GEEK Alert - Sonic Choking!

[lloyd-ss]

Some of us need to go back and complete our Bachelors first then maybe I will understand what I need to know to comprehend what you guys are disseminating here 
All kidding aside I have really enjoyed trying to follow along on this thread and I'm actually planning to attend the spring semester at my local SUNY
for some refresher math and mechanical engineering classes on fluid dynamics

Don, more power to you! Learning something new is about as healthy as it gets!  But at my age, I think going back of school school would be torture, LOL, so my hat is off to you.  Still, if I don't learn something new every day, I feel like I missed an opportunity. 
With a lot of this stuff, after you sift through what is presented, it boils down to common sense.  At least in hind-sight, LOL.  I am glad we have so many smart contributors and people asking questions.  The free flow of information is great.  I worked the last half of my career at a place that designed, built, and tested, high end inertial navigation systems.  I was fortunate enough t work with some REALLY smart people and whatever I needed to know, somebody there could help me. 
Lloyd 

[rsterne]

Lloyd, as always I am impressed by your ability to cut to the chase!.... It doesn't really matter if the transfer port is actually choked because the 1.9:1 DeltaP is reached or not, once the velocity reaches Mach 0.99, the flow rate through it flattens out at nearly the same value as if the port was choked, and can never exceed that value.... I also observed that with a 75% of bore transfer and barrel port, which is about as big as you can go without resulting to oval barrel porting or an axial flow design, that the transfer port velocity approaches Mach 1 at roughly the 5" point, which is a good point for the valve to close to maintain good efficiency.... It is interesting, however, how small an increase in port diameter is needed to push that distance much further down the barrel, and of course with bore-sized porting, you should never approach a choked condition even with the valve open to the muzzle if you keep the muzzle velocity in the mid 900s....

I'm going to do up a couple of diagrams showing where the velocity should, in theory, get to Mach 1 with some different size transfer ports so that everyone can compare them to the ones I already posted on the previous page.... That post compared WFO to choking occurring at 400 fps, which for a .22 cal Disco would require a smaller than stock transfer port of 1/8".... That value comes from: (0.125/0.217)^2 x 1207 fps = 400 fps.... I used 1207 as the average value for Mach 1 for a Disco because that is the speed of sound in air at 1500 psi.... Here are the graphs for a stock .22 cal Disco transfer port of 0.140", which would in theory choke at (0.140/0.217)^2 x 1207 = 502 fps....



You will notice that at 2000 psi, the valve closes before the pellet reaches 502 fps.... At 1600 psi, any choke would form just as the valve is closing.... but at 1200 psi, at least in theory, the choke would exist for about 1" of barrel travel.... I don't know how to properly represent what happens to the pressure in the barrel after the valve closes, but the flow through the port essentially stops at that point and all that is happening is that the air in the ports and barrel, between the valve seat and the pellet, expands to continue accelerating the pellet.... All the diagram represents is the time between a possible choke forming and the valve closing, and it would seem that during that time period (and possibly afterwards?) we have reduced acceleration occurring....

One other thing I just thought about.... A stock .22 cal Disco gets significantly louder about half way through the shot string.... Is it possible that increase in noise is partly the "sonic boom" from a choking event.... or is it just the increased residual muzzle pressure we are hearing?....

Bob

[rsterne]

Here is another way of thinking about the importance of the transfer port size.... I am going to be adding several examples to this post as I draw the graphs, for some typical PCPs, and will put a title above the graph and comments below it for each graph I add.... I'm going to start with a regulated .30 cal PCP designed for 90 FPE and using a 90 cc plenum with a 2000 psi setpoint, where the valve is closing at 50% of the 24" barrel....

.30 cal Regulated - 50% Closing point - 950 fps MV



The vertical green lines represent the theoretical point in the barrel where a choke would form at the transfer port for the sizes given (as % or boresize).... Note that an 80% or larger port should not choke in this gun....

