Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 42

■ Example 2.98
Problem: A tank 100 ft in diameter and 23 ft deep. If the flow into the tank is 1500
gpm and the flow of the tank is 325 gpm, how many hours will it take to fill the tank?

Solution: Calculate the volume in cubic feet.

Volume = 3.14 × (50 ft) 2 × 23 ft = 180,600 ft 3

or

Volume = 0.785 × (100 ft) 2 × 23 ft = 180,600 ft 3

Change cubic feet to gallons.

180,60 0 ft 3 × 7.48 gal l /f ft3 3 = 1,350,900 gal

Calculate the net inflow.

1500 gpm - 325 gpm = 1175 gpm

Calculate how long until full, or detention time, and change minutes to hours.

1,350,900 gal
1175 gpm

Detentiontime(min) = =1150 min

1150 min÷60 min/hr = 19.2 hr

C heMiCal a ddition C onversions
One of the most important water/wastewater operator functions is to make various
chemical additions to unit processes. In this section, we demonstrate how to calcu-
late the required amount of chemical (active ingredient and inactive ingredient), dry
chemical feed rate, and liquid chemical feed rate.

Required Amount of Chemical (Active Ingredient)

Chemical (lb/day) = Required dose (mg/L) × Flow (MGD) × 8.34 lb/MG/mg/L (2.11)

■ Example 2.99
Problem : A laboratory jar test indicates a required dose of 4.1 mg/L of ferric chlo-
ride. The flow rate is 5.15 MGD. How many pounds of ferric chloride will be needed
each day?

Solution :

Chemical = 4.1 mg/L × 5.15 MGD × 8.34 lb/MG/mg/L = 176.1 lb/day