# Second Law of Thermodynamics Alternative Statements

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Second Law of Thermodynamics Alternative Statements There is no simple statement that captures all aspects of the second law. Several alternative formulations of the second law are found in the technical literature. Three prominent ones are: Clausius Statement Kelvin-Planck Statement Entropy Statement

2 Aspects of the Second Law of Thermodynamics The second law of thermodynamics has many aspects, which at first may appear different in kind from those of conservation of mass and energy principles. Among these aspects are: predicting the direction of processes. establishing conditions for equilibrium. determining the best theoretical performance of cycles, engines, and other devices. evaluating quantitatively the factors that preclude attainment of the best theoretical performance level.

3 Clausius Statement of the Second Law It is impossible for any system to operate in such a way that the sole result would be an energy transfer by heat from a cooler to a hotter body.

4 Kelvin-Planck Statement of the Second Law It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving energy by heat transfer from a single thermal reservoir.

5 Kelvin Temperature Scale Consider systems undergoing a power cycle and a refrigeration or heat pump cycle, each while exchanging energy by heat transfer with hot and cold reservoirs: The Kelvin temperature is defined so that Q Q C H rev cycle = T T C H (Eq. 5.7)

6 Example: Power Cycle Analysis A system undergoes a power cycle while receiving 1000 kj by heat transfer from a thermal reservoir at a temperature of 500 K and discharging 600 kj by heat transfer to a thermal reservoir at (a) 200 K, (b) 300 K, (c) 400 K. For each case, determine whether the cycle operates irreversibly, operates reversibly, or is impossible. Hot Reservoir T H = 500 K Q H = 1000 kj Power Cycle Cold Reservoir W cycle Q C = 600 kj T C = (a) 200 K, (b) 300 K, (c) 400 K Solution: To determine the nature of the cycle, compare actual cycle performance (η) to maximum theoretical cycle performance (η max ) calculated from Eq. 5.9

7 Example: Power Cycle Analysis Actual Performance: Calculate η using the heat transfers: Q η = 1 Q C H 600 kj = 1 = kj Maximum Theoretical Performance: Calculate η max from Eq. 5.9 and compare to η: TC 200 K (a) ηmax = 1 = 1 = 0. 6 T 500 K H TC 300 K (b) ηmax = 1 = 1 = 0. 4 T 500 K H TC 400 K (c) ηmax = 1 = 1 = 0. 2 T 500 K H η η max 0.4 < 0.6 Irreversibly 0.4 = 0.4 Reversibly 0.4 > 0.2 Impossible

8 Clausius Inequality The Clausius inequality considered next provides the basis for developing the entropy concept in Chapter 6. The Clausius inequality is applicable to any cycle without regard for the body, or bodies, from which the system undergoing a cycle receives energy by heat transfer or to which the system rejects energy by heat transfer. Such bodies need not be thermal reservoirs.

9 Clausius Inequality The Clausius inequality is developed from the Kelvin-Planck statement of the second law and can be expressed as: δq T b = σ cycle (Eq. 5.13) where indicates integral is to be performed over all parts of the b boundary and over the entire cycle. subscript indicates integrand is evaluated at the boundary of the system executing the cycle.

10 Clausius Inequality The Clausius inequality is developed from the Kelvin-Planck statement of the second law and can be expressed as: δq T b = σ cycle (Eq. 5.13) The nature of the cycle executed is indicated by the value of σ cycle : σ cycle = 0 no irreversibilities present within the system σ cycle > 0 irreversibilities present within the system σ cycle < 0 impossible Eq. 5.14

11 Example: Use of Clausius Inequality A system undergoes a cycle while receiving 1000 kj by heat transfer at a temperature of 500 K and discharging 600 kj by heat transfer at (a) 200 K, (b) 300 K, (c) 400 K. Using Eqs and 5.14, what is the nature of the cycle in each of these cases? Solution: To determine the nature of the cycle, perform the cyclic integral of Eq to each case and apply Eq to draw a conclusion about the nature of each cycle.

12 Example: Use of Clausius Inequality Applying Eq to each cycle: δq T b = Q T in H Q T out C = σ cycle Hot Reservoir T H = 500 K Q H = 1000 kj Power Cycle W cycle Q C = 600 kj T C = (a) 200 K, (b) 300 K, (c) 400 K Cold Reservoir

13 Example: Use of Clausius Inequality Applying Eq to each cycle: (a) δq T 1000 kj 600 kj σ cycle = = 1kJ/K σ cycle = +1 kj/k > K 200 K b = Q T in H Q T out C = σ cycle Irreversibilities present within system (b) 1000 kj 600 kj σ cycle = = 0 kj/k σ cycle = 0 kj/k = K 300 K No irreversibilities present within system (c) 1000 kj 600 kj σ cycle = = 0.5 kj/k σ cycle = 0.5 kj/k < K 400 K Impossible

14 Chapter 6 Using Entropy

15 Learning Outcomes Demonstrate understanding of key concepts related to entropy and the second law... including entropy transfer, entropy production, and the increase in entropy principle. Evaluate entropy, evaluate entropy change between two states, and analyze isentropic processes, using appropriate property tables.

