Basic Units of Measurement and Conversions - Basic Mathematics for Water and Wastewater Operators - page 21

■ Example 2.25
Problem : A liquid chemical with a specific gravity of 1.22 is pumped at a rate of 40
gpm. How many pounds per day are being delivered by the pump?

Solution : Solve for pounds pumped per minute, then change to pounds/day.

8.34 lb/gal × 1.22 = 10.2 lb/gal

40 gal/min × 10.2 lb/gal = 408 lb/min

408 lb/min × 1440 min/day = 587,520 lb/day

■ Example 2.26
Problem : A cinder block weighs 70 lb in air. When immersed in water, it weighs 40
lb. What are the volume and specific gravity of the cinder block?

Solution: The cinder block displaces 30 lb of water; solve for cubic feet of water
displaced (equivalent to volume of cinder block).

30 lb waterdisplaced
62.4 lb/ft

3

= 0.48 ft

3

The cinder block volume is 0.48 ft 3 , which weighs 70 lb; thus,

70 lb ÷ 0.48 ft 3 = 145.8 lb/ft 3 density of cinder block

3

Densityofcinderblock
Den

145.8 lb/ft
62.4 lb/ft

Specificgravity = = = 234 .

3

sity of water

t eMperature C onversions
Most water/wastewater operators are familiar with the formulas used for Fahrenheit
and Celsius temperature conversions:

• °C = 5/9(°F - 32)

• °F = 9/5(°C) + 32

The difficulty arises when one tries to recall these formulas from memory.
Probably the easiest way to recall these important formulas is to remember these
basic steps for both Fahrenheit and Celsius conversions:

1. Add 40°.

2. Multiply by the appropriate fraction (5/9 or 9/5).

3. Subtract 40°.