Stormwater Calculations - Mathematics and Statistics for the Environment - page 808

The two methods for calculating the required size of the extended detention orifice allow for a
quick and conservative design (Method 2, above) and a similarly quick estimation with a routing
to verify the performance of the design (Method 1). Method 1, which uses the maximum hydraulic
head and maximum discharge in the calculation, results in a slightly larger orifice than the same
procedure using the average hydraulic head (Method 2). The routing allows the designer to verify
the performance of the calculated orifice size. As a result of the routing effect however, the actual
basin storage volume used to achieve the drawdown time will be less than the computed brim draw-
down volume. It should be noted that the routing of the extended detention of the runoff from storms
larger than the water quality storm (such as the 1-year frequency storm for channel erosion control)
will result in a proportionately larger reduction in the actual storage volume needed to achieve the
required extended detention. The procedure used to size an extended detention orifice includes the
first steps of the design of a multistage riser for a basin controlling water quality and/or channel
erosion, and peak discharge. These steps are repeated for sizing the 2-year and 10-year release open-
ings. Other design storms may be used as required by ordinance or downstream conditions.

25.7.9.2.1 Method 1. Water Quality Orifice Design Using Maximum
Hydraulic Head and Routing of Water Quality Volume
A water quality extended detention basin sized for two times the water quality volume will be used
here to illustrate the sizing procedure for an extended detention orifice.

1. Calculate the water quality volume ( V wq ) required for treatment. Assume the following:

V wq = 404,236 ft 2 × 1/2 in./12 in./ft = 16,843 ft 3

16,843 ft 3 ÷ 43,560 ft 2 /ac = 0.38 ac-ft

For extended detention basins, 2 × V wq = 2(0.38 ac-ft) = 0.76 ac-ft = 33,106 ft 3 .

2. Determine the maximum hydraulic head ( h max ) corresponding to the required water quality
volume. Assume from our example stage vs. storage curve (Figure 25.19) that 0.76 ac-ft
occurs at elevation 88 ft (approximate). Therefore, h max = 88 - 81 = 7 ft.

3. Determine the maximum discharge ( Q max ) resulting from the 30-hour drawdown require-
ment. The maximum discharge is calculated by dividing the required volume, in ft 3 , by
the required time, in seconds, to find the average discharge, and then multiplying by 2, to
determine the maximum discharge. Assume the following:

3

33,106 ft
(30hr)(3600 sec/hr)

Q avg = = 0.30 cf s

Q max =× = 2 0.30 cfs 0.60 cfs

4. Determine the required orifice diameter by rearranging the orifice equation, Equation
25.12, to solve for the orifice area (ft 2 ) and then diameter (ft). Insert the values Q max and
h max into the rearranged orifice equation, Equation 25.13, solve for the orifice area, and
then solve for the orifice diameter:

QCa h = 2 (25.12)

Q
C h

a = (25.13)

2