/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A diesel engine does not use spa... [FREE SOLUTION] | 91Ó°ÊÓ

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A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of \(21^{\circ} \mathrm{C}\) is compressed adiabatically to a temperature of \(688^{\circ} \mathrm{C}\) . Assume the air to be an ideal gas for which \(\gamma=\frac{7}{5} .\) Find the compression ratio, which is the ratio of the initial volume to the final volume.

Short Answer

Expert verified
The compression ratio is 18.27.

Step by step solution

01

Understand the Adiabatic Process

An adiabatic process is one in which no heat is transferred into or out of the system. For an ideal gas undergoing an adiabatic process, the equation \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \) holds true, where \( T_1 \) and \( T_2 \) are the initial and final temperatures, \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( \gamma \) is the heat capacity ratio of the gas.
02

Convert Temperatures to Kelvin

To use these equations, the temperatures must be in Kelvin. Convert the initial temperature \( T_1 = 21^{\circ} \text{C} \) to Kelvin: \( 21 + 273.15 = 294.15 \text{ K} \). Similarly, convert the final temperature \( T_2 = 688^{\circ} \text{C} \) to Kelvin: \( 688 + 273.15 = 961.15 \text{ K} \).
03

Apply the Adiabatic Equation

Using the adiabatic equation \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \), solve for the compression ratio \( \frac{V_1}{V_2} \). Substituting the values we have: \[ 294.15 \times V_1^{(7/5)-1} = 961.15 \times V_2^{(7/5)-1} \] Simplify this to find the volume ratio.
04

Solve the Equation for Compression Ratio

Rearranging the equation gives: \[ \left( \frac{V_1}{V_2} \right)^{\gamma-1} = \frac{T_2}{T_1} \] Substituting \( \gamma = \frac{7}{5} \):\[ \left( \frac{V_1}{V_2} \right)^{\frac{2}{5}} = \frac{961.15}{294.15} \] Calculate the right-hand side:\[ \left( \frac{V_1}{V_2} \right)^{\frac{2}{5}} = 3.268 \] Raise both sides to the power of \( \frac{5}{2} \) to solve for the compression ratio:\[ \frac{V_1}{V_2} = 3.268^{\frac{5}{2}} = 18.27 \]
05

Interpret the Result

The calculated compression ratio is \( 18.27 \). This means that the initial volume of air is compressed to a volume that is about 18.27 times smaller during the adiabatic process in the engine.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental concept in physics and chemistry. It provides the relationship between the pressure, volume, and temperature of a gas. The equation is written as: \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
The law assumes that gas molecules do not attract or repel each other, and that they occupy no space — both of which are idealizations. Despite these assumptions, it is accurate for many gases under a variety of conditions.
  • Used to calculate changes in gases when they undergo transformations.
  • While not directly used in the adiabatic process, forming the backdrop to understanding gas behavior.
Understanding this law is essential, especially as it lays the groundwork for more complex concepts, such as the heat capacity ratio and adiabatic processes.
Compression Ratio
The compression ratio is a key factor in the performance of engines. It describes how much the volume of gas within a cylinder is compressed. Mathematically, it’s expressed as \( \frac{V_1}{V_2} \), where \( V_1 \) is the initial volume, and \( V_2 \) is the final volume after compression.
A high compression ratio indicates that the gas has been significantly compressed, which is critical in diesel engines where the air is compressed so much that the temperature rises sufficiently to ignite the fuel spontaneously. In this example, the air is compressed from an initial to a much smaller volume (18.27 times smaller in the case study), achieving high efficiency.
  • Higher compression ratios generally lead to better thermal efficiency.
  • Important for engine designs, impacting power output and fuel efficiency.
Engines with higher compression ratios can convert more of the energy in fuel into work, which is one reason diesel engines are often more efficient than gasoline engines.
Heat Capacity Ratio
The heat capacity ratio, denoted by \( \gamma \), is critical for understanding the behavior of gases. It is defined as the ratio of the heat capacity at constant pressure (\( C_p \)) to the heat capacity at constant volume (\( C_v \)): \( \gamma = \frac{C_p}{C_v} \).
For an ideal gas, this ratio determines how its temperature changes during adiabatic processes. In our example, we used \( \gamma = \frac{7}{5} \), which is common for diatomic gases like nitrogen, which comprise most of the air.
  • This ratio influences the temperature and pressure changes within the engine.
  • Understanding \( \gamma \) helps in predicting how a gas behaves when compressed or expanded adiabatically.
Knowing the heat capacity ratio helps calculate the compression ratio accurately, which is important for determining engine performance and efficiency.
Adiabatic Equation
The adiabatic process is a key part of thermodynamics, particularly in understanding how engines work. In an adiabatic process, no heat is exchanged with the environment. The adiabatic equation for an ideal gas is \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \).
This equation connects the temperatures and volumes before and after the process. For an ideal gas undergoing adiabatic compression, even though no heat is gained or lost,
  • The internal energy can change, often observable as a temperature change.
  • Engines using adiabatic compression, like diesel engines, achieve high efficiency as they compress air without losing energy as heat.
In this problem, the adiabatic equation helps calculate the compression ratio needed to ignite the air-fuel mixture using the heat generated from compression, rather than a spark plug.

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Most popular questions from this chapter

Two Carnot air conditioners, \(A\) and \(B,\) are removing heat from different rooms. The outside temperature is the same for both rooms, 309.0 \(\mathrm{K}\) . The room serviced by unit A is kept at a temperature of 294.0 \(\mathrm{K}\) , while the room serviced by unit \(\mathrm{B}\) is kept at 301.0 \(\mathrm{K}\) . The heat removed from either room is 4330 \(\mathrm{J}\) J. For both units, find the magnitude of the work required and the magnitude of the heat deposited outside.

A power plant taps steam superheated by geothermal energy to 505 \(\mathrm{K}\) (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at 323 \(\mathrm{K}\) (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 84000 kilowatts. Determine (a) the maximum efficiency at which this plant can operate and \((\mathbf{b})\) the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

The inside of a Carnot refrigerator is maintained at a temperature of 277 \(\mathrm{K}\) , while the temperature in the kitchen is 299 \(\mathrm{K}\) . Using 2500 \(\mathrm{J}\) of work, how much heat can this refrigerator remove from its inside compartment?

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 228 \(\mathrm{J}\) when 166 \(\mathrm{J}\) of work is done on the system. In the second step, the internal energy of the system increases by 115 \(\mathrm{J}\) when 177 \(\mathrm{J}\) of work is done on the system. For the overall process, find the heat. What type of process is the overall process? Explain.

The temperature of a monatomic ideal gas remains constant during a process in which 4700 \(\mathrm{J}\) of heat flows out of the gas. How much work (including the proper \(+\) or \(-\) sign ) is done?

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