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Chapter 20: Problem 2

An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?

Short Answer

Expert verified
(a) Work output is 2600 J. (b) Thermal efficiency is approximately 28.89%.

Step by step solution

01

Understanding the Given Information

We are given that the heat absorbed by the engine, \( Q_1 \), is 9000 Joules. The heat discarded by the engine, \( Q_2 \), is 6400 Joules. We need to find both (a) the mechanical work output and (b) the thermal efficiency.
02

Mechanical Work Output Calculation

Mechanical work output, \( W \), of the engine can be calculated using the formula: \( W = Q_1 - Q_2 \). Substitute the given values: \( W = 9000 \, \text{J} - 6400 \, \text{J} = 2600 \, \text{J} \). Hence, the work output is 2600 Joules.
03

Thermal Efficiency Calculation

The thermal efficiency, \( \eta \), of an engine is calculated using the formula: \( \eta = \frac{W}{Q_1} \times 100 \% \). We already found \( W = 2600 \, \text{J} \) and \( Q_1 = 9000 \, \text{J} \). Substitute these values into the formula: \( \eta = \frac{2600}{9000} \times 100 \% = 28.89\% \). Hence, the thermal efficiency is approximately 28.89%.

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Key Concepts

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

Mechanical Work Output
Mechanical work output is a fundamental concept in thermodynamics, especially in the study of engines. In simple terms, it refers to the amount of energy converted from heat into mechanical energy or work. This transformation is central to the functioning of all engines and can be quantified using a straightforward formula.

For an engine, the mechanical work output (\( W \)) is calculated by the difference between the heat absorbed (\( Q_1 \)) and the heat discarded (\( Q_2 \)). In our example, this means:
  • The absorbed heat is 9000 Joules.
  • The discarded heat is 6400 Joules.
  • Thus, the mechanical work output is \( W = Q_1 - Q_2 = 9000 \, \text{J} - 6400 \, \text{J} = 2600 \, \text{J} \).
By understanding the output, we get insight into how efficiently a machine converts energy, helping engineers improve energy systems.
Thermal Efficiency
Thermal efficiency measures how well an engine converts heat into work. It's a vital concept for assessing engine performance. Simply put, thermal efficiency tells us what percentage of heat energy is converted into useful work.

To calculate thermal efficiency (\( \eta \)), use the formula:
  • \( \eta = \frac{W}{Q_1} \times 100 \% \)
  • Where \( W \) is the work output, and \( Q_1 \) is the heat absorbed.
  • For our engine: \( \eta = \frac{2600}{9000} \times 100 \% = 28.89\% \)
This calculation tells us that only about 28.89% of the heat energy is used to do work. The rest is lost, usually as waste heat. By understanding thermal efficiency, we can work towards designing engines that lose less energy.
Heat Engines
Heat engines are devices that convert heat energy into mechanical work. They play a crucial role in various technologies, from powering vehicles to generating electricity.

A heat engine operates by exploiting the flow of heat from a high-temperature source to a low-temperature sink, and in this process, it converts some of this heat into work. This process involves:
  • Absorbing heat from a high-temperature source (e.g., combustion in engines).
  • Performing work (e.g., moving pistons).
  • Releasing unused heat to a cold reservoir or sink.
The efficiency and work output of a heat engine are significant indicators of its effectiveness. By understanding these elements, engineers can innovate to create more sustainable and powerful engines, thereby advancing energy technology.

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Most popular questions from this chapter

20.17. A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 \(\mathrm{K}\) and 270 \(\mathrm{K}\) (a) If in each cycle the refrigerator receives 415 \(\mathrm{J}\) of heat energy from the reservoir at 270 \(\mathrm{K}\) , how many joules of heat energy does it deliver to the reservoir at 320 \(\mathrm{K} ?\) (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?

20.58. A \(0.0500-\mathrm{kg}\) cube of ice at an initial temperature of \(-15.0^{\circ} \mathrm{C}\) is placed in 0.600 \(\mathrm{kg}\) of water at \(T=45.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. (a) Calculate the final temperature of the water once the ice has melted. (b) Calculate the change in entropy of the system.

20.38. You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 \(\mathrm{atm}\) . With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 kg block of ice originally at \(0.0^{\circ} \mathrm{C}\) , using only the beat rejected by the engine?

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?

20.54. 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)?
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