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A tank is full of water. Find the work required to pump the water out of the spout. (Use 9.8 m/s^2 for g. Use 1000 kg/m^3 as the density of water. Assume r = 9 m and h = 3 m.)

The tank is a spherical shape with r as the radius(9m). On top of the tank is a spigot with h as the height (3m).

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2 answers
  1. How high is the top of the spout from the center of gravity of the tank ???
    radius + 3 = 9 + 3 = 12 meters

    So you have to lift the weight of the water 12 meters.
    mass of water in tank = m = (4/3) pi r^3 * 1000 Kg
    weight = w = m g = 9.8 * m
    work = weight * distance lifted = (4/3) pi 9^3 * 1000 Kg * 9.8 * 12 Joules

    Note, You will need more because the water has velocity when it exits but we do not have data for that
    (1/2) m v^2 :)

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  2. For now, assuming that the weight of water is 1 kg/m^2 (so we just have to worry about figuring the volume * distance)

    moving the center of mass to the height of the top of the sphere (forget the spout -- assume the water just runs out the top) requires

    4/3 π * 9^3 * 9 = 8748π J

    Doing the calculus, and figuring that the cross-section area of a slice of water at a distance of y units above the bottom of the sphere is π√(81-(9-y)^2) (this is correct, because even though 9-y goes from 9..-9, it is squared)

    ∫[0..18] πr^2h dy = ∫[0..18] π(81-(9-y)^2)(18-y) dy = 8748π J

    Now just multiply that by the correct density (N/m^3) and increase the height by the length of the spout, and you should get Damon's answer.

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