16 Learning Outcomes, cont. Represent heat transfer in an internally reversible process as an area on a temperature-entropy diagram. Apply entropy balances to closed systems and control volumes. Evaluate isentropic efficiencies for turbines, nozzles, compressors, and pumps.

17

18 Introducing Entropy Change and the Entropy Balance Mass and energy are familiar extensive properties of systems. Entropy is another important extensive property. Just as mass and energy are accounted for by mass and energy balances, entropy is accounted for by an entropy balance. Like mass and energy, entropy can be transferred across the system boundary.

19 Introducing Entropy Change and the Entropy Balance The entropy change and entropy balance concepts are developed using the Clausius inequality expressed as: δq T b = σ cycle (Eq. 5.13) where σ cycle = 0 no irreversibilities present within the system σ cycle > 0 irreversibilities present within the system σ cycle < 0 impossible Eq. 5.14

20 Defining Entropy Change Consider two cycles, each composed of two internally reversible processes, process A plus process C and process B plus process C, as shown in the figure. Applying Eq to these cycles gives, where σ cycle is zero because the cycles are composed of internally reversible processes.

21 Defining Entropy Change Subtracting these equations: Since A and B are arbitrary internally reversible processes linking states 1 and 2, it follows that the value of the integral is independent of the particular internally reversible process and depends on the end states only.

22 Defining Entropy Change Recalling (from Sec ) that a quantity is a property if, and only if, its change in value between two states is independent of the process linking the two states, we conclude that the integral represents the change in some property of the system. We call this property entropy and represent it by S. The change in entropy is (Eq. 6.2a) where the subscript int rev signals that the integral is carried out for any internally reversible process linking states 1 and 2.

23 Defining Entropy Change Equation 6.2a allows the change in entropy between two states to be determined by thinking of an internally reversible process between the two states. But since entropy is a property, that value of entropy change applies to any process between the states internally reversible or not. Entropy change is introduced by the integral of Eq. 6.2a for which no accompanying physical picture is given. Still, the aim of Chapter 6 is to demonstrate that entropy not only has physical significance but also is essential for thermodynamic analysis.

24 Entropy Facts Entropy is an extensive property. Like any other extensive property, the change in entropy can be positive, negative, or zero: By inspection of Eq. 6.2a, units for entropy S are kj/k and Btu/ o R. Units for specific entropy s are kj/kg K and Btu/ lb or.

25 Entropy Facts For problem solving, specific entropy values are provided in Tables A-2 through A-18. Values for specific entropy are obtained from these tables using the same procedures as for specific volume, internal energy, and enthalpy, including use of (Eq. 6.4) for two-phase liquid-vapor mixtures, and (Eq. 6.5) for liquid water, each of which is similar in form to expressions introduced in Chap. 3 for evaluating v, u, and h.

26 Entropy Facts For problem solving, states often are shown on property diagrams having specific entropy as a coordinate: the temperature-entropy and enthalpy-entropy (Mollier) diagrams shown here

27 Entropy and Heat Transfer By inspection of Eq. 6.2a, the defining equation for entropy change on a differential basis is (Eq. 6.2b) Equation 6.2b indicates that when a closed system undergoing an internally reversible process receives energy by heat transfer, the system experiences an increase in entropy. Conversely, when energy is removed by heat transfer, the entropy of the system decreases. From these considerations, we say that entropy transfer accompanies heat transfer. The direction of the entropy transfer is the same as the heat transfer.

28 Entropy and Heat Transfer In an internally reversible, adiabatic process (no heat transfer), entropy remains constant. Such a constantentropy process is called an isentropic process. On rearrangement, Eq. 6.2b gives Integrating from state 1 to state 2, (Eq. 6.23)

29 Entropy and Heat Transfer From this it follows that an energy transfer by heat to a closed system during an internally reversible process is represented by an area on a temperature-entropy diagram:

30 Entropy Balance for Closed Systems The entropy balance for closed systems can be developed using the Clausius inequality expressed as Eq and the defining equation for entropy change, Eq. 6.2a. The result is where the subscript b indicates the integral (Eq. 6.24) is evaluated at the system boundary. In accord with the interpretation of σ cycle in the Clausius inequality, Eq. 5.14, the value of σ in Eq adheres to the following interpretation σ: = 0 (no irreversibilities present within the system) > 0 (irreversibilities present within the system) < 0 (impossible)

31 Entropy Balance for Closed Systems That σ has a value of zero when there are no internal irreversibilities and is positive when irreversibilities are present within the system leads to the interpretation that σ accounts for entropy produced (or generated) within the system by action of irreversibilities. Expressed in words, the entropy balance is change in the amount of entropy contained within the system during some time interval net amount of entropy transferred in across the system boundary accompanying heat transfer during some time interval + amount of entropy produced within the system during some time interval

32 Entropy Balance for Closed Systems Example: One kg of water vapor contained within a piston-cylinder assembly, initially at 5 bar, 400 o C, undergoes an adiabatic expansion to a state where pressure is 1 bar and the temperature is (a) 200 o C, (b) 100 o C. Using the entropy balance, determine the nature of the process in each case. Since the expansion occurs adiabatically, Eq reduces to give S 2 0 δ = 2 Q S1 1 T b + σ m(s 2 s 1 ) = σ (1) Boundary where m = 1 kg and Table A-4 gives s 1 = kj/kg K.

