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Chapter 20: Problem 8

20.8. The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8 . (a) What is the ideal efficiency of the engine? Use \(\gamma=1.40 .\) (b) The engine in a Dodge Viper G.2 has a slightly higher compression ratio of 9.6 . How much increase in the ideal efficiency results from this increase in the compression ratio?

Short Answer

Expert verified
The ideal efficiency increases by 2.6% with the higher compression ratio.

Step by step solution

01

Understanding the Otto Cycle Efficiency Formula

The efficiency 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, \( \gamma = 1.40 \).
02

Calculating Ideal Efficiency for Mercedes-Benz

For the Mercedes-Benz SLK230 with a compression ratio of 8.8, substitute \( r = 8.8 \) and \( \gamma = 1.40 \) into the formula. Calculate:\[ \eta_{MB} = 1 - \frac{1}{{8.8^{1.40 - 1}}} \]This gives the ideal efficiency for the engine.
03

Substituting Values for Mercedes-Benz

First, calculate \( 8.8^{0.4} \). We get:\[ 8.8^{0.4} \approx 2.29 \]Now, substitute into the efficiency formula:\[ \eta_{MB} = 1 - \frac{1}{2.29} \approx 0.564 \text{ or } 56.4\% \]
04

Calculating Ideal Efficiency for Dodge Viper

For the Dodge Viper G.2 with a compression ratio of 9.6, substitute \( r = 9.6 \) and \( \gamma = 1.40 \) into the efficiency formula and calculate:\[ \eta_{DV} = 1 - \frac{1}{{9.6^{1.40 - 1}}} \]
05

Substituting Values for Dodge Viper

First, calculate \( 9.6^{0.4} \). We find:\[ 9.6^{0.4} \approx 2.44 \]Now, substitute into the efficiency formula:\[ \eta_{DV} = 1 - \frac{1}{2.44} \approx 0.590 \text{ or } 59.0\% \]
06

Calculating Increase in Efficiency

To find the increase in the ideal efficiency due to the change in compression ratio, calculate the difference:\[ \text{Increase in Efficiency} = \eta_{DV} - \eta_{MB} = 0.590 - 0.564 = 0.026 \text{ or } 2.6\% \]
07

Conclusion

The ideal efficiency of the Mercedes-Benz engine is 56.4%, and the Dodge Viper's ideal efficiency is 59.0%. The increase in ideal efficiency from the higher compression ratio is 2.6%.

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Key Concepts

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

Compression Ratio
The compression ratio is a crucial factor in internal combustion engines that significantly affects engine efficiency and performance. It is defined as the ratio of the maximum to minimum volume in the cylinder of an engine.
  • High compression ratios imply that the engine can compress the air-fuel mixture more efficiently, leading to better extraction of energy from the fuel.
  • In the case of the Mercedes-Benz SLK230, the compression ratio is 8.8.
  • The Dodge Viper G.2 features a slightly higher compression ratio of 9.6, contributing to its enhanced performance.
The relationship between the compression ratio and engine efficiency can be summarized as such: an increase in compression ratio generally leads to increased efficiency. This is because higher compression allows the engine to do more work with the same quantity of fuel, reducing heat losses.
Heat Capacity Ratio
The heat capacity ratio, often represented by the Greek letter \( \gamma \), plays an integral role in determining the efficiency of an Otto cycle engine. It is the ratio of the specific heat at constant pressure (\( C_p \)) to the specific heat at constant volume (\( C_v \)).
  • In thermal engines, the heat capacity ratio affects the expansion and compression of gases, thereby influencing the thermodynamic cycles.
  • A typical value for the heat capacity ratio for air (which is assumed for most Otto cycle calculations) is about 1.40, as used in both the Mercedes-Benz and Dodge Viper calculations.
  • The heat capacity ratio is vital because it dictates how much temperature will rise when a gas is compressed, which in turn affects efficiency.
Understanding this concept is fundamental because an optimal heat capacity ratio can maximize engine performance by balancing the heat input and extraction.
Ideal Efficiency Calculation
Ideal efficiency in an Otto cycle engine is a theoretical calculation that indicates the maximum efficiency the engine could achieve under perfect conditions. It is derived using the formula \( \eta = 1 - \frac{1}{r^{\gamma - 1}} \).
  • The term \( r \) is the compression ratio, and \( \gamma \) is the heat capacity ratio.
  • For the Mercedes-Benz SLK230, the efficiency calculated with a compression ratio of 8.8 and \( \gamma = 1.40 \) results in approximately 56.4%.
  • For the Dodge Viper G.2, with a higher compression ratio of 9.6, the efficiency increases to about 59.0%.
  • This demonstrates the importance of both the compression ratio and the heat capacity ratio in maximizing the efficiency of an internal combustion engine.
Performing these calculations allows engineers to predict how changes in engine design, such as an increased compression ratio, will impact performance.
Thermodynamics
Thermodynamics is the scientific study of heat transfer and work, particularly as it pertains to engines and machinery. In the context of the Otto cycle engine, thermodynamics explores how the energy from fuel is converted into mechanical work.
  • The Otto cycle is a fundamental thermodynamic cycle that describes the operation of spark-ignition engines.
  • It consists of four distinct processes: two adiabatic (no heat transfer) and two isochoric (constant volume) processes.
  • Understanding thermodynamics enables the calculation of crucial metrics like engine efficiency, which can be optimized by adjusting variables like compression and heat capacity ratios.
These principles are at the heart of designing more efficient engines, as they help predict how modifications will influence power output and fuel consumption. In essence, thermodynamics lays the groundwork for engineering advancements in engine technologies.

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Most popular questions from this chapter

A gasoline engine has a power output of 180 \(\mathrm{kW}\) (about 241 \(\mathrm{hp}\) ). Its thermal efficiency is 28.0\(\%\) . (a) How much heat must be supplied to the engine per second? (b) How much heat is discarded by the engine per second?

20.54. An air conditioner operates on 800 \(\mathrm{W}\) of power and has a performance coefficient of 2.80 with a room temperature of \(21.0^{\circ} \mathrm{C}\) and an outside temperature of \(35.0^{\circ} \mathrm{C}\) (a) Calculate the rate of heat removal for this unit. (b) Calculate the rate at which heat is discharged to the outside air. (c) Calculate the total entropy change in the room if the air conditioner runs for 1 hour. Calculate the total entropy change in the outside air for the same time period. (d) What is the net change in entropy for the system (room + outside air)?

20.21. A Carnot heat engine has a thermal efficiency of 0.600 , and the temperature of its hot reservoir is 800 \(\mathrm{K}\) . If 3000 \(\mathrm{J}\) of heat is rejected to the cold reservoir in one cycle, what is the work output of the engine during one cycle?

20.52. A typical coal-fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40\(\%\) (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of \(265 \times 10^{7} \mathrm{J} / \mathrm{kg}\) . How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river?(d) The river's temperature is \(18.0^{\circ} \mathrm{C}\) before it reaches the power plant and \(18.5^{\circ} \mathrm{C}\) after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

20.27 A 15.0-kg block of ice at \(0.0^{\circ} \mathrm{C}\) melts to liquid water at \(0.0^{\circ} \mathrm{C}\) inside a large room that has a temperature of \(20.0^{\circ} \mathrm{C}\) . Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. (a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to any equations. (b) Calculate the net entropy change of the system during this process. Explain whether or not this result is consistent with your answer to part (a).
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