A tank is full of oil weighing 30 lb/ft3. The tank is a right rectangular prism with a width of 2 feet, a depth of 2 feet, and a height of 3 feet. Find the work required to pump the water to a height of 2 feet above the top of the tank

the work will be W = 17373.96 lb*ft²/s²

Step-by-step explanation:

Neglecting the frictional losses in the pipe and the pump efficiency, then assuming a reversible process the work W will be

W = ∫pdV

from an energy balance (see note below), and assuming constant density of the oil, the pressures at pump output and pump input will be

P final = p₀ + ρ*g*h = p₀ + ρ*g*h + 1/2ρ*v²

P initial = p₀ + 1/2ρ*v²

where

p₀ = pressure at the surface

ρ = density

g = gravity

h = height of the oil level

the volume of pumped oil will be

V = A*h

dV=A*dh

then the pressure the pump has to work against is:

p = Pfinal - Pinitial = ρ*g*h

therefore

W=∫pdV = A∫ρ*g*h*dh = A*ρ*g*H²/2 = ρ*g*V*H/2

replacing values

V = 2 ft * 2 ft * 3 ft = 12 ft³

W = ρ*g*V*H/2 = 30 lb/ft³ * 32.174 ft/s² * 12 ft³ * 3 ft/2 = 17373.96 lb*ft²/s²

W = 17373.96 lb*ft²/s²

Note

- from mass conservation at the bottom of the tank and pump output

Apipe*vpipe = Apump*vpump

assuming that the discharge pipe has the same pipe diameter then Apump = Apipe → vpipe = vpump=v

- the final work is the average of the work for a completely charged tank and with almost any oil (H=0) (assuming no suction of air/vapor with low oil levels)