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
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)
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
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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