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To determine the entropy change for an irreversible process between states 1 and \(2,\) should the integral \(\int_{1}^{2} \delta Q / T\) be performed along the actual process path or an imaginary reversible path? Explain.

Short Answer

Expert verified
Answer: When calculating the entropy change for an irreversible process, we should use an imaginary reversible path between the initial and final states. This is because entropy is a state function that does not depend on the path taken, and using a reversible path allows us to properly evaluate the integral without dealing with the complexities involved in the heat exchange of an irreversible process.

Step by step solution

01

Definition of entropy change and reversible paths

Entropy change, denoted by ΔS, can be expressed in the formula \(\Delta S = \int_{1}^{2} \frac{\delta Q_{rev}}{T}\), where \(\delta Q_{rev}\) represents the heat exchanged during a reversible process and T is the absolute temperature. Notice that the formula specifies a reversible process, which implies that we need to determine the correct path involved (reversible or irreversible) to evaluate the entropy change.
02

Entropy as a state function

Entropy is a state function because it only depends on the initial and final states of the system and not the path taken to reach the final state. Since the entropy change has the same value regardless of whether the process is reversible or irreversible, it's crucial to consider this when deciding which path to use for the integral calculation.
03

Consequence of using an irreversible process path

When evaluating the integral for an irreversible process, we would have to account for the various factors causing irreversibility (such as friction, viscous dissipation, or chemical reactions) in the heat exchange \(\delta Q\). In general, this would make the integral difficult or even impossible to evaluate.
04

Using a reversible process path for calculating the entropy change

Since the actual process, in this case, is irreversible, it's not straightforward to determine the heat exchange and integrate it over the process path. However, because the entropy change is the same for both reversible and irreversible processes between the same states, we can design an imaginary reversible process between states 1 and 2. This will allow us to calculate ΔS using a reversible path, which should be more mathematically tractable.
05

Conclusion: Choosing the correct path

In conclusion, we should perform the integral \(\int_{1}^{2} \frac{\delta Q}{T}\) along an imaginary reversible path, not the actual irreversible process path. By doing so, we ensure that we can properly evaluate the entropy change without dealing with the complexities and uncertainties involved in the heat exchange resulting from an irreversible process.

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Key Concepts

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

Reversible Process
A reversible process in thermodynamics is one that can be reversed without leaving any trace on the surroundings. In other words, both the system and the environment return to their original states at the end of the reversible cycle. This idealized concept is essential because it provides a benchmark for calculating certain thermodynamic properties.

Mathematical Representation

For a process to be deemed reversible, the change must occur infinitesimally slowly, allowing the system to remain in a state of equilibrium with its surroundings. Mathematically, during a reversible process, the formula for entropy change, \( \Delta S \), could be represented as \( \Delta S = \int_{1}^{2} \frac{\delta Q_{rev}}{T} \), where \( \delta Q_{rev} \) is the small amount of heat exchanged reversibly and \( T \) is the absolute temperature.

This is an idealized concept, as all real processes are irreversible to some extent, but understanding reversible processes allows us to calculate maximum efficiency for systems like engines.
State Function
Entropy, symbolized as \( S \), is a fundamental thermodynamic quantity that, unlike heat or work, is a state function. State functions are properties that depend only on the current state of a system, not on how that state was reached.

Characteristic of State Functions

These functions are path-independent; that means the value of a state function at any given state is fixed, no matter the process the system underwent to arrive at that state. For example, when considering the elevation of a mountain peak, it doesn't matter whether you hiked up or took a helicopter, the elevation (state function) remains the same. Similarly, the change in entropy \( \Delta S \) from state 1 to state 2 is constant, irrespective of whether the process was reversible or irreversible.

This property allows us to use the reversible path to evaluate entropy changes in real-world, irreversible processes because the end value will be identical for the same initial and final states.
Irreversible Process
An irreversible process is one where the system cannot return to its initial state without a change in the external environment. In everyday life, most processes are irreversible due to factors like friction, inelastic collision, or natural phenomena such as spontaneous heat flow from hot to cold.

Complexity in Calculating Entropy

For these processes, calculating entropy change is not straightforward. It involves non-equilibrium states and often requires knowledge about the intricacies of the process, which might be unknown or complex. Moreover, because the state function should remain the same despite the path taken, using an irreversible process to calculate entropy change would not yield a general expression due to the irreversibilities.

To overcome this difficulty, we imagine a reversible process that takes the system between the same initial and final states. In doing so, we can use the simplified reversible process to accurately calculate the entropy change for the actual irreversible process, thereby simplifying the complex integration that would otherwise be required.

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Most popular questions from this chapter

Refrigerant-134a at \(140 \mathrm{kPa}\) and \(-10^{\circ} \mathrm{C}\) is compressed by an adiabatic \(1.3-\mathrm{kW}\) compressor to an exit state of \(700 \mathrm{kPa}\) and \(60^{\circ} \mathrm{C}\). Neglecting the changes in kinetic and potential energies, determine ( \(a\) ) the isentropic efficiency of the compressor, \((b)\) the volume flow rate of the refrigerant at the compressor inlet, in \(\mathrm{L} / \mathrm{min}\), and \((c)\) the maximum volume flow rate at the inlet conditions that this adiabatic \(1.3-\mathrm{kW}\) compressor can handle without violating the second law.

In a dairy plant, milk at \(4^{\circ} \mathrm{C}\) is pasteurized continuously at \(72^{\circ} \mathrm{C}\) at a rate of \(12 \mathrm{L} / \mathrm{s}\) for 24 hours a day and 365 days a year. The milk is heated to the pasteurizing temperature by hot water heated in a natural-gas-fired boiler that has an efficiency of 82 percent. The pasteurized milk is then cooled by cold water at \(18^{\circ} \mathrm{C}\) before it is finally refrigerated back to \(4^{\circ} \mathrm{C}\). To save energy and money, the plant installs a regenerator that has an effectiveness of 82 percent. If the cost of natural gas is \(\$ 1.30 /\) therm \((1 \text { therm }=105,500 \mathrm{kJ}),\) determine how much energy and money the regenerator will save this company per year and the annual reduction in entropy generation.

Combustion gases with a specific heat ratio of 1.3 enter an adiabatic nozzle steadily at \(800^{\circ} \mathrm{C}\) and \(800 \mathrm{kPa}\) with a low velocity, and exit at a pressure of 85 kPa. The lowest possible temperature of combustion gases at the nozzle exit is \((a) 43^{\circ} \mathrm{C}\) \((b) 237^{\circ} \mathrm{C}\) \((c) 367^{\circ} \mathrm{C}\) \((d) 477^{\circ} \mathrm{C}\) \((e) 640^{\circ} \mathrm{C}\)

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.

Can saturated water vapor at \(200 \mathrm{kPa}\) be condensed to a saturated liquid in an isobaric, closed system process while only exchanging heat with an isothermal energy reservoir at \(90^{\circ} \mathrm{C} ?\) (Hint: Determine the entropy generation.)

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