Adjusting the pressure of helium flow to account for smaller tube

  • #1
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TL;DR Summary
How to adjust bar to maintain the same helium flow rate from larger to smaller tubes
I’m not at all sure I’m posting this in the right forum. If not, please forgive me. Also, I have no background in physics. That said, here goes:

If 2bar of helium flowing through a 9.0mm inner diameter tube yields a flow rate of 15 liters per minute, what bar would one need to yield an identical flow rate through a 3.2mm inner diameter tube? Considering that the larger (9.0mm) tube is about 2.8 times larger than the smaller (3.2mm) one, would one just multiply 2bar by 2.8, which would equal about 9bar? I’m guessing not. So is there a formula one could use to determine what new bar one would need to yield an identical flow rate?
 
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  • #2
First, I need to make sure that I understand your question. Does the sketch below correctly show the problem?
Tube flow.jpg

Assumptions:
1) The flow rate is 15 liters per minute at STP - Standard Temperature and Pressure
2) The pressure is in bar of gauge pressure, not absolute pressure
3) The tube is discharging to atmosphere - 0 bar gauge, AKA 1.0 bar absolute
4) The tube is very long (several meters or longer) and smooth inside
5) The 3.2 mm tube is the same length, and has the same pressure at discharge end as the 9 mm tube

Please confirm or correct the above sketch and every one of the above assumptions. Then we can start the calculations.
 
  • #3
The sketch is correct, as are all assumptions less two:

1. The temperature is 70 F.

2. The length of the hose is only 1.5 meters.

Thank you.
 
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  • #4
What is the mass flow rate at 2 bars?
 
  • #5
The Helium flow through a 1.5m length of 9mm I.D. tubing with the described pressure conditions will be significantly (several orders) higher than 15 LPM. I'm not sure if this is an 'actual' situation or a 'thought experiment,' but the flow is being (mostly) limited by something other than the tubing; the description is wrong.
 
  • #6
The viscosity of helium at 20 C is 0.0002 Poise and, at 2 bars gauge, its density from the ideal gas law is 0.492 gm/liter. So 15 LPM, the mass flow rate is 7.39 gm/min = 0.123 gm/sec. At this mass flow rate, the Reynolds number for a 9 mm tube is $$Re=\frac{4\dot{m}}{\pi \mu D}=\frac{(4)(0.123)}{\pi (0.00020)(0.9)}=870$$This seems very low.
 
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  • #7
The equation for the pressure gradient in the tube goes like: $$\frac{dP}{dz}=-\left(\frac{4}{D}\right)\left(\frac{1}{2}\rho v^2\right)\left(\frac{0.0791}{Re^{0.25}}\right)$$where we have assumed that the flow is turbulent and the Reynolds number lies in the range 2000 < Re < 100000.

The term ##\rho v^2## can be rewritten as $$\rho v^2=\frac{\mu^2}{\rho D^2}\left(\frac{\rho v D}{\mu}\right)^2=\frac{\mu^2}{\rho D^2}(Re)^2$$and, from the ideal gas law, the density can be expressed as $$\rho=\frac{PM}{RT}$$If we combine the. previous equations, we obtain $$2P\frac{dP}{dz}=-\frac{0.3164RT\mu^2}{\rho D^3}(Re)^{1.75}$$For feed mass flow rate and viscosity, this takes the form $$2P\frac{dP}{dz}=\frac{k}{D^{4.75}}$$Integrating this over the length of the tube gives $$(P_{in}^2-P_{out}^2)D^{4.75}=Const$$

From this, it follows that, in the case of the 3.2mm tube, in inlet pressure will have to be ~32 barsg.
 
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