Environmental Engineering Reference
In-Depth Information
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Example 2.130
Problem
: A liquid chemical with a specific gravity of 1.22 is being pumped at a rate
of 50 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 water × 1.22 = 10.2 lb/gal liquid
50 gal/min × 10.2 lb/gal = 510 lb/min
510 lb/min × 1440 min/day = 734,400 lb/day
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Example 2.131
Problem
: A cinder block weighs 80 lb in air. When it is 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 water displaced ÷ 62.4 lb/ft
3
= 0.48 ft
3
water displaced
Cinder block volume is 0.48 ft
3
, which weighs 80 lb; thus,
80 lb ÷ 0.48 ft
3
= 166.7 lb/ft
3
3
Densityofcinderblock
Den
166.7 lb/ft
62.4 lb/ft
Specificgravity
=
=
= 267
.
3
sity of water
t
eMperature
C
onversion
e
xaMples
The two common methods for making temperature conversions are
•
°C = 5/9(°F - 32)
•
°F = 9/5(°C) + 32
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Example 2.132
Problem
: At a temperature of 4°C, water is at its greatest density. What is that tem-
perature in degrees Fahrenheit?
Solution:
(9/5 × °C) + 32 = (9/5 × 4) + 32 = 7.2 + 32 = 39.2°F
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:
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