Help get my head around basic air pressure /volume transference

[luben]

Hi Good Folks

 

I've been looking at Boyles laws but I‘m still having difficulty visualizing the basic rules of transference of compressed air / volume and how to apply Boyle Laws. I have created a scenario below can anyone help me towards the understanding.

 

Assume there are two compressed air cylinders. Cylinder A is volume 100m3 @ 2000psi. Cylinder B volume 10m3 and currently at normal atmospheric (14.7psi). Also Assume that the two tanks are connected and air can be moved from tank to another via necessary connecting pipe, valves and pumps etc.

 

So

 

A) If I were to open the connecting valve between the tanks I understand that the pressure would eventually equalize between both tanks as high pressure seeks low pressure. But what would the equalizing pressure now be in the two tanks / how to calculate ? I assume that the overall pressure is now lower than 2000psi as an extra volume containing lower pressure has had to be filled.

 

B) If wanted to increase the pressure in Tank B from atmospheric to 200psi using the 2000psi pressure from the large tank A. What would the new pressure / volume of air now be which remains in Tank A.

 

C) If tank A and B both had equal pressures, e.g 2000psi. If I wanted to pump the air from Tank B into tank A so that Tank B is reduced to atmospheric. Approximately what is the max pressure required from the pump. As there is equal force between the two tanks do you literally need only a few psi to begin with, then as more air is moved to tank A so a pressure bias is created thus requiring ever increasing pressure from the pump until 2000psi is required. Or is 2000psi+ required to move any extra air into tank A even though the starting pressure from the other tank is also 2000psi.

 

Apologies if this all sounds a bit simple or for any major flaws in my assumptions, but hey if don’t ask you ask you don’t learn huh. Greatly appreciate any help given.

 

Cheers

[Ndi]

Hum.

 

Well, since nobody tackled this, I'll give it a try.

 

I was always taught to take the easy path, so every time I see a pressure issue i simplify (to volume) - it can be added, unlike pressure.

 

Oh, and, also, I'll ignore imperial units since they make math too odd. I didn't even know physics was taught in Imperial. Your equations must be mighty odd. Anyway.

 

a) If one tank, A, has 100 m3 and one bar, that's 100 m3 of air at normal pressure (1). You can multiply by density to get actual air in grams but we'll skip that. Now, should we cram twice the air in there, we get twice the pressure. So at 2 bar, we have 200 m3 of air in an 100 m3 enclosure.

 

When you open the valve, you make the two tanks a single tank. Thus, sum of volumes over sum of contents. A is 100 m3 at 2 bar (200), the second is 10 m3 at 20 bar (200). So now you have 400 m3 of air on 110 m3, for a nice 3.63 bar. You see it's a lot closer to the large tank, which makes sense.

 

b) Again, volume. To push a 100 m3 from 1 to 2 bar, you need 100 m3 of air at standard pressure. That means that the first tube will lose that, meaning you now have 100 m3 at 10 m3 tank (pressure = 10), the new one will have 200 on 100 (pressure 2).

 

c) well, you get the point. Get actual quantity, do the math, re-apply into original.

 

Note, however, that none of this will work in real life as it should. Even in the classical interpretation of _ideal_ volumes, PV=k, and your tanks have different "k". Also, upon compression collisions go up and heat starts to really mess things up growing a tanks of air into a branch of physics. The above is an approximation of 1880's approximation of gases. Actual graphs are in 3d and quite curved, and that's ideals only.

 

But since you said you are now discovering Boyle's, I assume you just want a feel at the basics. Don't quote me on anything.

[luben]

Thanks for fast response ndi, that was useful, I've now got a much better visualisation of understanding whats going on.

 

Cheers L