Theoretical Maximum Velocity in a PCP

[rsterne]

Hammer momentum is mass x velocity....

Hammer energy is mass x velocity squared divided by two....

The hammer velocity at impact is determined by the mass of the hammer and the force exerted by the hammer spring on it, and the hammer travel while being accelerated by the spring.... This is quite complex, and depends on the preload and spring rate.... and any frictional losses in the system....

The force holding the valve closed is the diameter of the sealing surface of the poppet, squared, times PI/4, times the air pressure.... If your poppet diameter is 10mm, and the throat is 7 mm, the sealing diameter is likely somewhere around 8.5 mm, but could be 7 or 10.... The area (assuming 8.5mm) is (0.335 x 0.335 x PI/4) = 0.088 sq.in.... so at 3000 psi there is (0.088 x 3000) = about 264 lbf. holding the poppet closed, plus the valve spring force....

The hammer energy lost in lifting the valve off the seat is that force times the distance the poppet material is compressed, divided by two.... If your poppet is compressing 0.010" under that 264 lbf. load, then the energy required to "unstick" it would be (264 x 0.010) / 2 = 1.32 in.lb.... That amount of hammer energy is lost before the valve can flow any air, so only the residual energy and momentum is available to create lift and dwell....

Once the poppet is clear of the seat, the closing force is predominately the stem area time the pressure, plus the spring force.... but there is an additional unknown force caused by the drag of the air passing over the poppet and through the throat.... Since it is virtually impossible to calculate these variable forces, and apply them at the correct point in the poppet travel, it is equally impossible to calculate the total lift of the valve and its dwell.... at least it is well beyond my abilities....

Fortunately, even though the dynamics of the valve are hard to calculate, they are relatively easy to understand, as it their relationship to the hammer strike.... First, find the plateau velocity by increasing the hammer strike until you get no further increase in pellet velocity.... Then back off the velocity about 3-5% to get a reasonable tune....

Bob

[DomingoT]

Hi, this is my first post, just a comment on the max velocity of a PCP.
The theoretical max velocity of a PCP is tricky. The formula discussed above, v = F/m..., is an approximation that assumes isothermal expansion and neglects the kinetic energy of the gas behind the pellet (it also assumes infinite reservoir volume and neglects friction and nose work, but that is not fundamental). Notice that, per that formula, the velocity blows up when the pellet mass goes to zero, which is not physically meaningful. It also blows up when the barrel length goes to infinity, same problem.
The reason the max "theoretical" velocity is hard to visualize is that the reservoir discharge into the barrel is a non-stationary problem, and in a "non-stationary" problem the conversion of thermal energy into kinetic energy is counterintuitive (just about every grad student in gas dynamics trips on this one!). To get the right answer, you have to imagine there is an expansion (rarefaction) wave within the barrel, and the pellet is riding the front of that wave.  If you do this, you find a well-behaved "theoretical" maximum velocity which depends on the specific heats of the gas. For air, this max velocity corresponds to about Mach 5 (and much higher for lighter gases, such as He, reason why we use light-gas cannons to simulate meteorite impacts etc.) Emphasis on theoretical.

[rsterne]

The intent of my simplified formula is to supply a maximum FPE based on bore area x pressure x barrel length.... and only applies with an infinite reservoir (no difference between isothermal and adiabatic expansion)…. and with NO losses in the system.... As such it is the MAXIMUM energy that can be produced, and a great deal of that goes into accelerating the gas itself, which is why using Helium in a PCP can produce more FPE in the projectile than you can get with air.... Even if the projectile mass goes to zero, since we are still accelerating the gas itself, the formula doesn't "blow up", because there is still a mass that must be accelerated.... Trying to wrap my head around an "infinite barrel length" just makes me dizzy....  ….In practical terms, only the best of PCPs on air approach 50% of the maximum FPE, because of finite reservoirs, friction, and the mass of the air itself.... On Helium, they may approach 70%....

