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Actual pumping displacement of a 2-rotor
Does anyone know what the actual pumping displacement of a 2-rotor is? As I understand it, the 1.3L is the volume of the housing minus the volume of the rotor. I need the pumping displacement for some supercharger sizing calculations.
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That info (housing volume - rotor, etc.) is incorrect. Each rotor (180 degrees out of phase w/ respect to each other) pumps 654ml of fuel/air mix through the engine each rotation of the eccentric shaft. There's your 1.3L number - 654*2 = 1308. Now consider that the next revolution will also pump 1.3L. This would be equivalent to a 2.6L 4 cylinder 4 stroke, i.e., a 654ml power stroke each 180 degrees. The 4cyl 4stroke only pumps 1.3L each rotation due to 2 of the cylinders being on either a compression or power stroke which does not pump air into or out of the engine. Different reason, same result.
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If you are determining air pumping capability for supercharger sizing, you need to figure it out over time and in relation to a piston engine. Since a piston engine is figured out based on total displacement of all cylinders, this takes place in 2 crank revolutions. That's your time aspect. It's 2 revolutions. As Robert said the rotary displaces 1.3L per revolution so to equate it to a piston engine in the same amount of time, your calculations need to be based on 2.6L. It doesn't matter what Mazda calls the engine in terms of size or who argues what. For all sizing purposes, whether it's air or fuel, it's a 2.6.
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Amen.
-Mike |
Yes it is a 1.3 liter engine that breathes like a 2.6L 4stroke.
Barry |
Originally Posted by sniperstevedave
(Post 8954219)
Does anyone know what the actual pumping displacement of a 2-rotor is? As I understand it, the 1.3L is the volume of the housing minus the volume of the rotor. I need the pumping displacement for some supercharger sizing calculations.
^ Taken from here ^ *Note to summarize if you are using others data/formula set up for piston engines then you need to use either the 2 stoke equiv capacity (1.3) or the 4 stroke one (2.6), pretty simple* I decided to put this together for the new players who struggle with understanding what a wankel cycle is about and also the true capacity of the engine, a picture tells a thousands words, so I cut up a rotor and a shaft and marked them taking a photo at every 90 degree's of main shaft rotation, following a chamber from firing to firing or one full Wankel Combustion Cycle. 0 degree's TDC No1 chamber firing https://img401.imageshack.us/img401/839/0degso8.jpg 90 degree's https://img229.imageshack.us/img229/1204/90degxr0.jpg 180 degree's https://img229.imageshack.us/img229/4618/180degaw0.jpg 270 degree's https://img140.imageshack.us/img140/2852/270degym6.jpg 360 degree's (one revolution of crank) https://img440.imageshack.us/img440/2682/360deghi2.jpg 450 degree's https://img208.imageshack.us/img208/6769/450deglr0.jpg 540 degree's https://img212.imageshack.us/img212/5224/540degjo3.jpg 630 degree's https://img221.imageshack.us/img221/2944/630degtv4.jpg 720 degree's (two revolutions of crank) https://img380.imageshack.us/img380/2440/720degvr8.jpg 810 degree's https://img401.imageshack.us/img401/3505/810degho8.jpg 900 degree's https://img401.imageshack.us/img401/1644/900degxj4.jpg |
990 degree's
https://img87.imageshack.us/img87/6586/990degcj4.jpg 1080 degree's Wankel Cycle is complete ! (after 3 full revolutions of the crank shaft) No1 chamber firing again https://img87.imageshack.us/img87/3117/1080degbg5.jpg ^ continued as shit forum will not allow all pictures in one post :icon_tdow From the above you can see each individual separate chamber (3 per rotor) only fires after 1080 crank shaft degree's has elapsed,, this is why the Wankel is so different to ANY other type of engine, 2 strokes fire each individual chamber once every 360 degree's and 4 strokes fire every individual chamber every 720 degree's. If you look at a 13B with its 654cc per Individual chamber capacity (thus 1308cc) you can see it aspirates this ONCE every single revolution thus you can compare the 13B to a 2 stroke if you must do so on an equivalence basis (but remember you are not counting the other 2/3rd's of the combustion faces! Now if you compare it to the much more common 4 stroke engine you can see that 2 faces ONLY are being counted in the engine and thus it has aspirated a total of 2616cc over 720 degree's of crank shaft rotation....... nice little bit of info there but it still misses a whole 1/3rd of the engine! Finally the ONLY TRUE way to look at a Wankel Rotary is to view it in its own cycle! (and not comparing it to something that it is NOT!) this is only over 1080 degree's of crank shaft rotation, where ALL of the working faces can be accounted for (just as when you do a compression test to see if the poor little donk is healthy or not) :) For it is only when the entire engine has complete one full cycle of work can it thus be rated, be that as functional or in its true capacity sense. You will then see that the humble 13B is indeed 654cc x 3 working faces x number of rotors ! = 3924cc. Equivalence capacity to time scale (revolutions) for 13B engine, has one power pulse per 360 degree's per rotor 1308cc 360degree's (2 stroke) 2616cc 720 degree's (4 stroke) 3924cc 1080 degree's (Wankel Rotary) |
Originally Posted by sniperstevedave
(Post 8954219)
Does anyone know what the actual pumping displacement of a 2-rotor is? As I understand it, the 1.3L is the volume of the housing minus the volume of the rotor. I need the pumping displacement for some supercharger sizing calculations.
2616cc 720 degree's (4 stroke) is probably the number he needs to calc supercharger size and pulley ratio. |
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