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Height of fluid in cylinder

  1. Apr 22, 2014 #1
    Thanks in advance!

    I have problem that is causing me a headache to no end. In a cylinder, if one knows the total volume of cylinder and total volume of fluid in said cylinder, how would the height of this fluid be computed? I have no problems calculating volumes based on height, but the other way around and aargghhhh
     
  2. jcsd
  3. Apr 22, 2014 #2
    You need to know the specific shape of the cylinder, not just the total volume of it.
    Does the problem provide you with the horizontal radius of the cylinder, or maybe the height of the cylinder?
     
  4. Apr 22, 2014 #3
    The radius of the cylinder is 46 inches. The height of the fluid is below the total height if the cylinder. So, volume of cylinder would be 1000 ft^3, radius of cylinder is 46 inches, and volume of the fluid is 300 ft^3. What is the height of the fluid in the cylinder
     
  5. Apr 22, 2014 #4

    SteamKing

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    Do you know the formula for calculating the volume of a cylinder, given the radius and the height?
     
  6. Apr 22, 2014 #5

    Mark44

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    It also makes a difference how the cylinder is oriented. If the cylinder is vertical, the volume is proportional to the height. If the cylinder is lying horizontally, it's harder to determine the height of liquid from the volume that's present in the tank.
     
  7. Apr 23, 2014 #6
    Indeed, the answer comes easily if the cylinder is vertically oriented. Then pi*r^2*h is the quick solution. But, if the cylinder is on its side.... This is the problem I am trying to solve. I have no problem determining the volume of fluid in a partially filled tank when height of fluid is given and radius of cylinder is supplied. I am trying to work backwards, and am having no luck :(
     
  8. Apr 23, 2014 #7

    SteamKing

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    It seems like you have all the right information; what you are lacking is the geometry to make this information useful to you:

    http://en.wikipedia.org/wiki/Circular_segment

    You have a cylinder with diameter d = 46" and a total capacity of 1000 ft^3.

    If this cylinder is oriented so that its axis is horizontal, then the length of the cylinder will be:

    V = π*r[itex]^{2}[/itex]L

    1000 = π (23/12)[itex]^{2}[/itex]L

    L = 86.65 ft approx.

    Now, the cylinder is filled to 300 ft^3, what is the depth of the contents?

    In this case, we want to find the cross sectional area A so that A*L = 300 ft^3

    A = 300 / 86.65 = 3.46 ft^2 approx.

    We know from the article on circular segments that:

    A = (r[itex]^{2}[/itex]/2)*(α - sin(α)), where α is the central angle in radians

    So:

    3.46 = [(23/12)[itex]^{2}[/itex]/2]*(α - sin(α))

    (α - sin(α)) = 1.885 approx.

    This equation can be solved by iteration:
    Code (Text):

       α        α - sin(α)
      1.000     0.1585
      2.000     1.0907
      2.500     1.9015
      2.490     1.8835
      2.491     1.8853
     
    so α = 2.491 radians approx., which means the central angle Θ = 142.72 deg.

    From the article:

    h = r*(1 - cos(Θ/2))

    so

    h = (23)*(1 - cos(2.491/2)) = 15.65 inches

    For a capacity of 300 ft^3, the depth of the contents is about 15.65 inches.

    Calculations like this can be programmed into a spreadsheet, so that you can make a table of capacities for the cylinder at different depths. You can either measure the depth of fluid directly, or measure to the surface of the fluid, which distance is called the ullage.
     
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