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In a previous posting, I alluded to my suspicions
that a commonly accepted formula for calculating airflow for the rotary
engine might be flawed. I was asked off-line to expound on that
suspicion.
Well, I can think of no better audience to
present my suspicions to and see what you think. Last time I tried this on
another list I was berated, beat upon and chastised for daring to raise the
question - but, never one to be daunted for long and knowing this list has some
experienced and clear thinkers, here goes again.
1. The racing crowd was thrown
into a turmoil when the rotary appeared on the scene - what displacement class
should it be place in??
Well eventually the following was apparently (If
not agreed to was at least accepted) that the 1300 cc 13B rotary was equivalent
to a 160 CID 4 stroke engine. Then the standard air flow formula could be
applied - which is
Air Flow (CFM) = Displacement*rpm/(2
*1728)*Ve (Volumetric efficiency (Ve) is generally assume to be 100%
or 1, so we can drop it for this discussion as well as the effects of
compression ratios)
How did they arrive at 160 CID displacement, well
they apparently decided to reference the rotary to the standard 720 Deg 4
stroke rotation cycle even though for all six faces of the two rotors to go
through their cycle requires 1080 deg of e shaft rotation - well that is 360 deg
more than the standard of 720Deg. So if you limit consideration to only
720deg of rotation only 4 of the six rotor faces have occurred, so since the
displacement by a single face is 40 CID then 4* 40 = 160 CID. So that
appears to be how the decision of 160 CID equivalent displacement was arrived
at. No problem with that part - but, continuing on:
So taking the formula and assuming 6000 rpm, we
have Air Flow = 160*6000/(2*1728) = 277.77 CFM (assuming 100% Ve).
2. So what's the problem? Well,
unfortunately, I like to fully understand how and why something works rather
than just plugging in numbers to a formula (although, heaven knows I do enough
of that).
So here is how I started sliding down the
slipper slope {:>).
We know that the rotors turn at a speed 3 times
less than the E shaft. So if the e shaft is turning at 6000 rpm then the
rotors are turning at 6000/3 = 2000 rpm. Its the rotor of course that
actually suck in the air - not the eccentric shaft. So we should be able
to arrive at the same answer from either reference point (rotor or e
shaft). Well the formula above uses the e shaft reference. So I
looked at what the rotor reference would produce (expecting the same
answer).
If the rotors are spinning at 2000 rpm, we know
that 2 of the rotor faces have gone through their cycle for each 360 deg
revolution of the e shaft or 4 will have for 720 deg. Looking at it from
the rotor reference if the e shaft has turned 720 deg then the rotor has turned
720/3 = 240 degrees. 240/360 = 0.6666 * 6 faces = 4 faces. So that
confirms that 4 faces go through their cycle in 720 degs of e shaft
rotation. So if 4 faces of the two rotors complete their cycle in 720Deg
eshaft then we have
Air Flow = 4(number of faces) *40 (displacement for
each face) * rpm (rotor rpm)/(1728) note: we drop the division
by 2 which in the original 4 stroke reciprocating engine equation takes into
account that only 1/2 of the cylinders are sucking air on each revolution.
3. This Logic gives us
rotary Air Flow = 4*40*2000/(1728) = 185 CFM! This is ,of
course, NOT the 277 CFM arrived at by the approved
formula.
As you can see the implications of this (if
correct) are fairly significant and that is why even though I can't spot
the error, I must have made one.
I would greatly appreciate any assistance in
helping me understand where I have gone wrong in logic or math. Please
don't just quote an authority or formula - explain it and show me your
calculations on how you got there. Its things like this that keep me awake
at night {:>).
Help Meee..! 
Thanks
Ed
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