The Mach 5 maximum velocity will never be achieved with EXPANDING air (or Helium), as takes place in a PCP.... Light gas guns rely on rapid COMPRESSION of the gas (with heating), usually by a piston driven by an explosive charge.... so a completely different dynamic than a PCP.... In a PCP, the gas cools as it expands, at least to some degree.... You state that:

To get the right answer, you have to imagine there is an expansion (rarefaction) wave within the barrel, and the pellet is riding the front of that wave.  If you do this, you find a well-behaved "theoretical" maximum velocity which depends on the specific heats of the gas.

For a PCP, which relies on expansion of the gas, rather than compression like in a light gas gun.... it had been theorized that the maximum velocity possible was 1650 fps, because that was the average molecular speed of air at room temperature.... and hence the fastest rate at which the air in a parallel tube (barrel) could expand.... This has been left in the dust by both Lloyd and I, we have both recorded velocities using very light projectiles of over 2000 fps.... I agree with you that as the projectile gets very light, it acts like it is simply surfing along the front of the expanding gas.... What the actual "speed limit" is, is currently under debate.... My idea is that each "packet" of air released by the valve is accelerated by the one behind it, which allows the 1650 average molecular velocity to be exceeded.... Others have different theories....

With a very large reservoir, bore size porting, and finite barrel volume, there is little difference between the average effective pressure using adiabatic or isothermal expansion.... in both cases the residual muzzle pressure is not much less than the starting reservoir pressure.... In my 6mm, for the shots I conducted reaching 2092 fps with a 1.8 gr. Airsoft BB.... the calculated muzzle pressure with a 28" barrel and a 500 cc reservoir starting at 4200 psi is 4024 psi using isothermal expansion and 3955 psi using adiabatic.... The average barrel pressure during the shot is 97-98% of the starting pressure....

Welcome to the discussion, we look forward to your input....

Bob

[DomingoT]

In your derivation of your formula you made use of the isothermal expansion assumption, even if you didn't state it explicitly. That happened when you assumed the force was equal to P0*A for "any" position of the pellet in the bore.
A formula like yours may in fact give good estimates of max velocity, but that doesn't make it a theoretical maximum. Deriving the theoretical maximum is a lot more complicated because you must take into account the non-stationary of the expansion within the bore, and this is no longer an algebraic problem.
By the way, if I am not mistaken, implicit in your formula is an efficiency equal to [gamma - 1], where gamma is the ratio of specific heats, which in air is 1.4. This means that, per your formula, 40% of the thermal energy of the gas that enters the bore gets converted into kinetic energy of the pellet.
I almost forgot, the reason your formula blows up when the mass of the pellet approaches zero is because, in your formula, vel = srt(2 P0 A x/M), where M is the pellet mass, P0 the reservoir pressure, A the bore area, and x the bore length. As M approaches zero, your velocity goes to infinity like 1/sqr(M), and this is forbidden by the physics of gases (because in that case the gas molecules would not be able to keep up with the pellet!). Same with x, the bore length.

[rsterne]

You are correct, since the reservoir volume was specified as infinite, I used isothermal expansion.... because most people are familiar with that concept and Boyle's Law (but most don't understand adiabatic expansion).... My formula for maximum FPE is just that.... a way to show people that there is a maximum which cannot be exceeded (it can be simplified to pressure x barrel volume).... and in fact "real" PCPs do well to get to 50% of that.... In any "real" PCP, the expansion is somewhere between isothermal and adiabatic.... It cannot be purely adiabatic, because we can reach efficiencies beyond what that would permit.... once you take into account real world losses....

I don't understand what you mean by "must take into account the non-stationary of the expansion within the bore".... In addition, I do not know how to quantify the "thermal energy" of the gas that enters the bore (before the pellet exits the muzzle, after that it doesn't matter)…. so I can't comment on how close we are getting to the accuracy of your 40% assumption.... Thermodynamics is not my strong suit....

