# Gas flow in capillaries

1. Dec 7, 2015

### Damascenone

Can anyone help on this problem encountered in gas chromatography:
Two helium incoming gas lines A and B are connected together via a cross-shaped connector to two outlets C and D. All in and out lines are capillaries with diametre in the 0.1-1 mm range and lengths in the 10s of cm to 10s of metre range. Inlet pressure is in the 100s of kPa range (controlled by pressure regulators) for lines A and B. Outlet pressure is high vacuum for line C and 100 kPa (atmospheric pressure) for line D.
Inlet and outlet pressures and diametres and lengths of capillaries are known. How to calculate flow rate for each line? Thanks for your help!

2. Dec 7, 2015

### Staff: Mentor

What is the extent of your fluid dynamics background?

Chet

3. Dec 7, 2015

### Damascenone

Hi Chet,

I would like to say: I know a bit, but to be honest, I'm a chemist and my knowledge in fluid dynamics is zero!

Markus

4. Dec 7, 2015

### Staff: Mentor

OK. We're not supposed to solve this for you, but we can help by guiding you through the analysis. I'm guessing that this is a viscous laminar flow problem, because of the small capillary diameter and the constant tube diameter. Viscous laminar flow in a tube is governed by the Hagen Poiseuille law:
$$\frac{dp}{dx}=-\frac{128Q\mu}{\pi D^4}$$
where p is the pressure, x is the distance along the tube,Q is the volumetric flow rate, mu is the gas viscosity, and D is the tube diameter. The volumetric flow rate is related to the (local) gas density $\rho$ and the mass flow rate m by $Q=m/\rho$. The density is related to the pressure and the temperature (and molecular weight) by the ideal gas law. The mass flow rate in each tube is a constant from inlet to outlet.

Now it's your turn. If you substitute the equation for Q and the ideal gas law into the Hagen Poiseuille equation, what do you get?

Chet

5. Dec 7, 2015

### Damascenone

Thanks a lot Chet, I try to put the puzzle together... Considering, for the moment being, one capillary only, I get p1*Q1=p2*Q2 for the inlet and outlet. The viscosity of helium I was able to find, dp might be replaced by p2-p1, dx might be the length of my capillary, but then I run into trouble for Q which is certainly due to the fact that I don't know how to handle the differential equation. I would need another hint please!

Markus

6. Dec 7, 2015

### Staff: Mentor

From the ideal gas law, you have:

$$\rho=\frac{pM}{RT}$$
where M is the molecular weight, R is the gas constant, and T is the temperature. If we substitute this into the equation for Q, we obtain:
$$Q=\frac{mRT}{pM}$$
where the mass flow rate m is constant. What do you get if you substitute this into the differential equation?

Chet

7. Dec 9, 2015

### Damascenone

Yes I had been this far already but then I got confused by the dp/dx term, thinking I have to solve a differential equation but there is none to solve... Entering the formulas in a little spreadsheet now calculates me the flow rates as I vary the temperature. Great! Thanks again for your kind help!

Markus

8. Dec 9, 2015

### Staff: Mentor

Then you know you should be solving using p2, not p, correct?
$$\frac{dp^2}{dx}=-\frac{256mRT\mu}{M\pi D^4}$$
So $$p^2_{in}-p^2_{out}=\frac{256mRT\mu L}{M\pi D^4}$$
So, $$m=\frac{(p^2_{in}-p^2_{out})M\pi D^4}{256RT\mu L}$$
Your only unknown is p2 at the junction, and you should be solving for it under the constraint that the sum of the two mass flow rates into the junction are equal to the two mass flow rates out of the junction. This will be a linear equation in p2.

After getting your solution, you need to check to make sure that the Reynolds number in each of the 4 tubes does not exceed 2100, in which case the flow in that tube would be turbulent (and the analysis would have to be modified).

Chet