TM 5-814-9
Table A--5 Summary of Maximum Volumes
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).
(d) Total cell size. The calculated sediment
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
A-17
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