Environmental Engineering Reference
In-Depth Information
One cubic foot of water holds 7.48 gal. We can convert this volume in
cubic feet to gallons:
7.48 gal
1ft
3
198 ft
×
=
1481 gal
3
The test was done for 5 min. From this information, a pumping rate can
be calculated:
1481 gal
5min
296.2
1min
Pumping Rate
=
=
= 2gpm
296.
Example 5.2. Determining Pump Capacity with Influent
Problem: A wet well is 8.2 ft by 9.6 ft. The influent flow to the well, mea-
sured upstream, is 365 gpm. If the wet well rises 2.2 in. in 5 min, how
many gallons per minute is the pump discharging?
Solution:
Influent
=
Discharge
+
Accumulation
(5.2)
365 gal
1min
=
Discharge
+
Accumulation
We want to calculate the discharge. Influent is known, and we have
enough information to calculate the accumulation:
ft
12 in.
1
7.48 gal
1ft
Volume Accumulated .2 ft
=
×
9.6ft2.2 in.
×
×
×
=
108 gal
3
108 gal
5min
21.6 gal
1min
Accumulation
=
=
=
21.6gpm
Using Equation 5.2 (Influent = Discharge + Accumulation):
365 gpm = Discharge + 21.6
Subtracting from both sides:
365 gpm - 21.6 gpm = Discharge + 21.6 gpm - 21.6 gpm
343.4 gpm = Discharge
The wet well pump is discharging 343.4 gallons each minute.
5.3 PreliMinary TreaTMenT
The initial stage in the wastewater treatment process (following
collection and influent pumping) is preliminary treatment . Raw influ-
ent entering the treatment plant may contain many kinds of materials
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