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Time to Drain An Atmospheric Tank

  1. Dec 20, 2007 #1
    I am a little rusty on my differential equations as I haven't seen them in a few years!! I am looking for some direction on a question on a first order ODE. I am trying to find the time required to drain a tank that is open to atmosphere. Any help would be appreciated.

    Thanks!
     
  2. jcsd
  3. Dec 20, 2007 #2

    Astronuc

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    It depends on the volume (mass) and volumetric (mass) flow rate.

    Open to the atmosphere simply means that the surface of the liquid or the gas is at 1 atm of pressure. For liquids, the pressure will increase with depth, which could affect mass flow rate. The drain opening will also influence flow rate.
     
  4. Dec 20, 2007 #3
    I understand that the head pressure is going to decrease as the water level goes down which will affect the velocity of water leaving the tank....I am trying to develop an equation to get the time required to drain the tank.
     
  5. Dec 20, 2007 #4

    FredGarvin

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    This is directly from a PE exam review book:

    The general ODE covering the set up is:
    [tex]-A\frac{dH}{dt}=A_oC_o\sqrt{(2g \Delta H)}[/tex]

    Integrating gives:

    [tex]t=\frac{A\sqrt{\frac{2}{g}}}{A_o C_o}\left[H_i^{.5} - H_f^{.5}\right][/tex]

    Where:
    [tex]A[/tex]= the cross sectional area of the tank
    [tex]A_o[/tex]= area of the flow orifice
    [tex]g[/tex]= acceleration due to gravity
    [tex]C_o[/tex]= orifice discharge coefficient (usually about 0.6)
    [tex]H_i[/tex]= Initial tank head in ft.
    [tex]H_f[/tex]= Final head

    Give that a go and see how that does. Granted, if you have a very viscous fluid or the viscosity changes over time all bets are off since the Reynolds number is a function of that which then effects your discharge coefficient.
     
    Last edited: Dec 20, 2007
  6. Dec 20, 2007 #5
    Thanks. I assume this was derived from using the Bernoulli equation and an unsteady state material balance? I didn't get the same term on the right hand side....do you have the derivation as well?
     
  7. Dec 20, 2007 #6

    Astronuc

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    What is the geometry of the tank? Spherical? Cylindrical? If cylindrical, Vertical or horizontal?
     
  8. Dec 20, 2007 #7
    It is a rectangular concrete tank
     
  9. Dec 20, 2007 #8

    FredGarvin

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    It is definitely a product of the Bernoulli equation. It actually looks more like a plain square edge orifice flow equation.
     
  10. Jun 11, 2008 #9
    What would the formula be taking into account frictional losses due to pipe length and fittings etc. for a piped outlet (open to atmos)?
     
  11. Jun 12, 2008 #10

    FredGarvin

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    You can include those losses into the final head (which is what that variable is there for). If you notice, the flow is going to be proportional to the delta head number. If you increase the outlet head, the flow decreases.
     
  12. Jun 22, 2008 #11
    Fred,

    Since the flow is constantly changing as it's draining the frictional losses will also change. How do you determine what frictional losses to use in the final head?
     
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