Would not the velocity equation you state require the TOTAL mass to be used?.... The gas itself has significant mass, and cannot be ignored in calculating the velocity.... As an example, at 3000 psi air has a mass of 60 grains per CI.... At 3000 psi, the air in a .22 cal barrel that is 24" long weighs 55 gr.... which is heavier than most .22 cal bullets (and all .22 cal pellets).... Therefore, even if the mass of the bullet was zero, the velocity would not be infinite because that 3000 psi needs to accelerate at least half of that 55 gr. of air.... The mass of the air in the barrel therefore must limit the muzzle velocity for any given pressure, would it not?....

Likewise, as the barrel length increases, so does the mass of the air required to fill it.... which puts the conceptualization of that well beyond my mental abilities.... other than to logically and intuitively understand that infinite barrel length will not result in infinite velocity....

You hint that there is a maximum velocity that can be attained with a PCP, based on the gas molecules inability to keep up with the pellet.... We have heard that before (1650 fps at room temperature using air)…. If you have some other number in mind, I would love to hear it, and the reasoning behind it.... as this subject is much debated with no concrete answer so far....

Bob

[Scotchmo]

In your derivation of your formula you made use of the isothermal expansion assumption, even if you didn't state it explicitly. That happened when you assumed the force was equal to P0*A for "any" position of the pellet in the bore.
A formula like yours may in fact give good estimates of max velocity, but that doesn't make it a theoretical maximum. Deriving the theoretical maximum is a lot more complicated because you must take into account the non-stationary of the expansion within the bore, and this is no longer an algebraic problem.
By the way, if I am not mistaken, implicit in your formula is an efficiency equal to [gamma - 1], where gamma is the ratio of specific heats, which in air is 1.4. This means that, per your formula, 40% of the thermal energy of the gas that enters the bore gets converted into kinetic energy of the pellet.
I almost forgot, the reason your formula blows up when the mass of the pellet approaches zero is because, in your formula, vel = srt(2 P0 A x/M), where M is the pellet mass, P0 the reservoir pressure, A the bore area, and x the bore length. As M approaches zero, your velocity goes to infinity like 1/sqr(M), and this is forbidden by the physics of gases (because in that case the gas molecules would not be able to keep up with the pellet!). Same with x, the bore length.

Mach is a ratio. Mach 5 might be achievable when referencing the wave front. When referencing the reservoir air, that limit is about Mach 2.

The Mach limit at the reservoir-outlet/barrel-inlet is always 1. Couple that resulting flow velocity with the rate of expansion of the gas and that becomes the max velocity.

Based on Fanno flow equations, the absolute velocity limit of compressed air is determined solely by the Speed Of Sound (SOS) of the gas in the reservoir and it's gamma.

I still believe this to be the equation for the absolute velocity limit:
Vmax= SOS x sqrt((2/(k-1))+1)

https://www.gatewaytoairguns.org/GTA/index.php?topic=102604.msg1031163#msg1031163

FYI - at the reservoir pressures/temperatures we are dealing with, gamma is considerably greater than 1.4

[DomingoT]

You are almost there. The formula v = SOS sqr[2/(gamma -1)] applies to "steady state" flow. This fact is absolutely critical. In a PCP we don't have steady flow.
The correct formula is v = SOS 1/(gamma - 1), which applies to "unsteady" expansions (for gamma = 1.4 this corresponds to Mach about 5.) The speed of sound in this formula is the one in the reservoir.
You may ask, if the first formula represents energy conservation, does this mean the second formula violates energy conservation?  What people, even experienced fluid dynamic practitioners, tend to forget is that in unsteady flow the total temperature is NOT preserved. In an "unsteady" rarefaction wave (which is what you have in the bore) the energy of a fluid parcel is not constant (Bernoulli's equation is "not" valid in unsteady flow!)
Unfortunately, you very rarely find any discussion of the difference between these two regimes in fluid dynamic texts.
By the way, although gamma can be significantly higher in a PCP environment, this is not essential to the basic difference between steady and unsteady max velocity.

[DomingoT]

v = SOS 2 /(gamma - 1).
How does one edit a post?

