Thermodynamics

Thermodynamics Mr. M. F. Mohamed Hussain Assistant Professor Department of Aeronautical Engineering J.J. College of Engineering and Technology Ammapettai, Tiruchirappalli - 620009 Mr. Elphej Churchil. S. J Assistant Professor Department of Aeronautical Engineering J.J. College of Engineering and Technology Ammapettai, Tiruchirappalli - 620009 Mr. N. Prabakaran Assistant Professor Department of Aeronautical Engineering East West College of Engineering Bangalore, Karnataka – 560064 Mr. Nehru. K Assistant Professor Department of Aerospace Engineering SNS College of Technology Coimbatore - 641035 Mr. N. Shalom Assistant Professor Department of Mechanical Engineering Karpagam College of Engineering Coimbatore - 641032 Edition Details (I,II,III): I ISBN: 978-93-6786-035-9 Month & Year: October, 2025 Copyright @ Mr. M. F. Mohamed Hussain Mr. Elphej Churchil. S. J Mr. N. Prabakaran Mr. Nehru. K Mr. N. Shalom Pages: 463 Price: 1000/- About the Authors’ Mr. M. F. Mohamed Hussain works as an Assistant Professor in the department of Aeronautical Engineering, J.J. College of Engineering and Technology. He has 4 years of experience in teaching and 3 years of experience in research and also a member of AeSI. His research area includes Aerodynamics, Computational Fluid Dynamics, Aero Engineering Thermodynamics, Avionics, Flight Dynamics. Mr. Elphej Churchil. S. J is currently serving as an Assistant Professor in the Department of Aeronautical Engineering at J.J. College of Engineering and Technology, Tiruchirappalli, with more than 25 years of combined academic and industry experience. He has previously held leadership positions as Head of Department at Sathyabama Institute of Science and Technology, Chennai, and Park College of Engineering and Technology, Coimbatore, in addition to his industry role as a Design Engineer/Analyst at GCOL India, Mumbai. His academic and research expertise covers aircraft design, propulsion systems, aerodynamics, composites, and flight dynamics, with several publications, patents, and contributions to conferences and books. Mr. N. Prabakaran, graduated from Anna University, he has 12 Years of experience in teaching. Doing Ph.D. in Aerospace Engineering under Visvesvaraya Technological University, Karnataka. He has published 2 patents and 10 research papers in various journals and international conferences. His specialisations are in Aircraft Propulsion, Computational Fluid Dynamics, Thermodynamics, Thermal Engineering and Heat Transfer, currently working as an Assistant Professor in the Department of Aeronautical Engineering at East West College of Engineering, Bangalore – 560064, Karnataka. He is an active life time member of ISTE and IEI. Mr. Nehru. K, M.Tech., (Ph.D.) is an Assistant Professor in the Department of Aerospace Engineering at SNS College of Technology, Coimbatore. With more than eleven years of academic experience, he specializes in aircraft structures, vibration control, and fatigue life analysis. He earned his Master’s degree in Aeronautical Engineering from Hindustan University, graduating as a university topper, and is presently pursuing his Ph.D. in Mechanical Engineering at Anna University, Chennai. Mr. Nehru has published over 25 research papers in national and international journals and conferences. His research interests span vibration control in aircraft wings and fuselages, fatigue life estimation in gas turbine blades, and advanced composite structures. He has filed four patents one utility and three design patents and has successfully guided numerous student projects using CATIA, ANSYS, and CFD tools. He is a recipient of grants from CSIR and TNSCST and has organized several technical workshops, conferences, and industry–academia initiatives. A committed lifelong learner, he recently completed the AICTE–QIP postgraduate course on Advanced Aerospace Materials at IIT Bhilai, enhancing his expertise in cutting-edge aerospace technologies. Mr. Nehru is an Educator Associate Member of the American Institute of Aeronautics and Astronautics (AIAA) and a life member of several professional bodies. He is passionate about mentoring students, fostering innovation, and strengthening the link between academia and industry in aerospace engineering. Mr. N. Shalom is an accomplished academician and researcher, currently serving as an Assistant Professor in the Department of Mechanical Engineering at Karpagam College of Engineering, Coimbatore, Tamil Nadu, India. With a strong foundation in mechanical sciences, he holds a Bachelor of Engineering in Mechanical Engineering and a Master of Engineering in Thermal Engineering, both from Noorul Islam Centre for Higher Education, Kanyakumari. He is presently pursuing his Ph.D. at the same institution, with a focused interest in advanced materials and composites. With 3 years and 2 months of teaching experience, Mr. Shalom has consistently demonstrated excellence in both pedagogy and research. He has presented papers at leading national and international conferences, and his scholarly contributions include one Scopus-indexed and three SCI-indexed research articles published in reputed international journals. Mr. Shalom is also an innovator, having secured five patent grants, reflecting his commitment to applied research and technological advancement. He is a lifetime member of the International Association of Engineers (IAENG), actively engaging with global research communities. His core research interests lie in the field of Polymer Composites, with ongoing work aimed at developing next-generation materials with enhanced mechanical and thermal properties. Through his academic and research pursuits, Mr. Shalom continues to contribute meaningfully to the evolving landscape of engineering education and innovation. Preface Thermodynamics stands as one of the most fundamental and unifying branches of science and engineering. It provides the foundation for understanding the principles governing energy transformations, heat interactions, and the physical behavior of matter. From classical steam engines to modern power plants, from chemical reactions to biological systems, thermodynamics forms the core of how energy is utilized, conserved, and transformed in every aspect of nature and technology. This book on Thermodynamics has been prepared with the aim of offering a clear, comprehensive, and systematic understanding of the subject. It is designed to serve as a valuable reference for undergraduate and postgraduate students, as well as for educators, researchers, and professionals in various fields of science and engineering. The content is structured to provide a balanced blend of theoretical concepts, practical applications, and problem-solving techniques. Starting with the fundamental laws of thermodynamics, the book progresses through energy analysis, properties of pure substances, power and refrigeration cycles, and extends to contemporary topics such as thermodynamic efficiency, entropy generation, and the role of thermodynamics in sustainable energy systems. Special attention has been given to the clarity of explanation, logical flow of topics, and the inclusion of illustrative examples and diagrams to make complex concepts more accessible. Each chapter concludes with summary points and exercises, encouraging readers to apply their knowledge and strengthen their understanding through practice. This book also recognizes the growing importance of computational and experimental approaches in thermodynamic analysis. Where relevant, discussions have been enhanced with insights into modern simulation techniques, real-world applications, and interdisciplinary connections with fields such as materials science, environmental engineering, and renewable energy. We hope this work inspires curiosity, critical thinking, and a deeper appreciation for the elegance and universality of thermodynamic principles. If it assists even a few readers in grasping the beauty and relevance of this discipline, we will consider our efforts well rewarded. Finally, we welcome constructive feedback and suggestions from readers, which will help us improve future editions of this book. Mr. M. F. Mohamed Hussain Mr. Elphej Churchil. S. J Mr. N. Prabakaran Mr. Nehru. K Mr. N. Shalom Acknowledgement The completion of this book on Thermodynamics has been a deeply rewarding and enlightening journey, made possible through the constant support, encouragement, and contributions of numerous individuals and institutions. We extend our heartfelt gratitude to everyone who has played a part in bringing this work to realization. First and foremost, we express our sincere appreciation to our teachers, mentors, and colleagues, whose expert guidance, constructive feedback, and continuous encouragement have been instrumental in shaping the content and structure of this book. Their insights have enriched our understanding and enabled us to present both the fundamental principles and practical applications of thermodynamics with clarity and depth. We are profoundly thankful to our families for their unwavering love, patience, and moral support throughout this endeavor. Their understanding and motivation have been our strength during every stage of this writing process. Our deepest gratitude also goes to the scientific and academic communities whose remarkable research and contributions have expanded the horizons of thermodynamic science. Their innovative work and discoveries have been a source of inspiration in compiling and presenting this comprehensive study. We also acknowledge the assistance of modern computational tools and resources, which have greatly aided in data analysis, diagrammatic representation, and the overall refinement of this book. These technologies have helped demonstrate how thermodynamics continues to evolve with scientific and technological advancements. Above all, we offer our sincere thanks to Almighty God for His divine guidance, wisdom, and blessings, which have sustained us through challenges and enabled the successful completion of this work. We hope this book serves as a valuable resource for students, educators, researchers, and professionals, helping them understand, apply, and further explore the fascinating principles of Thermodynamics in science and engineering. Mr. M. F. Mohamed Hussain Mr. Elphej Churchil. S. J Mr. N. Prabakaran Mr. Nehru. K Mr. N. Shalom Thermodynamics 3 This relationship is expressed by the Knudsen Number (Kn): 𝐾𝑛=𝜆 𝐿 Where,  λ = Mean free path of the molecules  L = Characteristic length of the system Range of Kn Flow Regime Description Kn < 0.01 Continuum Flow Continuum assumption is fully valid; classical fluid mechanics applies. 0.01 < Kn < 0.1 Slip Flow Slight deviations from the continuum; surface slip effects begin to appear. 0.1 < Kn < 10 Transition Flow Continuum assumption partially breaks down; both molecular and continuum models may be needed. Kn > 10 Free Molecular Flow Continuum approach invalid; molecular (kinetic theory) approach required, such as in high vacuum or outer space. Table. 1.1 Range of Knudsen Number. Applications of the Continuum Concept  Valid for: Common engineering fluids like air, water, steam, and oil under normal temperature and pressure conditions.  Invalid for: Rarefied gases, high-altitude atmospheric layers, vacuum systems, and outer space environments. Problem 1.1: A gas is flowing through a small capillary tube of diameter 0.002 m at standard atmospheric conditions. The mean free path of the gas molecules under these conditions is 6.5×10−8 m. Determine whether the continuum assumption is valid for this flow. Solution: Given Data: Mean free path, 𝜆=6.5×10−8 m Thermodynamics 4 Characteristic length, 𝐿=0.002 m Formula Used: Knudsen Number (Kn)=𝜆 𝐿 Step 1: Substitute the given values 𝐾𝑛=6.5×10−8 0.002 Step 2: Perform the calculation carefully 𝐾𝑛=3.25×10−5 Step 3: Interpret the Knudsen Number From Table 1.1,  If Kn<0.01,→ Continuum Flow (continuum assumption valid)  If 0.01<Kn<0.1,→ Slip Flow  If 0.1<Kn<10, Transition Flow  If Kn>10, → Free Molecular Flow Since 𝐾𝑛=3.25×10−5 <0.01 the flow is within the continuum regime. Exercise 1.1: A spacecraft is moving through the upper atmosphere where the mean free path of air molecules is 0.02 m. The characteristic length of the spacecraft surface is approximately 0.5 m. Find: 1. Calculate the Knudsen number (Kn) for this situation. 2. Determine the flow regime (Continuum, Slip, Transition, or Free Molecular). 3. State whether the continuum assumption is valid. Thermodynamics 5 Exercise 1.2: Helium gas is flowing through a microchannel with a diameter of 5×10−4 m. The mean free path of helium molecules under given conditions is 6×10−7 m. Find: 1. Compute the Knudsen number (Kn). 2. Identify the flow regime using Table 1.1. 3. State whether the continuum approach or the molecular (kinetic theory) approach should be used to analyze the flow. 1.2 Macroscopic Approach The macroscopic approach is one of the two fundamental methods used in thermodynamics to study the behavior of matter the other being the microscopic (statistical) approach. In the macroscopic method, a system is analyzed based on measurable bulk properties without considering the molecular structure or behavior of individual particles. Definition The macroscopic approach in thermodynamics deals with the overall behavior of matter, considering the system as a continuous, homogeneous medium. It describes the state of the system using measurable macroscopic properties such as:  Pressure (P)  Volume (V)  Temperature (T)  Internal energy (U)  Enthalpy (H)  Entropy (S) These properties are averaged over a large number of particles and represent the state of the system as a whole. Key Features of the Macroscopic Approach 1. No molecular consideration: The internal structure and molecular motion of substances are not analyzed. Instead, the system is assumed to be uniform and continuous. 2. Measurable quantities: All properties used in this approach (e.g., pressure, temperature) can be experimentally measured with instruments. Thermodynamics 6 3. Few variables: The system’s condition is described using a small number of state variables, making analysis simple and practical. 4. Continuum assumption: The system is considered as a continuum, meaning that properties vary smoothly and continuously throughout the space without any molecular gaps or discontinuities. 5. Deterministic behavior: The system’s properties and changes follow deterministic laws (e.g., laws of thermodynamics), which predict the behavior of the system precisely under given conditions. Example Consider a gas in a cylinder fitted with a piston. Using the macroscopic approach, we analyze:  Pressure (P) acting on the piston head,  Volume (V) occupied by the gas, and  Temperature (T) of the gas. From these measurable quantities, we can determine the state of the system and apply thermodynamic laws (like PV=nRT) without considering the motion or collision of individual gas molecules. Fig. 1.2 Simple example of macroscopic approach. Thermodynamics 7 Advantages of the Macroscopic Approach  Simple and practical for engineering analysis.  Directly measurable properties make it experimentally verifiable.  Applicable to large-scale systems such as engines, turbines, refrigerators, etc.  Requires less computation than microscopic analysis. Limitations  Cannot explain microscopic phenomena, such as molecular motion or energy distribution among molecules.  Fails at very small scales, where molecular effects become significant (e.g., in rarefied gases, nanotechnology, or quantum systems).  Does not reveal the molecular basis of thermodynamic properties like entropy or temperature. Aspect Macroscopic Approach Microscopic Approach Basis Bulk behavior of matter Molecular and atomic behavior Variables used P, V, T, U, H, S Molecular velocity, energy distribution Analysis type Experimental, measurable Statistical and probabilistic Continuum assumption Assumed continuous Discrete molecular structure Ease of use Simpler and more practical More complex and theoretical Application Engineering thermodynamics Statistical thermodynamics Table. 1.2 Comparison with Microscopic Approach. 1.3 Thermodynamic systems – closed, open, and isolated In thermodynamics, a system refers to a specific quantity of matter or a particular region in space chosen for study. Everything outside the system is called the surroundings, and the imaginary boundary that separates the system from its surroundings is called the boundary. Thermodynamics 8 Fig. 1.3 Thermodynamics-System, Boundary and Surroundings. Depending on how energy and mass interact across this boundary, thermodynamic systems are classified into three main types:  Closed System  Open System  Isolated System 1. Closed System (Control Mass System) A closed system is one in which the mass of the system remains constant, but energy exchange with the surroundings is possible. The energy can be transferred in the form of heat, work, or both. In other words, no mass crosses the system boundary, but energy can flow in or out depending on the conditions. The boundary of the system separates it from its surroundings and can either be fixed (stationary) or movable (such as a piston). Characteristics of a Closed System 1. Constant Mass: The same set of particles remains within the system at all times. No matter enters or leaves. 2. Energy Interaction: The system can exchange heat energy (Q) or work (W) with its surroundings through the boundary. 3. Boundary Flexibility: The boundary may be fixed (like a rigid container) or movable (like a piston that can move up and down). 4. Continuum Assumption: The system is treated as a continuum, meaning its properties are uniform and continuous throughout. Thermodynamics 9 Illustration Consider a cylinder containing gas with a movable piston fitted at the top: Fig. 1.4 Closed System.  When heat is added to the gas, it expands and pushes the piston upward, doing work on the surroundings.  Conversely, when heat is removed, the gas contracts and the piston moves downward, with work being done on the gas.  Throughout this process, no gas particles escape or enter the cylinder only energy transfer takes place. Thus, the mass inside the cylinder remains constant, fulfilling the definition of a closed system. Thermodynamics 10 For a closed system: Δ𝑚=0 but 𝑄,𝑊≠0 Where,  𝚫𝐦= Change in mass of the system (zero for a closed system)  Q= Heat transfer across the boundary  𝐖= Work transfer across the boundary This means the mass of the system does not change, but energy interactions in the form of heat and work are possible. Examples 1. Piston-Cylinder Assembly in an Engine: When fuel burns in an internal combustion engine, the expanding gases push the piston. Although heat and work are exchanged, the gas itself does not cross the system boundary during the expansion stroke (if we ignore leakage). 2. Sealed, Heated Container of Water: When a sealed vessel containing water is heated, energy in the form of heat is added, causing some water to vaporize. The total mass (water + vapor) remains constant, but the internal energy changes due to heat addition. 3. Closed Pressure Cooker: A pressure cooker allows heat transfer from the stove to the food inside but does not allow steam to escape unless the safety valve opens. Hence, it is a practical example of a closed system under normal conditions. 2. Open System (Control Volume System) An open system, also known as a control volume, is one in which both mass and energy can cross the system boundary. Unlike a closed system where mass remains constant, an open system allows the continuous flow of matter into and out of the control region. This concept is commonly used in steady-flow devices, where the flow conditions at the inlet and outlet remain constant over time (such as in turbines, compressors, and pumps). Characteristics of an Open System 1. Mass Exchange: In an open system, mass can enter or leave through one or more openings. The system’s contents change continuously as new mass enters and old mass exits. 2. Energy Interaction: Energy can be transferred across the boundary in three ways:  Heat (Q): Thermal energy transfer due to temperature difference. Thermodynamics 11  Work (W): Mechanical energy transfer, such as shaft work or electrical work.  Mass Flow Energy: The energy associated with the flow of mass entering or leaving the system (includes kinetic and potential energy). 3. Boundary Nature: The boundary of an open system may be real (like the solid casing of a turbine or compressor) or imaginary (across a flow region, such as an inlet or outlet plane). The boundary can also change shape or position depending on the process. 4. Steady or Unsteady Operation:  In steady-flow systems, mass and energy entering and leaving the system remain constant over time.  In unsteady-flow systems, these quantities vary with time (e.g., filling or emptying of a tank). Illustration Consider a steam turbine, one of the best examples of an open system:  Steam enters the turbine at high pressure and high temperature.  Steam exits at a lower pressure after transferring some of its energy to rotate the turbine shaft.  The energy output from the system is in the form of mechanical work (W), and the mass flow (steam) continuously passes through the system. Thus, both mass and energy cross the system boundary, making it an open system. This behavior applies to many engineering devices designed to convert or transfer energy efficiently. For an open system: Δ𝑚≠0 and 𝑄,𝑊≠0 Where,  𝚫𝐦= Change in mass (not constant, since mass enters and exits)  Q= Heat transfer across the system boundary  𝐖= Work transfer (shaft work, flow work, etc.) This relationship shows that mass flow and energy transfer occur simultaneously in open systems. Thermodynamics 12 Fig. 1.5 Open System. Examples 1. Steam or Gas Turbine: Converts the enthalpy (thermal energy) of steam or gas into mechanical work. Mass enters and exits continuously. 2. Boiler: Water enters as a liquid and exits as steam. Both mass and heat energy are transferred through the boundaries. 3. Nozzle and Diffuser: In a nozzle, fluid accelerates (kinetic energy increases) while pressure decreases. In a diffuser, the reverse happens velocity decreases and pressure rises. Both involve mass and energy flow. 4. Air Compressor or Pump: These devices add energy to the fluid (work input) to increase its pressure. The mass (air or liquid) flows through continuously. Thermodynamics 19 Example: A 2-liter container of gas at 2 atm and 300 K is divided into two equal 1-liter containers:  Pressure (2 atm) and Temperature (300 K) → remain the same (Intensive).  Volume (1 L each) → halved (Extensive). Uses 1. Material Identification: Intensive properties like density and refractive index help identify substances. 2. System Design and Analysis: Engineers use both property types to analyze energy systems like turbines, compressors, and boilers. 3. Thermodynamic Calculations: Determining specific (per unit mass) quantities allows comparison between systems of different sizes. 4. Energy and Mass Balance: Extensive properties are used to calculate total energy or mass transfer in processes. 5. Process Optimization: Intensive properties help control system conditions such as pressure and temperature for maximum efficiency. Aspect Intensive Properties Extensive Properties Dependence Independent of the amount of matter Dependent on the amount of matter Effect of Subdivision Remain unchanged when divided Change proportionally when divided Additivity Not additive Additive in nature Usage Identify and characterize substances Measure total system behavior Relation to Matter Describe intrinsic nature Describe overall quantity Examples Temperature, Pressure, Density, Boiling Point Mass, Volume, Energy, Entropy, Enthalpy Type of Information Qualitative Quantitative Mathematical Relation Often derived as ratios of extensive properties Directly proportional to system size Table. 1.5 Difference between Intensive and Extensive Properties. Thermodynamics 20 1.5 Path and State Functions It is essential to distinguish between quantities that depend only on the state of a system and those that depend on the manner (path) in which the system changes from one state to another. These two categories are known as point (state) functions and path functions. Understanding their distinction forms the foundation of energy analysis in thermodynamic systems. 1. Path Functions A path function is a quantity whose value depends on the specific path or manner in which a system changes from an initial state to a final state. In other words, the process path the series of intermediate states determines the amount of the quantity, not just the starting and ending points. Path functions are not properties of the system but represent interactions at the boundary of the system, such as heat transfer (Q) and work (W). Consider a system that undergoes a change from state 1 to state 2. On a Pressure–Volume (P–V) diagram, this change can occur through different process paths such as A, B, or C: Fig. 1.7 Path Function.  Even though all paths start at state 1 and end at state 2, the area under each curve (representing work done) is different.  Hence, work and heat depend on the path followed, not merely the endpoints. Thermodynamics 21 This means the magnitude of energy transferred as work or heat varies with how the process occurs slow or fast, isothermal or adiabatic, reversible or irreversible. On a P–V diagram, the area under the curve between two states represents work. Different process curves (paths) between the same states yield different areas, confirming that work (and heat) are path-dependent quantities. Examples of Path Functions 1. Work (W): The work done by or on a system depends on how the process occurs. For a gas expanding or compressing, the P-V relationship determines the total work. Example: A gas expanding isothermally (constant temperature) does more work than when it expands adiabatically (no heat exchange), even if both processes start and end at the same states. 𝑊=∫ 2 1𝑃𝑑𝑉 The value of this integral depends on the path (i.e., how P changes with V ). 2. Heat Transfer (Q): The heat absorbed or rejected depends on how energy is added or removed. Example: Heating water slowly, rapidly, or under different pressures will result in different heat quantities, even if the final temperature is the same. Characteristics of Path Functions 1. Process Dependent: The value depends on the specific path or process taken between two states. 2. Different Paths, Different Values: Two systems reaching the same final state can have different heat or work interactions. 