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Detention Pond Design: Outlet Hydraulics and Storage-Indication Routing

A 2-PDH course on sizing a dry detention pond for peak-rate control: preliminary storage from TR-55, stage-storage by the average-end-area and conic methods, orifice and weir outlets, a combined rating curve, and storage-indication (Modified Puls) routing worked step by step. Graded quiz, certificate.

$29 · 2 PDH · the course text is free to read; enrollment unlocks the graded quiz and your certificate.

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Instructor: Michael Flynn, PE · Estimated time: 120 minutes including the quiz · Which states accept it?

Learning objectives

1. What a detention pond has to do

Development replaces pervious ground with roofs and pavement, so the same storm produces more runoff, sooner, with a higher peak. A dry detention pond stores part of that runoff and releases it through a restricted outlet, so that the peak flow leaving the site is no larger than a target, usually the pre-development peak for one or more design storms. The pond does not reduce the runoff volume; it spreads it out in time.

Before designing anything, find the criteria. A local stormwater ordinance or state manual normally sets:

A pond behind an embankment is also a dam. In most states, an embankment above a size threshold falls under the state dam safety program, with its own design storm and hazard classification. In North Carolina, for example, the Dam Safety Act applies to dams 15 ft or more in height with 10 acre-feet or more of storage, subject to exemptions. Check the thresholds early, because crossing one changes the spillway design storm and the review path.

2. A first estimate of storage

Routing needs a trial pond, so start with an estimate. TR-55 Chapter 6 relates the ratio of storage volume to runoff volume, Vs/Vr, to the ratio of peak outflow to peak inflow, qo/qi. For rainfall distributions Type II and III:

Vs/Vr = 0.682 − 1.43 (qo/qi) + 1.64 (qo/qi)² − 0.804 (qo/qi)³

(for Types I and IA the coefficients are 0.660, −1.76, 1.96 and −0.730). The relationship was fitted to many routed ponds and is meant for preliminary sizing only, with an error that TR-55 puts at up to about 25%: the routed design governs.

Example 1: preliminary storage. A 12-acre site produces 2.40 in of runoff in the 10-year storm after development, with a peak of 26 cfs. The pre-development 10-year peak, the allowable release, is 10 cfs.
qo/qi = 10/26 = 0.385
Vs/Vr = 0.682 − 1.43(0.385) + 1.64(0.385)² − 0.804(0.385)³ = 0.329
Vr = 2.40 in / 12 × 12 ac = 2.40 ac-ft
Vs = 0.329 × 2.40 = 0.79 ac-ft (34,380 ft³). Repeat for each controlled storm; the largest storage usually comes from the largest storm, but a tight low-frequency target can govern.

3. Stage-storage

Storage is computed from the area of the water surface at each contour or stage. Two formulas are in common use for the volume between stages 1 and 2, a height h apart:

Average end area: ΔS = h (A1 + A2) / 2      Conic: ΔS = (h/3) (A1 + A2 + √(A1A2))

The conic (frustum) formula is exact for a pond whose area grows smoothly with depth; average end area slightly overstates the volume, by more where the area changes quickly between stages. With 1-ft contours on a typical pond the difference is small, and the criteria manual or review agency may specify which to use.

Example 2: stage-storage table.
Stage (ft)Area (ft²)Storage, avg end area (ft³)Storage, conic (ft³)
100.06,20000
101.08,4007,3007,272
102.010,90016,95016,895
103.013,70029,25029,168
104.016,80044,50044,392
105.020,20063,00062,866
At the top contour the average-end-area total is 0.21% above the conic total. The rest of this course uses the conic values. The required storage from Example 1, 34,380 ft³, is reached between stages 103 and 104, so this grading is a reasonable trial for that site. The routing example below uses the same pond with a different inflow.

4. Outlet hydraulics

The outlet sets the stage-discharge relationship. Most detention outlets combine a small low-level orifice, which controls the frequent storms, with one or more higher openings or a weir for the larger storms, all inside a riser that discharges to a barrel through the embankment.