[DomingoT]

rsterne, your formula is valid for any polytropic process, you are right in this. But just to summarize, the max velocity attained in an expansion (no losses, etc.) when you properly account for unsteadiness is vmax = SOS 2/(gamma - 1), where the speed of sound is calculated at the reservoir temperature. This would only happen in an ideal world where the pellet is able to ride the front of the expansion wave.

[Scotchmo]

You are almost there. The formula v = SOS sqr[2/(gamma -1)] applies to "steady state" flow. This fact is absolutely critical. In a PCP we don't have steady flow.
The correct formula is v = SOS 1/(gamma - 1), which applies to "unsteady" expansions (for gamma = 1.4 this corresponds to Mach about 5.) The speed of sound in this formula is the one in the reservoir.
You may ask, if the first formula represents energy conservation, does this mean the second formula violates energy conservation?  What people, even experienced fluid dynamic practitioners, tend to forget is that in unsteady flow the total temperature is NOT preserved. In an "unsteady" rarefaction wave (which is what you have in the bore) the energy of a fluid parcel is not constant (Bernoulli's equation is "not" valid in unsteady flow!)
Unfortunately, you very rarely find any discussion of the difference between these two regimes in fluid dynamic texts.
By the way, although gamma can be significantly higher in a PCP environment, this is not essential to the basic difference between steady and unsteady max velocity.

I believe that your equation is based on the free expansion of a compressible gas:

Vmax = SOS x 2/(k - 1)

The equation that I'm using is based on Fanno flow which considers the adiabatic flow of a compressible gas through a constant area duct where the effect of friction is considered:

Vmax = SOS x sqrt((2/(k-1))+1)

For now, I'm sticking with that. If that velocity is exceeded in reality, then I'll reconsider.

[DomingoT]

Fanno flow is strictly "stationary" flow, this is "not" what you have in a PCP. Think about a situation that would apply to "stationary" 1-D flow and you will see under what conditions your conclusion applies. It is actually interesting that there is a difference between stationary and non-stationary maximum values. This has to do with the transmission of energy between non-stationary fluid parcels, which doesn't happen in stationary flow. It is subtle but important distinction.

[DomingoT]

Bob,
To address your questions,
1. Efficiency under isothermal process.
The thermal energy in the bore is TE = cv*T*density*A*x,  since you assume isothermal, T = T0 (in the reservoir). You can write this as TE = cv*P0/R*A*x, where R = gas  constant.
The kinetic energy of the pellet at position x (the basis of your derivation) is KE = P0*A*x
Efficiency = KE/(TH+KE) = P0*A*x/( cv*P0/R*A*x + P0*A*x)) = 1/(cv/R + 1) = R/(cv+R), now remember R = cp - cv, therefore Efficiency = (cp - cv)/cp = 1 -1/gamma = gamma - 1. Since for air gamma is approx 1.4, your efficiency is 1.4 - 1 = 0.4, or 40%.
2. Energy of the gas in the bore: not taken into account. To include the gas KE, you must account for the pressure change down the bore due to the gas inertia, and doing this invalidates your assumption that the energy imparted to the pellet is P0*A*x.
3. Non-stationary max velocity versus stationary max velocity: In a stationary flow, the total energy of a flow parcel (or particle) stays constant (Bernoulli's equation). In a non-stationary flow, the energy of flow "parcels" changes as they move along. If you convert all the energy of a flow "parcel" into KE, you get the maximum velocity that flow can reach. Since the energy contained in a stationary flow parcel is different than in a non-stationary one, the max velocity will differ. In a PCP airgun, the flow is non-stationary.

[rsterne]

I think I understand at least some of that....  …. Doesn't the value of the gamma of air change with pressure?....



Unless I am mistaken gamma is about 1.7 at 70*F and 3000 psi (maybe more as the air cools from expansion? ).... What does that do to the maximum efficiency?.... In addition, I am also confused about one other thing....

