CarmonColvin

First of all I am NOT an engineer so please forgive me if I use the wrong terminology or similar laymen mistake.

I am trying to calculate the BMEP (Brake Mean Effective Pressure) of a Wankel rotary engine. Searching the net I have found one formula that is slightly different for a 4-stroke and 2-stroke type of engine.

2-stroke BMEP = (HP * 6500) / (L * RPM)
4-stroke BMEP = (HP * 13000) / (L * RPM)
source: http://engineersedge.com/engine_formula_automotive.htm


The only difference between the two formulas are the numbers 6500 and 13000. The only correlation with a 2x difference between the two types of motors are the 2-stroke motor uses 100% of is displacement for combustion during one revolution where a 4-stroke motor uses 50% of its displacement for one revolution.

The only way I can make this work between those formulas is to look at it this way.


BMEP = (HP * (6500 / DU%) ) / (L * RPM)
Using the Displacement % used as DU%

OR

2-stroke BMEP = (HP * (6500 / 1.0)) / (L * RPM)
4-stroke BMEP = (HP * (6500 / .5)) / (L * RPM)

A Wankel Rotary engine uses 33% of its displacement for combustion for one revolution output from the motor.

Wankel Rotary BMEP (HP * (6500 / .333)) / (L * RPM)

IS THIS A CORRECT ASSUMPTION?

Data from an actual dyno run (measured at the wheels) of a slightly modified Mazda 13b (1308cc) non turbo motor put out 168 HP at 7000 RPM and 147 ft lbs of torque @ 4000 RPM.

BMEP = (168 * (6500 / .333)) / (1.308 * 7000)
BMEP = (168 * 19500) / 9156
BMEP = 3276000 / 9156
BMEP = 357.8
This number seems really high compared to other BMEP readings I have seen for 4 and 2 stroke engines.
And am I correct in assuming this is in pounds per liter?

If this is completely wrong please help me with a formula that will accurately give me BMEP of a Rotary Engine.

Thanks.

rotary emotions

Oh boy, this is complicated... Don't shoot me if I'm wrong, I'm just a poor normal person, not an engineer. But I think the 33% is wrong. The Wankel uses 100% of its displacement for combustion. The rotor will complete the cyclus on each side so has three ingition-moments per rev. This is then used to drive the excentric shaft by 3/1, so the output of the engine is, much like a 2-stroke: 1 combustioncycle for every rev of the engine (per rotor that is, of course). Since the output is like a 2-stroke the formula of a two stroke could apply. But I really don't think that's correct either, and the co-relation between two/fourstroke won't completely apply on a Wankel...

Rotortuner

I think you need to dounle the engine capacity of the rotary to 2.6 liters. Not sure, but i think that should bring the number down. ?
try that. remember roatarys use twice their displacement. am not an expert on exactly why, but rice racing and a few others on here know a lot more about it than me.

CJG

Rotortuner

I did that calculation you were using and with 2.6 liter i came up with 181.1. Does that sound mor beliviable?
BMEP = 181.1 ?

CarmonColvin

My thought behind the 33% is this.

Rotary:
For one revolution of the e-shaft you get 1 full combustion "stroke" from each rotor. So you get 2 combustions per revolution out of 6 total chambers. So only (2/6th = 1/3rd = 33%) of the measured volume of the motor has combustion for each revolution of the e-shaft.

Manually advance this movie slowly to see that for all 3 faces of a rotor to go through a cycle the e-shaft turns 3 times. Rotary Animation

2-stroke
Every time the piston goes down with the crank shaft it has had combustion. No matter how many pistions in the 2-stroke motor for one revolution of the crank shaft all of them will have a complete combustion cycle. So 100% of the measured volume is used.

4-stroke
It takes two revolutions of the crank shaft for a piston to go through a combustion stroke (one revolution is used on a exhaust and intake stroke). So 50% of its measured volume has a combustion stroke per revolution.

rotary emotions

Eh, you're not right there... You should see a single rotor. If you see the engine as two rotors and compare that to 1 cylinder you get wrong numbers. If you compare the crank with the E-shaft, you'll see that there is one combustion for every 360 degrees of E-shaft motion, and one combustion for every 360 degrees of crank shaft motion in a two stroke. When more rotors get involved, you'll get two combustions for every 360degrees, two cylinder two stroke just the same. Don't forget that the rotor isn't moving at the same speed as the E-shaft. Therefor: 3 rotor sides are doing the same as 3 pistons, but the E-shaft turns at 3/1 of the rotorspeed. So you get 1 cycle per rev for each rotor. You said it yourself:
.. to see that for all 3 faces of a rotor to go through a cycle the e-shaft turns 3 times...
So three faces mean three revs, or one rev/one face.
So: one combustion per rev.
The power output is very simular with a two-stroke, that's why a lot of people see it as two stroke. (it isn't, though)



Forget about the 2.6 liter thing, it's just something from tax and to get rotaries in higher race classes (they'd win every race otherwise, and some do not like that!!!).
The chamber volume of a 13B is 1308cc, and NOT 2.6 liters.

