



Posted by Carmon Colvin (12.166.72.98) on November 18, 2002 at 17:12:57:
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 4stroke and 2stroke type of engine.
2stroke BMEP = (HP * 6500) / (L * RPM)
4stroke BMEP = (HP * 13000) / (L * RPM)
source: /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 2stroke motor uses 100% of is displacement for combustion during one revolution where a 4stroke 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
2stroke BMEP = (HP * (6500 / 1.0)) / (L * RPM)
4stroke 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.
Carmon Colvin
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