> The actual displacement of the stock
> dual port engine is 1584 cc,
I'm sure you're right ... but I'd swear I've seen bugs running around with
an "1800" emblem on the back ... maybe that's a popular overbore kit ? More
likely just encroaching CRS.
> With this, and the actual multiplier for cfm to scfm at 100 psi
> (6.8),
Can't seem to derive this number : 100 psig = 114.7 psia. 114.7 / 14.7 =
7.8 ?
> the necessary speed for actual cfm @ 100 psi
> would be closer to 17630 rpm. This, of course, ignores heat of
> compression in both calculations, so the volume would be
> depressed some (about 15%, figuring a 100 degree F rise) if the
> compressed charge were allowed to cool to ambient before
> re-expansion.
You've also left out the effect of the compression ratio not being infinite.
In effect, at the end of each stroke, the combustion chamber represents
compressed air that is wasted. I guess you could say this is included in
VE, but then it's going to be a whole lot lower than 80% (which is decent
for a stock motor with only a muffler for backpressure). Another way to
look at it, we're talking about a huge backpressure here.
> There is no figure assumed for rate of use of air, and that is a
> missing variable.
Granted, the "worst case" consumption rate is a sub-problem.
> Also, compressor output cannot be
> assumed to be constant from 100 psi through 130 psi. If its
> maximum rated output is 58 scfm @ 100 psi, output at 130 psi
> will likely be lower.
Hey, I can _assume_ anything ! Whether the results mean anything in reality
is a different question <g> Besides, we're looking at a "worst case"
analysis for minimum time, and a lower output means a longer cycle time
(defined in this case as the interval between engine startings) so the error
is in our favor.
> If air use is greater than the rated output
> of the compressor, the compressor will not keep up,
> and the storage tank would have to be infinitely large to
> maintain a 60-second duty cycle.
Guess I wasn't clear, I was looking for how often the engine has to be
started. Sorry bout that. If the air use is equal to the output (can't be
greater for the steady state), then the engine only has to be started once
and the cycle time is infinite.
> Or is it assumed to be 60 seconds from shut-off at 130
> psi to start-up after draw-down to 100 psi? The latter
> case can be easily calculated (about 210 gallons capacity for air
> use slightly below the rated capacity of the
> compressor).
Well, you've just given us one point on the curve. One way to tackle the
problem without calculus is to pick several points, graph them and draw a
smooth line through the results ...
Randall
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