If Efficiency = 1 - 1/gamma …. for gamma = 1.4 we have.... Efficiency = 1 - 1/1.4 = 1 - 0.714 = 0.286 …. However you also state....

Efficiency = gamma - 1 …. so for gamma = 1.4 you get.... Efficiency = 1.4 - 1 = 0.4 …. How can you get 28.6% using one equation, and 40% using the other?.... Which one is correct?....

doing this invalidates your assumption that the energy imparted to the pellet is P0*A*x.

I never said that the energy imparted to the PELLET is P0*A*x.... That is the maximum energy that can be released by the expanding air, and INCLUDES that used to accelerate the gas itself.... That is the main reason we cannot get to more than about 50% of that value using air, or 70% using Helium.... Perhaps I should refer to that energy as "Energy In", rather than "Energy Out".... with the ratio of the two being the efficiency....

I am particularly intrigued by your statement #3, as you have put a "name" to a theory I proposed in this thread and others a couple of years ago.... I invisioned "packets" (parcels) of gas, each one being accelerated by the one behind it.... The average molecular speed within each packet may average 1650 fps, but the whole packet is moving down the barrel, and so can still exert force on the pellet even when the velocity exceeds 1650 fps.... Now I know that is called "non-stationary flow", and I sincerely thank you for that.... Can you put a "number" on that for air, starting at 70*F?.... I assume it varies with pressure and barrel length (minus losses)…. but I made a wild guess that at 70*F it was twice the average molecular velocity of 1650 fps.... ie 3300 fps would be the absolute maximum velocity limit in a PCP using air at 70*F.... Any validity to that?....

Bob

[DomingoT]

Yes, gamma varies significantly as a function of pressure, but assuming gamma is constant doesn't get in the way of the concepts. In fact, whenever one sets up a framework of analysis, it is best to keep it as simple as possible, and a constant gamma is one of the basic assumptions of simplicity without tearing up the physics.
More important, however, is your statement that P0*A*x includes the KE of the gas behind the pellet. It does NOT. Pressure * Area * distance is the work done by the gas behind the pellet ON THE PELLET ITSELF as it travels from the breech to position x. Because this is work done on the pellet, it turns into the pellet's KE, not the gas KE.
In order for the gas behind the pellet to get accelerated and acquire KE, you need a pressure gradient down the bore (pressure dropping along the bore), to produce the accelerating force. In the isothermal assumption, the pressure is uniform behind the pellet, and there is no pressure gradient - as a result, there is no acceleration of the gas behind the pellet. In the isothermal assumption, all there is behind the pellet is thermal energy, no kinetic energy. If you think there is something off here, you are right, an isothermal model doesn't have the flexibility to accommodate what actually happens.
Neglecting the KE of the gas behind the pellet (in a PCP) makes a difference of a couple of Joules in the pellet energy (for a 12 fpe, 4.5 mm gun).

[rsterne]

Neglecting the KE of the gas behind the pellet (in a PCP) makes a difference of a couple of Joules in the pellet energy (for a 12 fpe, 4.5 mm gun).

This seems to fly in the face of reality when dealing with powerful PCPs, as the mass of air used per shot is huge.... My 6mm produces 173 FPE (starting with 4200 psi in a 300 cc reservoir) with a 73.4 gr. bullet, but only 17.5 FPE with a 1.8 gr. airsoft BB.... The mass of air in the 28" barrel (21 cc vol.) at an average pressure of 4000 psi would be about 100 grains.... greater than the weight of the lead bullet.... and over 50 times the weight of the plastic BB....

I used this calculator for Adiabatic expansion to run the numbers for the above shot (scroll down)....