33 Entropy Balance for Closed Systems (a) Table A-4 gives, s 2 = kj/kg K. Thus Eq. (1) gives σ = (1 kg)( ) kj/kg K = kj/k Since σ is positive, irreversibilities are present within the system during expansion (a). (b) Table A-4 gives, s 2 = kj/kg K. Thus Eq. (1) gives σ = (1 kg)( ) kj/kg K = kj/k Since σ is negative, expansion (b) is impossible: it cannot occur adiabatically.

34 Entropy Balance for Closed Systems More about expansion (b): Considering Eq < 0 = < Since σ cannot be negative and For expansion (b) ΔS is negative, then By inspection the integral must be negative and so heat transfer from the system must occur in expansion (b).

35 Entropy Rate Balance for Closed Systems On a time rate basis, the closed system entropy rate balance is where (Eq. 6.28) ds dt Q T j j = = the time rate of change of the entropy of the system the time rate of entropy transfer through the portion of the boundary whose temperature is T j σ = time rate of entropy production due to irreversibilities within the system

36 Entropy Rate Balance for Closed Systems Example: An inventor claims that the device shown generates electricity at a rate of 100 kj/s while receiving a heat transfer of energy at a rate of 250 kj/s at a temperature of 500 K, receiving a second heat transfer at a rate of 350 kj/ s at 700 K, and discharging energy by heat transfer at a rate of 500 kj/s at a temperature of 1000 K. Each heat transfer is positive in the direction of the accompanying arrow. For operation at steady state, evaluate this claim. Q 1 = 250 kj/s T 1 = 500 K Q 2 = 350 kj/s T 2 = 700 K T 3 = 1000 K + Q 3 = 500 kj/s

37 Entropy Rate Balance for Closed Systems Applying an energy rate balance at steady state de dt 0 = Q + Q Q = W e Solving W e = 250 kj/s+ 350 kj/s 500 kj/s = 100kJ/s The claim is in accord with the first law of thermodynamics. Applying an entropy rate balance at steady state Solving 250 kj/s σ = 500 K σ = ( ) ds dt = 350 kj/s 500 kj/s K 1000 K kj/s kj/s = 0.5 K K Q 0 = T 1 1 Q + T 2 2 Q T Since σ is negative, the claim is not in accord with the second law of thermodynamics and is therefore denied σ

38 Entropy Rate Balance for Control Volumes Like mass and energy, entropy can be transferred into or out of a control volume by streams of matter. Since this is the principal difference between the closed system and control volume entropy rate balances, the control volume form can be obtained by modifying the closed system form to account for such entropy transfer. The result is (Eq. 6.34) m m e s where i s i and e account, respectively, for rates of entropy transfer accompanying mass flow at inlets i and exits e.

39 Entropy Rate Balance for Control Volumes For control volumes at steady state, Eq reduces to give (Eq. 6.36) For a one-inlet, one-exit control volume at steady state, Eq reduces to give (Eq. 6.37) where 1 and 2 denote the inlet and exit, respectively, and is the common mass flow rate at these locations. m

40 Entropy Rate Balance for Control Volumes Example: Water vapor enters a valve at 0.7 bar, 280 o C and exits at 0.35 bar. (a) If the water vapor undergoes a throttling process, determine the rate of entropy production within the valve, in kj/k per kg of water vapor flowing. (b) What is the source of entropy production in this case? p 1 = 0.7 bar T 1 = 280 o C p 2 = 0.35 bar h 2 = h 1 (a) For a throttling process, there is no significant heat transfer. Thus, Eq reduces to 0 0 = Q j + m s s σ T j j ( 1 2 ) + cv ( ) 1 2 cv 0 = m s s + σ

41 Entropy Rate Balance for Control Volumes Solving σ m cv = s 2 s 1 From Table A-4, h 1 = kj/kg, s 1 = kj/kg K. For a throttling process, h 2 = h 1 (Eq. 4.22). Interpolating in Table A-4 at 0.35 bar and h 2 = kj/kg, s 2 = kj/kg K. Finally σ cv m = ( ) kj/kg K = kj/kg K (b) Selecting from the list of irreversibilities provided in Sec , the source of the entropy production here is the unrestrained expansion to a lower pressure undergone by the water vapor.

