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Flow Rate of a Deflation Balloon

  1. Sep 6, 2009 #1
    A drug delivery infusion pump is made out of thick synthetic rubber and has a very small outlet. The balloon pump is filled with 500mL liquid drug and is now inflated into a ball shape. The drug will exit from the very small outlet at a nearly (but not always) constant flow rate of 5mL/hr (takes 100hr to deplete).

    How can I derive the flow rate as a function of time?
    I'd also like to know how the temperature and drug viscosity affect the flow rate.

    Someone suggested me to use Poiseuille-Hagen equation, but I need to know more details, especially how I can relate the flow rate to time and establish a model.

    Your insight will be highly appreciated!
     
  2. jcsd
  3. Sep 6, 2009 #2
    Let me know if i'm right: You have a liquid-filled ballon (spherical i guess) which has a little outlet through which liquid is verted outwards. So despite you give a value of a certain flow rate you'd like to know how it evolves with time?

    Thanks for clarifying :)
     
  4. Sep 6, 2009 #3
    That's exactly what I meant!
    The "constant flow rate" is only indicative (by product manufacturer). The actual flow rate seems decreasing as baloon deflating (from experimental data).
    Looking forward to your solution!
     
    Last edited: Sep 6, 2009
  5. Sep 6, 2009 #4
    Oh! thanks for your trust :) ya know its funny because I did something really similar to this yesterday helping another post.

    But i'm afraid I need more data because this is a very complex situation and if we want to obtain a practical solution (so that we can even calculate our own time of deplete, for example) we´ll have to do some simplifications in order to modelize the baloon.

    May you describe it to me a little more? thanks a lot
     
  6. Sep 6, 2009 #5
    can you tell me where that post (the similar question you answered yesterday) is located?

    sure we can simplify the model to start with, such as a perfect sphrical balloon shape, or even cylinder shape if that makes it easier, water as fluid, constant temperature, etc. I don't know what other information you need.
     
  7. Sep 7, 2009 #6
    Yes sure, the post is "Need formula/help with mass air flow" at Physics > Classical Physics, it was similar but not the same because there we dealed with a fixed-volume air tank. However some eqs. may help.

    The information I need is to modelize the baloon. Whether it decreases in volume as a sphere (or nearly does) or just wrinkle when water is released, or the description of the outlet in size, shape... are data that will allow us to make a good "practical" model of the pump.

    Thanks a lot.
     
  8. Sep 7, 2009 #7
    We can assume either
    (1) the balloon remains as a sphere when the volume decreases, or
    (2) the balloon keeps one diameter (along the axis of outlet) constant, while the radial dimension decreases.

    whichever is easier.

    The balloon is connected to a tubing, which has a very small outlet. The whole setup is more like the IV infusion bag, but instead of using a medication bag and the gravitation, we have an elastomeric baloon here.
     

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  9. Sep 7, 2009 #8
    Well, I see. The fact is that this is quite complicated because of the elastic nature of the baloon. Picking one of the options, it'd be far easier (1) rather than changing the very geometry of the baloon each second. I´m afraid I can't make a practical model of this if I consider it "realisticly" elastic.

    If i come up with an answer for anythig different I tell you.

    I'm sorry, :(
     
  10. Sep 8, 2009 #9

    Andy Resnick

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    It's not just a small hole (at least the devices I have seen)- there is a 'frit' inside that produces a constant resistance. The elastic energy of the balloon produces the driving force, pushing fluid through the porous frit.

    The Hagen-Poiseuille equation will not readily apply, due to the presence of the resistive element. A better approach would be Darcy's law (Or Poiseuille's law with a resistive term, sometimes called the Brinkmann equation).
     
  11. Sep 8, 2009 #10

    Andy Resnick

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    That's because the elastic energy of the balloon slowly dissipates as the balloon deflates- think of it as the driving force decreases over time.
     
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