What formula has been used for equating the amount of heat (btu's) our superchargers create at full boost, lets say 18 lbs?

 

22lbs here, I always felt if you really crank it the WIND will reduce the BTUs
Just adding a little levity
Mike

 

I have heard that the s/c is taking like 50 hp which would make sense looking at the belt.

50 hp = 50,000 watts or so

50,000 watts are about 150,000 btus most added to the air, some noise some mechanical losses to the case, more than youd think I bet cha....Woody

 

I found this on a BMW site. Don't know if this guy's right or not, but here it is.

Forced Induction in general produces heat. 1 psi of compressed air= 25 degrees F. Supercharger's are rated by adiabatic efficiency which is basically a measurement for how well it can compress air without heating it. As far as the gears are concerned, they sit in an oil bath to keep them cool and lubricated. You also have to take into effect what type of blower it is to determine where the compression of air takes place. As far as stock internals are concerned, you need to calculate your effective compression and determine wether or not you engine can handle the excess cylinder pressures. The formula is : boost/14.7 + 1 x static compression.

This is the forum:

How hot can a supercharger get???? - Bimmerforums - The Ultimate BMW Forum

 

Compressing a gas causes heating which in the extreme can cause the temps to rise enough to become incandescent, think of a blast wave.

THey were talking about temps which is a different question, the temps are probably in the 400 degree range in our engines, turbos are more efficient that our rotors and they will pull up into this range for a similar boost pressure. Out of the normal thinking for most people.

When you reexpand that air - let it ***** out, it will expand and cool, this is how you get liquid gasses like oxygen and liquid helium to form. Wdy

Enjoy the weekend

 

To get an estimate I would use the ideal gas law (PV=nRT). Assuming constant volume and mass, you can use T2=T1(P2/P1).. Assuming 60ºF inlet temp (288.7K), and a pressure ratio of 2.22 (32.7 PSIA/ 14.7 PSIA) the outlet temp would be 695ºF (641K). To figure out the heat generated, you would use Q=mCpdT (Heat = mass x specific heat x temp change.)
m = density x volume so 1.225kg/m^3 x .00416m^3 (assuming a 3" intake tube 3 ft long) = .0051 kg ..
Q = (.0051kg)(1 kJ/kg K)(641K)
Q=3269.1 J = 3.1 BTU
..Just an estimate

 

AHHhhhh THe volume is not constant, you assumed the temperature to raise the pressure to 2.22 times the ambient pressure, getting the temp.

Compression is WAY more difficult than the equation you selected, the pumping is also not adiabatic so there are inaccuracies.

I do love the 3 btus, sign me up for that one, though I stand by my number of 50,000 times 3.413 or about 150,000 BTUS.

Woody

 

I kept a constant v for simplicity, pretty much comparing the air in and out of the compressor if the inlet and exit manifold were equal in volume. Btu and watts are units of energy and power respectively.. Your 150,000 btu would actually be btu per hour. I didn't want to get into heat transfer lol.. Again just an estimate.

 

using..
(273K+ambientC)*(BoostkPaa/SC Inlet kPaa)^((Cp/Cv-1)/Cp/Cv)-273K = tempC post SCharger
- going into your cooler.....and assumes you dont loose many inWG through the cooler to the MAP

for 14.5psi sea level - stock pulley 74/155
my calcs get 90,000 BTU's/hr (26kW) added to 110,000 for a hot summers day.....
at 95F, 70%RH 673cfm intake snorkel at 6,000rpm...

see attach - cells shaded in yellow are variable you can change....
The volume post the Charge Air cooler is determined from the engine displacement x rpm and is hat I call "THE DOG"....so you can change various conditions (what I call THE TAIL) and the feed to the engine for a specific rpm stays the same...ie: the dog is waging the tail - not the other way 'round....
hence for various rpm, pulley configs etc...cfm intake changes, intake and TB drops change....etc...but feed to the motor...constant.
cfm should be 377...but 7cfm is fuel vapour.
Don't worry about the water side - that's a manual calc.....not linked (apologies)

I'm sure everyone will have a different take on this.....let it rip....

 

That's a nice spreadsheet! I was just coming up with a basic formula off the top of my head in about 5 minutes, but your model looks much more complete lol.

 

I love it the spread has 40.3 KW - about 40 hp.

BTUs were 137,000

I concur with the relative values, IE the s/c is making great heat and the I/C has a heck of a job to do.

I also love the comment that there is a boost loss in the I/C, that number is something I plan to do measurment on the losses this month. I expect that the results will be interesting.

Woody

 

137,000 is the BTU/hr required by the crank and at 65% eff. that's 90,000 input to the air coming into the SC.
122F AIT is an assumption that if you were running 100+mph in whatever gear it takes to keep 6,000rpm, your AIT's would be....122F with a stock IC, water pump etc.

65mm sheet comparison coming....

 

Thanks for the info, this supports my numbers as well.