42 Entropy Rate Balance for Control Volumes Comment: The value of the entropy production for a single component such as the throttling valve considered here often does not have much significance by itself. The significance of the entropy production of any component is normally determined through comparison with the entropy production values of other components combined with that component to form an integrated system. Reducing irreversibilities of components with the highest entropy production rates may lead to improved thermodynamic performance of the integrated system. p 1 = 0.7 bar T 1 = 280 o C p 2 = 0.35 bar h 2 = h 1

### Chapter 6: Using Entropy

Chapter 6: Using Entropy Photo courtesy of U.S. Military Academy photo archives. Combining the 1 st and the nd Laws of Thermodynamics ENGINEERING CONTEXT Up to this point, our study of the second law has

### Supplementary Notes on Entropy and the Second Law of Thermodynamics

ME 4- hermodynamics I Supplementary Notes on Entropy and the Second aw of hermodynamics Reversible Process A reversible process is one which, having taken place, can be reversed without leaving a change

### THE CLAUSIUS INEQUALITY

Part IV Entropy In Part III, we introduced the second law of thermodynamics and applied it to cycles and cyclic devices. In this part, we apply the second law to processes. he first law of thermodynamics

### Chapter 11. Refrigeration Cycles

Chapter 11 Refrigeration Cycles The vapor compression refrigeration cycle is a common method for transferring heat from a low temperature space to a high temperature space. The figures below show the objectives

### - Know basic of refrigeration - Able to analyze the efficiency of refrigeration system -

Refrigeration cycle Objectives - Know basic of refrigeration - Able to analyze the efficiency of refrigeration system - contents Ideal Vapor-Compression Refrigeration Cycle Actual Vapor-Compression Refrigeration

### Reversible & Irreversible Processes

Reversible & Irreversible Processes Example of a Reversible Process: Cylinder must be pulled or pushed slowly enough (quasistatically) that the system remains in thermal equilibrium (isothermal). Change

### The Second Law of Thermodynamics

The Second aw of Thermodynamics The second law of thermodynamics asserts that processes occur in a certain direction and that the energy has quality as well as quantity. The first law places no restriction

### Entropy. Objectives. MAE 320 - Chapter 7. Definition of Entropy. Definition of Entropy. Definition of Entropy. Definition of Entropy + Δ

MAE 320 - Chapter 7 Entropy Objectives Defe a new property called entropy to quantify the second-law effects. Establish the crease of entropy prciple. Calculate the entropy changes that take place durg

### FUNDAMENTALS OF ENGINEERING THERMODYNAMICS

FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant

### UNIT 5 REFRIGERATION SYSTEMS

UNIT REFRIGERATION SYSTEMS Refrigeration Systems Structure. Introduction Objectives. Vapour Compression Systems. Carnot Vapour Compression Systems. Limitations of Carnot Vapour Compression Systems with

### Chapter 6 Energy Equation for a Control Volume

Chapter 6 Energy Equation for a Control Volume Conservation of Mass and the Control Volume Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries,

### Esystem = 0 = Ein Eout

AGENDA: I. Introduction to Thermodynamics II. First Law Efficiency III. Second Law Efficiency IV. Property Diagrams and Power Cycles V. Additional Material, Terms, and Variables VI. Practice Problems I.

### SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition

SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide Problems -20 Steady

### Esystem = 0 = Ein Eout

AGENDA: I. Introduction to Thermodynamics II. First Law Efficiency III. Second Law Efficiency IV. Property Diagrams and Power Cycles V. Additional Material, Terms, and Variables VI. Practice Problems I.

### Lesson 5 Review of fundamental principles Thermodynamics : Part II

Lesson 5 Review of fundamental principles Thermodynamics : Part II Version ME, IIT Kharagpur .The specific objectives are to:. State principles of evaluating thermodynamic properties of pure substances

### Energy : ṁ (h i + ½V i 2 ) = ṁ(h e + ½V e 2 ) Due to high T take h from table A.7.1

6.33 In a jet engine a flow of air at 000 K, 00 kpa and 30 m/s enters a nozzle, as shown in Fig. P6.33, where the air exits at 850 K, 90 kpa. What is the exit velocity assuming no heat loss? C.V. nozzle.

### ME 201 Thermodynamics

ME 0 Thermodynamics Second Law Practice Problems. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F The maximum work a substance can do is given by its availablity.

### Thermodynamic Systems

Student Academic Learning Services Page 1 of 18 Thermodynamic Systems And Performance Measurement Contents Thermodynamic Systems... 2 Defining the System... 2 The First Law... 2 Thermodynamic Efficiency...

### 6.5 Simple Vapor Compression Refrigeration System:

6.5 Simple Vapor ompression Refrigeration System: A simple vapor compression refrigeration system consists of the following equipments: i) ompressor ii) ondenser iii) Expansion valve iv) Evaporator. B

### Thermodynamics - Example Problems Problems and Solutions

Thermodynamics - Example Problems Problems and Solutions 1 Examining a Power Plant Consider a power plant. At point 1 the working gas has a temperature of T = 25 C. The pressure is 1bar and the mass flow

### (b) 1. Look up c p for air in Table A.6. c p = 1004 J/kg K 2. Use equation (1) and given and looked up values to find s 2 s 1.