3. Not Properties: Since they depend on the path, they cannot describe the system’s condition at a point in time. 4. Inexact Differentials: Path functions are written as δQ and δW, not dQ or dW, indicating they are inexact differentials (not exact, not integrable without specifying a path). Thermodynamics 22 5. Nonzero in Cyclic Processes: The net value of a path function over a thermodynamic cycle is not zero it can be positive or negative (e.g., work output in engines). 2. Point Functions (State Functions) A point function, also known as a state function, is a quantity whose value depends only on the current state of the system and not on the path taken to reach that state. That means its value is determined solely by the state variables such as pressure (P), volume (V), temperature (T), and internal energy (U) and not on how the system arrived there. If a system moves from state 1 to state 2, the change in a point function is: Δ𝑍=𝑍2−𝑍1 This value is the same no matter how the system transitions between these states. Thus, point functions describe the condition or state of the system, while path functions describe the process that causes the state change. Fig. 1.8 Point (State) Function. Examples of Point Functions 1. Internal Energy (U): The change in internal energy depends only on the initial and final states, not on the process. Example: A gas expanding slowly or rapidly will have the same ΔU if the initial and final conditions are identical. Thermodynamics 23 Δ𝑈=𝑈2−𝑈1 2. Temperature (T): When water is heated from 20∘C to 80∘C, the temperature change ( ΔT=60∘C ) remains the same, regardless of how fast or slow the heating occurs. 3. Enthalpy (H): When water changes to steam at a given pressure, the enthalpy change depends only on the initial (liquid) and final (vapor) states, not on the heating method. 4. Entropy (S): For reversible or irreversible processes, the entropy change between two states depends only on those states. Characteristics of Point Functions 1. State Dependent: Depend only on the system's present condition, not on its history or process. 2. Exact Differentials: Represented as 𝐝𝐔,𝐝𝐇,𝐝𝐒, etc., meaning they are exact differentials that can be integrated directly. 3. Unique for Each State: Each thermodynamic state has a unique set of property values ( P,V,T,U, etc.). 4. Cyclic Processes: For a cyclic process, the net change in a point function is zero since the system returns to its initial state. ∮𝑑𝑈=0 Relationship between Path and Point Functions Path functions and point functions are interconnected in thermodynamics. The path functions (like heat and work) are responsible for changing the point functions (like internal energy or enthalpy). Key Relationships 1. Path Functions Cause Changes in Point Functions: Heat (Q) and Work (W) cause a change in the system's internal energy (U), enthalpy (H), or entropy (S). Δ𝑈=𝑄−𝑊 2. Different Paths → Same ΔU: For the same change of state, Q and W may vary with the process, but ΔU remains constant. 3. Point Functions Define Limits for Path Functions: The energy changes ( Δ𝑈 or ΔH ) set the boundaries for how much work or heat transfer can occur in a given process. Thermodynamics 24 Aspect Point Function (State Function) Path Function Definition Depends only on the state of the system Depends on the path taken between states Dependence Independent of path followed Dependent on the process path Examples Pressure (P), Temperature (T), Volume (V), Internal Energy (U), Enthalpy (H), Entropy (S) Work (W), Heat (Q) Value for Same States Same for all processes between two states Different for different processes Differentials Exact differentials (dU, dH, dS) Inexact differentials (δQ, δW) Additivity Additive and state-defined Non-additive and pathspecific Cyclic Process Net change over a cycle = 0 Net value over a cycle ≠ 0 Information Needed Only initial and final states Full path or process description Nature Property of the system Boundary interaction Table. 1.6 Difference between Point and Path Functions. Problem 1.2: A gas expands from an initial state of Pressure (𝐏𝟏)=𝟐𝟎𝟎 𝐤𝐏𝐚 and Volume (𝐕𝟏)=𝟎.𝟐𝐦𝟑 to a final volume of V2=0.6 m3. The process follows two different paths between the same states: 1. Path A: Isothermal expansion (Temperature constant) 2. Path B: Expansion following the law 𝑃=𝐶/𝑉2 The work done by the gas is given by 𝑊=∫ 𝑉2 𝑉1𝑃𝑑𝑉. Determine: 1. Calculate the work done for both paths. 2. Show that work (W) is a path function. 3. Comment on the change in internal energy (ΔU) between the same states. Thermodynamics 25 Assume: 𝑃1=200kPa=200×103 Pa 𝑉1=0.2 m3,𝑉2=0.6 m3 Solution: (A) Path A - Isothermal Process For an isothermal process, 𝑃𝑉= constant =𝐶 Hence, 𝑃=𝐶 𝑉 At state 1, 𝐶=𝑃1𝑉1=200×103×0.2=40,000 J Work done during isothermal expansion: 𝑊𝐴=∫ 𝑉2 𝑉1𝑃𝑑𝑉=∫ 𝑉2 𝑉1𝐶 𝑉𝑑𝑉=𝐶ln (𝑉2 𝑉1) Substitute values: 𝑊𝐴=40,000ln (0.6 0.2) 𝑊𝐴=40,000ln (3)=40,000×1.0986=43,944 J 𝑊𝐴=43.94 kJ (B) Path B - P = 𝑪 𝑽𝟐 At state 1, 𝐶=𝑃1𝑉12=200×103×(0.2)2=8,000 J.m Work done: 𝑊𝐵=∫ 𝑉2 𝑉1 𝑃𝑑𝑉=∫ 𝑉2 𝑉1 𝐶 𝑉2𝑑𝑉 Thermodynamics 26 𝑊𝐵=𝐶[−1 𝑉]𝑉1 𝑉2=𝐶(1 𝑉1−1 𝑉2) Substitute values: 𝑊𝐵=8,000(1 0.2−1 0.6) 𝑊𝐵=8,000(5−1.667)=8,000×3.333=26,664 J 𝑊𝐵=26.66 kJ (C) Comparison and Interpretation Path Process Type Expression Used Work Done (J) Work Done (kJ) A Isothermal 𝑊=𝐶ln (𝑉2/𝑉1) 43,944 43.94 B 𝑃=𝐶/𝑉2 𝑊=𝐶(1/𝑉1−1/𝑉2) 26,664 26.66 Result: Even though the initial and final states are the same, the work done differs for each path. Hence, Work (W) is a Path Function, not a property of the system. (D) Internal Energy Change ( ΔU ) Since the initial and final states are identical, all state properties (like internal energy) depend only on those states. Therefore, Δ𝑈=𝑈2−𝑈1=0 for both paths (since temperature and energy level depend only on the states, not on how the process occurred). Exercise 1.3: A gas expands from an initial state where 𝑃1=300kPa,𝑉1=0.1 m3 to a final volume 𝑉2=0.3 m3 Thermodynamics 27 according to the law 𝑃𝑉= constant (isothermal process). The same gas is then taken from the same initial to final state by another process defined by 𝑃=300− 500 V (where P is in kPa and V in m3 ). Determine: 1. Calculate the work done (W) for both processes. 2. Compare the results to show that work (W) is a path function. 3. Determine whether the change in internal energy (Δ𝑈) depends on the path or on the end states. Exercise 1.4: A gas is taken from state 1 to state 2 as follows: 𝑃1=100kPa, 𝑉1=0.4 m3 𝑃2=200kPa, 𝑉2=0.2 m3 Two different processes are used: 1. Path A: Straight-line process on a P-V diagram joining the two points. 2. Path B: Follows the law 𝑃𝑉1.2= constant. Determine: 1. Determine the work done (W) for both processes. 2. Compare and explain why the work done differs even though the initial and final states are the same. 3. Comment on whether the change in internal energy (ΔU) will be the same or different for both processes. 1.6 Quasi-Static Process A quasi-static process also called a quasi-equilibrium process is a theoretical thermodynamic process that occurs infinitesimally slowly so that the system remains infinitesimally close to equilibrium at all times. In other words, every intermediate state the system passes through can be considered as an equilibrium state. Although a perfectly quasi-static process is impossible in reality, it serves as an idealized model that allows accurate thermodynamic analysis. Thermodynamics 28 A quasi-static process is defined as: “A process that proceeds in such a manner that the system passes through a continuous sequence of equilibrium states, and the deviation from equilibrium is infinitesimally small.” Mathematically, it means the process occurs infinitely slowly, allowing uniform pressure, temperature, and density throughout the system at every instant. In a real process, when a system changes its state, there is usually an imbalance in pressure, temperature, or other properties between different parts of the system or between the system and surroundings. However, in a quasi-static process, these imbalances are assumed to be infinitesimally small, such that equilibrium conditions are practically maintained. Imagine a piston-cylinder containing gas: Fig. 1.9 Non-Quasi and Quasi Static Process. Thermodynamics 35 Even though no mass is lifted directly by the gas, the external effect the lifting of the weight indicates that work has been done by the system. Hence, the expansion of the gas is considered work done by the system on the surroundings. Key Concept: External Effects Only When evaluating thermodynamic work, only effects outside the system boundary are taken into account. Energy exchanges within the system do not constitute work from a thermodynamic perspective. Example: Consider a lift (elevator) as a thermodynamic system that includes a person and a suitcase. Fig. 1.13 Example of External Effects. If the person lifts the suitcase inside the lift, this energy transfer occurs within the system boundary. Therefore, no thermodynamic work is said to occur. Thermodynamics 36 However, if the entire lift moves upward due to a motor or external force, then energy is transferred across the boundary and this is considered thermodynamic work. Thus, only external mechanical effects are relevant in thermodynamic work analysis. In a general sense, work is the energy transfer that occurs when a force acts through a distance at the system boundary. 𝛿𝑊=𝐹𝑑𝑥 If a fluid element exerts a pressure (P) on a moving boundary of area A, the force exerted is: 𝐹=𝑃×𝐴 Since the volume change (dV) equals A×dx, the differential work done is: 𝛿𝑊=𝑃𝑑𝑉 Integrating between two states (1 and 2), we get the total work done: 𝑊=∫ 𝑉2 𝑉1𝑃𝑑𝑉 This is the boundary work, the most common type of thermodynamic work, which represents energy transfer due to volume change. Units of Work and Power Work In the SI system: Work = Force × Distance  Unit of force (F): Newton (N)  Unit of distance (d): metre (m) Therefore, the unit of work is: 1 N.m =1 Joule (J) Since thermodynamic systems often involve large energy values, work is expressed as: 1 kJ=1000 J Thermodynamics 37 Power The rate of doing work is called power. It is defined as work done per unit time: Power = Work Time In SI units: 1 Watt (𝑊)=1 Joule / second =1( N.m )/s For large-scale applications:  1 kW=1000 W  1MW=106 W Sign Convention for Work in Thermodynamics In thermodynamics, sign convention helps determine whether work is done by the system or on the system. Fig. 1.14 Sign convention of work. 1. Work Done by the System – Positive (+W): When the system expands and does work on the surroundings, work is considered positive. Example: Expansion of gas in a piston-cylinder arrangement. The gas pushes the piston upward, transferring energy outward → +W. 2. Work Done on the System – Negative (–W): When the surroundings compress the system or energy is transferred into the system, work is considered negative. Thermodynamics 38 Example: Compression of gas in a cylinder. The external pressure pushes the piston inward → –W. Process Description Sign of Work (W) Gas expansion System does work on surroundings Positive (+W) Gas compression Surroundings do work on system Negative (-W) No boundary movement No work done Zero (0) This sign convention ensures consistency when applying the First Law of Thermodynamics, which states: Δ𝑈=𝑄−𝑊 Where, Δ𝑈= change in internal energy, 𝑄= heat added to the system, 𝑊= work done by the system. Thermodynamic work always results in measurable mechanical effects external to the system boundary. It represents the energy interaction that can cause:  Movement (as in a piston)  Rotation (as in a turbine)  Electrical effects (as in generators)  Flow motion (as in pumps and compressors) It is important to note that heat transfer is distinguished from work because it is caused by a temperature difference, whereas work occurs due to a mechanical or force-related cause. Importance of Work 1. Energy Conversion: Work is the primary means of converting one form of energy into another, such as in engines and power plants. 2. System Analysis: Work quantifies mechanical energy transfer, which is essential for analyzing thermodynamic cycles. 3. Performance Evaluation: Engine and turbine efficiencies are based on the ratio of work output to energy input. Thermodynamics 39 4. Practical Applications: Work governs the operation of compressors, pumps, generators, and all mechanical systems involving energy exchange. Heat Heat is one of the two principal modes of energy transfer, the other being work. While work involves energy transfer due to a mechanical effect (like movement or force application), heat involves energy transfer due to a temperature difference between a system and its surroundings. Understanding heat transfer is fundamental for analyzing engines, refrigerators, power plants, and heat exchangers, as all these systems operate based on thermal energy exchange. Definition of Heat Heat is defined as a form of energy transfer between two systems (or between a system and its surroundings) that occurs solely due to a temperature difference. When two bodies at different temperatures are in contact:  Energy flows from the hotter body to the colder one.  This transfer continues until thermal equilibrium (equal temperature) is reached. Key Characteristics of Heat 1. Heat is not a property of a system: A body does not contain heat; it contains internal energy. Heat exists only during energy transfer due to temperature difference. 2. Heat is an energy in transit: It can only be recognized as it crosses the system boundary not stored inside a system. 3. Heat is a path function: The amount of heat transferred depends on the path followed during the process, not just the initial and final states. Thus, it is represented as an inexact differential (δQ), not an exact one (dQ). 4. Direction of Heat Flow: Always flows from higher temperature to lower temperature (according to the Second Law of Thermodynamics). 5. Units of Heat:  SI Unit: Joule (J)  Practical Unit: Kilojoule (kJ) or Calorie (cal) (1 cal = 4.186 J) Thermodynamics 40 Mechanism of Heat Transfer Heat transfer occurs in three distinct modes, depending on how energy moves through or between materials: 1. Conduction Conduction is the transfer of heat through a solid medium or between two bodies in direct contact, without any bulk motion of matter. Mechanism:  In solids, energy transfer occurs by vibrations of molecules and movement of free electrons (especially in metals).  Adjacent molecules pass on their energy, leading to a gradual transfer of heat through the material. Mathematical Relation (Fourier's Law): 𝑞=−𝑘𝐴𝑑𝑇 𝑑𝑥 Where,  𝑞= rate of heat transfer (W)  𝑘= thermal conductivity (W/m⋅K)  𝐴= area of heat transfer (m2)  𝑑𝑇/𝑑𝑥= temperature gradient (K/m) Characteristics:  Dominant in solids.  Requires physical contact between molecules.  The rate depends on the thermal conductivity ( 𝐤 ) of the material. Examples:  Heat transfer through the wall of a furnace.  A metal rod becoming hot at one end when the other end is heated.  Heat flow through the metal body of a cooking pot. Thermodynamics 41 2. Convection Convection is the transfer of heat through a fluid (liquid or gas) by the combined effect of molecular motion (conduction) and bulk fluid movement (advection). Mechanism: When a fluid is heated:  The region near the heat source becomes less dense and rises.  Cooler, denser fluid moves in to replace it.  This continuous circulation transfers heat within the fluid. Types of Convection 1. Natural (Free) Convection: Fluid motion is caused by density differences due to temperature variations. No external device (like a pump or fan) is used. Examples:  Cooling of hot water in an open vessel.  Air circulation in a heated room.  Wind movement in the atmosphere. 2. Forced Convection: Fluid motion is induced by external means, such as a pump, fan, or blower. Examples:  Cooling of an automobile radiator using a fan.  Air conditioning systems.  Circulation of coolant in power plants. Mathematical Relation (Newton's Law of Cooling): 𝑞=ℎ𝐴(𝑇𝑠−𝑇∞) Where,  𝑞= rate of convective heat transfer (W) Thermodynamics 42  ℎ= heat transfer coefficient (W/m2⋅ K)  𝐴= surface area (m2)  𝑇𝑠= surface temperature (K)  𝑇∞= ambient fluid temperature (K) Characteristics:  Important in liquids and gases.  Depends on fluid velocity, viscosity, and temperature difference.  Involves bulk movement of the fluid. 3. Radiation Radiation is the transfer of heat in the form of electromagnetic waves, primarily infrared radiation. It does not require any medium and can occur even through a vacuum. Mechanism: All bodies with temperature above absolute zero ( 0 K ) emit electromagnetic radiation. The intensity and wavelength of the emitted radiation depend on the temperature and surface characteristics. Mathematical Relation (Stefan-Boltzmann Law): 𝑞=𝜎𝐴𝑇4 Where,  𝑞= radiant heat transfer (W)  𝜎= Stefan-Boltzmann constant (5.67×10−8 W/m2⋅ K4)  𝐴= emitting area (m2)  𝑇= absolute temperature (K) Characteristics:  Does not require a medium (can occur in space).  Depends on the emissivity ( 𝜀 ) of the surface ( 0≤𝜀≤1 ).  Is mainly a surface phenomenon, but can be volumetric for gases containing CO2 or H2O vapor. Thermodynamics 43 Examples:  Solar radiation reaching Earth through space.  Heat radiating from a campfire or electric heater.  Thermal energy exchange between two parallel metal plates separated by vacuum. Sign Convention for Heat Transfer To ensure consistency in thermodynamic calculations, a standard sign convention is followed: Condition Description Sign of Heat (Q) Heat added to the system Energy enters the system Positive (+Q) Heat rejected by the system Energy leaves the system Negative (-Q) Fig. 1.15 Sign convention of heat. Examples:  When a gas in a piston-cylinder is heated →+Q (heat added).  When the gas cools and releases heat →−Q (heat rejected). This sign convention aligns with the First Law of Thermodynamics, expressed as: Δ𝑈=𝑄−𝑊 Where, Δ𝑈= change in internal energy, 𝑄= heat added to the system, Thermodynamics 44 𝑊= work done by the system. Importance of Heat in Thermodynamics 1. Energy Conversion: Heat transfer drives thermodynamic cycles like Carnot, Rankine, and Otto, converting heat energy into mechanical work. 2. Temperature Regulation: Understanding heat transfer enables efficient heating and cooling in industries and daily life. 3. Design Applications: Engineers design insulators, heat exchangers, and thermal systems using heat transfer principles. 4. Fundamental Law Applications: Heat is central to the First and Second Laws of Thermodynamics, which govern all energy interactions. Problem 1.4: A gas in a piston-cylinder device expands from an initial pressure of 200 kPa and volume of 0.1 m3 to a final volume of 0.25 m3. The pressure-volume relationship during the process follows the law: 𝑃𝑉= constant (Isothermal process) Find: 1. The work done (W) by the gas during the expansion. 2. State whether the work is positive or negative, based on the sign convention. Solution: Given Data: 𝑃1=200kPa=200×103 Pa 𝑉1=0.1 m3 𝑉2=0.25 m3 Process: 𝑃𝑉= constant Thermodynamics 51 Units of Internal Energy  In the SI system, internal energy is measured in Joules (J).  In the CGS system, it is measured in ergs ( 1 J=107 ergs). Problem 1.6: An ideal gas with a mass of 2 kg and specific heat at constant volume 𝐶𝑣= 718 J/kg⋅𝐾 is heated such that its temperature rises by 50 K. Find the change in internal energy. Solution: Δ𝑈=𝑚𝐶𝑣Δ𝑇 Δ𝑈=2×718×50=71,800𝐽 Result: The internal energy of the gas increases by 71.8 kJ. Significance of Internal Energy  Helps in understanding energy conversion during thermodynamic processes.  Used to calculate work, heat transfer, and efficiency in systems like engines, compressors, and turbines.  Acts as a thermodynamic property that determines system equilibrium. Exercise 1.7: A mass of 2 kg of air is heated in a closed, rigid container (constant volume) from an initial temperature of 30∘C to a final temperature of 130∘C. The specific heat of air at constant volume is given as 𝐶𝑣=0.718 kJ/kg.K. Find: 1. The change in internal energy ( ΔU ) of the air. 2. State whether the internal energy increases or decreases. 3. Determine whether any work is done by the system. 1.9 Enthalpy Enthalpy (H) is an important property used to describe the total heat content of a system. It is particularly useful when dealing with processes occurring at constant pressure such as in most chemical reactions, boilers, turbines, or heat exchangers. Enthalpy gives us a convenient way to express energy changes in systems that exchange heat with the surroundings under constant pressure. Thermodynamics 52 The enthalpy (H) of a system is defined as: 𝐻=𝑈+𝑃𝑉 Where,  𝐇= Enthalpy (Joules)  𝐔= Internal energy of the system (Joules)  𝐏= Pressure of the system (Pascals)  V= Volume of the system (m3) Enthalpy represents the total energy of the system, including:  The internal energy (𝐔) - energy due to molecular motion and interactions within the system.  The flow energy (PV) - energy required to make room for the system by displacing the surroundings at pressure 𝑃. So, enthalpy can be thought of as the total heat content or energy stored in a system at constant pressure. Derivation of Enthalpy Change Differentiating =𝑈+𝑃𝑉 : 𝑑𝐻=𝑑𝑈+𝑑(𝑃𝑉) If pressure is constant (a common case): 𝑑𝐻=𝑑𝑈+𝑃𝑑𝑉 From the First Law of Thermodynamics: 𝑑𝑈=𝛿𝑄−𝛿𝑊 At constant pressure, work done by the system is: 𝛿𝑊=𝑃𝑑𝑉 Substitute this into the equation: 𝑑𝑈=𝛿𝑄−𝑃𝑑𝑉 Hence, 𝑑𝐻=𝑑𝑈+𝑃𝑑𝑉=(𝛿𝑄−𝑃𝑑𝑉)+𝑃𝑑𝑉=𝛿𝑄 Thermodynamics 53 Therefore, at constant pressure: 𝑑𝐻=𝛿𝑄𝑝 That means the change in enthalpy equals the heat absorbed or released by the system at constant pressure. Enthalpy indicates the heat change in a process carried out at constant pressure.  If ΔH>0, the process absorbs heat from the surroundings → Endothermic process.  If ΔH<0, the process releases heat to the surroundings → Exothermic process. So, enthalpy change gives direct information about the energy exchange as heat in practical systems. Change in Enthalpy ( ΔH ) The change in enthalpy between two states is given by: Δ𝐻=𝐻2−𝐻1=(𝑈2+𝑃2𝑉2)−(𝑈1+𝑃1𝑉1) For a constant-pressure process: Δ𝐻=Δ𝑈+𝑃Δ𝑉 If the system is an ideal gas, using =𝑛𝑅𝑇 : Δ𝐻=Δ𝑈+Δ(𝑛𝑅𝑇) or for constant n : Δ𝐻=Δ𝑈+𝑛𝑅Δ𝑇 Enthalpy and Heat Capacity At constant pressure, the specific heat at constant pressure ( 𝐂𝐩 ) is defined as: 𝐶𝑝=(𝑑𝐻 𝑑𝑇)𝑃 For an ideal gas: Δ𝐻=𝑚𝐶𝑝Δ𝑇 Thermodynamics 54 Similarly, at constant volume: Δ𝑈=𝑚𝐶𝑣Δ𝑇 And we know: 𝐶𝑝−𝐶𝑣=𝑅 Where 𝐑 is the gas constant. Enthalpy in Different Processes Let's see how enthalpy behaves under different thermodynamic processes: (a) Constant Pressure Process  Heat added = change in enthalpy.  𝑄𝑝=Δ𝐻 (b) Constant Volume Process  No change in PV term.  𝑄𝑣=Δ𝑈  Enthalpy change is not equal to heat transfer here. (c) Isothermal Process (Constant Temperature)  For ideal gases, Δ𝑇=0⇒Δ𝐻=0 (d) Adiabatic Process (No Heat Exchange)  𝑄=0  Hence, Δ𝐻=0 only for reversible adiabatic processes involving ideal gases. Enthalpy of a System per Unit Mass Sometimes, we express enthalpy per unit mass, known as specific enthalpy (h): ℎ=𝑢+𝑃𝑣 Where:  𝐡= specific enthalpy ( kJ/kg )  𝐮= specific internal energy ( kJ/kg )  𝐯= specific volume (m3/kg) Thermodynamics 55 And the change: Δℎ=Δ𝑢+𝑃Δ𝑣 This form is commonly used in engineering thermodynamics, especially for steam and refrigerant tables. Problem 1.7: A gas is heated at constant pressure of 1.5 bar. The volume changes from 0.2 m3 to 0.25 m3, and the internal energy increases by 12 kJ. Find the change in enthalpy. Solution: Δ𝐻=Δ𝑈+𝑃Δ𝑉 Given: P=1.5 bar =1.5×105 Pa ΔU=12 kJ=12,000 J ΔV=0.25−0.2=0.05 m3 Δ𝐻=12,000+(1.5×105)(0.05) Δ𝐻=12,000+7,500=19,500𝐽 Result: The change in enthalpy is 19.5 kJ. Importance of Enthalpy Enthalpy is a crucial concept in both thermodynamics and engineering applications because:  It helps determine heat exchange in open systems (like turbines, compressors, boilers).  It simplifies energy analysis at constant pressure (common in natural conditions).  It is essential in chemical reactions to express heat of reaction, formation, and combustion.  It is used to calculate enthalpy of vaporization, enthalpy of fusion, etc. Exercise 1.8: A mass of 3 kg of air is heated at constant pressure from an initial temperature of 30∘C to a final temperature of 180∘C. The specific heat of air at constant pressure is given as 𝐶𝑝=1.005 kJ/kg.K. Find: 1. The change in enthalpy ( ΔH ) of the air. 2. State whether the process is endothermic or exothermic. Thermodynamics 56 1.10 Specific Heat Capacities When heat is supplied to a body, its temperature rises. However, the rate of temperature rise varies from one substance to another. For example:  Metals heat up quickly,  Water heats up slowly. This difference arises because each substance requires a different amount of heat to raise its temperature. This property is measured by Specific Heat Capacity. Fig. 1.16 Specific Heat Capacities. The Specific Heat Capacity (often simply called specific heat) of a substance is defined as the amount of heat required to raise the temperature of unit mass of a substance by 1∘C (or 1 K ). Thermodynamics 57 Mathematically: 𝐶= 𝑄 𝑚Δ𝑇 Where:  𝐶= specific heat capacity (J/kg⋅K)  𝑄= heat supplied (J)  𝑚= mass of the substance (kg)  Δ𝑇= change in temperature (K or ∘C) A higher specific heat means the substance needs more heat to increase its temperature. A lower specific heat means it heats up quickly with little energy. For example: Substance Specific Heat (J/kg•K) Water 4186 Copper 385 Iron 450 Air ≈1005 (at constant pressure) Thus, water has a very high specific heat - it can absorb or release large amounts of heat without changing its temperature quickly. That's why it's used as a coolant in engines and power plants. Heat Transfer Equation The amount of heat transferred to or from a substance can be calculated by: 𝑄=𝑚𝐶Δ𝑇 Where:  𝑄= heat gained or lost (J)  𝑚= mass ( kg )  𝐶= specific heat capacity (J/kg⋅K) Thermodynamics 58  Δ𝑇= temperature change (K or ∘C) 1. If heat is added, 𝑄 is positive. 2. If heat is removed, 𝑄 is negative. Types of Specific Heat Capacities In Thermodynamics, for gases, the specific heat depends on the conditions of heating - whether the volume or the pressure is kept constant. Hence, we define two types: (a) Specific Heat at Constant Volume (Cv) It is defined as the heat required to raise the temperature of unit mass of a gas by 𝟏K when volume is kept constant. Mathematically: 𝐶𝑣=(𝑑𝑄 𝑚𝑑𝑇)𝑣 At constant volume:  No work is done (since =0 ).  Hence, all heat supplied increases internal energy. From the first law: 𝑑𝑄=𝑑𝑈+𝑃𝑑𝑉⇒𝑑𝑄=𝑑𝑈 Thus, 𝐶𝑣=(𝑑𝑈 𝑚𝑑𝑇) (b) Specific Heat at Constant Pressure ( 𝐂𝐩 ) It is defined as the heat required to raise the temperature of unit mass of a gas by 𝟏K when pressure is kept constant. Mathematically: 𝐶𝑝=(𝑑𝑄 𝑚𝑑𝑇)𝑝 Thermodynamics 59 At constant pressure:  Part of the heat supplied increases internal energy.  