TM 5-814-9
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.
(i) Maximum pavement runoff volume, V9:
(b) Water volume. In addition to a sediment
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
(j) Maximum rainfall volume on basins, V10:
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
(k) Summary of maximum volumes. Table A-
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:
Calculate maximum wash volume for facility, V:
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
Sediment basin design. The effective water
(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
(a) Operation. A dual-celled sediment basin
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
A-16
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