Question

A pump is filling a cylindrical water tank 25 feet in diameter and 25 feet high. The pump flow rate is 300 GPM, after 2 hours of operation, the pressure exerted from the water on the bottom of the tank is _____ psi. (Assume that the tank was empty prior to the pumping process)

Answer 1

given

H = 25 feet = 25*0.3048 = 7.62 m

d = 25 feet = 25*0.3048 = 7.62 m

r = d/2 = 7.62/2 = 3.81 m

volume flow rate = 300 gallon/minute

= 300*3.785*10^-3 m^3/minute

= 1.1355 m^3/minute

volume of water in the tank after 2 hours, V = volume flowrate*time

= 1.1355*2*60

= 136.26 m^3

let h is the of water column in the tank.

use, V = pi*r^2*h

==> h = V/(pi*r^2)

= 136.26/(pi*3.81^2)

= 2.99 m

the pressure exerted from the water on the bottom of the tank,

P = Po + rho*g*h

= 1.013*10^5 + 1000*9.8*2.99

= 1.306*10^5 pa

= 1.306*10^5/6894.76

= 18.9 psi <<<<<<<<<----------------Answer