The remaining part does work in expanding the gas. From the first law: 𝑑𝑄=𝑑𝑈+𝑃𝑑𝑉 At constant pressure: 𝐶𝑝=1 𝑚(𝑑𝑈 𝑑𝑇+𝑃𝑑𝑉 𝑑𝑇) Relation Between 𝐂𝐩 and 𝐂𝐯 For an ideal gas, we know that: 𝑃𝑉=𝑅𝑇 Differentiating with respect to : 𝑃𝑑𝑉 𝑑𝑇+𝑉𝑑𝑃 𝑑𝑇=𝑅 At constant pressure, =0 : 𝑃𝑑𝑉 𝑑𝑇=𝑅 Hence: 𝐶𝑝−𝐶𝑣=𝑅 This is a fundamental thermodynamic relation. It shows that:  𝐶𝑝 is always greater than 𝐶𝑣,  Because at constant pressure, the gas does extra work during expansion. Thermodynamics 60 Ratio of Specific Heats ( 𝜸 ) The ratio of specific heats is an important thermodynamic constant, given by: 𝛾=𝐶𝑝 𝐶𝑣 This ratio (𝛾) appears in equations for adiabatic processes, speed of sound, and gas dynamics. Typical values: Gas 𝑪𝒑( 𝐉/𝐤𝐠⋅𝐊) 𝑪𝒗( 𝐉/𝐤𝐠⋅𝐊) 𝜸=𝑪𝒑/𝑪𝒗 Air 1005 718 1.4 Helium 5193 3120 1.66 CO2 846 657 1.29 Molar Specific Heat For gases, it's often convenient to express specific heat per mole instead of per kilogram. Fig. 1.17 Graphical representation of Specific Heat Capacities. Thermodynamics 67 𝑄 Heat added per unit mass J/kg 𝑊 Work done per unit mass J/kg 𝑚˙ Mass flow rate kg/s Final Form of SFEE: ℎ1+𝐶12 2+𝑔𝑧1+𝑄=ℎ2+𝐶22 2+𝑔𝑧2+𝑊 Problem 1.9: Steam enters a steam turbine steadily at a pressure of 3 MPa , temperature of 350∘C, and velocity of 60 m/s. It leaves the turbine at a pressure of 50 kPa , temperature of 100∘C, and velocity of 180 m/s. During the process, the heat loss to the surroundings is 30 kJ/kg of steam. Neglect potential energy changes. Find: 1. The work done per kilogram of steam (W). 2. The nature of work (whether output or input). Solution: Given Data: Property Inlet (1) Exit (2) Units Pressure 3 MPa 50 kPa - Temperature 350∘C 100∘C - Velocity 𝐶1=60 𝐶2=180 m/s Heat loss 𝑄=−30 - kJ/kg Neglect potential energy term (𝑔𝑧1≈𝑔𝑧2). From Steam Tables: At 3MPa,350∘C (superheated steam): ℎ1=3115 kJ/kg Thermodynamics 68 At 𝟓𝟎 𝐤𝐏𝐚, 𝟏𝟎𝟎∘𝐂 (saturated steam mixture): ℎ2=2676 kJ/kg Formula: Steady Flow Energy Equation (per kg of fluid): ℎ1+𝐶12 2×1000+𝑄=ℎ2+𝐶22 2×1000+𝑊 (All terms in kJ/kg ) Rearranging for work done: 𝑊=(ℎ1−ℎ2)+𝐶12−𝐶22 2000 +𝑄 Step 1: Substitute the known values 𝑊=(3115−2676)+(602−1802) 2000 −30 Step 2: Simplify step-by-step 𝑊=439+(3600−32400) 2000 −30 𝑊=439+−28800 2000 −30 𝑊=439−14.4−30 𝑊=394.6 kJ/kg Exercise 1.11: Air enters a single-stage air compressor steadily at a pressure of 100 kPa, temperature of 27∘C, and velocity of 15 m/s. It leaves the compressor at a pressure of 700 kPa , temperature of 200∘C, and velocity of 45 m/s. During the process, heat is rejected to the cooling water at a rate of 25 kJ/kg of air. Neglect potential energy changes. Find: 1. The work input (W) required per kilogram of air. 2. State whether the process is endothermic or exothermic. Exercise 1.12: Steam expands adiabatically in a convergent-divergent nozzle from an inlet pressure of 1 MPa and temperature of 250∘C to an exit pressure of 200 kPa . Assume that the heat loss is negligible ( Q=0 ) and potential energy changes are negligible. The enthalpy of steam at the inlet is h1=2940 kJ/kg, and at the exit h2=2650 kJ/kg. Thermodynamics 69 Find: 1. The exit velocity (𝐶2) of the steam. 2. The velocity increase (Δ𝐶) between inlet and outlet. 1.12 Application of the Steady Flow Energy Equation (SFEE) in Jet Engine Components A jet engine is an open system (or control volume) through which a working fluid (air or exhaust gases) flows continuously and steadily. At every stage intake, compression, combustion, expansion, and exhaust there is an exchange of heat, work, pressure, and velocity. The Steady Flow Energy Equation (SFEE) is used to analyze the energy transformation taking place in each component under steady-state conditions. Components of a Jet Engine The main components of a jet engine include: 1. Diffuser (Intake) 2. Compressor 3. Combustion Chamber 4. Turbine 5. Nozzle Let’s apply the SFEE to each of these. 1. Diffuser (Intake Section) Function The diffuser is the intake component of a jet engine or any steady-flow device where air enters at a high velocity and low pressure. Its main function is to slow down the incoming air before it enters the compressor. By reducing the velocity of the flow, the diffuser converts kinetic energy into pressure energy, which results in an increase in static pressure and a decrease in velocity of the air. In simple terms, the diffuser acts as a decelerating duct, preparing the air for efficient compression in the next stage. This process ensures smoother operation and higher overall efficiency of the jet engine. Thermodynamics 70 Fig. 1.19 Diffuser. Assumptions For the analysis of a diffuser using the Steady Flow Energy Equation (SFEE), the following assumptions are generally made: 1. No shaft work is done on or by the fluid, 𝑊=0 because the diffuser has no moving parts or mechanical components. 2. Negligible heat transfer, 𝑄=0 as the process is assumed adiabatic, meaning there is no significant heat exchange between the diffuser and its surroundings during the short time of air passage. Thermodynamics 71 3. Change in potential energy is very small and can be neglected since the height difference between inlet and outlet is minimal. Applying the SFEE The Steady Flow Energy Equation for a general open system is: ℎ1+𝐶12 2+𝑄=ℎ2+𝐶22 2+𝑊 Applying the assumptions 𝑄=0 and 𝑊=0, the equation simplifies to: ℎ1+𝐶12 2=ℎ2+𝐶22 2 This represents the energy balance between inlet and outlet of the diffuser. From the equation above, the total energy per unit mass (sum of enthalpy and kinetic energy) remains constant between the inlet and outlet of the diffuser. Since the diffuser slows down the flow: 𝐶2<𝐶1 Therefore, the kinetic energy decreases at the outlet. To maintain energy balance, the enthalpy (which represents internal and flow energy, closely linked to pressure and temperature) must increase: ℎ2>ℎ1 Thus, as air passes through the diffuser:  The velocity decreases,  The static pressure and enthalpy increase, and  The total energy remains constant (neglecting losses). This shows that the diffuser converts dynamic pressure (velocity head) into static pressure, which is crucial for efficient compression and combustion in subsequent engine stages. Result Kinetic Energy ↓, Pressure (Enthalpy) ↑ Thermodynamics 72 Hence, the diffuser is an essential component that:  Reduces the speed of incoming air,  Increases its pressure for the compressor stage, and  Improves overall engine performance and efficiency. 2. Compressor Function The compressor is one of the most critical components in a jet engine. Its primary function is to increase the pressure and temperature of the incoming air before it enters the combustion chamber. This is achieved by doing mechanical work on the air through the rotation of compressor blades connected to the turbine shaft. As the air passes through the compressor, it is compressed to a smaller volume, which raises its pressure and consequently its temperature due to the work input. This high-pressure, hightemperature air is essential for efficient combustion in the next stage of the engine. Fig. 1.20 Compressor. Assumptions For the analysis of a compressor using the Steady Flow Energy Equation (SFEE), the following assumptions are made, Thermodynamics 73 1. Negligible heat transfer: The compression process is assumed to be adiabatic, meaning 𝑄=0 because the time for air to pass through the compressor is very short, and the casing is generally well insulated. 2. Work input required: Mechanical energy is supplied to the air by the compressor shaft, so 𝑊>0 indicating work done on the fluid. 3. Change in potential energy is negligible due to minimal height difference. Applying the SFEE The Steady Flow Energy Equation for a general steady-flow process is given by: ℎ1+𝐶12 2+𝑄=ℎ2+𝐶22 2+𝑊 Since the process is adiabatic ( 𝑄=0 ), it simplifies to: ℎ1+𝐶12 2=ℎ2+𝐶22 2+𝑊 Rearranging: ℎ2−ℎ1=−𝑊 This shows that work is done on the air (since 𝑊 is positive for input), leading to an increase in enthalpy. From the above equation, it is clear that the enthalpy of air increases across the compressor because of the work supplied by the shaft. In practical terms: ℎ2>ℎ1 Since enthalpy (h) is directly related to temperature (T) for gases, an increase in enthalpy corresponds to a rise in temperature. Thermodynamics 74 Thus, as air flows through the compressor:  Pressure increases due to compression,  Temperature increases because of the work input, and  The kinetic energy change (𝐶2 2) is generally small compared to enthalpy change and is therefore neglected. Result Pressure ↑, Temperature ↑, Work Input Hence, in a compressor:  The mechanical energy supplied by the turbine is converted into pressure energy of the air.  The high-pressure, high-temperature air obtained at the outlet is then directed to the combustion chamber, ensuring efficient fuel burning and higher engine thrust. 3. Combustion Chamber (Burner) Function The combustion chamber, also known as the burner, is the section of the jet engine where the chemical energy of the fuel is converted into thermal energy. In this component, fuel is injected, mixed with highpressure air from the compressor, and then burned at nearly constant pressure. The primary purpose of the combustion chamber is to increase the enthalpy and temperature of the working fluid (air-fuel mixture) by adding a large amount of heat energy. The hightemperature gases produced in this stage provide the energy necessary to drive the turbine and, ultimately, generate thrust through the nozzle. Assumptions For the combustion chamber, the following assumptions are made while applying the Steady Flow Energy Equation (SFEE): 1. No shaft work The combustion chamber has no moving parts; therefore, no mechanical work is done by or on the system. 𝑊=0 Thermodynamics 75 Fig. 1.21 Combustion Chamber. 2. Significant heat addition Heat is added due to the combustion of fuel. Thus, 𝑄>0 The process is highly exothermic, and the added heat greatly increases the internal energy and enthalpy of the gases. 3. Negligible potential energy change The height difference between the inlet and outlet is small, so the potential energy term (𝑔𝑧) can be neglected. 4. Process occurs approximately at constant pressure Although minor pressure losses occur due to friction and flow resistance, the overall process is treated as constant pressure combustion for simplicity. Applying the SFEE The Steady Flow Energy Equation for a general open system is: ℎ1+𝐶12 2+𝑄=ℎ2+𝐶22 2+𝑊 Thermodynamics 76 Since the combustion chamber has no work interaction ( 𝐖=𝟎 ), the equation becomes: ℎ1+𝐶12 2+𝑄=ℎ2+𝐶22 2 The velocities of air entering and leaving the chamber are generally small compared to the change in enthalpy, so 𝐶12 and 𝐶22 terms can be neglected. Therefore: 𝑄=ℎ2−ℎ1 This means that the heat supplied by fuel combustion directly increases the enthalpy (and therefore temperature) of the air. From the simplified form of SFEE, we observe that the enthalpy of the fluid increases significantly due to the large heat addition from burning the fuel. Since ℎ∝𝑇 for gases, the increase in enthalpy results in a sharp rise in temperature, often ranging between 1000 K and 2000 K, depending on the fuel type and engine design. The combustion chamber thus serves as the energy source of the entire jet engine - converting chemical energy into high-temperature, high-energy gases that drive the turbine and produce thrust. Result Enthalpy ↑, Temperature ↑, Pressure ≈ Constant Hence, across the combustion chamber:  The enthalpy and temperature of the gas increase dramatically due to fuel combustion.  The pressure remains nearly constant, allowing smooth expansion of gases in the turbine.  The kinetic energy change is small and can be neglected compared to the large enthalpy rise. 4. Turbine Function The turbine is a vital component of the jet engine that extracts energy from the hightemperature, highpressure gases produced in the combustion chamber. As these gases expand through the turbine blades, part of their enthalpy energy is converted into mechanical work. Thermodynamics 83 The mathematical form of the First Law of Thermodynamics is given as: Δ𝑈=𝑞+𝑊 Where:  Δ𝑈= Change in internal energy of the system  𝑞= Heat transfer between the system and surroundings  𝑊= Work interaction between the system and its surroundings This equation signifies that the total energy change of a system equals the sum of the heat energy supplied and the work done on or by the system. Sign Conventions The sign of 𝑞 (heat) and 𝑊 (work) depends on the direction of energy transfer. The following conventions are generally used: Condition Heat (q) Work (W) Description Heat supplied to the system +ve - System absorbs heat Heat released by the system -ve - System loses heat Work done on the system - +ve Energy enters system Work done by the system - -ve Energy leaves system Table. 1.9 Sign conventions of first law of thermodynamics. Using this convention, the sign of Δ𝑈 will indicate whether the internal energy increases ( +ve ) or decreases (-ve). Let us analyze four different cases that combine compression, expansion, heating, and cooling of a gas inside a piston–cylinder system. Case I: Compression and Heating When a gas is heated and at the same time compressed (by placing additional weight on the piston), two types of energy transfer occur simultaneously: Thermodynamics 84 Fig. 1.25 Compression and Heating.  Heat is supplied to the system, therefore 𝑞=+𝑣𝑒  Work is done on the system by the surroundings due to compression, therefore 𝑊=+𝑣𝑒 According to the First Law of Thermodynamics, Δ𝑈=𝑞+𝑊 Since both 𝑞 and 𝑊 are positive, the change in internal energy ( Δ𝑈 ) is also positive. This means that the internal energy of the gas increases, as it receives energy in the form of both heat and compression work. Result: Internal Energy ↑ (Increases) Case II: Expansion and Cooling When a gas is cooled and simultaneously allowed to expand (the piston moves upward), both heat and work interactions occur but in the opposite direction: Thermodynamics 85 Fig. 1.26 Expansion and Cooling.  The gas loses heat to the surroundings, therefore 𝑞=−𝑣𝑒  The gas does work on the surroundings during expansion, therefore 𝑊=−𝑣𝑒 Applying the First Law: Δ𝑈=𝑞+𝑊 Since both 𝑞 and 𝑊 are negative, the total change in internal energy ( Δ𝑈 ) is negative. This indicates that the internal energy of the gas decreases, as energy leaves the system in both forms heat loss and work done by expansion. Result: Internal Energy ↓ (Decreases) Case III: Compression and Cooling In this case, the gas is cooled while being compressed (the piston is pushed downward). The process involves both heat rejection and work input: Thermodynamics 86 Fig. 1.27 Compression and Cooling.  Work is done on the system by the surroundings (due to compression), therefore 𝑊=+𝑣𝑒  The system loses heat to the surroundings (due to cooling), therefore 𝑞=−𝑣𝑒 Using the First Law: Δ𝑈=𝑞+𝑊 Here, 𝑞 is negative and 𝑊 is positive. The net change in internal energy ( Δ𝑈 ) depends on the relative magnitudes of the two quantities. If the work input is greater than the heat lost, Δ𝑈 will be positive (energy gain). If the heat loss is greater than the work input, Δ𝑈 will be negative (energy loss). Result: Internal Energy may Increase or Decrease depending on the values of 𝒒 and 𝑾 Case IV: Expansion and Heating When the gas is heated and simultaneously allowed to expand (the piston moves upward freely): Thermodynamics 87 Fig. 1.28 Expansion and Heating.  Heat is supplied to the system, therefore 𝑞=+𝑣𝑒  Work is done by the system on the surroundings during expansion, therefore 𝑊=−𝑣𝑒 Applying the First Law: Δ𝑈=𝑞+𝑊 In this case, 𝑞 is positive and 𝑊 is negative. The net change in internal energy ( Δ𝑈 ) depends on which energy effect is larger. If the heat added is greater than the work done by the gas, the internal energy will increase. If the work done exceeds the heat supplied, the internal energy will decrease. Result: Internal Energy may Increase or Decrease depending on the magnitude of 𝒒 and 𝑾 Case Process Description Heat (q) Work (W) Change in Internal Energy (ΔU) Result I Compression and Heating +ve +ve +ve Internal energy increases Thermodynamics 88 II Expansion and Cooling −ve −ve −ve Internal energy decreases III Compression and Cooling −ve +ve ±ve May increase or decrease IV Expansion and Heating +ve −ve ±ve May increase or decrease Table. 1.10 Summary of different cases of process. First Law of Thermodynamics for a Closed System In a closed system, mass remains constant - only energy crosses the boundary in the form of heat or work. The mechanical (pressure-volume) work done by or on the system is expressed as: 𝑊=−𝑃𝑒𝑥𝑡Δ𝑉 Where:  𝑃ext = constant external pressure  Δ𝑉= change in volume of the system Here, the negative sign appears because when the system expands, it does work on the surroundings, resulting in energy loss from the system. Substituting this into the First Law: Δ𝑈=𝑞+𝑊 Δ𝑈=𝑞−𝑃𝑒𝑥𝑡Δ𝑉 This is the First Law of Thermodynamics for a closed system undergoing pressure-volume work. First Law of Thermodynamics for an Open System In an open system, both mass and energy can cross the system boundary. Examples include turbines, compressors, nozzles, boilers, and pumps where fluid flows continuously through the device. Thermodynamics 89 For a steady-flow open system, the Steady Flow Energy Equation (SFEE) represents the First Law of Thermodynamics and is expressed as: 𝑄+(ℎ1+𝐶12 2+𝑔𝑧1)=𝑊+(ℎ2+𝐶22 2+𝑔𝑧2) or, rearranged, ℎ1+𝐶12 2+𝑔𝑧1+𝑄=ℎ2+𝐶22 2+𝑔𝑧2+𝑊 Where:  𝑄= heat supplied to the system per unit mass ( kJ/kg )  𝑊= work done by the system per unit mass ( kJ/kg )  ℎ=𝑢+𝑝𝑣= specific enthalpy (kJ/kg)  𝐶2 2= kinetic energy term (kJ/kg)  𝑔𝑧= potential energy term (kJ/kg) This equation states that: The total energy entering the system (as enthalpy, kinetic energy, potential energy, and heat) equals the total energy leaving the system (as enthalpy, kinetic energy, potential energy, and work). For many practical cases (like turbines or compressors), changes in potential energy are small, so the equation simplifies to: ℎ1+𝐶12 2+𝑄=ℎ2+𝐶22 2+𝑊  If work is produced by the system (e.g., turbine): 𝑊>0  If work is done on the system (e.g., compressor): 𝑊<0 The equation accounts for both energy transfer and mass flow in steady-flow devices. Limitations of the First Law of Thermodynamics (1) It does not indicate the direction of energy transfer The First Law tells us that energy can be exchanged between a system and its surroundings in the form of heat or work, but it does not specify the direction in which this transfer occurs. Thermodynamics 90 For example:  When two bodies at different temperatures are in contact, the First Law cannot tell us whether heat will flow from the hotter to the colder body or vice versa.  It only accounts for the amount of heat transferred, not the natural tendency of energy flow. In reality, heat always flows spontaneously from a higher temperature to a lower temperature, which is governed by the Second Law of Thermodynamics, not the first. (2) It does not tell about the possibility or spontaneity of a process The First Law only ensures that energy is conserved, but it cannot determine whether a process can occur naturally. For example:  The First Law does not prohibit heat from flowing from a cold body to a hot body it simply requires that energy be conserved.  However, we know such a process does not occur spontaneously without external work. Thus, the First Law gives no indication of feasibility or irreversibility of thermodynamic processes. (3) It provides no information about energy degradation or quality The First Law considers all forms of energy as equivalent in quantity. However, in practice, not all energy forms are equally useful. For instance:  Mechanical work can be completely converted into heat energy.  But heat energy cannot be completely converted back into work (some portion is always lost to surroundings). The First Law cannot distinguish between high-grade energy (like mechanical or electrical energy) and low-grade energy (like heat). It only considers their total amount, not their quality or usefulness. (4) It does not explain the concept of irreversibility In real systems, many processes are irreversible due to friction, unrestrained expansion, mixing, or heat transfer through finite temperature differences. Thermodynamics 91 The First Law cannot determine whether a process is reversible or irreversible it treats both in the same way as long as energy is conserved. Hence, it cannot explain why all natural processes are irreversible in practice. (5) It does not give efficiency limits for heat engines The First Law tells us that part of the heat supplied to a heat engine is converted into work, and the rest is rejected as waste heat. However, it gives no indication of the maximum efficiency that such a conversion process can achieve. For example:  The First Law can be satisfied even if 100% of heat is converted to work, but this never happens in real life.  The Second Law of Thermodynamics defines this limitation and gives the Carnot efficiency as the theoretical maximum. (6) It does not define entropy The First Law deals only with energy quantities and transformations. It does not introduce or explain the concept of entropy (S) a property that helps determine the direction, disorder, and irreversibility of processes. The Second Law is required to define entropy and explain its significance. Problem 1.10: A gas in a piston-cylinder device undergoes a thermodynamic process in which it expands from an initial volume of 0.25 m3 to a final volume of 0.75 m3 against a constant external pressure of 200 kPa. During the expansion, the gas receives 90 kJ of heat energy from the surroundings. Determine: 1. The work done ( W ) by the gas, 2. The change in internal energy ( ΔU ) of the gas, and 3. The type of process in terms of energy gain or loss. Solution: Given Data: 𝑃ext =200kPa=200×103 Pa Thermodynamics 92 𝑉1=0.25 m3 𝑉2=0.75 m3 𝑄=+90 kJ (heat supplied to the system) Formulae Used: 1. First Law of Thermodynamics: Δ𝑈=𝑄+𝑊 2. Work done by the system (constant external pressure): 𝑊=−𝑃𝑒𝑥𝑡(𝑉2−𝑉1) (The negative sign ensures the correct convention: Work done by the system →𝑊 negative and Work done on the system →𝑊 positive) Step 1: Calculate the Work Done (W) 𝑊=−𝑃𝑒𝑥𝑡(𝑉2−𝑉1) Substitute the given values: 𝑊=−(200×103)(0.75−0.25) 𝑊=−(200×103)(0.5) 𝑊=−100,000 J 𝑊=−100 kJ Interpretation: The work is negative, which means the system does work on the surroundings (expansion). Step 2: Apply the First Law of Thermodynamics Δ𝑈=𝑄+𝑊 Substitute values: Δ𝑈=(+90)+(−100) Δ𝑈=−10 kJ Interpretation: The change in internal energy is negative, meaning the internal energy of the gas decreases.  The system receives heat energy (+90 kJ).  But it does more work on the surroundings ( 100 kJ ). Thermodynamics 99 Heat Transfer 𝑄=𝑊=𝑚𝑅𝑇ln (𝑉2 𝑉1) Relation Between P, V, and T Since 𝑇= constant: 𝑃1𝑉1=𝑃2𝑉2 That is, Pressure is inversely proportional to Volume (Boyle's Law). 4. Adiabatic Process (No Heat Transfer Process) Definition An adiabatic process is one in which no heat is transferred to or from the system: 𝑄=0 Hence, the entire change in internal energy is due to work interaction. From the First Law: Δ𝑈=𝑄−𝑊 Δ𝑈=−𝑊 If the gas expands, it does work on the surroundings ( W positive) →Δ𝑈 decreases → temperature falls. If the gas is compressed, work is done on the gas →Δ𝑈 increases → temperature rises. Work Done For a reversible adiabatic process of an ideal gas: 𝑃𝑉𝛾= constant where 𝛾=𝐶𝑝 𝐶𝑣. The work done is given by: 𝑊=𝑃1𝑉1−𝑃2𝑉2 𝛾−1 Thermodynamics 100 Change in Internal Energy Since 𝑄=0, Δ𝑈=−𝑊=𝑚𝐶𝑣(𝑇2−𝑇1) Relation Between P, V, and T From 𝑃𝑉𝛾= constant, the following relations hold: 𝑇2 𝑇1=(𝑉1 𝑉2)𝛾−1 and 𝑇2 𝑇1=(𝑃2 𝑃1)(𝛾−1)/𝛾 5. Polytropic Process (Generalized Case) The polytropic process is a generalized form of all processes, represented by: 𝑃𝑉𝑛= constant Where, 𝑛 is called the polytropic index. Depending on the value of 𝑛, it can represent various processes: Process Condition Polytropic Index (n) Isobaric 𝑃= constant 𝑛=0 Isothermal 𝑇= constant 𝑛=1 Adiabatic 𝑄=0 𝑛=𝛾 Isochoric 𝑉= constant 𝑛=∞ Work Done in Polytropic Process 𝑊=𝑃2𝑉2−𝑃1𝑉1 1−𝑛 Heat Transfer 𝑄=𝑚𝐶𝑣(𝑇2−𝑇1)+𝑊 Thermodynamics 101 Problem 1.12: A mass of 1 kg of air (assumed ideal gas) is initially at 𝑃1=100kPa,𝑇1= 300 K and undergoes different thermodynamic processes until its final pressure becomes 𝑃2=200kPa. For air: 𝑅=0.287 kJ/kg.K, 𝐶𝑝=1.005 kJ/kg.K, 𝐶𝑣=0.718 kJ/kg.K, 𝛾=1.4 Determine for each process: 1. Final temperature 𝑇2 2. Final volume 𝑉2 3. Work done 𝑊 4. Change in internal energy Δ𝑈 5. Heat transfer 𝑄 Solution: Step 1: Find the initial specific volume ( 𝐕𝟏 ) From the ideal gas law: 𝑃1𝑉1=𝑅𝑇1 𝑉1=𝑅𝑇1 𝑃1 𝑉1=0.287×300 100 =0.861 m3/kg (a) Isobaric Process (Constant Pressure) 𝑃1=𝑃2=100kPa Let the temperature rise to 𝑇2=400 K (chosen for example). 1. Volume relation: 𝑉2 𝑉1=𝑇2 𝑇1=400 300=1.333 𝑉2=1.333×0.861=1.148 m3/kg 2. Work done: 𝑊=𝑃(𝑉2−𝑉1)=100(1.148−0.861)=28.7 kJ/kg 3. Change in internal energy: Δ𝑈=𝐶𝑣(𝑇2−𝑇1)=0.718(400−300)=71.8 kJ/kg Thermodynamics 102 4. Heat transfer: 𝑄=Δ𝑈+𝑊=71.8+28.7=100.5 kJ/kg (b) Isochoric Process (Constant Volume) 𝑉1=𝑉2=0.861 m3/kg Let pressure rise from 100 kPa to 200 kPa. From ideal gas law: 𝑃2 𝑇2=𝑃1 𝑇1 𝑇2=𝑇1𝑃2 𝑃1=300200 100=600 K 1. Work done: 𝑊=0 2. Change in internal energy: Δ𝑈=𝐶𝑣(𝑇2−𝑇1)=0.718(600−300)=215.4 kJ/kg 3. Heat transfer: 𝑄=Δ𝑈=215.4 kJ/kg (c) Isothermal Process (Constant Temperature) 𝑇1=𝑇2=300 K 𝑃1𝑉1=𝑃2𝑉2⇒𝑉2=𝑃1𝑉1 𝑃2=100×0.861 200 =0.4305 m3/kg 1. Work done: 𝑊=𝑅𝑇ln (𝑉2 𝑉1) 𝑊=0.287× 300ln (0.4305 0.861) 𝑊=86.1ln (0.5)=86.1(−0.693)=−59.6 kJ/kg 2. Change in internal energy: Δ𝑈=0 (since isothermal ideal gas) 3. Heat transfer: 𝑄=𝑊=−59.6 kJ/kg (negative → heat rejected) Thermodynamics 103 (d) Adiabatic Process (No Heat Transfer) 𝑄=0,𝑃1𝑉1𝛾=𝑃2𝑉2𝛾 1. Relation between temperatures: 𝑇2 𝑇1=(𝑃2 𝑃1)(𝛾−1)/𝛾 =2(0.4/1.4)=20.286 =1.22 𝑇2=1.22×300=366 K 2. Work done: 𝑊=𝐶𝑣(𝑇2−𝑇1) 1−𝐶𝑣 𝐶𝑝⇒𝑊=𝑚𝐶𝑣(𝑇2−𝑇1) 𝑊=0.718(366−300)=47.9 kJ/kg Since expansion (P↓,V↑)→ work done by system, =+47.9 kJ/kg. 3. Internal energy change: Δ𝑈=𝐶𝑣(𝑇2−𝑇1)=47.9 kJ/kg (Here 𝑄=0, so Δ𝑈=−𝑊 for compression or Δ𝑈=+𝑊 for expansion depending on direction.) (e) Polytropic Process (PV 𝐧= constant) Let 𝑛=1.3. 1. Temperature relation: 𝑇2 𝑇1=(𝑃2 𝑃1)(𝑛−1)/𝑛 =2(0.3/1.3) =20.231=1.175 𝑇2=1.175×300=352.5 K 2. Work done: 𝑊=𝑅(𝑇2−𝑇1) 1−𝑛 =0.287(352.5−300) 1−1.3 =0.287(52.5) −0.3 𝑊=−50.2 kJ/kg (Negative → work done on the system, i.e., compression.) 