4.1 Orifices

Q = Cd · A · √(2 g h)

with A the opening area (ft²), g = 32.2 ft/s² and h the head on the centroid of the opening for free discharge (or the difference between headwater and tailwater when the outlet is submerged). Cd = 0.60 is the usual design value for a sharp-edged plate; a rounded entrance is much higher, about 0.95, so match the coefficient to what will be built. Until the water covers the opening, it flows as a weir; this course's examples simply start the orifice flow when the water reaches the centroid, a common simplification that slightly understates outflow at very low stages.

4.2 Weirs

Q = Cw · L · H3/2

with L the crest length (ft) and H the head above the crest (ft). Cw is about 3.3 for a sharp-crested weir and 2.6 to 3.1 for broad-crested sections such as a riser wall or an emergency spillway crest (see the PE-Calc weir and orifice coefficient tables). Each end contraction reduces the effective length by about 0.1H. A weir in a riser also has an orifice limit: when the riser floods, the opening behaves as an orifice, and the lower of the weir and orifice discharges controls at each stage. Downstream, the barrel must carry the total, or it, not the riser openings, controls.

Example 3: combined rating. The pond in Example 2 gets a 12-in orifice with its invert at 100.0 (centroid 100.50, Cd = 0.6) and a rectangular weir 6 ft long in the riser wall with its crest at 103.0 (Cw = 3.1). The emergency spillway crest is at 104.5 and the top of the embankment at 105.0. At stage 104.0:
Orifice: Q = 0.6 × 0.785 × √(2 × 32.2 × 3.50) = 7.07 cfs
Weir: Q = 3.1 × 6 × 1.001.5 = 18.60 cfs
Total: 25.67 cfs. Below the weir crest, only the orifice discharges, which is what holds back the small storms. The last column is the storage-indication value used in Section 5, with Δt = 5 min.
Stage (ft)Storage (ft³)Orifice (cfs)Weir (cfs)Total O (cfs)2S/Δt + O (cfs)
100.0000.000.000.000.00
100.503,3710.000.000.0022.48
101.007,2722.670.002.6751.16
101.5011,7813.780.003.7882.32
102.0016,8954.630.004.63117.27
102.5022,6925.350.005.35156.63
103.0029,1685.980.005.98200.44
103.5036,4036.556.5813.13255.81
104.0044,3927.0718.6025.67321.62
104.5053,2147.5634.1741.73396.49

4.3 The barrel can take control

A riser drains through a barrel pipe under the embankment, and the rating built from the riser openings holds only while the barrel can carry what they pass. At higher stages the barrel flows full and the outflow becomes

Q = a √[ 2g H / (1 + Ke + kf) ],    kf = 29 n² L / R4/3

with a the barrel area, H the pond level above the outlet centroid (or above the tailwater, if higher), Ke the entrance loss, and kf the friction loss for a pipe of length L, Manning n and hydraulic radius R = D/4 when full. The true rating is the smaller of the riser rating and the barrel capacity at each stage.

Barrel check for Example 3. The riser discharges into an RCP barrel 60 ft long with its outlet invert at 97.4, n = 0.013, Ke = 0.5. For an 18-in barrel, kf = 29 × 0.013² × 60 / 0.3754/3 = 1.09; at the routed peak stage of Example 4 (103.89) it carries 21.1 cfs, less than the 22.5 cfs the riser passes. The barrel takes control at about elevation 103.83, so the routing of Example 4 would be wrong. A 24-in barrel carries 39.4 cfs at that stage and leaves the riser in control. Check the barrel at the highest stage the pond can reach, not only at the design peak.

5. Storage-indication (Modified Puls) routing

Routing applies continuity to the pond over a time step Δt: inflow minus outflow equals the change in storage. Using average flows over the step,

(I1 + I2)/2 − (O1 + O2)/2 = (S2 − S1)/Δt

The unknowns are O2 and S2, so collect them on one side:

2S2/Δt + O2 = (I1 + I2) + (2S1/Δt − O1)

Everything on the right is known at the start of the step. Because storage and outflow are both functions of stage, 2S/Δt + O is a known function of stage too (the last column of the rating table), so each step is: compute the right-hand side, look up the stage at which 2S/Δt + O equals it, read O2 and S2 at that stage, and carry 2S2/Δt − O2 = (2S2/Δt + O2) − 2O2 to the next step. This is "level-pool" routing: it assumes the water surface is flat, which is right for ponds and wrong for long, flat channels.

