Environmental Engineering Reference
In-Depth Information
■
Example 11.21
Problem
: A sedimentation basin is 60 ft long by 40 ft wide and 12 ft deep. What is
the volume of the tank in gallons?
Solution:
V
=
A
×
H
= 60 ft × 40 ft × 12 ft = 28,800 ft
3
28,800 ft
3
× 7.48 gal l /f ft3
3
= 215,425 gal
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Example 11.22
Problem:
A circular tank has a diameter of 40 ft and is 12 ft deep. How many gallons
will it hold?
Solution:
V
=
A
×
H
= (40 ft)
2
× 0.785 × 12 ft = 15,072 ft
3
15,072 ft
3
× 7.48 gal l /f ft3
3
= 112,739 gal
v
oluMe
of
p
ipes
The number of gallons contained in a 1-ft section of pipe can be determined by
squaring the diameter (in inches) and then multiplying by 0.048. To determine the
number of gallons in a particular length of pipe, multiply the gallons per foot by the
number of feet of pipe.
Volume in gallons =
D
2
(i (in.) × 0.0408 × Length (ft)
(11.10)
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Example 11.23
Problem:
A 12-in. line is 1200 ft long. How many gallons does the pipe hold?
Solution:
V
= (12 in.)
2
× 0.0408 × 1200 ft = 7050 gal
AREA EXAMPLES
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Example 11.24
Problem
: A basin has a length of 45 ft and a width of 14 ft. Calculate the area in
square feet.
Solution:
Area = Length × Width = 45 ft × 14 ft = 630 ft
2
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