/*! 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 approximately 26.43.

Step by step solution

01

Convert Temperatures to Kelvin

First, we need to convert the initial and final temperatures from Celsius to Kelvin. We do this by adding 273.15 to each temperature.- Initial temperature: \(21^{\circ} \text{C} + 273.15 = 294.15 \text{ K}\)- Final temperature: \(688^{\circ} \text{C} + 273.15 = 961.15 \text{ K}\)
02

Use the Adiabatic Process Formula

For an adiabatic process involving an ideal gas, the following relationship holds:\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]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 adiabatic index.
03

Plug in the Known Values

Given that \( \gamma = \frac{7}{5} = 1.4 \), and using the temperatures from Step 1, we plug in these values into the adiabatic formula:\[ \frac{961.15}{294.15} = \left( \frac{V_1}{V_2} \right)^{1.4 - 1} \]
04

Simplify the Expression

Calculate the left side of the equation:\[ \frac{961.15}{294.15} \approx 3.267 \]Thus, the equation simplifies to:\[ 3.267 = \left( \frac{V_1}{V_2} \right)^{0.4} \]
05

Solve for Compression Ratio

To find the compression ratio \( \frac{V_1}{V_2} \), raise both sides of the equation to the power of \( \frac{1}{0.4} \):\[ \left( 3.267 \right)^{\frac{1}{0.4}} = \frac{V_1}{V_2} \]Calculating gives:\[ \frac{V_1}{V_2} \approx 26.43 \]
06

Interpret the Results

The compression ratio, \( \frac{V_1}{V_2} \), is the factor by which the volume decreases during compression. In this case, the air is compressed to approximately 1/26.43 of its original volume.

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Key Concepts

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

Ideal Gas
In the realm of thermodynamics, an ideal gas is a theoretical gas that follows the ideal gas law: \[ PV = nRT \] This law relates the pressure (\( P \)), volume (\( V \)), amount of gas in moles (\( n \)), and temperature (\( T \)) of a gas, with \( R \) being the ideal gas constant. The assumptions made about ideal gases are:
  • The molecules are point particles with no volume.
  • The collisions between molecules and between molecules and the container walls are perfectly elastic.
  • There are no intermolecular forces present.
  • The average kinetic energy is proportional to the temperature in Kelvin.
By simplifying gas behavior, this concept helps us understand and predict how real gases behave under various conditions. In many practical applications, especially at high temperatures and low pressures, real gases behave nearly ideally. However, it's essential to remember this is a simplification and may not perfectly describe gases in all scenarios.
Compression Ratio
The compression ratio is a crucial part of understanding how diesel engines achieve efficient combustion without a spark plug. In essence, the compression ratio is the ratio of the volume of a cylinder when the piston is at the bottom (before compression) to the volume of the cylinder when the piston is at the top (after compression). This can be described as:\[ \text{Compression Ratio} = \frac{V_1}{V_2} \]where \( V_1 \) is the initial volume and \( V_2 \) is the final volume.In thermal applications, a high compression ratio means the air-fuel mix will be extensively compressed before combustion, leading to higher temperatures, which is necessary for igniting the fuel in a diesel engine. This means the cylinder's temperature rises enough to ignite the fuel naturally through compression, hence not needing a spark plug. In our example, a compression ratio of approximately 26.43 indicates a significant reduction in volume, enhancing the engine's capacity to perform efficiently.
Thermodynamics
Thermodynamics is the study of energy transformations, primarily dealing with how heat energy is converted and transferred. It revolves around several fundamental laws that govern these processes:
  • First Law of Thermodynamics: Often known as the law of energy conservation, it states that energy cannot be created or destroyed, only converted from one form to another.
  • Second Law of Thermodynamics: It asserts that energy transfers are never completely efficient and that entropy, or disorder in a system, tends to increase over time.
  • Third Law of Thermodynamics: It states that the entropy of a perfect crystal at absolute zero is exactly zero.
An adiabatic process, like the one in our combustion engine example, is a thermodynamic process in which no heat is transferred into or out of the system. Heat exchange is effectively isolated, meaning any change in temperature is purely due to work done on or by the system. This type of process is crucial for understanding engine cycles and calculations related to performance and efficiency. Understanding the principles of thermodynamics allows engineers to design systems that make the most efficient use of energy.

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Most popular questions from this chapter

A 52-kg mountain climber, starting from rest, climbs a vertical distance of \(730 \mathrm{m}\). At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| /\left|Q_{\mathrm{H}}\right|,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.

Three moles of neon expand isothermally to 0.250 from \(0.100 \mathrm{m}^{3}\). Into the gas flows \(4.75 \times 10^{3} \mathrm{J}\) of heat. Assuming that neon is an ideal gas, find its temperature.

A Carnot refrigerator is used in a kitchen in which the temperature is kept at \(301 \mathrm{K}\). This refrigerator uses \(241 \mathrm{J}\) of work to remove \(2561 \mathrm{J}\) of heat from the food inside. What is the temperature inside the refrigerator?

An engine does 18500 J of work and rejects 6550 J of heat into a cold reservoir whose temperature is \(285 \mathrm{K}\). What would be the smallest possible temperature of the hot reservoir?

An engine has an efficiency \(e_{1} .\) The engine takes input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) from a hot reservoir and delivers work of magnitude \(\left|W_{1}\right| .\)The heat rejected by this engine is used as input heat for a second engine, which has an efficiency \(e_{2}\) and delivers work of magnitude \(\left|W_{2}\right|\). The overall efficiency of this two- engine device is the magnitude of the total work delivered \(\left(\left|W_{1}\right|+\left|W_{2}\right|\right)\) divided by the magnitude \(\left|Q_{\mathrm{H}}\right|\) of the input heat. Find an expression for the overall efficiency \(e\) in terms of \(e_{1}\) and \(e_{2}\).

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