A Detention Basin That Will Not Fit Its Available Area
A 3.5-acre commercial redevelopment must provide on-site detention for the 25-year, 60-minute storm. Composite runoff coefficients are 0.30 pre-development and 0.72 post-development, and the 60-minute intensity is 3.9 in/hr. The only available basin area is a 0.20-acre rectangle between the rear parking row and the property line, and the outlet must be a gravity pipe to the existing storm main. The site is graded flat, so the basin can hold no more than 3 ft of water, with 1 ft of freeboard and 4:1 side slopes so maintenance crews can mow it. Soils are silty clay and no infiltration credit will be granted. The civil engineer has proposed a 9-in orifice at the bottom of the riser to hold the release to the pre-development peak. The landscape architect is asked to confirm that the basin fits and to comment on the outlet.
1. Compute the required detention storage and determine whether it fits in the available area at the stated depth and side slopes.
Read the worked answer
Rational method first: Q = CiA.
- Pre: 0.30 x 3.9 x 3.5 = 4.10 cfs.
- Post: 0.72 x 3.9 x 3.5 = 9.83 cfs.
- Excess: 5.73 cfs.
- Modified rational storage: 5.73 cfs x 3,600 s = 20,640 cf, or 0.47 ac-ft.
Now test the footprint. The 0.20 ac is 8,712 sf and must contain the basin at the top of the 4-ft excavation (3 ft of water plus 1 ft of freeboard). With 4:1 slopes each side climbs 16 ft, so (b + 32)(l + 32) is at most 8,712; a square gives an outside side of 93.3 ft and a bottom of about 61.3 ft, or 3,760 sf. The area at the 3-ft stage is (61.3 + 24) squared, about 7,280 sf. Prismoidal volume is roughly (3,760 + 7,280 + the square root of 3,760 x 7,280) / 3 x 3, about 16,300 cf.
Required is 20,640 cf, so the basin as drawn is about 4,300 cf, or 21 percent, short. It does not fit. Say so before the grading plan is set.
2. Check the proposed 9-in orifice. Assume a discharge coefficient of 0.6, and describe what else must be verified.
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Q = C A times the square root of 2gh, with C = 0.6 and h measured from the orifice centerline to the design water surface, 3 ft at the top of storage. The square root of (2 x 32.2 x 3) is 13.9 ft/s, so A = 4.10 / (0.6 x 13.9) = 0.49 sf and d = the square root of (4A/pi) = 0.79 ft, about 9.5 in. A 9-in orifice (A = 0.44 sf) releases only 3.7 cfs at the 3-ft stage, so the actual peak stage must rise above 3 ft to pass 4.10 cfs; the freeboard and emergency spillway must be sized for the full inflow, not just the release.
Two design cautions. A 9-in orifice at the bottom of a riser clogs easily, so add a trash rack, a micropool, and a minimum 6-in low-flow device. And because detention volume depends on release duration, do a stage-storage-discharge routing check rather than relying on a single-event rational volume. Confirm the 1-ft freeboard sits above the routed 100-year stage, not only the 25-year.
3. The basin is 4,300 cf short. Identify and rank the realistic options for closing that gap.
Read the worked answer
Rank by cost and schedule. First, reduce impervious cover: each 5 percent of the 3.5 acres converted from parking (C about 0.95) to landscape (C about 0.20) lowers the composite C by roughly 0.037 and the post peak by about 0.5 cfs, recovering about 1,800 cf. Converting about 10 percent of the site, 0.35 ac, nearly closes the gap and is usually cheaper than a vault; it is also the option the landscape architect can deliver.
Second, use vertical walls on the two long sides. A 0.20-ac rectangle with vertical walls and a 3-ft stage holds about 26,000 cf, but it needs structural design, a safety railing, and a maintenance plan for a drop-off.
Third, put 4,300 cf of arch chambers under the rear parking row with pretreatment. Do not count infiltration; the silty clay and the absence of credit make it unavailable. Whichever option is chosen, recheck the outlet, because a smaller basin empties faster and the release must still match the pre-development peak across the range of stages.