> I have a question about this theory...
> If I have a 350 Chevy at 6500 and 14lbs of boost with a 72% eff.
> Compressor and a VE 100% of I get 75Lbs per minute of air non intercooled.
Ok
> If I now use the same engine and add a air to water intercooler with a 45
>degree water and an Eff or 90% which is common I get 99 Lbs of air per minute
> and a corresponding increase in power. I would also probably need a larger
> turbo as well to provide the 32% increase in lbs of air that would be too
> large for the non-intercooled version.
Yeah, or more likely, you are just operating it a bit more inefficiently but
at higher flow. So the higher flow at a higher temp is cooled down to the
denser air at the conditions you specify. Using the same unit. Less
wastegate?
> If instead of using the intercooler I used the water injection and the
> turbo sized for the non intercooled version where do the pounds of air come
> from?? Don't I have 75 lbs per minute of colder air??
Yes plus you have the actual mass of water in the equation. But also the
turbo is not working as hard as in the first case you mentioned. IMHO.
> How would anyone propose to keep the larger turbo out of surge if the
> water injection had a bad day or were wrong on the btu calculation?
Because you are using the smaller turbo that meets the 75lbm flow you don't
have to worry about surge, If you do have surge (backpressure) then BOV.
> The change in charge air temp to go from non intercooled with 223 degree
> air to intercooled with 54 degree air is rather substanial. How much
> water would you need??
Look on the web site. I have an example and everything. And ways to
calculate it.
> The same engine at 30 lbs of boost is 75 dgree air intercooled and 342 non
> intercooled...
Yup, that's why you intercooled in the first place :^}And that's temperature
rise which need to be added to ambient temp, I suspect...
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