3. Change in internal energy: Δ𝑈=𝐶𝑣(𝑇2−𝑇1)=0.718(52.5)=37.7 kJ/kg Thermodynamics 104 4. Heat transfer: 𝑄=Δ𝑈+𝑊=37.7−50.2=−12.5 kJ/kg (Heat rejected.) Exercise 1.15: A mass of 𝟏 𝐤𝐠 of air initially at 𝑃1=200kPa,𝑇1=300 K undergoes an expansion process until its final pressure becomes 𝑃2=100kPa For air: 𝑅=0.287 kJ/kg.K, 𝐶𝑝=1.005 kJ/kg.K, 𝐶𝑣=0.718 kJ/kg.K, 𝛾=1.4 Find for each process: 1. Final temperature 𝑇2 2. Final volume 𝑉2 3. Work done 𝑊 4. Change in internal energy Δ𝑈 5. Heat transfer 𝑄 1.15 Zeroth Law of Thermodynamics Before the formulation of the First and Second Laws of Thermodynamics, scientists already needed a fundamental principle to define temperature and thermal equilibrium. This principle, established later, is known as the Zeroth Law of Thermodynamics. It was stated after the First and Second Laws, but because it forms the foundation of temperature measurement, it was placed before them in order and named the “Zeroth” Law. In simple terms, the Zeroth Law helps us understand temperature and provides the basis for comparing and measuring thermal equilibrium between different systems. Statement of the Zeroth Law of Thermodynamics If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This statement provides the logical foundation for defining temperature as a measurable property. Thermodynamics 105 Let’s consider three systems - System A, System B, and System C.  Suppose System A is in thermal equilibrium with System C.  And System B is also in thermal equilibrium with System C. Then, according to the Zeroth Law: System A and System B must also be in thermal equilibrium with each other. This means that when Systems A and B are brought into contact, no net heat transfer occurs between them both are at the same temperature. Fig. 1.29 Zeroth law of thermodynamics. Consider three bodies:  A: a hot metal block  B: a cup of water  C: a thermometer If the thermometer (System ) shows the same temperature reading when in contact with both the metal block (System A) and the water (System B), then: 𝑇𝐴=𝑇𝐵=𝑇𝐶 This indicates that A and B are at the same temperature, even though they are not in direct contact. Thus, the Zeroth Law allows us to use a thermometer as a device that can compare and measure the thermal states (temperatures) of different systems. Thermodynamics 106 Fig. 1.30 Simple example of Zeroth law of thermodynamics. Thermal Equilibrium Two bodies are said to be in thermal equilibrium when: 1. There is no net heat exchange between them when they are in contact. 2. Their temperatures remain equal. If heat flows between them, it means they are not in thermal equilibrium. The Zeroth Law forms the foundation for this concept - it gives meaning to the phrase "being at the same temperature." If:  System A is in thermal equilibrium with System C, 𝑇𝐴=𝑇𝐶  System B is in thermal equilibrium with System C, 𝑇𝐵=𝑇𝐶 Then, by the Zeroth Law: 𝑇𝐴=𝑇𝐵=𝑇𝐶 Thermodynamics 107 This mathematical relation shows that temperature ( T ) is a property that determines whether two systems are in thermal equilibrium. Importance and Significance The Zeroth Law of Thermodynamics is fundamental for several reasons: (1) Basis for Temperature Measurement It establishes temperature as a measurable and comparable property of systems. Without this law, it would be impossible to define or measure temperature accurately using thermometers. (2) Defines Thermal Equilibrium It explains that when two systems have the same temperature, there is no net heat flow between them a condition known as thermal equilibrium. (3) Foundation for Thermometers and Temperature Scales All temperature measuring devices (like thermometers) and temperature scales (Celsius, Kelvin, Fahrenheit) are based on this law. The Zeroth Law ensures that if a thermometer shows equal readings for two bodies, those bodies are at the same temperature. (4) Temperature as a State Variable It allows temperature (T) to be treated as a thermodynamic property or state variable, similar to pressure (P) and volume (V). This means temperature can define the state of a system in thermodynamic analysis. Limitations of the Zeroth Law Although the Zeroth Law is fundamental, it has a few limitations: 1. It applies only to systems in thermal contact and equilibrium, not to those still undergoing heat transfer. 2. It does not explain how or why thermal equilibrium is achieved that is covered by the mechanisms of heat transfer (conduction, convection, radiation). 3. It assumes temperature is a measurable property but does not explain how to measure it that is handled by thermometry and temperature scales. Thermodynamics 108 CHAPTER -2 SECOND LAW AND ENTROPY 2.1 Second Law of Thermodynamics The Second Law of Thermodynamics introduces a fundamental limitation on the conversion and direction of energy transformations. While the First Law ensures the conservation of energy, it does not indicate the direction in which energy changes occur or whether a process is feasible. The Second Law, therefore, governs the directionality and possibility of thermodynamic processes, establishing why some processes occur naturally while others cannot. It can be expressed in two equivalent but distinct forms:  Kelvin–Planck Statement  Clausius Statement Both statements describe limitations on cyclic devices such as heat engines, refrigerators, and heat pumps, and are interrelated a violation of one implies a violation of the other. 1. Kelvin-Planck Statement The Kelvin-Planck statement focuses on the limitations of heat engines, which are cyclic devices that convert heat energy into mechanical work. It establishes that even though energy is conserved (as per the First Law), there is a fundamental restriction on how efficiently this conversion can take place. Statement "It is impossible for any device that operates on a cyclic process to receive heat from a single reservoir and convert it completely into an equivalent amount of work." In simpler terms, No heat engine can ever achieve 100% efficiency, because a part of the heat absorbed from the hot source must always be rejected to a colder sink. Thermodynamics 115 However, according to the laws of thermodynamics, such machines are impossible. These laws particularly the First Law (conservation of energy) and the Second Law (direction of energy transformations) clearly prohibit perpetual motion. Let’s assume we design a wheel that keeps rotating forever using magnets, claiming it as a perpetual motion machine. Even if it starts spinning, friction at bearings and air resistance will gradually dissipate energy as heat, and the wheel will eventually stop. Thus, even in a vacuum and frictionless setup, energy would still redistribute, making perpetual motion impossible. Classification of Perpetual Motion Machines Perpetual motion machines are generally classified into two main types, based on which law of thermodynamics they violate: (a) Perpetual Motion Machine of the First Kind (PMM-I): A PMM of the First Kind is a machine that violates the First Law of Thermodynamics, which states that energy can neither be created nor destroyed; it can only change from one form to another. In other words, a PMM-I would create energy out of nothing. Thermodynamics 116 Example: Imagine a machine that keeps running forever and produces continuous mechanical work without any input of fuel, heat, or electrical energy. This means the output energy would exceed the input energy, directly violating the principle of energy conservation. Thermodynamic Representation: Energy Output > Energy Input Violation: The First Law requires that for any cyclic process: 𝑄=𝑊 Where,  𝑄= Heat supplied to the system  𝑊= Work done by the system A PMM-I would imply 𝑊>𝑄 or 𝑄=0,𝑊>0, which is impossible. Conclusion: PMM-I is impossible because energy cannot be produced without an equivalent input. Thermodynamics 117 (b) Perpetual Motion Machine of the Second Kind (PMM-II) A PMM of the Second Kind is a hypothetical device that violates the Second Law of Thermodynamics, which deals with the direction of spontaneous energy transfer and entropy. A PMM-II claims to spontaneously convert all available heat energy from a single heat reservoir into useful work - without any energy loss or heat rejection. Example: Imagine a heat engine that takes heat from a single heat source and converts it entirely into work, without rejecting any heat to a cooler sink. Thermodynamic Representation: 𝜂=100% Where, 𝜂 is the thermal efficiency of a heat engine. Thermodynamics 118 Violation: According to Kelvin-Planck's statement of the second law: "It is impossible to construct a device operating in a cycle that produces no other effect than the extraction of heat from a single reservoir and the performance of an equivalent amount of work." Hence, every real heat engine must reject some heat to a lower-temperature sink. Conclusion: PMM-II is also impossible because no engine can achieve 100% efficiency. Connection with Thermodynamic Laws Law of Thermodynamics Statement Violated by Type of PMM First Law Energy cannot be created or destroyed. Creates energy from nothing. PMM-I Second Law No engine can convert all heat into work. Converts all absorbed heat into work with 100% efficiency. PMM-II Why PMMs are Impossible? 1. Energy Conservation: Nature does not permit energy creation without input. Every form of work must come from some source of energy - mechanical, thermal, chemical, or electrical. 2. Entropy and Irreversibility: In real processes, there is always energy degradation due to friction, heat loss, or dissipation, increasing the entropy of the universe. 3. Practical Limitations: Mechanical losses (friction, air resistance), material imperfections, and heat transfer inefficiencies ensure that no process is perfectly reversible or loss-free. 4. Equivalence of the Kelvin-Planck and Clausius Statements Although the Kelvin-Planck and Clausius statements appear to describe different physical situations, they are actually equivalent. A violation of one would automatically lead to the violation of the other. Let's understand why. Thermodynamics 119 Proof of Equivalence 1. If the Kelvin-Planck statement is violated: Suppose a 100% efficient heat engine exists that converts all absorbed heat ( 𝑄1 ) into work ( 𝑊=𝑄1 ) with no heat rejection. This work could then drive a refrigerator that transfers heat from a cold body to a hot body without any net work input. This would violate the Clausius statement. 2. If the Clausius statement is violated: Suppose heat could flow from cold to hot without external work. Then, such a device could be combined with a normal heat engine to form a cycle that converts all absorbed heat into work, violating the Kelvin-Planck statement. Hence, both statements are mutually consistent and express the same fundamental principle of nature the impossibility of a Perpetual Motion Machine of the Second Kind (PMM-II). For a Cyclic Heat Engine: 𝑄1=𝑄2+𝑊 𝜂=𝑊 𝑄1=1−𝑄2 𝑄1 Here:  𝑄1= Heat absorbed from the source  𝑄2= Heat rejected to the sink  𝑊=𝑄1−𝑄2= Work done by the engine Since 𝑄2>0, efficiency 𝜂<1. For a Refrigerator or Heat Pump: 𝑄𝐻=𝑄𝐶+𝑊 Where:  𝑄𝐶= Heat extracted from the cold reservoir  𝑄𝐻= Heat delivered to the hot reservoir  𝑊= External work supplied Thermodynamics 120 Coefficient of Performance (COP) of a Refrigerator 𝐶𝑂𝑃𝑅=𝑄𝐶 𝑊=𝑄𝐶 𝑄𝐻−𝑄𝐶 Coefficient of Performance (COP) of a Heat Pump 𝐶𝑂𝑃𝐻𝑃 =𝑄𝐻 𝑊=𝑄𝐻 𝑄𝐻−𝑄𝐶 The two are related as: 𝐶𝑂𝑃𝐻𝑃 =𝐶𝑂𝑃𝑅+1 These relations demonstrate that external work is essential to make heat flow from cold to hot, and no cyclic device can operate without such work input - perfectly in line with the Second Law of Thermodynamics. Significance of the Second Law of Thermodynamics The Second Law is one of the most powerful principles in nature. It not only governs all thermodynamic systems but also dictates the irreversibility of natural processes. Key Significances: 1. Defines the Direction of Heat Flow: Heat always flows spontaneously from higher to lower temperature, never the reverse. 2. Sets Efficiency Limits: No heat engine can convert all absorbed heat into work; no refrigerator can operate without work input. 3. Introduces the Concept of Energy Quality: It differentiates between high-grade energy (like work) and low-grade energy (like heat). 4. Explains Irreversibility in Nature: All natural processes are irreversible and lead to an increase in entropy. 5. Forms the Basis for Entropy Concept: The Second Law introduces entropy (S) as a measure of disorder and the direction of spontaneous change. Thermodynamics 121 Problem 2.1: A heat engine operates between two thermal reservoirs - a hot reservoir at 600 K and a cold reservoir at 300 K. The engine absorbs 1000 kJ of heat from the hot reservoir during each cycle. 1. Determine the maximum possible efficiency of this engine (assuming it operates reversibly as a Carnot engine). 2. Calculate the maximum work output per cycle. 3. Find the amount of heat rejected to the cold reservoir. 4. If a refrigerator operates between the same two temperatures and removes 200 kJ of heat from the cold reservoir per cycle, determine the minimum work input required. Solution: Given Data: For both devices: 𝑇𝐻=600𝐾,𝑇𝐶=300𝐾 For the heat engine: 𝑄1=1000𝑘𝐽 For the refrigerator: 𝑄𝐶=200𝑘𝐽 1. Maximum (Carnot) Efficiency of Heat Engine For a reversible (Carnot) engine, efficiency is given by: 𝜂max=1−𝑇𝐶 𝑇𝐻 Substitute: 𝜂max =1−300 600=1−0.5=0.5 𝜂max =0.5 or 50% Thermodynamics 122 2. Maximum Work Output per Cycle Efficiency is the ratio of work done to heat absorbed: 𝜂=𝑊 𝑄1 Hence, 𝑊=𝜂𝑄1=0.5×1000=500𝑘𝐽 𝑊=500𝑘𝐽 3. Heat Rejected to the Cold Reservoir From the First Law: 𝑄1=𝑊+𝑄2 𝑄2=𝑄1−𝑊=1000−500=500 kJ 𝑄2=500 kJ This means 500 kJ of heat must be rejected to the cold reservoir, proving that 100% efficiency is impossible, consistent with the Kelvin-Planck statement. 4. Work Input for a Refrigerator Operating Between the Same Temperatures For a reversible refrigerator, the Coefficient of Performance (COP) is given by: 𝐶𝑂𝑃𝑅=𝑇𝐶 𝑇𝐻−𝑇𝐶 Substitute values: 𝐶𝑂𝑃𝑅=300 600−300=300 300=1 So, 𝐶𝑂𝑃𝑅=1=𝑄𝐶 𝑊 Hence, 𝑊= 𝑄𝐶 𝐶𝑂𝑃𝑅=200 1=200𝑘𝐽 𝑊=200𝑘𝐽 Thermodynamics 123 Exercise 2.1: A heat engine operates between a hot reservoir at 900 K and a cold reservoir at 300 K. During each cycle, it absorbs 1200 kJ of heat from the hot reservoir and rejects 500 kJ of heat to the cold reservoir. Determine: 1. The work output per cycle. 2. The actual thermal efficiency of the engine. 3. The maximum (Carnot) efficiency possible between these temperature limits. 4. The degree of perfection of the engine (ratio of actual efficiency to Carnot efficiency). Exercise 2.2: A refrigerator operates between a cold space at 260 K and the surrounding atmosphere at 310 K. The refrigerator extracts 400 kJ of heat from the cold space during each cycle. Determine: 1. The maximum (Carnot) COP of the refrigerator. 2. The minimum work input required to achieve this performance. 3. The heat rejected to the surrounding atmosphere. 2.2 Reversibility and Irreversibility According to the Second Law of Thermodynamics, every thermodynamic process can be classified into two categories: 1. Reversible (or Ideal) Process 2. Irreversible (or Actual or Natural) Process These classifications are based on whether the process can be exactly reversed to restore both the system and surroundings to their original states without leaving any change in the universe. Understanding reversibility is essential because it defines the theoretical limits of performance of thermodynamic devices such as engines, compressors, turbines, and refrigerators. Thermodynamics 124 1. Reversible Process (or Ideal Process) A reversible process is an idealized thermodynamic process that can be reversed completely without leaving any change either in the system or its surroundings. That means when the process is reversed, the system retraces the exact same path in the opposite direction on a pressure–volume (P–V) or temperature–entropy (T–S) diagram. Definition: A process that can be reversed in such a way that both the system and surroundings return to their original states without leaving any change in the universe is called a reversible process. In a reversible process, every intermediate state of the system is an equilibrium state. Therefore, the process proceeds infinitesimally slowly, allowing the system to remain in thermodynamic equilibrium throughout. Such a process is also called a quasi-static process, meaning that at every instant, the system and surroundings differ only by an infinitesimal amount of pressure, temperature, or other properties. Consider a system initially at state A that changes to state B during a thermodynamic process. This process is represented on a P–V diagram by a path A → B. If the process is reversed, the system exactly retraces the same path B → A, returning to its initial condition without any net effect on the surroundings. Fig. 2.3 Reversible process. Thermodynamics 131 Thus, even for a finite amount of heat transfer : Δ𝑇=𝑄 𝐶≈0 This means that no noticeable change in temperature occurs, regardless of the amount of heat exchanged. Although no real physical body can have infinite heat capacity, some natural and engineered systems have such a large mass or thermal inertia that their temperature changes are negligibly small during heat exchange. Therefore, in thermodynamics, such bodies are approximated as reservoirs. This assumption simplifies the analysis of energy interactions in various devices, especially in cyclic systems where temperature uniformity is required. Note: A thermal energy reservoir can exchange heat continuously with other systems without a measurable change in temperature. This makes it a constant-temperature heat source or sink - a crucial assumption for idealized cycles like the Carnot cycle, where the efficiency depends on two constant temperature levels. Examples of Thermal Energy Reservoirs Thermal reservoirs exist both in nature and in engineering systems. They can also be observed in phase change materials and everyday environments. (a) Natural Thermal Reservoirs 1. Atmosphere - absorbs or releases vast amounts of heat with negligible temperature change. 2. Oceans and large lakes - due to their enormous mass and specific heat capacity, they can act as stable heat sinks or sources. 3. Geothermal layers - Earth's crust maintains nearly constant temperatures, acting as a heat reservoir. 4. The Sun - supplies a continuous and practically inexhaustible flow of energy at an almost constant temperature. (b) Engineered Thermal Reservoirs 1. Industrial furnaces and boilers - serve as high-temperature reservoirs supplying heat to working fluids. Thermodynamics 132 2. Power plant condensers and cooling ponds - act as low-temperature reservoirs (heat sinks) for waste heat rejection. 3. Cryogenic tanks or liquid nitrogen systems - maintain constant low temperatures during heat exchange. (c) Phase-Change Systems Substances undergoing phase change (melting, freezing, boiling, condensation) at constant temperature behave as ideal thermal reservoirs. For example:  Melting ice absorbs a large amount of latent heat at 0∘C without increasing its temperature.  Boiling water at 100∘C absorbs heat while maintaining a constant temperature. Thus, phase-change materials (PCMs) are used in thermal energy storage systems because of their ability to store and release heat at nearly constant temperature. (d) Everyday Situations Even in small-scale systems, the surroundings can act as an approximate reservoir. For instance:  The air in a room can absorb heat from a running computer or lamp without a significant rise in room temperature.  A large metal tank of water used in laboratories can serve as a temporary heat sink for experiments. These are not perfect reservoirs but can be treated as such when the temperature change is negligibly small compared to the process being studied. Why Does the Temperature of a Reservoir Not Change? In practice, all real materials exhibit some temperature change during heat exchange. However, in thermodynamic analysis, a reservoir is considered ideal for simplification. Theoretical Reasoning For a reservoir: 𝐶→∞ ⇒ Δ𝑇=𝑄 𝐶→0 Thermodynamics 133 Thus, for any finite heat transfer 𝑄, the temperature change ( Δ𝑇 ) is practically zero. This assumption allows us to maintain constant-temperature boundaries for heat engines and refrigerators, making analytical solutions simpler and more accurate for idealized cycles such as:  Carnot cycle  Rankine cycle  Reversed Carnot cycle In real systems:  The reservoir's temperature changes slightly, depending on its mass and specific heat.  However, when the temperature change is very small relative to the process, it can be neglected without affecting accuracy. Hence, the constant temperature assumption ( ΔT≈0 ) is acceptable and forms the basis for the Second Law of Thermodynamics in cyclic processes. Heat Source and Heat Sink In a typical thermodynamic system, two types of reservoirs are required for cyclic operation: (a) Heat Source A heat source is a high-temperature reservoir that supplies heat energy (𝑄𝐻) to a working fluid or device. Examples:  Combustion chamber in a gas turbine  Boiler in a steam power plant  The Sun (for solar power systems) Mathematically, heat input from the source is expressed as: 𝑄𝐻=𝑚⋅𝑐⋅(𝑇𝐻−𝑇system ) (b) Heat Sink A heat sink is a low-temperature reservoir that receives or absorbs heat ( 𝑄𝐶 ) rejected by the system after useful work is extracted. Thermodynamics 134 Examples:  Cooling water in condensers  Air in radiators  Ocean or river used for waste heat disposal For a reversible heat engine operating between two reservoirs: 𝑄𝐻−𝑄𝐶=𝑊 and efficiency is given by: 𝜂=1−𝑄𝐶 𝑄𝐻 The presence of both source and sink is essential; without them, no cyclic heat engine can operate. If only one reservoir existed, the process would violate the Second Law of Thermodynamics, since heat must naturally flow from a higher to a lower temperature to produce work. Applications of Thermal Energy Reservoirs in Thermodynamic Systems (a) Heat Engines Heat engines convert thermal energy into mechanical work.  Source: High-temperature reservoir (e.g., furnace or combustion chamber)  Sink: Low-temperature reservoir (e.g., condenser, atmosphere) Example: In a steam power plant:  The boiler acts as the heat source.  The condenser acts as the heat sink. (b) Refrigerators and Heat Pumps These devices transfer heat from low to high temperature using external work input.  Refrigerator: Removes heat from a low-temperature space (source) and rejects it to surroundings (sink).  Heat Pump: Extracts heat from surroundings and delivers it to a high-temperature region (like a room). Thermodynamics 135 𝑄𝐻=𝑄𝐶+𝑊 Where, 𝑄𝐻= Heat delivered to hot region 𝑄𝐶= Heat absorbed from cold region 𝑊= Work input to compressor (c) Power Plants In power generation cycles:  The source is the steam or gas produced by burning coal, oil, or nuclear fuel.  The sink is the cooling water from rivers, cooling ponds, or cooling towers that absorb waste heat. Without both reservoirs, a power plant cannot function, since cyclic heat transfer requires two temperature levels. Significance of Thermal Energy Reservoirs 1. Simplify Thermodynamic Analysis: Reservoirs allow for constant-temperature assumptions in cycle studies like Carnot, Rankine, and Otto cycles. 2. Establish Limits of Efficiency: The performance of heat engines and refrigerators is determined by the temperatures of the two reservoirs: 𝜂Carnot =1−𝑇𝐿 𝑇𝐻 3. Provide Realistic Models for Energy Exchange: Large bodies like the atmosphere, oceans, or cooling towers behave approximately as thermal reservoirs in real-world systems. 4. Ensure Direction of Heat Transfer: They maintain the temperature gradient necessary for spontaneous heat flow - from high to low temperature, as dictated by the Second Law. Problem 2.2: A heat engine operates between two thermal energy reservoirs: 1. The high-temperature reservoir (heat source) is at 800 K , and 2. The low-temperature reservoir (heat sink) is at 300 K. During each cycle, the engine absorbs 2000 kJ of heat (𝑄𝐻) from the hot reservoir and rejects heat (𝑄𝐶) to the cold reservoir. Thermodynamics 136 Assuming the engine operates on a reversible (Carnot) cycle, determine: 1. The thermal efficiency of the engine. 2. The heat rejected to the cold reservoir (𝑄𝐶). 3. The work output of the engine per cycle. Solution: Given Data: 𝑇𝐻 =800 K 𝑇𝐿 =300 K 𝑄𝐻 =2000 kJ 1. Thermal Efficiency of the Carnot Engine For a reversible (Carnot) heat engine: 𝜂=1−𝑇𝐿 𝑇𝐻 Substitute values: 𝜂=1−300 800=1−0.375=0.625 𝜂=0.625 or 62.5% 2. Heat Rejected to the Cold Reservoir The efficiency is also defined as: 𝜂=𝑊 𝑄𝐻=1−𝑄𝐶 𝑄𝐻 Rearrange for 𝑄𝐶 : 𝑄𝐶=𝑄𝐻(1−𝜂) Substitute values: 𝑄𝐶=2000(1−0.625)=2000×0.375=750𝑘𝐽 𝑄𝐶=750𝑘𝐽 Thermodynamics 137 3. Work Output per Cycle From the First Law of Thermodynamics for cyclic processes: 𝑊=𝑄𝐻−𝑄𝐶 Substitute values: 𝑊=2000−750=1250𝑘𝐽 𝑊=1250𝑘𝐽 Exercise 2.3: A heat engine operates between two thermal reservoirs: 1. A high-temperature reservoir at 900 K, and 2. A low-temperature reservoir at 300 K. During each cycle, the engine absorbs 1500 kJ of heat from the hot reservoir. Determine: 1. The maximum possible efficiency of the engine. 2. The amount of heat rejected to the cold reservoir. 3. The work output per cycle for the reversible (Carnot) operation. Exercise 2.4: A refrigerator operates between two thermal energy reservoirs: 1. The cold reservoir (refrigerated space) is at 𝟐𝟕𝟎K, and 2. The hot reservoir (surrounding air) is at 310 K. During each cycle, the refrigerator removes 400 kJ of heat from the cold space. Find: 1. The maximum (Carnot) coefficient of performance (COP) of the refrigerator. 2. The minimum work input required to operate the refrigerator. 3. The heat rejected to the surrounding air. Thermodynamics 138 2.4 Carnot’s Theorem The Carnot’s Theorem, also known as Carnot’s Rule, was formulated in 1824 by the French engineer and physicist Nicolas Léonard Sadi Carnot. It is one of the most fundamental results of the Second Law of Thermodynamics, defining the maximum possible efficiency of a heat engine operating between two temperature limits. Carnot’s theorem provides a theoretical standard for all real heat engines by describing an ideal reversible engine (known as the Carnot engine) whose efficiency depends only on the temperatures of the heat source and sink, and not on the working substance or mechanism of operation. Statement of Carnot’s Theorem “No heat engine operating between two heat reservoirs can be more efficient than a reversible Carnot engine operating between the same reservoirs.” In simpler terms:  A Carnot engine sets the upper limit of efficiency for all heat engines.  Any irreversible (real) engine operating between the same hot and cold temperature limits will have lower efficiency. Thus, the Carnot theorem establishes an absolute thermodynamic benchmark no real engine, regardless of its design, can surpass the efficiency of the ideal Carnot engine. Fig. 2.6 Carnot’s Theorem. Thermodynamics 139 A Carnot engine is a reversible heat engine that operates on the Carnot cycle, consisting of four reversible processes: 1. Isothermal Expansion (at 𝑇𝐻 ) - Heat 𝑄𝐻 is absorbed from the high-temperature reservoir. 