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

Inputs: Pi = 28958 kPa (4200 psi), Vi - 0.0003 m^3 (300 cc), Vf = 0.000321 M^3 (321 cc), Ti = 294 *K (~70*F), gamma = 1.75 (from chart above).... This gives the following results....

n = 3.5539 moles
W = 573 J (423 FPE)
Pf = 25725 kPa (3731 psi)
Tf = 279.5 *K (a drop of only 14.5*C)

The predicted work done to the bullet is 423 FPE, but it ends up with only 173 FPE of energy for the lead bullet, and only 17.5 FPE for the plastic BB.... If a large portion of that 423 FPE does not go into accelerating the air, where is all the extra energy going?.... Incidently, my simple "back of the napkin" calculation gives a MAXIMUM of 454 FPE, assuming Isothermal expansion and an infinite reservoir.... and 423 is less than 454 (by only 7%)....

Bob

[DomingoT]

A great deal of the energy stays behind as heat.
Using the following arbitrary parameters, for illustration only, to get an idea of how the energy gets partitioned.

Pellet weight = 73.4 gr
Pressure = 290 bar
Barrel length = 700 mm
Transfer port diameter = 5 mm
Transfer channel volume = 1.6 cm3 (this is the wasted space between valve and TP)
Gamma = 1.5
Etc.

I get the following,
Muzzle energy = 231 J (170 fpe)
Thermal energy in the barrel = 214 J
Kinetic energy of air in the barrel = 24 J
Energy stuck in the transfer channel = 12 J

About 10% to 20% of the muzzle energy stays in the barrel as kinetic energy of the gas
In this case, about the same amount as the bullet energy stays in the barrel as thermal energy

What is the shot efficiency in the gun you refer to (fpe/in3)?

[rsterne]

I use full bore-area porting, 6mm or equivalent effective area on all ports.... Total port length from valve seat to base of bullet is roughly 1.5", so the total volume would be about 1.1-1.2 cc.... The shots were a "dump shot" as in the valve was still open when the pellet exited the barrel (no additional velocity available no matter how hard you hit the valve).... so no proper FPE/CI can be determined by a simple pressure drop on the reservoir, as some of the air is leaving the valve after Elvis has left the building.... I wasn't particularly fussy about recording the pressure drop, but it was about 300 psi.... That would be about 380 std. CI, so about 0.45 FPE/CI for the bullet, and about 0.045 FPE/CI for the plastic BB....

That 214 J seems like a huge amount of thermal energy left behind.... Would that not increase the temperature of the barrel significantly?.... In reality, PCPs get colder when you shoot them, not hotter.... One of our Forum members with a 20mm gets frost on the barrel after a few shots....

I am wondering about the idea of "packets" of air, each one being accelerated by the air exiting the valve.... Surely that must take energy?.... As the bullet progresses down the barrel, those packets must (somewhat) keep up with the bullet, and by the time it reaches the muzzle, the first packet must be doing the same as the muzzle velocity (1029 fps with the bullet).... In this gun, there is about 100 grains of air in the barrel, so if you divided that into 100 packets, each weighing 1 grain, the front one would have 2.35 FPE of energy.... If they were all going the same velocity, that is 235 FPE.... Even if you only used half the air mass, that works out to 118 FPE, a lot more than 24 J....

I haven't discussed it with you yet, but Lloyd has a PCP Internal Ballistics spreadsheet that adds the mass of the air released by the valve as the bullet moves down the barrel, in increments of 0.00001 sec.... You can toggle that on and off, and when I input the numbers for my 6mm, if I toggle the air mass off, the FPE of the bullet increases from 173 FPE to 289 FPE.... an increase of 116 FPE, pretty close to that guestimate of 118 FPE above.... Without using the air mass being accelerated along with the bullet, the spreadsheet makes NO sense, whether you use Isothermal or Adiabatic expansion (or something in between)…. It has inputs for wasted transfer port volume, bullet sliding and starting friction, barrel caliber and length, reservoir pressure and volume, gas (CO2, air, N2, He), bullet weight, valve dwell, and an "efficiency" (fudge) factor necessary to make it agree with test data.... Once the knowns are input, there is only one combination of dwell and efficiency that will match reality....

