Gas dynamics in the tuned exhaust of a high-performance 2T engine

I tried to start this discussion in another thread, but I was told that the question was inappropriate. Thus I open this new thread to discuss. For the original thread, see here: [Vous devez être inscrit et connecté pour voir ce lien]

I would like to treat everybody here as adults. Please do not respond with comments like "this question is too difficult" or "this question is uninteresting to other forum members". If you wish to refer to text books, please feel free to do so but please include the bibliographic information (*). If you wish to refer to journal papers or conference proceedings, please feel free to do so but again, please provide bibliographic. And also, please do not upload materials unless you are certain that the copyright allows you to upload the material.

In a forum contribution, the following is stated by an expert tuner:

Citation :

When the exhaust port opens, a pressure pulse starts moving through the exhaust pipe. It is reflected at the end cone and it should be back at the cylinder just before the exhaust port closes. Next a part of this reflected pulse bounces off the partly-closed exhaust port and a residual pulse starts moving down the exhaust pipe. This residual pulse too is reflected by the end cone and starts moving back to the cylinder. Ideally it will arrive at the exhaust port just when the port opens again. Then the cylinder pressure and the pressure of the residual pulse combine their energy and the resulting pulse will be stronger than the pulse from the previous cycle. And the combined pulse from the next cycle will be stronger still, and so on; we have achieved true resonance.

Some may argue that we want a low pressure in the exhaust pipe when the port opens because then the spent gases will experience less resistance while leaving the cylinder. But that is not true. Gas flow depends on a pressure ratio. But once that ratio reaches 2, the flow velocity will reach Mach 1, the speed of sound. Raising the pressure ratio any further will not raise the flow velocity any further. The cylinder pressure at exhaust opening can be as high as 11,7 bar and the pressure of the reflected pulse will then be about 2,6 bar. Thus the pressure ratio is well above 2, so lowering the pressure in the exhaust duct outside the cylinder will not do any good to the flow.

What has the exhaust timing got to do with the 'true resonance' I mentioned above? The initial pulse starts moving at Exhaust Opening and it has to be back at Exhaust Closing, or a little earlier. This pulse travels with the speed of sound and its journey up and down the exhaust pipe will take t seconds. The residual pulse starts moving at Exhaust Closing and it has to be back at the next Exhaust Opening. This pulse also travels with the speed of sound and its journey up and down the exhaust pipe will also take t seconds. So from EO to EC takes t seconds and from EC to EO also takes t seconds. In English: the exhaust port should be open just as long as it should be closed. Assuming that the crankshaft rotates with a uniform speed, this means that the crank angle during which the exhaust is open must be equal to the crank angle during which the port is closed. So both angles must be 180°.

I developed this line of thought some 40 years ago, but when I first published it in 1978 (in the motorcycle magazine Moto73 of which I was the technical editor) everybody called me crazy. Some people still do, but I got used to it.

I would like to hear from other experts about the above theory. I am especially perplexed with the sentence: "Then the cylinder pressure and the pressure of the residual pulse combine their energy and the resulting pulse will be stronger than the pulse from the previous cycle. And the combined pulse from the next cycle will be stronger still, and so on; we have achieved true resonance."

In my limited understanding of the exhaust tuning, the ideal case would be as follows:

- the exhaust port opens and a high-pressure wave starts travelling into the exhaust chamber
- the high-pressure wave reflects at the end of the chamber and the "echo" travels back to the cylinder
- the exhaust port should close just after the pressure wave has travelled into the cylinder for optimal "plugging effect".

In other words, the best filling of the cylinder is achieved when the port closes just __after__ the pressure wave has arrived in the cylinder. The pressure wave would be "trapped" in the cylinder, and the secondary echo would be small or even absent. Therefore I am confused about the need for the secondary echo, and I am confused about how the "energy combines"?

I guess my question is the following: in a tuned exhaust system for a high-performance 2T engine, what is the main objective of the tuned exhaust? Is the objective to reflect the pressure wave and then trap it in the cylinder, to increase the mass of air in the cylinder, or is the objective to set up a resonance? If the objective is to set up a resonance, then how does this cylinder - exhaust resonance behave with respect to the resonance of the air in the cylinder and crank case (and potentially, the air box)? On physical grounds, I would expect that the overall resonance frequency of the engine system (crank case - cylinder - exhaust) is not determined by the exhaust timing alone.

(*) bibliographic info: sufficient information so that anybody can find the material. For a book, list authors, title, publisher and year; for journal papers etc, list authors, title, journal title, volume number, page numbers, year.

Only one flaw in your thought experiment I can see after first reading: The closing exhaust port is more than halfway closed by the piston already during the plugging phase so the wave does not see an open end but a restricted pipe end and a large portion of the wave is reflected back out into the pipe. It does not get absorbed into the cylinder.

