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23 January, 17:20

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

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  1. 23 January, 17:27
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    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)
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