.30 cal Regulated - 33% Closing point - 909 fps MV



Reducing the dwell so that the valve closes at 33% of barrel length loses 41 fps, but now a 75% transfer port shouldn't choke.... This would probably be somewhere on the "knee" of the curve for this gun, and a pretty efficient tune, showing why the 75% transfer port is a good number for much of what we do....

.30 cal Regulated - 25% Closing point - 870 fps MV



Reducing the dwell further, so that the valve closes at 25% of barrel length, drops another 39 fps off the velocity, and now a 70% of boresize transfer port should be OK....

.30 cal Regulated - 100% Closing point - 979 fps MV



Going the other way, and keeping the valve open until the pellet has just left the barrel, only gains 29 fps, and would need a 90% transfer port to not choke.... Incidently, this tune uses about 75% more air than the 50% tune to produce that 29 fps gain.... an increase in FPE of just 6%....

I'm going to shift now to a .257 cal at 3000 psi with a 28" barrel, and the baseline will be a 72 gr. bullet at 960 fps with the valve closing at 50% of barrel length....

.257 cal 72 gr. - 50% Closing point - 960 fps MV



The higher pressure has increased the speed of sound to 1360 fps, allowing a 75% port to nearly remain unchoked at the point the valve closes (in theory).... Now, instead of changing the dwell, I'm going to change the bullet weight (a lot) to change the velocity, which because the bullet is moving faster or slower will change where the bullet is when the valve closes.... Added NOTE: Even though this gun is a lot different than the one above, with the valve closing at 50%, the diameter of the port when the choke occurs at the moment the valve closes is within a pretty narrow range (76-79%).... If the same velocity was used for the speed of sound in both examples, that difference would have disappeared....

.257 cal 116 gr. - 33% Closing point - 750 fps MV



Notice that because the velocity is so much lower, we shouldn't see choking even with a much smaller transfer port.... That doesn't mean that we wouldn't lose power by putting in a smaller port, only that choking is no longer an issue....

.257 cal 36 gr. - 84% Closing point - 1230 fps MV



Still the same valve dwell, but with half the bullet weight we started with, the gun is now shooting supersonic.... and the valve is open nearly until the bullet leaves the muzzle.... It now takes a 95% transfer port to avoid choking.... Wait a minute, we said the bullet was supersonic, how come the barrel isn't choked?.... That's because (in theory) the speed of sound in the 3000 psi air is 1360 fps....

I think that is enough examples, this was really only a side trip to look at what minimum transfer ports make sense.... and the discussion has moved on.... Added NOTE: It would appear that the primary factors in determining what size port is required to not choke are the muzzle velocity and the point in the barrel where the valve closes.... Of secondary importance is the operating pressure of the system, which is determined by both the starting reservoir pressure and it's volume, which determine the pressure in the barrel, and hence the speed of sound, at the moment the valve closes....


COMMENT: After perusing the above and trying some more examples, I have realized that one generic chart stands out as a point of interest....

Muzzle Velocity 950 fps - Valve closing at 33% - Port diameter 75% of boresize



While not exact for every gun, depending on the operating pressure and smoothness of flow, there is one useful piece of information to be had from this chart.... With the commonly used transfer and barrel port size of 75% of the caliber, any dwell over 33% of the barrel length will take a hit on efficiency in a gun shooting at 950 fps....

Bob

[Jim_Hbar]

 

the barrel is now filling at 1/4 the rate it was before M1 was reached. (I think that is correct, but please correct me if I am wrong.)

If upstream pressure is constant, the restriction determines what the maximum mass flow rate through the port can be..  As the velocity through the restriction increases, the delta P across the restriction also increases, and the mass flow rate through the port increases to the value given by density, mach1 and the port area.  It can never be higher, theoretically. (Btw, when people say theoretically, they usually really mean "simplistically - but that's another discussion.)

The stagnation pressure downstream (and thus the energy in the air) has been decreasing rapidly as the maximum mass flow rate approaches.

From my very first post in this thread!