2. Adiabatic Expansion - The system expands without heat exchange, lowering its temperature to 𝑇𝐶. 3. Isothermal Compression (at 𝑇𝐶 ) - Heat 𝑄𝐶 is rejected to the low-temperature reservoir. 4. Adiabatic Compression - The system is compressed without heat exchange, raising its temperature back to 𝑇𝐻. Because each step is reversible, the entire cycle represents the most efficient possible operation between two temperatures. The efficiency ( 𝜂 ) of any heat engine is defined as the ratio of the work output to the heat input: 𝜂=𝑊 𝑄𝐻 Since the work done is the difference between the heat absorbed from the source and heat rejected to the sink, 𝑊=𝑄𝐻−𝑄𝐶 Substituting, we get: 𝜂=1−𝑄𝐶 𝑄𝐻 Now, for a reversible (Carnot) engine, the ratio of heat transfers is proportional to the ratio of absolute temperatures of the reservoirs: 𝑄𝐶 𝑄𝐻=𝑇𝐶 𝑇𝐻 Therefore, the maximum possible (Carnot) efficiency becomes: 𝜂max =𝜂Carnot =1−𝑇𝐶 𝑇𝐻 Where:  𝜂max = Maximum or Carnot efficiency Thermodynamics 140  𝑇𝐻= Absolute temperature of the hot reservoir (in K )  𝑇𝐶= Absolute temperature of the cold reservoir (in K ) This equation shows that efficiency increases as the temperature difference between source and sink increases. From the equation: 𝜂Carnot =1−𝑇𝐶 𝑇𝐻 We can observe the following:  If 𝑇𝐶=𝑇𝐻, then 𝜂Carnot =0, meaning no work output (since there's no temperature difference).  If 𝑇𝐶 is very small compared to 𝑇𝐻, the efficiency approaches unity, but never reaches 100% in practice.  To increase efficiency, either increase 𝑇𝐻 (raise the source temperature) or decrease 𝑇𝐶 (lower the sink temperature). This demonstrates why no heat engine can be 100% efficient - some heat must always be rejected to the cold reservoir, as required by the Second Law of Thermodynamics. Entropy Relation and Reversibility Condition For a reversible Carnot cycle, the entropy change between the two reservoirs is given by: Δ𝑆=∫ 𝑏 𝑎𝑑𝑄𝑟𝑒𝑣 𝑇 Where:  Δ𝑆= Change in entropy  𝑑𝑄rev = Infinitesimal amount of reversible heat transfer  𝑇= Absolute temperature at which heat is transferred For a reversible cycle, the net entropy change of the universe is zero: Δ𝑆system +Δ𝑆surroundings =0 For the two reservoirs: 𝑄𝐻 𝑇𝐻=𝑄𝐶 𝑇𝐶 Thermodynamics 243 Problem 3.11: A Stirling air-standard cycle operates between 𝑇1=350 K and 𝑇3=900 K. The compression ratio ( V1/V2 ) is 8. The air used has 𝑅=0.287 kJ/kg.K and 𝐶𝑣= 0.718 kJ/kg.K. Assume perfect regeneration. Find: 1. The heat added per kg of air. 2. The heat rejected per kg. 3. The net work output. 4. The thermal efficiency. Solution: Given Data: Symbol Parameter 𝑇1 350 K 𝑇3 900 K 𝑉1/𝑉2 8 R 0.287 kJ/kg⋅K 𝐶𝑣 0.718 kJ/kg⋅K 𝛾 1.4 Step 1: Work During Isothermal Compression (1-2) 𝑊1−2=𝑚𝑅𝑇1ln (𝑉1 𝑉2) 𝑊1−2=0.287×350×ln (8) ln (8)=2.079 𝑊1−2=0.287×350×2.079=209.6 kJ/kg This is work input (negative), so 𝑊comp =−209.6 kJ/kg Thermodynamics 244 Step 2: Work During Isothermal Expansion (3-4) 𝑊3−4=𝑚𝑅𝑇3ln (𝑉4 𝑉3) 𝑉4/𝑉3=𝑉1/𝑉2=8 𝑊3−4=0.287×900×2.079=536.9 kJ/kg Step 3: Net Work Output 𝑊𝑛𝑒𝑡 =𝑊3−4+𝑊1−2 𝑊𝑛𝑒𝑡 =536.9−209.6=327.3 kJ/kg Step 4: Heat Added and Heat Rejected In the Stirling cycle (with perfect regeneration): 𝑄in =𝑄3−4 =𝑚𝑅𝑇3ln (𝑉4 𝑉3) 𝑄in =0.287×900×2.079=536.9 kJ/kg 𝑄out =𝑄1−2=𝑚𝑅𝑇1ln (𝑉1 𝑉2) 𝑄out =0.287×350×2.079=209.6 kJ/kg Step 5: Thermal Efficiency 𝜂=1−𝑄out 𝑄in 𝜂=1−209.6 536.9=1−0.390=0.61=61.0% or equivalently, 𝜂=1−𝑇1 𝑇3=1−350 900=1−0.389=0.611=61.1% Exercise 3.15: An air-standard Stirling cycle operates between temperature limits of 𝑇1= 310 K and 𝑇3=930 K. The volume ratio (V1/V2) is 5, and air is used as the working fluid. Take 𝑅=0.287 kJ/kg.K and 𝐶𝑣=0.718 kJ/kg.K. Assume perfect regeneration. Find: 1. The work done per kg of air during each isothermal process. 2. The net work output per kg of air. Thermodynamics 245 3. The thermal efficiency of the cycle. Exercise 3.16: An air-standard Stirling cycle operates with a compression ratio (𝑉1/𝑉2)=8. The minimum temperature is 300 K, and the maximum temperature varies as follows: (a) 𝑇3=900 K (b) 𝑇3=1200 K Assume 𝑅=0.287 kJ/kg.K. Determine: 1. The thermal efficiency for both temperature conditions. 2. The percentage improvement in efficiency when the maximum temperature increases from 900 K to 1200 K. Exercise 3.17: A Stirling air-standard cycle has a compression ratio (V1/V2)=7 and operates between T1=350 K and T3= 1050 K. The working fluid is air with = 0.287 kJ/kg.K. Assume perfect regeneration and 1 kg of air. Calculate: 1. The heat added per kg during isothermal expansion. 2. The heat rejected per kg during isothermal compression. 3. The net work output and efficiency of the cycle. 3.7 Brayton Cycle The Brayton Cycle, named after George Brayton, is the air-standard cycle that describes the operation of gas turbine engines. It is the ideal cycle for gas turbines, jet engines, and power plants operating with a compressor–combustion chamber–turbine arrangement.  Air is compressed,  Fuel is burned at constant pressure,  The hot gases expand in the turbine, producing work. In the air-standard Brayton cycle, the combustion process is replaced by external heat addition to simplify thermodynamic analysis. Thermodynamics 246 Definition The Brayton Air Standard Cycle is a reversible open thermodynamic cycle consisting of two isentropic and two constant-pressure processes: 1. Isentropic compression (in the compressor) 2. Constant-pressure heat addition (in the combustor) 3. Isentropic expansion (in the turbine) 4. Constant-pressure heat rejection (in the cooler) Components of the Cycle The basic components of a gas turbine engine working on the Brayton cycle are: 1. Compressor - compresses the air isentropically 2. Combustion chamber (heater) - adds heat at constant pressure 3. Turbine - expands the air isentropically to produce work 4. Cooler - rejects heat to close the cycle (in theoretical analysis) Air Standard Assumptions To analyze the cycle theoretically, the Air Standard Assumptions are made: 1. The working fluid is air, which behaves as an ideal gas. 2. The same air circulates in a closed loop. 3. All processes are internally reversible. 4. Combustion and exhaust are replaced by external heat addition and rejection. 5. Specific heats (Cₚ and Cᵥ) are constant. Processes of the Brayton Cycle The Brayton Air Standard Cycle consists of four reversible processes, as represented on P–V and T–S diagrams. Process 1-2: Isentropic Compression (in Compressor) Air is compressed adiabatically and reversibly (isentropic process). The piston or compressor raises the air pressure and temperature while volume decreases. No heat is exchanged. Equations: 𝑇2 𝑇1=(𝑃2 𝑃1)𝛾−1 𝛾=𝑟𝑝𝛾−1 𝛾 Thermodynamics 247 𝑉1 𝑉2=(𝑃2 𝑃1)1 𝛾 Where 𝑟𝑝=𝑃2 𝑃1→ Pressure ratio Fig. 3.7 P-V and T-S diagram for Brayton Cycle. Process 2-3: Constant Pressure Heat Addition Heat is added to the compressed air at constant pressure. Temperature increases from 𝑇2 to 𝑇3. This represents combustion in a gas turbine. Equation: 𝑄𝑖𝑛 =𝑚𝐶𝑝(𝑇3−𝑇2) Process 3-4: Isentropic Expansion (in Turbine) The high-pressure, high-temperature air expands adiabatically and reversibly in the turbine. The air does work on the turbine blades. Pressure and temperature drop as the air expands. Equations: 𝑇4 𝑇3=(𝑃4 𝑃3)𝛾−1 𝛾=(1 𝑟𝑝)𝛾−1 𝛾 𝑊turbine =𝑚𝐶𝑝(𝑇3−𝑇4) Thermodynamics 248 Process 4-1: Constant Pressure Heat Rejection Heat is rejected from the air at constant pressure, returning it to its initial temperature 𝑇1. Pressure remains constant, but volume decreases. Equation: 𝑄out =𝑚𝐶𝑝(𝑇4−𝑇1) Work and Heat Relations For 1 kg of air: Process Type Heat Work 1-2 Isentropic compression 0 𝑊𝑐=𝐶𝑝(𝑇2−𝑇1) 2-3 Constant pressure 𝑄in =𝐶𝑝(𝑇3−𝑇2) 0 3-4 Isentropic expansion 0 𝑊𝑡=𝐶𝑝(𝑇3−𝑇4) 4-1 Constant pressure 𝑄out =𝐶𝑝(𝑇4−𝑇1) 0 Net Work Done 𝑊𝑛𝑒𝑡 =𝑊𝑡−𝑊𝑐 𝑊𝑛𝑒𝑡 =𝐶𝑝[(𝑇3−𝑇4)−(𝑇2−𝑇1)] Thermal Efficiency 𝜂=1−𝑄out 𝑄in Substituting 𝑄in =𝐶𝑝(𝑇3−𝑇2) and 𝑄out =𝐶𝑝(𝑇4−𝑇1) : 𝜂=1−𝑇4−𝑇1 𝑇3−𝑇2 Using isentropic relations: 𝑇2 𝑇1=𝑟𝑝(𝛾−1)/𝛾,𝑇3 𝑇4=𝑟𝑝(𝛾−1)/𝛾 Thermodynamics 249 Simplifying, we get: 𝜂=1− 1 𝑟𝑝(𝛾−1)/𝛾 This is the ideal efficiency of the Brayton (Gas Turbine) Cycle. Key Observations Efficiency depends only on the pressure ratio (𝑟𝑝) and 𝛾. Higher 𝑟𝑝→ Higher efficiency. However, very high pressure ratios are limited by material and thermal constraints. Optimum Pressure Ratio for Maximum Work The net work output of a gas turbine does not always increase with 𝑟𝑝. There exists an optimum pressure ratio ( 𝐫𝐩, opt) for maximum specific work output. It is derived as: 𝑟𝑝(𝑜𝑝𝑡)=(𝑇3 𝑇1)𝛾 2(𝛾−1) Mean Effective Pressure (m.e.p.) m.e.p. =𝑊𝑛𝑒𝑡 𝑉1−𝑉2 A higher mean effective pressure indicates greater power output per unit volume of air. Applications The Brayton cycle is the basis of:  Jet engines (turbojet and turbofan)  Gas turbine power plants  Aircraft propulsion systems  Marine propulsion  Combined-cycle power plants (gas + steam turbine systems) Thermodynamics 250 Advantages  High power-to-weight ratio − suitable for aircraft.  Smooth and continuous combustion.  Less vibration (rotary motion).  Can use multiple fuels (diesel, natural gas, kerosene).  Compact and simple design. Limitations  Low thermal efficiency at small scales.  High temperature materials needed for turbine blades.  Efficiency drops at part load.  Difficult to start (needs external starter). Methods to Improve Efficiency To improve Brayton cycle efficiency, several modifications are used: Method Process Effect Regeneration Preheating of compressed air using turbine exhaust Increases efficiency Intercooling Multi-stage compression with cooling between stages Reduces compressor work Reheating Multi-stage expansion with heating between stages Increases turbine work Combined Cycle Brayton + Rankine (steam) Maximum overall efficiency Process Type Heat Work Entropy Change 1-2 Isentropic Compression 0 -ve 0 2-3 Constant Pressure Heat Addition +ve 0 +ve Thermodynamics 251 3-4 Isentropic Expansion 0 +ve 0 4-1 Constant Pressure Heat Rejection -ve 0 -ve Table. 3.7 Brayton Cycle with Entropy Change. Problem 3.12: A gas turbine plant operating on the Brayton air-standard cycle has a pressure ratio ( 𝐫𝐩 ) of 6. Air enters the compressor at 1 bar and 300 K, and the maximum cycle temperature is 1000 K . Assume 𝛾=1.4 and 𝐶𝑝=1.005 kJ/kg.K. Find: 1. The temperatures at all four points ( 𝑇2,𝑇3,𝑇4 ) 2. The cycle efficiency 3. The net work output per kg of air 4. The heat supplied per kg of air Solution: Given Data: Symbol Parameter Value 𝑃1 Inlet pressure 1 bar 𝑇1 Inlet temperature 300 K 𝑟𝑝=𝑃2 𝑃1 Pressure ratio 6 𝑇3 Max temperature 1000 K 𝛾 1.4 𝐶𝑝 1.005 kJ/kg⋅K Step 1: Process 1-2 (Isentropic Compression) 𝑇2 𝑇1=𝑟𝑝(𝛾−1)/𝛾 Thermodynamics 252 Substitute: 𝑇2=300×60.286 =300×1.668=500.4 K Step 2: Process 3-4 (Isentropic Expansion) 𝑇3 𝑇4=𝑟𝑝(𝛾−1)/𝛾 𝑇4=𝑇3 𝑟𝑝(𝛾−1)/𝛾 =1000 1.668=599.4𝐾 Step 3: Work Done by Compressor 𝑊𝑐=𝐶𝑝(𝑇2−𝑇1) 𝑊𝑐=1.005(500.4−300)=1.005×200.4=201.4 kJ/kg Step 4: Work Done by Turbine 𝑊𝑡=𝐶𝑝(𝑇3−𝑇4) 𝑊𝑡=1.005(1000−599.4)=1.005×400.6=402.6 kJ/kg Step 5: Net Work Output 𝑊𝑛𝑒𝑡 =𝑊𝑡−𝑊𝑐 𝑊𝑛𝑒𝑡 =402.6−201.4=201.2 kJ/kg Step 6: Heat Supplied 𝑄𝑖𝑛 =𝐶𝑝(𝑇3−𝑇2) 𝑄𝑖𝑛 =1.005(1000−500.4)=1.005×499.6=502.1 kJ/kg Step 7: Thermal Efficiency 𝜂=𝑊𝑛𝑒𝑡 𝑄𝑖𝑛 =201.2 502.1=0.400=40.0% Problem 3.13: A gas turbine plant works on a Brayton air-standard cycle with a pressure ratio of 8. The air enters the compressor at 1 bar and 290 K, and the maximum cycle temperature is 1100 K. Air mass flow rate is 𝟓kg/s. Assume 𝐶𝑝=1.005 kJ/kg.K and 𝛾= 1.4. Thermodynamics 259 (e) Stirling and Ericsson Cycles Both are external combustion regenerative cycles.  Stirling Cycle Efficiency: 𝜂Stirling =1−𝑇𝐿 𝑇𝐻  Ericsson Cycle Efficiency: 𝜂Ericsson =1−𝑇𝐿 𝑇𝐻 Key Point: Both have the same efficiency as the Carnot cycle, i.e., the maximum theoretical efficiency possible between two temperature limits. Relation to Carnot Efficiency The Carnot cycle represents the upper limit of efficiency for any heat engine operating between two temperature limits 𝑇𝐻 and 𝑇𝐿 : 𝜂Carnot =1−𝑇𝐿 𝑇𝐻 The air-standard efficiencies of Otto, Diesel, and Brayton cycles are always less than Carnot efficiency, but their trends approach it as irreversibilities are minimized. Factors Affecting Air Standard Efficiency Factor Effect on Efficiency Compression Ratio (r) Higher 𝑟→ higher efficiency Cut-off Ratio ( 𝛽 ) Higher 𝛽→ lower efficiency (in Diesel cycle) Pressure Ratio ( rp ) Higher 𝑟𝑝→ higher efficiency (in Brayton cycle) 𝛾 (Specific Heat Ratio) Higher 𝛾→ higher efficiency Temperature Ratio ( T1/T3 ) Larger temperature difference → higher efficiency Significance of Air Standard Efficiency  It provides a theoretical upper limit for real engine performance.  It helps in cycle comparison (Otto vs. Diesel vs. Brayton). Thermodynamics 260  It aids in design optimization of engines.  It shows how thermodynamic parameters influence performance. 3.9 Mean Effective Pressure (MEP) In the study of internal combustion engines and air-standard cycles, one of the most important parameters used to evaluate the performance of an engine is the Mean Effective Pressure (MEP). It serves as a useful measure to compare the power-producing capability of different engines, irrespective of their size or displacement. While the efficiency of a cycle indicates how effectively energy is converted from heat to work, it does not directly tell us how much power the engine develops. The Mean Effective Pressure bridges this gap it is a hypothetical constant pressure that, if it acted on the piston during the entire power stroke, would produce the same net work output as that actually produced by the cycle. In simpler terms, it translates the complex pressure variations during the cycle into a single representative pressure value that reflects the engine’s ability to perform work. In a real engine or a thermodynamic cycle, the pressure inside the cylinder is not constant - it continuously varies throughout the cycle as the piston moves between the top dead center (TDC) and bottom dead center (BDC). If we could replace this varying pressure by a constant average pressure that would produce the same work over one power stroke, that constant pressure is known as the Mean Effective Pressure. Mathematically, the Mean Effective Pressure (MEP) is defined as: MEP=𝑊net 𝑉𝑑 Where, 𝑊net = Net work output per cycle (in kJ or J) 𝑉𝑑= Displacement volume or swept volume of the cylinder (in m3 ) Thus, MEP can be interpreted as a measure of how effectively the engine uses its displacement volume to produce work. The higher the mean effective pressure, the greater the net work output for the same cylinder size. Thermodynamics 261 To understand the physical significance, imagine an engine cylinder in which the gas pressure remains constant throughout the expansion stroke. The work done by this gas on the piston would simply be the product of this constant pressure and the displacement volume: 𝑊=𝑃me×𝑉𝑑 In reality, the actual pressure fluctuates significantly during the cycle. However, if we consider the area enclosed by the indicator diagram (Pressure-Volume or P-V diagram), it represents the net work output. Dividing this area by the displacement volume gives us the mean effective pressure, i.e., the equivalent constant pressure that would yield the same work area. Therefore, MEP provides a clear and convenient way to express the output performance of an engine in terms of pressure - a parameter that is independent of engine size. Types of Mean Effective Pressure Depending on the type of engine and the process considered, different forms of mean effective pressure are defined: 1. Indicated Mean Effective Pressure (IMEP): This is based on the indicated power or the total work done by the gas on the piston as recorded on an indicator diagram. It represents the pressure corresponding to the gross work inside the cylinder before mechanical losses are considered. 2. Brake Mean Effective Pressure (BMEP): This corresponds to the useful power output available at the crankshaft after accounting for frictional and pumping losses. The brake mean effective pressure gives a realistic measure of the usable engine performance. 3. Friction Mean Effective Pressure (FMEP): This is the difference between IMEP and BMEP and represents the pressure loss due to mechanical friction and pumping effects within the engine. The relationship among these is expressed as: IMEP=BMEP+FMEP Thus, by comparing these values, engineers can determine how efficiently an engine converts the pressure developed in the cylinder into actual usable work at the shaft. Let the engine produce a net work per cycle 𝑊 and let the displacement volume be 𝑉𝑑. Thermodynamics 262 The mean effective pressure is then given by: 𝑃me =𝑊 𝑉𝑑 For a four-stroke engine, only one power stroke occurs in two revolutions of the crankshaft. If the engine runs at a speed of 𝑁 revolutions per minute (rpm), then the number of power strokes per minute is 𝑁/2. The indicated power (IP) can then be expressed as: 𝐼𝑃=𝑃me×𝐿×𝐴×𝑛 60 Where, 𝐿= Stroke length (m) 𝐴= Piston area ( m2 ) 𝑛= Number of power strokes per second On a Pressure–Volume (P–V) diagram, the area enclosed by the cycle represents the work output. If we replace this cycle by a rectangle having the same base (equal to the displacement volume) and the same area, the height of this rectangle would represent the mean effective pressure. Fig. 3.8 Graphical Representation of MEP. Thermodynamics 263 Hence, the MEP corresponds to the average height of the P–V diagram over the working stroke. This makes MEP a valuable tool for engine designers and performance analysts, as it converts the geometrical area of a complex diagram into a simple measurable quantity. Significance of Mean Effective Pressure Mean effective pressure serves as one of the most practical and fundamental parameters in engine performance analysis. It allows comparison between engines of different sizes because it relates work output to the displacement volume rather than the absolute work itself. For instance, two engines with different piston sizes or cylinder volumes may produce different total work, but if both have the same MEP, it indicates that both are equally effective in utilizing their swept volume. Moreover, MEP provides a direct indication of how much torque an engine can produce. In fact, for a given engine speed, torque is directly proportional to mean effective pressure. A high MEP value implies a greater torque output and better engine performance. Consequently, it is often used in design and testing to compare engine performance under various operating conditions or configurations. Derivation of Mean Effective Pressure (m.e.p.) for the Otto Cycle The mean effective pressure (m.e.p.), often denoted as 𝑝𝑚, is defined as the constant pressure that, if it acted on the piston during the power stroke, would produce the same net work output as that obtained in the actual cycle. 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 Where,  𝑊net = Net work done per cycle  𝑉1−𝑉2= Displacement volume (swept volume) This definition allows comparing the performance of different engines independent of their size. The net work output of the cycle is equal to the heat added minus the heat rejected: 𝑊𝑛𝑒𝑡 =𝑄𝑖𝑛−𝑄𝑜𝑢𝑡 Thermodynamics 264 For the Otto cycle, both 𝑄in and 𝑄out occur at constant volume, hence: 𝑄𝑖𝑛 =𝑚𝐶𝑣(𝑇3−𝑇2) 𝑄𝑜𝑢𝑡 =𝑚𝐶𝑣(𝑇4−𝑇1) Substitute these into the work equation: 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑣[(𝑇3−𝑇2)−(𝑇4−𝑇1)] We use the isentropic relations for compression (1-2) and expansion (3-4) processes: 𝑇2 𝑇1=𝑟(𝛾−1) and 𝑇3 𝑇4=𝑟(𝛾−1) From these, we get: 𝑇2=𝑇1𝑟(𝛾−1) and 𝑇4=𝑇3 𝑟(𝛾−1) Substitute 𝑇2 and 𝑇4 into the 𝑊net equation: 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑣[(𝑇3−𝑇1𝑟(𝛾−1))−( 𝑇3 𝑟(𝛾−1)−𝑇1)] Simplify: 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑣[𝑇3(1− 1 𝑟(𝛾−1))−𝑇1(𝑟(𝛾−1)−1)] Now recall: 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 Substitute 𝑊net from above: 𝑝𝑚=𝑚𝐶𝑣 𝑉1−𝑉2[𝑇3(1− 1 𝑟(𝛾−1))−𝑇1(𝑟(𝛾−1)−1)] We know: 𝑟=𝑉1 𝑉2⇒𝑉2=𝑉1 𝑟 Thermodynamics 265 Therefore: 𝑉1−𝑉2=𝑉1(1−1 𝑟) Substitute this in the equation for 𝑝𝑚 : 𝑝𝑚=𝑚𝐶𝑣 𝑉1(1−1 𝑟)[𝑇3(1− 1 𝑟(𝛾−1))−𝑇1(𝑟(𝛾−1)−1)] Using the ideal gas law for state 1: 𝑃1𝑉1=𝑚𝑅𝑇1⇒𝑚 𝑉1=𝑃1 𝑅𝑇1 Substitute into the m.e.p. equation: 𝑝𝑚=𝐶𝑣𝑃1 𝑅𝑇1(1−1 𝑟)[𝑇3(1− 1 𝑟(𝛾−1))−𝑇1(𝑟(𝛾−1)−1)] Since =𝐶𝑝−𝐶𝑣=𝐶𝑣(𝛾−1) : 𝐶𝑣 𝑅=1 (𝛾−1) Hence: 𝑝𝑚=𝑃1 (𝛾−1)(1−1 𝑟)[𝑇3 𝑇1(1− 1 𝑟(𝛾−1))−(𝑟(𝛾−1)−1)] The heat addition process (2-3) occurs at constant volume: 𝑇3 𝑇2=𝑄𝑖𝑛 𝑚𝐶𝑣𝑇2+1=𝑇3 𝑇2 Let's denote: 𝛼=𝑇3 𝑇2= temperature ratio during heat addition Now, since 𝑇2=𝑇1𝑟(𝛾−1) : 𝑇3 𝑇1=𝛼𝑟(𝛾−1) Thermodynamics 266 Substitute this into the expression for 𝑝𝑚 : 𝑝𝑚=𝑃1 (𝛾−1)(1−1 𝑟)[𝛼𝑟(𝛾−1)(1− 1 𝑟(𝛾−1))−(𝑟(𝛾−1)−1)] Simplify the bracket: 𝑝𝑚=𝑃1 (𝛾−1)(1−1 𝑟)[(𝛼− 1 𝑟(𝛾−1))−(𝑟(𝛾−1)−1)] 𝑝𝑚=𝑃1 (𝛾−1)(1−1 𝑟)[𝛼𝑟(𝛾−1)(1− 1 𝑟(𝛾−1))−(𝑟(𝛾−1)−1)] Where,  𝑝𝑚= Mean effective pressure  𝑃1= Initial pressure  𝑟= Compression ratio  𝛾= Specific heat ratio  𝛼=𝑇3/𝑇2= Temperature ratio during heat addition Derivation of Mean Effective Pressure (m.e.p.) for the Diesel Cycle The mean effective pressure (m.e.p.), denoted by 𝑝𝑚, is defined as the hypothetical constant pressure that, if acted on the piston during the power stroke, would produce the same net work as that obtained in the actual Diesel cycle. 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 Where,  𝑊net = Net work output per cycle (kJ/kg)  𝑉1−𝑉2= Displacement or swept volume (m3/kg) In the Diesel cycle: 𝑊𝑛𝑒𝑡 =𝑄𝑖𝑛−𝑄𝑜𝑢𝑡 Thermodynamics 267 From the heat addition and rejection processes:  Constant pressure heat addition (2-3): 𝑄𝑖𝑛 =𝑚𝐶𝑝(𝑇3−𝑇2)  Constant volume heat rejection (4-1): 𝑄𝑜𝑢𝑡 =𝑚𝐶𝑣(𝑇4−𝑇1) Hence: 𝑊𝑛𝑒𝑡 =𝑚[𝐶𝑝(𝑇3−𝑇2)−𝐶𝑣(𝑇4−𝑇1)] For one kg of air ( m=1 ): 𝑊𝑛𝑒𝑡 =𝐶𝑝(𝑇3−𝑇2)−𝐶𝑣(𝑇4−𝑇1) We know that 𝐶𝑝−𝐶𝑣=𝑅, and 𝐶𝑝 𝐶𝑣=𝛾. So: 𝐶𝑝=𝛾𝑅 𝛾−1,𝐶𝑣=𝑅 𝛾−1 Substitute these into 𝑊net : 𝑊𝑛𝑒𝑡 =𝑅 𝛾−1[𝛾(𝑇3−𝑇2)−(𝑇4−𝑇1)] We'll express all temperatures 𝑇2,𝑇3,𝑇4 in terms of 𝑇1, the compression ratio 𝑟, the cut-off ratio 𝜌, and the specific heat ratio 𝛾. (a) Process 1-2 (Isentropic Compression): 𝑇2 𝑇1=𝑟(𝛾−1) ⇒ 𝑇2=𝑇1𝑟(𝛾−1) (b) Process 2-3 (Constant Pressure Heat Addition): 𝑉3 𝑉2=𝜌 and 𝑇3 𝑇2=𝜌 Thus: 𝑇3=𝑇2𝜌=𝑇1𝑟(𝛾−1)𝜌 Thermodynamics 268 (c) Process 3-4 (Isentropic Expansion): 𝑇4 𝑇3=(𝑉4 𝑉3)(𝛾−1) But 𝑉4=𝑉1 and 𝑉3/𝑉2=𝜌, so: 𝑉4 𝑉3=𝑉1 𝑉3=𝑟 𝜌 Hence: 𝑇4=𝑇3(𝜌 𝑟)(𝛾−1)=𝑇1𝑟(𝛾−1)𝜌(𝜌 𝑟)(𝛾−1)=𝑇1𝜌𝛾 Therefore: 𝑇2=𝑇1𝑟(𝛾−1) 𝑇3=𝑇1𝑟(𝛾−1)𝜌 𝑇4=𝑇1𝜌𝛾 𝑊𝑛𝑒𝑡 =𝑅 𝛾−1[𝛾(𝑇3−𝑇2)−(𝑇4−𝑇1)] Substitute the expressions of 𝑇2,𝑇3,𝑇4 : 𝑊𝑛𝑒𝑡 =𝑅𝑇1 𝛾−1[𝛾(𝑟(𝛾−1)𝜌−𝑟(𝛾−1))−(𝜌𝛾−1)] Simplify: 𝑊net =𝑅𝑇1 𝛾−1[𝛾𝑟(𝛾−1)(𝜌−1)−(𝜌𝛾−1)] We know: 𝑟=𝑉1 𝑉2 ⇒ 𝑉2=𝑉1 𝑟 Therefore: 𝑉1−𝑉2=𝑉1(1−1 𝑟) Thermodynamics 275 At state 1, the air obeys: 𝑃1𝑉1=𝑚𝑅𝑇1 ⇒ 𝑚𝑅 𝑉1=𝑃1 𝑇1 Substitute into the m.e.p. expression: 𝑝𝑚=𝑃1(𝑇3−𝑇1) 𝑇1(1−1 𝑟)ln (𝑟) Simplify: 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)(𝑇3 𝑇1−1) 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)(𝑇3 𝑇1−1) Where, Symbol Meaning 𝑃1 Minimum (intake) pressure 𝑇1 Minimum (isothermal compression) temperature 𝑇3 Maximum (isothermal expansion) temperature 𝑟=𝑉1 𝑉2 Compression/expansion ratio 𝑝𝑚 Mean effective pressure Relation with Efficiency Since the Ericsson cycle has the same efficiency as the Carnot cycle: 𝜂Ericsson =1−𝑇1 𝑇3 We can also write: 𝑇3 𝑇1−1= 𝜂Ericsson 1−𝜂Ericsson Thermodynamics 276 Thus, substituting this into the m.e.p. expression gives an alternate form: 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)⋅𝜂Ericsson 1−𝜂Ericsson Derivation of Mean Effective Pressure (m.e.p.) for the Atkinson Cycle The mean effective pressure (m.e.p.), denoted 𝑝𝑚, is defined as: 𝑝𝑚=𝑊net 𝑉1−𝑉2 Where,  𝑊net =𝑄in −𝑄out is the net work per cycle  𝑉1−𝑉2 is the displacement volume (the swept volume during compression stroke) Process Type Relation Notes 1-2 Isentropic compression 𝑇2/𝑇1=𝑟𝑐(𝛾−1) Compression ratio 𝑟𝑐=𝑉1/𝑉2 2-3 Constant volume heat addition 𝑄𝑖𝑛 =𝑚𝐶𝑣(𝑇3−𝑇2) 𝑃3/𝑃2=𝑇3/𝑇2 3-4 Isentropic expansion 𝑇4/𝑇3=(𝑉4/𝑉3)(𝛾−1) =𝑟𝑒(𝛾−1) Expansion ratio 𝑟𝑒= 𝑉4/𝑉3 4-1 Constant pressure heat rejection 𝑄out =𝑚𝐶𝑝(𝑇4−𝑇1) 𝑃4=𝑃1 𝑊𝑛𝑒𝑡 =𝑄𝑖𝑛−𝑄𝑜𝑢𝑡 Substituting heat terms: 𝑊𝑛𝑒𝑡 =𝑚[𝐶𝑣(𝑇3−𝑇2)−𝐶𝑝(𝑇4−𝑇1)] Use 𝐶𝑝=𝛾𝐶𝑣 : 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑣[(𝑇3−𝑇2)−𝛾(𝑇4−𝑇1)] For one unit mass of air ( m=1 ): 𝑊𝑛𝑒𝑡 =𝐶𝑣[(𝑇3−𝑇2)−𝛾(𝑇4−𝑇1)] Thermodynamics 277 From the isentropic relations: (a) Compression process (1-2) 𝑇2=𝑇1𝑟𝑐(𝛾−1) (b) Expansion process (3-4) 𝑇4=𝑇3(1 𝑟𝑒)(𝛾−1) =𝑇3𝑟𝑒−(𝛾−1) 𝑊𝑛𝑒𝑡 =𝐶𝑣[(𝑇3−𝑇1𝑟𝑐(𝛾−1))−𝛾(𝑇3𝑟𝑒−(𝛾−1)−𝑇1)] Expand and simplify: 𝑊𝑛𝑒𝑡 =𝐶𝑣[𝑇3(1−𝛾𝑟𝑒−(𝛾−1))−𝑇1(𝑟𝑐(𝛾−1)−𝛾)] 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 We know: 𝑉1−𝑉2=𝑉1(1−1 𝑟𝑐) Substitute 𝑊net : 𝑝𝑚=𝐶𝑣 𝑉1(1−1 𝑟𝑐)[𝑇3(1−𝛾𝑟𝑒−(𝛾−1))−𝑇1(𝑟𝑐(𝛾−1)−𝛾)] From the ideal gas law at state 1: 𝑃1𝑉1=𝑅𝑇1 Hence: 𝐶𝑣 𝑉1=𝐶𝑣𝑅𝑇1 𝑅𝑉1𝑇1=𝑃1𝐶𝑣 𝑅𝑇1 But 𝑅=𝐶𝑝−𝐶𝑣=𝐶𝑣(𝛾−1)⇒𝐶𝑛 𝑅=1 𝛾−1 So: 𝐶𝑣 𝑉1=𝑃1 𝑇1(𝛾−1) Thermodynamics 278 Substitute into 𝑝𝑚 : 𝑝𝑚=𝑃1 (𝛾−1)𝑇1(1−1 𝑟𝑐)[𝑇3(1−𝛾𝑟𝑒−(𝛾−1))−𝑇1(𝑟𝑐(𝛾−1)−𝛾)] Divide the entire bracket by 𝑇1 : 𝑝𝑚=𝑃1 (𝛾−1)(1−1 𝑟𝑐)[𝑇3 𝑇1(1−𝛾𝑟𝑒−(𝛾−1))−(𝑟𝑐(𝛾−1)−𝛾)] 𝑝𝑚=𝑃1 (𝛾−1)(1−1 𝑟𝑐)[𝑇3 𝑇1(1−𝛾𝑟𝑒−(𝛾−1))−(𝑟𝑐(𝛾−1)−𝛾)] Symbol Meaning Description 𝑃1 Initial (intake) pressure At start of compression 𝑇1 Minimum temperature At start of compression 𝑇3 Maximum temperature After constant volume heat addition 𝑟𝑐=𝑉1 𝑉2 Compression ratio Determines compression stage 𝑟𝑒=𝑉4 𝑉3 Expansion ratio Greater than 𝑟𝑐 in Atkinson cycle 𝛾=𝐶𝑝 𝐶𝑣 Specific heat ratio ≈1.4 for air 𝑝𝑚 Mean effective pressure Average useful pressure per cycle Derivation of Mean Effective Pressure (m.e.p.) for the Stirling Cycle The mean effective pressure (m.e.p.), denoted 𝑝𝑚, is defined as the hypothetical constant pressure that would produce the same net work output as that obtained in the actual Stirling cycle during one power stroke. 