Force, Pressure, Head, and Velocity Calculations - Basic Mathematics for Water and Wastewater Operators - page 182

■ Example 12.13
Problem : Between the top of a reservoir and the watering point, the elevation is 125
ft. What will the static pressure be at the watering point?

Solution:

0.433 psi
1ft

125 ft × = 289 ft

■ Example 12.14
Problem : Find the pressure in psi in a tank 12 ft deep at a point 5 ft below the water
surface.

Solution:

Pressure = 0.433 × 5 ft = 2.17 psi

■ Example 12.15
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 ft/psi × 12.2 psi = 28.2 ft

■ Example 12.16
Problem : What is the static pressure 4 miles beneath the ocean surface?

Solution : Change miles to feet and then to pounds per square inch.

5280 ft/mile × 4 miles = 21,120 ft

21,120 ft ÷ 2.31 ft/psi = 9143 psi

■ Example 12.17
Problem: A 150-ft diameter cylindrical tank contains 2 MG of water. What is the
water depth?

Solution:

1. Change MG to cubic feet.

(2,000,000 gal)/(7.48 gal/ft 3 ) = 267,380 ft 3

2. Using volume, solve for depth.

Volume = 0.785 × (Diameter) 2 × Depth

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

Depth = 15.1 ft