Water topic 14 of 18 — free theory
Stormwater Management & BMPs
Detention vs retention, sizing storage from a hydrograph, first flush and water-quality volume, outlet structures, and LID practices.
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Detention vs retention
The two words get mixed up constantly, so nail the distinction now:
- Detention basin — dry between storms. It temporarily stores runoff and meters it out through a small outlet, shaving the flood peak so the downstream channel sees a gentler, delayed hydrograph.
- Retention basin (wet pond) — keeps a permanent pool between storms. Stormwater displaces the pool from the top; the permanent pool gives sediment time to settle, which is why retention is the water-quality workhorse.
Both flatten the outflow hydrograph. Only retention maintains a pool — if the stem mentions a permanent pool, "detention" is the wrong answer.
Sizing storage from the hydrograph
The required storage is simply the volume of inflow that arrives faster than the outlet can release it. For exam purposes the inflow is usually a triangular hydrograph and the allowable outflow is constant, which turns the whole problem into one triangle:
S = 12 (Ip − Qo) · Tb
| S | required storage volume |
| Ip | peak inflow rate (top of the triangle) |
| Qo | allowable constant outflow rate |
| Tb | base time of the hydrograph — convert to seconds before multiplying |
The half comes from the triangle area; the (Ip − Qo) is the triangle's height above the outflow line. 1 acre-ft = 1,233.5 m³ if the answer wants English units.
First flush and water-quality volume
The dirtiest water arrives first: the first flush washes the accumulated oil, metals, and sediment off pavements in the opening minutes of a storm. Water-quality design therefore targets that initial slug rather than the whole storm.
The water-quality volume (WQV) is commonly sized to capture the first 25 mm (1 in.) of runoff over the drainage area:
WQV = (0.025 m) × (drainage area)
| Capture the first flush, then drain it slowly | extended detention releases the WQV over roughly a day or two so particles settle |
WQV is a runoff depth times an area — don't confuse it with the full design-storm volume, which is much larger.
Outlet structures
The outlet is what makes a basin a basin. Two devices do almost all the work:
Orifice: Q = Cd · A · √(2gH)
| Cd | discharge coefficient, typically 0.60–0.65 |
| H | head measured to the centerline of the orifice |
Weir (overflow/emergency): Q = C · L · H1.5
| H | head measured above the crest |
Real basins use compound outlets — a small orifice low for the water-quality release, a weir higher up for floods — so the discharge grows in stages as the pool rises.
LID at a glance
Low-impact development (LID) treats stormwater at the source instead of piping it all to one big pond. The exam tests what each practice does, not detailed design:
- Bioretention (rain garden) — shallow ponding over mulch and engineered soil with an underdrain; captures the water-quality volume, filters pollutants, and infiltrates some flow.
- Permeable pavement — rainfall passes through the pavement surface into a stone reservoir below, then infiltrates into the soil; reduces runoff volume and peak right in the parking lot.
- Vegetated swales — broad, gently sloped grass channels that convey runoff slowly, filtering sediment and encouraging infiltration along the way (check dams flatten the slope further).
PE depth: routing and stage-storage-discharge
When the outflow isn't constant, the triangle shortcut won't do — use level-pool routing (the storage-indication method). Continuity on the basin reads:
dS/dt = I − O
Working from a stage-storage-discharge relationship (pool elevation → stored volume → outlet discharge), step through the inflow hydrograph interval by interval: each step's inflow minus outflow changes the storage, which sets the new stage and the next outflow. A sediment forebay at the inlet traps coarse material before it reaches the main pool, and the drawdown time (how long the WQV takes to drain) is checked so the basin is empty before the next storm.
Worked example Detention storage from a triangular hydrograph
Given:
- Triangular inflow hydrograph: peak Ip = 3.2 m³/s, base time Tb = 90 min.
- Allowable constant outflow Qo = 1.0 m³/s.
Solution:
- Convert the base time: Tb = 90 × 60 = 5,400 s.
- Triangle height above the outflow line: Ip − Qo = 3.2 − 1.0 = 2.2 m³/s.
- S = ½(2.2)(5,400) = 5,940 m³.
Answer: Required detention storage ≈ 5,940 m³.