Hydraulics & drainage · Open channels & culverts
Culvert outlet control headwater calculator
Headwater for a full-flowing culvert in outlet control: HW = H + ho − L S with entrance, friction and exit losses.
Standard: FHWA HDS-5 outlet control
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Inputs in the worked example
The live calculator starts with these values; change any of them and every result, check and diagram updates instantly.
| Input | Value | Unit |
|---|---|---|
| Flow | 4 | m³/s |
| Barrel width (box) / diameter | 1.8 | m |
| Barrel height | 1.2 | m |
| Length | 25 | m |
| Slope | 0.005 | m/m |
| Manning n | 0.013 | |
| Entrance loss ke | 0.5 square edge headwall | |
| Tailwater depth | 0.9 | m |
| Allowable headwater | 2 | m |
Worked example — results
| Result | Value | Unit |
|---|---|---|
| Area | 2.16 | m² |
| Hydraulic radius | 0.360 | m |
| Velocity | 1.85 | m/s |
| Losses (1 + ke + 2g n² L/R^(4/3)) V²/2g | 0.319 | m |
| ho = max(TW, (dc + D)/2) | 0.998 | m |
| HW = H + ho − L S | 1.19 | m |
- ✓ HW ≤ allowable
Method, formulas and assumptions
- Area = 1.8*1.2 = 2.16 m²
- Hydraulic radius = 2.16/(2*(1.8+1.2)) = 0.36 m
- Velocity = 4/2.16 = 1.8519 m/s
- Losses (1 + ke + 2g n² L/R^(4/3)) V²/2g = (1+0.5+2*9.81*0.013^2*25/0.36^(4/3))*1.8519^2/(2*9.81) = 0.31876 m
- ho = max(TW, (dc + D)/2) = MAX(0.9,((4/1.8)^2/9.81)^(1/3)/2+1.2/2) = 0.99775 m
- HW = H + ho − L S = 0.31876+0.99775-25*0.005 = 1.1915 m
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For preliminary design and education. Check every result against the current standard and your own engineering judgement before use.