Discharge

Flow Rate

Operating

Volume (V)

Parameter

Points

gpm

1pm

time, min

gal

L

V1 Bath

outflow

6395

24207

30

191862

726198

V2 Water cannons

12 cannons

80

303

3300

3168000

11990880

V3 Wash stations

24 hoses

30

114

3240

2332800

8829648

V4 Prep flushers

1 flusher

75

284

180

13500

51098

V5 Bath flushers

4 flushers

750

2840

60

180000

681300

V6 Trench flushers

3 flushers

75

284

3420

769500

2912558

V7 Interior wash

12 hoses

10

38

3240

388000

1471608

V8 Yard hydrants

6 nozzles

30

114

180

32400

122634

V9 Rainfall pavement

2.15 acres

1818

6881

60

109080

412868

V10 Rainfall basins

5.51 acres

4937

18686

60

296220

1121193

tions 6-3 and 6-4 with data from table A-2:

(10788045 L)

Sediment Volume = (Tracked vol. *+ *Wheeled

VEB =381044 cf (10699 CM) or;

vol.) x duration of wash season x sediment

VEB =Vmax = 724522 cf (20287 CM)

basin trap efficiency.

The designer uses a safety factor of 1.25 times the value of

Vmax or 905653 cf (25358CM to size the basin. The designer

Sediment volume

= (2.0 cu ft x 665 washed + 0.6 cu ft x 783 washes)

selects an effective water depth of 8 ft (2.4 M), 2.0 ft (0.6 M) of

wash

month

wash

month

sediment storage, 2.0 ft (0.6 M) of dead storage or a total depth

x 6 months x 0.82 percent

of 15 ft (3.7 M). The average area at the mid point of the

cleaning

equalization basin water zone is:

=8855 cf (248 CM)

The same equations that were used to size the cell based on

905653 cf/8 ft = 113207 sf (10528 SM)

water volume are used to size the cell for the sediment volume,

except that the volume of the ramp is assumed negligible. The

The designer selects to use a square basin (fig A-9) with side

width and length of the cell are known, and the depth is

slopes of 1 to 3.5. The basin is sized by taking the square root

calculated. Thus--

of the average area required to determine the average length and

width of the basin. This method gives the average basin size--

(Ls - Lr) xWx x Ds = Sediment volume

336 feet by 336 feet (102 meters by 102 meters). With the slope

(145 ft - 48 ft) x 48 ft x Ds = 8855 cf

of the walls at 1 to 3.5,7.0 ft (2.2 M) are added to the average

Ds = 1.9 ft. (0.58M)

side for each foot (meter) of depth above or subtracted below

say Ds = 2.0 ft. (0. 61M)

the average depth. The length of the basin at the water surface

would be 364 feet by 364 feet (110 meters by 110 meters).

With 3 feet (0.9 meters) of freeboard added, the overall top

basin dimensions are summarized in fig-

dimension of the basin will be 385 feet by 385 feet (117 meters

ure A-8.

by 117 meters). Subtracting 3.5 feet (1.1 meters) off of each

side for each foot (meter) below the average depth, the

(7) Equalization basin. The volume of water in the

dimensions of the bottom of the basin, with 2 feet (0.6 meters)

equalization basins is based on the greater of

allowed for sediment storage and 2 feet (0.6 meters) allowed for

Vavg for any assumed *5 *day wash period or

dead storage, would be 280 feet by 280 feet (86 meters by 86

Vmax:

meters).

VEB = 5xVavg=5x570042 gal= 2850210 gal