Detention Basin Storage Routing

\frac{d S}{d t} = I - O, I ˉ - O ˉ = \frac{Δ S}{Δ t}

Level-pool routing: average inflow minus average outflow equals storage change. Pond outflow is weir/orifice of HW. Peak O < peak I if there is storage.

$I$: Inflow
$O$: Outflow
$S$: Storage

Step-by-step solved example

Over 10 min, I goes 20→10 cfs (avg 15), O avg 6 cfs. ΔS?

Pond: inflow hydrograph vs outflow; ΔS = (I−O)Δt.

1. Δt
10 min=600 s.
2. ΔS
(15−6)×600=5400 ft³.

Answer: ΔS = 5400 ft³

10 practice questions

0/10 correct

1.If O=I always, storage is

2.A weir outlet O ∝

3.Retention vs detention

4.Trapezoidal rule on a hydrograph is

5.Peak reduction (attenuation) grows with

6.Emergency spillway is sized for

7.Δt=5 min, Ī=8 cfs, Ō=3 cfs: ΔS=

8.Stage-storage curve is

9.Pre- vs post-development: detention aims to

10.Pump station sump uses the same