ELECTRIC TURBO Fans are not designed for creating positive pressure! They can only put out 5 to 6 in H20 (.18-.22psi). They are designed to move air from one location to another not to pressurize it. OK, lets go with a 2001 celica gts for comparison because I'm tired of talking about Honda this and Honda that... specs: Displacement: 1.8 liters (1,796 cc) Bore x Stroke: 82.0 mm x 85.0 mm Compression ratio: 11.5:1 Horsepower: 180 hp @ 7,600 rpm Torque: 130 lb.-ft. @ 6,800 rpm Now that we have the specs of the motor, lets see how much air it consumes. v = (pi/4)*(bore)^2*(stroke)*(Ncylce)*(# Cylinders) v = 3.14/4*(3.228in)^2*(3.346in)*(7600/2)*(4) v = 416369in^3/min(1ft^3/1728ft^3) v = 241.0 cfm Mind you this is 100% VE, so say at a more reasonable 90% that leaves us with: v = 241.0cfm * 0.9 v = 216.9 cfm actual Now I am sure you are thinking that the fan is forcing the extra CFM into the motor right? Wrong. The air will actually just pass back between the blades and the housing, providing no benefit to power. Here is the thermodynamic method of finding the power a fan needs to move and slightly compress air (between 0.18-.22psi), but lets say it will magically create 5 psi boost for a "30hp" gain. Fan air power = w * Ht Fan air power = (216.9cfm)*(5psi)*(144in^3/1ft^2)(1hp/33000ftlb/min) Fan air power = 4.7 Hp 4.7hp = 3.5kw of power. Divide that by 13 volts (average operating battery voltage) and that leaves you with 270 amps of power. It leaves you to wonder if you would need a 270 amp inline fuse for this monster fan. I dunno about you, but I've yet to see anything with a 270 amp fuse in a car. Keep in mind, this is with 100% efficient motor as well, so it would actually have to consume more power than that. But wait....the factory rating on most alternators are around 80-90 amps! Hm, something seems wrong... seems this electric turbo is consuming 3 times as much as the alternator can put out. hmm... odd. Long story short an electric turbo just won't create more power than what they consume to run them. RAM AIR Operating Conditions Temperature = Tair = 20C = 293K Atm. pressure = Pair = 14.7PSIA Cpair = 1005J/KgK K = 1.4 The Cpair and K are constants for air. Case 1 @ 100Kmh (62Mph) = 27.78m/s Calculating temperature of the ram air Tramair = ((Vcar^2/2gc)/cp) + Tair Tramair = (((27.78m/s)^2/2(1kgm/Ns^2)/1005J/kg) + 293K Tramair = 293.4K The temperature increased by 0.4K or 0.4C. Pram = Pair (Tram/Tair)^(k/(k-1)) Pram = 14.7PSIA (293.4K/293K)^(1.4/(1.4-1)) Pram = 14.75PSIA - 14.7PSIA Pram = 0.05PSIG (gauge pressure) So as you can see driving 100kmh will only have a gain of 0.05 psi! now lets try for 200kmh. Case 2 @ 200Kmh (124Mph) = 55.5m/s Calculating temperature of the ram air Tramair = ((Vcar^2/2gc)/cp) + Tair Tramair = (((55.5m/s)^2/2(1kgm/Ns^2)/1005J/kg) + 293K Tramair = 294.5K The temperature increased by 1.5K or 1.5C. Pram = Pair (Tram/Tair)^(k/(k-1)) Pram = 14.7PSIA (294.5K/293K)^(1.4/(1.4-1)) Pram = 14.97PSIA - 14.7PSIA Pram = 0.27PSIG (gauge pressure) By seeing how the velocity of the car increases the ram air effect...it is barely anything! I'll conclude by saying that in racing circles where a 1/100th of a second counts, it's worth it, but don't let Pontiac fool you into thinking it works on the street.