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(a) Calculate the theoretical efficiency for an Otto-cycle engine with \(\gamma=1.40\) and \(r=9.50\) . (b) If this engine takes in \(10,000 \mathrm{J}\) of heat from burning its fuel, how much heat does it dis- card to the outside air?

Short Answer

Expert verified
(a) Efficiency is 59.3%. (b) Discarded heat is 4,070 J.

Step by step solution

01

Understanding the Otto Cycle Efficiency Formula

The efficiency \( \eta \) of an Otto cycle engine is given by the formula: \[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \]where \( r \) is the compression ratio and \( \gamma \) is the heat capacity ratio. In this problem, \( \gamma = 1.40 \) and \( r = 9.50 \).
02

Plug Values into the Efficiency Formula

Substitute the given values for \( r \) and \( \gamma \) into the efficiency formula: \[ \eta = 1 - \frac{1}{9.50^{1.40 - 1}} \]Simplify the exponent: \( \eta = 1 - \frac{1}{9.50^{0.40}} \).
03

Calculate the Numerical Efficiency

Calculate \( 9.50^{0.40} \), which approximately equals \( 2.46 \). Now compute the theoretical efficiency:\[ \eta = 1 - \frac{1}{2.46} \approx 1 - 0.407 \approx 0.593 \].Thus, the theoretical efficiency is approximately \( 59.3\% \).
04

Calculate Heat Discarded

The amount of heat discarded \( Q_C \) can be calculated using the formula:\[ Q_C = Q_H (1 - \eta) \]where \( Q_H = 10,000 \text{ J} \) is the heat taken in from the fuel. Substitute the known values:\[ Q_C = 10,000 \times (1 - 0.593) \].
05

Compute Heat Discarded

Compute \( Q_C \):\[ Q_C = 10,000 \times 0.407 = 4,070 \text{ J} \].Therefore, the engine discards approximately \( 4,070 \text{ J} \) of heat to the outside air.

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Key Concepts

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

Theoretical Efficiency of the Otto Cycle
Theoretical efficiency is a measure of how well an engine converts applied heat into mechanical energy. For the Otto cycle, which is common in gasoline engines, the theoretical efficiency is crucial as it defines the potential performance boundaries for the engine. The formula for calculating the theoretical efficiency, \( \eta \), of an Otto cycle engine is \[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \]. Here, \( r \) refers to the compression ratio, which indicates the volume change of the combustion chamber, while \( \gamma \) stands for the heat capacity ratio. By enhancing theoretical efficiency with a higher compression ratio and an optimal heat capacity ratio, engines can achieve greater performance and utilize fuel more effectively.In practical terms, while theoretical efficiency gives us the maximum possible efficiency, real-world factors like friction and heat losses ensure that the actual efficiency is always lower.
Understanding Compression Ratio
The compression ratio is a key factor in defining the performance of the Otto cycle engine. It is the ratio between the largest and smallest volume within the combustion chamber of an engine. Mathematically, it is expressed as:\[ r = \frac{V_{max}}{V_{min}} \] where \( V_{max} \) is the cylinder volume at the bottom of the stroke, and \( V_{min} \) is the volume when the piston is at the top of the stroke.The compression ratio impacts the efficiency directly; a higher compression ratio often results in a greater thermal efficiency. This is why high-performance engines frequently have higher compression ratios. However, there are practical limits, as overly high compression can cause engine knocking, potentially damaging the engine.
Role of Heat Capacity Ratio
The heat capacity ratio, denoted as \( \gamma \), is the specific heat capacity at constant pressure divided by the specific heat capacity at constant volume. Commonly for air and gasoline mixtures, this value is around 1.4 in an Otto cycle engine.The heat capacity ratio is crucial in efficiency calculations for an Otto cycle. It determines how the efficiency increases with changes in the compression ratio: a higher \( \gamma \) tends to increase the efficiency since it implies more available energy for work at a given temperature. In gas mixtures used in engines, \( \gamma \) remains fairly constant; hence, it acts as a constant factor in determining the potential efficiency of the cycle. However, alterations in the fuel mixture or engine design can lead to minor deviations in \( \gamma \), impacting the cycle's efficiency.
Heat Discarded from Otto Cycle Engines
Heat discarded represents the amount of heat energy expelled out of an engine without being utilized for work. In Otto cycle engines, understanding how much heat is discarded is important for designing more efficient systems and improving engine conservation concerns.The discarded heat \( Q_C \) from an Otto cycle engine can be calculated using the formula:\[ Q_C = Q_H (1 - \eta) \], where \( Q_H \) is the total heat input and \( \eta \) is the efficiency of the engine.In this context, if an Otto cycle engine receives \( 10,000 \text{ J} \) as proposed, and has an efficiency of \( 59.3\% \), it will discard \( 4,070 \text{ J} \) as unused energy to the atmosphere. Reducing heat discarded is pivotal for increasing power output and minimizing energy waste, critical for optimizing the engine's performance and environmental footprint.

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Most popular questions from this chapter

A diesel engine performs 2200 \(\mathrm{J}\) of mechanical work and discards 4300 \(\mathrm{J}\) of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?

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