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9 September, 05:05

A tank is in the shape of an inverted circular cone with height 500 ft and base radius 20 ft. It is filled with water to a height of 400 ft. Find the work required to empty the tank by pumping all of the water to the top of the tank. (Weight density of water rhog = 62.4 lb/ft3.)

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  1. 9 September, 05:17
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    1.338 * 10 ^ 9 lb-ft

    Step-by-step explanation:

    We have the following information:

    Water density 62.4 lb / ft ^ 3

    work equals force per distance, like this:

    W = F * d

    with the information in the statement we can deduce:

    20/500 = r / y

    r = y / 25

    now the differential volume would be:

    dv = pi * (r ^ 2) * dy

    dv = pi * ((y / 25) ^ 2) * dy

    dv = pi * (y ^ 2/625) * dy

    weight of slice would be:

    62.4 * pi * (y ^ 2/625) * dy

    = 0.3136 * y ^ 2 * dy

    work for the removal of slice:

    (500 - y) * 0.3136 * y ^ 2 * dy

    Now the total work is the integral of the previous expression that goes from y = 0 to y = 400

    Total work = integral (from 0 to 400) { (500 - y) * 0.3136 * y ^ 2 * dy}

    Total work = integral (from 0 to 400) {0.3136 * (500 * y ^ 2 - y ^ 3) dy}

    We integrate and we have:

    Total work = 0.3136 * ((500/3) * y ^ 3 - (y / 4) ^ 4)

    Total work = 0.3136 * ((500/3) * 400 ^ 3 - (400/4) ^ 4) - 0.3136 * ((500/3) * 0 ^ 3 - (0/4) ^ 4)

    Total work = 0.3136 * ((500/3) * 400 ^ 3 - (400 ^ 4) / 4)

    Total work = 1.338 * 10 ^ 9 lb-ft
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