Answer:
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)