𝑝𝑚=𝑊net 𝑉1−𝑉2 Thermodynamics 279 Where,  𝑊net = Net work done per cycle  𝑉1−𝑉2= Displacement (swept) volume This parameter represents the average pressure acting on the piston throughout one complete cycle. Process Type Description Heat/Work Relation 1-2 Isothermal compression At low temperature 𝑇1; heat rejected 𝑄12=𝑊12 =𝑚𝑅𝑇1ln 𝑉1 𝑉2 2-3 Constant volume (isochoric) heat addition Regenerative heating 𝑄23 =𝑚𝐶𝑣(𝑇3−𝑇2) 3-4 Isothermal expansion At high temperature 𝑇3; heat absorbed 𝑄34=𝑊34 =𝑚𝑅𝑇3ln 𝑉4 𝑉3 4-1 Constant volume (isochoric) heat rejection Regenerative cooling 𝑄41 =𝑚𝐶𝑣(𝑇4−𝑇1) For a perfect regenerator, 𝑄23 =𝑄41 Hence, only the isothermal processes (1-2 and 3-4) contribute to the net external heat and work interactions. The total net work done is the algebraic sum of the isothermal works: 𝑊𝑛𝑒𝑡 =𝑊34−𝑊12 Substituting the isothermal work relations: 𝑊𝑛𝑒𝑡 =𝑚𝑅𝑇3ln (𝑉4 𝑉3)−𝑚𝑅𝑇1ln (𝑉1 𝑉2) In the Stirling cycle, the expansion ratio equals the compression ratio: 𝑉4 𝑉3=𝑉1 𝑉2=𝑟 Thermodynamics 280 Therefore: 𝑊𝑛𝑒𝑡 =𝑚𝑅(𝑇3−𝑇1)ln (𝑟) where 𝑟=𝑉1 𝑉2 is the volume ratio. By definition: 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 Substitute 𝑊net : 𝑝𝑚=𝑚𝑅(𝑇3−𝑇1)ln (𝑟) 𝑉1−𝑉2 We know: 𝑉2=𝑉1 𝑟 Thus: 𝑉1−𝑉2=𝑉1(1−1 𝑟) Substitute into the m.e.p. equation: 𝑝𝑚=𝑚𝑅(𝑇3−𝑇1)ln (𝑟) 𝑉1(1−1 𝑟) At state 1 (beginning of isothermal compression): 𝑃1𝑉1=𝑚𝑅𝑇1 ⇒𝑚𝑅 𝑉1=𝑃1 𝑇1 Substitute into the m.e.p. expression: 𝑝𝑚=𝑃1(𝑇3−𝑇1) 𝑇1(1−1 𝑟)ln (𝑟) Thermodynamics 281 Simplify the temperature ratio: 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)(𝑇3 𝑇1−1) 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)(𝑇3 𝑇1−1) Where, Symbol Meaning 𝑃1 Initial pressure (intake/compression start) 𝑇1 Minimum temperature (cold reservoir) 𝑇3 Maximum temperature (hot reservoir) 𝑟=𝑉1/𝑉2 Volume ratio (compression/expansion ratio) 𝑝𝑚 Mean effective pressure (average pressure during cycle) Alternative Expression in Terms of Temperature Ratio Let the temperature ratio be: 𝜏=𝑇3 𝑇1 Then: 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)(𝜏−1) Relation to Thermal Efficiency The thermal efficiency of the Stirling cycle equals that of the Carnot cycle: 𝜂Stirling =1−𝑇1 𝑇3=1−1 𝜏 Thermodynamics 282 We can rearrange this: 𝜏−1= 𝜂Stirling 1−𝜂Stirling Substituting into the 𝑝𝑚 expression: 𝑝𝑚=𝑃1ln (𝑟) (1−1 𝑟)𝜂Stirling 1−𝜂Stirling Derivation of Mean Effective Pressure (m.e.p.) for the Brayton Cycle The mean effective pressure (m.e.p.), denoted 𝑝𝑚, is the hypothetical constant pressure that, if acted on the piston during one power stroke, would produce the same net work as the actual Brayton cycle. 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 Where,  𝑊net = Net work output per cycle (kJ/kg)  𝑉1−𝑉2= Displacement or swept volume (m3/kg) Process Type Description Equations 1-2 Isentropic compression 𝑇2/𝑇1=(𝑃2/𝑃1)(𝛾−1)/𝛾 =𝑟𝑝(𝛾−1)/𝛾 𝑉2/𝑉1=(𝑃1/𝑃2)1/𝛾 =1 /𝑟𝑝1/𝛾 2-3 Constant pressure heat addition 𝑄in =𝑚𝐶𝑝(𝑇3−𝑇2) 𝑃3=𝑃2 3-4 Isentropic expansion 𝑇4/𝑇3=(𝑃4/𝑃3)(𝛾−1)/𝛾 =1 /𝑟𝑝(𝛾−1)/𝛾 4-1 Constant pressure heat rejection 𝑄out =𝑚𝐶𝑝(𝑇4−𝑇1) 𝑃4=𝑃1 Here, 𝑟𝑝=𝑃2 𝑃1 is the pressure ratio, and 𝛾=𝐶𝑝 𝐶𝑣. Thermodynamics 283 𝑊net =𝑊turbine −𝑊compressor For isentropic processes: 𝑊turbine =𝑚𝐶𝑝(𝑇3−𝑇4) 𝑊compressor =𝑚𝐶𝑝(𝑇2−𝑇1) Hence: 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑝[(𝑇3−𝑇4)−(𝑇2−𝑇1)] From the isentropic relations: 𝑇2 𝑇1=𝑟𝑝(𝛾−1)/𝛾,𝑇3 𝑇4=𝑟𝑝(𝛾−1)/𝛾 Hence: 𝑇2=𝑇1𝑟𝑝(𝛾−1)/𝛾,𝑇4=𝑇3/𝑟𝑝(𝛾−1)/𝛾 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑝[(𝑇3−𝑇3 𝑟𝑝(𝛾−1)/𝛾)−(𝑇1𝑟𝑝(𝛾−1)/𝛾−𝑇1)] Simplify terms: 𝑊𝑛𝑒𝑡 =𝑚𝐶𝑝[𝑇3(1− 1 𝑟𝑝(𝛾−1)/𝛾)−𝑇1(𝑟𝑝(𝛾−1)/𝛾−1)] 𝑝𝑚=𝑊𝑛𝑒𝑡 𝑉1−𝑉2 Substitute 𝑊net : 𝑝𝑚=𝑚𝐶𝑝[𝑇3(1− 1 𝑟𝑝(𝛾−1)/𝛾)−𝑇1(𝑟𝑝(𝛾−1)/𝛾−1)] 𝑉1−𝑉2 For the isentropic compression process 1−2 : 𝑉1 𝑉2 =(𝑃2 𝑃1)1/𝛾 =𝑟𝑝1/𝛾 ⇒𝑉2=𝑉1 𝑟𝑝1/𝛾 Thermodynamics 284 Thus: 𝑉1−𝑉2=𝑉1(1− 1 𝑟𝑝1/𝛾) 𝑝𝑚=𝑚𝐶𝑝 𝑉1(1− 1 𝑟𝑝1/𝛾)[𝑇3(1− 1 𝑟𝑝(𝛾−1)/𝛾)−𝑇1(𝑟𝑝(𝛾−1)/𝛾−1)] From the ideal gas law at state 1: 𝑃1𝑉1=𝑚𝑅𝑇1 Hence: 𝑚 𝑉1=𝑃1 𝑅𝑇1 Since 𝐶𝑝=𝛾𝑅 𝛾−1 : 𝑚𝐶𝑝 𝑉1=𝑃1𝛾 (𝛾−1)𝑇1 Substitute into 𝑝𝑚 : 𝑝𝑚=𝑃1𝛾 (𝛾−1)𝑇1(1− 1 𝑟𝑝1/𝛾)[𝑇3(1− 1 𝑟𝑝(𝛾−1)/𝛾)−𝑇1(𝑟𝑝(𝛾−1)/𝛾−1)] 𝑝𝑚=𝑃1𝛾 (𝛾−1)(1− 1 𝑟𝑝1/𝛾)[𝑇3 𝑇1(1− 1 𝑟𝑝(𝛾−1)/𝛾)−(𝑟𝑝(𝛾−1)/𝛾−1)] 𝑝𝑚=𝑃1𝛾 (𝛾−1)(1− 1 𝑟𝑝1/𝛾)[𝑇3 𝑇1(1− 1 𝑟𝑝(𝛾−1)/𝛾)−(𝑟𝑝(𝛾−1)/𝛾−1)] Thermodynamics 291 0.6117 kPa. This point is extremely important, as it serves as a reference for defining the thermodynamic temperature scale and calibrating temperature measurement instruments. Problem 4.1: A closed vessel contains 2 kg of water at a pressure of 400 kPa. The quality (dryness fraction) of the mixture is 𝟎.𝟔. Determine the following properties of the mixture: 1. The temperature of the mixture 2. The specific volume (v) 3. The enthalpy (h) 4. The entropy (s) 5. The total volume of the mixture Solution: Given Data: 𝑃=400kPa,𝑥=0.6,𝑚=2 kg Step 1: Identify the Phase Region The problem states that the water has a quality ( 𝑥=0.6 ). This means that the mixture lies in the two-phase region (liquid + vapor mixture). Step 2: Obtain Saturation Properties from Steam Tables (at 400 kPa ) At P = 𝐤𝐏𝐚, from standard saturated steam tables: Property Symbol Saturated Liquid (f) Saturated Vapor (g) Temperature ( ∘C ) T5 143.6∘C - Specific volume ( m3/kg ) v 0.001093 0.4624 Enthalpy (kJ/kg) h 604.7 2556.3 Entropy ( kJ/kg⋅K ) s 1.733 7.127 Thermodynamics 292 Step 3: Apply the Mixture Relations In the two-phase region, any property of the mixture is given by: 𝑦=𝑦𝑓+𝑥(𝑦𝑔−𝑦𝑓) Where, 𝑦 can be 𝑣,ℎ,𝑠, etc. Step 4: Calculate Properties (a) Temperature Since the mixture is saturated at 400 kPa , 𝑇=𝑇𝑠𝑎𝑡 =143.6∘C (b) Specific Volume (v) 𝑣=𝑣𝑓+𝑥(𝑣𝑔−𝑣𝑓) 𝑣=0.001093+0.6(0.4624−0.001093) 𝑣=0.001093+0.6(0.461307) 𝑣=0.001093+0.2767842=0.2779 m3/kg (c) Enthalpy (h) ℎ=ℎ𝑓+𝑥(ℎ𝑔−ℎ𝑓) ℎ=604.7+0.6(2556.3−604.7) ℎ=604.7+0.6(1951.6) ℎ=604.7+1170.96=1775.66 kJ/kg (d) Entropy (s) 𝑠=𝑠𝑓+𝑥(𝑠𝑔−𝑠𝑓) 𝑠=1.733+0.6(7.127−1.733) 𝑠=1.733+0.6(5.394) 𝑠=1.733+3.236=4.969 kJ/kg.K (e) Total Volume (V) 𝑉=𝑚×𝑣=2×0.2779=0.5558 m3 Thermodynamics 293 Problem 4.2: A sample of steam has the following measured properties: Pressure =1MPa and Temperature =250∘C. Determine: 1. The phase or region in which the steam exists (saturated, superheated, or compressed liquid). 2. The specific enthalpy (h), specific volume (v), and entropy (s) of the steam. Solution: Given Data: 𝑃=1MPa,𝑇=250∘𝐶 Step 1: Determine Saturation Temperature at 1 MPa From the saturated steam table, At =1MPa : 𝑇sat =179.9∘C Since the actual temperature 𝑇=250∘𝐶 is greater than the saturation temperature, the state of the substance is superheated vapor. Step 2: Refer to Superheated Steam Tables ( 𝐏=𝟏 𝐌𝐏𝐚 ) From the superheated steam table at 1 MPa: Temperature ( ∘𝐂 ) Specific Volume (𝐦𝟑/𝐤𝐠 ) Enthalpy (kJ/kg) Entropy ( 𝐤𝐉/𝐤𝐠⋅𝐊 ) 200 0.2060 2875.4 7.127 250 0.2138 2943.0 7.245 300 0.2218 3012.4 7.359 Step 3: Read Off the Properties At 𝑃=1MPa and 𝑇=250∘𝐶, the properties are directly available: 𝑣=0.2138 m3/kg ℎ=2943.0 kJ/kg 𝑠=7.245 kJ/kg.K Thermodynamics 294 Exercise 4.1: A rigid container of volume 𝟎.𝟒𝐦𝟑 contains 𝟏.𝟓 𝐤𝐠 of water at a pressure of 𝐤𝐏𝐚. Determine: 1. The temperature of the mixture 2. The quality (dryness fraction) of steam 3. The enthalpy and entropy of the mixture 4. The phase condition (saturated, subcooled, or superheated) of the substance Exercise 4.2: Steam enters a turbine at a pressure of 𝟏.𝟓 𝐌𝐏𝐚 and a temperature of 𝟑𝟓𝟎∘. Determine: 1. The specific enthalpy, specific volume, and entropy of the steam at this state 2. The degree of superheat above the saturation temperature 3. Sketch the state on a T-s diagram indicating its region 4.2 Phase Rule The Phase Rule is a fundamental principle in thermodynamics and physical chemistry that establishes a relationship between the number of phases, components, and degrees of freedom (independent variables) in a system at equilibrium. It is primarily used to analyze multiphase systems, such as mixtures of solids, liquids, and vapors, and to understand how temperature, pressure, and composition affect equilibrium conditions. This rule was first developed by Josiah Willard Gibbs in 1876 and is often referred to as Gibbs’ Phase Rule. In a thermodynamic system, several phases can coexist in equilibrium for example, ice, liquid water, and water vapor can all exist together under specific conditions. Each phase has its own intensive properties such as temperature, pressure, and composition. The Phase Rule provides a mathematical relationship that tells us how many independent intensive variables can be changed without disturbing the number of phases in equilibrium. Statement of the Phase Rule The Gibbs Phase Rule is expressed as: 𝐹=𝐶−𝑃+2 Thermodynamics 295 Where,  𝐹= Number of degrees of freedom (or variance)  𝐶= Number of components  𝑃= Number of phases in equilibrium (a) Components (C) A component is the minimum number of independent chemical constituents required to describe the composition of all the phases present in the system. It represents the distinct substances that can combine or separate to form all the existing phases. For example:  In a system of ice, water, and water vapor, there is only one component (H2O) because all three phases are chemically identical.  In a system containing CaCO3 (solid), CaO (solid), and CO2 (gas), there are two components because chemical reactions relate these substances (CaCO3⇋CaO+ CO2). (b) Phases (P) A phase is a homogeneous and physically distinct part of a system that is mechanically separable from other parts of the system. Each phase is uniform in composition, temperature, and pressure. Phases are separated by definite boundaries called interfaces. Examples:  Ice, liquid water, and steam are three distinct phases of a one-component system ( H2O ).  Liquid water and oil form two immiscible liquid phases. (c) Degrees of Freedom (F) The degrees of freedom represent the number of independent intensive variables (such as temperature, pressure, or concentration) that can be varied without changing the number of phases in equilibrium. It indicates how many conditions can be independently changed while maintaining phase equilibrium.  If 𝐹=0, the system is invariant (no freedom to change conditions). Thermodynamics 296  If 𝐹=1, the system is univariant (one variable can change independently).  If 𝐹=2, the system is bivariant (two variables can change independently). Derivation of Gibbs Phase Rule Let us derive the Phase Rule equation, 𝐹=𝐶−𝑃+2, in a logical way. Step 1: Define Variables Each phase in a system has:  Temperature (T)  Pressure (P)  Composition (expressed by the mole fractions of components) For a system with C components and P phases: Each phase has C mole fractions, but since their total must be 1, only ( C−1 ) are independent. Hence, the total number of variables for all phases =𝑃(𝐶−1)+2. (The extra " +2 " corresponds to T and P , which are common to all phases in equilibrium.) Step 2: Apply Equilibrium Conditions For equilibrium, the temperature and pressure must be the same in all phases, and the chemical potential of each component must be the same in all phases. For each component, there are (𝑃−1) equilibrium relations because chemical potential must be equal in all P phases. Hence, total number of equilibrium relations =𝐶(𝑃−1). Step 3: Calculate Degrees of Freedom (F) Now, 𝐹= (Number of variables) − (Number of equilibrium relations) 𝐹=[𝑃(𝐶−1)+2]−[𝐶(𝑃−1)] 𝐹=𝑃𝐶−𝑃+2−𝐶𝑃+𝐶 𝐹=𝐶−𝑃+2 Thus, the Phase Rule is: 𝐹=𝐶−𝑃+2 Thermodynamics 297 This equation gives the number of degrees of freedom (F) for any system in equilibrium based on the number of components ( C ) and phases ( P ). It tells us how many intensive variables (temperature, pressure, composition) can be changed independently without altering the number of phases. Applications of Phase Rule Let us apply the Phase Rule to some common thermodynamic systems to understand its practical meaning. (a) One-Component System ( C=1 ) Case 1: One Phase ( 𝐏=1 ) 𝐹=1−1+2=2 Hence, 𝐅=2→ Bivariant system. Example: Pure liquid water. Here, both pressure and temperature can be varied independently. Case 2: Two Phases ( P=2 ) 𝐹=1−2+2=1 Hence, 𝐅=𝟏→ Univariant system. Example: Water coexisting with steam (liquid-vapor equilibrium). If the pressure is fixed, the temperature is automatically determined (and vice versa). Case 3: Three Phases ( P=3 ) 𝐹=1−3+2=0 Hence, 𝐅=0→ Invariant system. Example: Triple Point of water, where ice, liquid water, and vapor coexist. At this point, neither temperature nor pressure can be changed without disturbing equilibrium. (b) Two-Component System ( C=2 ) For example, a water-salt ( NaCl−H2O ) mixture:  If one phase (liquid solution) exists: 𝐹=2−1+2=3 Thermodynamics 298 Three degrees of freedom (temperature, pressure, and concentration can be varied independently).  If two phases (solution + solid salt) coexist: 𝐹=2−2+2=2 Two degrees of freedom remain.  If three phases (solution + ice + salt) coexist: 𝐹=2−3+2=1 One degree of freedom. Importance and Applications The Phase Rule is significant because it: 1. Predicts the number of variables required to describe a system's state. 2. Helps identify equilibrium conditions among phases. 3. Guides experimental determination of phase diagrams. 4. Applies to metallurgy, geology, materials science, and chemical engineering - wherever phase equilibria occur (e.g., alloys, solutions, and crystallization processes). Limitations of the Phase Rule While the Gibbs Phase Rule is highly useful, it has certain limitations: 1. It assumes the system is in thermodynamic equilibrium. 2. It applies only to non-reactive systems (unless extended to include chemical reactions). 3. It does not specify the nature of the phases - only the number. 4. It neglects external fields (e.g., electric, magnetic, or gravitational). Problem 4.3: At a certain condition, ice, liquid water, and water vapor coexist in equilibrium. Using Gibbs' Phase Rule, determine: 1. The number of components ( 𝐶 ), 2. The number of phases (P), 3. The degrees of freedom (F), and 4. The nature of the system (invariant, univariant, or bivariant). Thermodynamics 299 Also, explain physically what this condition represents. Solution: Given Data System: Water (H2O) Phases present: Ice (solid), liquid water, and water vapor. Step 1: Identify the Number of Components ( C ) In this system, all phases - solid, liquid, and vapor - consist of the same chemical substance (H2O). Hence, 𝐶=1 Step 2: Identify the Number of Phases (P) The system contains three distinct phases: 1. Solid (Ice) 2. Liquid (Water) 3. Vapor (Steam) Thus, 𝑃=3 Step 3: Apply Gibbs' Phase Rule The general form of the Phase Rule is: 𝐹=𝐶−𝑃+2 Substitute the values: 𝐹=1−3+2=0 The system has 𝐅=0, which means zero degrees of freedom. Therefore, it is an invariant system - no independent variable (temperature or pressure) can be changed without disturbing the equilibrium among the three phases. This condition is unique and fixed. The state described corresponds to the Triple Point of Water, where the three phases - ice, liquid water, and vapor - coexist in equilibrium. Thermodynamics 300 At this condition: 𝑇=0.01∘𝐶,𝑃=0.6117kPa If either temperature or pressure is altered, one of the phases will disappear, and the equilibrium will be destroyed. Problem 4.4: Consider the equilibrium system: CaCO3(𝑠)⇋CaO( s)+CO2(𝑔) Determine: 1. The number of components (C), 2. The number of phases (P), 3. The degrees of freedom (F), and describe the type of system using Gibbs' Phase Rule. Solution: Step 1: Identify the Phases Present In this reaction:  Solid phase 1: Calcium carbonate ( CaCO3 )  Solid phase 2: Calcium oxide (CaO)  Gaseous phase: Carbon dioxide (CO2) Thus, the system contains: 𝑃=3 Step 2: Identify the Number of Components (C) To determine components, we must consider the chemical relationship among the substances. Here, one chemical equation relates the species: CaCO3⇋CaO+CO2 Since the system has 3 species but 1 independent chemical reaction, the number of components is given by: 𝐶=( number of species )−( number of independent reactions ) 𝐶=3−1=2 Thermodynamics 307 3. The temperature, enthalpy, and entropy of the substance. Solution: Given Data 𝑃=200kPa,𝑣=0.885 m3/kg,𝑚=2 kg Step 1: Locate the Given Pressure on Steam Tables At =𝟐𝟎𝟎 𝐤𝐏𝐚, from the saturated water-steam table, the following properties are obtained: Property Symbol Saturated Liquid (f) Saturated Vapor (g) Temperature ( ∘C ) Ts 120.2 - Specific volume ( m3/kg ) v 0.001061 0.8858 Enthalpy (kJ/kg) h 504.5 2706.7 Entropy ( kJ/kg⋅K ) s 1.5302 7.1271 Step 2: Compare Given Specific Volume with Saturated Values Given 𝑣=0.885 m3/kg At 200 kPa : 𝑣𝑓=0.001061,𝑣𝑔=0.8858 Now, compare: 𝑣𝑓<𝑣<𝑣𝑔 0.001061<0.885<0.8858 Since the given specific volume lies between the saturated liquid and saturated vapor values, the substance exists in the two-phase (liquid-vapor mixture) region. Step 3: Determine the Quality ( x ) For a mixture: 𝑣=𝑣𝑓+𝑥(𝑣𝑔−𝑣𝑓) Thermodynamics 308 Rearranging: 𝑥= 𝑣−𝑣𝑓 𝑣𝑔−𝑣𝑓 Substitute: 𝑥= 0.885−0.001061 0.8858−0.001061 𝑥=0.883939 0.884739=0.9989 Hence, 𝑥=0.999 (approximately) The quality (dryness fraction) is 99.9%, which means the water is almost dry saturated vapor. Step 4: Determine the Temperature For a mixture at a given pressure, the temperature equals the saturation temperature at that pressure. 𝑇=𝑇sat =120.2∘C Step 5: Determine the Enthalpy (h) The enthalpy of the mixture is: ℎ=ℎ𝑓+𝑥(ℎ𝑔−ℎ𝑓) Substitute the values: ℎ=504.5+0.9989(2706.7−504.5) ℎ=504.5+0.9989(2202.2) ℎ=504.5+2199.8=2704.3 kJ/kg Step 6: Determine the Entropy (s) 𝑠=𝑠𝑓+𝑥(𝑠𝑔−𝑠𝑓) Substitute: 𝑠=1.5302+0.9989(7.1271−1.5302) Thermodynamics 309 𝑠=1.5302+0.9989(5.5969) 𝑠=1.5302+5.5907=7.1209 kJ/kg.K Step 7: Determine the Total Volume 𝑉=𝑚×𝑣=2×0.885=1.77 m3  At 𝐤𝐏𝐚, the saturation dome (bounded by the saturated liquid and vapor lines) defines the twophase region.  Since the calculated specific volume (v=0.885 m3/kg) lies almost equal to 𝑣𝑔= 0.8858 m3/kg, the state lies very close to the right-hand boundary of the dome, representing dry saturated vapor.  On the p-v diagram, this point would be located just inside or at the saturated vapor line, indicating that the substance is nearly completely vaporized. Exercise 4.5: A closed vessel contains 1.8 kg of water at a pressure of 300 kPa and a specific volume of 0.5 m3/kg. Determine the following: 1. Identify the phase region in which the substance exists (compressed liquid, mixture, saturated vapor, or superheated vapor). 2. If it is in the two-phase region, determine the dryness fraction (x). 3. Estimate the enthalpy and entropy of the mixture using appropriate relations. 4. Indicate the location of this state on a 𝐩−𝐯 diagram, and describe qualitatively how the state would move if the substance is heated at constant pressure. Thermodynamics 310 Exercise 4.6: At a pressure of 20 MPa , the temperature of water is 360∘C. From the given data:  Critical temperature of water =374.15∘C  Critical pressure of water =22.12MPa Determine: 1. Determine the phase or region of water at these conditions (compressed liquid, twophase, or supercritical fluid). 2. Explain what happens if the temperature is slightly increased while keeping the pressure constant. 3. On a qualitative p-v diagram, indicate where this state lies with respect to the critical point. 4. State one engineering application where supercritical fluids are used. 4.4 p–T and T–v Diagrams 1. p–T Diagram The p–T diagram (pressure–temperature diagram) shows the equilibrium states of a substance when pressure and temperature vary. It is divided into distinct regions corresponding to solid, liquid, and vapor phases, separated by characteristic curves (Figure 4.3). 1. Sublimation Curve: Separates the solid and vapor regions. Along this line, the solid phase changes directly into vapor a process called sublimation. 2. Fusion Curve: Represents equilibrium between solid and liquid phases.  For most substances, it slopes upward, indicating contraction during freezing.  For water, it slopes backward, showing expansion upon freezing. 3. Vaporization Curve: Separates the liquid and vapor regions. Boiling occurs along this curve, which terminates at the critical point. 4. Triple Point: The intersection of the three curves sublimation, fusion, and vaporization where solid, liquid, and vapor coexist in equilibrium. Thermodynamics 311 For water: 1. Temperature = 273.16 K (0.01°C) 2. Pressure = 0.6113 kPa Fig. 4.3 p-T diagram for pure substance. 5. Critical Point: The endpoint of the vaporization curve, beyond which the liquid and vapor phases become indistinguishable, forming a supercritical fluid. 2. T–v Diagram The T–v diagram (temperature–specific volume diagram) illustrates how the specific volume of a pure substance changes with temperature at constant pressures. It resembles the p–v diagram but includes isobars (constant-pressure lines), as shown in Figure 4.4. Regions in the T–v Diagram 1. Compressed (Subcooled) Liquid Region: Left of the saturated liquid line, the substance exists as a liquid below its saturation temperature. Thermodynamics 312 2. Wet Region (Liquid–Vapor Mixture): Between the saturated liquid and saturated vapor lines, the substance is a mixture of liquid and vapor. The dryness fraction (x) defines the proportion of vapor. 3. Superheated Vapor Region: Right of the saturated vapor line, the vapor exists above its saturation temperature the superheated region used extensively in power and refrigeration cycles. Fig. 4.4 T-v diagram for pure substance. Critical Point and Supercritical Fluid As pressure increases, the saturated liquid and vapor lines converge at the critical point, where both phases merge into a single homogeneous state the supercritical fluid. Above this point, there is no distinct phase boundary between liquid and vapor. Significance of p-T and T-v Diagrams  The p–T diagram helps in understanding phase equilibrium, the triple point, and critical conditions. Thermodynamics 313  The T–v diagram aids in analyzing heating, vaporization, and condensation processes at constant pressures.  Together, these diagrams are fundamental for designing boilers, condensers, turbines, and for interpreting thermodynamic cycles in power and refrigeration systems. Problem 4.6: A pure substance (water) exists at a temperature of 120∘C and a pressure of 200 kPa. Using the 𝐩−𝐓 and 𝐓−𝐯 diagrams, determine: 1. The phase region in which the substance exists (compressed liquid, saturated mixture, or superheated vapor). 2. The saturation temperature and saturation pressure corresponding to these conditions. 3. Explain the state of the substance on both the p−T and T−v diagrams. Solution: Given Data 𝑇=120∘𝐶,𝑃=200kPa Step 1: Obtain Saturation Properties at 200 kPa From the saturated water-steam tables: Property Symbol Saturation Value Saturation Temperature 𝑇sat 120.2∘C Saturation Pressure 𝑃sat 200 kPa Step 2: Compare Given Conditions with Saturation Data Given: 𝑇=120∘𝐶,𝑃=200kPa and from tables: 𝑇sat =120.2∘𝐶,𝑃sat =200kPa Thermodynamics 314 Now compare:  The given temperature (120∘C)≈ saturation temperature (120.2∘C).  The given pressure (200kPa)= saturation pressure. Hence, the system lies on the saturation line. This means the water exists at saturation conditions - i.e., it is about to vaporize or condense, depending on the direction of heat transfer. Step 3: Identify the Phase Region Since the given condition matches the saturation values:  The substance is neither a subcooled liquid nor a superheated vapor.  It exists as a mixture of liquid and vapor in equilibrium. Therefore, the system lies in the two-phase (wet region). Step 4: Representing the State on the Diagrams (a) On the 𝒑−𝑻 Diagram The point lies exactly on the vaporization curve that separates the liquid and vapor regions. At this line, liquid and vapor coexist in equilibrium at 𝑇sat =120.2∘𝐶 and 𝑃sat =200kPa. Thermodynamics 315 Any small increase in temperature (at constant pressure) will cause complete vaporization, moving the point into the superheated vapor region. (b) On the T-v Diagram The state point lies within the dome-shaped region bounded by the saturated liquid line and the saturated vapor line. The exact position within the dome depends on the dryness fraction ( 𝐱 ), which defines how much of the mixture is vapor.  If =0 : Saturated liquid.  If =1 : Saturated vapor.  If 0<𝑥<1 : Mixture of both phases. Step 5: Interpretation At 𝑃=200kPa and 𝑇=120∘𝐶, water exists in a state of phase equilibrium.  If heat is added, the mixture will gradually convert into vapor (boiling).  If heat is removed, vapor will condense into liquid. Thus, the point corresponds to the liquid-vapor equilibrium line on the 𝐩−𝐓 diagram and the two-phase dome on the T-v diagram. Exercise 4.7: A sample of water exists at a pressure of 400 kPa and a temperature of 200∘C. Using the 𝐩−𝐓 and 𝐓−𝐯 diagrams, determine: 1. The phase region (compressed liquid, mixture, superheated vapor, or supercritical fluid). 2. The saturation temperature and saturation pressure corresponding to these conditions. 3. Sketch the approximate location of the state on both the 𝑝−𝑇 and 𝑇−𝑣 diagrams. 4. Explain what happens if the temperature is slightly increased while keeping the pressure constant. Exercise 4.8: For a pure substance, the following data are known:  Triple point temperature =0.01∘C, pressure =0.6113kPa  Critical temperature =374.15∘C, pressure =22.12MPa Answer the following: 1. Explain the physical significance of the triple point and the critical point on the 𝐩−𝐓 diagram. 2. Identify what happens to the phase of the substance if: (a) Pressure is reduced below the triple point pressure. Thermodynamics 316 (b) Temperature is raised above the critical temperature. 3. Sketch the 𝐩−𝐓 diagram, clearly marking the sublimation, fusion, and vaporization curves along with the triple point and critical point. 4.5 T-s Diagram The T–s diagram is an essential thermodynamic chart that represents the phase-change behavior of pure substances, particularly water and steam. In this diagram, temperature (T) is plotted on the vertical axis and entropy (s) on the horizontal axis. It is especially useful for analyzing steam power cycles such as the Rankine, reheat, and regenerative cycles, as it provides a direct visual understanding of thermodynamic processes. Fig. 4.5 T-s diagram for water. Thermodynamics 323 Step 1: Obtain Properties at Inlet (State 1) From superheated steam tables at 𝑃1=3MPa and 𝑇1=350∘C : ℎ1=3115.3 kJ/kg,𝑠1=6.743 kJ/kg.K Step 2: Determine the Final State (State 2) The expansion is isentropic, so: 𝑠2=𝑠1=6.743 kJ/kg.K At 𝑃2=0.1MPa, from saturated steam tables: 𝑠𝑓=1.303,𝑠𝑔=7.355,ℎ𝑓=417.5,ℎ𝑔=2675.4 Step 3: Check Whether Steam is Wet or Superheated Since 𝑠2=6.743 lies between 𝑠𝑓 and 𝑠𝑔 : 𝑠𝑓<𝑠2<𝑠𝑔⇒ Wet region Thus, the steam at the exit is wet. Step 4: Find the Dryness Fraction ( 𝐱𝟐 ) 𝑥2=𝑠2−𝑠𝑓 𝑠𝑔−𝑠𝑓 𝑥2=6.743−1.303 7.355−1.303=5.44 6.052=0.899 𝑥2=0.899 or 89.9% Step 5: Find Enthalpy at Exit ( h2 ) ℎ2=ℎ𝑓+𝑥2(ℎ𝑔−ℎ𝑓) ℎ2=417.5+0.899(2675.4−417.5) ℎ2=417.5+0.899(2257.9) ℎ2=417.5+2030.3=2447.8 kJ/kg Step 6: Determine Turbine Work Output For an isentropic expansion, the work done per kg of steam is equal to the enthalpy drop: 𝑊=ℎ1−ℎ2 𝑊=3115.3−2447.8=667.5 kJ/kg Thermodynamics 324 Exercise 4.11: Steam enters a turbine at a pressure of 2.5 MPa and a temperature of 400°C. It expands isentropically to a pressure of 0.15 MPa. Determine: 1. The enthalpy at inlet and outlet of the turbine using the Mollier chart or steam tables. 2. The dryness fraction of the steam at the exit. 3. The work output (per kg of steam). 4. Represent this process on a h–s diagram, clearly showing the isentropic line and phase regions. Exercise 4.12: Steam at a pressure of 1.2 MPa and a temperature of 250°C is throttled to a pressure of 0.2 MPa. Assuming the process is isenthalpic (constant enthalpy), determine: 1. The enthalpy before and after throttling. 2. The final state of the steam (wet, saturated, or superheated). 3. The dryness fraction of the steam after throttling, if in the wet region. 