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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 \(\times\) 10\(^5\) Pa), density \(\rho\) = 1.23 kg/m\(^3\), and temperature 15.0\(^\circ\)C. It is compressed adiabatically to 1.45 \(\times\) 10\(^5\) Pa. In the intercooler, the air is cooled to the original temperature of 15.0\(^\circ\)C at a constant pressure of 1.45 \(\times\) 10\(^5\) Pa. (a) Draw a \(pV\)-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 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger alone?

Short Answer

Expert verified
Using both turbocharger and intercooler increases power by about 24.2%. Using only the turbocharger increases power by about 18.0%.

Step by step solution

01

Understanding the Problem

We need to solve a multi-part physics problem involving a turbocharger and intercooler in a car engine system. The components require us to create a pV-diagram, calculate masses of air in engine cylinders under different conditions, and determine the percentage increase in engine power.
02

Analyzing Adiabatic Compression

For adiabatic processes, we use the equation \( p_1 V_1^\gamma = p_2 V_2^\gamma \), where \( \gamma = 1.4 \) for air. We know the initial pressure and final pressure and want to understand the volume changes. Initially: \( p_1 = 1.01 \times 10^5 \) Pa. Finally: \( p_2 = 1.45 \times 10^5 \) Pa.
03

Drawing the pV-Diagram

To draw the pV-diagram, note that the processes include an adiabatic compression (increasing pressure and decreasing volume without heat exchange) followed by an isobaric cooling (constant pressure and decreasing temperature). Start from an initial state with pressure \( p_1 \), density \( \rho_1 \), and temperature \( 15.0^{\circ} \mathrm{C} \), compress adiabatically to \( p_2 \), then cool isobarically back to \( 15.0^{\circ} \mathrm{C} \).
04

Calculating Volume Exiting Intercooler

With constant temperature and pressure at the intercooler exit, apply the ideal gas law \( pV = nRT \). Assume \( R = 287 \) J/kg·K for air, calculate the final density \( \rho \) at \( p_2 \). Using \( V = 575 \) cm\(^3\) (converted to cubic meters), determine \( m = \rho V \).
05

Calculating Initial Mass

At atmospheric pressure, calculate the initial mass of air using initial conditions (\( \rho = 1.23 \) kg/m\(^3\) and \( V = 575 \) cm\(^3\)). Use \( m = \rho V \) to obtain mass \( m_1 \).
06

Percentage Increase in Power Using Intercooler

Power increase is proportional to mass. Calculate the percentage increase by \( \frac{m_{intercooler}}{m_1} \times 100 - 100\).
07

Without Intercooler (Adiabatic Only)

If the intercooler is not used, repeat similar steps for an adiabatic process with new density \( \rho_a \) determined by adiabatic conditions and calculate the mass \( m_a \).
08

Percentage Increase Using Turbocharger Alone

Calculate \( \frac{m_a}{m_1} \times 100 - 100\). This gives the percentage increase in power if only the turbocharger is used without the intercooler.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Turbocharger
A turbocharger is a device used in engines to increase efficiency and power output by compressing the air that enters the engine's cylinders. This compression process allows a greater mass of air to mix with the fuel, maximizing combustion. The principle behind a turbocharger is to utilize engine exhaust gases to drive a turbine, which in turn powers a compressor.
This compressor draws in outside air, compresses it, and forces it into the combustion chambers of the engine.
  • Increased air density leads to more oxygen available for combustion.
  • The result is improved engine performance and often better fuel economy.
Turbochargers are especially beneficial in engines operating at high altitudes where air pressure is lower, as they provide the necessary boost to maintain engine power. Understanding turbochargers involves examining both mechanical and thermodynamics principles, as they rely on high-speed rotational motion and efficient energy conversion processes.
Adiabatic process
An adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings. This means that any change to the system's internal energy is solely due to work done on or by the system. Key traits of an adiabatic process include:
  • Temperature changes are a result of internal energy shifts, not external heating or cooling.
  • For gas, it often involves compression or expansion where pressure and volume changes are linked via the equation: \[ p_1 V_1^\gamma = p_2 V_2^\gamma \]
Here, \(\gamma\) is the adiabatic index, which is 1.4 for air. Adiabatic processes are critical in understanding how turbochargers work, as the air is compressed rapidly causing a rise in temperature without heat exchange with the environment. It's important to note that for compression, the temperature of the gas increases, which can further be lowered using an intercooler before entering the engine.
pV-diagram
A pV-diagram is a graphical representation of the changes in pressure (p) and volume (V) of a system. It's an essential tool in thermodynamics for visualizing different processes like adiabatic, isothermal, isobaric, and others. Typically, a pV-diagram for a turbocharged engine cycle includes:
  • An upward sloped curve for adiabatic compression, indicating increased pressure and decreased volume with no heat exchange.
  • A horizontal line for isobaric cooling, which shows constant pressure as the volume decreases due to temperature reduction when passing through the intercooler.
  • The axes represent pressure on the vertical and volume on the horizontal.
By analyzing these diagrams, one can see how the volume of air decreases during compression and how any heat interaction stabilizes or changes the pressure afterwards. This visual representation helps in understanding the efficiency and performance adjustments made by devices like turbochargers and intercoolers.
Ideal gas law
The ideal gas law is a fundamental equation in thermodynamics that relates the pressure, volume, and temperature of a gas. It's expressed as:\[ pV = nRT \]where \(p\) is pressure, \(V\) is volume, \(n\) is the amount of substance in moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. This equation assumes that the gas behaves ideally, meaning the particles are in constant motion and the interactions between them are negligible.
Understanding the ideal gas law is crucial when dealing with processes like those in a turbocharged engine where:
  • The compressed air exiting a turbocharger can be analyzed by the changes in pressure and temperature.
  • It allows for calculations of changes in state, such as displayed in the problem where air pressure and temperature affect density.
Knowing these relations helps in determining the mass of air, the key to estimating the engine's power output.

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Most popular questions from this chapter

The engine of a Ferrari F355 F1 sports car takes in air at 20.0\(^\circ\)C and 1.00 atm and compresses it adiabatically to 0.0900 times the original volume. The air may be treated as an ideal gas with \(_\Upsilon\) = 1.40. (a) Draw a \(pV\)-diagram for this process. (b) Find the final temperature and pressure.

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

During an adiabatic expansion the temperature of 0.450 mol of argon (Ar) drops from 66.0\(^\circ\)C to 10.0\(^\circ\)C. The argon may be treated as an ideal gas. (a) Draw a \(pV\)-diagram for this process. (b) How much work does the gas do? (c) What is the change in internal energy of the gas?

On a warm summer day, a large mass of air (atmospheric pressure 1.01 \(\times\) 10\(^5\) Pa) is heated by the ground to 26.0\(^\circ\)C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 \(\times\) 10\(^5\) Pa. Assume that air is an ideal gas, with \(\Upsilon\) = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C\(^\circ\) per 100 m of altitude, is called the dry \(adiabatic\) \(lapse\) \(rate\).)

Starting with 2.50 mol of N\(_2\) gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0\(^\circ\)C, a chemist first heats the gas at constant volume, adding 1.36 \(\times\) 10\(^4\) J of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. Calculate (a) the final temperature of the gas; (b) the amount of work done by the gas; (c) the amount of heat added to the gas while it was expanding; (d) the change in internal energy of the gas for the whole process.

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