Yes, I also thought about that: the port is closing when the pressure peak returns at the cylinder.

I have no experience, but I expect the following: when the exhaust port opens, the pressure inside the cylinder is high. When the port opens, a steep pressure wave will start moving into the exhaust. The pressure in the cylinder decreases rapidly, so that the pressure wave has a steep front and a slightly "tapered" back. This "steep front" will remain because the high pressure causes high propagation velocity. Thus the "echo" will have the same "steep front, tapered back."

So how should I interpret the conventional story about the "plugging"? I would say that the high pressure pulse should (at least partly) enter into the cylinder so that there can be a mass flow from the exhaust back into the cylinder, and the exhaust should close when the maximum amount of mixture is pushed back into the cylinder. Or perhaps more specifically: as long as the pressure in the exhaust is higher than in the cylinder, mass flow will go into the cylinder. Thus the best moment to close the exhaust is when the pressure in the cylinder and exhaust are identical.

Is the above the correct way of thinking?

ReinanRacing a écrit:

the exhaust should close when the maximum amount of mixture is pushed back into the cylinder.

That is correct.

Citation :

as long as the pressure in the exhaust is higher than in the cylinder, mass flow will go into the cylinder.

As long as the pipe pressure is higher than the cylinder pressure, the gas mass in the pipe will experience an acceleration in the direction of the cylinder. But that does not tell us anything about the actual flow direction, which may still be outward from the cylinder, in which case this flow velocity will just be slowed down until it comes to a stop, and only then will the flow direction be reversed.

Citation :

Thus the best moment to close the exhaust is when the pressure in the cylinder and exhaust are identical.
Is the above the correct way of thinking?

It is not. Mass flow from the pipe into the cylinder will cause the cylinder pressure to rise and the pipe pressure to drop.
But as mentioned above, as long as the pipe pressure exceeds the cylinder pressure, the flow will experience an acceleration in the direction of the cylinder, steadily raising the flow velocity. When pipe pressure and cylinder pressure are equal, this velocity has reached its maximum. Next, the mass entering the cylinder will raise the cylinder pressure above the pipe pressure, slowing the flow down. And only when the flow has come to a stop, is it sensible to close the exhaust port.

I see, I see.

When the exhaust opens, the hot exhaust gas enters into the exhaust pipe and reaches a high propagation velocity, perhaps even sonic conditions.

To "charge" mixture back into the cylinder, one needs to first decelerate the flow into the pipe, and then reverse the flow, so that the mass flow is in the direction of the cylinder.

In my "imagination", when the exhaust opens the pressure pulse looks like a "spike" with a steep "front" and a tapered "rear". When this pressure pulse is reflected in the exhaust, what is the shape of the reflected pulse? Is it a spike, or is it more or less "flattened"? From the above discussion, it would be good to have a "flattened" shape, because there would be more time to accelerate the flow into the cylinder? I guess the "flattening" depends on the taper angle of the reflector section of the exhaust?

Most text books use the image of a "plugging pulse" but if I read the comments from Frits Overmars, it seems better to have a "charging wave".

ReinanRacing a écrit:

what is the main objective of the tuned exhaust?
Is the objective to reflect the pressure wave and then trap it in the cylinder, to increase the mass of air in the cylinder,
or is the objective to set up a resonance?

The answer is the first sentence, no doubt at all.
To show this I come back with images where is resonance and where is no resonance.
For the moment you can see this: [Vous devez être inscrit et connecté pour voir ce lien]

The only time the particles and the wave move at the same speed is during sonic flow - that is the definition of sonic flow.

The fact that the fresh charge move only a short distance into the pipe system is why there is such a big drive currently to cool the exhaust port passage to keep this charge cool, and to smooth the flow by the shape of the auxiliary ports and to reduce the duct diameter to prevent as far as possible the mixing between the fresh charge and the burnt gas. That way the plugging pulse pushes cool and pure gas back into the cylinder. This makes more power and decreases detonation.

Vannik sent me a very nice video showing the pressure pulses in the exhaust. We are currently trying to figure out a way to share that video.

In Vannik's video, the following phenomena occur: At Exhaust Port Open (EPO), a very strong pressure pulse enters into the exhaust. The pulse then "flattens" due to the shape of the pipe, and when the pulse is reflected and starts traveling back to the cylinder, the pressure increase at the exhaust port is rather gradual. At the same time, the piston is moving up so that indeed, some of the returning high-pressure wave actually reflects from the piston and start traveling back into the exhaust. Once the exhaust port closes, the pressure waves in the exhaust "die away" rather quickly, or so it seems.

