Please help us figure out an engineering quandry involving a compressor-and-tank system!

May 26, 2011 11:43 AM

Can you help us analyze some of the thermodynamic and mechanical properties of a hypothetical compressor-and-tank system? Diagram and explanation provided.

My coworkers and I have been puzzling over a thought experiment involving the compressor and tank system in the following diagram:

http://www.inactivex.net/compressor-diagram.jpg

Note: Valves I1, I2, O1, and O2 are check valves; they only permit fluid to flow in the direction shown by their arrow. In this diagram, that direction is always toward the right.

At t=0, the system is completely filled with a gas and the valve at the top is open so that all parts of the system are at equal pressure.

To begin, the operator starts operating the compressor by pushing and pulling on the compressor handle, which is attached to a piston-like device inside the compressor. The operator does this at a steady pace and establishes a (relatively - except for when he reverses directions) consistent flow of gas through the system, where the pressure difference between gas cylinder 1 and gas cylinder 2 remains small.

Then, an assistant closes the valve at the top of the system. Our question is what happens to the amount of work needed to move the compressor handle? Does it get easier to move, more difficult to move, or is there no change in difficulty? Further, if there is a change in difficulty, does the change continue in the same direction as the compressor continues to operate, or might the difficulty decrease at first and later increase (or vice-versa)?

We all agree that the difficulty of ejecting gas through the compressor's outputs would seem to increase as the pressure on the output side increases, and that the difficulty of drawing gas in through the compressor's inputs would seem to increase as the suction on the input side increases. However, it also seems to us that as the pressure on the input side decreases, fewer molecules of gas would be drawn into the compressor on each stroke and hence less work would be required to exhaust those molecules, even against a higher pressure on the output side.

We are all computer science / I.T. guys, not mechanical or chemical engineers, so none of us know how to correctly analyze the situation.

Use the following assumptions:
-the gas is an ideal gas
-no part of the system leaks
-there is no friction between any moving parts
-there is no backward flow through any check valves
-everything that is supposed to be sealed is perfectly sealed
-etc

posted by Juffo-Wup to science & nature (8 answers total)

You can use the ideal gas law (PV/T=PV/T) to determine how much the pressure of cylinder 2 will increase with each stroke of the handle. Then you can look at the pressure differential between the two cylinders, and multiply that by the area of the piston to get the force required for each stroke. As the pressure differential will be changing with each stroke, it will require more force each time to move the handle.

Eventually, you will have enough of a vacuum inside cylinder 1 for it to collapse.
posted by bajema at 11:57 AM on May 26, 2011

We all agree that the difficulty of ejecting gas through the compressor's outputs would seem to increase as the pressure on the output side increases, and that the difficulty of drawing gas in through the compressor's inputs would seem to increase as the suction on the input side increases. However, it also seems to us that as the pressure on the input side decreases, fewer molecules of gas would be drawn into the compressor on each stroke and hence less work would be required to exhaust those molecules, even against a higher pressure on the output side.

think pressure differential. Then you see that your first sentence is correct and just stop there.
posted by caddis at 12:24 PM on May 26, 2011

I think it would get harder quickly at first, then more slowly, and then the difficulty would remain constant. This is because the pipes between the pump and output valves have a non-zero volume, and in order for either of the output check valves to open, the pressure on the pump side of the valve has to exceed the pressure on the Gas#2 side. After a certain amount of pumping, you'd reach a point where there was so little gas on the pump side of each output valve that even when it was fully compressed, it's pressure would not be high enough to force the valve open, so moving the pump handle back and forth would just compress and relax the same gas over and over again but never move it through the valve.
posted by jon1270 at 12:26 PM on May 26, 2011 [1 favorite]

Speaking as a chemical engineer who's worked with check valves, but I'm no mechanical and am pretty rusty... you guys forgot that vacuum exerts a force too.

With a finite amount of air, the system would eventually equalize, with the piston generating both the max pressure at full compression (because it can no longer generate pressure greater than the compressed air upstream) and the max vacuum at full extension (well really, just low pressure probably) which is equal to the pressure downstream.

At equalization, all of the air to the RHS will have reached its peak pressure, and the force of the high-pressure side is equal to the eventual max amount of pressure generated by the piston-side, meaning the check valves will remain shut - you can't force anything else through. To the LHS of the piston, a vacuum is being created, which also exerts a force on the check valves, sucking them shut. You can't suck any more through.

So you reach a stalemate, and the difficulty of moving the piston ultimately becomes proportional to the amount of high-pressure and vaccuum that you generate within the system. But yes, I would think it becomes harder and harder as the pressure differential grows.
posted by lizbunny at 12:29 PM on May 26, 2011

lizbunny just to be a pedant the vacuum doesn't exert a force. The check valves aren't being sucked shut so much as gas on the other side of the check valve is exerting a greater pressure than the vacuum and pushing them shut.
posted by selenized at 12:38 PM on May 26, 2011 [2 favorites]

Nice tricky little problem.

For simplicity assume that the volumes of the 2 cylinders and the compressor are equal, and equal to about a mole of the gas at the temperature and pressure of the beginning of the process.

Each forward or reverse stroke will transfer half of the gas which was remaining in cylinder #1 at the end of the previous stroke to cylinder #2.

Therefore, in less than 100 strokes you will be down to your final molecule of the gas that was originally in cylinder #1. The next stroke will transfer that molecule to cylinder #2. After that, there will be a total vacuum in cylinder #1 and the compressor, and the work of each stroke will drop to zero.

Before that, things are more complicated. The resistance to the stroke can never be greater than the difference in pressures of the two cylinders times the area of one face of the piston, and will only reach that value when the compressed gas side of the compressor reaches the pressure of cylinder #2. this will happen later in every stroke than it did on the previous stroke, and will make the rest of the stroke more difficult than the corresponding part of the previous stroke was. However, the beginning of each stroke will meet less resistance than the beginning of the previous stroke, and the resistance will continue to be lower up until a point near--but not identical to-- the point where the pressures between the two cylinders are equal.

Actual calculation of the total work involved is complicated by the fact that the gas in cylinder #2 and the gas in the compression side of the compressor will heat up with every stroke and during every stroke, thereby changing the pressures in those two locations.
posted by jamjam at 1:48 PM on May 26, 2011

Not "the point where the pressures between the two cylinders are equal", but 'the point where the pressure in cylinder #2 and the compression side of the compressor are equal'.
posted by jamjam at 1:55 PM on May 26, 2011

Good grief, "Each forward or reverse stroke will transfer half of the gas which was remaining in cylinder #1 at the beginning of the previous stroke to cylinder #2."
posted by jamjam at 2:04 PM on May 26, 2011