4. Indicate the process on the Mollier chart, labeling the constant-enthalpy (horizontal) line. 4.7 p–v–T Surface In thermodynamics, the behavior of a pure substance can be completely described by the three fundamental properties pressure (p), specific volume (v), and temperature (T). When these three variables are plotted in a three-dimensional coordinate system, the resulting representation is known as the p–v–T surface. In this plot:  Specific volume (v) and temperature (T) form the horizontal plane (independent variables).  Pressure (p) is represented on the vertical axis (dependent variable). Each point on this surface corresponds to a unique equilibrium state of the substance. Characteristics of the p–v–T Surface 1. Single-Phase Regions: Solid, liquid, and vapor phases appear as smooth, continuous curved surfaces. Thermodynamics 325 2. Two-Phase Regions: Areas where two phases coexist (solid–liquid, liquid–vapor, solid– vapor) form surfaces perpendicular to the p–T plane, representing saturation conditions. 3. Triple Point Line: The line where solid, liquid, and vapor phases coexist in equilibrium.  For water: T=273.16 K, P=0.6113 kPa.  Along this line, temperature and pressure remain constant while volume changes. 4. Critical Point: Located at the end of the saturated liquid and vapor surfaces. Beyond this point, there is no distinction between liquid and vapor phases the substance becomes supercritical. Substances Expanding and Contracting on Freezing 1. Water (Expands on Freezing): Due to its open crystalline ice structure, water expands on freezing. The solid–liquid surface slopes backward. 2. CO₂ (Contracts on Freezing): Most substances, such as carbon dioxide, contract upon freezing. Their solid–liquid surface slopes forward. (a) 2-D Diagram. Thermodynamics 326 (b) 3-D Diagram. Fig. 4.7 p-v-T surface for water and CO2. Relation to Two-Dimensional Diagrams The p–v–T surface is the complete thermodynamic representation, while commonly used 2D diagrams are its projections: Thermodynamics 327 Diagram Projection Plane Representation p–v Diagram p–v plane Pressure vs. Specific Volume p–T Diagram p–T plane Pressure vs. Temperature T–v Diagram T–v plane Temperature vs. Specific Volume Thus, each 2D diagram provides a simplified view of the full 3D surface. Significance  Provides a complete thermodynamic map of a pure substance.  Helps visualize phase changes, the triple point, and the critical point.  Clarifies relationships between pressure, volume, and temperature.  Though comprehensive, it is used mainly for theoretical understanding; engineers prefer 2D projections or steam tables for practical analysis. Problem 4.9: A sample of water exists at a pressure of 0.1 MPa and a temperature of 120∘C. Using the concept of the p−v−T surface, determine: 1. The phase region in which the water exists. 2. The saturation temperature corresponding to the given pressure. 3. The location of this state on the 𝑝−𝑣−𝑇 surface. 4. Explain what happens to the state of the water if: (a) Temperature is increased at constant pressure. (b) Pressure is increased at constant temperature. Solution: Given Data 𝑃=0.1MPa,𝑇=120∘C Step 1: Obtain Saturation Data from Steam Tables At 𝑃=0.1MPa(100kPa), from the saturated steam table: 𝑇𝑠𝑎𝑡 =99.6∘C Thermodynamics 328 Step 2: Compare Given Temperature with Saturation Temperature Given 𝑇=120∘𝐶 and 𝑇sat =99.6∘C 𝑇>𝑇sat Hence, the water is above the saturation temperature for the given pressure. Step 3: Identify the Phase Region When 𝑇>𝑇sat at a fixed pressure, the substance exists entirely as superheated vapor. Therefore, the state lies in the single-phase vapor region of the p−v−T surface, above the vaporization dome. Step 4: Locate the State on the p−v−T Surface At 0.1 MPa , the liquid-vapor equilibrium (saturation line) occurs at 99.6∘C. Since the actual temperature is higher (120∘C), the state lies on the vapor surface above the two-phase dome in the superheated vapor region. If plotted on the 3D surface:  The temperature axis moves forward (increasing T),  The pressure remains constant, and Thermodynamics 329  The point lies above the vaporization surface on the p−v−T diagram. Step 5: Analyze the Changes (a) If Temperature Increases at Constant Pressure: The state moves further into the superheated region, increasing the specific volume. (b) If Pressure Increases at Constant Temperature: The state moves toward the saturation surface. If pressure exceeds the saturation pressure, the vapor condenses into liquid - the point crosses into the liquid region. Exercise 4.13: A sample of water is observed at a pressure of 0.5 MPa and a temperature of 150°C. Using the concept of the p–v–T surface and standard steam data, determine: 1. The phase region in which the substance lies (liquid, mixture, or vapor). 2. The saturation temperature corresponding to 0.5 MPa and compare it with the given temperature. 3. The direction of change on the p–v–T surface if: (a) The pressure is increased at constant temperature. (b) The temperature is increased at constant pressure. 4. Sketch the approximate position of this state on the p–v–T surface diagram. Exercise 4.14: A pure substance exhibits the following properties:  Triple point: 0.01°C and 0.6113 kPa  Critical point: 374.15°C and 22.12 MPa Answer the following: 1. Explain the significance of the triple point line and critical point on the p–v–T surface. 2. What happens to the phase behavior if the temperature is below the triple point value? 3. Describe how the saturation dome changes as the pressure increases toward the critical point. 4. Sketch the p–v–T surface showing the triple point line and critical point for this substance. Thermodynamics 330 4.8 Thermodynamic Properties of Steam Steam, or water vapor, is one of the most commonly used working fluids in thermodynamic systems such as steam turbines, boilers, condensers, and power plants. Its behavior and performance depend on several thermodynamic properties, which describe its condition and energy content under different states of temperature and pressure. Understanding these properties is essential for analyzing energy conversion processes and for using steam tables and thermodynamic charts effectively. 1. Pressure (p) Pressure is the normal force exerted by the steam molecules per unit area of the container walls. It determines the boiling temperature of water; higher pressure increases the boiling point. Measured in kPa, bar, or MPa. Example: At 100 kPa , water boils at 100∘C, while at 700 kPa , it boils at about 165∘C. 2. Temperature (T) Temperature is the measure of the thermal state of the steam, indicating how hot or cold it is. It defines the energy level of the molecules. Measured in ∘C,K, or ∘F. It can exist as:  Saturation Temperature ( Ts ): Temperature at which water starts to boil for a given pressure.  Superheated Temperature: Temperature above the saturation temperature at the same pressure. 3. Specific Volume (v) Specific volume is the volume occupied by one kilogram of steam, expressed as: 𝑣=𝑉 𝑚 ( m3/kg) It varies with temperature and pressure:  Increases greatly during vaporization.  Steam occupies a much larger volume than liquid water at the same temperature and pressure. Thermodynamics 331 4. Enthalpy (h) Enthalpy represents the total heat content or total energy of steam per unit mass. ℎ=𝑢+𝑝𝑣 where,  𝑢= internal energy,  𝑝𝑣= flow energy (work of pushing fluid). Measured in kJ/kg. Steam enthalpy includes:  hf: Enthalpy of saturated liquid (water).  hfg: Latent heat of vaporization.  hg: Enthalpy of saturated vapor =ℎ𝑓+ℎ𝑓𝑔. This property is crucial for evaluating heat addition, work output, and efficiency in power cycles. 5. Internal Energy (u) Internal energy is the energy stored within the substance due to molecular motion and interactions. For steam: 𝑢=ℎ−𝑝𝑣 It represents the energy change during heating, evaporation, or condensation at constant volume. Measured in kJ/kg. 6. Entropy (s) Entropy is a measure of the degree of disorder or the energy unavailable for work in a system. Δ𝑆=𝑄𝑟𝑒𝑣 𝑇  Measured in 𝐤𝐉/𝐤𝐠⋅𝐊.  Increases during vaporization since molecules become more disordered.  Used for analyzing reversible and adiabatic (isentropic) processes, especially in turbines and compressors. Thermodynamics 332 7. Quality or Dryness Fraction (x) In the wet region, steam exists as a mixture of liquid water and vapor. The dryness fraction (x) indicates the mass fraction of vapor in the mixture: 𝑥= 𝑚vapor 𝑚liquid +𝑚vapor  𝑥=0 : Saturated liquid (no vapor)  𝑥=1 : Dry saturated vapor  0<𝑥<1 : Wet steam mixture Thermodynamic properties in the wet region are calculated as: 𝑦=𝑦𝑓+𝑥(𝑦𝑔−𝑦𝑓) Where, 𝑦 can be ℎ,𝑣,𝑢, or 𝑠. 8. Specific Heat (Cp and Cv) Cp (Specific Heat at Constant Pressure): Heat required to raise the temperature of 1 kg of steam by 1∘C at constant pressure. Cv (Specific Heat at Constant Volume): Heat required to raise the temperature of 1 kg of steam by 1∘C at constant volume. In superheated steam: ℎ2−ℎ1=𝐶𝑝(𝑇2−𝑇1) 9. Specific Weight (w) Specific weight is the weight per unit volume of steam: 𝑤=1 𝑣 Measured in kg/m3, it is inversely proportional to the specific volume. 10. Latent Heat (hfg) Latent heat is the amount of heat energy required to convert saturated liquid into saturated vapor without a temperature change at constant pressure. ℎ𝑓𝑔 =ℎ𝑔−ℎ𝑓 Thermodynamics 339 From the First Law: 𝑄=𝑊+Δ𝑈 𝑄=𝑝1𝑣1−𝑝2𝑣2 𝑛−1 +(𝑢2−𝑢1) 𝑄= 𝑛 𝑛−1(𝑝1𝑣1−𝑝2𝑣2)+(ℎ2−ℎ1) In steam systems:  𝑛=1.13 (for wet steam)  𝑛=1.3 (for superheated steam) This process represents irreversible adiabatic or practical expansion in steam turbines. (vii) Throttling Process In throttling, the steam expands through a restriction such as a valve or orifice, resulting in a pressure drop with no work interaction and no heat exchange. Thus, the enthalpy remains constant: ℎ1=ℎ2 If both states are within the wet region: ℎ𝑓1+𝑥1ℎ𝑓𝑔1=ℎ𝑓2+𝑥2ℎ𝑓𝑔2 If the exit state is superheated: ℎ𝑓1+𝑥1ℎ𝑓𝑔1=ℎ𝑔2+𝐶𝑝(𝑇𝑠𝑢𝑝2−𝑇𝑠2) This process is isenthalpic and forms the basis for calorimeter experiments and refrigeration throttling valves. Problem 4.10: A quantity of steam initially at a pressure of 1 MPa and dryness fraction (𝐱𝟏)=0.9 is expanded or heated through several non-flow processes. Calculate the work done (W) and heat transfer (Q) for each of the following processes, assuming 1 kg of steam: 1. Constant Volume Process 2. Constant Pressure Process 3. Constant Temperature Process 4. Hyperbolic Process ( pv= constant) 5. Reversible Adiabatic Process Thermodynamics 340 6. Polytropic Process ( n=1.2 ) 7. Throttling Process Steam data is to be taken from standard tables at 1 MPa and 0.1 MPa where applicable. Solution: Given Data Property Symbol Value Pressure at state 1 𝑃1 1 MPa Dryness fraction 𝑥1 0.9 Mass of steam 𝑚 1 kg From steam tables at 𝑃1=1MPa : Symbol Value ℎ𝑓 761.7 kJ/kg ℎ𝑓𝑔 2015.3 kJ/kg 𝑣𝑓 0.001127 m3/kg 𝑣𝑔 0.1944 m3/kg Hence: 𝑣1=𝑣𝑓+𝑥1(𝑣𝑔−𝑣𝑓)=0.001127+0.9(0.1944−0.001127) 𝑣1=0.1751 m3/kg ℎ1=ℎ𝑓+𝑥1ℎ𝑓𝑔 =761.7+0.9(2015.3)=2576.5 kJ/kg (i) Constant Volume Process Steam at 1 MPa and 𝑥1=0.9 is heated at constant volume until the pressure rises to 1.6 MPa. At 𝑃2=1.6MPa: From tables: 𝑣𝑓=0.00113 m3/kg,𝑣𝑔=0.1238 m3/kg,ℎ𝑓=798.5,ℎ𝑓𝑔 =1890.7 Thermodynamics 341 Since 𝑣1=𝑣2 : 𝑥2=𝑣2−𝑣𝑓 𝑣𝑔−𝑣𝑓=0.1751−0.00113 0.1238−0.00113=1.42 Since 𝑥2>1, the steam is superheated. From superheated table at =1.6MPa,𝑇=300∘𝐶 : ℎ2=3050 kJ/kg,𝑣2=0.145 m3/kg Work done: 𝑊=0 Heat transfer: 𝑄=Δ𝑈=(ℎ2−ℎ1)−(𝑝2𝑣2−𝑝1𝑣1) 𝑄=(3050−2576.5)−(1.6×103×0.145−1.0×103×0.1751) 𝑄=473.5−(232−175.1)=416.6 kJ/kg (ii) Constant Pressure Process The same steam ( 1MPa,x1=0.9 ) is heated at constant pressure until it becomes superheated at T2=250∘C. From superheated tables: At 1MPa,250∘C→ℎ2=2945.3 kJ/kg,𝑣2=0.2579 m3/kg Work done: 𝑊=𝑝(𝑣2−𝑣1)=1×103(0.2579−0.1751)=82.8 kJ/kg Heat transfer: 𝑄=ℎ2−ℎ1=2945.3−2576.5=368.8 kJ/kg (iii) Constant Temperature Process At 1 MPa , saturation temperature =179.9∘C. Steam expands isothermally (and hence at constant pressure in wet region) until it becomes dry saturated. Final enthalpy: ℎ2=ℎ𝑔=2778.1 kJ/kg Work done: 𝑊=𝑄=ℎ2−ℎ1=2778.1−2576.5=201.6 kJ/kg Thermodynamics 342 (iv) Hyperbolic Process ( = constant) Steam expands from 𝑃1=1MPa to 𝑃2=0.5MPa, obeying 𝑝𝑣=𝐶. 𝑝1𝑣1=𝑝2𝑣2⇒𝑣2=𝑝1𝑣1 𝑝2=1×0.1751 0.5 =0.3502 Work done: 𝑊=𝑝1𝑣1ln (𝑣2 𝑣1) 𝑊=1000(0.1751)ln (0.3502 0.1751)=121.5 kJ/kg Since Δ𝑈≈0, 𝑄=𝑊=121.5 kJ/kg (v) Reversible Adiabatic (Isentropic) Process In this process: 𝑄=0 𝑊=−Δ𝑈=(ℎ1−ℎ2)−(𝑝1𝑣1−𝑝2𝑣2) Assume expansion from 𝑃1=1MPa to 𝑃2=0.1MPa. From steam tables at 0.1MPa:ℎ𝑔= 2675.4,𝑠𝑔=7.3546,𝑠𝑓=1.303. Since it is isentropic, 𝑠2=𝑠1=6.743 kJ/kg.K 𝑥2=𝑠2−𝑠𝑓 𝑠𝑔−𝑠𝑓=6.743−1.303 7.355−1.303=0.90 ℎ2=ℎ𝑓+𝑥2ℎ𝑓𝑔 =417.5+0.9(2257.9)=2449.6 kJ/kg 𝑊=(ℎ1−ℎ2)−(𝑝1𝑣1−𝑝2𝑣2) 𝑊=(2576.5−2449.6)−(1000×0.1751−100×0.1944) 𝑊=126.9−(175.1−19.4)=−28.8 kJ/kg Negative sign → work done by system =28.8 kJ/kg. Thermodynamics 343 (vi) Polytropic Process ( 𝐧=𝟏.𝟐 ) 𝑊=𝑝1𝑣1−𝑝2𝑣2 𝑛−1 𝑄= 𝑛 𝑛−1(𝑝1𝑣1−𝑝2𝑣2)+(ℎ2−ℎ1) Assuming expansion from 1 MPa to 0.1 MPa , with 𝑛=1.2. 𝑝1𝑣1=175.1,𝑝2𝑣2=19.4 𝑊=175.1−19.4 0.2 =778.5 kJ/kg 𝑄=1.2 0.2(175.1−19.4)+(2449.6−2576.5) 𝑄=6(155.7)−126.9=934.2−126.9=807.3 kJ/kg (vii) Throttling Process During throttling: ℎ1=ℎ2 If steam at 1MPa,𝑥1=0.9 is throttled to 0.1 MPa : ℎ1=ℎ2=2576.5 kJ/kg At 0.1 MPa : ℎ2=ℎ𝑓+𝑥2ℎ𝑓𝑔 ⇒2576.5=417.5+𝑥2(2257.9) 𝑥2=0.96 Thus, the steam becomes drier after throttling, and 𝑄=𝑊=0,ℎ1=ℎ2 Exercise 4.15: A mass of 1 kg of wet steam at 1.2 MPa and dryness fraction 0.8 is heated at constant pressure until it becomes superheated steam at 250∘C. Determine: 1. The work done during the process. 2. The heat supplied to the steam. 3. The increase in internal energy. 4. Represent this process on 𝐩−𝐯 and 𝐓−𝐬 diagrams. Thermodynamics 344 (Use appropriate values from the steam tables at 1.2 MPa and 250∘C.) Exercise 4.16: Steam initially at 1 MPa and 250∘C expands polytropically according to the law 𝑝𝑣1.3= constant until the pressure falls to 0.2 MPa . Determine: 1. The work done per kilogram of steam. 2. The change in internal energy. 3. The heat transfer during the process. 4. Represent this process on the 𝐩−𝐯 and 𝐓−𝐬 diagrams, clearly indicating the direction of change. (Take the required properties of superheated steam from tables and assume ideal gas behavior if necessary.) 4.10 Calculation of Work Done and Heat Transfer in Flow Processes Flow processes (also known as steady-flow processes) are those in which mass continuously enters and leaves a control volume, such as in turbines, compressors, boilers, condensers, and nozzles. In contrast to non-flow processes (where the system boundary is fixed), flow processes involve energy transfer due to both mass flow and work interactions. The analysis of these processes is based on the Steady Flow Energy Equation (SFEE) derived from the First Law of Thermodynamics for open systems. Steady Flow Energy Equation (SFEE) For a control volume operating under steady-state conditions: 𝑄˙−𝑊 ˙=𝑚˙[(ℎ2−ℎ1)+𝐶22−𝐶12 2+𝑔(𝑧2−𝑧1)] Where,  𝑄˙= Rate of heat transfer to the system ( kJ/kg⋅s )  𝑊 ˙= Rate of work done by the system (kJ/kg.s)  𝑚˙ = Mass flow rate ( kg/s )  ℎ= Specific enthalpy (kJ/kg) Thermodynamics 345  𝐶= Velocity ( m/s )  𝑧= Elevation (m)  𝑔= Acceleration due to gravity (9.81 m/s2) For per unit mass of fluid, the steady-flow energy equation becomes: 𝑄−𝑊=(ℎ2−ℎ1)+𝐶22−𝐶12 2+𝑔(𝑧2−𝑧1) This is the fundamental equation used to calculate work done and heat transfer in various steady-flow devices. Components of Energy in Flow Processes A flowing fluid possesses three forms of energy per unit mass: 1. Internal Energy (u): Energy due to molecular motion and bonding. 2. Flow Energy (pv): Energy required to push the mass into or out of the control volume. Flow Energy =𝑝×𝑣 (where, 𝑝= pressure, 𝑣= specific volume) 3. Kinetic and Potential Energy: Due to velocity and elevation: 𝐶2 2 and 𝑔𝑧 The total energy per unit mass of flowing fluid is: 𝐸=ℎ+𝐶2 2+𝑔𝑧 Steady Flow Energy Equation in Differential Form For infinitesimal changes: 𝛿𝑄−𝛿𝑊=𝑑ℎ+𝑑(𝐶2 2)+𝑔𝑑𝑧 In many engineering devices, changes in kinetic and potential energy are negligible compared to enthalpy change. Hence, 𝛿𝑄−𝛿𝑊≈𝑑ℎ Thermodynamics 346 Application of SFEE to Different Flow Devices Let us now apply the Steady Flow Energy Equation to different practical devices to calculate work done and heat transfer. (i) Boiler (Constant Pressure Heat Addition) Function: Converts water into steam by heat addition at constant pressure. For boilers: 𝑊 ˙=0 (no work done by the fluid) 𝑄=ℎ2−ℎ1 Hence, heat added = increase in enthalpy of steam. This process corresponds to the constant pressure heating of a fluid. (ii) Condenser (Constant Pressure Heat Rejection) Function: Condenses exhaust steam from the turbine into water by rejecting heat to the cooling water. At constant pressure: 𝑊 ˙=0 𝑄=ℎ2−ℎ1 Here, heat rejected = decrease in enthalpy of steam. (iii) Nozzle (Velocity Increase) Function: Converts the enthalpy of fluid into kinetic energy. Applying SFEE between inlet (1) and exit (2): 𝑄−𝑊=(ℎ2−ℎ1)+𝐶22−𝐶12 2 For an adiabatic nozzle, 𝑄=0 and =0 : ℎ1−ℎ2=𝐶22−𝐶12 2 Hence, drop in enthalpy = increase in kinetic energy. Thermodynamics 347 The exit velocity can be calculated as: 𝐶2=√2(ℎ1−ℎ2)×103 (Enthalpy in kJ/kg is converted to J/kg by multiplying by 1000.) (iv) Diffuser (Velocity Decrease) Function: Converts kinetic energy into pressure energy. For adiabatic diffuser: ℎ2−ℎ1=𝐶12−𝐶22 2 Hence, increase in enthalpy = decrease in kinetic energy. (v) Steam Turbine (Work-Producing Device) Function: Expands steam to produce mechanical work. For an adiabatic turbine: 𝑄=0 𝑊=ℎ1−ℎ2 Hence, turbine work done per kg of steam = enthalpy drop across the turbine. This is the most important flow process in power plants. (vi) Compressor or Pump (Work-Absorbing Device) Function: Compresses a gas or liquid by doing external work on the fluid. From SFEE: 𝑊=ℎ2−ℎ1 If the process is isentropic (adiabatic reversible): 𝑄=0 𝑊=ℎ2−ℎ1 This represents minimum work input. Thermodynamics 348 (vii) Throttling Process Function: Expansion through a restriction (e.g., valve or orifice) where pressure decreases abruptly and no heat or work is exchanged. 𝑄=0,𝑊=0 Hence, from SFEE: ℎ1=ℎ2 Enthalpy remains constant. This is called an isenthalpic process, used in calorimeter experiments and refrigeration systems. Neglecting Kinetic and Potential Energy Changes In most practical devices (turbines, boilers, etc.), 𝐶22−𝐶12 2,𝑔(𝑧2−𝑧1)≪(ℎ2−ℎ1) Therefore, they are neglected for simplicity, and SFEE reduces to: 𝑄−𝑊=ℎ2−ℎ1 Engineering Significance The Steady Flow Energy Equation is the foundation for analyzing all thermal and fluid systems. It helps determine:  Work output in turbines,  Work input in compressors and pumps,  Heat transfer in boilers and condensers,  Velocity and enthalpy changes in nozzles and diffusers. It connects energy conservation with real engineering processes. Problem 4.11: Steam enters a steam turbine steadily at a pressure of 3 MPa and temperature of 350∘C with a velocity of 60 m/s, and leaves at a pressure of 0.1 MPa with a velocity of 180 m/s. The mass flow rate of steam is 𝐤𝐠/𝐬, and heat loss to the surroundings is estimated to be 𝟑𝟎 𝐤𝐉/𝐤𝐠 of steam. Neglecting changes in potential energy, determine: 1. The work done per kilogram of steam by the turbine. Thermodynamics 355 Thus, efficiency improves by:  Increasing turbine inlet temperature and pressure.  Decreasing condenser pressure.  Implementing superheating or reheating. Temperature-Entropy (T-s) Diagram The T-s diagram provides a clear visualization of the Rankine cycle's thermodynamic processes. It helps identify the heat addition and rejection zones, turbine expansion, and pump work. Figure 4.9 illustrates the Rankine cycle on a T-s diagram. Fig. 4.9 T-s diagram for Rankine cycle. Processes on T-s Diagram 1. A-B: Isentropic Compression (Pump) Thermodynamics 356 Vertical line representing compression of liquid at constant entropy. 2. B-C-D: Isobaric Heat Addition (Boiler)  B-C: Heating of compressed liquid.  C-D: Phase change (vaporization) at constant temperature and pressure. 3. D-E: Superheating (Optional)  Steam is heated beyond the saturation temperature to prevent condensation in the turbine. 4. E-F: Isentropic Expansion (Turbine)  Steam expands adiabatically through the turbine producing work.  Represented by a vertical line (Δ𝑆=0) in the ideal case. 5. F-A: Isobaric Heat Rejection (Condenser)  Steam condenses to saturated liquid at constant pressure. Superheating and Reheating Superheating To minimize moisture at the turbine exit and prevent blade erosion, steam is superheated before expansion. This improves efficiency and ensures drier steam at the outlet. Reheating In large power plants, steam after partial expansion is reheated and then expanded again in another turbine stage. This is called the Reheat Rankine Cycle, and it provides:  Higher efficiency,  Reduced moisture content,  Better turbine performance. Importance of the Rankine Cycle  Forms the theoretical basis of all steam power plants.  Allows quantitative evaluation of plant performance.  Demonstrates the conversion of thermal energy → mechanical energy → electrical energy. Thermodynamics 357 Forms the foundation for improved cycles such as: 1. Reheat Rankine Cycle 2. Regenerative Rankine Cycle 3. Organic Rankine Cycle Problem 4.12: In a Rankine cycle, steam enters the turbine at a pressure of 3 MPa and a temperature of 350∘C, and it exhausts to the condenser at a pressure of 𝟏𝟎 𝐤𝐏𝐚. The condensate leaves the condenser as saturated liquid, and the pump delivers this water to the boiler. Assume isentropic expansion in the turbine and isentropic compression in the pump. Determine: 1. The enthalpy at all key points ( 1,2,3, and 4 ). 2. The work done by the turbine. 3. The work input to the pump. 4. The net work output per kg of steam. 5. The thermal efficiency of the cycle. 6. The specific steam consumption (SSC). Use steam tables where required. Solution: Given Data Property Symbol Boiler pressure 𝑃3=3MPa Condenser pressure 𝑃1=10kPa Turbine inlet temperature 𝑇3=350∘C Process type Turbine: Isentropic, Pump: Isentropic Step 1: Determine Properties at Key Points At Point 1 (Condenser Outlet / Pump Inlet) At 𝑃1=10kPa, steam is a saturated liquid. Thermodynamics 358 From steam tables: ℎ1=ℎ𝑓=191.83 kJ/kg 𝑣𝑓=0.001010 m3/kg At Point 2 (Pump Outlet / Boiler Inlet) For isentropic compression, 𝑊pump =𝑣𝑓(𝑃2−𝑃1) 𝑊pump =0.001010×(3000−10)=3.02 kJ/kg ℎ2=ℎ1+𝑊pump =191.83+3.02=194.85 kJ/kg At Point 3 (Boiler Outlet / Turbine Inlet) At 𝑃3=3MPa and 𝑇3=350∘C : From superheated steam tables: ℎ3=3115.3 kJ/kg,𝑠3=6.743 kJ/kg.K At Point 4 (Turbine Exit / Condenser Inlet) At 𝑃4=10kPa, and assuming isentropic expansion, 𝑠4=𝑠3=6.743 From saturation data at 10 kPa : 𝑠𝑓=0.6492,𝑠𝑔=8.151,ℎ𝑓=191.83,ℎ𝑔=2584.7 Dryness fraction: 𝑥4=𝑠4−𝑠𝑓 𝑠𝑔−𝑠𝑓=6.743−0.6492 8.151−0.6492=0.833 ℎ4=ℎ𝑓+𝑥4(ℎ𝑓𝑔)=191.83+0.833(2392.9)=2187.5 kJ/kg Step 2: Calculate Work and Heat Interactions (a) Turbine Work 𝑊𝑡=ℎ3−ℎ4=3115.3−2187.5=927.8 kJ/kg Thermodynamics 359 (b) Pump Work 𝑊𝑝=ℎ2−ℎ1=3.02 kJ/kg (c) Net Work Output 𝑊net =𝑊𝑡−𝑊𝑝=927.8−3.02=924.8 kJ/kg Step 3: Heat Addition and Rejection (a) Heat Added in Boiler 𝑄in =ℎ3−ℎ2=3115.3−194.85=2920.45 kJ/kg (b) Heat Rejected in Condenser 𝑄out =ℎ4−ℎ1=2187.5−191.83=1995.67 kJ/kg Step 4: Thermal Efficiency 𝜂=𝑊net 𝑄in =924.8 2920.45=0.3167 𝜂=31.7% Step 5: Specific Steam Consumption (SSC) SSC=3600 𝑊net =3600 924.8=3.89 kg/kWh Exercise 4.19: Steam enters the turbine of an ideal Rankine cycle at a pressure of 3 MPa and a temperature of 400∘C, and expands to a condenser pressure of 10 kPa . The condensate leaves the condenser as saturated liquid and is pumped back to the boiler. Assuming isentropic expansion in the turbine and isentropic compression in the pump: Determine: 1. The enthalpy and entropy at all key points ( 1,2,3, and 4). 2. The turbine work, pump work, net work output, and heat supplied per kilogram of steam. 3. The thermal efficiency of the cycle. 4. The specific steam consumption (SSC) in kg/kWh. 5. Sketch the T-s diagram for the process. Thermodynamics 360 (Use superheated and saturated steam tables as necessary.) Exercise 4.20: In a steam power plant operating on the Rankine cycle, steam enters the turbine at 2.5 MPa and 350∘C and leaves at 0.05 MPa . The condensate is then pumped back to boiler pressure, where it is again heated and vaporized. Assuming ideal (reversible adiabatic) expansion and compression: Find: 1. The heat added in the boiler per kg of steam. 2. The heat rejected in the condenser. 3. The work done by the turbine and work input to the pump. 4. The net work done and cycle efficiency. 5. Indicate all the processes on a T-s diagram. (Use relevant property data from the steam tables for accurate results.) 4.12 Reheat and Regenerative Rankine Cycles While the simple Rankine cycle provides the basic concept of steam power generation, its practical efficiency can be improved through certain modifications. Two such major improvements are: 1. Reheating of steam - To reduce moisture content and increase turbine work. 2. Regeneration - To improve efficiency by preheating feedwater using extracted steam. Both methods enhance thermal performance, reduce losses, and make the cycle more suitable for modern power plants. Reheat Rankine Cycle Principle In a Reheat Rankine cycle, steam is expanded in two stages:  High-Pressure (HP) Turbine  Low-Pressure (LP) Turbine Between these two stages, the steam is returned to the boiler or a reheater, where it is reheated at constant pressure before entering the LP turbine. This reheating process increases the Thermodynamics 361 average temperature of heat addition and improves both efficiency and steam quality at the turbine exit. Components and Process Description Fig. 4.10 Reheat Cycle. 1. Process 1-2: Isentropic expansion of steam through the HP turbine from boiler pressure to reheat pressure. 2. Process 2-3: The partially expanded steam is sent back to the reheater, where it is heated at constant pressure up to nearly the same temperature as before expansion. 3. Process 3-4: The reheated steam expands isentropically in the LP turbine, producing additional work. 4. Process 4-5: The exhaust steam enters the condenser, where it is condensed to saturated liquid. 5. Process 5-1: The condensate is pumped back to boiler pressure by the feed pump, completing the cycle. Thermodynamics 362 T-s and h-s Diagrams The T-s diagram of the reheat cycle shows two expansion processes (1-2 and 3-4) separated by a constantpressure reheating process (2-3). The 𝐡-s diagram similarly illustrates increased turbine work due to reheat. Fig. 4.11 T-s and h-s Diagrams. Thermodynamics 363 Work and Heat Calculations 1. Turbine Work: 𝑊𝑡=(ℎ1−ℎ2)+(ℎ3−ℎ4) 2. Pump Work: 𝑊𝑝=ℎ6−ℎ5 3. Heat Supplied: 𝑄𝑖𝑛 =(ℎ1−ℎ6)+(ℎ3−ℎ2) 4. Net Work Output: 𝑊𝑛𝑒𝑡 =𝑊𝑡−𝑊𝑝 5. Thermal Efficiency: 𝜂reheat =𝑊net 𝑄in =(ℎ1−ℎ2)+(ℎ3−ℎ4)−(ℎ6−ℎ5) (ℎ1−ℎ6)+(ℎ3−ℎ2) Effect of Reheating  Increases turbine work output: Due to additional expansion in the LP turbine.  