SpudFiles

If pressure was not multiplied by each stage, then a paintball 3 stage compressor would be good for maybe 600 PSI instead of 4500.

Maybe my 1/2 stroke example was confusing. Let's redesign for an example. Using a low pressure example to keep from blowing out a head gasket and providing high volume instead, lets do a 2 stage compressor. Each stage is to compress it's inlet pressure 5X by compressing the volume of absolute pressure into 1/5th the volume and then the outlet valve opens. We will look at this from a power required standpoint.

First stage takes in atmospheric air. Assume no dead space such as a perfect piston fit. 1 atm in and 1 atm on the piston backside.. It works as follows. 1/2 way through the compression stroke the piston pressure is 2 atm, at 3/4 stroke the pressure is 4 atm and at 4/5ths stoke the outlet valve opens and the 1 unit of air is expelled on the remaining 1/5th of the stoke until the entirety of the volume is expelled. Force on the piston is not a difficult calculation. Most of the piston stoke delivers NO air to the outlet as the outlet valve is closed for 4/5th of the pressure stroke distance. Pressure delivered is at 5 Atm and is 1/5 the original volume.

Second stage receives this air, but to get the inlet full for the stoke, the first stage needed 5 stokes to provide this 1 volume at the higher pressure. It could have either a displacement 5 times as large, have 5 in parallel, or run 5 times faster. Note the HP requirement to provide the first stage to keep up with the input requirement of the 2nd stage.

The second stage takes in an easy cylinder full of air at 5 Atm and has 5 Atm of air on the piston backside. Intake stroke energy requirement is the same as the first stage. No pressure differential. Things change on the second stage on the compression stroke. At 1/2 way up the cylinder pressure is now 10 Atm in pressure. Please note the first stage has a 1 atm increase at this position and the second stage is pushing 5 Atm increase at 1/2 the piston travel. At 3/4 the piston travel the cylinder volume is 1/4 it's original volume at 4X the initial pressure or 20 Atm. The outlet valve is still closed. At this point in time we have 20 Atm in the cylinder and 5 Atm still in the compressor shell and piston backside. Simple math will show the crankshaft force is pushing a 15 Atm difference. At 4/5ths the piston travel, the outlet valve opens again just like the first stage and this volume of air compressed again to 1/5th its volume is delivered at 5X it's inlet pressure at 25 Atm. In this case the last 1/5th the piston stroke is delivering the cylinder of air out the valve at 25 Atm pressure.

Please note at the work portion where our perfect pump delivers the air on the last 1/5th of the piston stroke, the outlet is 25 Atm and the inlet is 5 Atm so it is pushing an increase of 20 Atm. The first stage only had a differential of 4 Atm.

Simple math will convert the absolute pressures to relative that you would read on a normal pressure gauge.

The first stage is pumping into about 30 PSI and the second stage is pumping into 360 PSI using a rounded Atm pressure of 15. For this to work the first stage needs 5X the displacement, or using matched fridges, 5 compressors in parallel feeding into one second stage. The single second stage has about the same HP requirement as the combined HP of the 5 compressors feeding into the second stage.

I hope that example works better for you. This is ignoring all leaks, dead space, compression heating, and other system losses. The real world would have to account for those factors.

A bit late but I just read this last post today. I was looking back to this thread because I did a little experiment on one of my compressors that usually slows down at 650psi. I fed some low pressure into the input (not sure of the exact pressure, less than 70psi I think) and the compressor delivered 1200psi in about the same amount of time (I shut down the compressor at that point).

Although I still don't agree with the multiplying effect of a pressurised input.

I understand that reducing volume to 1/5th will multiply pressure by 5, but what I am getting at is that the first stage producing 5atm gives a force differential of 4xArea whereas the second stage gives a 20xArea force, so will require more power to do the same stroke in the same amount of time.
In multi stage compressors they tend to reduce the piston diameter of the next stage to compensate (also as Tech said, so that 5 first stages aren't required so that the second stage can keep up).
But upping the inlet pressure on one same compressor won't multiply its max pressure as the motor can only provide that same power. If it usually stalls with a 4xArea force differential, a 5atm input will make the compressor stall at 10atm.

Also the opening pressure of the valve isn't fixed it depends on the pressure in the reservoir. There will be an initial minimum opening pressure (spring, friction etc...) but after that it is once again a force differential that will make the valve open.

It all depends on what the motor and flywheel inertia are capable of.