Choose Δt small compared with the rise of the hydrograph. A step of about one-fifth to one-tenth of the time to peak is usual; too long a step clips the peak and, with a small outlet, can make the computation oscillate.

Example 4: routing a hydrograph. An NRCS triangular inflow hydrograph with a peak of 36 cfs at 30 min and a base of 2.67 × 30 = 80 min enters the pond of Example 3, empty at the start. Δt = 5 min = 300 s. Rows every 10 min, plus the peak (bold).
t (min)I (cfs)2S/Δt + O (cfs)O (cfs)Stage (ft)Storage (ft³)
00.000.000.00100.000
1012.0024.000.54100.523,519
2024.0089.423.98101.6112,816
3036.00191.495.86102.9027,844
4028.81285.6918.54103.7440,073
5021.63305.8322.47103.8942,504
6014.44290.1919.40103.7740,619
707.26261.1114.04103.5437,060
800.07225.348.44103.2432,536
900.00195.515.91102.9528,439
1000.00172.175.59102.6824,988
1100.00150.165.24102.4221,738
1200.00129.564.88102.1618,702
1300.00110.444.48101.9115,893
1400.0092.914.07101.6613,326
1500.0077.073.63101.4211,016
1600.0063.023.16101.208,980
1700.0050.882.66101.007,233
1800.0040.762.13100.825,794
1900.0032.771.58100.684,678
2000.0027.011.01100.573,899
2100.0023.530.43100.513,466
One step in full, from t = 25 to 30 min: I1 + I2 = 30.00 + 36.00 = 66.00 cfs; 2S1/Δt − O1 = 2 × 19,571/300 − 4.99 = 125.49 cfs; so 2S2/Δt + O2 = 191.49 cfs, which the rating gives at stage 102.90 with O2 = 5.86 cfs.
Result: peak outflow 22.47 cfs at t = 50 min, a 38% reduction from the 36-cfs inflow peak, at a peak stage of 103.89 ft, 0.61 ft below the emergency spillway crest. The inflow volume is 86,519 ft³; the volume released plus the volume still in the pond at the end differs from it by +0.00%, a check worth making on every routing.

Two properties of level-pool routing are useful checks on any result. First, the outflow peak lies on the falling limb of the inflow hydrograph, at the moment the two curves cross: before that instant inflow exceeds outflow and the pond is still filling, so storage and stage, and therefore outflow, are still rising. In Example 4 the peak outflow at 50 min comes well after the inflow peak at 30 min; at that step the falling inflow is 21.63 cfs and the outflow 22.47 cfs, so the two curves crossed within the last step. A shorter time step would place the crossing more exactly. Second, the peak stage and the peak storage occur at the same time as the peak outflow, because outflow is a single-valued function of stage.

6. Finishing the design

6.1 Sizing the emergency spillway

An open-channel emergency spillway is usually rated as a broad-crested weir, Q = C L H1.5, with C about 2.6 to 3.1 depending on the crest shape and head. The governing case is set by the criteria: the spillway design storm, often with the principal outlet assumed blocked, and the required freeboard above the spillway water surface.

Example 5: spillway length. Suppose the criteria require the spillway alone to pass the full inflow peak of 36 cfs (no credit for storage or the principal outlet), with C = 2.8. With the crest at 104.5 and the top of embankment at 105.0, keeping 0.1 ft of freeboard leaves H = 0.40 ft, so L = 36 / (2.8 × 0.401.5) = 50.8 ft, an impractically long crest. Raising the embankment to 106.0 with 0.5 ft of freeboard gives H = 1.00 ft and L = 12.9 ft. The distance between the principal-outlet peak, the spillway crest and the top of the embankment is usually decided by this check, so make it early, before the grading is fixed.

