Power output of a intercooler and turbocharger

AI Thread Summary
The power output of an automobile engine increases with the mass of air entering the cylinders, which is enhanced by a turbocharger that compresses air before it enters the engine. This compression raises the air temperature, necessitating an intercooler to cool the air back to the original temperature while maintaining pressure. Calculations for the mass of air exiting the intercooler and the resulting power output increase are essential for understanding performance gains. The discussion also highlights the need to differentiate between calculations with and without the intercooler, as the power ratios will vary. Understanding these dynamics is crucial for optimizing engine performance through forced induction systems.
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Homework Statement


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 (like a 2008 Mini Cooper S) have a turbocharger, which compresses the air before it enters the engine, giving it a greater mass of air per unit 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 drawn into the turbocharger at atmospheric pressure (1.01 × 10^5 Pa), density = 1.23 kg/m^3 , and temperature 15.0 C. It is compressed adiabatically to 1.45 × 10^5 Pa. In the intercooler, the air is cooled to the original temperature of 15.0 C at a constant pressure of 1.45 × 10^5 Pa.
a) Draw a − diagram for this sequence of processes.
b) 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 × 10^5 Pa? Compared to the power output of an engine that takes in air at 15.0 C and atmospheric pressure, what percentage increase in power is obtained by using the turbocharger plus intercooler?
c) Repeat the previous part just using the turbocharger (no intercooler). What percentage increase in power do we get this time?

Homework Equations


P=W/t
PV=nRT
m=PV/RT*(0.29)
T=298
R=8.314

The Attempt at a Solution


I have drawn the p-v diagram for part a.
For part b) I assume you would find the mass needed by using m=PV/RT=9.74*10^(-4), for the second part however how would you find the power obtained?
For part c) what would change in your calculation?
 
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You don't find the power obtained. You find the ratio of the powers obtained.

I haven't checked your result, but isn't the equation you used for the number of moles, not the mass? How much mass is taken into the cylinder without the turbocharger.
 
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