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Consider a process that involves no irreversibilities. Will the actual useful work for that process be equal to the reversible work?

Short Answer

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Answer: Yes, the actual useful work for a process with no irreversibilities will be equal to the reversible work.

Step by step solution

01

Understanding reversible and irreversible processes

A reversible process is a process that occurs in such a way that both the system and its surroundings can be returned to their initial states without any net work or heat transfer between them. It is a hypothetical concept, and no real process is completely reversible due to internal and external irreversibilities. In contrast, an irreversible process is a real process that occurs with some degree of irreversibility.
02

Understanding useful work

Useful work is the output work that can be used to perform a specific function or task without being changed into any other form. In thermodynamics, useful work is typically defined as the work done by the system on the surroundings during a process, excluding any work done to overcome irreversibilities.
03

Relationship between reversible and irreversible work

Since a reversible process is essentially a process with no irreversibilities, the reversible work is the maximum amount of work that can be done by a system during a process. It serves as a benchmark to compare the actual work done by a system during an irreversible process, which is always less than the reversible work due to the existence of irreversibilities.
04

Evaluating the actual useful work and reversible work for a process with no irreversibilities

As the given process has no irreversibilities, it can be considered as a reversible process. As a result, the actual useful work will be equal to the reversible work since there are no irreversibilities to reduce the amount of useful work. So, the answer to the exercise question is yes, the actual useful work for a process with no irreversibilities will be equal to the reversible work.

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Key Concepts

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

Irreversible Process
An irreversible process in thermodynamics is one that cannot return both the system and its surroundings to their initial states without leaving a net change. Such processes are the norm in the real world; perfect reversibility is an idealized concept. Irreversibilities can occur due to friction, inelastic deformations, non-quasi-static changes, or thermal gradients, among others.

Understanding the concept of irreversibility is crucial in thermodynamics because it influences the efficiency of engines and other thermodynamic systems. For example, in an engine, irreversible processes result in losses due to exhaust heat or mechanical friction, thereby reducing the system's efficiency. In essence, irreversibilities are associated with wasted energy that cannot be fully recovered.

Characteristics of Irreversible Processes

  • They generate entropy, leading to a net increase in the entropy of the universe.
  • They are non-quasi-static, meaning they cannot be described by a series of equilibrium states.
  • They include natural processes such as mixing, combustion, and diffusion.
Useful Work in Thermodynamics
When examining useful work in thermodynamics, we focus on the work that can be harnessed and utilized to achieve a desired function or productivity. It is essentially the portion of energy transferred from a system to its surroundings that can be employed for practical purposes.

In a thermodynamic cycle, not all work produced by the system qualifies as useful work. For example, the work expended on irreversible processes, like overcoming friction or dealing with thermal resistances, is not considered useful because it does not contribute to the output of the system. Therefore, a clear distinction between the total work done and the useful work is important when assessing the performance and efficiency of thermodynamic systems.

Examples of Useful Work

  • Electric Work: The work done by a power plant to generate electricity.
  • Mechanical Work: The work done by an engine to propel a vehicle.
Relationship Between Reversible and Irreversible Work
Understanding the relationship between reversible and irreversible work is fundamental for grasping the limits of a system's performance. A reversible process represents a hypothetical ideal where no energy is wasted. It defines the upper boundary of the amount of work a system can theoretically deliver.

However, since all real processes are inherently irreversible to some extent, the actual work obtained is always less than what would be achieved in a perfectly reversible process. This is because irreversible processes involve dissipative effects, such as friction and heat loss, that are not present in reversible processes.

Measuring the Impact of Irreversibilities

  • The difference between reversible and irreversible work serves as a measure of the irreversibilities within a system.
  • Efficiency metrics, like the Carnot efficiency, relate to the concept of reversible work and set a benchmark for comparison.
In the case of a process with no irreversibilities, the actual useful work would indeed equal the work from a reversible process, as both would be operating at the theoretical maximum efficiency.

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Most popular questions from this chapter

Keeping the limitations imposed by the second law of thermodynamics in mind, choose the wrong statement below: (a) A heat engine cannot have a thermal efficiency of \(100 \%\) (b) For all reversible processes, the second-law efficiency is \(100 \%\) (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. (d) The second-law efficiency of a process is \(100 \%\) if no entropy is generated during that process. \((e)\) The coefficient of performance of a refrigerator can be greater than 1

Can a system have a higher second-law efficiency than the first-law efficiency during a process? Give examples.

Steam at \(7 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) enters a two-stage adiabatic turbine at a rate of \(15 \mathrm{kg} / \mathrm{s}\). Ten percent of the steam is extracted at the end of the first stage at a pressure of \(1.8 \mathrm{MPa}\) for other use. The remainder of the steam is further expanded in the second stage and leaves the turbine at 10 kPa. If the turbine has an isentropic efficiency of 88 percent, determine the wasted power potential during this process as a result of irreversibilities. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\).

A piston-cylinder device contains 8 kg of refrigerant134a at \(0.7 \mathrm{MPa}\) and \(60^{\circ} \mathrm{C}\). The refrigerant is now cooled at constant pressure until it exists as a liquid at \(20^{\circ} \mathrm{C}\). If the surroundings are at \(100 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\), determine (a) the exergy of the refrigerant at the initial and the final states and (b) the exergy destroyed during this process.

An adiabatic steam nozzle has steam entering at \(300 \mathrm{kPa}, 150^{\circ} \mathrm{C},\) and \(45 \mathrm{m} / \mathrm{s},\) and leaving as a saturated vapor at \(150 \mathrm{kPa}\). Calculate the actual and maximum outlet velocity. Take \(T_{0}=25^{\circ} \mathrm{C}\).

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