Geoscience Reference
In-Depth Information
EXAMPLE 23.24
Problem: A flocculation basin is 40 ft long by 12 ft wide with water to a depth of 9 ft. What is the
volume of water (in gallons) in the basin?
Solution:
Volume = Length (ft) × Width (ft) × Depth (ft) × 7.48 ga l /f (ft) 3
= 40 ft × 12 ft × 9 ft × 7.48 gal l /f ft 3 = 32,314 gal
EXAMPLE 23.25
Problem: A flocculation basin is 50 ft long by 22 ft wide and contains water to a depth of 11 ft, 6 in.
How many gallons of water are in the tank?
Solution: First convert the 6-in. portion of the depth measurement to feet:
(6 in.)/(12 in./ft) = 0.5 ft
Then calculate basin volume:
Volume = Length (ft) × Width (ft) × Depth (ft) × 7.48 ga l /f (ft) 3
= 50 ft × 22 ft × 11.5 ft × 7.48 gal l /f ft 3 = 94,622 gal
23.4.4.2 Detention Time
Because coagulation reactions are rapid, detention time for flash mixers is measured in seconds,
whereas the detention time for flocculation basins is generally between 5 and 30 min. The equation
used to calculate detention time is shown below:
Detention time (min) = Volume of tank (gal)/Flow rate (gpm)
(23.24)
EXAMPLE 23.26
Problem: The flow to a flocculation basin that is 50 ft long by 12 ft wide by 10 ft deep is 2100 gpm.
What is the detention time in the tank (in minutes)?
Solution:
Tank volume (gal) = 50 ft × 12 ft × 10 ft × 7.48 gal l /f ft 3 = 44,880 gal
Detention time = Volume of tank (gal)/Flow rate (gpm) = 44,880 gal/2100 gpm = 21.4 min
EXAMPLE 23.27
Problem: A flash mix chamber is 6 ft long by 4 ft with water to a depth of 3 ft. If the flow to the flash
mix chamber is 6 MGD, what is the chamber detention time in seconds (assuming that the flow is
steady and continuous)?
Solution: First, convert the flow rate from gpd to gps so the time units will match:
6,000,000/(1440 min/day × 60 sec/min) = 69 gps
Then calculate detention time:
Detention time = Volume of tank (gal)/Flow rate (gpm)
= (6 ft × 4 ft × 3 ft × 7.48 ga l /f ft 3 )/69 gps = 7.8 sec
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