rkostolni

I want to replace my Greddy intake filter with a K&N intake filter and the location where I want to put it requires it to be a small filter. I am trying to figure out just how small I can go without introducing a restriction into the system. K&N provides this formula for figuring out the required filter area

A=(CID * RPM)/20839.

Now my question is with the CID. I've read Rotary God's post on the true CID of a 1.3L rotary engine being 2.6L. In case anyone hasn't read it, its basically because we have 3 rotor faces so 1.3L *3 =3.9L, but then the rotary engine is a 6 stroke engine instead of a 4 stroke, so 3.9L *4/6 = 2.6L.

But I was thinking about that and that doesn't quiet make sense to me.
-A piston engine makes 1 intake stroke for 1 revolution of the crank shaft.
-The displacement on a piston engine is the volume swept by all of the cylinders of one bank. So on a 4 cyl, the volume swept by two pistons determines the CID.


-A rotary engine makes 1 intake stroke for 1 revolution of the eccentric shaft, just like a piston engine does
-The displacement of a rotary motor is determined by the volume swept by one face of each rotor.
-Each rotor sweeps 1.3L/2, so two rotors sweep the total 1.3L volume.

-So, the engine breathes 1.3L worth of air for each rotation of the eccentric shaft.

So it seems to me the CID I would use here (without compensating for the turbo) is 1.3L = 79.3 cu in. The fact that there are 3 rotor faces and the motor being a 6 stroke is compensated for in the fact that the rotors are spinning at 1/3 the rate of the eccentric shaft.

Once I compensate for the turbo, which I have done by:
(6psi + 14.7psi)/14.7psi = 40% increase. Then guessing at the efficiency of the turbo being 60%, (Best guess since I don't have a compresor map) I find the turbo gives a 24% increase in air flow.

So I think I should use roughly a 100CID to enter into the formula for the filter area. I'm sure Rotary god is correct though that I should have started with 2.6L instead of 1.3L, but I would like to know why.

If I used a 2.6L displacement then I would end up with a 200CID in the formula.

HD-Paschke

A Wankel rotary engine is a real four stroke engine and no 6 stroker!
HKM = 720° for 4 strokes, one stroke 180°.
KKM57P (Wankel) = 1080° for 4 strokes, one stroke 270°.
The stroke duration is 50% longer, CID= 2,6ltrI

rkostolni

The rotary engine is a 4 cycle engine, but it is a 6 stroke engine. I say this because it requires 3 revolutions of the eccentric shaft to complete 1 full combustion cycle whereas a 4 stroke piston engine requires only 2 revolutions of the crank shaft to complete the 4 cycles. A 2 stroke engine requires only 1 revolution of the crank shaft to complete all 4 cycles of combustion.

HD-Paschke

So, then designate these 6 Strokes.
Draw the pV-Diagramm.
I recommend Richard F. Ansdale The Rotary Engine.
(Der Wankelmotor Konstruktion und Wirkungsweise)

HDP
Experince in KKK and KKM57 since 1951!

rkostolni

I think you are confusing cycle and stroke. Actually, I guess stroke is a very ambiguous word anyway. I totally agree that a rotary has 4 cycles. Intake, compression, power, exhaust, all engine have 4 cycles.

By saying it is 6 strokes, I mean refering to the difference in the number of revolutions of the shaft to accomplish those cycles. A 2 stroke weedwacker engine still accomplishes the 4 cycles, but it can do all 4 cycles with only 1 up and 1 down of the piston, or 1 revolution of the crank - 360 degrees of rotation. A 4 stroke 4 cycle piston engine can accomplish all 4 cycles in 1 up 1 down 1 up 1 down, or 2 revolutions of the crank - 720 degrees of rotation. A rotary requires 1080 degrees of rotation to accomplish all 4 cycles. This means 3 rotations of the eccentric shaft.

HD-Paschke

I think you are confusing strokes and revolutions!
Strokes and cycle (Takt) are in the Otto cycle terminus the same.
(You should not forget a rotary has no stroke. No ups and downs)
Each stroke lasts 270° or 3/4 revolutions.
Your rotor have three sides and we have every revolutions one working cycle per chamber.
Your CID = 1,3ltr x 2 = 2,6ltr for a 13BR compare to a four stroke piston engine.
Why? You have no empty stroke like a piston engine (four stroke).

This is wrong for a four stroke piston engine.

You' have only ever second revolution 1 intake stroke per cylinder ( 4 stroke).
You have an empty stroke.
http://en.wikipedia.org/wiki/Four_Stroke
Wrong

You have only ever second revolution 1 intake stroke per cylinder
Right Terminus: A rotary engine makes 1 intake stroke for 1 revolution of the eccentric shaft per chamber, just like a 2 piston 4 stroke engine with the double displacement does!
Right. Vc= Vmax-Vmin
Without comment

Right Terminus:
The displacment per Chamber (13BR) is Vc=3 x ( Qudratwurzel aus 3 ) x Rges x e x b
13BR e=15mm R=103mm a=2mm B=80mm
Rges=R+a
Vc= 3x 1,73205x 10,5x 1,5 x 8 =654,7ccm per Chamber
Vcges= Vc x (Number of Chambers)
Vcges = 654,7 x 2 = 1309,4ccm ~ 1,3ltr
CID = Vcges x 2 compare to a four stroke engine. Why factor 2? We have 2 maximas per Chamber!
Why? We have no empty stroke.