When the flow is choked, that restriction determines the mass flow rate of the system, until the pressure ratio is reached.
(deleted)

When the ports are big enough - What this calculated mass flow rate will do, is put an upper limit on the acceleration possible from the pellet - and that's how I likely will use it.
That's all - pretty small potatoes..

Thinking through it a bit more and crunching a few numbers, I agree that the choke has very little practical effect at the velocities we are talking about, if the ports are reasonable.  If they are too small, I'm sure it does have some effect.

And of course, it applies after the pellet leaves the barrel, but that's not interesting - jokes over by then!

A dump shot should be affected by the choked flow if the barrel is long enough, but perhaps not as much some would think.
If you look at the last four panels on that Wiki diagram, the velocity downstream of the orfice is super-sonic in what could be considered the barrel.  Intuitively, people under stand case A & B, and perhaps will go as far as C & D, but rarely E, and the fact that the flow can be super-sonic well beyond the nozzle as in F & G just goes beyond intuition.
Especially when considering that the flow is only sonic at the restriction.
To go through an orfice and then speed up?
But that's what happens, and the flow is choked.

BTW, as upstream pressure is increased relative to the downstream pressure, the flow regimes move down that chart.
The standing waves shown at the outlet (in the last panel), are related to the shock diamonds in the outlet of a jet or rocket engine.

Lloyd, the way I would use all this "choke" information, is calculate what the maximum mass flow rate (Mmax) into the barrel can be, as determined by the restriction area, or the caliber of the barrel. 
Then I would go through the time iterations/numerical method calculations, and calculate the massflow rate (Mr) at the restriction/barrel entrance.  Once Mach "C" is reached in the port (a tweak point, start at perhaps .85??), assume that the calculated mass makes it through the restriction, but the pressure drop across the port is increasing by a linear function of the velocity in the port. 
Say (Pu-Pd)=(1-.53)(Vn-C)/(1-C)
Where Pu = Pressure Upstream, Pd = Pressure downstream, Vn is velocity in the nozzle expressed as Mach#, C is the Mach number to be tweeked where things start to get messy.
Once Vn = 1, Pu = 1.893Pd, Mr = Mmax there on out..

Jim

[rsterne]

Jim, thanks for confirming that the mass flow rate limit is the governing factor, and even more thanks for suggesting a way to predict the downstream pressure once Mach is approached....

This begs the question, how do we calculate the maximum mass flow rate?.... I assume the critical factors are the reservoir pressure and the area of the restriction?.... Other than environmental factors (temperature/others?) likely the molecular weight of the gas (eg. He, N2, Air would be different) and/or the speed of sound for that gas at the pressure used?....

And then, of course, the other question is what happens to the pressure in the barrel after the valve closes.... I assume it then simply expands, starting from the last value used before the valve closed, as the pellet continues towards the muzzle?....

Bob

[Jim_Hbar]

Bob - For the maximum mass flow rate, I would just start with reservoir pressure, air density at that pressure, Mach 1 at that pressure, and the area of the restriction.  Mass flow simply = density x velocity x area x (some constants to make the units work out)  Then de-rate that mass flow by 5 or 10% or so to account for pressure drops to the restriction, sharp edge orifice effects, boundary layers, etc... 
It's going to be very close!

[lloyd-ss]

the barrel is now filling at 1/4 the rate it was before M1 was reached. (I think that is correct, but please correct me if I am wrong.)

If upstream pressure is constant, the restriction determines what the maximum mass flow rate through the port can be..  As the velocity through the restriction increases, the delta P across the restriction also increases, and the mass flow rate through the port increases to the value given by density, mach1 and the port area.  It can never be higher, theoretically.........