So: the question remains: how important is the "180 degree timing"? From Vannik's video, I completely understand the hypothesis that one would have the "cleanest" pressure waves if the timing is 180 degrees, but my physical intuition says that this depends (by and large) on the dispersion and dissipation of the pressure wave. That is: if the wave in the exhaust basically flattens out and/or dies out between EPC and EPO, then you're really not that much concerned about the 180-degree timing.

A very nice description on the resonance effect vs the plain tuned pipe can be found in Blair's first book:

The Basic Design of Two-Stroke Engines, Gordon P Blair, 1990, SAE

Para 5.4 - Single cylinder high specific output two-stroke engine, Starting on page 243

I have uploaded the video of the wave animation to the following web page, it is at the bottom of the page:

Gasdynamic wave animation

Vannik a écrit:

I have uploaded the video of the wave animation to the following web page, it is at the bottom of the page:

Gasdynamic wave animation

Vannik, thanks very much! I think it is a great video which is very instructive!

When the exhaust port opens, a very short high pressure pulse starts traveling into the exhaust (the "Right Moving Wave"). Due to the shape of the exhaust chamber, the pulse flattens out, and finally reflects from the end of the chamber. The "Right Moving Wave" now becomes the "Left Moving Wave" (LMW). At the cylinder, the LMW is not strongly peaked, rather, a very gradual increase of the pressure occurs, pushing mixture back into the cylinder. Just before the exhaust port closes, there seems to be a bit of a RMW, it appears that the port closing is just a bit too late. At any rate, this video is very instructive and makes the discussion much more clear. Thank you.

Vannik was gracious enough to send me three more graphs, and he has allowed me to post those graphs here. All graphs pertain to the Aprilia RSA engine at 13,500 RPM.

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The above figure shows the purity during the cycle. The red line shows the purity in the cylinder: combustion starts just before 0 degree, and purity goes down to 0 at 30 degrees (combustion complete). When the transfer ports open, the purity rapidly goes to 1.0 (all fresh mixture), and when the exhaust port closes, a bit of exhaust gas is also trapped, giving a purity of 0.95 just before ignition. The black line shows the purity in the transfer ports (?). I cannot explain why the purity dips when the transfer ports open, perhaps the pressure in the cylinder is still too high and some exhaust gas is pushed into the transfer ports. The green line shows the purity in the exhaust (I think). When the exhaust port is closed, the purity in the exhaust is 0.75. I guess that fresh air is pulled in through the tail pipe.

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This figure shows the Mach number in the exhaust port and in the exhaust. When the exhaust port opens, the flow reaches Mach 1 and stays at Mach 1 for about 30 degrees (longer than I expected). Propagation speed then goes down, but just before the INLET port opens, the flow reaches Mach 1 again. This is the effect of the low pressure in the exhaust, pulling mixture through the engine. When the transfer port closes, the flow velocity switches sign: mixture is now moving back into the cylinder (and at high propagation speed). As indicated earlier, just before the exhaust port closes, some mixture escapes into the exhaust.

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The above figure shows the mass flow. The black line shows the mass flow at the exhaust port. When the port opens, mass flows out, into the exhaust. When the transfer port closes, the flow reverses, and mass enters back into cylinder. The total mass from the cylinder into the exhaust is the integral over the positive part of the graph; call this MCylToExh. The mass from the exhaust into the cylinder is the integral over the negative part of the graph; call the MExhToCyl. In the ideal case, one should have:

MCylToExh - MExhToCyl = MExhaustGas

but looking at the graph, it seems that the flow from the cylinder into the exhaust is larger, and that would mean that some fresh air goes into the exhaust and never comes back. Maybe Vannik can explain.

The mass flow graph shows mass flow rate (dM/dt) in g/s and if the actual mass that is flowed is summed over 3 full cycles the error is less than 2%. I am investigating ways to decrease the error further.

I haven't studied the equations of GPB in detail, but from the look of it, it appears that he uses an explicit differencing in time and a non-conservation form for the gas flow. In GPB's text book, he explicitly mentions that you need to "connect" the solution from one mesh to the solution in the next mesh, in other words, the coupling between the meshes does not necessarily converse the mass flux and total mass.

I have no experience with engine modeling software, but for helium-cooled nuclear reactors, the error on the mass is on the order of 1.0e-6 - 1.0e-8. But then again, in nuclear reactors one is dealing with large mass flow rate but no shocks or waves, so the calculation is much more benign.

The GPB method connects the meshes in a mass conservative way so that is not the cause of the small error. I suspect the following:
1. Numerical resolution - maybe changing to double or quad precision will improve it.
2. Numerical accuracy or method accuracy - for best accuracy you need to choose the timestep so that the wave travels just less than the mesh length, but because the wave speed differs dramatically, including shocks and contact discontinuities, the timestep is chosen for the mesh with the fastest wave speed (The so called Levy-Courant condition) which results in a large number of meshes where the wave only travels partially through the mesh.