# Pressure calculation for compressible gas and variable 'g'

1. Dec 19, 2015

### ozmac

I have what seems like a straight forward question, but am unable to find a formula.

The question originates from my previous topic which was trolled: https://www.physicsforums.com/threads/blowing-up-a-balloon-using-centrifugal-force.848892/

Basically I have a 1.5m hollow length of 10mm tube and due to centrifugal/centripetal force, at one closed end of the tube the acceleration is 24g, and at the other open end it is zero. The gradient is linear (Ac=w^2*r)

I'm trying to calculate the pressure of the gas at the closed end of the tube.

As the air is compressible, you can't use p = h ρ g because the density of the air is a variable dependent upon the pressure. But the pressure calculation here uses the density. There must be a different formula which uses the molecular mass of air. But even then, the amount of molecules of air in the tube is dependent upon the pressure, as with the air that is accelerated towards the closed end of the tube, this results in additional air being sucked in the open end of the tube.

There are so many things going on dependent on each other, all the while with a hugely variable 'gravity' so I just don't know where to begin. The easiest way would be to put a pressure switch at the end of the tube, but I'd prefer if there was also a way to calculate this also.

Thanks

2. Dec 19, 2015

### Staff: Mentor

If p(r) is the pressure at radial location r, p(r+Δr) is the gas pressure at r + Δr (these pressures act on the cross sectional area A), and if the amount of mass between r and r + Δr is ρAΔr, what is the force balance on this mass?

3. Dec 19, 2015

### ozmac

Density is not constant though, the gas is compressible, I'm not sure how that helps me

It seems most literature is based on assumptions of incompressible fluids and constant acceleration/gravity, none of which come close to a suitable approximation in this case.

4. Dec 19, 2015

### Staff: Mentor

Well, the analysis I was trying to help you through takes into account the gas compressibility. But, if you are not even willing to try to proceed, then I can't help you. I have the entire analysis worked out already, including the effect of gas compressibility. But Physics Forums rules prevent us from just giving you the solution. There has to be some effort on your part.

Chet

5. Dec 19, 2015

### ozmac

Ahh ok, fair enough (and awesome btw!).
I presume for your force balance you can ignore the forces on the cylindrical part of the wall as they all balance each other, and just look at forces along the direction of the tube.
So you have the force on one end p(r)*A and on the other end p(r+Δr)*A (where A = π(0.005)^2)
Hmmm but that's not balanced, as you have a change in force of p(Δr)*A... Which means I'm wrong, or that this is not static and the change in force results in flow, i.e the compression. But I'm not sure what to do next.

6. Dec 20, 2015

### Staff: Mentor

You've done well so far. What you're missing is that there is a radial acceleration, because the parcel of air in your force balance is traveling in circular motion. You already wrote down what that radial acceleration was in your first post. If ω is the angular velocity of your small air parcel and r is its radial location, what is its radial acceleration? What does your force balance look like now?

Chet

7. Dec 22, 2015

### Staff: Mentor

It doesn't look like you are going to respond to this, so I can tell you that I've done the calculation, and there is a good reason why you did not see any effect at 120 rpm. It's just now a high enough rotational speed. The density of air is very low, even when compressed at one end of the tube at 24 g. You are going to need speeds 10x as high before you start to see some effect on your balloon.

chet