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If a closed system would undergo an internally reversible process and an irreversible process between the same end states, how would the changes in entropy for the two processes compare? How would the amounts of entropy produced compare?

Short Answer

Expert verified
Entropy change is the same for both processes, but entropy production is greater in the irreversible process.

Step by step solution

01

Understand the Problem

The problem involves comparing entropy changes and entropy production in a closed system undergoing two types of processes: an internally reversible process and an irreversible process between the same end states.
02

Define Entropy Change

For both processes, the change in entropy \(\triangle S\) depends only on the initial and final states, since entropy is a state function. Thus, \(\triangle S \) will be the same for both processes.
03

Internally Reversible Process

In an internally reversible process, the system can return to its initial state without leaving any changes in the surroundings. There is no internal entropy production \(\triangle S_{gen} = 0 \).
04

Irreversible Process

An irreversible process, however, involves entropy generation within the system due to factors like friction, unrestrained expansion, rapid mixing, etc. Thus, the entropy generation \(\triangle S_{gen} > 0 \) for an irreversible process.
05

Compare Entropy Changes

Since the entropy change depends only on the initial and final states, the change in entropy for both processes will be the same, \(\triangle S_{reversible} = \triangle S_{irreversible}\).
06

Compare Entropy Produced

Entropy production in an internally reversible process is zero \(\triangle S_{gen} = 0 \), whereas entropy production in an irreversible process is positive \(\triangle S_{gen} > 0 \). Therefore, the amount of entropy produced in the irreversible process will be greater.

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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 is an idealized concept referring to a process that can be reversed without leaving any trace on the surrounding. In a reversible process, the system and its surroundings can be returned to their original states without any net change. This means that there are no losses, no friction, and everything happens infinitely slowly allowing the system to stay in equilibrium throughout the process.
For example, think of a very slow compression or expansion of a gas in a piston. If done infinitely slowly, each tiny adjustment allows the surroundings and the system to adapt, maintaining equilibrium. No entropy is generated within the system during this type of process because it avoids dissipative factors like friction.
It's important to note that in real life, truly reversible processes do not exist; they are theoretical constructs. However, they are useful for understanding the limits of performance and efficiency in thermodynamic systems.
irreversible process
An irreversible process, as opposed to a reversible one, involves factors that generate entropy and result in dissipative phenomena. These factors can include friction, rapid mixing, unrestrained expansions, and other forms of resistances or gradients. Because of these factors, irreversible processes cannot be exactly reversed to restore both the system and environment to their original states without net changes.
Consider stirring cream into coffee. The process happens quickly and results in a uniform mixture, but you cannot un-stir the mixture to separate the cream from the coffee again. This is a basic example of the irreversibility in thermodynamic processes.
In these processes, entropy is produced within the system. Mathematically, if \(\triangle S_{gen} > 0\), this indicates that there is entropy generation due to irreversibilities. Understanding this helps in analyzing the real-world performance of engines, refrigerators, and other thermodynamic systems which are always less efficient than their ideal reversible counterparts.
entropy generation
Entropy generation is a concept deeply tied to the second law of thermodynamics. It quantifies the amount of entropy produced within a system due to irreversible processes. Unlike reversible processes where \( \triangle S_{gen} = 0 \), in irreversible processes \( \triangle S_{gen} > 0 \), meaning there is a net increase in entropy.
Entropy generation is a measure of the 'lost work' or inefficiencies within a system. Higher entropy generation often correlates with greater inefficiencies. For example, in an engine, minimizing friction, turbulence, and other resistive forces can reduce entropy generation, leading to greater efficiency.
It's useful to understand that while \( \triangle S \) (change in entropy) between two states is the same for both reversible and irreversible processes because entropy is a state function, \( \triangle S_{gen} \) differs. In the case of the exercise, this means that the entropy change for both the internally reversible and irreversible processes might be identical, but entropy generation is zero for the reversible process and positive for the irreversible one. This is crucial for designing efficient thermodynamic systems.

