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Chapter 6: Problem 38

A closed system undergoes a process in which work is done on the system and the heat transfer \(Q\) occurs only at temperature \(T_{\mathrm{b}}\). For each case, determine whether the entropy change of the system is positive, negative, zero, or indeterminate. (a) internally reversible process, \(Q>0\). (b) internally reversible process, \(Q=0\). (c) internally reversible process, \(Q<0\). (d) internal irreversibilities present, \(Q>0\). (e) internal irreversibilities present, \(Q=0\). (f) internal irreversibilities present, \(Q<0\).

Short Answer

Expert verified
(a) positive, (b) zero, (c) negative, (d) indeterminate, (e) positive, (f) indeterminate.

Step by step solution

01

Understanding Entropy Change

The entropy change of a system is a measure of disorder or randomness. For a closed system undergoing a process, the entropy change can be influenced by heat transfer (Q) and whether the process is internally reversible or has internal irreversibilities.
02

Case (a): Internally Reversible Process, Q > 0

For an internally reversible process with heat transfer Q > 0, the entropy change (螖S) of the system is positive. The system absorbs heat, resulting in an increase in entropy.
03

Case (b): Internally Reversible Process, Q = 0

For an internally reversible process with no heat transfer Q = 0, the entropy change (螖S) of the system is zero. Since there is no entropy generation due to heat transfer, the entropy remains unchanged.
04

Case (c): Internally Reversible Process, Q < 0

For an internally reversible process with heat transfer Q < 0, the entropy change (螖S) of the system is negative. The system releases heat, resulting in a decrease in entropy.
05

Case (d): Internal Irreversibilities Present, Q > 0

For a process with internal irreversibilities and heat transfer Q > 0, the entropy change (螖S) of the system is indeterminate. Internal irreversibilities can generate entropy, and the total entropy change depends on an additional component from these irreversibilities.
06

Case (e): Internal Irreversibilities Present, Q = 0

For a process with internal irreversibilities and no heat transfer Q = 0, the entropy change (螖S) is positive. Internal irreversibilities generate entropy even in the absence of heat transfer.
07

Case (f): Internal Irreversibilities Present, Q < 0

For a process with internal irreversibilities and heat transfer Q < 0, the entropy change (螖S) of the system is indeterminate. The entropy reduction due to heat loss may be countered by entropy generation due to irreversibilities.

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Key Concepts

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

entropy
Entropy is a fundamental concept in thermodynamics that measures the amount of disorder or randomness in a system. It's denoted by the symbol \( S \). When heat is added to a system, the molecules gain energy and move more randomly, increasing the entropy. Conversely, removing heat decreases the entropy as the system becomes more ordered. The change in entropy, \( \text{螖}S \), is given by the equation \( \text{螖}S = \frac{Q}{T} \) for a reversible process, where \( Q \) is the heat transfer and \( T \) is the absolute temperature. Understanding entropy helps in determining the feasibility and direction of a thermodynamic process.
internally reversible process
An internally reversible process is one where the system changes state in such a way that it can be reversed by an infinitesimal change without leaving any net effect on the system or its surroundings. These processes are idealized, meaning no real-world processes are perfectly reversible due to friction, unrestrained expansion, and other internal dissipative effects. When considering entropy change:
  • For \( Q > 0 \): The system absorbs heat, resulting in an increase in entropy. \( \text{螖}S > 0 \).
  • For \( Q = 0 \): No heat transfer means the entropy change is zero. \( \text{螖}S = 0 \).
  • For \( Q < 0 \): The system releases heat, resulting in a decrease in entropy. \( \text{螖}S < 0 \).
These relationships help predict the direction of entropy change for various processes.
internal irreversibilities
Internal irreversibilities refer to factors within the system that prevent it from returning to its original state without changes in its surroundings. These include friction, non-quasi-static processes, and heat transfer through a finite temperature difference. Presence of internal irreversibilities often leads to generation of entropy, denoted by \( S_{\text{gen}} \). Key points related to internal irreversibilities are:
  • When \( Q > 0 \): The entropy change is indeterminate since it also includes entropy generated from irreversibilities (\( \text{螖}S = \frac{Q}{T} + S_{\text{gen}} \)).
  • When \( Q = 0 \): Entropy change is positive due to entropy generation from irreversibilities \( \text{螖}S = S_{\text{gen}} > 0 \).
  • When \( Q < 0 \): The process could lead to a decrease in entropy due to heat loss, but internal irreversibilities might result in an indeterminate total entropy change (\( \text{螖}S = \frac{Q}{T} + S_{\text{gen}} \)).
This explains why internal irreversibilities complicate the prediction of entropy change.
heat transfer
Heat transfer is the movement of thermal energy from one object or system to another. It's a central concept in thermodynamics and can significantly influence the entropy change in a system. There are three main modes of heat transfer: conduction, convection, and radiation. Key points concerning entropy changes with heat transfer:
  • When heat is added to a system (\( Q > 0 \)), the increase in energy causes an increase in entropy.
  • When no heat is transferred (\( Q = 0 \)), there is no change in entropy due to thermal energy.
  • When heat is removed from a system (\( Q < 0 \)), it results in a decrease in entropy.
Understanding heat transfer helps in analyzing how energy moves through systems and how it affects entropy.

