Water Treatment Processes - Mathematics and Statistics for the Environment - page 637

■ EXAMPLE 23.68
Problem: During a 70-hr filter run, a total of 22.4 million gal of water are filtered. What is the aver-
age flow rate through the filter in gpm during this filter run?
Solution:
Flow rate = Total gallons produced/Filter run (min)
= 22,400,000 gal/(70 hr × 60 min/hr) = 5333 gpm
■ EXAMPLE 23.69
Problem: At an average flow rate of 4000 gpm, how long of a filter run (in hours) would be required
to produce 25 MG of filtered water?
Solution: Write the equation as usual, filling in known data:
Flow rate = Total gallons produced/filter run (min)
4000 gpm = 25,000,000 gal/( x hr × 60 min/hr)
Then solve for x:
x = 25,000,000 gal/(4000 gpm × 60 min/hr) = 104 hr
■ EXAMPLE 23.70
Problem: A filter box is 20 ft by 30 ft (including the sand area). If the influent valve is shut, the water
drops 3 in./min. What is the rate of filtration in MGD?
Solution:
Given:
Filter box = 20 ft × 30 ft
Water drops = 3 in./min
Step 1. Find the volume of water passing through the filter:
Volume = Area × Height
Area = Width × Length
Note: The best approach to performing calculations of this type is a step-by-step one, breaking
down the problem into what is given and what is to be found.
Area = 20 ft × 30 ft = 600 ft 2
Convert 3.0 in. into feet: 3/12 = 0.25 ft
Volume = 600 ft 2 × 0.25 ft = 150 ft 3 of water passing through the filter in 1 min
Step 2. Convert cubic feet to gallons:
150 ft 3 × 7.48 ga l /f ft3 3 = 1122 gpm
Step 3. The problem asks for the rate of filtration in MGD. To find MGD, multiply the number of
gallons per minute by the number of minutes per day:
1122 gpm × 1440 min/day = 1.62 MGD