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Work is entropy free, and sometimes the claim is made that work will not change the entropy of a fluid passing through an adiabatic steady-flow system with a single inlet and outlet. Is this a valid claim?

Short Answer

Expert verified
Answer: Work is entropy-free in an adiabatic steady-flow system with a single inlet and outlet only if the process is reversible, meaning no irreversible entropy generation occurs within the system. In real-world scenarios and irreversible processes, this claim does not hold, as work is not entropy-free for irreversible processes in adiabatic systems.

Step by step solution

01

Understanding adiabatic processes and entropy

Adiabatic processes are those where no heat is exchanged between a system and its surroundings. This is a thermodynamically idealized scenario, but it can be closely approximated in well-insulated systems or when a process happens quickly enough that heat transfer is negligible. Entropy is a thermodynamic property that measures the amount of energy dispersion or disorder in a system. A system's entropy can change when heat is added or removed, but it can also change due to irreversible processes.
02

Examining the work-energy theorem

The work-energy theorem states that the work done on a system is equal to the change in its internal energy. In an adiabatic steady-flow system, we have energy conservation such that the work done equals the change in enthalpy. Here we should note that entropy and enthalpy are related, but they are not the same thing.
03

Evaluating the claim about work, entropy, and adiabatic systems

In adiabatic systems, there is no heat transfer, so the only way for the system to exchange energy with its surroundings is through the inlet and outlet via work. If the process were reversible, no entropy would be generated within the system, and the claim that work will not change the entropy of the fluid would be valid. However, this claim does not hold for irreversible processes.
04

Understanding entropy generation in irreversible processes

In real-life scenarios, adiabatic processes can be irreversible, meaning that entropy can still be generated within the system, even if there is no heat transfer. Sources of irreversibility include friction, flow resistance, and shock. In these cases, even though the system is adiabatic and work is being done, there is still a change in entropy due to the irreversibility of the process.
05

Conclusion

The claim that work will not change the entropy of a fluid passing through an adiabatic steady-flow system with a single inlet and outlet is only valid if the process is reversible, meaning no irreversible entropy generation occurs within the system. When dealing with real-world scenarios and irreversible processes, this claim does not hold, as work is not entropy-free for irreversible processes in adiabatic systems.

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Key Concepts

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

Adiabatic Steady-Flow System
When understanding adiabatic steady-flow systems, it's critical to grasp the concept of adiabatic processes. In an adiabatic steady-flow system, there is no heat transferred to or from the fluid as it moves through the system, which typically involves components like turbines, compressors, or nozzles. Given these systems have a single inlet and outlet, the flow is consistent, with no accumulation of mass or energy inside the system—making them a staple in thermodynamic analysis for engines and refrigeration cycles.

An important aspect of these systems is that while there is no heat exchange, they can perform or absorb work. This might lead to a misconception that work cannot alter the entropy of a fluid undergoing such a process. However, while this could be true under ideal, reversible conditions, real-world applications often involve irreversible factors that indeed affect entropy, which we'll explore further.
Work-Energy Theorem
The work-energy theorem connects the work done on a system to its kinetic energy. For adiabatic steady-flow systems, when applying this theorem, we're mostly concerned with the internal energy changes that result from work done on or by the system. If work is done on the system, such as in a compressor, it increases the internal energy, often reflected in higher temperature or pressure. Conversely, if the system does work, as in the case of a turbine, its internal energy decreases.

It is crucial to differentiate between enthalpy and entropy here—while enthalpy is a measure of total energy inclusive of flow work, entropy measures disorder or randomness. Linking these concepts can be challenging, so when interpreting the work-energy theorem, one must be cautious not to simplify enthalpy and entropy as interchangeable.
Entropy Generation
Entropy generation is the hallmark of real-world, irreversible processes and is indicative of energy dispersion that cannot be completely converted back into work. While an idealized process may boast no entropy change, every actual process is subject to inefficiencies that lead to entropy generation. These may arise due to various factors such as friction, unanticipated heat transfer, chemical reactions, or mixing of different substances.

Understanding that irreversible processes lead to a net increase in entropy can be complex. Thus, while one may intuitively believe work to be 'entropy-free' since it is organized energy, the practical generation of entropy in adiabatic systems due to irreversibilities must be acknowledged. This further implies that despite no heat exchange, the entropy of a fluid can indeed increase as a result of work done in real-life systems.
Irreversible Processes
In thermodynamics, an irreversible process breaks the idealized symmetry of time; you cannot simply reverse the process and restore the system to its initial state without changes in the surrounding environment. Factors like friction, turbulence, and inelastic deformation are commonly associated with real-world irreversibility. These processes generate entropy because they entail energy dispersion in forms that cannot be harnessed to do useful work.

When connecting irreversibility to our adiabatic steady-flow systems, the key takeaway is that even in the absence of heat transfer, work done due to these irreversible processes still increases the system’s entropy. The initial claim that work in adiabatic processes could not change entropy is thus refuted by recognizing the inherent irreversibility present in actual scenarios. This recognition demystifies the seemingly paradoxical idea that work can indeed alter entropy, despite its 'organized' nature compared to 'disorganized' heat energy.

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Most popular questions from this chapter

Air is compressed steadily by a compressor from \(100 \mathrm{kPa}\) and \(17^{\circ} \mathrm{C}\) to \(700 \mathrm{kPa}\) at a rate of \(5 \mathrm{kg} / \mathrm{min}\). Determine the minimum power input required if the process is (a) adiabatic and ( \(b\) ) isothermal. Assume air to be an ideal gas with variable specific heats, and neglect the changes in kinetic and potential energies.

Liquid water at \(200 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\) is heated in a chamber by mixing it with superheated steam at \(200 \mathrm{kPa}\) and \(150^{\circ} \mathrm{C} .\) Liquid water enters the mixing chamber at a rate of \(4.3 \mathrm{kg} / \mathrm{s},\) and the chamber is estimated to lose heat to the surrounding air at \(20^{\circ} \mathrm{C}\) at a rate of \(1200 \mathrm{kJ} / \mathrm{min}\). If the mixture leaves the mixing chamber at \(200 \mathrm{kPa}\) and \(80^{\circ} \mathrm{C}\) determine \((a)\) the mass flow rate of the superheated steam and \((b)\) the rate of entropy generation during this mixing process.

Refrigerant-134a is expanded adiabatically from 100 psia and \(100^{\circ} \mathrm{F}\) to a saturated vapor at 10 psia. Determine the entropy generation for this process, in Btu/lbm-R.

Steam enters an adiabatic turbine at \(8 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{kg} / \mathrm{s}\), and exits at \(0.2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) The rate of entropy generation in the turbine is \((a) 0 \mathrm{kW} / \mathrm{K}\) \((b) 7.2 \mathrm{kW} / \mathrm{K}\) \((c) 21 \mathrm{kW} / \mathrm{K}\) \((d) 15 \mathrm{kW} / \mathrm{K}\) \((e) 17 \mathrm{kW} / \mathrm{K}\)

For an ideal gas with constant specific heats show that the compressor and turbine isentropic efficiencies may be written as $$\eta_{C}=\frac{\left(P_{2} / P_{1}\right)^{(k-1) / k}}{\left(T_{2} / T_{1}\right)-1} \text { and } \eta_{T}=\frac{\left(T_{4} / T_{3}\right)-1}{\left(P_{4} / P_{3}\right)^{(k-1) / k}-1}$$ The states 1 and 2 represent the compressor inlet and exit states and the states 3 and 4 represent the turbine inlet and exit states.

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