V8

= yard hydrant flow rate x clean-

one cell to be closed for cleaning while

up period x number clean-ups

the other is treating wastewater; the

wastewater will not have the designed

V8

= Q6 x K x 3=10800 gal x 60 min

detention time, but the wash facility will

x3

be operable. The installation has indicated

that the basins will be cleaned every 6

V8

= 32400 gal (122634 L)

months and it will take approximately 5

days to drain and clean each cell.

V9

= pavement runoff rate x storm

volume, each cell in the basin must be

duration

able to hold the greater of the overflow

and bath volume or the stormwater vol-

V9

= Q9 x TS = 1818 gpm x 60 min

ume; therefore, the bath and the overflow

V9

= 109080 gal (412868 L)

volumes are compared with the

stormwater volume to determine which

value will control the basin site. The total

volume of water required in each cell is

V10 = rainfall rate x storm duration

computed:

V10 = Q10 x TS = 4937 gpm x 60 min

Vw

=

Bath overflow volume + bath volume =

time + V

V10 = 296220 gal (1121193 L)

Qover detention

B

min + 202633 gal

Vw

=

1581 gpmx 120

ns (1485056 L)

5 summarizes the computed maximum

Vw

=

392353 gallo

volumes for the stated conditions at Fort

(1469 CM)

Vw

=

52453 cf

Swampy.

Calculate storm water volume:

Vst

=

paved area storm runoff

Vmax = U1V1+U2V2+U3V3+U4V4+

V

gpcf

U5V5 +U6V6 +U7V7 +U8V8

Vst

=

o/7.48

Vmax = 1.0(191862 ga1) + 0.9

14582 cf (408 CM)

Vst

=

109080/7.48 =

(3168000gal) *+ *0.8(2332800

gal) *+ *0.2(13500 gal) *+*

Since the overflow plus the bath volume is larger than the

0.2(180000 gal) +0.2(769500

stormwater runoff volume, the ovefflow plus bath volume is the

gal) +0.8(388800 gal) +

controlling factor and each cell is sized to detain at least 52453

0.2(32400 gal)

cf (1504 CM) of water. The designer chooses an effective water

depth of 8 ft. (2.4 M), an access ramp slope of 1 on 6, a ramp

Vmax = 54319422 gal (20512512L) or

width of 18 ft. and a basin length-to-width ratio of 3 to 1. Figure

A-7 shows the general geometry of each cell. The dimensions

Vmax = 724,522 cf (20287 CM)

of the basin are then calculated using the information given and

noting that the volume occupied by the ramp must be

(6)

considered:

depth of the basin will be 8 feet (2.4 meters)

and free board is 2 feet (0.6 meters). The

Vsw= LS x WS x DW - l/2 x Lr x Dw x Wr = 52453

sediment depth will be determined by calcula-

W

x 8 x 18 = 52453

tions. Settling tests performed on soils found in

3Ws x S x 8 - l/2 x 48

the training areas showed that 82 percent of the

042

WS = 2

suspended solids will settle in less than 2 hours

(13. 7M)

(120 minutes). This can be checked

WS = 45 ft.

theoretically by using Stokes Law and the

. (41.2M)

and Ls = 135 ft

nondispersed soil particle gradation test re-

sults, after the basin size has been determined.

(c) Sediment volume. The designer estimates that

the installation has average-to-heavy soiling

conditions. By applying Stokes Law, using

is required. One cell will be used to settle

Qover with the basin size previously calculated,

solids during continuous washing and the

and checking the soil particle size information,

adjacent basin will detain the surge

removal of 82 percent of the soil, as deter-

volume created when the bath is flushed.

mined by settling tests, is verified. Using equa-

A dual-cell arrangement will also allow

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