Problem 1 Given: Air cooled where: T 1 = 858K, P 1 = P = 4.5 MPa gage, T = 15 o C = 88K Find: (a) Show process on a T-s diagram (b) Calculate change in specific entropy if air is an ideal gas (c) Evaluate

### Chapter 5 The Second Law of Thermodynamics

Chapter 5 he Second aw of hermodynamics he second law of thermodynamics states that processes occur in a certain direction, not in just any direction. Physical processes in nature can proceed toward equilibrium

### The Second Law of Thermodynamics

Objectives MAE 320 - Chapter 6 The Second Law of Thermodynamics The content and the pictures are from the text book: Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill,

### Entropy and The Second Law of Thermodynamics

The Second Law of Thermodynamics (SL) Entropy and The Second Law of Thermodynamics Explain and manipulate the second law State and illustrate by example the second law of thermodynamics Write both the

### Physics 2326 - Assignment No: 3 Chapter 22 Heat Engines, Entropy and Second Law of Thermodynamics

Serway/Jewett: PSE 8e Problems Set Ch. 22-1 Physics 2326 - Assignment No: 3 Chapter 22 Heat Engines, Entropy and Second Law of Thermodynamics Objective Questions 1. A steam turbine operates at a boiler

### APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES

APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES INTRODUCTION This tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more

### AC 2011-2088: ON THE WORK BY ELECTRICITY IN THE FIRST AND SECOND LAWS OF THERMODYNAMICS

AC 2011-2088: ON THE WORK BY ELECTRICITY IN THE FIRST AND SECOND LAWS OF THERMODYNAMICS Hyun W. Kim, Youngstown State University Hyun W. Kim, Ph.D., P.E. Hyun W. Kim is a professor of mechanical engineering

### Chapter 15: Thermodynamics

Chapter 15: Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat

### 1.5. Which thermodynamic property is introduced using the zeroth law of thermodynamics?

CHAPTER INTRODUCTION AND BASIC PRINCIPLES. (Tutorial). Determine if the following properties of the system are intensive or extensive properties: Property Intensive Extensive Volume Density Conductivity

### Exergy: the quality of energy N. Woudstra

Exergy: the quality of energy N. Woudstra Introduction Characteristic for our society is a massive consumption of goods and energy. Continuation of this way of life in the long term is only possible if

### Second Law of Thermodynamics

Thermodynamics T8 Second Law of Thermodynamics Learning Goal: To understand the implications of the second law of thermodynamics. The second law of thermodynamics explains the direction in which the thermodynamic

### Availability. Second Law Analysis of Systems. Reading Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.

Availability Readg Problems 10.1 10.4 10.59, 10.65, 10.66, 10.67 10.69, 10.75, 10.81, 10.88 Second Law Analysis of Systems AVAILABILITY: the theoretical maximum amount of reversible work that can be obtaed

### = Q H Q C Q H Q C Q H Q C. ω = Q C W =

I.D The Second Law The historical development of thermodynamics follows the industrial olution in the 19 th century, and the advent of heat engines. It is interesting to see how such practical considerations

### Refrigeration and Air Conditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Refrigeration and Air Conditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No. # 18 Worked Out Examples - I Welcome back in this lecture.

### Mass and Energy Analysis of Control Volumes

MAE 320-Chapter 5 Mass and Energy Analysis of Control Volumes Objectives Develop the conservation of mass principle. Apply the conservation of mass principle to various systems including steady- and unsteady-flow

### THE SECOND LAW OF THERMODYNAMICS

1 THE SECOND LAW OF THERMODYNAMICS The FIRST LAW is a statement of the fact that ENERGY (a useful concept) is conserved. It says nothing about the WAY, or even WHETHER one form of energy can be converted

### Lesson 10 Vapour Compression Refrigeration Systems

Lesson 0 Vapour Compression Refrigeration Systems Version ME, II Kharagpur he specific objectives of the lesson: his lesson discusses the most commonly used refrigeration system, ie Vapour compression

### ESO 201A/202. End Sem Exam 120 Marks 3 h 19 Nov Roll No. Name Section

ESO 201A/202 End Sem Exam 120 Marks 3 h 19 Nov 2014 Roll No. Name Section Answer Questions 1 and 2 on the Question Paper itself. Answer Question 3, 4 and 5 on the Answer Booklet provided. At the end of

### Process Integration Prof. Bikash Mohanty Department of Chemical Engineering Indian Institute of Technology, Roorkee

Process Integration Prof. Bikash Mohanty Department of Chemical Engineering Indian Institute of Technology, Roorkee Module - 6 Integration and placement of equipment Lecture - 4 Placement of Heat Engine,

