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The Outo-cycle engine in a Mercedes-Benz SL.K230 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 GT2 has a slightly higher compression ratio of 9.6 . How much increase in the ideal efficicncy results from this increase in the compression ratio?

Short Answer

Expert verified
The ideal efficiency for the Outo-cycle engine is computed in step 1. The ideal efficiency for the Dodge Viper GT2 engine is computed in step 2. The difference between these two values, computed in step 3, represents the increase in efficiency resulting from the higher compression ratio in the Dodge Viper GT2 engine.

Step by step solution

01

Compute the efficiency of the Outo-cycle engine

First, use the formula \[ \eta=1-(\frac{1}{r^{\gamma -1}}) \] for the Outo-cycle engine with a compression ratio (r) of 8.8 and heat capacity ratio ( \( \gamma \) ) of 1.4. Substitute the given values into the formula to find the efficiency.
02

Compute the efficiency of the Dodge Viper GT2 engine

Next, use the same formula for the Dodge Viper GT2 engine, which has a different compression ratio of 9.6. Substitute these values into the formula to find the efficiency.
03

Calculate the difference in efficiencies

Subtract the efficiency of the Outo-cycle engine calculated in step 1 from the efficiency of the Dodge Viper GT2 engine calculated in step 2 in order to calculate the increase in efficiency due to the larger compression ratio.

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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 aspect of understanding the Otto cycle, which describes the process used in standard internal combustion engines. It is defined as the ratio of the total volume of the cylinder when the piston is at its lowest point (bottom dead center) to the volume when the piston is at its highest point (top dead center). The formula for compression ratio \( r \) is given by:
  • \( r = \frac{V_{max}}{V_{min}} \)
where \( V_{max} \) is the maximum cylinder volume, and \( V_{min} \) is the minimum cylinder volume.
An engine with a higher compression ratio compresses the air-fuel mixture more tightly, leading to more efficient fuel burning and higher power output. For instance, in the exercise provided, the Mercedes-Benz SL.K230 has a compression ratio of 8.8, while the Dodge Viper GT2 has a slightly higher ratio of 9.6.
A greater compression ratio usually means increased power and efficiency but can also lead to complications such as engine knocking. Thus, manufacturers carefully balance the compression ratio to optimize engine performance.
Thermal Efficiency
Thermal efficiency in the context of the Otto cycle refers to how effectively an engine converts the energy from fuel into work. It is a measure of the engine's performance and is affected by parameters like the compression ratio and the specific heat ratio.The thermal efficiency \( \eta \) of an Otto cycle engine can be calculated using the formula:
  • \( \eta = 1 - \left( \frac{1}{r^{\gamma - 1}} \right) \)
where \( r \) is the compression ratio and \( \gamma \) is the specific heat ratio.
In our case, the specific heat ratio \( \gamma \) is given as 1.40. By substituting the respective compression ratios for the Mercedes-Benz SL.K230 (8.8) and Dodge Viper GT2 (9.6), we can calculate the corresponding thermal efficiencies.
It is important to understand that higher compression ratios result in higher calculated thermal efficiencies, meaning more efficient energy conversion.The thermal efficiency is idealized and does not account for real-world losses such as friction or heat dissipation, but it provides a good benchmark for comparing different engines.
Specific Heat Ratio
The specific heat ratio, also known as the adiabatic index (\( \gamma \)), is the ratio of the specific heat at constant pressure \( C_p \) to the specific heat at constant volume \( C_v \). It plays a vital role in calculating the performance and efficiency of engines using the Otto cycle. The equation for the specific heat ratio is:
  • \( \gamma = \frac{C_p}{C_v} \)
In the context of our exercise, \( \gamma \) is 1.40, which is typical for air in standard conditions and critical for determining the thermal efficiency of the Otto cycle.
The specific heat ratio affects how volume and pressure change within the engine's cylinders during the compression and power strokes. A higher specific heat ratio means that the working gas is more likely to expand and contract quickly when heated or cooled, which influences the engine's performance characteristics.
Engine design considers this ratio to optimize the combustion process, minimize energy losses, and achieve desired power outputs. Therefore, knowing \( \gamma \) is essential for understanding and comparing engine efficiency and behavior in real-world scenarios.

