Compressed gas flow rate calculation

Click For Summary

Discussion Overview

The discussion revolves around calculating the flow rate of compressed gas for a mechanical design involving a plunger and a critical orifice. Participants explore the implications of gas behavior under pressure, the governing equations, and the practical considerations for achieving a low constant pressure over an extended period.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • One participant describes their design involving a compressed gas cylinder and seeks guidance on calculating the gas flow rate needed to fill a 3 mL chamber over 16 hours at a low pressure of 1 psi.
  • Another participant suggests using the ideal gas law and Boyle's law to derive the necessary flow rate, providing a specific calculation for the volume of gas needed to achieve the desired pressure.
  • There is a discussion about whether the flow rate can be simply calculated as volume divided by time, with a clarification that this depends on whether a specific pressure-time curve is required.
  • A participant raises the issue of the changing volume of the reservoir as fluid is pushed out, indicating that this complicates the flow rate calculation.
  • Concerns are expressed about controlling the flow rate, with suggestions including charging a gas cartridge to maintain pressure or using a capillary tube, although space limitations are noted.

Areas of Agreement / Disagreement

Participants express varying opinions on the best approach to calculate and control the gas flow rate, indicating that multiple competing views remain. There is no consensus on a single method or solution.

Contextual Notes

Participants acknowledge limitations related to the assumptions made about gas behavior, the need for a specific pressure-time curve, and the physical constraints of the design, such as space for regulators or capillary tubes.

spiri
Messages
27
Reaction score
0
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.
 
Engineering news on Phys.org
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
 
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?
 
spiri said:
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?

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!
 
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.
 

Attachments

  • photo.jpg
    photo.jpg
    20.9 KB · Views: 2,462
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.
 
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?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
598
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
6K
Replies
13
Views
3K
  • · Replies 45 ·
2
Replies
45
Views
7K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
0
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K