Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 16

Problem: How many gallons of water are required to fill a tank that holds 7540 lb
of water?

Solution:

Pounds
8.34 lb/gal

7540 lb
8.34 lb/gal

= = 904 ga l

■ Example 2.10
Convert milligrams per liter to pounds.
Note: For plant operations, concentrations in milligrams per liter or parts per mil-
lion determined by laboratory testing must be converted to quantities of pounds,
kilograms, pounds per day, or kilograms per day.

Pounds = Concentration (mg/L) × Volume (MG) × 8.34 lb/MG/mg/L

Problem: The solids concentration in the aeration tank is 2580 mg/L. The aeration
tank volume is 0.95 MG. How many pounds of solids are in the tank?

Solution:

2580 mg/L × 0.95 MG × 8.34 lb/MG/mg/L = 20,441.3 lb

■ Example 2.11
Convert milligrams per liter to pounds per day.

Pounds/day = Concentration (mg/L) × Flow (MGD) × 8.34 lb/MG/mg/L

Problem: How many pounds of solids are discharged per day when the plant effluent
flow rate is 4.75 MGD and the effluent solids concentration is 26 mg/L?

Solution:

26 mg/L × 4.75 MGD × 8.34 lb/mg/L/MG = 1030 lb/day

■ Example 2.12
Convert milligrams per liter to kilograms per day.

Kilograms/day = Concentration (mg/L) × Volume (MG) × 3.785 L/gal

Problem: The effluent contains 26 mg/L of BOD 5 . How many kilograms per day of
BOD 5 are discharged when the effluent flow rate is 9.5 MGD?
Solution:

26 mg/L × 9.5 MG × 3.785 L/gal = 934 kg/day