Water Resources HubWater topic 13 of 18 — free theory
Culverts & Spillways
Inlet vs outlet control in culverts, the weir equation, spillway flow, and tailwater effects.
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Inlet control vs outlet control
A culvert is a short conduit carrying a stream under a road embankment. What limits its discharge depends on where the choke point is, and the exam loves asking you to tell the two regimes apart:
- Inlet control — the entrance is the bottleneck. Capacity depends on the inlet geometry and the headwater depth, and on nothing else: not the barrel roughness, not the barrel length, not the slope, not the tailwater. Flow typically shoots through the barrel at supercritical, part-full depth.
- Outlet control — the barrel is flowing full (or the outlet is submerged), so the whole barrel plus the tailwater set the capacity. Now everything counts: entrance loss, barrel friction, exit loss, and the tailwater elevation.
How do you tell which one governs? Compute the headwater for both regimes at the given discharge — the regime that demands the higher headwater (or passes the lower discharge at a given headwater) is the one in charge. Quick rules of thumb: a steep barrel with low tailwater usually means inlet control; a mild slope, a long barrel, or a high tailwater usually means outlet control.
Outlet control (barrel full): HW = TW + (Ke + 1) V²2g + hf
| HW | headwater depth above the outlet invert |
| TW | tailwater depth above the outlet invert |
| Ke | entrance loss coefficient (≈ 0.5 for a square-edged entrance) |
| hf | barrel friction loss = SfL, with Manning Sf = (Vn/R2/3)² |
| The "1" is the exit loss — one full velocity head is dissipated at the outlet. |
Under inlet control, skip this entirely — barrel friction and tailwater do not change the answer.
The weir equation
Weirs and spillway crests both follow the same relationship: discharge grows with the head to the three-halves power.
Q = C · L · H1.5
| C | weir coefficient — broad-crested ≈ 1.6–1.7 (SI) / ≈ 3.0 (English); sharp-crested ≈ 1.84 (SI) / 3.33 (English); ogee spillway ≈ 2.2 (SI) |
| L | crest length (perpendicular to flow) |
| H | head above the crest — not above the channel bottom |
A sharp-crested weir passes more flow per metre of crest than a broad-crested one (higher C), but it is fragile and never used as a dam spillway — spillway crests are broad-crested or ogee-shaped.
Spillways and tailwater
A dam's spillway is just a weir writ large. The service spillway (often gated or ogee-shaped) handles everyday floods; the emergency spillway (usually a broad-crested crest) only runs in rare events. Both get a rating curve — a table or plot of discharge versus head — built straight from Q = C·L·H1.5.
Tailwater matters when it climbs high enough to interfere. If the downstream water surface rises above the culvert outlet crown and the barrel runs full, the regime flips to outlet control. For weirs, high tailwater submerges the crest and trims the discharge below the free-flow value — a submerged weir passes less than the equation says, never more.
PE depth: culvert performance curves and overtopping
On the PE side, a culvert is analysed with a performance curve: headwater depth (often plotted as HW/D) against discharge, with separate curves for inlet control and outlet control. The governing curve is whichever sits higher — the culvert follows the envelope of the two. For multi-barrel culverts, divide the total discharge by the number of barrels before entering the curves.
Overtopping: Qroad = C · L · H1.5 with H above the roadway
If the headwater climbs above the road surface, the roadway itself becomes a broad-crested weir and passes flow over the top — the total discharge is the culvert flow plus the overtopping flow. High-velocity culvert outlets also need energy dissipation (stilling basins, riprap aprons) so the jet does not scour the downstream channel.
Worked example Barrel losses under outlet control
Given:
- 1.2 m diameter concrete culvert, n = 0.013, length 25 m, square-edged entrance (Ke = 0.5).
- Q = 2.5 m³/s, barrel flowing full under outlet control.
Solution:
- A = π(1.2)²/4 = 1.131 m², so V = 2.5/1.131 = 2.21 m/s. R = D/4 = 0.30 m.
- Friction slope: Sf = (Vn/R2/3)² = (2.21×0.013/0.302/3)² = (0.06413)² = 0.004113. hf = 0.004113×25 = 0.103 m.
- Velocity head: V²/2g = 2.21²/19.62 = 0.249 m. Entrance loss = 0.5×0.249 = 0.125 m; exit loss = 1.0×0.249 = 0.249 m.
- Total barrel loss = 0.103 + 0.125 + 0.249 = 0.48 m.
Answer: The barrel consumes ≈ 0.48 m of head (entrance + friction + exit), which must be added to the tailwater to get the headwater.