Low pressure dump chambers and efficiency

[FatMan]

I think I get it now

Steve,

I think I'm getting you now. No diss to Doc Nickel, but I don't see that charging a dump chamber is any more of an issue than charging the area behind the ball when the poppet opens, or charging the chamber behind the poppet after a shot. Any time the air is transferred from one place to another friction and heat loss result in some energy loss, but that's going to be true for just about any marker, poppet or dump chamber.

OK on dropping point 2.

The pressure spike is later, but why does that imply a loss of energy - rather than just a timing of when the energy is applied? I think its that totally ungrounded assumption that I take issue with.

I'll give you that a dump chamber might be a little less efficient - though I'm not entirely sold on that. On the whole though, the complete system seems to be very efficient.

I guess what I've always wondered is how much energy is being wasted out the end of the barrel after the ball has left. What I think is GOOD about the dump chamber arrangement is that there is a single burst, and once the ball has left the barrel there isn't more pressure comming down the tube, where as it seems like a poppet valve which is "open to the bottle" more or less, is still pumping pressure in behind the first pulse - pressure which isn't really effecting the ball, only being spent down the barrel.

This is sort of related to the stuff AGD posted over the weekend - though not exactly I admit.

Anyway, I'm glad we've managed to communicate on this - you're right this isn't the most efficient communication, but then if it weren't for this we wouldn't HAVE this conversation and noone else would have heard it (and potentially learned something).

FatMan

[steveg]

English Canadian joke:
How do you gag a Frenchman?
Tie his hands behind his back.

[steveg]

Going back to the Tippmann (model 98 valve)

The velocity adjuster is a setscrew placed in the airflow
after the poppet valve.
By turning the screw in and out the velocity lowers or raises

More obstruction=lower velocity
less obstruction=higher velocity

Soooo the question is; Is the lower velocity a product of
the same volume(quantity) of air being slowed down
and therefore imparting less energy or momentum to the ball
or is it a product if an obstruction in the path of the
air allowing less air out of the valve?

[FatMan]

It lowers the volume of air

I think clearly it slows the air going past the valve, thus less air gets past the valve.
Less air past the valve is less energy expended. I don't think the rate of the air itself is much of an issue (within a range). Unless you slow the rate of increase WAY down, that's not what's going to control the velocity. Its the total energy transfered, not the rate of transfer that is critical.

FatMan

[bjjb99]

I agree, the set screw in the Tippmann restricts the airflow, so less air gets released in the forward direction. Incidentally, some fraction of the air that doesn't make it towards the ball due to the restriction actually goes towards throwing the hammer back for the next shot. This can be seen when pressures start to drop in a Tippmann, screwing in the velocity adjuster can cause the gun to recock when leaving it open would result in the blowback's characteristic low-gas runaway burp.

You see the same effect in some blowbacks when firing with no paint in the barrel. The slight difference in airflow when a ball is not there results in the hammer not being thrown back far enough to catch the sear.

BJJB

[FreshmanBob]

someone needs to convert a shocker(or any dump valve) into a pump and compare it to a well tuned (*cough*Palmer) sheridan pump. Using co2 and N2 may also produce slightly different results because the dump valve has to completely refill each shot, which might need more expansion energy(im not sure, but its possible). That means less liquid co2 is expanded an you may shoot more liquid than with a poppet valve.

[Doc Nickel]

Hey guys? Please don't put words into my mouth, okay?

The loss of energy of the compressed gas is minimal, but it does exist. I posted that in the Guild as an example of a factor, not as an explanation for the entire cause.

And yes, the expansion/recompression of the gas does indeed cause a loss of energy. Josh stated it himself in the first post when he mentioned the RT's chamber heating problem- that heat came from somewhere

If you compress a gas, it warms, if you allow it to expand, it cools. The energy that does the work we're doing- firing a paintball- comes from the potential energy created when the gas was compressed. At any time after that, if the gas is allowed to expand and at least partially recompress, as in filling a "dump" type chamber, some of that energy is irrevocably lost.