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Most popular questions from this chapter

As a budding mechanical engineer, you are called upon to design a Carnot engine that has \(2.00 \mathrm{~mol}\) of a monatomic ideal gas as its working substance and operates from a high-temperature reservoir at \(500^{\circ} \mathrm{C}\). The engine is to lift a \(15.0 \mathrm{~kg}\) weight \(2.00 \mathrm{~m}\) per cycle, using \(500 \mathrm{~J}\) of heat input. The gas in the cngine chamber can have a minimum volume of \(5.00 \mathrm{~L}\) during the cycle. (a) Draw a \(p V\) -diagram for this cycle. Show in your diagram where heat enters and leaves the gas. (b) What must be the temperature of the cold reservoir? (c) What is the thermal efficicncy of the cngine? (d) How much heat cnergy does this engine waste per cycle? (e) What is the maximum pressure that the gas chamher will have to withstand?

You are conducting experiments to study prototype heat engines. In one test, \(4.00 \mathrm{~mol}\) of argon gas are taken around the cycle shown in Fig. \(\mathrm{P} 20.55 .\) The pressure is low cnough for the gas to be treated as ideal. You measure the gas temperature in states \(a, b, c,\) and \(d\) and find \(T_{a}=250.0 \mathrm{~K}\) \(T_{h}=300.0 \mathrm{~K}, \quad T_{c}=380.0 \mathrm{~K}, \quad\) and \(T_{d}=316.7 \mathrm{~K}\) (a) Calculate the efficiency \(e\) of the cycle. (b) Disappointed by the cycle's low efficicncy, you consider doubling the number of moles of gas while kecping the pressure and volume the same. What would \(e\) be then? (c) You remember that the efficiency of a Carnot cycle increases if the temperature of the hot reservoir is increased. So, you return to using \(4.00 \mathrm{~mol}\) of gas but double the volume in states \(c\) and \(d\) while keeping the pressures the same. The resulting temperatures in these states are \(T_{c}=760.0 \mathrm{~K}\) and \(T_{d}=633.4 \mathrm{~K}\). \(T_{a}\) and \(T_{b}\) remain the same as in part (a). Calculate \(e\) for this cycle with the new \(T_{c}\) and \(T_{d}\) values. (d) Encouraged by the increase in efficiency, you raise \(T_{c}\) and \(T_{d}\) still further. But \(e\) doesn't increase very much; it scems to be approaching a limiting valuc. If \(T_{a}=250.0 \mathrm{~K}\) and \(T_{b}=300.0 \mathrm{~K}\) and you keep volumes \(V_{a}\) and \(V_{b}\) the same as in part (a), then \(T_{d} T_{d}=T_{b} / T_{a}\) and \(T_{c}=1.20 T_{d}\). Derive an expression for \(e\) as a function of \(T_{d}\) for this cycle. What value docs \(e\) approach as \(T_{d}\) becomes very large?

An ideal Carnot cngine opcrates between \(500^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\) with a heat input of \(250 \mathrm{~J}\) per cycle. (a) How much heat is delivered to the cold reservoir in each cycle? (b) What minimum number of cycles is necessary for the engine to lift a \(500 \mathrm{~kg}\) rock through a height of \(100 \mathrm{~m}^{\prime} ?\)

A heat engine uses a large insulated tank of ice water as its cold reservoir. In 100 cycles the engine takes in \(8000 \mathrm{~J}\) of heat energy from the hot reservoir and the rejected heat melts \(0.0180 \mathrm{~kg}\) of ice in the tank. During these 100 cycles, how much work is performed by the engine?

Three moles of an ideal gas undergo a reversible isothermal compression at \(20.0^{\circ} \mathrm{C}\). During this compression, \(1850 \mathrm{~J}\) of work is done on the gas. What is the change of cntropy of the gas?

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