Circumference, Area, and Volume - Basic Mathematics for Water and Wastewater Operators - page 171

■ Example 11.25
Problem : Calculate the surface area of a basin that is 90 ft long, 20 ft wide, and 12
ft deep.
Solution:
Area = Length × Width = 90 ft × 20 ft = 1800 ft 2
■ Example 11.26
Problem : Calculate the cross-sectional area in square feet for a 3-ft-diameter main
that has just been laid.
Solution:
Area = 0.785 × D 2 = 0.785 × (3 ft) 2 = 7.1 ft 2
■ Example 11.27
Problem : Calculate the cross-sectional area in square feet for a 36-in.-diameter main
that has just been laid.
Solution:
36 in. = 3 ft
A = 0.785 × D 2 = 0.785 × (3 ft) 2 = 7.1 ft 2
■ Example 11.28
Problem : Calculate the area in square feet for a 48-in. interceptor line that has just
been laid.
Solution:
48 in. = 4 ft
A = 0.785 × D 2 = 0.785 × (4 ft) 2 = 12.6 ft 2
■ Example 11.29
Problem : Calculate the cross-sectional area in square feet for a 3-in. pipe.
Solution:
3 in. = 0.25 ft
A = 0.785 × D 2 = 0.785 × (0.25 ft) 2 = 0.049 ft 2