For the plastic BB, with a MV of 2092 fps, the first 1 gr. packet of air behind it would have 9.7 FPE of energy.... The last packet leaving the valve could not exceed the SOS at the pressure at the valve throat (~3900 psi), which is about 1470 fps (see chart below), so the maximum FPE of the last packet would be 4.8 FPE (and probably less)…. Some (or all) packets must be expanding as they move from valve to muzzle at these velocities.... 



How would your calculation work with the 1.8 gr. plastic BB at 2092 fps?.... Why did you choose a gamma of 1.5 at ~4000 psi, instead of about 1.75?....

I am immensely enjoying this discussion.... Please don't think that my questioning your numbers is a put-down, I merely want to understand the concepts, and make sure that they agree with our observed data.... If not, I need to understand why....

Bob

[Scotchmo]

Fanno flow is strictly "stationary" flow, this is "not" what you have in a PCP. Think about a situation that would apply to "stationary" 1-D flow and you will see under what conditions your conclusion applies. It is actually interesting that there is a difference between stationary and non-stationary maximum values. This has to do with the transmission of energy between non-stationary fluid parcels, which doesn't happen in stationary flow. It is subtle but important distinction.

"stationary" flow is an oxymoron.

Fanno flow is 2-d flow. It varies radially and along the flow axis.

[DomingoT]

Scotchmo,
In fluid dynamics, "stationary" means steady-state, it doesn't mean quiescent - it is part of the jargon. It comes from the idea that if the flow is in steady state and you look at the streamlines, the streamlines don't move, even though the flow itself moves.  Like any jargon, the word "stationary" entails a bit of idiomatic abuse.
The flow in an airgun is not steady-state, and that is an essential difference in that total enthalpy behaves very differently in steady and unsteady flows. As I said earlier, I have seen experienced people trip on this one.
Fanno flow is steady-state, "one-dimensional" flow. You may be thinking, I suppose, that it has a radial component because Fanno flow refers to flow in a duct with wall friction, and the wall friction induces a radial component of movement.  But Fanno flow refers to a one-dimensional "equivalent" of duct flow with friction - there no radial component. The streamlines of Fanno flow and the streamlines of flow in a duct with friction do NOT coincide - even though both have the same cross-sectional mass, momentum and energy fluxes.
That said, some people may refer to Fanno flow "quasi" one-dimensional, but this is a misnomer; there are no "radial" dependencies in Fanno's relationships.

[DomingoT]

Bob, it seems my last reply to your last post dissolved in cyberspace, don't see it anywhere...
The lift and dwell of the valve are the linchpin in capturing the internal dynamics. If you know these, computing the mass and energy transfers is straightforward. From your comments much earlier (which I hadn't fully read when I first started posting), it is clear you have a solid understanding of where the difficulty lies - we don't have a "simple" way to compute the pressure on the poppet downstream face (the face away from the reservoir). And, as it turns out, the whole thing depends crucially on this pressure!
There is something called the instantaneous mixing assumption. If you assume that the gas entering the transfer channel mixes instantaneously with the air that is already stationed there, you can use the pressure in the transfer channel as a  proxy for the pressure on the downstream face of the poppet. This assumption is open to criticism, but it allows you to close the problem, and get the dwell and lift as part of the solution.
In addition to this difficulty, there is the issue of the precise mechanism whereby the hammer transfers momentum and energy to the valve. Here again, to make the problem manageable, we have to introduce assumptions. The assumption I use is that the time scale of the elastic deformation of the valve stem (the "give" of the stem) under the impact of the hammer is "much smaller" than the dwell time and the time it takes the poppet to lift itself from its deflected seating area. There is a bit of math in justifying these assumptions, you can find the details in my paper "Internal Ballistics of a PCP Airgun", which I posted a few days ago on Researchgate and Scribd (search "internal ballistics PCP airguns researchgate" and it should pop up) - I don't write the URL in case that was the reason my previous post vanished =)
If I run my calculations with a 1.8 gr pellet (!) I get a velocity of about 2,300 f/sec, but these are just "toy" numbers, really, because I am not using the right parameters for your gun.
One other question: Does your gun experience hammer bounce? I am intrigued by the amount of air you quote.