Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 41

■ Example 2.94
Problem: Find the detention time in a 55,000-gal reservoir if the flow rate is 75 gpm.

Solution:

55,000 gal
75 gpm

min
60 min/hr

733

Detentiontime = = 733 min; = 12.2 hr

■ Example 2.95
Problem: If the fuel consumption of a boiler is 30 gpd, how many days will the
1000-gal tank last?

Solution:

1000 gal ÷ 30 gpd = 33.3 days

■ Example 2.96
Problem : A 60,000 gal tank receives 250,000 gpd flow. What is the detention in
hours?

Solution:

60,000 gal
250,000 gpd

0.24 days
24 hr/day

Detentiontime = = 0.24 d ays; = 5.8 hr

■ Example 2.97
Problem: A tank is 70 ft by 60 ft by 10 ft, and the flow is 3.0 MGD. What is the
detention time in hours?

Solution: Find the volume in cubic feet.

70 ft × 60 ft × 10 ft = 42,000 ft 3

Change cubic feet to gallons.

42,000 ft 3 × 7.48 gal l /f ft3 3 = 314,160 gal

Change MGD to gpd.

3 MGD = 3,000,000 gpd

Find the detention time in days and change days to hours.

314,160 gal
3,000,000 gpd

0.10 days
24 hr/day

Detentiontime = = 00days; 0.1 = 2.4 hr