Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 33

■ Example 2.66
Problem: Find the flow in cubic feet per second in an 8-in. line if the velocity is 3 ft/sec.
Solution:
1. Determine the cross-sectional area of the line in square feet. Start by con-
verting the diameter of the pipe to inches.
2. The diameter is 8 in.; therefore, the radius is 4 in. = 4/12 of a foot, or 0.33
ft.
3. Find the area ( A ) in square feet:
A = π × r 2
A = π × (0.33 ft) 2 = 3.14 × 0.109 ft 2 = 0.342 ft 2
4. Q = V × A :
Q = 3 ft/sec × 0.342 ft 2 = 1.03 cfs
■ Example 2.67
Problem: Find the flow in gallons per minute when the total flow for the day is
75,000 gpd.
Solution:
75,500 gpd ÷ 1440 min/day = 52 gpm
■ Example 2.68
Problem: Find the flow in gpm when the flow is 0.45 cfs.
Solution:
0.45 cfs × 449 gpm/cfs = 202 gpm
■ Example 2.69
Problem: A system uses 3 MGD. How many gallons per minute does it use?
Solution: Convert MGD to gpm.
3 MGD × 700 = 2100 gpm
■ Example 2.70
Problem: A pipeline has a carrying capacity of 2 cfs. How many gallons per minute
can it handle?
Solution : Convert cfs to gpm.
2 cfs × 449 gpm/cfs = 898 gpm