Book cover for Thermodynamics: An Engineering Approach

Thermodynamics: An Engineering Approach

Yunus A. Cengel, Michael A. Boles

ISBN #9781259822674

9th Edition

2,694 Questions

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59,300 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This chapter bridges the fundamental concepts of entropy and exergy in thermodynamics. Key topics include the measurement and significance of entropy, the idealized concept of isentropic processes, and the application of T ds relations to compute entropy changes in both closed systems and control volumes. Furthermore, by introducing exergy, reversible work, and second?law efficiency, the chapter provides engineers with a robust framework for diagnosing and reducing irreversibilities in devices like turbines, compressors, nozzles, and pumps. Ultimately, exergy analysis serves as a powerful tool in optimizing energy systems by highlighting areas where energy quality is degraded.

Learning Objectives

1

Explain the concept of exergy and its relation to the second law of thermodynamics in engineering systems.

2

Describe how entropy changes in pure substances and during isentropic processes, and apply the T ds relations.

3

Define reversible work, exergy destruction, and second?law efficiency, and demonstrate their roles in evaluating device performance.

4

Analyze both closed systems and control volumes using exergy balances to identify and minimize irreversibilities.

5

Apply the principles of entropy and exergy analysis to optimize the design and operation of turbines, compressors, nozzles, and pumps.

Key Concepts

CONCEPT

DEFINITION

Exergy

A measure of the quality of energy, representing its potential to perform work. It degrades due to irreversibilities in real processes.

Entropy

A thermodynamic property that measures the degree of disorder or randomness in a system, and indicates irreversibility in processes.

Reversible Work

The maximum work that can be extracted from a system if a process is conducted in a completely reversible manner.

Exergy Destruction

The loss of potential work due to irreversibilities, usually manifested as an increase in entropy in real processes.

Second-Law Efficiency

The ratio of the actual work output to the maximum possible (reversible) work output, serving as a measure of process performance.

Isentropic Process

A process that occurs at constant entropy, representing an idealized reversible process with no entropy generation.

T ds Relations

Fundamental thermodynamic relations that link temperature (T), entropy (s), and the differential changes in energy, forming the basis for many entropy calculations.

Example Problems

Example 1

Does a cycle for which $\oint \delta Q>0$ violate the Clausius inequality? Why?

Example 2

Does the cyclic integral of heat have to be zero (i.e. does a system have to reject as much heat as it receives to complete a cycle)? Explain.

Example 3

Is a quantity whose cyclic integral is zero necessarily a property?

Example 4

Is an isothermal process necessarily internally reversible? Explain your answer with an example.

Example 5

Is the value of the integral $\int_{1}^{2} \delta Q / T$ the same for all reversible processes between states 1 and $2 ?$ Why?
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Step-by-Step Explanations

QUESTION

Calculate the change in entropy (ΔS) for 1 kg of an ideal gas with a constant specific heat (c_p = 1.005 kJ/kg·K) when it is heated from 300 K to 400 K at constant pressure.

STEP-BY-STEP ANSWER:

Step 1: Identify the relevant formula for constant pressure: ΔS = c_p * ln(T2/T1).
Step 2: Substitute the provided temperatures into the formula: ΔS = 1.005 * ln(400/300).
Step 3: Compute the logarithmic term: ln(400/300) = ln(1.3333) ≈ 0.28768.
Step 4: Multiply the specific heat by the logarithm: ΔS = 1.005 * 0.28768 ≈ 0.289 kJ/kg·K.
Final Answer: The entropy change is approximately 0.289 kJ/kg·K.

Entropy Change of a Pure Substance (Constant Specific Heats)

QUESTION

Explain the steps to determine the exergy destruction during a heat transfer process if given the actual work output and the reversible work potential.

STEP-BY-STEP ANSWER:

Step 1: Define the reversible work potential for the process under ideal conditions.
Step 2: Measure or calculate the actual work output from the process.
Step 3: Compute the difference between the reversible work and the actual work; this difference represents the work lost due to irreversibilities.
Step 4: Relate this lost work to exergy destruction, acknowledging that increased entropy correlates directly with a reduction in available work.
Final Answer: Exergy destruction is quantified by the difference between the reversible work potential and the actual work output, highlighting the impact of irreversibilities and entropy generation.

Exergy Destruction in a Reversible vs. Irreversible Process

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Common Mistakes

  • Confusing energy with exergy; remembering that exergy measures the quality or work potential of energy, not just its quantity.
  • Assuming that a process with no heat transfer automatically has zero entropy generation; real processes often involve irreversibilities.
  • Misapplying the T ds relation by neglecting the variations in specific heats, especially when using constant specific heat approximations in situations where variable specific heats are more accurate.
  • Overlooking the fact that even processes labeled as 'isentropic' are idealizations; in real systems, small irreversibilities always result in some entropy generation.