/quote]

Lloyd, the way I would use all this "choke" information, is calculate what the maximum mass flow rate (Mmax) into the barrel can be, as determined by the restriction area, or the caliber of the barrel. 
Then I would go through the time iterations/numerical method calculations, and calculate the massflow rate (Mr) at the restriction/barrel entrance.  Once Mach "C" is reached in the port (a tweak point, start at perhaps .85??), assume that the calculated mass makes it through the restriction, but the pressure drop across the port is increasing by a linear function of the velocity in the port. 
Say (Pu-Pd)=(1-.53)(Vn-C)/(1-C)
Where Pu = Pressure Upstream, Pd = Pressure downstream, Vn is velocity in the nozzle expressed as Mach#, C is the Mach number to be tweeked where things start to get messy.
Once Vn = 1, Pu = 1.893Pd, Mr = Mmax there on out..

Jim

Jim,
Thanks for the insight you have provided!  Your statement that the max mass flow is dependent upon density, Mach 1, and the port area is very helpful in explaining the rapidly decreasing efficiencies of high velocity shots.   I was a bit surprised at how quickly Mach1 could be reached in the T-port.

I am having a little trouble with the concept of Pu/Pd=1.893.  I could see where this might be true in an open system, but the PCPs we are working with are closed systems.  Does the Pu/Pd=1.893 apply for closed systems as well as open systems?  In a closed system, the application might be different.  Here is my reasoning.  Until the pellet leaves  the barrel, the PCP airgun is a closed by the pellet in the barrel.  The flow of the air that is released by the valve is basically pushing against a cork in the barrel.  That cork consists of the friction of the bullet along the barrel walls, the inertia of the bullet, and the steadily increasing inertia of the column of air that is accelerating behind the pellet. It should be noted that the mass of the air column can be significant percentage of the weight of the pellet and must be included in the calculation.

As the air vel in the T-port approaches M1, Pu/Pd = 1, and delta P is close to zero.  Even when M1 is first reached, delta P =0.  And as the mass air flow stays at the steady, pressure dependent  maximum rate, the bullet keeps the air from  expanding to much. The amount of "make-up" passing into the barrel can be calculated, and its expansion against the restriction of the bullet can also be calculated.  I  think the ration of Pu/Pd will be closer to 1 than to 1.893, but I could be wrong.  Could that be the case?
Lloyd

[rsterne]

OK, well using the Metric system that looks pretty easy.... Density in kg/m^3.... Velocity in m/sec.... Area in m^2.... multiply them all together and you get kg/sec.... So at 2000 psi, the air density is 164.4 kg/m^3, the speed of sound is 1252 fps (381.6 m/sec), and let's use a 1" diameter, which gives an area of 5.067 cm^2.... and there are 10,000 cm^2 in one m^2.... so we have....

164.4 x 381.6 x 5.067 / 10,000 = 31.79 kg/sec.... That is the mass flow limit at 2000 psi through a 1" diameter.... Did I get the math right (other than the derating)?....

Assuming I did, then multiplying that number by the caliber squared (in inches) should give the mass flow limit at 2000 psi.... so for a .250" cal (or port) you would have....  31.79 x 0.250 x 0.250 =1.987 kg/sec at 2000 psi....

Bob

[Cal]

the barrel is now filling at 1/4 the rate it was before M1 was reached. (I think that is correct, but please correct me if I am wrong.)

If upstream pressure is constant, the restriction determines what the maximum mass flow rate through the port can be..  As the velocity through the restriction increases, the delta P across the restriction also increases, and the mass flow rate through the port increases to the value given by density, mach1 and the port area.  It can never be higher, theoretically.........

/quote]

Lloyd, the way I would use all this "choke" information, is calculate what the maximum mass flow rate (Mmax) into the barrel can be, as determined by the restriction area, or the caliber of the barrel. 
Then I would go through the time iterations/numerical method calculations, and calculate the massflow rate (Mr) at the restriction/barrel entrance.  Once Mach "C" is reached in the port (a tweak point, start at perhaps .85??), assume that the calculated mass makes it through the restriction, but the pressure drop across the port is increasing by a linear function of the velocity in the port. 
Say (Pu-Pd)=(1-.53)(Vn-C)/(1-C)
Where Pu = Pressure Upstream, Pd = Pressure downstream, Vn is velocity in the nozzle expressed as Mach#, C is the Mach number to be tweeked where things start to get messy.
Once Vn = 1, Pu = 1.893Pd, Mr = Mmax there on out..

