Culvert Inlet / Orifice Control

Q = C_{d} A 2 g H (inlet submerged, orifice-like)

Inlet control: barrel can carry more than the inlet will admit; HW and inlet geometry govern (orifice or weir). Outlet control: barrel friction/tailwater — energy equation along the barrel. FHWA nomographs in spirit; FE uses orifice/weir/energy.

$H$: Head on the inlet (HW minus centroid/invert per figure)
$C d$: Orifice coefficient (~0.5–0.7)

Step-by-step solved example

Box 4 ft × 4 ft, Cd=0.62, H=9 ft of water on centroid. Q?

Submerged inlet as an orifice: Q = Cd A √(2gH).

1. A and √(2gH)
A=16 ft², √(64.4×9)=√580=24.1 ft/s.
2. Q
0.62×16×24.1=239 cfs.

Answer: Q ≈ 240 cfs

10 practice questions

0/10 correct

1.Unsubmerged inlet often behaves as

2.Outlet control uses

3.Increasing roughness n in outlet control

4.Square-edged inlet Cd is nearer

5.HW/D is

6.A projecting thin-wall inlet is worse than

7.If tailwater is very high, control tends to

8.A=10 ft², H=4 ft, Cd=0.6, g=32.2: Q nearest

9.Multiple barrels: Q total is

10.Energy dissipators at the outlet address