Basic Math Operations - Mathematics and Statistics for the Environment - page 72

■ EXAMPLE 2.62
Problem: Find the pressure (psi) in a tank 12 ft deep at a point 5 ft below the water surface.

Solution:

Pressure (psi) = 0.433 × 5 ft = 2.17 psi

■ EXAMPLE 2.63
Problem: A pressure gauge at the bottom of a tank reads 12.2 psi. How deep is the water in the tank?

Solution:

Head (ft) = 2.31 × 12.2 psi = 28.2 ft

■ EXAMPLE 2.64
Problem: What is the pressure (static pressure) 4 miles beneath the ocean surface?

Solution: Change miles to feet, then to psi:

5280 ft/mile × 4 = 21,120 ft

21,120 ft
2.31 ft/psi

= 9143 psi

■ EXAMPLE 2.65
Problem: A 150-ft-diameter cylindrical tank contains 2.0 MG water. What is the water depth? At
what pressure would a gauge at the bottom read in psi?

Solution:

1. Change MG to cubic feet:

2 000 000
748

, ,

.

gal

ft 3 = 267 380 ,

2. Using volume, solve for depth:

Volume = 0.785 × D 2 × depth

267,380 f ft3 3 = 0.785 × (150) 2 × depth

Depth = 15.1 ft

■ EXAMPLE 2.66
Problem: The pressure in a pipe is 70 psi. What is the pressure in feet of water? What is the pres-
sure in psf?

Solution:

1. Convert pressure to feet of water:

70 psi × 2.31 ft/psi = 161.7 ft of water