Jim

Jim,
Thanks for the insight you have provided!  Your statement that the max mass flow is dependent upon density, Mach 1, and the port area is very helpful in explaining the rapidly decreasing efficiencies of high velocity shots.   I was a bit surprised at how quickly Mach1 could be reached in the T-port.

I am having a little trouble with the concept of Pu/Pd=1.893.  I could see where this might be true in an open system, but the PCPs we are working with are closed systems.  Does the Pu/Pd=1.893 apply for closed systems as well as open systems?  In a closed system, the application might be different.  Here is my reasoning.  Until the pellet leaves  the barrel, the PCP airgun is a closed by the pellet in the barrel.  The flow of the air that is released by the valve is basically pushing against a cork in the barrel.  That cork consists of the friction of the bullet along the barrel walls, the inertia of the bullet, and the steadily increasing inertia of the column of air that is accelerating behind the pellet. It should be noted that the mass of the air column can be significant percentage of the weight of the pellet and must be included in the calculation.

As the air vel in the T-port approaches M1, Pu/Pd = 1, and delta P is close to zero.  Even when M1 is first reached, delta P =0.  And as the mass air flow stays at the steady, pressure dependent  maximum rate, the bullet keeps the air from  expanding to much. The amount of "make-up" passing into the barrel can be calculated, and its expansion against the restriction of the bullet can also be calculated.  I  think the ration of Pu/Pd will be closer to 1 than to 1.893, but I could be wrong.  Could that be the case?
Lloyd

Great contributions!

It is difficult to pull the important details from the possibilities.

The pell must see pressure at it's base in order to satisfy acceleration equations.  Propellent velocity does no good.
Have you ever seen an internal ballistics calculation for PB based on propellent velocity? 
F=ma and all that.....


Still,  The concept of "choking", seems to mean different things to different people in this discussion.

If the mass flow is set by "choked conditions", what effect will that have if the pell velocity is a function of propellant velocity?

Can you see the disconnect?

Why do light gasses propel with more velocity?  because they show up with higher pressures more rapidly!

[Jim_Hbar]

Lloyd:

The air won't flow without a pressure drop (speaking in stagnation pressure terms, as I always do) - in other words, there is a constant loss of energy, (or work done by the air), to pass from the high pressure area to the low(er) pressure area.  If there is no pressure drop, there is no flow....  The greater the flow, the greater the pressure drop required to sustain that flow, everything else being equal.  Until you hit the wall and can't push any more through the hole......

Here's a web page to look at - for simplicity sake, they use the value of 2 instead of 1.893, (and 50% instead of 53%)...  Cv is measure of the flow capacity of a valve or component in a system, and is how industrial pneumatic flow capacities are rated. 
But the chart illustrates the point I'm be-laboring.

[lloyd-ss]

First off, I want to thank everyone for their patience with me on this.  I know it is a pain when someone comes late to the discussion and then presents ideas and questions that have already been flogged to death.  My apologies.
I am sure that I am misapplying some terms and that I have some semantics issues, but my guess is that we are not as far apart as it seems.

Cal, yes, I realize that pellet velocity (or pellet acceleration) is dependent on the force from the pressure on its cross sectional area, and not the velocity of the air behind it.  I might not have stated that very clearly.  BTW, where I live, the VT at the bottom of your posts can only mean Virginia Tech. Is that by any chance the case?

Jim, My background is more straight mechanical so the fluid mechanics are a little less intuitive to me and I am sure my terminology is a bit off, so I appreciate your persistence and patience.  I am familiar with the C sub V factor (at least for solenoids), but, like in calculus, there is a difference in being able to work the problems and actually "understanding" the concepts. I can fumble through the problems, but you understand what is happening.