Reduces moisture content: Steam quality at the turbine exhaust improves, reducing erosion of blades.  Slight improvement in efficiency: Due to higher mean temperature of heat addition. However, the gain in efficiency is marginal because additional reheating increases cost and system complexity. Optimum Reheat Pressure In most modern power plants, the reheat pressure is maintained between 0.2 to 0.25 of the boiler pressure. This ensures a good balance between turbine work and moisture control. Advantages of Reheating  Increases turbine output.  Reduces moisture at turbine exit (improves blade life).  Improves overall efficiency of the cycle.  Enhances nozzle and blade efficiency. Thermodynamics 364 Disadvantages of Reheating  Increases plant cost and maintenance.  Efficiency gain is relatively small compared to added complexity. Regenerative Rankine Cycle Principle In the Regenerative Rankine cycle, part of the steam from the turbine is extracted (bled) at intermediate stages and used to preheat the feedwater before it enters the boiler. This preheating reduces the temperature difference between the feedwater and the boiler, minimizing irreversibilities and improving overall thermal efficiency. Purpose of Regeneration  Reduces the amount of external heat required in the boiler.  Raises the average temperature of heat addition.  Improves thermal efficiency.  Reduces condenser load and size. Principle of Operation 1. Steam is extracted from the turbine at one or more intermediate pressures. 2. The extracted steam transfers heat to the feedwater in feedwater heaters (FWHs). 3. The feedwater then enters the boiler at a higher temperature, reducing the boiler's fuel requirement. Types of Feedwater Heaters 1. Open (Direct-Contact) Heaters:  Steam and feedwater mix directly.  Simple and inexpensive.  Pressure of extracted steam = pressure of feedwater. 2. Closed Heaters:  Steam condenses on one side of a heat exchanger, transferring heat to feedwater flowing on the other side.  No mixing occurs.  Allows separate pressures for feedwater and bleed steam. Thermodynamics 371 Assume all expansion and compression processes are isentropic, and neglect pump work. Determine: 1. The fraction of steam extracted for regeneration. 2. The specific work output of the cycle. 3. The heat added in the boiler and reheater. 4. The cycle thermal efficiency. 5. Draw a neatly labeled T−s diagram showing both reheating and regeneration processes. (Use appropriate enthalpy and entropy values from superheated steam and saturation tables.) 4.13 Heat Rate The Heat Rate is a key performance parameter used to measure the efficiency of power plants, particularly steam power plants operating on the Rankine cycle. It indicates how effectively a power plant converts heat energy (thermal input) into useful electrical energy (output). In simple terms, it expresses the amount of heat energy required to produce one unit of electrical energy. The heat rate is defined as: Heat Rate = Heat Supplied to the Cycle Net Work Output of the Cycle It can also be expressed in terms of thermal efficiency ( 𝜂 ): Heat Rate =3600 𝜂 Where,  Heat Rate is in kJ/kWh,  𝜼= Thermal efficiency of the plant (decimal form),  3600 represents the number of kilojoules in one kilowatt-hour ( 1kWh=3600 kJ ). The heat rate represents the energy input required by a power plant to generate one kilowatthour (kWh) of electricity. Thermodynamics 372  A lower heat rate means the plant is more efficient - it requires less fuel energy per unit of electricity generated.  A higher heat rate indicates poorer efficiency and higher fuel consumption. Thus, the heat rate serves as an inverse measure of efficiency. Relation Between Heat Rate and Thermal Efficiency Since efficiency ( 𝜂 ) is given by: 𝜂= Net Work Output Heat Supplied Rearranging gives: Heat Rate =3600 𝜂 Hence,  When 𝜼 increases, the heat rate decreases.  For example, if a plant has an efficiency of 36%, Heat Rate =3600 0.36 =10,000 kJ/kWh This means the plant needs 10,000 kJ of heat to generate 1 kWh of electricity. Units of Heat Rate System Typical Unit Equivalent SI System kJ/kWh 1kWh=3600 kJ Imperial (USA) Btu/kWh 1kWh=3412Btu Example: A modern steam turbine power plant may have a heat rate of 9,000− 10,000 kJ/kWh, corresponding to efficiencies of 𝟑𝟔−𝟒𝟎%. Factors Affecting Heat Rate The heat rate of a power plant is influenced by several design and operational factors: 1. Boiler Efficiency: Lower flue gas losses and better heat transfer improve overall efficiency, reducing heat rate. Thermodynamics 373 2. Turbine Efficiency: Higher turbine efficiency means more mechanical work per unit of steam, reducing heat rate. 3. Condenser Pressure: A lower condenser pressure increases turbine expansion ratio and improves efficiency. 4. Superheat and Reheat Temperature: Increasing these temperatures raises the average temperature of heat addition, improving cycle performance. 5. Regeneration: Preheating the feedwater reduces fuel input and thus the heat rate. 6. Auxiliary Power Consumption: Lower internal power consumption (e.g., pumps, fans) reduces total energy input per kWh output. Importance of Heat Rate The heat rate is a practical measure of power plant performance used by engineers and operators because:  It connects fuel consumption directly to electricity generation.  It helps evaluate economic operation of the plant.  It identifies areas for improvement in thermal efficiency.  It provides a basis for comparing different power plants or cycles. Typical Values of Heat Rate Type of Power Plant Thermal Efficiency (%) Heat Rate (kJ/kWh) Old Steam Power Plant 30 12,000 Modern Steam Plant (Reheat + Regeneration) 38 9,470 Supercritical Steam Plant 42 8,570 Combined Cycle Gas Plant 55 6,550 Ultra-Supercritical Plant 45 8,000 Thermodynamics 374 Specific Steam Consumption (SSC) and Heat Rate There is a close relationship between Specific Steam Consumption (SSC) and Heat Rate: Heat Rate =SSC× Enthalpy Drop per kg of Steam Also, since: SSC=3600 Work Output per kg Therefore, improvement in turbine performance (reducing SSC) directly reduces the heat rate. Methods to Improve Heat Rate 1. Increasing Boiler Pressure and Temperature: Improves efficiency and reduces fuel input. 2. Using Reheating and Regeneration: Raises the average temperature of heat addition and reduces losses. 3. Reducing Condenser Pressure: Increases turbine expansion and net work output. 4. Improving Heat Exchanger Performance: Reduces temperature drops and energy wastage. 5. Periodic Maintenance and Optimization: Keeps turbine and boiler efficiency at design levels. Problem 4.14: A steam power plant operates on a Rankine cycle between the following states:  Steam enters the turbine at 40 bar and 400∘C.  It expands isentropically to a condenser pressure of 0.05 bar.  The condensate from the condenser is pumped back to boiler pressure. Neglect the pump work. Determine: 1. The turbine work output ( kJ/kg ). 2. The heat supplied in the boiler (kJ/kg). 3. The thermal efficiency of the cycle (%). 4. The heat rate (kJ/kWh). Use the following steam table data: Thermodynamics 375 Solution: Property Symbol 40 bar, 𝟒𝟎𝟎∘𝐂 0.05 bar Enthalpy of superheated steam ℎ1 3232.0 kJ/kg - Entropy of superheated steam 𝑠1 6.769 kJ/kg⋅K - Saturated liquid enthalpy ℎ𝑓 - 191.8 kJ/kg Saturated vapour enthalpy ℎ𝑔 - 2584.7 kJ/kg Saturated liquid entropy 𝑠𝑓 - 0.649 kJ/kg⋅K Saturated vapour entropy 𝑠𝑔 - 8.396 kJ/kg⋅K Step 1: Determine the Dryness Fraction at Turbine Exit Since expansion is isentropic, 𝑠2=𝑠1=6.769 kJ/kg.K At 0.05 bar, 𝑠𝑓=0.649,𝑠𝑔=8.396 Dryness fraction, 𝑥2=𝑠2−𝑠𝑓 𝑠𝑔−𝑠𝑓=6.769−0.649 8.396−0.649=6.120 7.747=0.790 𝑥2=0.79 Step 2: Determine Enthalpy at Turbine Exit ℎ2=ℎ𝑓+𝑥2ℎ𝑓𝑔 At 0.05 bar: ℎ𝑓=191.8,ℎ𝑓𝑔 =ℎ𝑔−ℎ𝑓=2584.7−191.8=2392.9 ℎ2=191.8+0.79(2392.9)=191.8+1890.4=2082.2 kJ/kg ℎ2=2082.2 kJ/kg Step 3: Turbine Work Output 𝑊𝑡=ℎ1−ℎ2=3232.0−2082.2=1149.8 kJ/kg 𝑊𝑡=1149.8 kJ/kg Thermodynamics 376 Step 4: Heat Supplied in the Boiler At boiler outlet, ℎ1=3232.0,ℎfeed ≈ℎ𝑓( at 40 bar )=1087.3 kJ/kg (For simplicity, pump work is neglected.) 𝑄𝑖𝑛 =ℎ1−ℎfeed =3232.0−1087.3=2144.7 kJ/kg 𝑄𝑖𝑛 =2144.7 kJ/kg Step 5: Thermal Efficiency 𝜂=𝑊𝑡 𝑄𝑖𝑛 =1149.8 2144.7=0.536 𝜂=53.6% Step 6: Heat Rate Heat Rate =3600 𝜂 Heat Rate =3600 0.536=6716.4 kJ/kWh Heat Rate =6716.4 kJ/kWh Exercise 4.23: A steam power plant (Plant A) operates on a simple Rankine cycle with a thermal efficiency of 34%, while another plant (Plant B) operates on an improved ReheatRegenerative Rankine cycle with a thermal efficiency of 42%. 1. Calculate the heat rate for both plants in kJ/kWh. 2. Determine how much fuel energy is saved per kWh of electricity produced by the more efficient plant. 3. If both plants produce 200 MW of power, find the difference in total heat input (MJ/s) between them. 4. Comment on the effect of efficiency improvement on fuel consumption and operational cost. Exercise 4.24: A thermal power station produces 𝟐𝟓𝟎MW of electrical power and consumes 𝟐,𝟐𝟓𝟎,𝟎𝟎𝟎MJ of heat energy per hour from the fuel. 1. Determine the thermal efficiency of the plant. 2. Calculate the heat rate in kJ/kWh. Thermodynamics 377 3. If the fuel cost is ₹ 4 per MJ, estimate the fuel cost per kWh of electricity generated. 4. Suggest two methods by which the heat rate of this plant can be improved. 4.14 Specific Steam Consumption (SSC) In steam power plants, one of the most important performance indicators of a turbine or cycle is the Specific Steam Consumption (SSC). It indicates how much steam mass flow is required to produce one unit of useful power output. While thermal efficiency and heat rate express how effectively energy is converted, the SSC expresses how economically the working fluid (steam) is utilized to generate power. The Specific Steam Consumption (SSC) is defined as: "The mass of steam required to produce one unit of power output (usually 1 kWh ) in a steam turbine or power plant." Mathematically, SSC= Mass flow rate of steam Power output or, SSC=3600 𝑊net Where,  SSC = Specific Steam Consumption (kg/kWh)  𝐖net = Net work output per kg of steam (kJ/kg)  3600= Conversion factor (1kWh=3600 kJ) The SSC indicates the quantity of steam required to generate one unit of electricity.  A lower SSC means the plant or turbine is more efficient - less steam is needed for the same power output.  A higher SSC indicates poor efficiency, requiring more steam per kWh of power. Thermodynamics 378 Relation Between SSC, Work Output, and Heat Rate There is a direct relationship between SSC, specific work output, and heat rate. SSC=3600 𝑊net and Heat Rate =SSC×(ℎ1−ℎ2) Also, since 𝜂=𝑊net 𝑄in =3600 Heat Rate it follows that: SSC∝1 Efficiency Hence,  Higher efficiency → Lower SSC  Lower efficiency → Higher SSC Units of SSC System Unit Equivalent Meaning SI System kg/kWh kilograms of steam per kilowatt-hour CGS/Other kg/MJ or kg/h per kW rarely used variants Thus, if a turbine has an SSC of .𝟎 𝐤𝐠/𝐤𝐖𝐡, it means 𝟒 𝐤𝐠 of steam are required to produce 𝟏 𝐤𝐖𝐡 of power. Factors Affecting Specific Steam Consumption 1. Boiler Pressure and Temperature: Higher pressures and superheating increase turbine work output, reducing SSC. 2. Condenser Pressure: Lower condenser pressure improves turbine expansion ratio, increasing efficiency and lowering SSC. Thermodynamics 379 3. Turbine Efficiency: Better turbine design (high isentropic efficiency) produces more work per kg of steam, lowering SSC. 4. Cycle Modifications: Using Reheating and Regeneration increases mean temperature of heat addition, thus reducing SSC. 5. Auxiliary Power Consumption: Higher internal power use (e.g., pumps, fans) effectively increases SSC. Typical Values of SSC Type of Cycle / Power Plant Specific Steam Consumption (kg/kWh) Simple Rankine Cycle 5.0-6.0 Rankine with Reheat 3.5-4.0 Rankine with Reheat & Regeneration 3.0-3.5 Supercritical Power Plant 2.5−3.0 Modern supercritical and ultra-supercritical power plants achieve SSC values as low as 2.5 kg/kWh, indicating very high efficiency. Importance of SSC  It provides a direct measure of cycle performance.  Useful for comparing turbines or plants operating under different conditions.  Helps in fuel and cost estimation for plant operation.  Aids in identifying areas of efficiency improvement. Methods to Reduce SSC 1. Increase boiler pressure and temperature (superheating). 2. Employ reheat and regenerative feedwater heating. 3. Maintain low condenser pressure. 4. Improve turbine and pump efficiencies. 5. Ensure proper maintenance to minimize mechanical and thermal losses. Thermodynamics 380 Problem 4.15: A steam turbine operates on an ideal Rankine cycle. Steam enters the turbine at 30 bar and 400∘C and expands isentropically to a condenser pressure of 0.1 bar. Neglect the pump work. Determine: 1. The specific work output of the turbine ( kJ/kg ). 2. The thermal efficiency of the cycle (%). 3. The specific steam consumption (SSC) in kg/kWh. 4. Comment on the result. Use the following steam table data: Property Symbol 30 bar, 400∘C 0.1 bar Enthalpy of superheated steam ℎ1 3230.5 kJ/kg - Entropy of superheated steam 𝑠1 6.769 kJ/kg⋅K - Saturated liquid enthalpy ℎ𝑓 - 191.8 kJ/kg Saturated vapour enthalpy ℎ𝑔 - 2584.7 kJ/kg Saturated liquid entropy 𝑠𝑓 - 1.302 kJ/kg⋅K Saturated vapour entropy 𝑠𝑔 - 8.151 kJ/kg⋅K Solution: Step 1: Find the Dryness Fraction at the Turbine Exit Since the expansion is isentropic, 𝑠2=𝑠1=6.769 kJ/kg.K At 𝑃2=0.1bar, 𝑠𝑓=1.302,𝑠𝑔=8.151 Thermodynamics 387 CHAPTER -5 BASICS OF PROPULSION AND HEAT TRANSFER 5.1 Classification of Jet Engines Jet engines are a type of air-breathing propulsion system designed to produce thrust by expelling a high-velocity jet of gases from the rear of the engine. The operation of a jet engine is fundamentally based on Newton’s Third Law of Motion, which states that “For every action, there is an equal and opposite reaction.” When gases are expelled at high speed in one direction, a reaction force of equal magnitude acts in the opposite direction, propelling the engine and therefore the aircraft forward. Depending on the method of air intake, compression, combustion, and expansion, jet engines can be classified into various categories. The main classification is based on whether the engine uses atmospheric air or carries its own oxidizer. 1. Based on the Type of Jet Propulsion System Jet engines can be broadly divided into two main types: (a) Air-Breathing Engines Air-breathing engines derive the oxygen required for combustion from the surrounding atmosphere. The incoming air is compressed, mixed with fuel, and burnt to produce hightemperature and high-pressure gases, which expand through a nozzle to generate thrust. These engines are commonly used in aircraft operating within the Earth’s atmosphere. They include the following types: 1. Turbojet Engine 2. Turbofan Engine 3. Turboprop Engine 4. Ramjet Engine 5. Scramjet Engine Thermodynamics 388 (b) Non-Air-Breathing Engines Unlike air-breathing engines, non-air-breathing engines carry both fuel and oxidizer onboard, making them capable of operating even outside the atmosphere such as in space. They do not rely on external air and thus can function in a vacuum. These engines are primarily used in rockets and space vehicles, where atmospheric oxygen is unavailable. Example: Rocket Engines. 2. Types of Air-Breathing Jet Engines The air-breathing jet engines vary mainly in how they compress air, extract energy, and generate thrust. Each type has its own operational speed range, efficiency characteristics, and applications. (a) Turbojet Engine A turbojet engine is the simplest form of a gas turbine engine. In this type, all the air entering the engine passes through the core, where it undergoes compression, combustion, and expansion. The high-energy exhaust gases are then expelled through a convergent nozzle to produce thrust. Fig. 5.1 Turbojet Engine. Working Principle: Air is first decelerated and compressed by the compressor, after which fuel is injected and burnt in the combustion chamber. The resulting high-pressure gases pass through a turbine, which extracts some energy to drive the compressor. The remaining energy in the exhaust gases produces a high-velocity jet that generates thrust. Thermodynamics 389 Major Components: Diffuser → Compressor → Combustion Chamber → Turbine → Exhaust Nozzle Characteristics: Turbojets perform best at high speeds and altitudes, making them suitable for supersonic aircraft. However, at low speeds, they are relatively inefficient and produce high noise levels. Applications: Commonly used in fighter jets and military aircraft that operate at supersonic speeds. (b) Turbofan Engine The turbofan engine is an advanced version of the turbojet that includes a large fan at the front of the engine. This fan pushes a large volume of air around the engine core (known as the bypass air) and combines it with the thrust produced by the jet exhaust. Fig. 5.2 Turbofan Engine. Working Principle: Part of the air passes through the engine core and undergoes combustion, while the rest bypasses the core through an outer duct. The bypass ratio the ratio of the mass flow rate of bypass air to that of the core air determines the engine’s efficiency and performance characteristics. Advantages: Turbofans are more fuel-efficient, quieter, and better suited for subsonic flight speeds compared to turbojets. They produce smoother operation and lower emissions. Applications: Widely used in commercial airliners (such as Boeing and Airbus aircraft) and transport planes, where efficiency and noise reduction are critical. Thermodynamics 390 (c) Turboprop Engine A turboprop engine combines the principles of a gas turbine with a propeller system. In this engine, the turbine extracts a large portion of the energy from the combustion gases to drive the propeller through a reduction gearbox. The propeller provides the majority of the thrust, while only a small portion comes from the jet exhaust. Fig. 5.3 Turboprop Engine. Features and Performance: Turboprops are extremely fuel-efficient at lower flight speeds (below about 800 km/h) and at low altitudes. However, their performance decreases at higher speeds due to propeller drag. Applications: Commonly used in cargo planes, regional transport aircraft, and maritime patrol or surveillance aircraft, where fuel economy and reliability are essential. (d) Ramjet Engine The ramjet is a simple air-breathing jet engine with no moving parts such as compressors or turbines. Instead, it relies entirely on the ram effect created by the forward motion of the aircraft to compress incoming air. Thermodynamics 391 Fig. 5.4 Ramjet Engine. Working Principle: As the aircraft moves at high speed, air is forced into the inlet and compressed. Fuel is injected and burned, and the high-pressure gases expand through the nozzle to produce thrust. Since there is no compressor, the engine cannot generate thrust at zero speed it requires an initial boost to reach operating speed. Operating Range: Ramjets are effective only at supersonic speeds (Mach 2–6), as they depend on the dynamic pressure of incoming air for compression. Applications: Primarily used in missiles, target drones, and high-speed experimental aircraft, where simplicity and compactness are advantageous. (e) Scramjet Engine (Supersonic Combustion Ramjet) A scramjet engine is an advanced form of ramjet designed for hypersonic flight. The key difference is that combustion occurs while the airflow remains supersonic within the combustion chamber. Fig. 5.5 Scramjet Engine. Thermodynamics 392 Working Principle: In scramjets, the incoming air is compressed by the vehicle’s high-speed motion, but it is not slowed down to subsonic speeds before entering the combustor. This allows efficient combustion at speeds exceeding Mach 5, resulting in extremely high thrustto-weight ratios. Advantages: Scramjets are highly efficient at hypersonic speeds, offering enormous potential for space launch systems and reusable spaceplanes. However, they cannot operate from rest and require another propulsion system to reach operational speed. Applications: Used in hypersonic research programs, space launch vehicles, and nextgeneration aerospace systems. 3. Non-Air-Breathing Engines (a) Rocket Engine The rocket engine represents a class of non-air-breathing propulsion systems that carry both fuel and oxidizer onboard. This enables them to operate in environments devoid of atmospheric oxygen, such as outer space. Fig. 5.6 Rocket Engine. Working Principle: Fuel and oxidizer are combusted in a combustion chamber to produce high-temperature, high-pressure gases, which are then expelled through a nozzle to generate thrust. Because all reactants are self-contained, rockets can function in a vacuum. Thermodynamics 393 Characteristics: Rocket engines generate extremely high thrust and are capable of propelling vehicles to very high velocities. However, they have high fuel consumption and typically limited reusability due to thermal and structural stresses. Applications: Widely used in spacecraft, satellite launch vehicles, ballistic missiles, and space exploration missions. 5.2 Basic Jet Propulsion Arrangement Jet propulsion is a system used to produce forward thrust for aircraft by expelling mass (gases) at high velocity in the opposite direction. The essential concept behind jet propulsion is derived from Newton’s Third Law of Motion: “For every action, there is an equal and opposite reaction.” When a mass of gas is accelerated rearward through a nozzle, the aircraft experiences an equal and opposite forward reaction force known as thrust. The basic jet propulsion arrangement includes a sequence of components that enable air intake, compression, combustion, expansion, and exhaust each contributing to the production of thrust. Basic Components of a Jet Propulsion System A jet engine operates as a continuous thermodynamic cycle (an open Brayton cycle) consisting of five main stages: 1. Intake (Diffuser Section) 2. Compression (Compressor Section) 3. Combustion (Combustion Chamber) 4. Expansion (Turbine Section) 5. Exhaust (Nozzle Section) Each of these components has a specific function, and together they form the basic jet propulsion arrangement. Thermodynamics 394 Fig. 5.7 Basic arrangement of Jet Propulsion System. (1) Air Intake (Diffuser Section) The air intake or inlet diffuser is the first component of the propulsion system. Its primary purpose is to capture and decelerate the incoming air efficiently before it enters the compressor.  The air entering the engine is slowed down to increase its static pressure, allowing for better compression efficiency.  The diffuser is designed aerodynamically to minimize energy losses and prevent turbulence.  The pressure at the end of the diffuser should be as high as possible since it forms the starting point for compression. Function: Converts high-velocity, low-pressure air into low-velocity, high-pressure air suitable for the compressor. (2) Compressor Section After air passes through the diffuser, it enters the compressor, which increases the air’s pressure and temperature before combustion. There are two main types of compressors used in jet engines:  Centrifugal Compressors: Air is accelerated outward by a rotating impeller and then diffused to raise pressure.  Axial-Flow Compressors: Air flows parallel to the engine axis through alternating rows of rotating (rotor) and stationary (stator) blades. Function: To compress the air to 10–40 times the atmospheric pressure (depending on engine type). This pressurized air is essential for efficient fuel combustion. Thermodynamics 395 (3) Combustion Chamber The combustion chamber (or combustor) is where the compressed air is mixed with fuel and ignited. The chemical energy of the fuel is converted into thermal energy.  The burning process must be stable, efficient, and produce a uniform high-temperature gas stream.  Only about one-third of the compressed air is used for combustion; the rest is used to cool the chamber walls and dilute the hot gases to prevent turbine damage.  The temperature at the outlet of the combustion chamber typically ranges from 1000°C to 1500°C. Function: Converts chemical energy of fuel into high-temperature, high-energy gases. (4) Turbine Section The turbine extracts a portion of the energy from the hot gases exiting the combustion chamber. This energy is used to drive the compressor and other accessories (like fuel pumps).  The turbine consists of alternating rows of stationary nozzles and rotating blades.  The nozzle guide vanes direct the gas flow at an optimum angle onto the turbine blades, causing them to rotate.  The turbine must balance between extracting sufficient energy to drive the compressor and leaving enough energy in the exhaust gases to produce thrust. Function: Converts part of the gas energy into mechanical energy to operate the compressor. (5) Exhaust Nozzle The nozzle is the final component of the propulsion system. It converts the thermal and pressure energy of the exhaust gases into kinetic energy, producing a high-velocity jet at the exit.  The gas expands through the nozzle, increasing its velocity to supersonic speeds in some designs.  The difference between the exhaust velocity and the incoming air velocity produces the net thrust. Function: Accelerates exhaust gases to generate the reaction force (thrust) that propels the aircraft forward. Thermodynamics 396 Working Principle of Jet Propulsion The working of a basic jet propulsion engine can be explained as follows: 1. Atmospheric air enters the intake diffuser, where its velocity is reduced and pressure increased. 2. The compressor then compresses the air to a much higher pressure. 3. Fuel is injected and burnt in the combustion chamber, raising the temperature and energy of the gases. 4. These high-pressure gases expand through the turbine, providing the necessary power to drive the compressor. 5. Finally, the gases exit through the nozzle, where they are accelerated to high velocity, producing forward thrust due to the reaction principle. This entire process occurs continuously during engine operation, generating a smooth and sustained thrust. Energy Conversion in a Jet Engine Stage Type of Energy Conversion Air Intake Kinetic → Pressure Energy Compressor Mechanical → Pressure Energy Combustion Chemical → Thermal Energy Turbine Thermal → Mechanical Energy Nozzle Pressure/Thermal → Kinetic Energy Thus, the total energy transformation from fuel to thrust involves multiple intermediate conversions, with thrust being the final useful output. Applications The basic jet propulsion arrangement is the foundation for all modern gas turbine engines, including:  Turbojet engines (military and supersonic aircraft)  Turbofan engines (commercial airliners)  Turboprop engines (propeller-driven aircraft)