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How to predict the volume of a tank of water at any time? (fluid motion)

  1. Jul 29, 2011 #1
    We have a cylindrical tank of known dimensions. Attached at the bottom is a pipe perpendicular to the tank of known dimensions. We fill the tank with a liquid and let it drain out through the pipe. What formula will predict the volume of the tank at any time t? I believe the Hagen–Poiseuille equation (involves the viscosity of the liquid) and Pascal's law (involves the density of the liquid) are involved, but i cannot derive an equation for the volume at any time. All working and any explanation where ever necessary would be appreciated.

    Heres my attempt at it:
    ΔP = 8μLQ / πr^4

    ΔP is the pressure drop
    L is the length of pipe
    μ is the dynamic viscosity
    Q is the volumetric flow rate
    r is the radius

    ΔP = ρgΔh

    g is acceleration due to gravity
    ρ is the fluid density
    Δh is the height of fluid

    ΔP = 8μLQ / πr^4
    ΔP = ρgΔh
    Volume of tank = πr^2h
    8μLQ / πr^4 = ρgΔh
    Q = ρgΔh ✕ πr^4 / 8μL

    This is essentially equal to Q = k Δh where k is = ρg ✕ πr^4 / 8μL
    Im fairly sure is is redundant however as of course the rate of flow is proportional to the change in height of water as when the Δh is multiplied by the area of the circular face of the tank it will give the volume of the cylinder of water which has left through the pipe during the Δh. There has to be an equation for the relationship im looking for as the results of the test will be consistent.
    Last edited: Jul 29, 2011
  2. jcsd
  3. Jul 29, 2011 #2
    Is you tank horizontal or vertical? This will govern the method to apply.
  4. Jul 29, 2011 #3
    Its a vertical standing tank, on its circular face.
  5. Jul 30, 2011 #4
  6. Jul 30, 2011 #5
    Thanks edgepflow for the link. However that websites seems to be focused more on a Torricelli's law situation, which was created for holes. It does appear to take into account an oriface discharge constant but only relates it to the type of hole. As a large part of my experiment will be testing different viscosity and lengths of the pipe, a formula (such as the Hagen–Poiseuille equation) which takes these both into account is something that should be part of the answer.
  7. Jul 31, 2011 #6
    I think you can 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.

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

    Then, as was done in the link: Integrate Q(h) = -A dh/dt.
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