Basic Math Operations - Mathematics and Statistics for the Environment - page 66

To determine the volume of a cone and sphere, we use the following equations and examples.

Volume of cone

π
12

Volume of cone = × Diameter × Diameter × Height (2.10)

Note that

π
12

314
12

.

= = 0 262 .

Key Point: The diameter used in the formula is the diameter of the base of the cone.

■ EXAMPLE 2.51
Problem: The bottom section of a circular settling tank has the shape of a cone. How many cubic
feet of water are contained in this section of the tank if the tank has a diameter of 120 ft and the
cone portion of the unit has a depth of 6 ft?

Solution:

Volume (ft 3 ) = 0.262 × 120 ft × 120 ft × 6 ft = 22,637 ft 3

Volume of sphere

π
6

Volume of sphere = × Diameter × Diameter × Diame ter (2.11)

Note that

π
6

314
6

.

= = 0 524 .

■ EXAMPLE 2.52
Problem: What is the volume (ft 3 ) of a spherical gas storage container with a diameter of 60 ft?

Solution:

Volume (ft 3 ) = 0.524 × 60 ft × 60 ft × 60 ft = 113,184 ft 3

Circular process and various water and chemical storage tanks are commonly found in water/
wastewater treatment. A circular tank consists of a circular floor surface with a cylinder rising
above it (see Figure 2.9). The volume of a circular tank is calculated by multiplying the surface area
times the height of the tank walls.

■ EXAMPLE 2.53
Problem: If a tank is 20 feet in diameter and 25 feet deep, how many gallons of water will it hold?

Hint: In this type of problem, calculate the surface area first, multiply by the height, and then con-
vert to gallons.

Solution:

r = D ÷ 2 = 20 ft ÷ 2 = 10 ft

A = π × r 2 = π × 10 ft × 10 ft = 314 ft 2

V = A × H = 314 ft 2 × 25 ft = 7850 ft 3 × 7.48 gal l /f ft 3 = 58,718 gal