rotarygod

When I wrote that article, I used the terms "stroke" and "cycle" only to clarify what I was trying to say from an example standpoint. In the technical sense that qualifies an engine as a 2 stroke or a 4 stroke in regards to crankshaft rotation, the rotary is a 6 stroke. However if we consider how the combustion chamber moves and goes through it's phases, it is still just an Otto cycle 4 stroke engine. It's another one of those strange rotary only things that makes it difficult to compare to a piston engine. For all intents and purposes a rotary is a 4 stroke engine that takes 50% longer to do everything in terms of crank rotation.

I like to use a general rule that you should use the biggest air filter that you can in the space available and no smaller. You can't have too much filter but you can have too little. It doesn't really matter if you place the filter in spot A in the desired location or if you place the filter in spot B and run a sealed duct over to spot A. It works the same way. If you can't get enough filter area you should think about a new intake pickup location. At the very least stick a larger filter where it fits and then duct to that spot.

rkostolni

I agree 1 cylinder makes an intake stroke every other revolution, but I was talking about the engine as a whole. The engine has 2 banks of cylinders, from a black box perspective, the engine is making an intake "stroke" for every revolution of the crank. Granted it is done by opposing cylinders.



Again see above. From the perspective of the air filter, both the rotary and the piston engine make require one intake pulse per revolution of their shaft.


Why do you multiply 1.3L by 2?

rkostolni

I think what we're really disagreeing about here is my use of the word "stroke", you are correct in that "stroke" is not the right word to use. I was just trying to point out the difference that a piston engine requires 2 rotations of the crank whereas a rotary requires 3 rotations to complete a combustion cycle.


To help with figuring out the answer to my original question I will break it up into parts.

Part 1.

How much air would the Renesis breath in NA form if it had 100% VE for one revolution of the eccentric shaft. I think it would breath 1.3L, is that correct?

HD-Paschke

http://www.rotaryengineillustrated.com/re101/cycle.php#
We you can see we have two BDC (chamber maximum) and two TDC (Chamber minimum).



For a 6 Stroke Engine need you three BDC and three TDC.

Spider

The engine should give 2 chuffs for one crank rotation if the apex seals are working.

The swept volume is 654cc. 2 chuffs indicates two presure releases. And I assume what comes out must have gone in. So 1308cc would have been breathed.

Yes 1.3L.

Andrew

rkostolni

Okay, so then if the Renesis breathes 1.3L worth of air for 1 turn of the eccentric shaft, then it seems to me that 1.3L is what I should use for engine displacement to figure out the area needed for my intake filter. But why does Rotary God say that whenever sizing parts for a rotary use 2.6L as the displacement.

rotarygod

Because if you size everything for a 1.3L engine everything will come out insanely small and be only half of what you need it to be. The rotary displaces 1.3L for every 1 revolution of the eccentric shaft. However a piston engine that displaces 2.6L will do it in 2 full revolutions of the crank and not 1. We need to take revolutions into account so 1.3L X 2 rotations = 2.6

rkostolni

I thought piston engine displacement was rated as the volume swept per bank of cylinders. So on a 4 cyl. it is the volume swept by 2 of the cylinders.

In which case a 2.3L engine means 2 of the 4 cylinder is 2.3L, or all 4 cylinders is 4.6L worth of displacement. Maybe I have that wrong though.

But if that is right then, 1 revolution of the crank fill 2 of the 4 cylinders which is 2.3L worth of displacement. The next revolution fills the other 2 cylinders, so another 2.3L worth of displacement. And then it repeats. So that would mean every revolution fills 2.3L worth of displacement on a piston engine.

rotarygod

A piston engine size is based on total swept volume of all cylinders. This just happens to take 2 crankshaft rotations on piston engines. This means a 2.6 liter piston engine displaces 1.3L per revolution which is that same as a 13B Renesis. This is where the confustion lies. A rotary displacement is not figured out according to total displacement of all rotor faces. Instead it is rate in how much air it displaces per eccentric shaft revolution. If a 2.6 liter piston engine were actually rated per crank rotation as the rotary is, it too would be rated at 1.3 liters.

rkostolni

Ahh, okay, that is what was wrong with my line of thinking. I always though piston displacements were rated per bank of cylinders.

So then next question. If I assume the Renesis is a 2.6L engine which gives about 160 cu in. Then to account for the turbo I added 25% onto that, based on the line of thinking stated above. So would 200 cu in. as the displacement be a reasonable value to use for CID in K&N's filter area formula?

rotarygod

For forced induction you need to figure out how much more air you are using. If you are running 7 psi, you will need about 50% more airflow over stock all things being equal. Another things which is going to affect it is the fact that the rotary uses about 40% more air to make the same amount of horsepower as a typical piston engine. For this reason alone I use 4 liters as a good intake sizing chart but forced induction will add to it. You're going to find that K&N's formula isn't very helpful for what you are trying to do. As a general rule, use the largest filter you can in the space available. You can't go wrong this way.