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Can anyone figure our how to predict the volume in a tank at any time?

  1. Aug 14, 2011 #1
    I have a tank (actually a PVC tube with an end cap) that has an interior diameter of 86 mm. It stands on its circular face. Coming out near the bottom of the tank at a right angle is a metal pipe 11 mm in diameter (interior) and 312 mm long (in total that is, note that some of the pipe is protruding within the tank and not just out of the tank to stabilise it). Here is a picture similar to what I did on page 3 http://seniorphysics.com/physics/ejp_sianoudis.pdf (I’m not doing the same experiment).

    I fill the tank with water until enough water is in the tank so that it starts to flow out of the pipe (there is some space beneath the exit pipe's hole). I do this so then I know if I add another litre of water now, 1 litre should come out (no space inside the tank beneath the pipe hole now). My experiment is aimed at investigating the effect of viscosity on the time the tank takes to empty (not really empty because when the water level is beneath the top of the pipe the theory does not stand, so about there). I need a formula that can predict the volume of fluid in the tank at any time and takes into account viscosity and the pipe length. I created one based off the Hagen–Poiseuille equation, but it did not match up with the data i collected and it also suggested that a viscosity that was 100 greater would take 100 times as long to flow, which is not the case.

    If anyone knows anything that could help me, such as if this is even predictable and why/why it isn't, I would appreciate it.
  2. jcsd
  3. Aug 14, 2011 #2
    Start with the treatment in (probably close to what you tried):


    And modify this treatment to include your other effects as follows:

    DP = rho * g * h = KT * rho * v^2/2 + f (L/D) * rho * v^2/2

    K = sum of all resistance coefficients (including opening)
    L = length of pipe
    and the rest are the usual meening.

    Note that you can capture your viscosity effects with correlations for f that depend on Reynold's Number.

    You can adjust your K for loss factors for friction as follows: K = f * Leq / D.

    So solve this above equation for Q as a function h.

    Then, as was done in the link: Integrate Q(h) = -A dh/dt.
  4. Aug 15, 2011 #3
    First of all I cant see where the equation at http://www.lmnoeng.com/Tank/TankTime.htm
    takes pressure into account so I dont know how to "modify this treatment to include the other effects as follows:

    DP = rho * g * h = KT * rho * v^2/2 + f (L/D) * rho * v^2/2"
    Or where this came from or what T is.

    Secondly i dont know how to calculate The Darcy friction factor for turbulent flow (as is the case here initially) for K = f * Leq / D. Also what does Leq represents.

    I'm afraid I have no idea about most of what you say. It might help to tell you that before now I have had no experience with fluid mechanics, but i can integrate functions like where the flow rate is proportional to the volume of water etc. Really what would help the most if you could give some kind of example of doing what you are talking about. Either way thanks for helping.
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