Pipe Pressure Drop Calculator

Enter flow, size, length and material roughness to get head loss and pressure drop — the numbers that decide whether the far shower gets decent pressure.

Pipe Pressure Drop Calculator inputs

Copper/PEX ≈ 0.0015 · new steel ≈ 0.045 · old steel ≥ 1.

Last updated: 2026-09-28

How the calculation works

  • Reynolds number from velocity and internal diameter picks the regime: laminar below 2300, turbulent above.
  • Turbulent friction comes from the Swamee-Jain explicit equation (≈1% vs Colebrook); laminar is the exact 64/Re.
  • Head loss converts to pressure through ρ·g — about 9.8 kPa per metre of water.

Formula

h = f · (L/D) · V²/2g
Δp = ρ·g·h
f: Swamee-Jain (turbulent), 64/Re (laminar)
Formula variables
SymbolMeaningUnit
hHead lossm
fDarcy friction factor—
ReReynolds number—

Worked example

30 L/min in 20 mm ID copper over 30 m: V = 1.59 m/s, Re ≈ 31,900 (turbulent), f ≈ 0.0246. h = 0.0246 × 1500 × 0.129 = 4.76 m → 46.6 kPa (6.8 psi) lost. Fittings add 20–50% on top — a real design pressure budget.

Interpreting the result

Pressure loss is the budget every distribution system spends: the available pressure at the street minus everything the path consumes must still satisfy the worst fixture (typically 100 kPa / 15 psi for showers). Loss scales with V² and L/D — velocity is the lever, which is why upsizing one trade size transforms long runs. Old steel pipe's rising roughness quietly strangles flow: a 1 mm roughness multiplies friction several-fold versus new copper.

Assumptions

  • Water at 20 °C, straight pipe, full bore.
  • No elevation change or fitting losses included.
  • Smooth-wall Darcy-Weisbach model: published copper tables (Hazen-Williams, C = 145) read ~10% higher for small tubes at domestic flows.

Limitations

  • Fittings, valves and bends dominate short runs — add equivalent length or 20–50% margin.
  • For other fluids, adjust viscosity and density (this model is water-specific).

Frequently asked questions

How much pressure do I lose per metre of pipe?

It depends on velocity: at 1.5 m/s in 20 mm copper, roughly 1.5 kPa per metre. Loss scales with V², so 3 m/s costs 4× as much.

What is the Swamee-Jain equation?

An explicit approximation of the Colebrook friction factor: f = 0.25/[log₁₀(ε/3.7D + 5.74/Re⁰·⁹)]². Accurate to about 1% for turbulent water mains without iterating.

Related tools

Technical references