# Turbocharger Question

1. Mar 18, 2009

### Physics_Kid

i have a question about a turbocharger (air compressor).

a turbo's function is to compress the exit air so that more air can fill the cylinders. let me give an example which will help setup my question:

XYZ motor pumps 10000cfm normally aspirated (1ATM) at max power (lets says 6500 rpm). to double max power the motor needs to combust twice the fuel along with twice the air, so we need to process 20000cfm @1ATM at 6500 rpm. so we need a compressor and the turbo will do that for us and we need 2ATM on the turbo exit to get 2x(na) into the motor.

so my question is, if i stick a flow meter on the inlet and exit of the turbo and run the turbo at 2ATM (aka 2PR or +14.7psi boost) do the flow meters read the same value (given area of inlet and exit tubing is equal). my initial thoughts are no, but some others are saying flow is the same. for a liquid pump i see flow rate is preserved because of incompressible fluid, but the turbo compresses the gas.

i said no because the turbo exit is attached to the inlet of another pump which has fixed physical pumping volume at 6500 rpm (and the compressor itself needs to have mass air preservation across the compressor, mass air(in) = mass air(out)). the turbo has to suck in 20000cfm (2x the NA motor) and compress that air 2x (2ATM) on the exit side. this equates to 2x(na) mass air exiting the turbo. i mentioned "mass air" but i'm only asking about volume rates.

so does P1V1=P2V2 hold true for the turbo where constant P1=1ATM(turbo inlet), V1=Xcfm, constant P2=2*P1, and V2=Ycfm. so if the motor can pump 10000cfm(@6500rpm) i need to fill that same space with a gas that has been compressed 2ATM, doubling the mass air being pumped into the motor. so does 20000cfm @1ATM = 10000cfm @2ATM ?

i may be wrong, just need a explanation.

thanks.

Last edited: Mar 19, 2009
2. Mar 19, 2009

### KLoux

Mass is conserved. So considering the flow rate, (I'll assume you're talking about a mass flow rate, units would be mass/time), the mass entering the turbo must be equal to the mass exiting the turbo. This means that if the area of the opening is the same on the inlet and the outlet, $$P_{1}V_{1} = P_{2}V_{2}$$ holds. Or with different inlet and outlet areas, you have $$A_{1}P_{1}V_{1} = A_{2}P_{2}V_{2}$$.

It sound like you might be confused about how a turbo increases the amount of air flowing into the engine if the mass of air entering the turbo is equal to the mass of air exiting the turbo... Think of it like this: Turbo's don't just push air, they also suck it in. Adding a turbo will increase the amount of air that is coming through your air filter (way back at the beginning of your intake system) compared to a car without a turbo. You have to look at it on a system level rather than a component level.

And believe it or not, the same equations are true for water and air... A substance doesn't need to be compressible in order for the pressure to change.

-Kerry

3. Mar 19, 2009

### Physics_Kid

i am ok with mass flow preservation and the fact that if the turbo is sucking in more air over the na setup (it has to suck in 2x(na) in order to get 2x motor power).

my question is strictly about volume rates. what's inside that volume is not relevant in my question.

i have not yet done my own experiment with flow meters, hence why i am asking before going to the lab, etc. i'm only interested in the volume or volume rate on the inlet and exit while compressing the exit to 2ATM and then filling a motor that can physically only pump a volume of 10000cfm, so the volume rate on the turbo exit side cannot be more or less than whatever the engine is pumping, etc. does this fit the governing equations?

4. Mar 19, 2009

### KLoux

Ahha, now I understand your question - Yes: the volumetric flow rate can be different at the inlet and the outlet.

This still goes back to mass preservation. If mass is preserved, but the pressure of the gas changes (you were right about the compressibility bit, too), the volume will change, hence, the volumetric flow rate changes. Volumetric flow rate is area times velocity, right? And mass flow rate is density times volumetric flow rate. And the density is a function of pressure and temperature. So your volumetric flow rate is a function of pressure and temperature, also, if you hold everything else constant. This means that even if the pressure changes, the volumetric flow rate is not necessarily different (if temperature changes in just the right way), but it certainly could be.

Hope this is what you were looking for...

-Kerry