Let me try another slightly different explanation of  my visualization of what is happening as the air reaches M1 in the T-port. Let me know where I am going off course. Thank you in advance for the help.
For simplicity sake and to help me better understand what is happening I would like to use some real numbers. Let's assume a reservoir of infinite volume filled to 1000 psi.  The T-port is the main restriction and is one half the diameter of the barrel.  The pellet is light enough to achieve high velocity at this pressure.
1) After the valve opens, and the pellet starts accelerating, the velocity of the moving column of air quadruples as it passes through the T-port, and then drops back down on the pellet side of the T-port. As long as the vel through the T-port stays below M1, the reservoir is able to keep the barrel filled behind the pellet such that the barrel and reservoir are essentially the same pressure. 
(Is that explanation correct?)
2) As the pellet continues to accelerate, the air passing through the T-Port continues to fill the barrel faster and faster, and quickly reaches M1 and thus its mass flow limit. This is where I am getting hung up.  Just before reaching M1, the pressure on either side of the T-port is close to being the same. (Correct??) 
3) When M1 is reached the air flow doesn't stop, but continues to fill the barrel at a constant rate equal to the mass flow limit. (Correct?)
4) The pellet continues to accelerate, but because the fill rate is now a constant and cannot keep the barrel filled to 1000psi, the pressure in the barrel starts to drop.  But the pressure drop is gradual and doesn't just plummet to 53%, or 530psi. (Correct?)
5) Or, does the air flow through the T-port basically stop until the pellet has moved far enough to let the barrel pressure drop to 530psi?

I am confused. Please help! Thanks,
Lloyd

[Sfttailrdr46]

  Lloyd let me say that your confusion and hard work with Cal and Bob working to clarify it is helping me to finally grasp what the significance of the thread and you guys goals achieve. Scarily it is beginning to make sense to me and the Impact on High performance tuning of our favorite toys

[Jim_Hbar]

I'm going to be busy for the next couple of days, except in the evening.  Have an appointment in an hour, so this is going to blunt, with little explanation.

1) After the valve opens, and the pellet starts accelerating, the velocity of the moving column of air quadruples as it passes through the T-port, and then drops back down on the pellet side of the T-port. As long as the vel through the T-port stays below M1, the reservoir is able to keep the barrel filled behind the pellet such that the barrel and reservoir are essentially the same pressure. 
(Is that explanation correct?) No - intially 1000 psi gets there quickly, as there is no flow, but starts dropping as flow increases
2) As the pellet continues to accelerate, the air passing through the T-Port continues to fill the barrel faster and faster, and loose more and more pressure and quickly reaches M1 and thus its mass flow limit. This is when the pressure downstream of the port is 530 psi I am getting hung up.  Just before reaching M1, the pressure on either side of the T-port is close to being the same. (Correct??)  No
3) When M1 is reached the air flow doesn't stop, but continues to fill the barrel at a constant rate equal to the mass flow limit. (Correct?) Yes And there will actually be a pressure drop from the port to the pellet, if the pellet is a significant distance down the barrel.
4) The pellet continues to accelerate, but because the fill rate is now a constant and cannot keep the barrel filled to 530psi, the pressure in the barrel starts to drop even more.  But the pressure drop is gradual  (and doesn't just plummet to 53%, or 530psi.Correct?) It was already at that level at step 3

Study the page and the table I linked to in my last post - It just happens to be the first table that popped up in a Google, that had the info I was looking for - I'll look for a better explanation/link tonight.

[rsterne]

Lloyd, you are struggling with EXACTLY the same problems I have (and still am).... I agree 100% with your first point, and most of the second, right up to the point where you say the pressure downstream of the restricition is basically equal to the upstream pressure right up to the velocity at the restriction reaching Mach 1.... That seems inconsistent with the requirement post choke for it to (suddenly) be only 53% of the upstream pressure (sudden discontinuous change in pressure seems unrealistic).... I agree 100% with your point 3, and I would love to believe that point 4 is what happens, but I think Jim will shoot it down as the 53% drop across the restriction is apparently a condition of the velocity reaching Mach 1.... Your point 5 seems unrealistic as well, of course, not only counter-intuitive, but involves stopping and starting the flow of a large mass of air....

