Fluid Mechanics & Hydraulics

Darcy–Weisbach Pipe Head Loss

Pressure-pipe friction by Darcy–Weisbach, hL = f·(L/D)·V²/2g plus ΣK·V²/2g minor losses, with f from 64/Re (laminar) or the Colebrook equation (Swamee–Jain seeded). Solves for head loss, flowrate, or required diameter; roughness by value or by material (steel/cast iron/concrete/PVC/glass); viscosity direct or from water temperature. Reports f, Reynolds number, velocity, and regime. Use for: pipe friction loss, 'what diameter is needed', pump-system loss terms, laminar vs turbulent checks. Not for: partially full pipes (use Manning) or Hazen–Williams-specified problems.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

solve forThe unknown: head loss from the flow, flowrate from an allowed head loss, or required diameterlengthPipe length [dimension: length]diameter (optional)Pipe inside diameter (required unless solving for it) [dimension: length]flowrate (optional)Volumetric flowrate (give this OR velocity; required unless solving for flowrate) [dimension: flowrate]velocity (optional)Mean pipe velocity — alternative to flowrate [dimension: velocity]head loss (optional)Head loss across the pipe run (required when solving for flowrate or diameter) [dimension: length]roughness (optional)Absolute wall roughness ε (give this OR material) [dimension: length]material (optional)Pipe material for a standard roughness: commercial steel ε = 0.045 mm, cast iron 0.26 mm, concrete 1.0 mm, PVC/glass 0.0015 mmkinematic viscosity (optional)Fluid kinematic viscosity ν in m²/s (e.g. 1.004e-6 for water at 20 °C). Give this OR temperature; omit both for water at 20 °C.temperature (optional)Water temperature — ν is looked up from the standard water property table [dimension: temperature]minor loss k (optional)Sum of minor-loss coefficients ΣK (entrance, bends, valves, exit); omitted = 0

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

laminar oil line: V = 1 m/s, D = 50 mm, L = 10 m, ν = 1e-4 m²/s

Given

solve for: headLoss
length: 10 m
diameter: 50 mm
velocity: 1 m/s
kinematic viscosity: 0.0001

Assumptions

Steady incompressible full-pipe flow; the pipe is circular and flowing full.
Laminar regime: f = 64/Re — the wall roughness does not enter.

Solution steps

Reynolds number
Re classifies the regime: below about two thousand laminar, above about four thousand turbulent.
Re = V·D/ν
Re = 1 m/s × 0.05 m / 0.0001 m²/s = 500
= 500
Friction factor (laminar)
Laminar pipe flow has the exact solution f = 64/Re, independent of wall roughness.
f = 64/Re
f = 64 / 500 = 0.128
= 0.128
Friction head loss
Darcy–Weisbach: the friction loss scales with the length-to-diameter ratio and the velocity head.
hf = f·(L/D)·V²/2g
h_f = 0.128 × (10 m / 0.05 m) × 0.051 m = 1.305 m
= 1.305 m

Results

Total head loss hL

1.305m

Darcy friction factor f

0.128

Reynolds number Re

500

Mean pipe velocity V

1m/s

Flowrate Q

0.00196m^3/s

Where this answer was checked

source: Laminar pipe flow closed form f = 64/Re (NCEES FE Reference Handbook) — Re = VD/ν = 500 < 2000; hL = f·(L/D)·V²/2g
verified by: hand-recomputed
derivation: Re = 1·0.05/1e-4 = 500. f = 64/500 = 0.128. hL = 0.128·(10/0.05)·1²/(2·9.80665) = 25.6·0.0509858 = 1.30524 m. No roughness needed in laminar flow.

Example 2

SI commercial steel: Q = 10 L/s, D = 0.1 m, L = 50 m, water at 20 °C

Given

solve for: headLoss
length: 50 m
diameter: 0.1 m
flowrate: 10 L/s
roughness: 0.045 mm
temperature: 20 degC

Assumptions

The fluid is water: ν was taken from the standard water property table at the given temperature.
Steady incompressible full-pipe flow; the pipe is circular and flowing full.

