ok so wierd question.

if i have a 6.2L motor. that means every time my crank spins once. it sucks in and blows out 6.2L of air.

so if i bolt say a 2.3L supercharger(whipple style) and that means in one crank spin i am feeding 2.3L of air.

so does that mean in thereo i am consumming 8.5L of air per crank spin?

and if so how do i figure this with a turbo?

thanks

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http://www.whipplesuperchargers.com/...sp?ProdID=1162

^^^ info i found. so if i spin my motor 6000 rpm for a min and the blower spins 18,000 rpm for a min. then the blow spins 3 times for my cranks every 1. so that means. that it is feeding 2.3L/rev so 2.3 x 3 = 6.9 + 6.2(my motor) = 13.1L of air for every crank spin?

 

Your theory assumes a few items which aren't necessarily true. At exactly 100% VE, your motor only consumes 1/2 your displacement, so only 3.1L per revolution. Your 2.3L blower only pushes almost 2.3L of air at peak efficiency. The best designs still cannot achieve 100% efficiency. In reality you just have to figure how much air the blower or turbo is providing at X rpm's, and that is what your motor is eating. This involves a detailed knowledge of compressor maps.

 

In a four stroke engine, the engine will fire each cylinder once in every two revolutions. At 100% volumetric efficiency (WOT in a perfect world) that would mean that for every revolution of the crank, your engine fires four cyl (half of its displacement) moving 3.1L of air.

For theory's sake, a 2.3L blower will displace 2.3L of air into the engine per BLOWER revolution. Due to the different sized pullies, the blower and the crank are NOT revolving at the same speed (rpm). The blower will be revolving more quickly than the engine. You would need to know this "step up" ratio to determine the "volume" of air that it it will displace crank revolution.

Your "step up" ratio is 3:1 as referenced in your original post. That means for every revolution of the crank, the blower will (in a perfect world) take in 6.9L of air (2.3X3). What your theory does not account for is that the VOLUME of air that moves into the engine each revolution does not change. Even though the blower is able to "inhale" 6.9L of air per engine revolution, due to the unchanging cyl volume (3.1L per revolution), it must "exhale" the MASS of the initial 6.9L of air into a smaller volume (3.1L). The volumetric differential between the blower's intake and exhaust tracts creates PRESSURE. More simply even though the blower on your engine can "inhale" 6.9L per crank revolution (13.8L per complete combustion of 8 cyl) it is "exhale" is limited by the cylinder volume of 3.1L per crank revolution (6.2L per complete combustion of 8 cyl.). The volume of air is the same for both the NA and supercharged combos (in a perfect world assuming no other variables). The MASS of air, on the other hand is different due to it being at a higher PRESSURE (density).

Now to answer your question, each time the engine spins once, the blower takes in 6.9L of air. You do not add the volume of the engine as the blower is now the ONLY source of air into the engine. Given that the blower takes in 6.9L of air, the engine takes in that same MASS of the 6.9L of air, but in the smaller 3.1L volume (1/2 of total engine displacement) making the pressure higher (Boost).

 

thank you

 

So are you trying to say that a 4.6 cobra motor with a 2.3 whipple (popular mod) is equal to a 11.5 L?

 

Read my post

 

Read the thread, newb.

 

Only if it makes 1100 hp+

Don't they all?


Superchargers are really "density increasers" as ryarbrough suggested. Picture more oxygen molecules stuffed into the cylinders along with more fuel to burn with the extra oxygen. Not all of those molecules coming thru the blower get into the cylinder. Their brothers are taking up much of the space and causing a back pressure which you could call "boost".

It's kind of like packing people into subway cars like they often do in Japan. Not everyone who wants to get in can fit, so they wait for the next car. Just because you get into the station door (blower intake) doesn't mean you'll get into the next car.


Jon

 

So with all those people jamed in there the heat goes up.

 

Yepper!

 

...and after the cycle is complete (train gets to the next station)...it's harder for them all to get back out because the train still has the same sized doors as it did off-peak hours (out of boost)...part of the reason why VE tends to drop as boost comes up...because if they don't all make it out the door...now there's less room for the next group of people to fit inside the train.

 

Mike, did you forget about all the milling around and fighting the passengers did from stop to stop? They got very agitated during their trip and when the doors opened they shoved the folks in front of them out with vigor!

It's all about "delta p". Getting out is far easier than getting in with many times the delta p.

Assuming the passengers are going out the opposite doors from the incoming folks (cross-flow train cars), I think it would be important when you opened the exit doors. It may be before the train completely stops.

Are we getting too carried away here?


Jon

 

haha...yes

 

No, that actually worked out to be a great analogy.

I never really thought to hard about this stuff, but it makes sense now.

Even though im N/A.

Thanks guys.

 

The only real difference between forced induction and a well-designed NA engine is the amount of increased density. NA engines get "boost" from intake harmonics or "inertia ramming". The same type of packing in of the molecles occurs for the same reason...pressure differential between the cylinder and the intake tract. It's called "Delta p". Delta refers to a difference and p refers to pressure.

Visualizing compressible fluid (air) flow in an operating engine is very difficult. Sometimes analogies help.


Jon

 

I love this section of the forum. You guys are pretty damn smart...This was stuff that I thought that i already knew, but you guys really cleared it up with the "subway" analogy...Good stuff!!!