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Math Help - Work Problem: Hemispherical Tank

  1. #1
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    Work Problem: Hemispherical Tank

    A tank is a hemisphere of radius 12 feet with flat base on the ground and is filled to a height of 9 feet with liquid nitrogen which has a density of 50.45lb/ft^3. How much work is needed to pump all of the hydrogen to the top of the tank?

    MY WORK:

    Work = volume*density*distance

    The equation of the circle is x^2 +y^2 = r^2
    Given radius = 12, therefore x^2+y^2=144; x^2=144-y^2

    Force = volume * density

    Density of nitrogen is 50.45 lbs/ft^3.
    Volume = pi*x^2 dy;

    Force = [pi*x^2 dy]*[50.45]
    Force = 50.45pi[x^2 dy]

    Distance = (-y) because all y-coordinates below origin is negative.


    So,
    Work = [(50.45pi)(x^2)dy]*(-y)

    Plugging in 144-y^2 for x^2 by equation of circle:

    Work = [(50.45pi)(144 - y^2)dy]*(-y)
    Work = (50.45pi)[y^3 - 144y]dy

    Nitrogen = 9 feet deep, therefore integrate by dy from y = -12 to y = -3.


    Work = (50.45pi)INT.(-12, -3)[y^3 -144y]dy
    Work = (50.45pi)[y^4/4 - 72y^2]|(-12, -3)
    Work = (50.45pi)[{(1/4)(-3)^4 - 72(-3)^2} - {(1/4)(-12)^4 - 72(-12)^2}]
    Work = (50.45pi)[{81/4 - 648} - {5184 - 10368}]
    Work = (50.45pi)[-2511/4 + 5184]
    Work = (50.45pi)[4556.25]
    Work = APPROX 722135.3231 ft-lbs.
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  2. #2
    Member Greengoblin's Avatar
    Joined
    Feb 2008
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    UK
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    I don't think your formulas are correct for force and work:

    W=Fs

    Where:
    W = work
    F = force
    s = distance force is applied over

    F=ma

    Where:
    m = mass
    a = acceleration

    m=\rho V

    Where:

    \rho = density
    V = Volume

    Putting this together:

    W=\rho Vas

    Where the acceleration will be 9.8ms^{-2} needed to overcome gravity.
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