π × 0.05²/4) = 1.528 m/s.
3. Re = 998 × 1.528 × 0.05 ÷ 0.001 = 76,242, so turbulent; ε/D = 0.0009.
4. Colebrook iteration converges to f = 0.02248.
5. Δp = 0.02248 × (100/0.05) × (998 × 1.528²/2) = 0.02248 × 2000 × 1164.9 = 52,380 Pa = 52.38 kPa.
6. Head loss: 52,380 ÷ (998 × 9.80665) = 5.352 m.

How to read the result

Here a pump must supply about 5.4 m of head just to overcome 100 m of pipe, before any lift or fittings. In turbulent flow Δp rises roughly with the square of flow, and at constant flow it rises steeply as the diameter shrinks (roughly 1/D⁵), so one size larger often cuts losses by more than half.

Common mistakes

• Using nominal size instead of inner diameter.
• Forgetting minor losses from elbows, valves and tees, which can be significant in short systems.
• Confusing the Darcy factor with the Fanning factor, which is four times smaller.

Frequently asked questions

Is there an explicit alternative to Colebrook?

Yes. The Haaland (1983) and Swamee-Jain (1976) formulas give f directly, usually within a few percent of Colebrook.

Why is laminar 64/Re independent of roughness?

In laminar flow the slow layer near the wall covers the roughness, so friction depends only on viscosity.

How do I convert head loss to bar?

For water, 10.2 m of head is about 1 bar, so 5.35 m is about 0.52 bar.

Sources: C. F. Colebrook, Journal of the Institution of Civil Engineers 11 (1939) 133–156 · L. F. Moody, Transactions of the ASME 66 (1944) 671–684 · F. M. White, Fluid Mechanics, McGraw-Hill