Temperature gain = (Pressure Ratio^0.28-1) x Absolute ambient air temp°
Then, you have to divide the above result by the efficiency % of whatever blower you are running. The nicest vortech units I believe were near 80%.
For 6psi of boost at 100°F ambient temperature for example:
((14.7psi + 6psi/14.7psi)^0.28-1) x 560°
Sweet, Rankine temperature scale. For when you want Absolute Zero = 0, but aren't snobby enough to use Celsius-based Kelvin.
I wonder how greatly boost juice (50/50 water meth) would affect this math.
Note that I'm not a thermodynamicist or HVAC engineer, so some of this is Scientific Wild-Assed Guess, but it should just add more layers of calculation around latent heat of vaporization.
Here's a list that includes methyl alcohol and water. It's probably at standard pressure and tempertaure instead of 140F and 6psig, but gives you a starting point.
Latent heat of vaporization for fluids like alcohol, ether, nitrogen, water and more.
www.engineeringtoolbox.com
This page lets you calculate water for different temperatures
Online calculator, figures and tables showing heat of vaporization of water, at temperatures from 0 - 370 °C (32 - 700 °F) - SI and Imperial units.
www.engineeringtoolbox.com
If you know how many pounds you're injecting per hour, you can get the BTU cooling per hour due to vaporization from the BTU/lb number (assuming complete vaporization). At 5000rpm (2500 4-stroke intake cycles per minute) a 302ci engine is pulling in roughly 755,000ci or 437 cubic feet of air per minute. This page has calculations for "sensible heat" that we can reverse for temperature drop. Near the bottom it gets into advanced examples that account for humidity, but I'll use the simple equation.
Latent and sensible cooling and heating equations - imperial units.
www.engineeringtoolbox.com
The Imperial unit formula
BTU/hr = 1.08 * cfm * deltaT
converts to
BTU/hr / (1.08 * cfm) = deltaT
Back-of-napkin calculation
50/50 water/meth latent heat of vaporization is roughly 720 BTU/lb. I'll call this mix 8lb/gal.
Snow's methanol FAQ says "For most engines in the 200-500hp range, the standard 3qt tank will last around a tank of gasoline." Say you go through 1/2 tank gas in a 30-minute session (SWAG) that gives you 1.5 gal/hr, IF the wm injection flow tracks the fuel injection flow (which there's a good chance it doesn't, but I'll leave that as an exercise for the reader). So that's 12lb/hr at 720 BTU/lb = 8640 BTU/hr.
8640 BTU/hr / (1.08 * 437 cfm) = 18.3F temperature drop
YMMV, operate at your own risk, objects are closer than they appear, no user serviceable parts inside, professional driver on a closed course, etc.
Anyone care to check my math?