Geoscience Reference
In-Depth Information
EXAMPLE 24.78
Problem: A calibration test is conducted for a solution chemical feeder. During 5 min, the solution
feeder delivers a total of 700 mL. The polymer solution is a 1.3% solution. What is the lb/day feed
rate? (Assume that the polymer solution weighs 8.34 lb/gal.)
Solution: The mL/min flow rate is calculated as
700 mL ÷ 5 min = 140 mL/min
Then convert the mL/min flow rate to a gpd flow rate:
140 mL/min 1440 min/day
3785 mL/gal
×
=
53 gpd
Now calculate the lb/day feed rate:
Chemical (mg/L) × Flow (MGD) × 8.34 lb/day = Chemical (lb/day)
13,000 mg/L × 0.000053 MGD × 8.34 lb/day = 5.7 lb/day polymer
Actual pumping rates can be determined by calculating the volume pumped during a specified time
frame; for example, if 120 gal are pumped during a 15-min test, the average pumping rate during
the test is 8 gpm. The gallons pumped can be determined by measuring the drop in tank level dur-
ing the timed test:
Volume pumped(gal)
Duration of t
Flow (gpm)
=
(24.107)
est(min)
The actual flow rate (gpm) is then calculated using
2
3
0.785
×
(Diameter)
×
Drop in l
level (ft)
×
7.48 gal/ft
Flow rate (gpm)
=
(24.108)
Duration of test (mi
n)
EXAMPLE 24.79
Problem: A pumping rate calibration test is conducted for a 5-min period. The liquid level in the
4-ft-diameter solution tank is measured before and after the test. If the level drops 0.4 ft during the
5-min test, what is the pumping rate in gpm?
Solution:
2
3
0.785
×
(Diameter)
×
Drop i
nnlevel (ft)
×
7.48 gal/ft
Pumpingrate(gpm)
=
Duration of test
(min)
3
0.785
×××
(
4ft4ft)
0.4 ft
×
7.48 gal/ft
=
5min
=
38 gpm
 
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