Engine Turbochargers and Intercoolers

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The discussion focuses on the mechanics of turbochargers and intercoolers in automotive engines, emphasizing the relationship between air mass and engine power output. Turbochargers compress air before it enters the engine, increasing the air mass available for combustion, while intercoolers cool the compressed air to maintain efficiency. The calculations provided aim to determine the mass of air that can fill an engine cylinder at a specific pressure and volume. The user arrives at a result of 9.75 x 10^-4 m^3 but expresses uncertainty about its accuracy. The conversation highlights the importance of unit conversions and correct application of gas laws in engine performance calculations.
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The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine's cylinders to react chemically with gasoline. Many cars have a turbocharger which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an intercooler in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure 1.01 x 10^5Pa, density rho = 1.23 , kg/m^3, and temperature 15.0 ^C. It is compressed adiabatically to 1.45x 10^5 Pa. In the intercooler, the air is cooled to the original temperature of 15.0 C at a constant pressure of 1.45x 10^5Pa.

If the volume of one of the engine's cylinders is 575 , cm^3, what mass of air exiting from the intercooler will fill the cylinder at 1.45 x10^5Pa?


PV=nrt
m=nM, where M is the molar mass

So i solved for n and got PV/RT=n


m=(PV/RT)M, where M is .029

P=1.45 x 10^5
V=575 x 10^-6
R=8.314
T=298


plug it in and i got 9.75 x10^-4, does that seem reasonable for an engine?


Im thinking its wrong but I am not sure why
 
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Without showing your units who knows?
 
Everything was converted to standard units


P=1.45 x 10^5 P
V=575 x 10^-6 m^3
R=8.314
T=298 K


so my answer would be 9.75 x10^-4 m^3
 
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