globodj

Can anyone tell me if this is correct?

I am currently running a 85mm pulley on my S/C. If I reduce the pulley size to a 70mm it represents a 18% reduction in diameter.

After a dynotune, with the 85mm pulley my output is 178 hp.

Without any other changes, can I calculate the potential increase in hp like this:

178hp x 18% = 32 hp increase

Is this a close calculation or is there other factors that will determine the actual output?

Thank you.

richyvrlimited

iTrader: (1)

You're assuming 100% efficiency, which it isn't, the faster spun charger will create much more heat, at absolute best an M45 supercharger is 67% efficient. If you're already out of that efficiency zone then the output will be even less.

the MP62 is slightly more efficent at 68% but moves more air for the same pulley ratio.

http://www.eaton.com/Eaton/ProductsS...Maps/index.htm

a 9% increase in power from that pulley change is much more realistic IMO.

globodj

The car is a 94R with a 99 low mileage engine.
S/C - cold side UB2 sold about 7 years ago by FM
BEGI air intercooler
600cc injectors
high capacity aluminum radiator
FM coolant re-route kit
RB 4/1 header, high flow cat and RB catback
MS PLP
NGK Iridium IX-4 plugs
upgraded ignition wires

With the setup listed above, how much more boost/hp can I induce safely without internal mods?

richyvrlimited

iTrader: (1)

Whatever ratio will net 250~ ft/lbs of torque. That's when rods give up.

globodj

Good to know.
Thanks.

skidude

iTrader: (3)

Is that still true with superchargers? Turbos use power from the exhaust to create boost, and are pretty efficient at it. Supercharges actually use the horsepower at the crank to make boost, so they are directly stealing it from the wheels.

richyvrlimited

iTrader: (1)

?? 250ftlbs is 250ftlbs.

If anything a supercharger can make more torque safely as it's much linear in its torque delivery than a turbo i.e. no sudden hit as the thing spools at xxxxrpm

ianferrell

The problem is you're having to make some large chunk of power to turn the blower... so out of my ***, if the blower takes 50 ft/lbs of torque to move enough air to make 200 ft/lbs at the wheels, its actually making 250 ft/lbs of torque and the rods are seeing that much force, even though you're only at 230 whp 200 wtq or something.

skidude

iTrader: (3)

That is what I was trying to ask, but ianferrell said it much better. This is all ignoring the drivetrain losses, of course.

globodj

Yes, there will be about a 4% loss. Corky Bell as usual was kind enough to give me a "guesstimation" based on what info he has. Here's his figures:

No, not quite right. Two things; the added power must take some of the added power to run the blower faster and at a higher pressure. Second, the air will be heated more because of more compression to a higher pressure. Heating the air will move the air molecules a little farther apart, thus a few less per CFM.

An approximation would be about 8% of the intended gain power the SC, or: 8% 0f 18 = 1.5% That drops your expected gain to 16.5%.

The heating of additional compression will reduce the air density proportional to the absolute temperature. “Absolute” means all the way down to no heat, which is about 460 F below or normally referred to “0” which is just 32F below freezing.

Guessing that your SC produced 6 psi boost, the boost change would be about 4.5 psi. This will create a temp increase of approximately 17F for every psi, or about 75 F.

The temp change of the first 6 psi would be about 6 x 17 = 102F. That drops the air density by approx;

On an 80 F day:
Change = (460 = 80 + 102 )/(460 + ambient of 80) = 12%

For the next 4.5 psi, with the additional temp of 75F:

Change will be about (642 + 75)/ 642 = another 12% of the expected gain.

So, of the 18%,, down to 16.5% due to powering the blower, another 12% drops due to the air density change.

Or: 16.5% x (1 - .12 ) or 16.5% x .88 = 14.5%

So, instead of expecting 18%, or 32 hp, you should see something closer to 14.5% or about 27hp.