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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 J of heat from burning its fuel, how much heat does it discard to the outside air?

Short Answer

Expert verified
(a) Efficiency is 61.5%. (b) Heat discarded is 3,850 J.

Step by step solution

01

Formula for Otto Cycle Efficiency

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

Substitute the Given Values

Substitute \( \gamma = 1.40 \) and \( r = 9.50 \) into the efficiency formula:\[\eta = 1 - \frac{1}{9.50^{1.40 - 1}}\]
03

Calculate the Efficiency

Carry out the calculation:\[9.50^{0.40} \approx 2.60\]Therefore, the efficiency \( \eta \) is:\[\eta = 1 - \frac{1}{2.60} \approx 1 - 0.385 = 0.615\]So, the efficiency is approximately 61.5%.
04

Calculate Heat Discarded

The efficiency of an engine is also defined as the ratio of work output to heat input. If \( Q_{input} = 10,000 \) J and the efficiency \( \eta = 0.615 \), the work done \( W \) is:\[W = \eta \times Q_{input} = 0.615 \times 10,000 = 6,150 \text{ J}\]The heat discarded to the surrounding \( Q_{output} \) is the remaining energy:\[Q_{output} = Q_{input} - W = 10,000 - 6,150 = 3,850 \text{ J}\]

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Key Concepts

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

Understanding Heat Capacity Ratio
The heat capacity ratio, often symbolized by \( \gamma \), is a crucial factor in thermodynamics, particularly for the Otto cycle which is common in internal combustion engines. It is defined as the ratio of the heat capacity at constant pressure \( C_p \) to the heat capacity at constant volume \( C_v \). Mathematically, it's expressed as:\[ \gamma = \frac{C_p}{C_v} \]This ratio provides insight into how a gas behaves when subjected to changes in pressure and temperature. A higher \( \gamma \) typically indicates a larger change in pressure or temperature for a given energy change. In the context of an Otto cycle, a higher heat capacity ratio means that the gas can do more work on the piston, leading to higher efficiency. For most diatomic gases, \( \gamma \) is around 1.4. This is why, in our problem, \( \gamma = 1.40 \) is utilized to find the efficiency.
The Role of Compression Ratio
The compression ratio \( r \) in an Otto cycle is defined as the ratio of the total volume when the piston is at the bottom of its stroke to the volume when the piston is at the top of its stroke. It is a critical parameter for determining engine efficiency:\[ r = \frac{V_{bottom}}{V_{top}} \]A higher compression ratio means the gas is compressed into a smaller volume before combustion. This increase in compression can significantly improve the engine's thermal efficiency since it allows for more complete combustion and better use of the air-fuel mixture. It is important to note that while higher compression ratios can yield better efficiency, they also require engines that can handle higher pressures. For our problem, the given compression ratio is \( r = 9.50 \), which indicates a relatively high level of efficiency suitable for gasoline engines, as evidenced in the calculated efficiency of approximately 61.5%.
Heat Engine Calculations Simplified
Heat engine calculations aim to quantify how effectively an engine converts heat from fuel into work. In the Otto cycle, the efficiency \( \eta \) is expressed as:\[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \]This formula highlights how the heat capacity ratio and compression ratio influence efficiency. When an engine receives a specific amount of heat energy, the work output can be calculated by multiplying the efficiency by the heat energy intake:\[ W = \eta \times Q_{input} \]For example, with an input of 10,000 J and an efficiency of 61.5%, the work done is 6,150 J. The heat discarded, or the heat not converted to work, is simply the difference between input energy and work output:\[ Q_{output} = Q_{input} - W \]In our scenario, this is 3,850 J, indicating the amount of energy lost to the environment. Understanding these calculations is crucial for designing more efficient engines and reducing energy waste in real-world applications.

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Most popular questions from this chapter

A Carnot engine whose high-temperature reservoir is at 620 K takes in 550 J of heat at this temperature in each cycle and gives up 335 J to the low- temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal efficiency of the cycle?

You decide to use your body as a Carnot heat engine. The operating gas is in a tube with one end in your mouth (where the temperature is 37.0\(^\circ\)C) and the other end at the surface of your skin, at 30.0\(^\circ\)C. (a) What is the maximum efficiency of such a heat engine? Would it be a very useful engine? (b) Suppose you want to use this human engine to lift a 2.50-kg box from the floor to a tabletop 1.20 m above the floor. How much must you increase the gravitational potential energy, and how much heat input is needed to accomplish this? (c) If your favorite candy bar has 350 food calories (1 food calorie = 4186 J) and 80% of the food energy goes into heat, how many of these candy bars must you eat to lift the box in this way?

You make tea with 0.250 kg of 85.0\(^\circ\)C water and let it cool to room temperature (20.0\(^\circ\)C). (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all of the heat lost by the water goes into the air. What is the total entropy change of the system tea + air?

A Carnot engine operates between two heat reservoirs at temperatures \(T_H\) and \(T_C\) . An inventor proposes to increase the efficiency by running one engine between \(T_H\) and an intermediate temperature \(T'\) and a second engine between \(T'\) and \(T_C\) , using as input the heat expelled by the first engine. Compute the efficiency of this composite system, and compare it to that of the original engine.

A refrigerator has a coefficient of performance of 2.10. In each cycle it absorbs 3.10 \(\times\) 10\(^4\) J of heat from the cold reservoir. (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

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