I'll offer up a better way to think about it . . . it's not mathematically "perfect", but it does not need to be - it will help anyone get their head wrapped around this. The fact that air is not an ideal gas is not critical to understand this - yes, it does affect things at the higher pressures, but since we are not going into the deep calculations here, we can set it aside to better understand what is going on. So, I will refer to "air" moving forward in this with the understanding that I am not talking about the exact molar number of molecules involved, as we would be if we calculated taking into account the full van der Waals forces and equation, but rather a simplified set of units to get people thinking closer to how this all works. More can be done later if desired, but it really is not needed. I hope this helps
@Misguided understand this, along with others too . . .
First, since air is compressible, we must take that into account as we think through this - the amount of air we have on hand (and use per shot) is a function of both the volume and the pressure of the air. With that in mind, think of the air not in terms of volume OR pressure, but together in a unit I will call the "bar-cc" of air. We all know how much a cc is, and of course a bar is ~14.5 psi, or 1 standard atmosphere of pressure. With this unit, we can calculate the amount of air in both the reservoir and the plenum in a regulated gun (by the way, this thinking also works for unregulated guns too and is the basis for most efficiency calculations too).
In the OP's original example, the example started with a 600cc tank and 5cc plenum, but importantly did not include any reference to pressure - thus it only talked about the volume and not the amount of air in the system. We need the pressure to know how much air we have on hand. To complete the example, and to make the math easy, let’s assume the reservoir starts at 200 bar and plenum is set to run at 100 bar. With that, we now know that the amount of air we have in each is:
Reservoir: 600cc x 200bar = 120,000 bar-cc of air
Plenum: 5cc x 100bar = 500 bar-cc of air
OK. Now we know where we are starting. Now we must consider what happens when we shoot a pellet. It is important to understand that all the air in the plenum is not used - only part of it is, unless the valve is a "dump valve", which are never used with regulated guns (they are used on multi pump guns). There are ways to measure and figure this amount out but let’s just assume that this gun uses half of it on the shot as it is set, and we assume that the shot cycle is so much faster than the regulator that the pressure drops 50% in the plenum to 50 bar before any meaningful amount of air flows through the regulator.
Thus, we use 5cc x 50bar = 250 bar-cc of air on each shot. Now with that, we can calculate the number of shots we would get out of a full reservoir before it falls to the regulator set point. Since the regulator is set at half the initial tank pressure, that means that only 50% of the air in the tank can be used before we hit that set point, at which point the plenum will no longer fully refill after each shot, and each progressive shot will likely slow down (depending on how the gun is tuned).
So, we have 60,000 bar-cc of air to use, and we use 250 bar-cc per shot, so we get 240 shots on the fill. Note - this value is an approximation, and it is the first point in all this discussion so far where the non-ideal gas nature of "air" has an impact. Specifically, it means that we won't get the full 240 shots because of the non-linearity of the molar amount of air at the higher pressures, so in all likelihood we would get somewhere around 230 shots on that fill - not a big deal in my book, and not worth complicating the discussion to "get it fully correct" on this point . . .
Now we can finally look at what happens when we increase that plenum to 15cc of volume. There are two cases we can build on this - I'll do one, and then explain what happens in most cases with the other. The first case is one where the reservoir volume does not change as we increase the plenum. This is common on guns that have a separate changeable plenum, like the many of the FX guns and others that are similar. The second case is one with a tube gun in which there is one combined space for air and the regulator and plenum takes up part of that space - so increasing the regulator volume will decrease the amount of high pressure air available for use.
If we keep the 600cc tank unchanged and increase the regulator to 15ccs, and leave the regulated at the same setting of 100 bar, we now have the following:
Reservoir: 600cc x 200bar = 120,000 bar-cc of air
Plenum: 15cc x 100bar = 1500 bar-cc of air
Now we must understand what happens when we fire the pellet. Before we used half of the air that we had in the plenum when the gun was shot, but what happens now? If we use half of it, we will be using 3 times as much air (750 bar-cc instead of 250). Is that even possible? Well, this will depend on the gun itself, and how we choose to set it up. Even without changing anything, in almost all cases, the gun will use a bit less air, at least as measured in bar-ccs because the gun will be running much more efficiently - the larger plenum (in this case, three times larger) leads to less pressure drop during the firing cycle and thus a faster closing valve (shorter dwell time) - and also a bit more power in terms of a faster pellet flying out the barrel. Because the gun is more efficient, it will get one of the three things happen either directly or with tuning:
- More shots at the about same power level
- More power over about the same number of shots (or possible less if tuned for even more power afterwards)
- Some blend of the two, typically decided by whoever is tuning the gun after the change
With no changes to the regulated pressure, the larger plenum only added a small bit of air in the system from the extra 10 ccs at 100 bar (less than 1% more air) but it completely changed the operating window of the gun - meaning the efficiency of air usage as it is fired.
Furthermore, the opportunity now exists to lower the regulated pressure and get similar performance to before - as a broad example, remember the gun original shot using only 250 bar-cc of air. We could potentially lower the regulator from 100 bar to 50 bar and still have 765 bar-cc of air in the plenum for the shot. Now if we retune and assume that it gives the same power as before using 300 bar-cc used per shot (since it will be less efficient at the lower reg pressure), we have a whole different case. But now we also have more usable air in the 600cc tank since now we get to use 150 bar of it instead of just 100 bar on a fill, so we have 600cc x 150bar = 90,000 bar-cc of usable air per fill, and at 300 bar-cc per shot we now get 300 shots per fill compared to the 240 before! But do know that we are now using 50 percent more air in each refill of the gun and that air will come from a tank or compressor so is not free . . . especially if hand pumping, as the refill will take at least 50% more strokes.
Now things do become more complicated in the second scenario (trading off reservoir volume for plenum volume), such that each case likely needs to be looked at individually if the swing is significant. But even in this case, the efficiency gains often lead to more shots at higher power with a larger plenum - I have seen and measured it directly in high power tunes with fixed tube on Marauders, where a larger plenum gave more usable shots even though the reservoir space became smaller (again, still needing to put more air in the gun at each refill due to the impact of regulator set point changes). But that is more nuanced discussion, as understanding it means fully understanding the shifts in efficiency that take place, and whether they can overcome the changes to the amount of usable air that we have after the change, but the basic concepts are the same . . . .
Hope this helps the understanding - if not, I just wasted a bit over an hour.
Note: edited to clean up a few things, and to include the OP's link in the thread to let him know this is here to read . . .
