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**

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