### An introduction to thermodynamics applied to Organic Rankine Cycles

An introduction to thermodynamics applied to Organic Rankine Cycles By : Sylvain Quoilin PhD Student at the University of Liège November 2008 1 Definition of a few thermodynamic variables 1.1 Main thermodynamics

### where V is the velocity of the system relative to the environment.

Exergy Exergy is the theoretical limit for the wor potential that can be obtaed from a source or a system at a given state when teractg with a reference (environment) at a constant condition. A system

### = T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T

hermodynamics: Examples for chapter 3. 1. Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c,

### In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0

Chapter 4 Entropy and second law of thermodynamics 4.1 Carnot cycle In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0 since the internal energy U is a

### 4.1 Introduction. 4.2 Work. Thermodynamics

Thermodynamics 4-1 Chapter 4 Thermodynamics 4.1 Introduction This chapter focusses on the turbine cycle:thermodynamics and heat engines. The objective is to provide enough understanding of the turbine

### Laws of Thermodynamics

Laws of Thermodynamics Thermodynamics Thermodynamics is the study of the effects of work, heat, and energy on a system Thermodynamics is only concerned with macroscopic (large-scale) changes and observations

### Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson

Thermochemistry r2 d:\files\courses\1110-20\99heat&thermorans.doc Ron Robertson I. What is Energy? A. Energy is a property of matter that allows work to be done B. Potential and Kinetic Potential energy

### Entropy and the Second Law of Thermodynamics. The Adiabatic Expansion of Gases

Lecture 7 Entropy and the Second Law of Thermodynamics 15/08/07 The Adiabatic Expansion of Gases In an adiabatic process no heat is transferred, Q=0 = C P / C V is assumed to be constant during this process

### The First Law of Thermodynamics

The First Law of Thermodynamics (FL) The First Law of Thermodynamics Explain and manipulate the first law Write the integral and differential forms of the first law Describe the physical meaning of each

### We will try to get familiar with a heat pump, and try to determine its performance coefficient under different circumstances.

C4. Heat Pump I. OBJECTIVE OF THE EXPERIMENT We will try to get familiar with a heat pump, and try to determine its performance coefficient under different circumstances. II. INTRODUCTION II.1. Thermodynamic

### Clausius Inequality (contd )

Module 7 Entropy Suppose we have an engine that receives from several heat reservoirs and rejects heat to several reservoirs, we still have the equation valid. Clausius Inequality Clausius Inequality (contd

### ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

Chapter 20: ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 1. In a reversible process the system: A. is always close to equilibrium states B. is close to equilibrium states only at the beginning and end

### CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS. Blank

CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS Blank SONNTAG/BORGNAKKE STUDY PROBLEM 7-1 7.1 A car engine and its fuel consumption A car engine produces 136 hp on the output shaft with a thermal efficiency

### Expansion and Compression of a Gas

Physics 6B - Winter 2011 Homework 4 Solutions Expansion and Compression of a Gas In an adiabatic process, there is no heat transferred to or from the system i.e. dq = 0. The first law of thermodynamics

### Heat Engines, Entropy, and the Second Law of Thermodynamics

2.2 This is the Nearest One Head 669 P U Z Z L E R The purpose of a refrigerator is to keep its contents cool. Beyond the attendant increase in your electricity bill, there is another good reason you should

### The First Law of Thermodynamics: Closed Systems. Heat Transfer

The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy gained

### Physics 5D - Nov 18, 2013

Physics 5D - Nov 18, 2013 30 Midterm Scores B } Number of Scores 25 20 15 10 5 F D C } A- A A + 0 0-59.9 60-64.9 65-69.9 70-74.9 75-79.9 80-84.9 Percent Range (%) The two problems with the fewest correct

### Chapter 11, Problem 18.

Chapter 11, Problem 18. Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and 10 C at a rate of 0.12 kg/s, and it leaves at 0.7 MPa and 50 C. The refrigerant is

### Basic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Basic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 10 Second Law and Available Energy - I Good morning, I welcome you to this

### UNIT 2 REFRIGERATION CYCLE

UNIT 2 REFRIGERATION CYCLE Refrigeration Cycle Structure 2. Introduction Objectives 2.2 Vapour Compression Cycle 2.2. Simple Vapour Compression Refrigeration Cycle 2.2.2 Theoretical Vapour Compression

### Sustainable Energy Utilization, Refrigeration Technology. A brief history of refrigeration. Björn Palm

Sustainable Energy Utilization, Refrigeration Technology Björn Palm bpalm@energy.kth.se This email message was generated automatically. Please do not send a reply. You are registered for the following

### A HEAT ENGINE: PRODUCES WORK FROM HEAT BY WASTING A FRACTION OF HEAT INPUT TO A LOW TEMPERATURE RESERVOIR

RANKINE POWER GENERAION CYCE A EA ENGINE: PRODUCES WORK FROM EA BY WASING A FRACION OF EA INPU O A OW EMPERAURE RESERVOIR o C CARACERISICS s (kj/kg-k. Rankine cycle is a eat engine comprised of four internally