I'm surely no expert on physics, but that's the root concept of entropy at work.

Poppet type valves see only a small drop in pressure, so there's minimal recompression at work. Dump chambers see a far larger drop, so there's more loss. But again, that's just one factor, not the entire explanation.

As for dump vs. poppet, keep in mind dwell times and the pressure rise times have a lot to do with it as well.

Carry on.

Doc.

[FatMan]

Hey Doc!

Hey Doc,

Nice to see you here. I didn't intend to drag you into this, your name was thrown around and I hadn't seen your post - so I wasn't arguing with you. Sorry.

Anyway, I'm also not a physicist, but I do know something about it. From what I do know there's nothing inherent in compressing or expanding a gas that causes energy loss. In a perfect universe with perfect insulators and no friction you should be able to compress, expand, and recompress gas all you want without energy loss - the energy is simply transferred from place to place. Or am I missing something (wouldn't be the first time )

Of course, we don't live in a perfect universe. When we compress a gas we lose some of the energy in heat loss as the gas cools. When the gas escapes it has to overcome some friction - which can also cause heat that is lost, thus in practice the effect you noted.

I guess what I'm not seeing is why transferring a gas from one chamber to another, and then out the barrel is all that different for a poppet valve versus a dump chamber? In both we will see a cavity that has its pressure reduce and then increase. Granted the rates are a little different, the specific pressures and volumes are a little different, but I still don't see an *inherent* difference.

Anyway, I realize the discussion may be academic, as these losses probably pale in comparison to other factors that affect efficiency, but I'm actually interested in the subject - and since we're discussing it ... well ...

So, if you want to just let this go, that's fine but if you have some insight I'm missing I'd like to hear about it - as long as we're on the subject.



FatMan

[BlackVCG]

Originally posted by pbjosh
The air comes out a .25" dia hole, which has .196 in2 of area.

The pressure of the air drops to 60-70 psi or so AT the ball. Because of restrictions in the system. Designed in to make a Blow-Forward marker work. If there were no restricions, then the bolt WOULDN'T go forward.

Your math is wrong. The I.D. of the PT tip is .25" so you need the radius squared times pi. Not the diameter squared times pi. A = piR^2 gives us an area of .049in^2.

How does a "restriction" in the system drop the pressure? I'd have to argue it simply relates to the relationship of pressure and volume. You're releasing a fixed volume of air at 375psi in a controlled manner and due to volumetric expansion, the pressure decreases. Also, you say if there were no restrictions, then the bolt wouldn't go forward. I'm assuming you're saying that the power tube is the restriction. I guess if you want to look at it that way, then the bolt certainly wouldn't move because you have to have some way to build up pressure behind the bolt. You have to have that "restriction" there in order to make it practical. The only way you couldn't have that "restriction" there is to make an erroneously large bolt the size of the dump chamber. It's just not practical and pointless.

Furthermore, I don't think anyone has really discussed it yet, but the way you'd lose energy in a system from the dynamic flow is due to turbulent flow. When you have turbulent flow making sharp turns and entering smaller orfices, it forms small eddies at the transition point, which equates to a lose of pressure downstream and therfore a lose of energy. This is more applicable when you're dealing with liquids in a constant flow system, but I'd have to argue that in a pneumatic valve for a paintball gun, which is a non-constant flowing system, there is no energy lose do to turbulent flow created by sharp bends and reductions in orfice sizes throughout the airflow path of the valve. In aerodynamics it's a whole different issue. You're dealing with constant flow, so losses do to turbulent flow are an issue. Which is why they design planes and such to get the Nr below 2000 and thus have laminar flow.

Originally posted by FatMan

Anyway, I'm also not a physicist, but I do know something about it. From what I do know there's nothing inherent in compressing or expanding a gas that causes energy loss. In a perfect universe with perfect insulators and no friction you should be able to compress, expand, and recompress gas all you want without energy loss - the energy is simply transferred from place to place. Or am I missing something (wouldn't be the first time )

You are correct, this is known as an adiabatic system. In an adiabatic system the Work (W) = 0; Energy transfer by heat (Q) = 0; and so therefore the change in Internal Energy is = 0.

