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Compressed gas flow rate calculation

  1. Jul 31, 2014 #1
    Folks, I'm new to this thread. I'm a mechanical engineer, but I haven't done much fluids work since I graduated, 20 years ago. I have a question about calculating gas flow rate, so any help would be greatly appreciated. I have a design that I'm working on that will use a compressed gas cylinder to push a plunger that pushes a fluid on the other side of the plunger out of a critical orifice. I need a very low constant pressure (about 1 psi), the cylinder manufacturers are telling me that I need a flow rate. I've told them that the gas will fill a chamber with a piston and the total volume of that chamber is 3 mL. This filling will occur over a period of 16 hours, so that flow rate is 0.188 mL/hr if it were a liquid. I'm not sure how if that's the flow rate they are looking for (and neither do they) since it's a compressed gas. Any ideas on a governing equation(s) that I can use.
     
  2. jcsd
  3. Jul 31, 2014 #2
    Welcome to Physics Forums! I'm not exactly sure what you're describing. However, since you're working with a gas, you should use the equation of state. If you are using a ideal gas such as air, nitrogen, oxygen, or a noble gas you should use the ideal gas law (PV=nRT). You may have to use one of the many derivations that arise from the ideal gas law.

    You can assume: Isothermal flow so T is constant. R is an ideal gas constant (for your specific gas), it will not change. What you might use is Boyle's law: P1V1=P2V2. Assuming the tank is initially at atmospheric pressure and your chamber's volume remains constant, I calculated (using Boyle's law) that you need to add 0.204 mL of your air to pressurize it to 1 psi gage. (assuming atmospheric pressure is 101325 Pa) If you don't care about the rate at which it is pressurized you can just use .01275 mL/hr flow rate.

    Equation I used: Patm*Vneeded=(Pgiven+Patm)*Vgiven
    Where Vneeded=Vgoing in+Vgiven
     
  4. Jul 31, 2014 #3
    Thank you. So, I think that's what I used to calculate the volume. So is the flow rate that volume divided by the time?
     
  5. Jul 31, 2014 #4
    It depends if you need a specific pressure-time curve or not. If you don't care about the pressure-time curve you can just divide by time. If you want a steady (constant slope) increase in pressure you will need to do some more analysis, let me know and we can work on it!
     
  6. Jul 31, 2014 #5
    So, I think I need to clarify, because the volume of the tank is changing. As you can see in the attachment, the volume of the reservoir changes as the fluid is pushed out of it and replaced by the gas. V1 is the initial volume where the fluid volume is maximum and the gas is minimum and V2 is the final volume where the fluid is empty and the gas volume is maximum. The reservoir total volume is 3 mL.
     

    Attached Files:

  7. Aug 1, 2014 #6
    So what is controlling your flow rate?

    You mentioned the liquid and a critical orifice - you also want a 1 psi pressure to force the liquid through the orifice. In that case, one option is to charge a large enough cartridge to 1 psi so that the increase in volume of gas into the reservoir does not change the initial pressure, keeping temperature constant for 16 hours.

    The other option is to regulate the amount of air that enters into the reservoir regardless of cartridge pressure with something such as a very long small ID capillary tube, or something similar, whatever is the standard technology in industry or laboratory for such a low flowrate.
     
  8. Aug 1, 2014 #7
    Thank you for your help, first of all.

    The flow rate of the liquid is controlled by a critical orifice and the force pushing it. The temperature will be held constant throughout the process, so that's helpful. I have identified a cartridge that is charged at something like 800 psi and holds 075g of CO2, so I think that it's overkill, but I believe it will work. I'm not sure how I can control the flow rate, the problem is I don't have the space to use a regulator. I don't think I have the room for a long capillary tube, but maybe. Any other thoughts?
     
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