### AN INTRODUCTION TO THE CONCEPT OF EXERGY AND ENERGY QUALITY. Truls Gundersen

AN INRODUION O HE ONEP OF EXERGY AND ENERGY QUALIY by ruls Gundersen Department of Energy and Process Engineering Norwegian University of Science and echnology rondheim, Norway Version 4, March 211 ruls

### APPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction.

APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas expansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend

### THERMODYNAMICS NOTES - BOOK 2 OF 2

THERMODYNAMICS & FLUIDS (Thermodynamics level 1\Thermo & Fluids Module -Thermo Book 2-Contents-December 07.doc) UFMEQU-20-1 THERMODYNAMICS NOTES - BOOK 2 OF 2 Students must read through these notes and

### Chapter 22. Heat Engines, Entropy, and the Second Law of Thermodynamics

Chapter 22 Heat Engines, Entropy, and the Second Law of Thermodynamics C HAP T E R O UTLI N E 221 Heat Engines and the Second Law of Thermodynamics 222 Heat Pumps and Refrigerators 223 Reversible and Irreversible

### Lecture 36 (Walker 18.8,18.5-6,)

Lecture 36 (Walker 18.8,18.5-6,) Entropy 2 nd Law of Thermodynamics Dec. 11, 2009 Help Session: Today, 3:10-4:00, TH230 Review Session: Monday, 3:10-4:00, TH230 Solutions to practice Lecture 36 final on

### Chapter 6 The first law and reversibility

Chapter 6 The first law and reversibility 6.1 The first law for processes in closed systems We have discussed the properties of equilibrium states and the relationship between the thermodynamic parameters

### Problem Set 1 3.20 MIT Professor Gerbrand Ceder Fall 2003

LEVEL 1 PROBLEMS Problem Set 1 3.0 MIT Professor Gerbrand Ceder Fall 003 Problem 1.1 The internal energy per kg for a certain gas is given by U = 0. 17 T + C where U is in kj/kg, T is in Kelvin, and C

### Process chosen to calculate entropy change for the system

Chapter 6 Exaple 6.3-3 3. ---------------------------------------------------------------------------------- An insulated tank (V 1.6628 L) is divided into two equal parts by a thin partition. On the left

### C H A P T E R T W O. Fundamentals of Steam Power

35 C H A P T E R T W O Fundamentals of Steam Power 2.1 Introduction Much of the electricity used in the United States is produced in steam power plants. Despite efforts to develop alternative energy converters,

### There is no general classification of thermodynamic cycle. The following types will be discussed as those are included in the course curriculum

THERMODYNAMIC CYCLES There is no general classification of thermodynamic cycle. The following types will be discussed as those are included in the course curriculum 1. Gas power cycles a) Carnot cycle

### Practice Problems on Conservation of Energy. heat loss of 50,000 kj/hr. house maintained at 22 C

COE_10 A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kj/hr is maintained at 22 C at all times during a winter night for 10 hr. The house is to be heated by 50 glass

### ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6

Spring 2003 Page 1 of 6 2-30 Steam Tables Given: Property table for H 2 O Find: Complete the table. T ( C) P (kpa) h (kj/kg) x phase description a) 120.23 200 2046.03 0.7 saturated mixture b) 140 361.3

### Isentropic flow. Wikepedia

Isentropic flow Wikepedia In thermodynamics, an isentropic process or isoentropic process (ισον = "equal" (Greek); εντροπία entropy = "disorder"(greek)) is one in which for purposes of engineering analysis

### N 2 C balance: 3a = = 9.2 a = H CO = CO H 2. theo. A-F ratio =

15.26 Liquid propane is burned with dry air. A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis: 8.6% CO 2, 0.6% CO, 7.2% O 2 and 83.6% N

### Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 10 Steam Power Cycle, Steam Nozzle Good afternoon everybody.

### Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator

Lecture. Real eat Engines and refrigerators (Ch. ) Stirling heat engine Internal combustion engine (Otto cycle) Diesel engine Steam engine (Rankine cycle) Kitchen Refrigerator Carnot Cycle - is not very

### Introduction to Thermodynamics

Introduction to Thermodynamics Training Objectives The participant will be introduced to: 1.1 basic concepts and definitions. 1.2 the properties of a pure substance. 1.3 work and heat. 1.4 the fist law

### 20 m neon m propane 20

Problems with solutions:. A -m 3 tank is filled with a gas at room temperature 0 C and pressure 00 Kpa. How much mass is there if the gas is a) Air b) Neon, or c) Propane? Given: T73K; P00KPa; M air 9;

### ENGINEERING COUNCIL CERTIFICATE LEVEL THERMODYNAMIC, FLUID AND PROCESS ENGINEERING C106 TUTORIAL 5 STEAM POWER CYCLES

ENGINEERING COUNCIL CERTIFICATE LEVEL THERMODYNAMIC, FLUID AND PROCESS ENGINEERING C106 TUTORIAL 5 STEAM POWER CYCLES When you have completed this tutorial you should be able to do the following. Explain

### Engineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 919-9670 Web Site:

Engineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 919-9670 E-Mail: info@engineering-4e.com Web Site: http://www.engineering-4e.com Brayton Cycle (Gas Turbine) for Propulsion Application

### THERMODYNAMICS TUTORIAL 5 HEAT PUMPS AND REFRIGERATION. On completion of this tutorial you should be able to do the following.