Cal, I likewise realize that the propellant velocity is NOT what drives the pellet, it is pressure at the base of the pellet that does that, (which in the end is just rapidly, and mostly randomly, moving air molecules bouncing around), and the applicable formula is indeed F=ma.... That is EXACTLY where the disconnect in our reasoning is.... I think we can all grasp (and agree with) the concept of the mass flow reaching a limit when the flow at the restriction reaches Mach 1.... What Lloyd and I are having trouble with is the huge and sudden drop in pressure that accompanies that.... As I stated in one of my earlier posts, if that drop was sudden, like turning off a switch, then at that instant, the volume of air at the original pressure between the restriction and the pellet would simply expand (backwards) until the pressure was equalized.... and that would instantly cancel the choke.... Yet the choke must occur once the velocity reaches Mach 1....

Jim, once again thanks for your input.... in particular this sentence in response to Lloyd's point 1....

No - intially 1000 psi gets there quickly, as there is no flow, but starts dropping as flow increases

Reading this, to me, was the lightbulb moment.... At the instant the valve opens, but the pellet is stationary, there is virtually NO flow, those randomly bouncing around air molecules are just raising the pressure between the valve and the pellet until it starts to move, reaching 1000 psi very quickly.... Then the pellet begins to accelerate down the barrel, expanding the volume behind it, and the air from the reservoir keeps that volume full, but with enough pressure gradient to keep the air flowing through the barrel and ports as fast as the pellet is moving away from the valve.... When we place a restriction between the valve and the pellet, we begin to generate an additional pressure gradient across that restriction, which we call DeltaP.... This pressure gradient increases as the velocity through the restriction approaches Mach 1, eventually reaching 1.893:1 when the velocity reaches Mach 1, and the flow is then fully choked.... I add the word "fully" because this has been an exponential process, becoming noticeable at about Mach 0.8 (where the transonic region begins) and quickly building from there to Mach 1, when the flow is officially "choked"....

I do have a question, and that is in regards to the statement that the pressure at the pellet will be even lower than 530 psi.... IIRC, you stated previously that it is the DeltaP over the entire system that must be >1.893:1, so if that is the case, would not the pressure at the pellet be 530 psi when the choke first occurs?.... As the pellet continues to accelerate, and with constant mass flow input, why does it have to drop further?.... There is still air being supplied by the valve, in abundance, and the downstream pressure has no affect over the flow rate in a choked port.... Could not the pressure just downstream of the restriction be, say, 600 psi at the moment the choke forms, and increasing to, say, 650 psi as the pellet moves away from it, to keep the pressure gradient in the barrel, and yet maintain the DeltaP across the system because the air at the pellet is still only 530 psi?.... I know this is probably insignificant, but I want to understand as it is important that we can calculate the pressure, hence the force, accelerating the pellet....

Bob

[rsterne]

I had a look at the page that Jim linked to, and the chart clearly shows the mass flow (SCFM) limit when the outlet pressure drops below half the inlet pressure.... The formula given intrigues me, however....

Q = 0.6875 x (P1-P2)^0.5 x (P1-P2)^0.5.... which simplifies to Q= 0.6875 x (P1-P2).... which doesn't agree with the chart, as any difference of pressure of (for example) 10 psi should have the same flow in SCFM.... I therefore believe that equation must have been misprinted.... If the correct formula can be found, then would it be possible, using a mass flow calculation, to work it in reverse and calculate the pressure difference across the restriction as the velocity increases?.... We need to get a feel for how the DeltaP builds as the flow through the restriction approaches Mach 1.... Linear?, Exponential? (and to what power?)....

Bob

[lloyd-ss]

Bob,
I noticed that problem with the formula, too.
Jim,
Thank you for the answers. They are tremendously helpful.  The understanding that the entire air circuit is a continuum of pressure gradients that vary based on air velocity and air volume and passage and port shape, etc, etc, is the lightbulb moment for me.  Very helpful, now I just need to apply it.

