I hope this isnt but i saw a vid were a dsm smoked a z06 and some one said that that for every bar (14.7psi) of boost the displacement of a motor doubles hows true is that.?? is it that simple? is it only based on psi or does turbo out put/size play a roll? because if so there is a replacement for displacement lol happy NEW YRS BTW!

 

As you well guessed its more than just pressure, but based on the math yeah it would seem that way wouldnt it. While it is true to some extent there are so many variables in play that you cant say it is in one case and not the other. The reason most of us think in terms of the output based on the airmass of the engine instead.

For instance the old rule of thumb is that for every lb/min the engine is consuming (or turbo is flowing more accurately) it will generate approx 10whp.

 

thnx!

 

It is true, but you have to remember that volumetric airflow doesn't equal mass airflow. Mass airflow includes temperature as well. It would be true for mass airflow if the turbo didn't heat the air and the VE of the engine wasn't changing. But, as we all know, turbos heat the air while pressurizing it (why we run intercoolers) and that is why turbo compressor maps have efficieny islands. It shows how much the air is being heated for a given mass airflow and boost pressure. The more efficient the engine (higher VE) and the more efficient the turbo, the closer you will get to that ideal.

Here are a couple quick equations that should help you out:

Volumetric airflow through an engine:

Airflow (CFM) = PR[RPM*V.E.*Cid/3456]
PR=Pressure ratio=(boost in psi+atmos(psi))/atmos(psi)
RPM = RPM of engine
V.E. = volumetric efficiency at RPM being measured
Cid=cubic inch displacement= 122 for our 2.0L engines

As you can see, doubling the atmospheric pressure (boost of 14.7psi at seal level) will double the volumetric airflow. However, as mentioned, we aren't doubling the mass airflow and thus the HP. You need to take the temperature into effect.

As you can see in the equation below, mass airflow is inversely proportional to temperature. To get the mass airflow from the volumetric airflow from the above equation:
n (lbs/min) = P (psia) x V (CFM) x 29 / (10.73[ft3·psi· °R-1·lb-mol-1] x T)

The temperature increase from a turbo's coimpressor can be found by:
Tout = Tin + Tin x [-1+(Pout/Pin)0.263]/(compressor efficiency)

Tin - compressor inlet temp
Pout - compressor outlet pressure
Pin - compressor inlet pressure (usually a little below ambient due to vacuum in intake pipe, etc)
compressor efficiency - the efficieny of the compressor taken from the compressor map for your pressure ratio and volumetric airflow

A quick example from the web here shows an example where the inlet temperature is 70 deg F, the suction pressure is -0.5 psig (a slight vacuum), the discharge pressure is 19 psig, and the efficiency is 72%. The outlet temperature is 257.8F.

This is where our intercoolers come into play. There are even more mathematical equations for intercooling, with heat transfer rates, pressure drops, etc, etc, but basically an intercooler has a percent efficiency, meaning how close the intercooler can cool the air back down to the original turbo inlet temp. It seems as if 80% is a good number for a good intercooler.

So, anyway, to answer your question, doubling the atmospheric pressure with boost (14.7psi) will double the volumetric airfllow, but will not double the mass airflow and corresponding horsepower due to temperature increase. The better the turbo's and intercooler's efficiency, the closer you will get.

Making more power out of a turbocharged car really is simple...just keep VE as high as possible and temp as low as possible. Easier said than done sometimes, though.


Eric

 

lol you totally annihilated that answer. Bravo.

 

Eric , perhaps you could elaborate

 

hey eric THANKS u definitely killed it!. surprisingly i understood the math (glad i took physics) , and makes more sence now . so i have new data to show off and use for proof when i get into arguments lol.