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On 9/26/2011 9:02 AM, stevei@carey.asn.au wrote:
Hi Guys
Could you educate me here.
Q1. If Dave is boosting to 45"Hg then am I right that that is ~22psi.(now I take it that is referenced to static at whatever altitude Dave quoted)
45" mmHg ~ 22 PSI... if both have a common zero point.. I dont do cars, so I dont know what the motorheads use for zero - ambient or true vacuum. But your numbers are correct presuming 0" = 0 psi.
Q2. Now lets say a 13B with a wonderful intake has 100%VE and makes x horsepower at 6K RPM.
If we boost that same 13B at 14.7psi after a really nice intercooler and spray bar that drops the temp to ambient (hypothetically speaking).
Will this then double the charge mass and therefore the potential energy and so produce 2x horsepower at 6K RPM?
If you increase the intake charge pressure from 14.7 psi/30" mmHg (ambient sea level pressure) and double it to almost 30 psi/60" mmhg, (and for theory sake, have a hypothetically perfect intake and intercooler, with no losses) and add fuel proportional to the mass of the air (which in this case is doubled) then your HP would be 2x. You seem to grasp the academic concept.
Now.. reality check time....
And you will make twice as much heat that needs to be rejected out of the cooling system, but twice the hp wont give you twice the speed. Twice the fuel cost - at a minimum... you likely will have to run rich to avoid detonation in a real world installation, and intake charge temps may be close to 200 degrees after the intercooler. Drag increases in an exponential fashion, so you will make twice as much heat (linear), generate more power (linear), Generate more drag (logarithmic), and get decreasing amounts of speed for every increased unit of power generated (negative logarithm)
If this is really dumb thinking. My apologies.
I deeply appreciate the great minds and practice on this list.
Cheers
Steve Izett
Dave
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