Randall Young wrote:
>>What do you need the air tank for?
> Even assuming you just vent the excess air rather than starting and stopping
> the motor, a tank still allows you to operate tools that take more air than
> the compressor supplies, and also provides a buffer for tools that have
> large instantaneous air draws (like jackhammers and chisels).
>> If you have 58 SCFM (or CFM,
>>I have not found anything to clarify on this for this setup)
> Should I say it again ? Compressors are *always* in scfm, even when they
> don't say so.
> Lessee, ISTR a typical bug motor is 1800cc, we should get nearly half that
> per crank revolution. A cubic foot is .0283 cubic meters, or 28300 cc, so
> it would have to spin at 58 * 28300 / 900 rpm to deliver 58 scfm @ 0 psi
> (ignoring VE), or a little over 1800 rpm. However, even if the kit has
> components to raise the compression ratio through the roof, the efficiency
> won't be very good at 100 psi, so I'd guess it turns more like 3600 rpm to
> deliver the rated 58 scfm @ 100 psi.
> Throw in the multiplier from scfm to cfm, which is almost 8 @ 100 psi, and
> the engine would have to turn some 14000 rpm just to get the free air
> delivery. Somehow, I doubt that's what they're doing ...
Ummm, let's do the math a bit more simply, and rightly assume SCFM for the
output. The actual displacement of the stock
dual port engine is 1584 cc, so, with the cam timing adjusted to open #2 and #4
valves once each rev, the displacement
per rev is 792 cc, or 48.32 cu. in/rev (@16.39 cc/cu. in.). This translates to
0.028 std. cu. ft/rev., or 58 SCFM @ 2074
rpm. Assuming the manufacturers of the kit have done their homework, and the
compressor side of the engine attains a
volumetric efficiency of 80% at working rpm, this would translate to rated
output @ 2592 rpm.
With this, and the actual multiplier for cfm to scfm at 100 psi (6.8), 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
>>The compressor is going to run the same amount of
>>time whether you have the tank or not, since your air usage requirements
>>will still be the same.
> How long do you think the starter would last, if you started the engine
> every 10 seconds ? Can you recharge the battery in 5 seconds ?
> Exercise for the student : how big does the tank have to be to guarantee a
> minimum 60 second cycle time (assuming the motor is started when the
> pressure falls to 100 psi, and shut off when it rises to 130 psi. For
> simplicity, assume compressor output is constant at 58 scfm.)
There is no figure assumed for rate of use of air, and that is a missing
variable. 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. 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. Also is the 60-second duty
cycle assumed to be starting at 130 psi shut-off, draw-down to 100 psi, and
then recharge to 130 psi, with air used all
the while? 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). The former case requires some calculus, I think.
Michael D. Porter
Never let anyone drive you crazy when you know it's within walking distance.