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Most popular questions from this chapter

An insulated, rigid tank is divided into two compartments by a frictionless, thermally conducting piston. One compartment initially contains \(1 \mathrm{~m}^{3}\) of saturated water vapor at \(4 \mathrm{MPa}\) and the other compartment contains \(1 \mathrm{~m}^{3}\) of water vapor at \(20 \mathrm{MPa}, 800^{\circ} \mathrm{C}\). The piston is released and equilibrium is attained, with the piston experiencing no change of state. For the water as the system, determine (a) the final pressure, in MPa. (b) the final temperature, in \({ }^{\circ} \mathrm{C}\). (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

One-tenth kilogram of water executes a Carnot power cycle. At the beginning of the isothermal expansion, the water is a saturated liquid at \(160^{\circ} \mathrm{C}\). The isothermal expansion continues until the quality is \(98 \%\). The temperature at the conclusion of the adiabatic expansion is \(20^{\circ} \mathrm{C}\). (a) Sketch the cycle on \(T-s\) and \(p-v\) coordinates. (b) Determine the heat added and net work, each in kJ. (c) Evaluate the thermal efficiency.

Air enters a compressor operating at steady state at \(17^{\circ} \mathrm{C}\), 1 bar and exits at a pressure of 5 bar. Kinetic and potential energy changes can be ignored. If there are no internal irreversibilities, evaluate the work and heat transfer, each in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of air flowing, for the following cases: (a) isothermal compression. (b) polytropic compression with \(n=1.3\). (c) adiabatic compression. Sketch the processes on \(p-v\) and \(T-s\) coordinates and associate areas on the diagrams with the work and heat transfer in each case. Referring to your sketches, compare for these cases the magnitudes of the work, heat transfer, and final temperatures, respectively.

Air is compressed in an axial-flow compressor operating at steady state from \(27^{\circ} \mathrm{C}, 1\) bar to a pressure of \(2.1\) bar. The work input required is \(94.6 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing through the compressor. Heat transfer from the compressor occurs at the rate of \(14 \mathrm{~kJ}\) per \(\mathrm{kg}\) at a location on the compressor's surface where the temperature is \(40^{\circ} \mathrm{C}\). Kinetic and potential energy changes can be ignored. Determine (a) the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the rate at which entropy is produced within the compressor, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of air flowing.

At steady state, work at a rate of \(25 \mathrm{~kW}\) is done by a paddle wheel on a slurry contained within a closed, rigid tank. Heat transfer from the tank occurs at a temperature of \(250^{\circ} \mathrm{C}\) to surroundings that, away from the immediate vicinity of the tank, are at \(27^{\circ} \mathrm{C}\). Determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), (a) for the tank and its contents as the system. (b) for an enlarged system including the tank and enough of the nearby surroundings for the heat transfer to occur at \(27^{\circ} \mathrm{C}\).
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