[FreshmanBob]

When it comes to turbulance, your talking such a small space that the air is moving through i really doubt it could make that large of an impact. If we were dealing with much larger systems, not channels through guns 1/8th of an inch wide i could see your point but i doubt the energy loss is substantial

[FatMan]

Puzzle

OK, so all of this really has had me thinking a lot about pressure and energy and all that. So here is an interesting questions:

Suppose I have a large steel tank of N2 at 5000 psi with an even temperature throughout the gas and tank. I also have a steel bottle at 0 psi (OK, its actually at sea-level atmospheric pressure, but you know what I mean) and it is also at the same temperature. Now, I connect the two with a hose, and I open the valve long enough to fill the bottle to 2500 psi. The tank is much larger, so its pressure drops, but not so much.

Now, the interesting thing is this: the N2 in the bottle is now at a lower pressure than it was a moment ago. All the same, the gas and the bottle have heated up. The tank should be cooler than it was a moment ago - which makes sense as the gas is at lower pressure. Given time, the excess heat in the bottle dissipates into the room, and some heat from the room warms the tank, with corresponding adjustments in pressure - in the ideal case no energy is lost.

Why did the bottle heat up? yes the air in the bottle is under higher pressure than before, but most of the air in the bottle came from the tank and was at a still higher pressure before. If anything I would expect the heating and cooling to cancel out.

So, I don't know the answer to this question, but I suspect it has something to do with something AGD brought up in another thread - so called dynamic pressure. Could it be that when gas is released, more than just gas molecules is lost? Does some of the engergy in the remaining molecules transfer to the leaving molecules in the form of motion - and dynamic pressure - that converts back to stored energy in the form of heat in the bottle? Thus the resulting cooling of the tank and heating of the bottle?

If this is the case, then one would expect that the faster you fill the bottle, the warmer it would get (and the cooler the tank would get). My experience says this is true, though I don't know this any better than the original question.

Now, if all of this is true, then to answer my own question from a couple posts back - the inherent loss in a dump chamber comes in the fact that by storing the charge momentarily we convert the dynamic energy to heat, which can be lost through dissipation. Of course, that will depend on the recharge speed which would imply it is more of an issue on an RT (which recharges fast) than a standard AIR valve (which recharges slower) and thus represents a trade-off (which can only be evaluated in terms of the whole system).

So - am I out to lunch on this? Is this even close? Gee, I ought to go back and get another PhD in Physics!

Until next time,

FatMan

[steveg]

Ah-ha starting to come around arn't you
By the way which of you Guy's are putting words into Doc's mouth

[AGD]

The little bottle heats up for two reasons. First when the air is flowing through the hose to the small tank it is at much lower pressure. The lower pressure automatically means lower temp (below ambient) so during transistion the air picks up heat from the suroundings and when recompressed to the same pressure, has more BTU's per cubic inch = hotter. The second thing that happens is the flowing air has turbulence and friction which also serves to heat the air. THe faster you transfer it the more heat it generates. The RT valve gets most of it's heat from friction in the recharge flow.

AGD

[FatMan]

Umm ... well ...

Originally posted by steveg
Ah-ha starting to come around arn't you
By the way which of you Guy's are putting words into Doc's mouth

Well, actually Steve, you were the one who invoked his name way back at the beginning of the thread. I don't know anything about his mouth - I try to stay away from that!

AGD, so I was dead and completely wrong! Cool!

OK, so heat picked up from outside the system during transfer and later lost in the dump chamber doesn't count as lost energy.

The turbulence and friction - where does that energy come from? Is that energy previously stored in the system as pressure that is converted to heat and lost, or is it energy picked up from outside the system that temporarily increases the energy in the system until it is dissipated?

FatMan

[FreshmanBob]

because the air it's self is causing the friction and turbulance its comming from the energy needed to compress the gas into the large bottle to 5000psi in the first place.