The two main limits on these compressors are compression ratio and flywheel torque.

ramses wrote:Working in absolute bar, the theoretical maximum pressure is the product of the compression ratios of all stages.

The force required is the on a given stage with pressure on both sides is the pressure differential across that stage times the cross section of that cylinder.

The volume moved by a stage is equal to the absolute pressure in bar (atmospheres, really) times the cross sectional area times the stroke of that stage.


So if the case never bursts, the motor will always be the limiting factor. That is to say that you can keep adding inlet pressure until the motor stalls or the case blows up.

You can't predict stall pressure simply by comparing the linear force available from the motor at a given displacement with the force on the piston at a given displacement (which increases somewhat exponentially, neglecting deadspace)


Because the pump has a flywheel with inertia, force required to not stall can (in the short term) exceed the force available at the motor at some displacements (which is NOT constant)

Thus, if flow through the inlet check valve is great enough, and there is only atmospheric pressure on the "back" of the piston, the inlet can actually transfer energy to the piston, accelerating it, driving the flywheel, and, if the motor can spin faster, ease SOME of the motor load. This will never be as good as if the "back" of the piston was connected to the inlet with a large buffer volume and near infinite flow, but it's better than nothing.

Similarly, if the back IS exposed to inlet pressure, and the inlet check valve in the cylinder head has insufficient flow, the motor will have to pull a pseudo-vacuum to prepare for the next stroke. If flow is poor enough, the pressure in the cylinder at the start of the next pump will be less than inlet pressure (and the pressure in the case), and the pressure in the case will drive the piston until the pseudo-vacuum has collapsed. This will drive the motor, but will also decrease the actual displaced volume and actual compression ratio.

ramses wrote:It all depends on what the motor and flywheel inertia are capable of.

The two main limits on these compressors are compression ratio and flywheel torque.

And motor power. If the flywheel doesn't come up to speed between strokes, it will progressively turn slower each turn until it stalls.

Power is still required.

The 4 limits are
Horsepower
Compression Ratio
Inertia
Dead space

This assumes all components are rated to handle the pressure required without breaking.

I've been looking for a cheap but safe way to get my chamber Pumped to 800 PSI. Would it be possible to pump up a large tank to 100 or so PSI with a hand pump, and have this tank output to the inlet of my fridge compressor?

(oops)

i know this thread has been dominated my recipricating compressor so i figured i will put add input on rotary compressors.

with my rotary the motor is the limiting factor. this means that my max pressure is reached without a presssureised input. (a less than atmosheric pressure might get me higher but pumping would be extremly slow then the limiting factor would switch over to deadspace as the pressre went down)

i do use a pressurized input becuase its way to slow for me. i typicly run a 40~50psi ( 2.75~3.5bar) in as it inceases the output volume considerably and the motor doesnt stall out too early. its not uncommon for me to run it till stall and it stall at inceasingly lower preesures as the inlet pressure is inceased

if i want to get near the max out of my compressor i start with the inlet reg set at 40psi and the ball valve fully open. as i near 500bar (34.5bar) i start closing off the ballvalve limiting the flow. with this technique i can get around 750psi(52bar) (i dont know exactly becuase my gauge is only a 600psi [41bar] one)

anouther difference between rotary and recipricating is the inlet and outlet configurations. i do not think that it is mounted on springs insidethe case. the inlet goes into the case then directly to the pump head. the outlet is pump directly into the case. so withy mine the case is at the same pressure as the oulet is. which means the case has to be built to withstand a much higher pressure

on a side note i got the chance and fittings to test the massive fridgy ( 3hp 90lb recipricating comprssor) it quickly and easily maxed out my gauge at 600psi. im gonna need a much higher gauge... also i looked at the gauge on a similar compressor that is still in use for refrigeration and it runs and slighly under 450psi so its no wonder that 600psi is no problem

Technician1002 wrote:

ramses wrote:It all depends on what the motor and flywheel inertia are capable of.

The two main limits on these compressors are compression ratio and flywheel torque.

And motor power. If the flywheel doesn't come up to speed between strokes, it will progressively turn slower each turn until it stalls.

Power is still required.

The 4 limits are
Horsepower
Compression Ratio
Inertia
Dead space

This assumes all components are rated to handle the pressure required without breaking.

compression ratio ~= dead space.

you are correct that the flywheel would slow down, but when I said flywheel torque, I meant (not said, yet again) the maximum sustainable peak torque.

anyone one got any decent high pressurs yet ? havent touched this kinda stuf in a while might have to start again

i think a french guy got to like 1000+, cant find the thread.