Calculate head loss, flow rate, or required diameter for a full circular water pipe using Hazen-Williams, with SI or US units and worked steps.
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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
SI head loss: C = 130, D = 0.3 m, Q = 0.1 m³/s, L = 500 m
Given
solve for
headLoss
hazen williams c
130
length
500 m
diameter
0.3 m
flowrate
0.1 m^3/s
Assumptions
Hazen–Williams is an empirical fit for WATER at ordinary temperatures flowing full in a pressure pipe — it does not apply to other fluids or partially full pipes.
Turbulent flow in the range the correlation was calibrated for.
Solution steps
Velocity from continuity
Mean velocity from the flowrate and the full-pipe area.
V = Q/A = 4Q/(πD²)
V = 0.1 m^3/s / 0.0707 m^2 = 1.41 m/s
= 1.41 m/s
Friction slope from Hazen–Williams
Invert the velocity form for the friction slope; a full circular pipe has hydraulic radius R = D/4.
V = 0.849·C·R^(0.63)·S^(0.54) with R = D/4 ⇒ S = (V/(0.849·C·R^(0.63)))^(1/0.54)
R = 0.075 m; C = 130; S = 0.006433
= 0.006433
Head loss over the run
The loss is the slope times the length.
hL = S·L
h_L = 0.006433 × 500 m = 3.216 m
= 3.216 m
Results
Friction head loss hL
3.216m
Mean pipe velocity V
1.41m/s
Friction slope S = hL/L
0.006433
Where this answer was checked
source
Hazen–Williams SI velocity form V = 0.849·C·R^0.63·S^0.54 (NCEES FE Reference Handbook) — solve S from V, then hL = S·L
verified by
hand-recomputed
derivation
A = π·0.3²/4 = 0.0706858 m²; V = 0.1/0.0706858 = 1.414711 m/s; R = D/4 = 0.075 m. S = (V/(0.849·130·0.075^0.63))^(1/0.54) = 0.0064326. hL = 0.0064326·500 = 3.2163 m. Cross-check with the monomial hL = 10.67·L·Q^1.852/(C^1.852·D^4.87) = 3.2102 m (−0.19%, the rounded 10.67 constant).
Example 2
US flowrate: C = 100 cast iron, D = 12 in, hL = 10 ft over L = 1000 ft
Given
solve for
flowrate
hazen williams c
100
length
1000 ft
diameter
12 in
head loss
10 ft
Assumptions
Hazen–Williams is an empirical fit for WATER at ordinary temperatures flowing full in a pressure pipe — it does not apply to other fluids or partially full pipes.
Turbulent flow in the range the correlation was calibrated for.
Solution steps
Friction slope
Available head loss per unit length.
S = hL/L
S = 10 ft / 1000 ft = 0.01
= 0.01
Hazen–Williams velocity
Velocity from the SI form with R = D/4 for a full circular pipe.
V = 0.849·C·R^(0.63)·S^(0.54) with R = D/4
R = 0.25 ft; C = 100; V = 4.576 ft/s
= 4.576 ft/s
Flowrate from continuity
Multiply by the full-pipe area.
Q = V·A = V·πD²/4
Q = 4.576 ft/s × 0.785 ft^2 = 3.594 cfs
= 3.594 cfs
US-customary form (footnote)
The identical equation in US units — the 1.318 constant is just 0.849 carried through the ft↔m conversion of R and V.
V(ft/s) = 1.318·C·R(ft)^(0.63)·S^(0.54)
Results
Flowrate Q
3.594cfs
Mean pipe velocity V
4.58ft/s
Friction slope S = hL/L
0.01
Where this answer was checked
source
Hazen–Williams US velocity form V = 1.318·C·R^0.63·S^0.54 — S = 0.01; V then Q = V·A
verified by
hand-recomputed
derivation
R = 1/4 ft; R^0.63 = 0.417544; S^0.54 = 0.01^0.54 = 0.0831764. V = 1.318·100·0.417544·0.0831764 = 4.57739 ft/s. A = π/4 = 0.785398 ft²; Q = 4.57739·0.785398 = 3.59507 cfs. (SI cross-check: 0.849·0.3048^−0.37 = 1.31772, within 0.02% of the 1.318 constant.)
Example 3
SI diameter: C = 150 PVC, Q = 0.05 m³/s, hL = 5 m over L = 1000 m
Given
solve for
diameter
hazen williams c
150
length
1000 m
flowrate
0.05 m^3/s
head loss
5 m
Assumptions
Hazen–Williams is an empirical fit for WATER at ordinary temperatures flowing full in a pressure pipe — it does not apply to other fluids or partially full pipes.
Turbulent flow in the range the correlation was calibrated for.
Solution steps
Friction slope
Available head loss per unit length.
S = hL/L
S = 5 m / 1000 m = 0.005
= 0.005
Required diameter from the discharge form
Substituting A = πD²/4 and R = D/4 into the velocity form collapses Hazen–Williams to a single monomial in D, solved directly.
Q = 0.278·C·D^(2.63)·S^(0.54) ⇒ D = (Q/(0.278·C·S^(0.54)))^(1/2.63)
discharge constant = 0.2784; C = 150; D = 0.2299 m
= 0.2299 m
Results
Required inside diameter D
0.2299m
Mean pipe velocity V
1.2m/s
Friction slope S = hL/L
0.005
Notes
A computed diameter is a hydraulic minimum — real designs round UP to the next commercial pipe size.
Where this answer was checked
source
Hazen–Williams discharge form Q = 0.2784·C·D^2.63·S^0.54 — invert the monomial for D
verified by
hand-recomputed
derivation
S = 5/1000 = 0.005; S^0.54 = 0.0572063. Discharge constant 0.849·(π/4)·4^−0.63 = 0.2784196. D = (0.05/(0.2784196·150·0.0572063))^(1/2.63) = (0.0209653)^(0.380228) = 0.229877 m ≈ 230 mm.