Perry Casson wrote:

OK Perry - we will use your own procedure for the caculations
First - each rotor of the Wankel fires and expands a full one
cavity charge for each turn of the E shaft - I think we can all
agree on that, So if it burns and expands that charge - it
had to pump it - so potentially it pumped about 40 cu in
for each rotor for one turn. The 13B with two rotors thus
80 times the rpm 7000 = 560,000 cu in or 324 cu ft min
With sls air at ,0023769 slugs per cu ft is 1488 lbs per
hour. Notice that the new numbers are quite different
than yours - I do not Know what you used but there is
no smoke nor mirrors to the method I have just described
and it will work equally well for any positive displacement
heat engine Using this flow and your same 12.5 and .47
we would be entitled to 253 hp, A good, modified NA 13B
does about 180HP/253 for an equivalent V.E. of 0.71

Vance - The problem I have with your numbers is as I understand things
a 13B does not displace 40 cu in for each rotor.  Here's  how I
understand things, the swept volume of a 13B rotor cavity is 327 cc
(26.7 cu in) it takes 3 full turns of the e-shaft to cycle all 6
chambers  so each rotation of the e-shaft only pumps 2/3 of the rated 80
cu in.  The reason the engine displacement is called 1.3 liters ( 80 cu
in) is because a four stroke piston engine pumps it's complete
displacement in two turns of the crankshaft.  In two turns of a e-shaft
a rotary will have a swept volume on 80 cu in so to make things more
relative to the 4 stroke centric world it's call 1.3L engine.  Have I
been living a lie all this time?

I did have a mistake in my first posting, at 7000 rpm  the volume pumped
would  be 216 cfm  or  972 lbs / hr if VE = 1.0 which should make 151 HP
in the world as I know it.

Perry Casson

It is equivalent to a 2 cyl 1.3 L if you consider it a 2 cycle and a 4 cyl 
2.6 L if you treat it as a 4 cycle. Despite what Vance says :) the amount 
of air pumped per rev is all that counts.

Paul Lamar
 
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