Find pipe head loss, flow rate, or diameter using Darcy-Weisbach. Include minor losses and see friction factor, Reynolds number, and flow regime.
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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
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.