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Most popular questions from this chapter

Steam at \(0.7 \mathrm{MPa}, 355^{\circ} \mathrm{C}\) enters an open feedwater heater operating at steady state. A separate stream of liquid water enters at \(0.7 \mathrm{MPa}, 35^{\circ} \mathrm{C}\). A single mixed stream exits as saturated liquid at pressure \(p\). Heat transfer with the surroundings and kinetic and potential energy effects can be ignored. (a) If \(p=0.7 \mathrm{MPa}\), determine the ratio of the mass flow rates of the incoming streams and the rate at which entropy is produced within the feedwater heater, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of liquid exiting. (b) Plot the quantities of part (a), each versus pressure \(p\) ranging from \(0.6\) to \(0.7 \mathrm{MPa}\).

At steady state, a device receives a stream of saturated water vapor at \(210^{\circ} \mathrm{C}\) and discharges a condensate stream at \(20^{\circ} \mathrm{C}, 0.1 \mathrm{MPa}\) while delivering energy by heat transfer at \(300^{\circ} \mathrm{C}\). The only other energy transfer involves heat transfer at \(20^{\circ} \mathrm{C}\) to the surroundings. Kinetic and potential energy changes are negligible. What is the maximum theoretical amount of energy, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of steam entering, that could be delivered at \(300^{\circ} \mathrm{C} ?\)

Steam enters a horizontal \(15-\mathrm{cm}\)-diameter pipe as a saturated vapor at 5 bar with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and exits at \(4.5\) bar with a quality of \(95 \%\). Heat transfer from the pipe to the surroundings at \(300 \mathrm{~K}\) takes place at an average outer surface temperature of \(400 \mathrm{~K}\). For operation at steady state, determine (a) the velocity at the exit, in \(\mathrm{m} / \mathrm{s}\). (b) the rate of heat transfer from the pipe, in \(\mathrm{kW}\). (c) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for a control volume comprising only the pipe and its contents. (d) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for an enlarged control volume that includes the pipe and enough of its immediate surroundings so that heat transfer from the control volume occurs at \(300 \mathrm{~K}\). Why do the answers of parts (c) and (d) differ?

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) enters a nozzle operating at steady state at 28 bar, \(267^{\circ} \mathrm{C}\), and \(50 \mathrm{~m} / \mathrm{s}\). At the nozzle exit, the conditions are \(1.2\) bar, \(67^{\circ} \mathrm{C}, 580 \mathrm{~m} / \mathrm{s}\), respectively. (a) For a control volume enclosing the nozzle only, determine the heat transfer, in \(\mathrm{kJ}\), and the change in specific entropy, in \(\mathrm{kJ} / \mathrm{K}\), each per \(\mathrm{kg}\) of carbon dioxide flowing through the nozzle. What additional information would be required to evaluate the rate of entropy production? (b) Evaluate the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of carbon dioxide flowing, for an enlarged control volume enclosing the nozzle and a portion of its immediate surroundings so that the heat transfer occurs at the ambient temperature, \(25^{\circ} \mathrm{C}\).

An ideal gas with constant specific heat ratio \(k\) enters a nozzle operating at steady state at pressure \(p_{1}\), temperature \(T_{1}\), and velocity \(\mathrm{V}_{1} .\) The air expands isentropically to a pressure of \(p_{2}\) (a) Develop an expression for the velocity at the exit, \(\mathrm{V}_{2}\), in terms of \(k, R, \mathrm{~V}_{1}, T_{1}, p_{1}\), and \(p_{2}\), only. (b) For \(\mathrm{V}_{1}=0, T_{1}=1000 \mathrm{~K}\), plot \(\mathrm{V}_{2}\) versus \(p_{2} / p_{1}\) for selected values of \(k\) ranging from \(1.2\) to \(1.4\).

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