Power output of a intercooler and turbocharger

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SUMMARY

The discussion focuses on the power output of an automobile engine enhanced by a turbocharger and intercooler, specifically in a 2008 Mini Cooper S. The turbocharger compresses air to increase its mass per unit volume, while the intercooler cools the compressed air to maintain efficiency. Calculations reveal that the mass of air exiting the intercooler at 1.45 × 10^5 Pa is crucial for determining the percentage increase in power output compared to atmospheric conditions. The discussion also highlights the need to differentiate between power calculations with and without the intercooler.

PREREQUISITES
  • Understanding of thermodynamics, specifically adiabatic processes
  • Familiarity with the ideal gas law (PV=nRT)
  • Knowledge of automotive engineering concepts, particularly turbocharging
  • Basic proficiency in calculating mass flow rates and power output
NEXT STEPS
  • Calculate the mass of air entering the engine without a turbocharger
  • Explore the effects of intercooling on engine efficiency and performance
  • Investigate the thermodynamic principles behind adiabatic compression
  • Learn about different types of turbochargers and their impact on power output
USEFUL FOR

Automotive engineers, performance tuners, and students studying thermodynamics or automotive engineering will benefit from this discussion, particularly those interested in enhancing engine performance through forced induction systems.

Anne3
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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?
 
Last edited:
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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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