Psi vs Cfm Options

[KOU In3]

Ok, I've always been led to believe that CFM was more important a measure than psi and that the larger turbos flow grossly more air at a given psi than their smaller brethren. I understand the basic part that the hot air produced when the turbo is overworked leads to a higher psi reading from the same quantity of actual air.

Here's where I start to get confused though: if we keep the intake charge's temperature constant through effecient intercooling shouldn't these values be the same? Basicly if you pressurize the manifold to 20psi using 120 degree air from a 14b or the same temp and psi from say a mammoth BR580 shouldn't this be using the same cfm of air?

Right when I think it all makes sense I talk myself back into the classic fallacy of 'dude it's 20 psi!'. Is the Someone straighten me out here please.

[ICGerms]

...I understand the basic part that the hot air produced when the turbo is overworked leads to a higher psi reading from the same quantity of actual air...

I think that you're confusing yourself on this...hot air isn't going to produce more psi. An "overworked turbo" will produce more air, albeit hot air. A turbo that is not being overworked (e.g. a larger one) will produce the same amount of air (in cfm) but that air will be at a lower temperature. In other words, the larger turbo is more efficient in producing the larger amount of air. Also, the larger turbo will be better able to produce more cfm at lower psi than the smaller turbo. Say turbo X produces 100cfm @ 15psi and turbo Y produces 200cfm @ 15psi...turbo Y is the bigger turbo. Sure, turbo x could probably produce 200cfm, but it would have to do so at a higher psi...30psi, for example. Since turbo X is most likely out of it's efficiency range, it is producing those 200cfm as hot air now, while turbo Y is able to produce it's efficient (15psi) 200cfm as nice, engine-friendly cool air.

[KOU In3]

think that you're confusing yourself on this...hot air isn't going to produce more psi

Hot air is going to produce more psi. Not useful for performance but still more psi nonetheless.

Imagine a sealed 10 pound nitrous tank filled with air. Resting at 1100 psi at 70 degrees farenheit. Now since it's sealed, the amount of air in the tank isn't changing. Crank the ambient temperature up to 100 degrees farenheit though and the pressure (psi) of the air in that tank is going to rise. I think one of the laws governing gas expansion applies actually. PV=NRT or one of those. If I try to quote it I'll definate make an even bigger @ss of myself though.

Should be the same principle when pressurizing a manifold. Not saying that that's going to net a performance gain just that the hot air will result in a higher psi. You are right that I've confused myelf though.

I think I'm getting confused at how the turbo is moving more air at the same psi. If it's not due to the lower charge temp. Is the reading of what is being pressurized taken somewhere where I'm not getting this? Is the cubic feet of air measurement still a temperature sensative unit or an absolute?

Thanks for humoring me here.

[AWD DSM 1]

Would it make any sense to say that although they may be pressurizing the same (20 psi say) that the bigger turbo will create more velocity? Basically it gains the cfm by moving the air through the intake at a faster pace? I don't have anything to back this up other than that would be one way to explain why a bigger turbo could produce more air at the same psi.

The temp would also have a big part in it. 100 deg air is much more dense than 200 deg air, thus more cfm.... right? One thing I don't quite understand is how turbo efficency works. Why is it that a larger turbo can produce cooler pressurized air than a small one? I know that no matter how you do it, air will increase in temp as you pressurize it some.

Maybe it's just been too long since my last physics class...

[natedogg]

PV=NRT

temperature (T) and pressure (P) are dynamic variables in our system, but volume (V) is a constant unless we change it so a better name for it might be static variable. R is a constant correction factor and N is a variable based on the amount or mass of gas molecules we are moving.

N=(PV)/(RT) I write the equation in terms of N because that is really what we are concerned with...the mass of air. Its obvious by looking at the equation this way that we want high pressure and volume, but low temperature. Pressure is dependent upon temperature, but temperature is not dependent upon pressure. Imagine an air compressor. The air inside an air compressor charged to 90 psi isn't any warmer than the ambient air outside the compressor, but if you start heating the air inside the compressor its pressure will increase. Apply this to a turbo...increasing the pressure is not increasing the temperature of the air. But the higher pressure fighting the compressor blades causes a lot more friction and this is where your heat and inefficiency comes from. Now this increasing temperature is, like Travis said increasing the pressure within the system and the converging feed back of the increasing temps and pressures leads to very hot air. So with pressure and temperature fighting our efficiency we need to find a different way to increase the mass of air going to our engine. Lets increase the volume. This allows more air at the same efficient pressure and temp. Of course this is done with larger IC and piping. This explains the ideal gas law in our application.

Now where does the turbo come in at? Well, the best way I can think of to explain this is with two fans. A big one and a small one. The small fan has to spin twice as fast to flow as much air as the large one at the same pressure. Ala a small turbo has to spin twice as fast to flow as much air as a larger turbo at the same pressure. Since the small turbo as to spin twice as fast it is heating the air much faster than the larger turbo and therefore becoming inefficient much sooner. The larger turbo is capable of flowing more air at the same pressure because it has larger fins and can just plain move more air with each revolution of the fins.

That's how I understand it anyway. Hope this helps.

[AWD DSM 1]

All bow to the great source of knowledge and wisdom.....

Seriously though, that was a pretty good explination that I think I understood. I think....