Solution steps

Velocity from continuity
The mean velocity follows from the flowrate and the pipe area.
V = Q/A = 4Q/(πD²)
V = 0.01 m^3/s / 0.00785 m^2 = 1.27 m/s
= 1.27 m/s
Reynolds number
Re classifies the regime: below about two thousand laminar, above about four thousand turbulent.
Re = V·D/ν
Re = 1.27 m/s × 0.1 m / 1.00e-6 m²/s = 126800
= 126800
Friction factor (Colebrook)
The implicit Colebrook equation is seeded with the explicit Swamee–Jain estimate and converged by fixed-point iteration.
1/√f = −2·log10( (ε/D)/3.7 + 2.51/(Re·√f) )
ε/D = 0.00045; Swamee–Jain seed f = 0.0196; converged f = 0.01951
= 0.01951
Friction head loss
Darcy–Weisbach: the friction loss scales with the length-to-diameter ratio and the velocity head.
hf = f·(L/D)·V²/2g
h_f = 0.01951 × (50 m / 0.1 m) × 0.0827 m = 0.8064 m
= 0.8064 m

Results

Total head loss hL

0.8064m

Darcy friction factor f

0.01951

Reynolds number Re

126800

Mean pipe velocity V

1.27m/s

Flowrate Q

0.01m^3/s

Where this answer was checked

source: Colebrook equation hand-iterated (fixed point on 1/√f), Swamee–Jain cross-check — ε = 0.045 mm steel; hL = f·(L/D)·V²/2g
verified by: hand-recomputed
derivation: V = 0.01/(π·0.1²/4) = 1.273240 m/s; Re = 1.273240·0.1/1.004e-6 = 126 817; ε/D = 4.5e-4. Swamee–Jain: f₀ = 0.25/log₁₀(4.5e-4/3.7 + 5.74/126817^0.9)² = 0.0195987 → x₀ = 1/√f₀ = 7.14310. Fixed point xₙ₊₁ = −2log₁₀(1.21622e-4 + 2.51xₙ/126817): x₁ = 7.16009, x₂ = 7.15898, x₃ = 7.15905 → converged x = 7.159045, f = 1/x² = 0.0195115 (Swamee–Jain +0.45% high). hL = 0.0195115·(50/0.1)·1.273240²/(2·9.80665) = 0.80636 m.

Example 3

US cast iron, solve Q: D = 6 in, L = 500 ft, hL = 5.8987 ft

Given

solve for: flowrate
length: 500 ft
diameter: 6 in
head loss: 5.89875 ft
material: cast-iron
kinematic viscosity: 0.00000121

Assumptions

Wall roughness taken from the standard Moody-chart table for the given material (new, clean pipe; aged pipe can be far rougher).
Steady incompressible full-pipe flow; the pipe is circular and flowing full.

Solution steps

Solve the head-loss equation for Q
Head loss grows monotonically with flowrate, so the flowrate giving exactly the target loss is found by a bracketed root search, seeded with the explicit Swamee–Jain flowrate formula. The friction factor is re-evaluated at every trial flowrate.
hL = (f·L/D + ΣK)·V²/2g with V = Q/A
Target h_L = 5.9 ft → Q = 0.7854 cfs (V = 4 ft/s)
= 0.7854 cfs
Reynolds number
Re classifies the regime: below about two thousand laminar, above about four thousand turbulent.
Re = V·D/ν
Re = 4 ft/s × 0.5 ft / 1.21e-6 m²/s = 153600
= 153600
Friction factor (Colebrook)
The implicit Colebrook equation is seeded with the explicit Swamee–Jain estimate and converged by fixed-point iteration.
1/√f = −2·log10( (ε/D)/3.7 + 2.51/(Re·√f) )
ε/D = 0.001706; Swamee–Jain seed f = 0.02391; converged f = 0.02372
= 0.02372
Back-substitution check
Recompute the head loss at the solution — it must reproduce the target.
hf = f·(L/D)·V²/2g
h_f = 0.02372 × (500 ft / 0.5 ft) × 0.249 ft = 5.899 ft
= 5.899 ft

Results

Flowrate Q

0.7854cfs

Darcy friction factor f

0.02372

Reynolds number Re

153600

Mean pipe velocity V

4ft/s

Head loss hL (given)

5.899ft

Where this answer was checked

source: Forward Colebrook hand computation inverted (constructed): V = 4 ft/s produces this exact head loss — cast iron ε = 0.26 mm; recover Q from the head loss
verified by: hand-recomputed
derivation: Forward: D = 0.5 ft = 0.1524 m, V = 4 ft/s = 1.2192 m/s, ν = 1.21e-6 → Re = 153 559, ε/D = 2.6e-4/0.1524 = 1.70604e-3. Swamee–Jain f₀ = 0.0239148 → x₀ = 6.46647; fixed point: x₁ = 6.49316, x₂ = 6.49249, x₃ = 6.49251 → f = 0.0237233 (Swamee–Jain +0.81%). hL = 0.0237233·(152.4/0.1524)·1.2192²/(2·9.80665) = 1.797938 m = 5.89875 ft. So solving for Q from hL = 5.8987 ft must recover Q = V·A = 4·π·0.5²/4 = 0.785398 cfs.