For what its worth, a single shot from a .22 pcp might use 22 cuin of air at std pressure, delivered in 0.002 seconds of valve open-time. (The pellet might stay int he barrel for .004 seconds, though.)  That converts to about 390 scfm.  And a valve or orifice in the circuit might have a Cv of 0.2 (big guess).  Maybe a pressure drop of 300 psi is fully acceptable. I do not know if that pressure drop is reasonable or achievable.

Lloyd

[rsterne]

Since nobody disputed my formula for the Mass Flow Limit in Post #107 above, I decided to work with it and came up with the following.... First, a repeat of the Speed of Sound chart I came up with earlier in this thread.... http://www.gatewaytoairguns.org/GTA/index.php?topic=66866.0 .... Note that the values for Air were extrapolated using the geometric average for N2 an O2 in the ratio 78:21....



Next, using the same calculator at http://www.peacesoftware.de/einigewerte/einigewerte_e.html I determined and graphed the Density at 20*C vs. Pressure as follows....



I then used the formula in Post #107 above to calculate the Mass Flow Limit through a 1" diameter and plotted it vs. Pressure as follows....



What a nice surprise that it turned out to be so close to a linear function that we can consider it as such, particularly since we will have to derate the value slighty to allow for turbulence, etc., which will induce a greater uncertainty than any non-linearity.... We can now safely state that for our purposes the Mass Flow Limit through any given orifice is directly proportional to the upstream pressure.... I took this one step further, and divided the Mass Flow Limit by the Pressure and converted to grams instead of kg. and came up with the following chart to show the slight variation in the value of the "constant"....



Over the range of interest (1K-4K) it varies from 15 to 17.4, with an average value over that range of 16.3.... If we derate that by 8%, we get a value of 15 g/sec/psi, in other words the Mass Flow Limit is 15 grams/second for each psi of upstream pressure on a 1" orifice.... I think that should be a usable value for further calculations, unless somebody can blow a hole in my math....

ADDITION: Using that figure, at 2000 psi, in a time of 1 millisecond in a .22 caliber barrel, we would see about....
15 g/sec/psi x 2000 psi x 0.001 sec. x (0.22")^2 = 15 x 2 x 0.0484 = 1.45 grams/millisecond for the MAXIMUM flow rate possible....
Using Lloyd's example above, 22CI of air at std. pressure is 360 cc, and has a mass of 360 x 0.001204 = 0.433 grams spread over 2 milliseconds, for an AVERAGE flow rate of 0.22 grams/millisecond.... It certainly seems possible to reconcile these two figures....

Bob

[PakProtector]

from the 'E' panel...Mach Cones are the visible effects of the shockwaves... One of my favourite visuals.
http://www.aerospaceweb.org/question/propulsion/q0224.shtml
cheers,
Douglas

[Jim_Hbar]

The understanding that the entire air circuit is a continuum of pressure gradients that vary based on air velocity and air volume and passage and port shape, etc, etc,

That's a very good description! I wish I would of thought to express it like that earlier.

Bob - I wouldn't trust that link I put up.  It's just the first table that I found that demonstrated the point I was trying to make at the time....  It makes no sense to express a formula in that form (two square roots of the same term multiplied together?) , and therefore, I suspect it must be in error.....  The 40PSIG Column also has goofy numbers, btw.  It was copyrighted in 1990, so you would think they would had gotten around to fixing the boo-boos by now.

Unfortunately, any books that I may have still are at our other place, so I can't even go looking for information..
I used to rely on a particular binder of information from a certain manufacturer for my calculations, however, they now have converted that approach and information to an application that is sold for a price..

Try this one for formulas - it looks better.  And it is the explanation for this calculator http://www.generant.com/cvcalc.aspx

And here's another to read http://www.swagelok.com/downloads/webcatalogs/EN/MS-06-84.PDF

[lloyd-ss]

Jim,
That Swagelock tech bulletin is pulling it all together for me.  Thank you!
Lloyd