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H2O and CO2 Pressure Imbalance

  1. Apr 8, 2013 #1
    H2O and CO2 must mix in an open line (not closed pressurized vessel) via stainless steel membrane; the dilemma is:

    H2O incoming line pressure is approximately 50psi and CO2 line pressure is approximately 100psi. The problem arises when both elements collide at the point of the membrane where some pressure is created and at which time the pressure from the CO2 attempts to back pressure the H2O back down the line causing a significant loss in H2O volume.

    I believe that the pressure of the CO2 should remain high in an attempt to force a carbonation mixture with the H2O in an open environment.

    These are the two concepts that I’ve devised:

    a) Increase the pressure of the H2O so that it will remain consistent in any environment, and if so what is the best solution for do so?

    b) Intermittently inject the CO2 into the flow of H2O, and if so what is the best solution for doing so?

    Thanks in advance for any help with this problem.
  2. jcsd
  3. Apr 8, 2013 #2


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    The pressures have to match, so you need a regulator on the higher pressure line.
  4. Apr 8, 2013 #3


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    Or a pump on the water supply.
  5. Apr 8, 2013 #4

    Without using a pump per say, would a restrictive devise such as a Venturi act as a balancer between the pressures?
  6. Apr 8, 2013 #5


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    Maybe you're thinking of something like an eductor? They're also called injectors, jet pumps and other things. In your case, the motive fluid would need to be the CO2 but I'm assuming that's a gas in your system, in which case it won't be able to pull much liquid into the flow stream. If you only need a relatively small amount of water to CO2, you could contact a supplier and see if they could be made to work for your system.
  7. Apr 9, 2013 #6
    Thanks for the input! The concept looks appealing, not sure if the gas side of CO2 will indeed pull the correct volume of water needed to achieve the desired results, but I am certainly going to work with this idea until I prove that it will not work. Thanks again.
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