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Math Help - Emptying a water tank (Inverted circular cone)

  1. #1
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    Emptying a water tank (Inverted circular cone)

    ok, i drew out a diagram and I have no idea how or where to start this problem. Please help guide me so I can learn how to do it...

    A water tank has the shape of an inverted circular cone with a diameter of 4 feet and a height of 6 feet. If the tank is full, find the work required to empty the tank through a pipe that extends 3 feet above the top of the tank. Use 62.5 as the weight-density of water. Give your answer in foot-pounds.

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    just found a really helpful sample problem online. thanks anyway.

    might be posting another problem here in a few
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  3. #3
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    Quote Originally Posted by Grimey View Post
    ok, i drew out a diagram and I have no idea how or where to start this problem. Please help guide me so I can learn how to do it...

    A water tank has the shape of an inverted circular cone with a diameter of 4 feet and a height of 6 feet. If the tank is full, find the work required to empty the tank through a pipe that extends 3 feet above the top of the tank. Use 62.5 as the weight-density of water. Give your answer in foot-pounds.
    work = integral of WALT

    W = weight-density in lbs per cubic foot

    A = cross-sectional area of a representative horizontal "slice" of liquid in terms of y

    L = lift distance of the representative "slice" in terms of y

    T = "slice" thickness , dy.

    let the vertex of the cone be at the origin ... one side of the cone is modeled by the linear equation ... y = \frac{6}{2} x = 3x

    "slice" radius = x = \frac{y}{3} , so the cross-sectional area of a representative "slice" is , A = \pi \left(\frac{y}{3}\right)^2

    lift distance, L = 9-y

    the liquid resides between y = 0 and y = 6 ... your limits of integration.

    set up the integral and find the work required.
    Last edited by skeeter; September 20th 2009 at 12:57 PM. Reason: fixed lift distance expression
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    thanks, ill be sure to remember WALT.

    This is how I solved it, not 100% if its right but it seems it might be...

    Weight of slab = density * volume
    weight of slab = 500 pi dx

    Distance = x + 3

    Work = (x + 3)500pi dx

    W = integral from 0 to 6 [(x+3)(500pi)]dx

    W = 56,548.7 ft lbs
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  5. #5
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    Quote Originally Posted by Grimey View Post
    thanks, ill be sure to remember WALT.

    This is how I solved it, not 100% if its right but it seems it might be...

    Weight of slab = density * volume
    weight of slab = 500 pi dx

    Distance = x + 3

    Work = (x + 3)500pi dx

    W = integral from 0 to 6 [(x+3)(500pi)]dx

    W = 56,548.7 ft lbs
    this would not be correct for the described cone.

    62.5 \pi \int_0^6 \left(\frac{y}{3}\right)^2 (9-y) \, dt = 2250\pi ft-lbs
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  6. #6
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    Quote Originally Posted by skeeter View Post
    this would not be correct for the described cone.

    62.5 \pi \int_0^6 \left(\frac{y}{3}\right)^2 (9-y) \, dt = 2250\pi ft-lbs

    Thanks.

    So with finding A, the cross section area...
    If its a disc, it will always be A = pi r^2?
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