6.2 A water-quality orifice

Where a water-quality volume (WQV) must drain slowly, a small orifice at the bottom of the pond controls it. A quick first size uses the average flow WQV/T and the average head, taken here as half the maximum head; the drawdown is then checked stage by stage, as in the drawdown note above.

Example 6: sizing for a 48-hour drawdown. A WQV of 0.25 ac-ft = 10,890 ft³ fills the pond of Example 2 to elevation 101.41. Average flow over 48 hours: 10,890 / (48 × 3600) = 0.063 cfs. With half the maximum head, 0.70 ft, the orifice area is 0.063 / [0.60 √(2 × 32.2 × 0.70)] = 0.0156 ft², a diameter of 1.69 in; use 1.75 in. The stage-by-stage check for that orifice gives a drawdown of 54.3 hours, longer than the target. The average-head estimate is only approximate because the pond is widest, and the head greatest, at the start. Trying the next size, 2.00 in, gives 41.4 hours, inside a 24-to-48-hour window, so that is the orifice to specify. Always check the first size by the stage-by-stage computation. An orifice this small needs a trash rack or a protected inlet, and many manuals set a minimum diameter.

7. Common mistakes

8. Summary

A detention design is a loop: estimate storage, grade a pond, choose outlets, build the rating and the storage-indication curve, route each design storm, and adjust until every release target and freeboard requirement is met. The routing itself is a short, mechanical calculation; the engineering lies in the criteria, the outlet details that will actually be built, and the checks around the result.

References: USDA NRCS TR-55, Urban Hydrology for Small Watersheds (1986), Ch. 6; USDA NRCS National Engineering Handbook Part 630, Ch. 17 (flood routing); FHWA HEC-22, Urban Drainage Design Manual, 3rd ed., Ch. 8 (detention and retention facilities); Brater & King, Handbook of Hydraulics; ASCE/WEF MOP 77.

Final quiz

23 questions. You need 70% to pass. You can retake it as many times as you need.

1. A dry detention pond designed for peak-rate control mainly changes:

2. With the TR-55 Type II/III relation, qo/qi = 20/50 gives Vs/Vr most nearly:

3. A 10-ac site has 3.0 in of runoff and qo/qi = 0.5. The TR-55 preliminary storage (ac-ft) is most nearly:

4. Contours 1 ft apart enclose 5,000 and 9,000 ft². The conic-method volume between them (ft³) is most nearly:

5. Compared with the conic method, the average-end-area method:

6. An 8-in sharp-edged orifice (Cd = 0.60) has 4.0 ft of head on its centroid. Q (cfs) is most nearly:

7. A 5-ft rectangular weir (Cw = 3.1) runs with 0.8 ft of head. Q (cfs) is most nearly:

8. A 12-in orifice with its invert at 100.0 (Cd = 0.60, free discharge) at a pond stage of 102.0. Q (cfs) is most nearly:

9. Orifice head for free discharge is measured from the water surface to:

10. When a riser floods, its crest opening:

11. In storage-indication routing with I1 = 10, I2 = 16 cfs, S1 = 9,000 ft³, O1 = 4 cfs and Δt = 300 s, 2S2/Δt + O2 (cfs) is:

12. In level-pool routing, the peak outflow occurs:

13. With no inflow, a pond of 6,000 ft² surface area falls 0.5 ft while releasing an average 2.0 cfs. The time (min) is most nearly:

14. 1.8 acre-feet of storage, in cubic feet, is most nearly:

15. A routing time step of about this fraction of the time to peak is usual:

16. A rounded (bell-mouth) orifice entrance has a discharge coefficient of about:

17. Level-pool routing assumes:

18. A pond meets its 100-year release target. The reviewer will still expect:

19. Water-quality volumes are commonly required to drain over about:

20. A good check on any routing computation is:

21. A pond outlet's rating is built from the riser openings. It remains valid only while:

22. An emergency spillway (C = 2.8) must pass 50 cfs with 1.0 ft of head. The crest length (ft) is most nearly:

23. A 2.0 ac-ft water-quality volume must drain in 40 hours. The average release rate (cfs) is most nearly:

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