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Pipe head loss calculator (Darcy–Weisbach)

Head loss and pressure drop in a full pipe from flow, size, length and roughness — Darcy–Weisbach with Colebrook.

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gpm
in
ft
ft

The Handbook gives ε as ranges; these presets are mid-range values. Use the value a problem gives.

°F

IAPWS-95 properties. The PE Civil Handbook's water table prints ν = 1.217 × 10⁻⁵ ft²/s at 60 °F; this gives 1.2085 × 10⁻⁵ (0.7% lower).

Straight pipe only — add fitting and valve losses separately.

Result

Head loss, h_f
11.34ft of fluid
Pressure drop
4.911psi
Velocity
5.04ft/s
Reynolds number
139,900
Friction factor, f
0.01927
Loss per 100 ft
2.268ft
  1. 1.Q = 200 gpm ÷ 448.8 = 0.4456 ft³/s; D = 4.026/12 = 0.3355 ft
  2. 2.A = πD²/4 = 0.0884 ft²; V = Q/A = 5.04 ft/s
  3. 3.Water at 60 °F (IAPWS-95): ν = 1.208 × 10⁻⁵ ft²/s, ρ = 62.37 lbm/ft³
  4. 4.Re = VD/ν = 139,900 → turbulent
  5. 5.ε/D = 0.0004471; Colebrook → f = 0.01927
  6. 6.h_f = f (L/D)(V²/2g) = 0.01927 × (500/0.3355) × 0.3948 = 11.34 ft
  7. 7.Δp = γ h_f / 144 = 62.37 × 11.34 / 144 = 4.911 psi

How it's tested on the exam

Shows up on: PE Mechanical (Thermal & Fluid Systems) · PE Mechanical (HVAC & Refrigeration) · PE Civil (Water Resources & Environmental)

  • Head loss is in feet of the flowing fluid. Convert to pressure with that fluid's specific weight — for water, about 2.31 ft per psi.
  • Minor losses are separate. Fittings and valves add K·V²/2g (or equivalent lengths) on top of the straight-pipe loss.
  • Velocity from flow: V = Q/A with Q in ft³/s. 1 ft³/s is 448.8 gpm.
  • Doubling the flow roughly quadruples the loss in turbulent flow (h_f ∝ V² with f nearly constant) — a quick sanity check on your answer.

Worked example

200 gpm of 60 °F water flows through 500 ft of 4-in Schedule 40 steel pipe (inside diameter 4.026 in, ε = 0.00015 ft). Find the head loss and pressure drop.

  1. 1.Q = 200 / 448.8 = 0.4456 ft³/s; A = π(0.3355)²/4 = 0.08840 ft²; V = 5.04 ft/s
  2. 2.Re = VD/ν = 5.04 × 0.3355 / 1.2085 × 10⁻⁵ = 139,900 → turbulent
  3. 3.ε/D = 0.000447; Colebrook gives f = 0.01927
  4. 4.h_f = f (L/D)(V²/2g) = 0.01927 × (500/0.3355) × 0.3948 = 11.34 ft
  5. 5.Δp = γ h_f = 62.37 × 11.34 / 144 = 4.91 psi

h_f ≈ 11.3 ft of water; Δp ≈ 4.9 psi.

Where it is in the Handbook

  • Hydraulics, Fluids, and Pipe Flow — Darcy-Weisbach Equation — PE Mechanical Reference Handbook
  • Hydraulics — Head Loss Due to Flow (Darcy-Weisbach Equation) — PE Civil Reference Handbook

Section names as printed in the current NCEES PE Reference Handbooks (Mechanical 2.1, Civil 2.2); numbering can shift between versions. On exam day you'll use your own approved calculator — here's which models NCEES allows.