CarmonColvin

I am not comparing a 2 rotor to a 1 cyl engine. That is not what the formula is doing either.

I am comparing an engine design that has 100% of its measured volume go through a combustion cycle for ever 1 revolution of its crank shaft. To a motor design that has 33% of its measured volume go through a combustion cycle for each e-shaft revolution.

For a 2-stroke it does not matter if it has 1, 2, 300 cyls. For one revolution of crank rotation every cyl completes a full cycle.
For a rotary it does not matter if it has 1, 2, 3 or more rotors. For one revolution of the e-shaft rotation each rotor goes through only 1 combustion cycle of its possible 3. So 33% of its chambers fires. If it is a 1 rotor that would be 1 of 3 chambers, for a 2 rotor it would be 2 of 6 chambers for a 3 rotor it would be 3 of 9 chambers, etc. etc.. either way it is 1/3.

My original question had nothing to do with the accuracy of the 33%. My question was my theory behind applying the 33% in the correct position of the given formula.

RICE RACING

Use 2.6lt for 13B, my BMEP on my engine is over 2MPa, well over !

I developed my own formula and it checks out with SAE papers that specify BMEP and Peak chamber pressures in wankels so I am confident that it is right

I will look at my HP (all my formulas are stored in that) and I will post it.

RICE RACING

BMEP,Pa= HP/00.1341/(RPM/60/2)/(Capacity,Lt/1000)

is the formula I use.....

Cap =2.616
HP=566
RPM=7500
BMEP=2581486Pa or 2.581Mpa

zenkiFC

so rotary displacement is measured by all 3 rotor faces, right? 654cc per rotor, 218cc for each face? sorry if this sounds stupid i just need to understand.

Maxer

No..... 654 cc for each face of the rotor , or 654cc of mixture drawn in for one rev of the e-shaft (one third rotor rev).
hence , two rotors is 1308 cc of "power" for each rev of the e-shaft.

zenkiFC

oh I see so it uses all of its displacement for each rotation of the e-shaft, as opposed to a piston engine using 1/2 for each rotation of the crankshaft?

RICE RACING

Basically yes, if you want to use all of the displacement of a wankel rotary you need to do three full crank shaft revolutions, this works out to 654cc X 3 X number of rotors, which works out to 3.9 odd liters for a 13B ! but this is over 3 crank revs !

Then you got a 4 cycle piston engine that completes it's power cycle for all chambers in two crank revs !

And you got a 2 stroke piston engine that completes it's power cycle for all chambers in one crank rev !

You gotta compare an apple to an apple and a good mean base line is out put shaft RPM, If you se the std as 720deg of rotation (two crank rpm's "complete 4 stroke piston cycle = most common std") then a Wankel 13B will complete 654cc X 2 crank revs X 2 rotors, in a 13B's case this = 2.616lt, its as simple as that. And for the two stroke piston engine you double it's capacity cause it has completed a power cycle in only one rpm.

The moral ? A 13B can be called a 1.308lt engine, or a 2.616lt engine or a 3.924lt engine, it is ALL of these as it does complete ONE power cycle across both rotors in one engine crank rev, though it does take three crank revolutions for ALL three rotor faces to do one power cycle.

To enable us to use common formule developed for 4 cycle piston engines a 13B needs to be regarded as a 2.616lt engine, cause this is what it is in this term of reference.

CarmonColvin

I was misinfomred how the displacement of a rotary was calculated. I was under the impression that 1.3L was the total of all 6 chambers when each of them were at their largest capacity. And it used 1/3 of that each revolution.

Using your updated theory with comparing it to a 2.616L 4 stroke engine and my original formula for a 4 stroke. (168 HP @ 7000 RPM, measured at the wheels)

4-stroke BMEP = (HP * 13000) / (L * RPM)

BMEP = (168 * 13000) / (2.616 * 7000)
BMEP = 2184000 / 18312
BMEP = 119.266 (lbs per L? if so that is 8.22 BAR)


Using your formula for the same specs looks like this.
BMEP,Pa = HP / 00.1341 / (RPM / 60 / 2) / (Capacity,Lt / 1000)
BMEP,Pa = 168 / 00.1341 / (7000 / 60 / 2) / (2.616 / 1000)
BMEP,Pa = 168 / 00.1341 / 58.3333 / 0.002616
BMEP,Pa = 8209.675102

And if my Math is correct 8209.675 is 8.209 BAR. Amazingly close to the 8.22 bar above. BMEP,bar of around 10.0 seems to be average for a mild modified engine (of any type). So it sounds right to me when you consider that the 168HP was measured at the rear wheels.

And applying your HP figure (566 @ 7500) to my formula I also get 25.86 bar. This number is very high but probably very reasonable for a turbocharged engine.


Do you agree that both formulas are getting the same numbers and are correct?

now.. what to do with these numbers? hmmmm..

RICE RACING

Yup Spot on !

Yes the BMEP is very high not above mechanical limits though, touch wood !