THERMODYNAMICS TUTORIAL 5 HEAT PUMPS AND REFRIGERATION On completion of this tutorial you should be able to do the following. Discuss the merits of different refrigerants. Use thermodynamic tables for

### THE UNIVERSITY OF TRINIDAD & TOBAGO

THE UNIVERSITY OF TRINIDAD & TOBAGO FINAL ASSESSMENT/EXAMINATIONS JANUARY/APRIL 2013 Course Code and Title: Programme: THRM3001 Engineering Thermodynamics II Bachelor of Applied Sciences Date and Time:

### Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 7 Ideal Gas Laws, Different Processes Let us continue

### UNIT-III PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE

UNIT-III PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE Pure Substance A Pure substance is defined as a homogeneous material, which retains its chemical composition even though there may be a change

### Final Exam Review Questions PHY Final Chapters

Final Exam Review Questions PHY 2425 - Final Chapters Section: 17 1 Topic: Thermal Equilibrium and Temperature Type: Numerical 12 A temperature of 14ºF is equivalent to A) 10ºC B) 7.77ºC C) 25.5ºC D) 26.7ºC

### ME 239: Rocket Propulsion. Nozzle Thermodynamics and Isentropic Flow Relations. J. M. Meyers, PhD

ME 39: Rocket Propulsion Nozzle Thermodynamics and Isentropic Flow Relations J. M. Meyers, PhD 1 Assumptions for this Analysis 1. Steady, one-dimensional flow No motor start/stopping issues to be concerned

### Mohan Chandrasekharan #1

International Journal of Students Research in Technology & Management Exergy Analysis of Vapor Compression Refrigeration System Using R12 and R134a as Refrigerants Mohan Chandrasekharan #1 # Department

### Fundamentals of Thermodynamics Applied to Thermal Power Plants

Fundamentals of Thermodynamics Applied to Thermal Power Plants José R. Simões-Moreira Abstract In this chapter it is reviewed the fundamental principles of Thermodynamics aiming at its application to power

### Thermodynamics Answers to Tutorial # 1

Thermodynamics Answers to Tutorial # 1 1. (I) Work done in free expansion is Zero as P ex = 0 (II) Irreversible expansion against constant external pressure w = P ex (V 2 V 1 ) V 2 = nrt P 2 V 1 = nrt

### PG Student (Heat Power Engg.), Mechanical Engineering Department Jabalpur Engineering College, India. Jabalpur Engineering College, India.

International Journal of Emerging Trends in Engineering and Development Issue 3, Vol. (January 23) EFFECT OF SUB COOLING AND SUPERHEATING ON VAPOUR COMPRESSION REFRIGERATION SYSTEMS USING 22 ALTERNATIVE

### VAPOUR-COMPRESSION REFRIGERATION SYSTEM

VAPOUR-COMPRESSION REFRIGERATION SYSTEM This case study demonstrates the modeling of a vapour compression refrigeration system using R22 as refrigerant. Both a steady-state and a transient simulation will

### The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.

The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed

### EXPERIMENT 21. Thermal Efficiency Apparatus. Introduction. Experiment 21. Thermal Efficiency Apparatus

EXERIMENT 21 Thermal Efficiency Apparatus Introduction The Thermal Efficiency Apparatus can be used as a heat engine or a heat pump. hen used as a heat engine, heat from the hot reservoir is used to do

### Chapter 6: Entropy and the Laws of Thermodynamics

Chapter 6: Entropy and the Laws of Thermodynamics Goals of Period 6 Section 6.1: To examine order, disorder and entropy Section 6.2: To discuss conservation of energy and the first law of thermodynamics

### Sheet 5:Chapter 5 5 1C Name four physical quantities that are conserved and two quantities that are not conserved during a process.

Thermo 1 (MEP 261) Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles 7 th Edition, McGraw-Hill Companies, ISBN-978-0-07-352932-5, 2008 Sheet 5:Chapter 5 5 1C Name four physical

### ME 6404 THERMAL ENGINEERING. Part-B (16Marks questions)

ME 6404 THERMAL ENGINEERING Part-B (16Marks questions) 1. Drive and expression for the air standard efficiency of Otto cycle in terms of volume ratio. (16) 2. Drive an expression for the air standard efficiency

### a) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L

hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal