Student Study Guide for 5. th. edition of Thermodynamics by Y. A. Çengel & M. A. Boles Chapter_ 1 to 8 and 10

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-1

Study Guide for Thermodynamics: an Engineering Approach By Michael A. Boles Department of Mechanical and Aerospace Engineering NC State University Raleigh, NC 2795-7910

Chapter 1-1

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

A CP CV COP d E e F g H h h K k k kt M M m N n η ηth P Pa Qnet qnet R Ru S s T U

2

1-2

Nomenclature

area (m ) specific heat at constant pressure (kJ/(kg⋅K)) specific heat at constant volume (kJ/(kg⋅K)) coefficient of performance exact differential stored energy (kJ) stored energy per unit mass (kJ/kg) force (N) acceleration of gravity ( 9.807 m/s2) enthalpy (H= U + PV) (kJ) specific enthalpy (h= u + Pv) (kJ/kg) convective heat transfer coefficient (W/(m2⋅K) Kelvin degrees specific heat ratio, CP/CV 103 thermal conductivity (W/(m-°C)) molecular weight or molar mass (kg/kmol) 106 mass (kg) moles (kmol) polytropic exponent (isentropic process, ideal gas n = k) isentropic efficiency for turbines, compressors, nozzles thermal efficiency (net work done/heat added) pressure (kPa, MPa, psia, psig) Pascal (N/m2) net heat transfer (∑Qin - ∑Qout) (kJ) Qnet /m, net heat transfer per unit mass (kJ/kg) particular gas constant (kJ/(kg⋅K)) universal gas constant (= 8.314 kJ/(kmol⋅K) ) entropy (kJ/K) specific entropy (kJ/(kg⋅K)) temperature ( °C, K, °F, R) internal energy (kJ) Chapter 1-2

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

u V V G V

v v

X X x Z Wnet wnet Wt δ ε φ ρ ω

1-3

specific internal energy (kJ/(kg ⋅K)) volume (m3 ) volume flow rate (m3/s) velocity (m/s) specific volume (m3/kg) molar specific volume (m3/kmol) distance (m) exergy (kJ) quality elevation (m) net work done [(∑Wout - ∑Win)other + Wb] (kJ) 2 where Wb = 1 PdV for closed systems and 0 for control volumes Wnet /m, net work done per unit mass (kJ/kg) weight (N) inexact differential regenerator effectiveness relative humidity density (kg/m3) humidity ratio

z

Subscripts, superscripts a actual b boundary f saturated liquid state g saturated vapor state fg saturated vapor value minus saturated liquid value gen generation H high temperature HP heat pump L low temperature net net heat added to system or net work done by system other work done by shaft and electrical means P constant pressure Chapter 1-3

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

REF rev s sat v 1 2 i e ⋅

1-4

refrigerator reversible isentropic or constant entropy or reversible, adiabatic saturation value constant volume initial state finial state inlet state exit state per unit time

REFERENCE Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed., New York, McGraw-Hill: 2006.

Chapter 1-4

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-5

Chapter 1: Introduction and Basic Concepts INTRODUCTION The study of thermodynamics is concerned with the ways energy is stored within a body and how energy transformations, which involve heat and work, may take place. One of the most fundamental laws of nature is the conservation of energy principle. It simply states that during an energy interaction, energy can change from one form to another but the total amount of energy remains constant. That is, energy cannot be created or destroyed. This review of thermodynamics is based on the macroscopic approach where a large number of particles, called molecules, make up the substance in question. The macroscopic approach to thermodynamics does not require knowledge of the behavior of individual particles and is called classical thermodynamics. It provides a direct and easy way to obtain the solution of engineering problems without being overly cumbersome. A more elaborate approach, based on the average behavior of large groups of individual particles, is called statistical thermodynamics. This microscopic approach is rather involved and is not reviewed here and leads to the definition of the second law of thermodynamics. We will approach the second law of thermodynamics from the classical point of view and will learn that the second law of thermodynamics asserts that energy has quality as well as quantity, and actual processes occur in the direction of decreasing quality of energy. Closed, Open, and Isolated Systems A thermodynamic system, or simply system, is defined as a quantity of matter or a region in space chosen for study. The region outside the system is called the surroundings. The real or imaginary surface that

Chapter 1-5

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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separates the system from its surroundings is called the boundary. The boundary of a system may be fixed or movable. Surroundings are physical space outside the system boundary.

Systems may be considered to be closed or open, depending on whether a fixed mass or a fixed volume in space is chosen for study. A closed system consists of a fixed amount of mass and no mass may cross the system boundary. The closed system boundary may move. Examples of closed systems are sealed tanks and piston cylinder devices (note the volume does not have to be fixed). However, energy in the form of heat and work may cross the boundaries of a closed system.

Chapter 1-6

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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An open system, or control volume, has mass as well as energy crossing the boundary, called a control surface. Examples of open systems are pumps, compressors, turbines, valves, and heat exchangers.

An isolated system is a general system of fixed mass where no heat or work may cross the boundaries. An isolated system is a closed system with no energy crossing the boundaries and is normally a collection of a main system and its surroundings that are exchanging mass and energy among themselves and no other system.

Isolated System Boundary Heat = 0 Work = 0 Mass = 0 Across Isolated Boundary

Work Mass Surr 1

Surr 4

System Heat

Surr 3 Mass

Surr 2 Chapter 1-7

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-8

Since some of the thermodynamic relations that are applicable to closed and open systems are different, it is extremely important that we recognize the type of system we have before we start analyzing it. Properties of a System Any characteristic of a system in equilibrium is called a property. The property is independent of the path used to arrive at the system condition. Some thermodynamic properties are pressure P, temperature T, volume V, and mass m. Properties may be intensive or extensive. Extensive properties are those that vary directly with size--or extent--of the system. Some Extensive Properties a. mass b. volume c. total energy d. mass dependent property Intensive properties are those that are independent of size. Some Intensive Properties a. temperature b. pressure c. age d. color e. any mass independent property Chapter 1-8

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Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Extensive properties per unit mass are intensive properties. For example, the specific volume v, defined as

Volume V v= = mass m and density ρ, defined as

mass m ρ= = volume V

FG m IJ H kg K 3

FG kg IJ Hm K 3

are intensive properties.

Units An important component to the solution of any engineering thermodynamic problem requires the proper use of units. The unit check is the simplest of all engineering checks that can be made for a given

Chapter 1-9

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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solution. Since units present a major hindrance to the correct solution of thermodynamic problems, we must learn to use units carefully and properly. The system of units selected for this course is the SI System, also known as the International System (sometimes called the metric system). In SI, the units of mass, length, and time are the kilogram (kg), meter (m), and second (s), respectively. We consider force to be a derived unit from Newton's second law, i.e.,

Force = (mass)(acceleration) F = ma In SI, the force unit is the newton (N), and it is defined as the force required to accelerate a mass of 1 kg at a rate of 1 m/s2. That is,

m 1N = (1kg )(1 2 ) s This definition of the newton is used as the basis of the conversion factor to convert mass-acceleration units to force units. The term weight is often misused to express mass. Unlike mass, weight Wt is a force. Weight is the gravitational force applied to a body, and its magnitude is determined from Newton's second law.

Wt = mg where m is the mass of the body and g is the local gravitational acceleration (g is 9.807 m/s2 at sea level and 45 °latitude). The weight of a unit volume of a substance is called the specific weight w and is determined from w = ρ g, where ρ is density.

Chapter 1-10

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1-11

Oftentimes, the engineer must work in other systems of units. Comparison of the United States Customary System (USCS), or English System, and the slug system of units with the SI system is shown below. SI mass kilogram (kg) time second (s) length meter (m) force newton (N)

USCS pound-mass (lbm) second (s) foot (ft) pound-force (lbf)

Slug slug-mass (slug) second (s) foot (ft) pound-force (lbf)

Sometimes we use the mole number in place of the mass. In SI units the mole number is in kilogram-moles, or kmol. Newton’s second law is often written as

ma F = gc where gc is called the gravitational constant and is obtained from the force definition. In the SI System 1 newton is that force required to accelerate 1 kg mass 1 m/s2. The gravitational constant in the SI System is

m (1kg)(1 2 ) ma kg m s = =1 gc = F 1N N s2 In the USCS 1 pound-force is that force required to accelerate 1 poundmass 32.176 ft/s2. The gravitational constant in the USCS is Chapter 1-11

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

ma gc = = F

(1lbm)(32.2 1lbf

1-12

ft ) 2 s = 32.2 lbm ft lbf s2

In the slug system, the gravitational constant is

ft (1slug)(1 2 ) lbm ft ma s gc = = =1 lbf s2 F 1 lbf Example 1-1 An object at sea level has a mass of 400 kg. a) Find the weight of this object on earth. b) Find the weight of this object on the moon where the local gravitational acceleration is one-sixth that of earth. (a)

Wt = mg

F I mIG 1 N J F W = (400 kg )G 9.807 J G H s K G kg m JJ H sK t

2

2

= 3922.8 N Chapter 1-12

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-13

Note the use of the conversion factor to convert mass-acceleration units into force units.

(b)

Wt = mg

F I 9.807 m I G 1 N J F = (400 kg )G H 6 s JK GG kg m JJ H sK 2

2

= 6538 . N Example 1-2E An object has a mass of 180 lbm. Find the weight of this object at a location where the local gravitational acceleration is 30 ft/s2.

Wt = mg

= (180 lbm)(30

1 lbf ft )( ) 2 ft s 32.2 lbm s2

= 167.7 lbf

Chapter 1-13

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-14

State, Equilibrium, Process, and Properties State Consider a system that is not undergoing any change. The properties can be measured or calculated throughout the entire system. This gives us a set of properties that completely describe the condition or state of the system. At a given state all of the properties are known; changing one property changes the state. Equilibrium A system is said to be in thermodynamic equilibrium if it maintains thermal (uniform temperature), mechanical (uniform pressure), phase (the mass of two phases, e.g., ice and liquid water, in equilibrium) and chemical equilibrium.

Chapter 1-14

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-15

Process Any change from one state to another is called a process. During a quasiequilibrium or quasi-static process the system remains practically in equilibrium at all times. We study quasi-equilibrium processes because they are easy to analyze (equations of state apply) and work-producing devices deliver the most work when they operate on the quasi-equilibrium process.

In most of the processes that we will study, one thermodynamic property is held constant. Some of these processes are

Process Property held constant isobaric pressure isothermal temperature isochoric volume isentropic entropy (see Chapter 7)

System Boundary F Water

Constant Pressure Process

We can understand the concept of a constant pressure process by considering the above figure. The force exerted by the water on the face of the piston has to equal the force due to the combined weight of the piston and the bricks. If the combined weight of the piston and bricks is constant, then F is constant and the pressure is constant even when the water is heated. Chapter 1-15

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-16

We often show the process on a P-V diagram as shown below.

Steady-Flow Process Consider a fluid flowing through an open system or control volume such as a water heater. The flow is often defined by the terms steady and uniform. The term steady implies that there are no changes with time. The term uniform implies no change with location over a specified region. Engineering flow devices that operate for long periods of time under the same conditions are classified as steady-flow devices. The processes for these devices is called the steady-flow process. The fluid properties can change from point to point with in the control volume, but at any fixed point the properties remain the same during the entire process. State Postulate As noted earlier, the state of a system is described by its properties. But by experience not all properties must be known before the state is Chapter 1-16

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-17

specified. Once a sufficient number of properties are known, the state is specified and all other properties are known. The number of properties required to fix the state of a simple, homogeneous system is given by the state postulate: The thermodynamic state of a simple compressible system is completely specified by two independent, intensive properties. Cycle A process (or a series of connected processes) with identical end states is called a cycle. Below is a cycle composed of two processes, A and B. Along process A, the pressure and volume change from state 1 to state 2. Then to complete the cycle, the pressure and volume change from state 2 back to the initial state 1 along process B. Keep in mind that all other thermodynamic properties must also change so that the pressure is a function of volume as described by these two processes. P

2

Process B Process A

1

V

Pressure Force per unit area is called pressure, and its unit is the pascal, N/m2, in the SI system and psia, lbf/in2 absolute, in the English system.

Chapter 1-17

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

P=

1-18

Force F = Area A

N m2 N 1 MPa = 106 2 = 103 kPa m 1 kPa = 103

The pressure used in all calculations of state is the absolute pressure measured relative to absolute zero pressure. However, pressures are often measured relative to atmospheric pressure, called gage or vacuum pressures. In the English system the absolute pressure and gage pressures are distinguished by their units, psia (pounds force per square inch absolute) and psig (pounds force per square inch gage), respectively; however, the SI system makes no distinction between absolute and gage pressures.

These pressures are related by

Pgage = Pabs − Patm Pvac = Patm − Pabs Chapter 1-18

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-19

Or these last two results may be written as

Pabs = Patm ± Pgage Where the +Pgage is used when Pabs > Patm and –Pgage is used for a vacuum gage. The relation among atmospheric, gage, and vacuum pressures is shown below.

Some values of 1 atm of pressure are 101.325 kPa, 0.101325 MPa, 14.696 psia, 760 mmHg, and 29.92 inches H2O.

Small to moderate pressure differences are measured by a manometer and a differential fluid column of height h corresponds to a pressure difference between the system and the surroundings of the manometer.

Chapter 1-19

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-20

This pressure difference is determined from the manometer fluid displaced height as

∆P = ρ g h

( kPa )

The text gives an extensive review of the manometer pressure relations. For further study of the manometer pressure relations, see the text. Other devices for measuring pressure differences are shown below.

Chapter 1-20

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-21

Example 1-3 A vacuum gage connected to a tank reads 30 kPa at a location where the atmospheric pressure is 98 kPa. What is the absolute pressure in the tank?

Pabs = Patm − Pgage = 98 kPa − 30 kPa = 68 kPa

Chapter 1-21

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-22

Example 1-4 A pressure gage connected to a valve stem of a truck tire reads 240 kPa at a location where the atmospheric pressure is 100 kPa. What is the absolute pressure in the tire, in kPa and in psia?

Pabs = Patm + Pgage = 100 kPa + 240 kPa = 340 kPa The pressure in psia is

14.7 psia = 49.3 psia Pabs = 340 kPa 1013 . kPa What is the gage pressure of the air in the tire, in psig?

Pgage = Pabs − Patm = 49.3 psia − 14.7 psia = 34.6 psig Check the side walls of the tires on your car or truck. What is the maximum allowed pressure? Is this pressure in gage or absolute values?

Chapter 1-22

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-23

Example 1-5 Both a gage and a manometer are attached to a gas tank to measure its pressure. If the pressure gage reads 80 kPa, determine the distance between the two fluid levels of the manometer if the fluid is mercury, whose density is 13,600 kg/m3.

∆P h= ρg

103 N / m2 80 kPa kPa h= kg m 1N 13600 3 9.807 2 m s kg m / s2 = 0.6 m Temperature Although we are familiar with temperature as a measure of “hotness” or “coldness,” it is not easy to give an exact definition of it. However, temperature is considered as a thermodynamic property that is the measure of the energy content of a mass. When heat energy is transferred to a body, the body's energy content increases and so does its temperature. In fact it is the difference in temperature that causes energy, called heat transfer, to flow from a hot body to a cold body. Two bodies are in thermal equilibrium when they have reached the same temperature. If two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. This simple fact is known as the zeroth law of thermodynamics.

Chapter 1-23

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

1-24

The temperature scales used in the SI and the English systems today are the Celsius scale and Fahrenheit scale, respectively. These two scales are based on a specified number of degrees between the freezing point of water ( 0°C or 32°F) and the boiling point of water (100°C or 212°F) and are related by

9 T ° F = T ° C + 32 5 Example 1-6 Water boils at 212 °F at one atmosphere pressure. At what temperature does water boil in °C.

T = (T ° F − 32)

5 5° C = (212 − 32) ° F = 100° C 9 9° F

Like pressure, the temperature used in thermodynamic calculations must be in absolute units. The absolute scale in the SI system is the Kelvin scale, which is related to the Celsius scale by

T K = T ° C + 273.15 In the English system, the absolute temperature scale is the Rankine scale, which is related to the Fahrenheit scale by

T R = T ° F+ 459.67 Also, note that

T R = 1.8 T K Chapter 1-24

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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Below is a comparison of the temperature scales. °C 99.975

0.01

-273.15

K 373.125

273.16

0

°F 211.955

32.02

-273.15

R 671.625

Boiling point of water at 1 atm

491.69

Triple point of water

0

Absolute zero

This figure shows that that according to the International Temperature Scale of 1990 (ITS-90) the reference state for the thermodynamic temperature scale is the triple point of water, 0.01 °C. The ice point is 0°C, but the steam point is 99.975°C at 1 atm and not 100°C as was previously established. The magnitude of the kelvin, K, is 1/273.16 of the thermodynamic temperature of the triple point of water. The magnitudes of each division of 1 K and 1°C are identical, and so are the magnitudes of each division of 1 R and 1°F. That is,

Chapter 1-25

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

°

°

1-26

∆T K = (T2 C + 273.15) - (T1 C + 273.15) °

°

= T2 C - T1 C = ∆T ° C ∆T R = ∆T ° F

Chapter 1-26

2-1

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Chapter 2: Energy, Energy Transfer, and General Energy Analysis We will soon learn how to apply the first law of thermodynamics as the expression of the conservation of energy principle. But, first we study the ways in which energy may be transported across the boundary of a general thermodynamic system. For closed systems (fixed mass systems) energy can cross the boundaries of a closed system only in the form of heat or work. For open systems or control volumes energy can cross the control surface in the form of heat, work, and energy transported by the mass streams crossing the control surface. We now consider each of these modes of energy transport across the boundaries of the general thermodynamic system. Energy G

Consider the system shown below moving with a velocity V at an elevation Z relative to the reference plane.

General System

G V

CM

Z Reference Plane, Z=0

The total energy E of a system is the sum of all forms of energy that can exist within the system such as thermal, mechanical, kinetic, potential, electric, magnetic, chemical, and nuclear. The total energy of the system is normally thought of as the sum of the internal energy, kinetic energy,

Chapter 2-1

2-2

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

and potential energy. The internal energy U is that energy associated with the molecular structure of a system and the degree of the molecular activity (see Section 2-1 of text for more detail). The kinetic energy KE exists as a result of the system's motion relative to an external reference G frame. When the system moves with velocity V the kinetic energy is expressed as

G2 V KE = m 2

( kJ )

The energy that a system possesses as a result of its elevation in a gravitational field relative to the external reference frame is called potential energy PE and is expressed as

PE = mgZ

( kJ )

where g is the gravitational acceleration and z is the elevation of the center of gravity of a system relative to the reference frame. The total energy of the system is expressed as

E = U + KE + PE

( kJ )

or, on a unit mass basis,

E U KE PE = + + m m m m G2 V = u+ + gZ 2

e=

(

kJ ) kg

where e = E/m is the specific stored energy, and u = U/m is the specific internal energy. The change in stored energy of a system is given by Chapter 2-2

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

∆E = ∆U + ∆KE + ∆PE

2-3

( kJ )

Most closed systems remain stationary during a process and, thus, experience no change in their kinetic and potential energies. The change in the stored energy is identical to the change in internal energy for stationary systems.

If ∆KE = ∆PE = 0,

∆E = ∆U

( kJ )

Energy Transport by Heat and Work and the Classical Sign Convention Energy may cross the boundary of a closed system only by heat or work. Energy transfer across a system boundary due solely to the temperature difference between a system and its surroundings is called heat. Energy transferred across a system boundary that can be thought of as the energy expended to lift a weight is called work. Heat and work are energy transport mechanisms between a system and its surroundings. The similarities between heat and work are as follows: 1. Both are recognized at the boundaries of a system as they cross the boundaries. They are both boundary phenomena. 2. Systems possess energy, but not heat or work.

Chapter 2-3

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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3. Both are associated with a process, not a state. Unlike properties, heat or work has no meaning at a state. 4. Both are path functions (i.e., their magnitudes depends on the path followed during a process as well as the end states. Since heat and work are path dependent functions, they have inexact differentials designated by the symbol δ. The differentials of heat and work are expressed as δQ and δW. The integral of the differentials of heat and work over the process path gives the amount of heat or work transfer that occurred at the system boundary during a process. 2

∫

δ Q = Q12

(not ∆Q)

δ W = W12

(not ∆W )

1, along path 2

∫

1, along path

That is, the total heat transfer or work is obtained by following the process path and adding the differential amounts of heat (δQ) or work (δW) along the way. The integrals of δQ and δW are not Q2 – Q1 and W2 – W1, respectively, which are meaningless since both heat and work are not properties and systems do not possess heat or work at a state. The following figure illustrates that properties (P, T, v, u, etc.) are point functions, that is, they depend only on the states. However, heat and work are path functions, that is, their magnitudes depend on the path followed.

Chapter 2-4

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-5

700 kPa

100 kPa

0.01 m3

0.03 m3

A sign convention is required for heat and work energy transfers, and the classical thermodynamic sign convention is selected for these notes. According to the classical sign convention, heat transfer to a system and work done by a system are positive; heat transfer from a system and work a system are negative. The system shown below has heat supplied to it and work done by it. In this study guide we will use the concept of net heat and net work.

System Boundary Chapter 2-5

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-6

Energy Transport by Heat Recall that heat is energy in transition across the system boundary solely due to the temperature difference between the system and its surroundings. The net heat transferred to a system is defined as

Qnet = ∑ Qin − ∑ Qout Here, Qin and Qout are the magnitudes of the heat transfer values. In most thermodynamics texts, the quantity Q is meant to be the net heat transferred to the system, Qnet. Since heat transfer is process dependent, the differential of heat transfer δQ is called inexact. We often think about the heat transfer per unit mass of the system, Q.

Q q= m Heat transfer has the units of energy measured in joules (we will use kilojoules, kJ) or the units of energy per unit mass, kJ/kg. Since heat transfer is energy in transition across the system boundary due to a temperature difference, there are three modes of heat transfer at the boundary that depend on the temperature difference between the boundary surface and the surroundings. These are conduction, convection, and radiation. However, when solving problems in thermodynamics involving heat transfer to a system, the heat transfer is usually given or is calculated by applying the first law, or the conservation of energy, to the system. An adiabatic process is one in which the system is perfectly insulated and the heat transfer is zero. Chapter 2-6

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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Introduction to the Basic Heat Transfer Mechanisms For those of us who do not have the opportunity to have a complete course in heat transfer theory and applications, the following is a short introduction to the basic mechanisms of heat transfer. Those of us who have a complete course in heat transfer theory may elect to omit this material at this time. Heat transfer is energy in transition due to a temperature difference. The three modes of heat transfer are conduction, convection, and radiation. Conduction through Plane Walls Conduction heat transfer is a progressive exchange of energy between the molecules of a substance.

Fourier's law of heat conduction is

Chapter 2-7

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-8

dT  Qcond = − A k t dx here Q cond

kt A dT dx

= heat flow per unit time (W) = thermal conductivity (W/m⋅K) = area normal to heat flow (m2) = temperature gradient in the direction of heat flow (°C/m)

Integrating Fourier's law

∆T Q cond = kt A ∆x Since T2>T1, the heat flows from right to left in the above figure.

Chapter 2-8

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Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Example 2-1 A flat wall is composed of 20 cm of brick having a thermal conductivity kt = 0.72 W/m⋅K. The right face temperature of the brick is 900°C, and the left face temperature of the brick is 20°C. Determine the rate of heat conduction through the wall per unit area of wall.

Tright = 900°C Tleft = 20°C

20 cm

∆T Q cond = k t A ∆x ∆T Q cond W (900 − 20) K = kt = 0.72 ∆x m⋅ K A 0.2m W = 3168 2 m

FG H

IJ K

Convection Heat Transfer Convection heat transfer is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion.

Chapter 2-9

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

The rate of heat transfer by convection law of cooling, expressed as

Q conv

is determined from Newton's

Q conv = h A (Ts − Tf ) here = heat transfer rate (W) = heat transfer area (m2) = convective heat transfer coefficient (W/m2⋅K) = surface temperature (K)

Q conv

A h Ts

Chapter 2-10

2-10

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-11

Tf = bulk fluid temperature away from the surface (K) The convective heat transfer coefficient is an experimentally determined parameter that depends upon the surface geometry, the nature of the fluid motion, the properties of the fluid, and the bulk fluid velocity. Ranges of the convective heat transfer coefficient are given below. h W/m2⋅K free convection of gases free convection of liquids forced convection of gases forced convection of liquids convection in boiling and condensation

2-25 50-100 25-250 50-20,000 2500-100,000

Radiative Heat Transfer Radiative heat transfer is energy in transition from the surface of one body to the surface of another due to electromagnetic radiation. The radiative energy transferred is proportional to the difference in the fourth power of the absolute temperatures of the bodies exchanging energy.

Chapter 2-11

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-12

The net exchange of radiative heat transfer between a body surface and its surroundings is given by

c

4 Q rad = ε σ A Ts4 − Tsurr

h

here Q rad

A σ

= heat transfer per unit time (W) = surface area for heat transfer (m2) = Stefan-Boltzmann constant, 5.67x10-8 W/m2K4 and 0.1713x10-8 BTU/h ft2 R4 = emissivity

ε Ts = absolute temperature of surface (K) Tsurr = absolute temperature of surroundings (K)

Example 2-2 A vehicle is to be parked overnight in the open away from large surrounding objects. It is desired to know if dew or frost may form on the vehicle top. Assume the following: •

• • •

Convection coefficient h from ambient air to vehicle top is 6.0 W/m2⋅°C. Equivalent sky temperature is -18°C. Emissivity of vehicle top is 0.84. Negligible conduction from inside vehicle to top of vehicle.

Determine the temperature of the vehicle top when the air temperature is 5 oF. State which formation (dew or frost) occurs. Chapter 2-12

2-13

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Tsky = -

Tair = 5 C

Qrad

Qconv Ttop

Under steady-state conditions, the energy convected to the vehicle top is equal to the energy radiated to the sky.

Q conv = Q rad The energy convected from the ambient air to the vehicle top is

Q conv = Atop h (Tair − Ttop ) The energy radiated from the top to the night sky is

d

4 4 Q rad = ε σ Atop Ttop − Tsky

i

Setting these two heat transfers equal gives

d

4 4 Atop h (Tair − Ttop ) = ε σ Atop Ttop − Tsky

d

4 4 h (Tair − Ttop ) = ε σ Ttop − Tsky

Chapter 2-13

i

i

2-14

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6.0

W (5 + 273) − Ttop K m2 K

FG H

= (0.84) 5.67 x10−8

IJ K

W 4 4 4 T ( 18 273 ) K − − + top m2 K 4

Write the equation for Ttop in °C (T K = T°C + 273)

d5 − T i top

(0.84)(5.67) = 6.0

LMF T + 273I GMH 100 JK N top

4

O − (2.55) P PQ 4

Using the EES software package

Ttop = −3.38 ° C Since Ttop is below the triple point of water, 0.01°C, the water vapor in the air will form frost on the car top (see Chapter 14).

Extra Problem Explore what happens to Ttop as you vary the convective heat transfer coefficient. On a night when the atmosphere is particularly still and cold and has a clear sky, why do fruit growers use fans to increase the air velocity in their fruit groves?

Chapter 2-14

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Chapter 2-15

2-15

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-16

Energy Transfer by Work Electrical Work The rate of electrical work done by electrons crossing a system boundary is called electrical power and is given by the product of the voltage drop in volts and the current in amps.

We = V I

(W)

The amount of electrical work done in a time period is found by integrating the rate of electrical work over the time period.

2

We = ∫ V I dt 1

(kJ)

Mechanical Forms of Work Work is energy expended by a force acting through a distance. Thermodynamic work is defined as energy in transition across the system boundary and is done by a system if the sole effect external to the boundaries could have been the raising of a weight.

Mathematically, the differential of work is expressed as

Chapter 2-16

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-17

G G δW = F ⋅ d s = F ds cosΘ here Θ is the angle between the force vector and the displacement vector. As with the heat transfer, the Greek symbol δ means that work is a pathdependent function and has an inexact differential. If the angle between the force and the displacement is zero, the work done between two states is

z z 2

2

W12 = δ W = F ds 1

1

Work has the units of energy and is defined as force times displacement or newton times meter or joule (we will use kilojoules). Work per unit mass of a system is measured in kJ/kg. Common Types of Mechanical Work Energy (See text for discussion of these topics) • • • • •

Shaft Work Spring Work Work done of Elastic Solid Bars Work Associated with the Stretching of a Liquid Film Work Done to Raise or to Accelerate a Body

Net Work Done By A System The net work done by a system may be in two forms other work and boundary work. First, work may cross a system boundary in the form of a rotating shaft work, electrical work or other the work forms listed above. We will call these work forms “other” work, that is, work not associated Chapter 2-17

2-18

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

with a moving boundary. In thermodynamics electrical energy is normally considered to be work energy rather than heat energy; however, the placement of the system boundary dictates whether to include electrical energy as work or heat. Second, the system may do work on its surroundings because of moving boundaries due to expansion or compression processes that a fluid may experience in a piston-cylinder device. The net work done by a closed system is defined by

Wnet =

d∑ W

out

− ∑ Win

i

other

+ Wb

Here, Wout and Win are the magnitudes of the other work forms crossing the boundary. Wb is the work due to the moving boundary as would occur when a gas contained in a piston cylinder device expands and does work to move the piston. The boundary work will be positive or negative depending upon the process. Boundary work is discussed in detail in Chapter 4.

e j

Wnet = Wnet

other

+ Wb

Several types of “other” work (shaft work, electrical work, etc.) are discussed in the text. Example 2-3 A fluid contained in a piston-cylinder device receives 500 kJ of electrical work as the gas expands against the piston and does 600 kJ of boundary work on the piston. What is the net work done by the fluid? Chapter 2-18

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Wb=600 kJ

Wele =500 kJ

Wnet = (Wnet

)

other

+ Wb

Wnet = (Wout − Win , ele )

other

+ Wb

Wnet = ( 0 − 500 kJ ) + 600 kJ Wnet = 100 kJ

Chapter 2-19

2-19

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-20

The First Law of Thermodynamics The first law of thermodynamics is known as the conservation of energy principle. It states that energy can be neither created nor destroyed; it can only change forms. Joule’s experiments lead to the conclusion: For all adiabatic processes between two specified states of a closed system, the net work done is the same regardless of the nature of the closed system and the details of the process. A major consequence of the first law is the existence and definition of the property total energy E introduced earlier. The First Law and the Conservation of Energy The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat or work. Energy may cross a system boundary (control surface) of an open system by heat, work and mass transfer. A system moving relative to a reference plane is shown below where z is G the elevation of the center of mass above the reference plane and V is the velocity of the center of mass.

Energyin

System

G V

CM

z

Energyout

Reference Plane, z = 0

For the system shown above, the conservation of energy principle or the first law of thermodynamics is expressed as

Chapter 2-20

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-21

FG Total energy IJ − FG Total energy IJ = FG The change in total IJ H entering the systemK H leaving the systemK H energy of the systemK or

Ein − Eout = ∆E system Normally the stored energy, or total energy, of a system is expressed as the sum of three separate energies. The total energy of the system, Esystem, is given as

E = Internal energy + Kinetic energy + Potential energy E = U + KE + PE Recall that U is the sum of the energy contained within the molecules of the system other than the kinetic and potential energies of the system as a whole and is called the internal energy. The internal energy U is dependent on the state of the system and the mass of the system. For a system moving relative to a reference plane, the kinetic energy KE and the potential energy PE are given by G V G V =0

KE = ∫ PE = ∫

z

z =0

G G G mV 2 mV dV = 2

mg dz = mgz

Chapter 2-21

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-22

The change in stored energy for the system is

∆E = ∆U + ∆KE + ∆PE

Now the conservation of energy principle, or the first law of thermodynamics for closed systems, is written as

Ein − Eout = ∆U + ∆KE + ∆PE If the system does not move with a velocity and has no change in elevation, it is called a stationary system, and the conservation of energy equation reduces to

Ein − Eout = ∆U Mechanisms of Energy Transfer, Ein and Eout The mechanisms of energy transfer at a system boundary are: Heat, Work, mass flow. Only heat and work energy transfers occur at the boundary of a closed (fixed mass) system. Open systems or control volumes have energy transfer across the control surfaces by mass flow as well as heat and work. 1. Heat Transfer, Q: Heat is energy transfer caused by a temperature difference between the system and its surroundings. When added to a system heat transfer causes the energy of a system to increase and heat

Chapter 2-22

2-23

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

transfer from a system causes the energy to decrease. Q is zero for adiabatic systems. 2. Work, W: Work is energy transfer at a system boundary could have caused a weight to be raised. When added to a system, the energy of the system increase; and when done by a system, the energy of the system decreases. W is zero for systems having no work interactions at its boundaries. 3. Mass flow, m: As mass flows into a system, the energy of the system increases by the amount of energy carried with the mass into the system. Mass leaving the system carries energy with it, and the energy of the system decreases. Since no mass transfer occurs at the boundary of a closed system, energy transfer by mass is zero for closed systems. The energy balance for a general system is

Ein − Eout = ( Qin − Qout ) + (Win − Wout ) + ( Emass , in − Emass , out ) = ∆Esystem

Expressed more compactly, the energy balance is

Ein − Eout =  

Net energy transfer by heat, work, and mass

∆Esystem   

(kJ )

Change in internal, kinetic, potential, etc., energies

or on a rate form, as

E in − E out  

Rate of net energy transfer by heat, work, and mass

=

∆E system   

Rate change in internal, kinetic, potential, etc., energies

Chapter 2-23

( kW )

2-24

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

For constant rates, the total quantities during the time interval ∆t are related to the quantities per unit time as

Q = Q ∆t , W = W ∆t , and

∆E = ∆E ∆t ( kJ )

The energy balance may be expressed on a per unit mass basis as

e in − eout = ∆esystem

(kJ / kg )

and in the differential forms as

δ Ein − δ Eout = δ Esystem δ e in − δ eout = δ esystem

(kJ ) (kJ / kg )

First Law for a Cycle A thermodynamic cycle is composed of processes that cause the working fluid to undergo a series of state changes through a process or a series of processes. These processes occur such that the final and initial states are identical and the change in internal energy of the working fluid is zero for whole numbers of cycles. Since thermodynamic cycles can be viewed as having heat and work (but not mass) crossing the cycle system boundary, the first law for a closed system operating in a thermodynamic cycle becomes

Chapter 2-24

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

0

Qnet − Wnet = ∆Ecycle Qnet = Wnet

Chapter 2-25

2-25

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-26

Example 2-4 A system receives 5 kJ of heat transfer and experiences a decrease in energy in the amount of 5 kJ. Determine the amount of work done by the system. Qin =5 kJ

∆E=-5 kJ kJ ∆E = -5

Wout=?

System Boundary

We apply the first law as

Ein − Eout = ∆Esystem Ein = Qin = 5 kJ Eout = Wout

∆Esystem = −5 kJ Eout = Ein − ∆Esystem Wout = 5 − ( −5 )  kJ Wout = 10 kJ The work done by the system equals the energy input by heat plus the decrease in the energy of the working fluid.

Chapter 2-26

2-27

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Example 2-5 A steam power plant operates on a thermodynamic cycle in which water circulates through a boiler, turbine, condenser, pump, and back to the boiler. For each kilogram of steam (water) flowing through the cycle, the cycle receives 2000 kJ of heat in the boiler, rejects 1500 kJ of heat to the environment in the condenser, and receives 5 kJ of work in the cycle pump. Determine the work done by the steam in the turbine, in kJ/kg. The first law requires for a thermodynamic cycle

Qnet − Wnet = ∆Ecycle0 Qnet = Wnet Qin − Qout = Wout − Win Wout = Qin − Qout − Win W Q and q = m m = qin − qout + win

Let w = wout

wout = ( 2000 − 1500 + 5 ) wout = 505

kJ kg

Chapter 2-27

kJ kg

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-28

Example 2-6 Air flows into an open system and carries energy at the rate of 300 kW. As the air flows through the system it receives 600 kW of work and loses 100 kW of energy by heat transfer to the surroundings. If the system experiences no energy change as the air flows through it, how much energy does the air carry as it leaves the system, in kW? System sketch:

Q out E mass , in

Open System

E mass , out

Win Conservation of Energy:

E in − E out = ∆E system E mass ,in + Win − E mass , out − Q out = ∆E system = 0 E mass ,out = E mass , in + Win − Q out E mass ,out = ( 300 + 600 − 100 ) kW = 800 kW

Chapter 2-28

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-29

Energy Conversion Efficiencies A measure of performance for a device is its efficiency and is often given the symbol η. Efficiencies are expressed as follows:

η=

Desired Result Required Input

How will you measure your efficiency in this thermodynamics course? Efficiency as the Measure of Performance of a Thermodynamic cycle A system has completed a thermodynamic cycle when the working fluid undergoes a series of processes and then returns to its original state, so that the properties of the system at the end of the cycle are the same as at its beginning. Thus, for whole numbers of cycles

Pf = Pi , Tf = Ti , u f = ui , v f = vi , etc. Heat Engine A heat engine is a thermodynamic system operating in a thermodynamic cycle to which net heat is transferred and from which net work is delivered. The system, or working fluid, undergoes a series of processes that constitute the heat engine cycle.

Chapter 2-29

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-30

The following figure illustrates a steam power plant as a heat engine operating in a thermodynamic cycle.

Thermal Efficiency, ηth The thermal efficiency is the index of performance of a work-producing device or a heat engine and is defined by the ratio of the net work output (the desired result) to the heat input (the cost or required input to obtain the desired result).

η th =

Desired Result Required Input

For a heat engine the desired result is the net work done (Wout – Win) and the input is the heat supplied to make the cycle operate Qin. The thermal efficiency is always less than 1 or less than 100 percent.

Chapter 2-30

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

η th =

where

2-31

Wnet , out Qin

Wnet , out = Wout − Win Qin ≠ Qnet

Here, the use of the in and out subscripts means to use the magnitude (take the positive value) of either the work or heat transfer and let the minus sign in the net expression take care of the direction.

Example 2-7 In example 2-5 the steam power plant received 2000 kJ/kg of heat, 5 kJ/kg of pump work, and produced 505 kJ/kg of turbine work. Determine the thermal efficiency for this cycle. We can write the thermal efficiency on a per unit mass basis as:

Chapter 2-31

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

ηth =

2-32

wnet , out qin

kJ w − win kg = out = kJ qin 2000 kg = 0.25 or 25%

( 505 − 5 )

Combustion Efficiency Consider the combustion of a fuel-air mixture as shown below.

Fuel CnHm

CO2 H2O N2

Combustion Chamber

Air Reactants TR, PR

Qout = HV

Products PP, TP

Fuels are usually composed of a compound or mixture containing carbon, C, and hydrogen, H2. During a complete combustion process all of the carbon is converted to carbon dioxide and all of the hydrogen is converted to water. For stoichiometric combustion (theoretically correct amount of air is supplied for complete combustion) where both the reactants (fuel plus air) and the products (compounds formed during the combustion process) have the same temperatures, the heat transfer from the combustion process is called the heating value of the fuel.

Chapter 2-32

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-33

The lower heating value, LHV, is the heating value when water appears as a gas in the products.

LHV = Qout with H 2Ovapor in products The lower heating value is often used as the measure of energy per kg of fuel supplied to the gas turbine engine because the exhaust gases have such a high temperature that the water formed is a vapor as it leaves the engine with other products of combustion.

The higher heating value, HHV, is the heating value when water appears as a liquid in the products.

HHV = Qout with H 2Oliquid in products The higher heating value is often used as the measure of energy per kg of fuel supplied to the steam power cycle because there are heat transfer processes within the cycle that absorb enough energy from the products of combustion that some of the water vapor formed during combustion will condense. Combustion efficiency is the ratio of the actual heat transfer from the combustion process to the heating value of the fuel.

ηcombustion

Qout = HV

Chapter 2-33

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-34

Example 2-8 A steam power plant receives 2000 kJ of heat per unit mass of steam flowing through the steam generator when the steam flow rate is 100 kg/s. If the fuel supplied to the combustion chamber of the steam generator has a higher heating value of 40,000 kJ/kg of fuel and the combustion efficiency is 85%, determine the required fuel flow rate, in kg/s.

ηcombustion m fuel =

m fuel

m fuel

Qout m steam qout to steam = = HV m fuel HHV

m steam qout to steam ηcombustion HHV

kg steam   kJ     100   2000 s kg   steam  =  kJ  ( 0.85 )  40000  kg fuel   kg fuel = 5.88 s

Chapter 2-34

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-35

Generator Efficiency:

η generator

Welectrical output = W

mechanical input

Power Plant Overall Efficiency:

ηoverall ηoverall ηoverall

 Q in , cycle   Wnet , cycle   Wnet , electrical output =  m fuel HHV fuel   Q in , cycle   Wnet , cycle    = ηcombustionηthermalη generator Wnet , electrical output = m fuel HHV fuel

Motor Efficiency:

ηmotor

Wmechanical output = W electrical input

Chapter 2-35

  

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

2-36

Lighting Efficacy:

Lighting Efficacy =

Amount of Light in Lumens Watts of Electricity Consumed

Type of lighting Ordinary Incandescent Ordinary Fluorescent

Efficacy, lumens/W 6 - 20 40 - 60

Effectiveness of Conversion of Electrical or chemical Energy to Heat for Cooking, Called Efficacy of a Cooking Appliance:

Cooking Efficacy =

Useful Energy Transferred to Food Energy Consumed by Appliance

Chapter 2-36

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 1

Chapter 3: Properties of Pure Substances We now turn our attention to the concept of pure substances and the presentation of their data. Simple System A simple system is one in which the effects of motion, viscosity, fluid shear, capillarity, anisotropic stress, and external force fields are absent. Homogeneous Substance A substance that has uniform thermodynamic properties throughout is said to be homogeneous. Pure Substance A pure substance has a homogeneous and invariable chemical composition and may exist in more than one phase. Examples: 1. 2. 3. 4. 5.

Water (solid, liquid, and vapor phases) Mixture of liquid water and water vapor Carbon dioxide, CO2 Nitrogen, N2 Mixtures of gases, such as air, as long as there is no change of phase.

State Postulate Again, the state postulate for a simple, pure substance states that the equilibrium state can be determined by specifying any two independent intensive properties.

Chapter 3-1

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

The P-V-T Surface for a Real Substance ♦ P-V-T Surface for a Substance that contracts upon freezing

♦ P-V-T Surface for a Substance that expands upon freezing

Chapter 3-2

3- 2

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 3

Real substances that readily change phase from solid to liquid to gas such as water, refrigerant-134a, and ammonia cannot be treated as ideal gases in general. The pressure, volume, temperature relation, or equation of state for these substances is generally very complicated, and the thermodynamic properties are given in table form. The properties of these substances may be illustrated by the functional relation F(P,v,T)=0, called an equation of state. The above two figures illustrate the function for a substance that contracts on freezing and a substance that expands on freezing. Constant pressure curves on a temperature-volume diagram are shown in Figure 311. These figures show three regions where a substance like water may exist as a solid, liquid or gas (or vapor). Also these figures show that a substance may exist as a mixture of two phases during phase change, solid-vapor, solid-liquid, and liquid-vapor. Water may exist in the compressed liquid region, a region where saturated liquid water and saturated water vapor are in equilibrium (called the saturation region), and the superheated vapor region (the solid or ice region is not shown). Let's consider the results of heating liquid water from 20°C, 1 atm while keeping the pressure constant. We will follow the constant pressure process shown in Figure 3-11. First place liquid water in a piston-cylinder device where a fixed weight is placed on the piston to keep the pressure of the water constant at all times. As liquid water is heated while the pressure is held constant, the following events occur.

Process 1-2:

Chapter 3-3

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 4

The temperature and specific volume will increase from the compressed liquid, or subcooled liquid, state 1, to the saturated liquid state 2. In the compressed liquid region, the properties of the liquid are approximately equal to the properties of the saturated liquid state at the temperature.

Process 2-3: At state 2 the liquid has reached the temperature at which it begins to boil, called the saturation temperature, and is said to exist as a saturated liquid. Properties at the saturated liquid state are noted by the subscript f and v2 = vf. During the phase change both the temperature and pressure remain constant (according to the International Temperature Scale of 1990, ITS90, water boils at 99.975°C ≅ 100°C when the pressure is 1 atm or 101.325 kPa). At state 3 the liquid and vapor phase are in equilibrium and any point on the line between states 2 and 3 has the same temperature and pressure.

Chapter 3-4

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 5

Process 3-4: At state 4 a saturated vapor exists and vaporization is complete. The subscript g will always denote a saturated vapor state. Note v4 = vg.

Thermodynamic properties at the saturated liquid state and saturated vapor state are given in Table A-4 as the saturated temperature table and Table A-5 as the saturated pressure table. These tables contain the same information. In Table A-4 the saturation temperature is the independent property, and in Table A-5 the saturation pressure is the independent property. The saturation pressure is the pressure at which phase change will occur for a given temperature. In the saturation region the temperature and pressure are dependent properties; if one is known, then the other is automatically known. Process 4-5: If the constant pressure heating is continued, the temperature will begin to increase above the saturation temperature, 100 °C in this example, and the volume also increases. State 5 is called a superheated state because T5 is greater than the saturation temperature for the pressure and the vapor is not about to condense. Thermodynamic properties for water in the superheated region are found in the superheated steam tables, Table A-6.

Chapter 3-5

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 6

This constant pressure heating process is illustrated in the following figure.

99.975 ≅

Figure 3-11 Consider repeating this process for other constant pressure lines as shown below.

Chapter 3-6

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 7

If all of the saturated liquid states are connected, the saturated liquid line is established. If all of the saturated vapor states are connected, the saturated vapor line is established. These two lines intersect at the critical point and form what is often called the “steam dome.” The region between the saturated liquid line and the saturated vapor line is called by these terms: saturated liquid-vapor mixture region, wet region (i.e., a mixture of saturated liquid and saturated vapor), two-phase region, and just the saturation region. Notice that the trend of the temperature following a constant pressure line is to increase with increasing volume and the trend of the pressure following a constant temperature line is to decrease with increasing volume.

Chapter 3-7

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

P2 = 1000 kPa P1 = 100 kPa 179.88oC 99.61oC

Chapter 3-8

3- 8

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 9

The region to the left of the saturated liquid line and below the critical temperature is called the compressed liquid region. The region to the right of the saturated vapor line and above the critical temperature is called the superheated region. See Table A-1 for the critical point data for selected substances. Review the P-v diagrams for substances that contract on freezing and those that expand on freezing given in Figure 3-21 and Figure 3-22. At temperatures and pressures above the critical point, the phase transition from liquid to vapor is no longer discrete.

Figure 3-25 shows the P-T diagram, often called the phase diagram, for pure substances that contract and expand upon freezing.

Chapter 3-9

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

The triple point of water is 0.01oC, 0.6117 kPa (See Table 3-3). The critical point of water is 373.95oC, 22.064 MPa (See Table A-1).

Chapter 3-10

3- 10

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 11

Plot the following processes on the P-T diagram for water (expands on freezing) and give examples of these processes from your personal experiences. 1. process a-b: liquid to vapor transition 2. process c-d: solid to liquid transition 3. process e-f: solid to vapor transition

Chapter 3-11

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 12

Property Tables In addition to the temperature, pressure, and volume data, Tables A-4 through A-8 contain the data for the specific internal energy u the specific enthalpy h and the specific entropy s. The enthalpy is a convenient grouping of the internal energy, pressure, and volume and is given by

H = U + PV The enthalpy per unit mass is

h = u + Pv We will find that the enthalpy h is quite useful in calculating the energy of mass streams flowing into and out of control volumes. The enthalpy is also useful in the energy balance during a constant pressure process for a substance contained in a closed piston-cylinder device. The enthalpy has units of energy per unit mass, kJ/kg. The entropy s is a property defined by the second law of thermodynamics and is related to the heat transfer to a system divided by the system temperature; thus, the entropy has units of energy divided by temperature. The concept of entropy is explained in Chapters 6 and 7. Saturated Water Tables Since temperature and pressure are dependent properties using the phase change, two tables are given for the saturation region. Table A-4 has temperature as the independent property; Table A-5 has pressure as the independent property. These two tables contain the same information and often only one table is given.

Chapter 3-12

3- 13

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

For the complete Table A-4, the last entry is the critical point at 373.95 oC. TABLE A-4 Saturated water-Temperature table Temp., T °C 0.01 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 ٠ ٠ 360 365 370 373.95

Sat. Press., Psat kPa 0.6117 0.8725 1.228 1.706 2.339 3.170 4.247 5.629 7.385 9.595 12.35 15.76 19.95 25.04 31.20 38.60 47.42 57.87 70.18 84.61 101.42 ٠ ٠ 18666 19822 21044 22064

Specific volume, m3/kg Sat. Sat. vapor, liquid, vf vg 0.001000 206.00 0.001000 147.03 0.001000 106.32 0.001001 77.885 0.001002 57.762 0.001003 43.340 0.001004 32.879 0.001006 25.205 0.001008 19.515 0.001010 15.251 0.001012 12.026 0.001015 9.5639 0.001017 7.6670 0.001020 6.1935 0.001023 5.0396 0.001026 4.1291 0.001029 3.4053 0.001032 2.8261 0.001036 2.3593 0.001040 1.9808 0.001043 1.6720 ٠ ٠ ٠ ٠ 0.001895 0.006950 0.002015 0.006009 0.002217 0.004953 0.003106 0.003106

Internal energy, kJ/kg Sat. Sat. Evap., liquid, vapor, uf ufg ug 0.00 2374.9 2374.9 21.02 2360.8 2381.8 42.02 2346.6 2388.7 62.98 2332.5 2395.5 83.91 2318.4 2402.3 104.83 2304.3 2409.1 125.73 2290.2 2415.9 146.63 2276.0 2422.7 167.53 2261.9 2429.4 188.43 2247.7 2436.1 209.33 2233.4 2442.7 230.24 2219.1 2449.3 251.16 2204.7 2455.9 272.09 2190.3 2462.4 293.04 2175.8 2468.9 313.99 2161.3 2475.3 334.97 2146.6 2481.6 355.96 2131.9 2487.8 376.97 2117.0 2494.0 398.00 2102.0 2500.1 419.06 2087.0 2506.0 ٠ ٠ ٠ ٠ ٠ ٠ 1726.16 625.7 2351.9 1777.22 526.4 2303.6 1844.53 385.6 2230.1 2015.8 0 2015.8

Chapter 3-13

Enthalpy, kJ/kg Sat. liquid, hf 0.00 21.02 42.02 62.98 83.91 104.83 125.74 146.64 167.53 188.44 209.34 230.26 251.18 272.12 293.07 314.03 335.02 356.02 377.04 398.09 419.17 ٠ ٠ 1761.53 1817.16 1891.19 2084.3

Evap., hfg 2500.9 2489.1 2477.2 2465.4 2453.5 2441.7 2429.8 2417.9 2406.0 2394.0 2382.0 2369.8 2357.7 2345.4 2333.0 2320.6 2308.0 2295.3 2282.5 2269.6 2256.4 ٠ ٠ 720.1 605.5 443.1 0

Entropy, kJ/kg⋅K Sat. vapor, hg 2500.9 2510.1 2519.2 2528.3 2537.4 2546.5 2555.6 2564.6 2573.5 2582.4 2591.3 2600.1 2608.8 2617.5 2626.1 2634.6 2643.0 2651.4 2659.6 2667.6 2675.6 ٠ ٠ 2481.6 2422.7 2334.3 2084.3

Sat. liquid, sf 0.0000 0.0763 0.1511 0.2245 0.2965 0.3672 0.4368 0.5051 0.5724 0.6386 0.7038 0.7680 0.8313 0.8937 0.9551 1.0158 1.0756 1.1346 1.1929 1.2504 1.3072 ٠ ٠ 3.9165 4.0004 4.1119 4.4070

Evap., sfg 9.1556 8.9487 8.7488 8.5559 8.3696 8.1895 8.0152 7.8466 7.6832 7.5247 7.3710 7.2218 7.0769 6.9360 6.7989 6.6655 6.5355 6.4089 6.2853 6.1647 6.0470 ٠ ٠ 1.1373 0.9489 0.6890 0

Sat. vapor, sg 9.1556 9.0249 8.8999 8.7803 8.6661 8.5567 8.4520 8.3517 8.2556 8.1633 8.0748 7.9898 7.9082 7.8296 7.7540 7.6812 7.6111 7.5435 7.4782 7.4151 7.3542 ٠ ٠ 5.0537 4.9493 4.8009 4.4070

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Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

TABLE A-5 Saturated water-Pressure table Press. P kPa

Sat. Temp., Tsat °C

0.6117 1.0 1.5 2.0 2.5 3.0 4.0 5.0 7.5 10 15 20 25 30 40 50 75 100 125 ٠ ٠ 20,000 21,000 22,000 22,064

0.01 6.97 13.02 17.50 21.08 24.08 28.96 32.87 40.29 45.81 53.97 60.06 64.96 69.09 75.86 81.32 91.76 99.61 105.97 ٠ ٠ 365.75 369.83 373.71 373.95

Sat. liquid, vf 0.001000 0.001000 0.001001 0.001001 0.001002 0.001003 0.001004 0.001005 0.001008 0.001010 0.001014 0.001017 0.001020 0.001022 0.001026 0.001030 0.001037 0.001043 0.001048 ٠ ٠ 0.002038 0.002207 0.002703 0.003106

Specific volume, m3/kg Sat. vapor, vg 206.00 129.19 87.964 66.990 54.242 45.654 34.791 28.185 19.233 14.670 10.020 7.6481 6.2034 5.2287 3.9933 3.2403 2.2172 1.6941 1.3750 ٠ ٠ 0.005862 0.004994 0.003644 0.003106

Sat. liquid, uf 0.00 29.30 54.69 73.43 88.42 100.98 121.39 137.75 168.74 191.79 225.93 251.40 271.93 289.24 317.58 340.49 384.36 417.40 444.23 ٠ ٠ 1785.84 1841.62 1951.65 2015.8

Internal energy, kJ/kg Evap., Sat. vapor, ufg ug 2374.9 2374.9 2355.2 2384.5 2338.1 2392.8 2325.5 2398.9 2315.4 2403.8 2306.9 2407.9 2293.1 2414.5 2282.1 2419.8 2261.1 2429.8 2245.4 2437.2 2222.1 2448.0 2204.6 2456.0 2190.4 2462.4 2178.5 2467.7 2158.8 2476.3 2142.7 2483.2 2111.8 2496.1 2088.2 2505.6 2068.8 2513.0 ٠ ٠ ٠ ٠ 509.0 2294.8 391.9 2233.5 140.8 2092.4 0 2015.8

Sat. liquid, hf 0.00 29.30 54.69 73.43 88.42 100.98 121.39 137.75 168.75 191.81 225.94 251.42 271.96 289.27 317.62 340.54 384.44 417.51 444.36 ٠ ٠ 1826.59 1887.97 2011.12 2084.3

Enthalpy, kJ/kg Evap., hfg 2500.9 2484.4 2470.1 2459.5 2451.0 2443.9 2432.3 2423.0 2405.3 2392.1 2372.3 2357.5 2345.5 2335.3 2318.4 2304.7 2278.0 2257.5 2240.6 ٠ ٠ 585.5 450.4 161.5 0

Sat. vapor, hg 2500.9 2513.7 2524.7 2532.9 2539.4 2544.8 2553.7 2560.7 2574.0 2583.9 2598.3 2608.9 2617.5 2624.6 2636.1 2645.2 2662.4 2675.0 2684.9 ٠ ٠ 2412.1 2338.4 2172.6 2084.3

Sat. liquid, sf 0.0000 0.1059 0.1956 0.2606 0.3118 0.3543 0.4224 0.4762 0.5763 0.6492 0.7549 0.8320 0.8932 0.9441 1.0261 1.0912 1.2132 1.3028 1.3741 ٠ ٠ 4.0146 4.1071 4.2942 4.4070

Entropy, kJ/kg⋅K Evap., sfg 9.1556 8.8690 8.6314 8.4621 8.3302 8.2222 8.0510 7.9176 7.6738 7.4996 7.2522 7.0752 6.9370 6.8234 6.6430 6.5019 6.2426 6.0562 5.9100 ٠ ٠ 0.9164 0.7005 0.2496 0

For the complete Table A-5, the last entry is the critical point at 22.064 MPa. Saturation pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature. Saturation temperature is the temperature at which the liquid and vapor phases are in equilibrium at a given pressure. In Figure 3-11, states 2, 3, and 4 are saturation states.

Chapter 3-14

Sat. vapor, sg 9.1556 8.9749 8.8270 8.7227 8.6421 8.5765 8.4734 8.3938 8.2501 8.1488 8.0071 7.9073 7.8302 7.7675 7.6691 7.5931 7.4558 7.3589 7.2841 ٠ ٠ 4.9310 4.8076 4.5439 4.4070

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 15

The subscript fg used in Tables A-4 and A-5 refers to the difference between the saturated vapor value and the saturated liquid value region. That is,

u fg = ug − u f h fg = hg − h f s fg = sg − s f The quantity hfg is called the enthalpy of vaporization (or latent heat of vaporization). It represents the amount of energy needed to vaporize a unit of mass of saturated liquid at a given temperature or pressure. It decreases as the temperature or pressure increases, and becomes zero at the critical point. Quality and Saturated Liquid-Vapor Mixture Now, let’s review the constant pressure heat addition process for water shown in Figure 3-11. Since state 3 is a mixture of saturated liquid and saturated vapor, how do we locate it on the T-v diagram? To establish the location of state 3 a new parameter called the quality x is defined as

x=

masssaturated vapor masstotal

=

mg m f + mg

The quality is zero for the saturated liquid and one for the saturated vapor (0 ≤ x ≤ 1). The average specific volume at any state 3 is given in terms of the quality as follows. Consider a mixture of saturated liquid and saturated vapor. The liquid has a mass mf and occupies a volume Vf. The vapor has a mass mg and occupies a volume Vg.

Chapter 3-15

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

We note

V = V f + Vg m = m f + mg V = mv , V f = m f v f , Vg = mg v g mv = m f v f + mg v g v=

mf v f m

+

mg v g m

Recall the definition of quality x

x=

mg m

=

mg m f + mg

Then

mf m

=

m − mg m

Chapter 3-16

= 1− x

3- 16

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 17

Note, quantity 1- x is often given the name moisture. The specific volume of the saturated mixture becomes

v = (1 − x )v f + xv g

The form that we use most often is

v = v f + x (v g − v f ) It is noted that the value of any extensive property per unit mass in the saturation region is calculated from an equation having a form similar to that of the above equation. Let Y be any extensive property and let y be the corresponding intensive property, Y/m, then

Y = y f + x( yg − y f ) m = y f + x y fg

y=

where y fg = y g − y f The term yfg is the difference between the saturated vapor and the saturated liquid values of the property y; y may be replaced by any of the variables v, u, h, or s. We often use the above equation to determine the quality x of a saturated liquid-vapor state. The following application is called the Lever Rule:

x=

y − yf

Chapter 3-17

y fg

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 18

The Lever Rule is illustrated in the following figures.

Superheated Water Table A substance is said to be superheated if the given temperature is greater than the saturation temperature for the given pressure. State 5 in Figure 3-11 is a superheated state. In the superheated water Table A-6, T and P are the independent properties. The value of temperature to the right of the pressure is the saturation temperature for the pressure. The first entry in the table is the saturated vapor state at the pressure.

Chapter 3-18

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 19

Compressed Liquid Water Table A substance is said to be a compressed liquid when the pressure is greater than the saturation pressure for the temperature. It is now noted that state 1 in Figure 3-11 is called a compressed liquid state because the saturation pressure for the temperature T1 is less than P1. Data for water compressed liquid states are found in the compressed liquid tables, Table A-7. Table A-7 is arranged like Table A-6, except the saturation states are the saturated liquid states. Note that the data in Table A-7 begins at 5 MPa or 50 times atmospheric pressure.

Chapter 3-19

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 20

At pressures below 5 MPa for water, the data are approximately equal to the saturated liquid data at the given temperature. We approximate intensive parameter y, that is v, u, h, and s data as

y ≅ y f @T The enthalpy is more sensitive to variations in pressure; therefore, at high pressures the enthalpy can be approximated by

h ≅ h f @T + v f ( P − Psat )

For our work, the compressed liquid enthalpy may be approximated by

h ≅ h f @T

Chapter 3-20

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 21

Saturated Ice-Water Vapor Table When the temperature of a substance is below the triple point temperature, the saturated solid and liquid phases exist in equilibrium. Here we define the quality as the ratio of the mass that is vapor to the total mass of solid and vapor in the saturated solid-vapor mixture. The process of changing directly from the solid phase to the vapor phase is called sublimation. Data for saturated ice and water vapor are given in Table A-8. In Table A-8, the term Subl. refers to the difference between the saturated vapor value and the saturated solid value.

The specific volume, internal energy, enthalpy, and entropy for a mixture of saturated ice and saturated vapor are calculated similarly to that of saturated liquid-vapor mixtures.

yig = y g − yi y = yi + x yig

Chapter 3-21

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 22

where the quality x of a saturated ice-vapor state is

x=

mg mi + mg

How to Choose the Right Table The correct table to use to find the thermodynamic properties of a real substance can always be determined by comparing the known state properties to the properties in the saturation region. Given the temperature or pressure and one other property from the group v, u, h, and s, the following procedure is used. For example if the pressure and specific volume are specified, three questions are asked: For the given pressure,

Is v < v f ? Is v f < v < v g ? Is v g < v ? The answer to one of these questions must be yes. If the answer to the first question is yes, the state is in the compressed liquid region, and the compressed liquid tables are used to find the properties of the state. If the answer to the second question is yes, the state is in the saturation region, and either the saturation temperature table or the saturation pressure table is used to find the properties. Then the quality is calculated and is used to calculate the other properties, u, h, and s. If the answer to the third

Chapter 3-22

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 23

question is yes, the state is in the superheated region and the superheated tables are used to find the other properties. Some tables may not always give the internal energy. When it is not listed, the internal energy is calculated from the definition of the enthalpy as

u = h − Pv

Chapter 3-23

3- 24

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Example 2-1 Find the internal energy of water at the given states for 7 MPa and plot the states on T-v, P-v, and P-T diagrams. Steam

700

7000 kPa

600 500

T [C]

400 300 200 100 0 10-4

10-3

10-2

10-1

100

101

102

103

3

v [m /kg]

Steam

105

104

285.9 C 3

P [kPa]

10

374.1 C

102

101

100 10-4

10-3

10-2

10-1 3

v [m /kg]

Chapter 3-24

100

101

102

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

P

CP

Steam

7 MPa Triple Point

0.01

285.8

373.95

1. P = 7 MPa, dry saturated or saturated vapor Using Table A-5,

u = u g = 2581.0

kJ kg

Locate state 1 on the T-v, P-v, and P-T diagrams.

2. P = 7 MPa, wet saturated or saturated liquid Using Table A-5,

u = u f = 1258.0

kJ kg

Locate state 2 on the T-v, P-v, and P-T diagrams.

Chapter 3-25

T, °C

3- 25

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 26

3. Moisture = 5%, P = 7 MPa let moisture be y, defined as

y= then, the quality is

mf m

= 0.05

x = 1 − y = 1 − 0.05 = 0.95

and using Table A-5,

u = u f + x(u g − u f ) = 1258.0 + 0.95(2581.0 − 1257.6) = 2514.4

kJ kg

Notice that we could have used

u = u f + x u fg Locate state 3 on the T-v, P-v, and P-T diagrams. 4. P = 7 MPa, T = 600°C For P = 7 MPa, Table A-5 gives Tsat = 285.83°C. Since 600°C > Tsat for this pressure, the state is superheated. Use Table A-6.

u = 3261.0

Chapter 3-26

kJ kg

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 27

Locate state 4 on the T-v, P-v, and P-T diagrams. 5. P = 7 MPa, T = 100°C Using Table A-4, At T = 100°C, Psat = 0.10142 MPa. Since P > Psat, the state is compressed liquid. Approximate solution:

u ≅ u f @ T =100C = 419.06

kJ kg

Solution using Table A-7: We do linear interpolation to get the value at 100 °C. (We will demonstrate how to do linear interpolation with this problem even though one could accurately estimate the answer.) P MPa 5 7 10

u kJ/kg 417.65 u=? 416.23

The interpolation scheme is called “the ratio of corresponding differences.” Using the above table, form the following ratios.

Chapter 3-27

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

5−7 417.65 − u = 5 − 10 417.65 − 416.23 kJ u = 417.08 kg Locate state 5 on the T-v, P-v, and P-T diagrams. 6. P = 7 MPa, T = 460°C Since 460°C > Tsat = 385.83°C at P = 7 MPa, the state is superheated. Using Table A-6, we do a linear interpolation to calculate u. T °C 450 460 500

u kJ/kg 2979.0 u=? 3074.3

Using the above table, form the following ratios.

u − 2979.0 460 − 450 = 500 − 450 3074.3 − 2979.0 kJ u = 2998.1 kg Locate state 6 on the T-v, P-v, and P-T diagrams.

Chapter 3-28

3- 28

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 29

Example 2-2 Determine the enthalpy of 1.5 kg of water contained in a volume of 1.2 m3 at 200 kPa. Recall we need two independent, intensive properties to specify the state of a simple substance. Pressure P is one intensive property and specific volume is another. Therefore, we calculate the specific volume. . m3 Volume 12 m3 = = 0.8 v= 15 . kg mass kg

Using Table A-5 at P = 200 kPa, vf = 0.001061 m3/kg ,

vg = 0.8858 m3/kg

Now,

Is v < v f ? No Is v f < v < v g ? Yes Is v g < v ? No Locate this state on a T-v diagram. T

v

Chapter 3-29

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 30

We see that the state is in the two-phase or saturation region. So we must find the quality x first.

v = v f + x (v g − v f )

x=

v − vf vg − v f

0.8 − 0.001061 0.8858 − 0.001061 = 0.903 (What does this mean?) =

Then,

h = h f + x h fg = 504.7 + (0.903)(2201.6) = 2492.7

kJ kg

Example 2-3 Determine the internal energy of refrigerant-134a at a temperature of 0°C and a quality of 60%. Using Table A-11, for T = 0°C, then,

uf = 51.63 kJ/kg

Chapter 3-30

ug =230.16 kJ/kg

3- 31

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

u = u f + x (u g − u f ) = 51.63 + (0.6)(230.16 − 51.63) = 158.75

kJ kg

Example 2-4 Consider the closed, rigid container of water shown below. The pressure is 700 kPa, the mass of the saturated liquid is 1.78 kg, and the mass of the saturated vapor is 0.22 kg. Heat is added to the water until the pressure increases to 8 MPa. Find the final temperature, enthalpy, and internal energy of the water. Does the liquid level rise or fall? Plot this process on a P-v diagram with respect to the saturation lines and the critical point. P mg, Vg Sat. Vapor

mf, Vf Sat. Liquid

v

Let’s introduce a solution procedure that we will follow throughout the course. A similar solution technique is discussed in detail in Chapter 1. System: A closed system composed of the water enclosed in the tank Property Relation: Steam Tables Process: Volume is constant (rigid container)

Chapter 3-31

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 32

For the closed system the total mass is constant and since the process is one in which the volume is constant, the average specific volume of the saturated mixture during the process is given by

v=

V = constant m

or

v2 = v1

Now to find v1 recall that in the two-phase region at state 1

x1 =

mg 1 m f 1 + mg1

=

0.22 kg = 0.11 (1.78 + 0.22) kg

Then, at P = 700 kPa

v1 = v f 1 + x1 (vg1 − v f 1 ) = 0.001108 + (0.11)(0.2728 − 0.001108) m3 = 0.031 kg

State 2 is specified by: P2 = 8 MPa, v2 = 0.031 m3/kg At 8 MPa = 8000 kPa, vf = 0.001384 m3/kg at 8 MPa, v2 = 0.031 m3/kg.

Chapter 3-32

vg = 0.02352 m3/kg

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 33

Is v2 < v f ? No Is v f < v2 < v g ? No Is v g < v2 ? Yes Therefore, State 2 is superheated. Interpolating in the superheated tables at 8 MPa, v = 0.031 m3/kg gives, T2 = 361 °C h2 = 3024 kJ/kg u2 = 2776 kJ/kg Since state 2 is superheated, the liquid level falls. Extra Problem What would happen to the liquid level in the last example if the specific volume had been 0.001 m3/kg and the pressure was 8 MPa?

Chapter 3-33

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 34

Equations of State The relationship among the state variables, temperature, pressure, and specific volume is called the equation of state. We now consider the equation of state for the vapor or gaseous phase of simple compressible substances.

b

g

F P, T , v ≡ 0

Ideal Gas Based on our experience in chemistry and physics we recall that the combination of Boyle’s and Charles’ laws for gases at low pressure result in the equation of state for the ideal gas as

TI F P = RG J H vK where R is the constant of proportionality and is called the gas constant and takes on a different value for each gas. If a gas obeys this relation, it is called an ideal gas. We often write this equation as

Pv = RT The gas constant for ideal gases is related to the universal gas constant valid for all substances through the molar mass (or molecular weight). Let Ru be the universal gas constant. Then,

R=

Chapter 3-34

Ru M

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 35

The mass, m, is related to the moles, N, of substance through the molecular weight or molar mass, M, see Table A-1. The molar mass is the ratio of mass to moles and has the same value regardless of the system of units.

M air = 28.97

g kg lbm = 28.97 = 28.97 gmol kmol lbmol

Since 1 kmol = 1000 gmol or 1000 gram-mole and 1 kg = 1000 g, 1 kmol of air has a mass of 28.97 kg or 28,970 grams.

m=NM The ideal gas equation of state may be written several ways.

Pv = RT

V P = RT m PV = mRT

b g

m PV = MR T M PV = NRu T V P = Ru T N Pv = Ru T Chapter 3-35

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 36

Here P = absolute pressure in MPa, or kPa 3 v = molar specific volume in m /kmol T = absolute temperature in K Ru = 8.314 kJ/(kmol⋅K) Some values of the universal gas constant are Universal Gas Constant, Ru 8.314 kJ/(kmol⋅K) 8.314 kPa⋅m3/(kmol⋅K) 1.986 Btu/(lbmol⋅R) 1545 ft⋅lbf/(lbmol⋅R) 10.73 psia⋅ft3/(lbmol⋅R) The ideal gas equation of state can be derived from basic principles if one assumes 1. Intermolecular forces are small. 2. Volume occupied by the particles is small. Example 2-5 Determine the particular gas constant for air and hydrogen.

Chapter 3-36

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

R=

3- 37

Ru M

Rair =

kJ kmol − K = 0.287 kJ kg kg − K 28.97 kmol

8.314

Rhydrogen =

kJ kmol − K = 4.124 kJ kg kg − K 2.016 kmol

8.314

The ideal gas equation of state is used when (1) the pressure is small compared to the critical pressure or (2) when the temperature is twice the critical temperature and the pressure is less than 10 times the critical pressure. The critical point is that state where there is an instantaneous change from the liquid phase to the vapor phase for a substance. Critical point data are given in Table A-1.

Compressibility Factor To understand the above criteria and to determine how much the ideal gas equation of state deviates from the actual gas behavior, we introduce the compressibility factor Z as follows.

Chapter 3-37

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 38

Pv = Z Ru T or

Z=

Pv Ru T

For an ideal gas Z = 1, and the deviation of Z from unity measures the deviation of the actual P-V-T relation from the ideal gas equation of state. The compressibility factor is expressed as a function of the reduced pressure and the reduced temperature. The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure, which are defined as

TR =

T Tcr

and

PR =

P Pcr

where Pcr and Tcr are the critical pressure and temperature, respectively. The critical constant data for various substances are given in Table A-1. This is known as the principle of corresponding states. Figure 3-51 gives a comparison of Z factors for various gases and supports the principle of corresponding states.

Chapter 3-38

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 39

When either P or T is unknown, Z can be determined from the compressibility chart with the help of the pseudo-reduced specific volume, defined as

vR =

vactual R Tcr Pcr

Figure A-15 presents the generalized compressibility chart based on data for a large number of gases.

Chapter 3-39

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

3- 40

These charts show the conditions for which Z = 1 and the gas behaves as an ideal gas: 1. 2.

PR < 10 and TR > 2 or P < 10Pcr and T > 2Tcr PR > V1

G V1

G G V2 TL, to the working fluid in a pistoncylinder device that does some boundary work. Process 2-3 Reversible adiabatic expansion during which the system does work as the working fluid temperature decreases from TH to TL. Process 3-4 The system is brought in contact with a heat reservoir at TL < TH and a reversible isothermal heat exchange takes place while work of compression is done on the system. Process 4-1 A reversible adiabatic compression process increases the working fluid temperature from TL to TH

TL =const.

Chapter 6-15

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-16

You may have observed that power cycles operate in the clockwise direction when plotted on a process diagram. The Carnot cycle may be reversed, in which it operates as a refrigerator. The refrigeration cycle operates in the counterclockwise direction.

TL =const.

Carnot Principles The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the Kelvin-Planck and Clausius statements. A heat engine cannot operate by exchanging heat with a single heat reservoir, and a refrigerator cannot operate without net work input from an external source. Consider heat engines operating between two fixed temperature reservoirs at TH > TL. We draw two conclusions about the thermal efficiency of reversible and irreversible heat engines, known as the Carnot principles. (a)The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.

η th < η th , Carnot

(b) The efficiencies of all reversible heat engines operating between the same two constant-temperature heat reservoirs have the same efficiency. Chapter 6-16

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-17

As the result of the above, Lord Kelvin in 1848 used energy as a thermodynamic property to define temperature and devised a temperature scale that is independent of the thermodynamic substance. The following is Lord Kelvin's Carnot heat engine arrangement.

Since the thermal efficiency in general is

η th = 1 −

QL QH

For the Carnot engine, this can be written as

η th = g ( TL , TH ) = 1 − f ( TL , TH ) Chapter 6-17

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Considering engines A, B, and C

Q1 Q1 Q2 = Q3 Q2 Q3 This looks like

f ( T1 , T3 ) = f ( T1 , T2 ) f ( T2 , T3 ) One way to define the f function is

f ( T1 , T3 ) =

θ ( T2 ) θ ( T3 ) θ ( T3 ) = θ ( T1 ) θ ( T2 ) θ ( T1 )

The simplest form of θ is the absolute temperature itself.

f ( T1 , T3 ) =

T3 T1

The Carnot thermal efficiency becomes

η th , rev = 1 −

TL TH

This is the maximum possible efficiency of a heat engine operating between two heat reservoirs at temperatures TH and TL. Note that the temperatures are absolute temperatures.

Chapter 6-18

6-18

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-19

These statements form the basis for establishing an absolute temperature scale, also called the Kelvin scale, related to the heat transfers between a reversible device and the high- and low-temperature heat reservoirs by

QL T = L QH TH Then the QH/QL ratio can be replaced by TH/TL for reversible devices, where TH and TL are the absolute temperatures of the high- and lowtemperature heat reservoirs, respectively. This result is only valid for heat exchange across a heat engine operating between two constant temperature heat reservoirs. These results do not apply when the heat exchange is occurring with heat sources and sinks that do not have constant temperature. The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows:

η th

R|< η S|= η T> η

th , rev

irreversible heat engine

th , rev

reversible heat engine

th , rev

impossible heat engine

Reversed Carnot Device Coefficient of Performance If the Carnot device is caused to operate in the reversed cycle, the reversible heat pump is created. The COP of reversible refrigerators and heat pumps are given in a similar manner to that of the Carnot heat engine as

Chapter 6-19

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-20

QL 1 = QH − Q L QH − 1 QL TL 1 = = TH − TL TH − 1 TL

COPR =

QH QH QL = = QH − Q L QH − 1 QL TH TH TL = = TH − TL TH − 1 TL

COPHP

Again, these are the maximum possible COPs for a refrigerator or a heat pump operating between the temperature limits of TH and TL. The coefficients of performance of actual and reversible (such as Carnot) refrigerators operating between the same temperature limits compare as follows:

R|< COP COP S= COP |T> COP R

R , rev

irreversible refrigerator

R , rev

reversible refrigerator

R , rev

impossible refrigerator

Chapter 6-20

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-21

A similar relation can be obtained for heat pumps by replacing all values of COPR by COPHP in the above relation. Example 6-2 A Carnot heat engine receives 500 kJ of heat per cycle from a hightemperature heat reservoir at 652oC and rejects heat to a low-temperature heat reservoir at 30oC. Determine (a) The thermal efficiency of this Carnot engine. (b) The amount of heat rejected to the low-temperature heat reservoir. a. TH = 652oC

η th , rev = 1 −

QH

( 30 + 273) K ( 652 + 273) K = 0.672 or 67.2% = 1−

WOUT

HE

TL TH

QL

TL = 30oC

b.

QL T = L QH TH ( 30 + 273) K = 0.328 ( 652 + 273) K Q L = 500 kJ ( 0.328) =

= 164 kJ Chapter 6-21

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-22

Example 6-3 An inventor claims to have invented a heat engine that develops a thermal efficiency of 80 percent when operating between two heat reservoirs at 1000 K and 300 K. Evaluate his claim. TH = 1000 K

η th , rev = 1 −

QH

300 K 1000 K = 0.70 or 70% = 1−

WOUT

HE

TL TH

QL

TL = 300 K

The claim is false since no heat engine may be more efficient than a Carnot engine operating between the heat reservoirs. Example 6-4 An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2oC while operating in a room where the temperature is 25 oC and has a COP of 13.5. Is there any truth to his claim? TH = 25oC

Win

R QL

TL = 2oC

QL TL = QH − Q L TH − TL ( 2 + 273) K = ( 25 − 2 ) K = 11.96

COPR =

QH

The claim is false since no refrigerator may have a COP larger than the COP for the reversed Carnot device. Chapter 6-22

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

6-23

Example 6-5 A heat pump is to be used to heat a building during the winter. The building is to be maintained at 21oC at all times. The building is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5 oC. Determine the minimum power required to drive the heat pump unit for this outside temperature.

Q Lost

Win

o

21 C

Q H

Q L HP

-5 oC

The heat lost by the building has to be supplied by the heat pump.

kJ Q H = Q Lost = 135000 h COPHP

Q H TH = = Q H − Q L TH − TL ( 21 + 273) K = ( 21 − ( −5)) K = 11.31 Chapter 6-23

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Using the basic definition of the COP

COPHP =

Q H Wnet , in

Wnet , in =

Q H COPHP

135,000 kJ / h 1 h 1 kW . 1131 3600s kJ / s = 3.316 kW =

Chapter 6-24

6-24

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-1

Chapter 7: Entropy: A Measure of Disorder

Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic disorder for a system. Entropy is a measure of energy that is no longer available to perform useful work within the current environment. To obtain the working definition of entropy and, thus, the second law, let's derive the Clausius inequality. Consider a heat reservoir giving up heat to a reversible heat engine, which in turn gives up heat to a piston-cylinder device as shown below.

Chapter 7-1

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-2

We apply the first law on an incremental basis to the combined system composed of the heat engine and the system.

Ein − Eout = ∆Ec δQR − (δWrev + δWsys ) = dEc where Ec is the energy of the combined system. Let Wc be the work done by the combined system. Then the first law becomes

δWc = δWrev + δWsys δQR − δWc = dEc If we assume that the engine is totally reversible, then

δQR δQ = TR T δQ δQR = TR T The total net work done by the combined system becomes

δWc = TR

δQ − dEc T

Now the total work done is found by taking the cyclic integral of the incremental work. δQ − dEc Wc = TR T If the system, as well as the heat engine, is required to undergo a cycle, then

z z

Chapter 7-2

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

z

7-3

dEc = 0

and the total net work becomes

Wc = TR

z

δQ T

If Wc is positive, we have a cyclic device exchanging energy with a single heat reservoir and producing an equivalent amount of work; thus, the Kelvin-Planck statement of the second law is violated. But Wc can be zero (no work done) or negative (work is done on the combined system) and not violate the Kelvin-Planck statement of the second law. Therefore, since TR > 0 (absolute temperature), we conclude

Wc = TR

z

δQ ≤0 T

or

z

δQ ≤0 T

Here Q is the net heat added to the system, Qnet.

z

δQnet ≤0 T

This equation is called the Clausius inequality. The equality holds for the reversible process and the inequality holds for the irreversible process.

Chapter 7-3

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-4

Example 7-1 For a particular power plant, the heat added and rejected both occur at constant temperature and no other processes experience any heat transfer. The heat is added in the amount of 3150 kJ at 440oC and is rejected in the amount of 1950 kJ at 20oC. Is the Clausius inequality satisfied and is the cycle reversible or irreversible?

z

δQnet ≤0 T

Fz G δQ IJ + z FG δQ IJ ≤ 0 H T K H T K FG Q IJ + FG Q IJ ≤ 0 HTK HTK FG Q IJ + FG −Q IJ ≤ 0 HT K H T K FG 3150 kJ IJ + FG −1950 kJ IJ ≤ 0 H (440 + 273) K K H (20 + 273) K K net

net

in

out

net

net

in

in

in

out

out

out

kJ ≤0 K kJ −2.237 ≤ 0 K Calculate the net work, cycle efficiency, and Carnot efficiency based on TH and TL for this cycle. (4.418 − 6.655)

Wnet = Qin − Qout = (3150 − 1950) kJ = 1200 kJ

Chapter 7-4

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

ηth =

7-5

Wnet 1200 kJ = = 0.381 or 38.1% Qin 3150 kJ

ηth , Carnot = 1 −

TL (20 + 273) K = 1− = 0.589 or 58.9% TH (440 + 273) K

The Clausius inequality is satisfied. Since the inequality is less than zero, the cycle has at least one irreversible process and the cycle is irreversible. Example 7-2 For a particular power plant, the heat added and rejected both occur at constant temperature; no other processes experience any heat transfer. The heat is added in the amount of 3150 kJ at 440oC and is rejected in the amount of 1294.46 kJ at 20oC. Is the Clausius inequality satisfied and is the cycle reversible or irreversible?

z

δQnet ≤0 T

Fz G δQ IJ + z FG δQ IJ ≤ 0 H T K H T K FG Q IJ + FG −Q IJ ≤ 0 HT K H T K FG 3150 kJ IJ + FG −1294.46 kJ IJ ≤ 0 H (440 + 273) K K H (20 + 273) K K net

net

in

out

in

in

out

out

(4.418 − 4.418)

Chapter 7-5

kJ =0 K

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-6

The Clausius inequality is satisfied. Since the cyclic integral is equal to zero, the cycle is made of reversible processes. What cycle can this be? Calculate the net work and cycle efficiency for this cycle.

Wnet = Qin − Qout = (3150 − 1294.46) kJ = 1855.4 kJ η th =

. kJ Wnet 185554 = = 0.589 or 58.9% 3150 kJ Qin

Definition of Entropy Let’s take another look at the quantity

z

δQnet ≤0 T

If no irreversibilities occur within the system as well as the reversible cyclic device, then the cycle undergone by the combined system will be internally reversible. As such, it can be reversed. In the reversed cycle case, all the quantities will have the same magnitude but the opposite sign. Therefore, the work WC, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case. Then it follows that WC,int rev = 0 since it cannot be a positive or negative quantity, and therefore

Fz G δQ IJ H T K

=0

net

int rev

for internally reversible cycles. Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and Chapter 7-6

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-7

the inequality for the irreversible ones. To develop a relation for the definition of entropy, let us examine this last equation more closely. Here we have a quantity whose cyclic integral is zero. Let us think for a moment what kind of quantities can have this characteristic. We know that the cyclic integral of work is not zero. (It is a good thing that it is not. Otherwise, heat engines that work on a cycle such as steam power plants would produce zero net work.) Neither is the cyclic integral of heat. Now consider the volume occupied by a gas in a piston-cylinder device undergoing a cycle, as shown below.

When the piston returns to its initial position at the end of a cycle, the volume of the gas also returns to its initial value. Thus the net change in volume during a cycle is zero. This is also expressed as

zb

g

dV = 0

Chapter 7-7

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-8

We see that the cyclic integral of a property is zero. A quantity whose cyclic integral is zero depends only on the state and not on the process path; thus it is a property. Therefore the quantity (δQnet/T)int rev must be a property. In-class Example Consider the cycle shown below composed of two reversible processes A and B. Apply the Clausius inequality for this cycle. What do you conclude about these two integrals?

δQ I F G JK zHT 

2

net

1

and

int rev

δQ I F G JK zHT 

2

1

along path A

P

net

int rev

along path B

2

B A

1

V A cycle composed of two reversible processes.

Apply the Clausius inequality for the cycle made of two internally reversible processes: δQnet =0 T int rev

Fz G H

IJ K

Chapter 7-8

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-9

FG δQ IJ = z FG δQ IJ z You should find: H T K H T K

 

  2

2

net

1

net

1

int rev

along path A

int rev

along path B

Since the quantity (δQnet/T)int rev is independent of the path and must be a property, we call this property the entropy S. The entropy change occurring during a process is related to the heat transfer and the temperature of the system. The entropy is given the symbol S (kJ/K), and the specific entropy is s (kJ/kg⋅K). The entropy change during a reversible process, sometimes called an internally reversible process, is defined as

dS =

δQnet T

S2 − S1 =

z

int rev 2

1

δQnet T

int rev

Consider the cycle 1-A-2-B-1, shown below, where process A is arbitrary that is, it can be either reversible or irreversible, and process B is internally reversible. P

2

B A

1

V A cycle composed of reversible and irreversible processes. Chapter 7-9

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Fz G δQ IJ H T K

≤0

net

z

int rev

δQnet + 1   T 

2

along A

7-10

δQ I F G JK ≤ 0 zHT 

1

net

2

int rev

along B

The integral along the internally reversible path, process B, is the entropy change S1 –S2. Therefore,

z

2

1

δQnet + S1 − S2 ≤ 0 T

or

S2 − S1 ≥

z

2

1

δQnet T

In general the entropy change during a process is defined as

δQnet dS ≥ T where

= holds for the internally reversible process > holds for the irreversible process

Consider the effect of heat transfer on entropy for the internally reversible case.

Chapter 7-10

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-11

δQnet dS = T Which temperature T is this one? If

δQnet > 0,

then dS > 0

δQnet = 0, δQnet < 0,

then dS = 0 then dS < 0

This last result shows why we have kept the subscript net on the heat transfer Q. It is important for you to recognize that Q has a sign depending on the direction of heat transfer. The net subscript is to remind us that Q is positive when added to a system and negative when leaving a system. Thus, the entropy change of the system will have the same sign as the heat transfer in a reversible process. From the above, we see that for a reversible, adiabatic process

dS = 0 S2 = S1 The reversible, adiabatic process is called an isentropic process. Entropy change is caused by heat transfer and irreversibilities. Heat transfer to a system increases the entropy; heat transfer from a system decreases it. The effect of irreversibilities is always to increase the entropy. In fact, a process in which the heat transfer is out of the system may be so irreversible that the actual entropy change is positive. Friction is one source of irreversibilities in a system.

Chapter 7-11

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-12

The entropy change during a process is obtained by integrating the dS equation over the process:

∆S sys = S2 − S1 ≥

z

2

1

FG kJ IJ H KK

δQnet T

Here, the inequality is to remind us that the entropy change of a system 2 during an irreversible process is always greater than 1δQ / T , called the entropy transfer. That is, some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibilities. The entropy generated during a process is called entropy generation and is denoted as Sgen.

z

We can remove the inequality by noting the following

∆S sys = S2 − S1 =

z

2

1

δQnet + S gen T

FG kJ IJ H KK

Sgen is always a positive quantity or zero. Its value depends upon the process and thus it is not a property. Sgen is zero for an internally reversible process.

z

2

The integral 1δQ / T is performed by applying the first law to the process to obtain the heat transfer as a function of the temperature. The integration is not easy to perform, in general.

Chapter 7-12

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-13

Definition of Second Law of Thermodynamics Now consider an isolated system composed of several subsystems exchanging energy among themselves. Since the isolated system has no energy transfer across its system boundary, the heat transfer across the system boundary is zero.

Applying the definition of entropy to the isolated system

∆Sisolated ≥

z

2

1

δQnet 0, adiabatic T

The total entropy change for the isolated system is

∆Sisolated ≥ 0 This equation is the working definition of the second law of thermodynamics. The second law, known as the principle of increase of entropy, is stated as The total entropy change of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant. Chapter 7-13

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-14

Now consider a general system exchanging mass as well as energy with its surroundings.

S gen = ∆Stotal = ∆S sys + ∑ ∆S surr ≥ 0 where

= holds for the totally reversible process > holds for the irreversible process

Thus, the entropy generated or the total entropy change (sometimes called the entropy change of the universe or net entropy change) due to the process of this isolated system is positive (for actual processes) or zero (for reversible processes). The total entropy change for a process is the amount of entropy generated during that process (Sgen), and it is equal to the sum of the entropy changes of the system and the surroundings. The entropy changes of the important system (closed system or control volume) and its surroundings do not both have to be positive. The entropy for a given system (important or surroundings) may decrease during a process, but the

Chapter 7-14

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-15

sum of the entropy changes of the system and its surroundings for an isolated system can never decrease. Entropy change is caused by heat transfer and irreversibilities. Heat transfer to a system increases the entropy, and heat transfer from a system decreases it. The effect of irreversibilities is always to increase the entropy. The increase in entropy principle can be summarized as follows:

S gen = ∆STotal

R|> 0 S|= 0 T< 0

Irreversible processes Reversible processes Impossible processes

Some Remarks about Entropy 1. Processes can occur in a certain direction only, not in just any direction, such that Sgen ≥ 0. 2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. The entropy of the universe is continuously increasing. 3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities present during that process.

Chapter 7-15

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-16

Heat Transfer as the Area under a T-S Curve For the reversible process, the equation for dS implies that

δQnet dS = T δQnet = TdS or the incremental heat transfer in a process is the product of the temperature and the differential of the entropy, the differential area under the process curve plotted on the T-S diagram.

z

2

Qnet = TdS 1

In the above figure, the heat transfer in an internally reversible process is shown as the area under the process curve plotted on the T-S diagram.

Chapter 7-16

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-17

Isothermal, Reversible Process For an isothermal, reversible process, the temperature is constant and the integral to find the entropy change is readily performed. If the system has a constant temperature, T0, the entropy change becomes

∆S = S2 − S1 =

z

2

1

δQnet Qnet = T To

For a process occurring over a varying temperature, the entropy change must be found by integration over the process. Adiabatic, Reversible (Isentropic) Process For an adiabatic process, one in which there is no heat transfer, the entropy change is

∆S = S2 − S1 ≥

z

2

1

δQnet T

0, adiabatic

∆S = S2 − S1 ≥ 0 If the process is adiabatic and reversible, the equality holds and the entropy change is

∆S = S2 − S1 = 0 S2 = S1

or on a per unit mass basis

S m s2 = s1 s=

Chapter 7-17

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-18

The adiabatic, reversible process is a constant entropy process and is called isentropic. As will be shown later for an ideal gas, the adiabatic, reversible process is the same as the polytropic process where the polytropic exponent n = k = Cp/Cv. The principle of increase of entropy for a closed system exchanging heat with its surroundings at a constant temperature Tsurr is found by using the equation for the entropy generated for an isolated system. Surroundings Tsurr A general closed system (a cup of coffee) exchanging heat with its surroundings

Qout, sys System Boundary

S gen = ∆Stotal = ∆Ssys + ∑ ∆Ssurr ≥ 0 ∆Ssys = ( S2 − S1 ) sys

∑ ∆S

surr

=

Qnet , surr Tsurr

S gen = ∆Stotal = m( s2 − s1 ) sys +

Qnet , surr Tsurr

≥0

where

Qnet , surr = −Qnet , sys = −(0 − Qout , sys ) = Qout , sys Chapter 7-18

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-19

Effect of Heat Transfer on Entropy Let's apply the second law to the following situation. Consider the transfer of heat from a heat reservoir at temperature T to a heat reservoir at temperature T - ∆T > 0 where ∆T > 0, as shown below. HR at T

Q

Two heat reservoirs exchanging heat over a finite temperature difference

HR at T-∆T

The second law for the isolated system composed of the two heat reservoirs is

S gen = ∆Stotal = ∆Ssys + ∑ ∆Ssurr ≥ 0

S gen = ∆Stotal = ∆S HR @T + ∆S HR @T − ∆T In general, if the heat reservoirs are internally reversible

∆Ssys = ∆S HR @T = ∆S HR @T − ∆T = Chapter 7-19

z

δQnet 1 T −Q 2

T +Q T − ∆T

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

−Q

S gen = ∆STotal =

S gen = ∆STotal

T

+

7-20

+Q T − ∆T

∆T O L = T MN (T − ∆T ) PQ Q

Now as ∆T → 0, Sgen → 0 and the process becomes totally reversible. Therefore, for reversible heat transfer ∆T must be small. As ∆T gets large, Sgen increases and the process becomes irreversible. Example 7-3 Find the total entropy change, or entropy generation, for the transfer of 1000 kJ of heat energy from a heat reservoir at 1000 K to a heat reservoir at 500 K. HR at T=1000 K

Areas = 1000 kJ

T 1000 K

Q=1000 kJ HR at T-∆T = 500K

500 K 0

1

The second law for the isolated system is

Chapter 7-20

2 S, kJ/K

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

S gen = ∆STotal =

−Q

+

7-21

+Q

T T − ∆T −1000kJ 1000kJ = + 1000 K 500 K kJ = ( −1 + 2) K kJ =1 K

What happens when the low-temperature reservoir is at 750 K? The effect of decreasing the ∆T for heat transfer is to reduce the entropy generation or total entropy change of the universe due to the isolated system and the irreversibilities associated with the heat transfer process. Third Law of Thermodynamics The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero. This law provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is called absolute entropy. Entropy as a Property Entropy is a property, and it can be expressed in terms of more familiar properties (P,v,T) through the Tds relations. These relations come from the analysis of a reversible closed system that does boundary work and has heat added. Writing the first law for the closed system in differential form on a per unit mass basis

Chapter 7-21

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

δQint rev

7-22

δWint rev, out

System used to find expressions for ds

δQint rev − δWint rev, out = dU δQint rev = T dS δWint rev, out = P dV TdS − P dV = dU On a unit mass basis we obtain the first Tds equation, or Gibbs equation, as

Tds = du + Pdv Recall that the enthalpy is related to the internal energy by h = u + Pv. Using this relation in the above equation, the second Tds equation is

T ds = dh − v dP These last two relations have many uses in thermodynamics and serve as the starting point in developing entropy-change relations for processes. The successful use of Tds relations depends on the availability of property relations. Such relations do not exist in an easily used form for a general pure substance but are available for incompressible substances (liquids, Chapter 7-22

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-23

solids) and ideal gases. So, for the general pure substance, such as water and the refrigerants, we must resort to property tables to find values of entropy and entropy changes. The temperature-entropy and enthalpy-entropy diagrams for water are shown below.

Chapter 7-23

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-24

Shown above are the temperature-entropy and enthalpy-entropy diagrams for water. The h-s diagram, called the Mollier diagram, is a useful aid in solving steam power plant problems. Chapter 7-24

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-25

Example 7-4 Find the entropy and/or temperature of steam at the following states: P 5 MPa

T

1 MPa

50oC

1.8 MPa 40 kPa

Region

s kJ/(kg K)

120o C

400o C Quality, x = 0.9

40 kPa

7.1794

(Answers are on the last page of Chapter 7.) Example 7-5 Determine the entropy change of water contained in a closed system as it changes phase from saturated liquid to saturated vapor when the pressure is 0.1 MPa and constant. Why is the entropy change positive for this process? System: The water contained in the system (a piston-cylinder device) T Steam

Chapter 7-25

s

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-26

Property Relation: Steam tables Process and Process Diagram: Constant pressure (sketch the process relative to the saturation lines) Conservation Principles: Using the definition of entropy change, the entropy change of the water per mass is

∆s = s2 − s1 = sg − s f = s fg = 6.0562

kJ kg ⋅ K

The entropy change is positive because: (Heat is added to the water.) Example 7-6 Steam at 1 MPa, 600oC, expands in a turbine to 0.01 MPa. If the process is isentropic, find the final temperature, the final enthalpy of the steam, and the turbine work. System: The control volume formed by the turbine 1

T Wout

Control surface

2

Chapter 7-26

s

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-27

Property Relation: Steam tables Process and Process Diagram: Isentropic (sketch the process relative to the saturation lines on the T-s diagram) Conservation Principles: Assume: steady-state, steady-flow, one entrance, one exit, neglect KE and PE Conservation of mass:

m 1 = m 2 = m First Law or conservation of energy: The process is isentropic and thus adiabatic and reversible; therefore Q = 0. The conservation of energy becomes

Ein = E out m 1h1 = m 2 h2 + Wout Since the mass flow rates in and out are equal, solve for the work done per unit mass

Chapter 7-27

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-28

E in = E out Wout = m 1h1 − m 2 h2 = m (h1 − h2 ) Wout = h1 − h2 w= m Now, let’s go to the steam tables to find the h’s.

kJ  = 3698.6 h P1 = 1MPa   1 kg   kJ T1 = 600o C   s1 = 8.0311  kg ⋅ K The process is isentropic, therefore; s2 = s1 = 8.0311 kJ/(kg K ) At P2 = 0.01 MPa, sf = 0.6492 kJ/kg⋅K, and sg = 8.1488 kJ/(kg K); thus, sf < s2 < sg. State 2 is in the saturation region, and the quality is needed to specify the state.

Chapter 7-28

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-29

s2 = s f + x2 s fg x2 =

s2 − s f s fg

8.0311 − 0.6492 = 0.984 7.4996 h2 = h f + x2 h fg =

= 191.8 + (0.984)(2392.1) kJ = 2545.6 kg Since state 2 is in the two-phase region, T2 = Tsat at P2 = 45.81oC.

w = h1 − h2 kJ = (3698.6 − 2545.6) kg kJ = 1153 kg Entropy Change and Isentropic Processes The entropy-change and isentropic relations for a process can be summarized as follows: 1. Pure substances: Any process: ∆s = s2 − s1 (kJ/kg⋅K)

Chapter 7-29

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-30

Isentropic process: s2 = s1 2. Incompressible substances (Liquids and Solids):

ds =

du P + dv T T

The change in internal energy and volume for an incompressible substance is

du = C dT dv ≅ 0 The entropy change now becomes

C dT +0 T 2 C ( T ) dT ∆s = 1 T ds =

z

If the specific heat for the incompressible substance is constant, then the entropy change is Any process:

s2 − s1 = Cav ln

T2 T1

Isentropic process: T2 = T1 3. Ideal gases:

Chapter 7-30

(kJ/kg⋅K)

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-31

a. Constant specific heats (approximate treatment): Any process: (can you fill in the steps?)

s2 − s1 = Cv , av ln

T2 v + R ln 2 T1 v1

(kJ/kg⋅K)

and (can you fill in the steps?)

s2 − s1 = C p , av ln

T2 P − R ln 2 T1 P1

(kJ/kg⋅K)

Or, on a unit-mole basis, s2 − s1 = Cv , av ln

T2 v + Ru ln 2 T1 v1

(kJ/kmol⋅K)

s2 − s1 = C p , av ln

T2 P − Ru ln 2 T1 P1

(kJ/kmol⋅K)

and

Chapter 7-31

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-32

Isentropic process: (Can you fill in the steps here?)

FG T IJ HT K 2

1

FG T IJ HT K

k −1

1

s = const .

2

1

Fv I =G J Hv K 2

FPI =G J HPK

( k −1)/ k

2

s = const .

1

 P2  v  = 1     P1  s = const .  v2 

k

For an isentropic process this last result looks like Pvk = constant which is the polytropic process equation Pvn = constant with n = k = Cp/Cv.

b. Variable specific heats (exact treatment): From Tds = dh - vdP, we obtain

∆s = ∫

2

C p (T ) T

1

dT − R ln

P2 P1

The first term can be integrated relative to a reference state at temperature Tref..

z

2

1

C p (T ) T

dT = =

z z

Tref

C p (T )

T2

T C p (T )

Tref

T

T1

Chapter 7-32

dT +

dT −

z z

T2

C p (T )

T1

T C p (T )

Tref

T

Tref

dT

dT

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-33

The integrals on the right-hand side of the above equation are called the standard state entropies, so, at state 1, T1, and state 2, T2; so is a function of temperature only.

s1o = s = o 2

z z

T1

C p (T )

T2

T C p (T )

Tref

T

Tref

dT dT

Therefore, for any process: s2 − s1 = s2o − s1o − R ln

P2 P1

(kJ/kg⋅K)

or s2 − s1 = s2o − s1o − Ru ln

P2 P1

(kJ/kmol⋅K)

The standard state entropies are found in Tables A-17 for air on a mass basis and Tables A-18 through A-25 for other gases on a mole basis. When using this variable specific heat approach to finding the entropy change for an ideal gas, remember to include the pressure term along with the standard state entropy terms--the tables don’t warn you to do this. Isentropic process: ∆s = 0

Chapter 7-33

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

s2o = s1o + R ln

P2 P1

7-34

(kJ/kg⋅K)

If we are given T1, P1, and P2, we find so1 at T1, calculate so2, and then determine from the tables T2, u2, and h2. When air undergoes an isentropic process when variable specific heat data are required, there is another approach to finding the properties at the end of the isentropic process. Consider the entropy change written as

∆s = ∫

2

1

C p (T ) T

dT − R ln

P2 P1

Letting T1 = Tref, P1 = Pref = 1atm, T2 = T, P2 = P, and setting the entropy change equal to zero yield

F PI GH P JK ref

s = const

F1 = EXPG z HR

T

C p (T ' )

Tref

T'

IJ K

dT '

We define the relative pressure Pr as the above pressure ratio. Pr is the pressure ratio necessary to have an isentropic process between the reference temperature and the actual temperature and is a function of the actual temperature. This parameter is a function of temperature only and is found in the air tables, Table A-17. The relative pressure is not available for other gases in this text.

b g Pr

s = const

F1 = EXPG z HR Chapter 7-34

T

C p (T ' )

Tref

T'

IJ K

dT '

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-35

The ratio of pressures in an isentropic process is related to the ratio of relative pressures.

 P2 / Pref  P2  =   P  1  s = const .  P1 / Pref

 P = r2   s = const . Pr1

There is a second approach to finding data at the end of an ideal gas isentropic process when variable specific heat data are required. Consider the following entropy change equation set equal to zero. From Tds = du + Pdv, we obtain for ideal gases

∆s =

z

2

1

v Cv (T ) dT + R ln 2 T v1

Letting T1 = Tref, v1 = vref, T2 = T, v2 = v, and setting the entropy change equal to zero yield

FvI GH v JK ref

s = const

F 1 = EXPG − z H R

T

Tref

IJ K

Cv (T ') dT ' T'

We define the relative volume vr as the above volume ratio. vr is the volume ratio necessary to have an isentropic process between the reference temperature and the actual temperature and is a function of the actual temperature. This parameter is a function of temperature only and is found in the air tables, Table A-17. The relative volume is not available for other gases in this text.

bv g

r s = const

F 1 = EXPG − z H R

T

Tref

Chapter 7-35

IJ K

Cv (T ') dT ' T'

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

FG v IJ Hv K 2

1

s = const .

F v /v I =G H v / v JK 2

ref

1

ref

7-36

= s = const .

vr 2 vr 1

Extra Assignment For an ideal gas having constant specific heats and undergoing a polytropic process in a closed system, Pvn = constant, with n = k, find the heat transfer by applying the first law. Based on the above discussion of isentropic processes, explain your answer. Compare your results to this problem to a similar extra assignment problem in Chapter 4. Example 7-7 Aluminum at 100oC is placed in a large, insulated tank having 10 kg of water at a temperature of 30oC. If the mass of the aluminum is 0.5 kg, find the final equilibrium temperature of the aluminum and water, the entropy change of the aluminum and the water, and the total entropy change of the universe because of this process. Before we work the problem, what do you think the answers ought to be? Are entropy changes going to be positive or negative? What about the entropy generated as the process takes place? System: Closed system including the aluminum and water. Tank insulated boundary

Water AL

Chapter 7-36

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-37

Property Relation: ? Process: Constant volume, adiabatic, no work energy exchange between the aluminum and water.

Conservation Principles: Apply the first law, closed system to the aluminum-water system.

Q − W = ∆U system 0 − 0 = ∆U water + ∆U AL Using the solid and incompressible liquid relations, we have

mwater Cwater ( T2 − T1 ) water + m AL C AL (T2 − T1 ) AL = 0 But at equilibrium, T2,AL = T2,water = T2

T2 = =

mwater Cwater (T1 ) water + mAL C AL (T1 ) AL mwater Cwater + mAL C AL

10kg water (4.18kJ / kg water ⋅ K )(303K ) + 0.5kg AL (0.941kJ / kg AL ⋅ K )(373K ) 10kg water (4.18kJ / kg water ⋅ K ) + 0.5kg AL (0.941kJ / kg AL ⋅ K )

= 3038 . K

The second law gives the entropy production, or total entropy change of the universe, as

Chapter 7-37

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-38

S gen = ∆Stotal = ∆Swater + ∆S AL ≥ 0 Using the entropy change equation for solids and liquids,

∆S AL = m AL C AL ln

T2 T1, AL

= 0.5kg ( 0.941 = −0.0966

FG H

kJ 303.8 K ) ln kg ⋅ K (100 + 273) K

IJ K

kJ K

∆S water = mwater Cwater ln = 10kg ( 4.177 = +0.1101

T2 T1,water

FG H

kJ 303.8 K ) ln kg ⋅ K ( 30 + 273) K

kJ K

Why is ∆SAL negative? Why is ∆Swater positive?

S gen = ∆Stotal = ∆S water + ∆S AL . = (01101 − 0.0966) = +0.0135

Why is Sgen or ∆STotal positive?

Chapter 7-38

kJ K

kJ K

IJ K

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-39

Example 7-8 Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa and 800 K, respectively. Assuming ideal gas behavior, find the entropy change of the carbon dioxide by first assuming constant specific heats and then assuming variable specific heats. Compare your results with the real gas data obtained from the EES software. a. Assume the Table A-2(a) data at 300 K are adequate; then Cp = 0.846 kJ/kg⋅K and R = 0.1889 kJ/kg-K.

s2 − s1 = C p , av ln

T2 P − R ln 2 T1 P1

FG H

IJ K

FG H

kJ kJ 800 K 2000kPa ln . ln − 01889 kg ⋅ K kg ⋅ K 400 K 50kPa kJ . = −01104 kg ⋅ K = 0.846

IJ K

b. For variable specific heat data, use the carbon dioxide data from Table A-20.

F s − s I − R ln P s −s =G H M JK P F (257.408 − 225.225) kJ / kmol ⋅ K IJ − 0.1889 kJ lnFG 2000kPa IJ =G 44 kg / kmol kg ⋅ K H 50kPa K H K o T2

2

o T1

2

1

CO2

= +0.0346

1

kJ kg ⋅ K

Chapter 7-39

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-40

c. Using EES for carbon dioxide as a real gas: Deltas

=ENTROPY(CarbonDioxide,T=800,P=2000)ENTROPY(CarbonDioxide,T=400,P=50) = +0.03452 kJ/kg⋅K

d. Repeat the constant specific heat calculation assuming Cp is a constant at the average of the specific heats for the temperatures. Then Cp = 1.054 kJ/kg⋅K (see Table A-2(b)). s2 − s1 = C p , av ln

T2 P − R ln 2 T1 P1

FG H

IJ K

FG H

800 K 2000kPa kJ kJ ln ln − 0.1889 400 K 50kPa kg ⋅ K kg ⋅ K kJ = +0.0337 kg ⋅ K = 1.054

IJ K

It looks like the 300 K data give completely incorrect results here. If the compression process is adiabatic, why is ∆s positive for this process?

Chapter 7-40

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-41

Example 7-9 Air, initially at 17oC, is compressed in an isentropic process through a pressure ratio of 8:1. Find the final temperature assuming constant specific heats and variable specific heats, and using EES. a. Constant specific heats, isentropic process

FG T IJ HT K 2

1

FPI =G J HPK

( k −1)/ k

2

s = const .

1

For air, k = 1.4, and a pressure ratio of 8:1 means that P2/P1 = 8.

FPI T =TG J HPK 2

( k −1)/ k

2

1

1

bg

= (17 + 273) K 8

(1.4 −1)/1.4

= 525.3K = 252.3o C

b. Variable specific heat method

FG P IJ HPK

=

2

1

s = const .

FP /P I GH P / P JK 2

ref

1

ref

= s = const .

Pr 2 Pr 1

Using the air data from Table A-17 for T1 = (17+273) K = 290 K, Pr1 = 1.2311.

Pr 2 = Pr 1

P2 P1

= 12311 . (8) = 9.8488 Interpolating in the air table at this value of Pr2, gives T2 = 522.4 K = 249.4oC Chapter 7 - 41

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-42

c. A second variable specific heat method. Using the air table, Table A-17, for T1 = (17+273) K = 290 K, soT1 = 1.66802 kJ/kg⋅K. For the isentropic process s2o = s1o + R ln

P2 P1

bg

kJ kJ + 0.287 ln 8 kg ⋅ K kg ⋅ K kJ = 2.26482 kg ⋅ K

= 1.66802

At this value of soT2, the air table gives T2 = 522.4 K= 249.4oC. This technique is based on the same information as the method shown in part b. d. Using the EES software with T in oC and P in kPa and assuming P1 = 100 kPa. s_1 = ENTROPY(Air, T=17, P=100) s_2 = s_1 T_2 = TEMPERATURE(Air, P=800, s=s_2) The solution is: s_1 = 5.668 kJ/kg⋅K s_2 = 5.668 kJ/kg⋅K T_2 = 249.6oC

Chapter 7 - 42

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-43

Example 7-10 Air initially at 0.1 MPa, 27oC, is compressed reversibly to a final state. (a) Find the entropy change of the air when the final state is 0.5 MPa, 227oC. (b) Find the entropy change when the final state is 0.5 MPa, 180oC. (c) Find the temperature at 0.5 MPa that makes the entropy change zero. Assume air is an ideal gas with constant specific heats. Show the two processes on a T-s diagram. a.

s2 − s1 = C p , av ln = 1.005

T2 P − R ln 2 T1 P1

FG H

IJ K

FG H

IJ K

FG H

0.5 MPa kJ ( 227 + 273) K kJ ln ln − 0.287 . MPa 01 kg ⋅ K ( 27 + 273) K kg ⋅ K

= +0.0507

IJ K

kJ kg ⋅ K

b.

s2 − s1 = C p , av ln = 1.005

T2 P − R ln 2 T1 P1

FG H

kJ (180 + 273) K kJ 0.5 MPa − 0.287 ln ln kg ⋅ K ( 27 + 273) K kg ⋅ K 0.1 MPa

= −0.0484

kJ kg ⋅ K

Chapter 7 - 43

IJ K

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

c.

FPI T =TG J HPK 2

7-44

( k −1)/ k

2

1

1

F 0.5 MPa IJ = ( 27 + 273) K G H 01. MPa K

(1.4 −1)/1.4

= 475.4 K = 202.4 o C The T-s plot is T

2

c

P2

a

b P1 1 s

Give an explanation for the difference in the signs for the entropy changes.

Chapter 7 - 44

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-45

Example 7-11 Nitrogen expands isentropically in a piston cylinder device from a temperature of 500 K while its volume doubles. What is the final temperature of the nitrogen, and how much work did the nitrogen do against the piston, in kJ/kg? System: The closed piston-cylinder device T

W

P1

1

Nitrogen system boundary

P2

2

s

P

1

P1 s = const. 2

P2

v

Property Relation: Ideal gas equations, constant properties Process and Process Diagram: Isentropic expansion Conservation Principles: Second law:

Chapter 7 - 45

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-46

Since we know T1 and the volume ratio, the isentropic process, ∆s = 0, allows us to find the final temperature. Assuming constant properties, the temperatures are related by

F v I T =TG J Hv K 1I F = 500 K G J H 2K k −1

2

1

1

2

1.4 −1

= 378.9 K = 105.9 o C Why did the temperature decrease? First law, closed system: Note, for the isentropic process (reversible, adiabatic); the heat transfer is zero. The conservation of energy for this closed system becomes

Ein − E out = ∆E −W = ∆U W = − ∆U Using the ideal gas relations, the work per unit mass is

Chapter 7 - 46

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-47

W = − mCv (T2 − T1 ) W w = = − Cv (T2 − T1 ) m kJ = −0.743 (378.9 − 500) K kg ⋅ K kJ = 90.2 kg Why is the work positive? Extra Assignment For the isentropic process Pvk = constant. Use the definition of boundary work to show that you get the same result as the last example. That is, determine the boundary work and show that you obtain the same expression as that for the polytropic boundary work.

Example 7-12 A Carnot engine has 1 kg of air as the working fluid. Heat is supplied to the air at 800 K and rejected by the air at 300 K. At the beginning of the heat addition process, the pressure is 0.8 MPa and during heat addition the volume triples. (a) Calculate the net cycle work assuming air is an ideal gas with constant specific heats. (b) Calculate the amount of work done in the isentropic expansion process.

Chapter 7 - 47

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-48

(c) Calculate the entropy change during the heat rejection process. System: The Carnot engine piston-cylinder device. W

Air system boundary Q

Property Relation: Ideal gas equations, constant properties. Process and Process Diagram: Constant temperature heat addition. QH

T TH

1

2

TL 4

QL

3

s Carnot Cycle

Conservation Principles: a. Apply the first law, closed system, to the constant temperature heat addition process, 1-2.

Chapter 7 - 48

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

Qnet ,12 − Wnet ,12 = ∆U12 = mCv (T2 − T1 ) = 0 = Wnet ,12

Qnet ,12

So for the ideal gas isothermal process,

Wnet ,12 = Wother ,12 + Wb ,12

z z

2

= PdV 1

dV 1 V V2 = mRT ln V1 =

2

mRT

FG IJ H K

= 1kg (0.287

kJ )(800 K ) ln(3) kg ⋅ K

= 252.2 kJ But

Qnet ,12 = QH QH = 252.2 kJ

Chapter 7 - 49

7-49

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-50

The cycle thermal efficiency is

η th =

Wnet , cycle QH

For the Carnot cycle, the thermal efficiency is also given by

TL 300 K ηth = 1 − = 1− TH 800 K = 0.625 The net work done by the cycle is

Wnet , cycle = η th QH = 0.625(252.2 kJ ) = 157.6 kJ b. Apply the first law, closed system, to the isentropic expansion process, 23. But the isentropic process is adiabatic, reversible; so, Q23 = 0.

Chapter 7 - 50

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-51

Ein − E out = ∆E −W = ∆U W23 = − ∆U 23 Using the ideal gas relations, the work per unit mass is

W23 = − mCv (T3 − T2 ) kJ )(300 − 800) K = − (1kg )(0.718 kg ⋅ K = 359.0 kJ This is the work leaving the cycle in process 2-3. c. Using equation (6-34)

∆s34 = C p ln

FG T IJ − R lnFG P IJ HT K H P K 4

4

3

3

But T4 = T3 = TL = 300 K, and we need to find P4 and P3. Consider process 1-2 where T1 = T2 = TH = 800 K, and, for ideal gases

Chapter 7 - 51

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

m1 = m2 PV PV 1 1 = 2 2 T1 T2 P2 = P1

V1 3V1

= (800 kPa )

1 3

= 266.7 kPa Consider process 2-3 where s3 = s2.

FG IJ H K F TI P = PG J HT K

T3 P3 = T2 P2 3

2

( k −1)/ k

k /( k −1)

3

2

F 300K IJ = 266.7 kPa G H 800k K

1.4 /(1.4 −1)

Chapter 7 - 52

= 8.613 kPa

7-52

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-53

Now, consider process 4-1 where s4 = s1.

T  P4 = P1  4   T1 

k /( k −1)

1.4 /(1.4 −1)

 300 K  = 8000 kPa    800k  = 25.834 kPa Now,

L F F TI P IO ∆S = mMC lnG J − R lnG J P N H T K H P KQ F PI = − mR lnG J HPK kJ . kPa I F 25834 = (1 kg )(0.287 ) lnG J kg ⋅ K H 8.613kPa K p

4

4

3

3

4

3

= −0.315

kJ K

Extra Problem Use a second approach to find ∆S34 by noting that the temperature of process 3-4 is constant and applying the basic definition of entropy for an internally reversible process, dS = δQ/T.

Chapter 7 - 53

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7-54

Reversible Steady-Flow Work Isentropic, Steady Flow through Turbines, Pumps, and Compressors Consider a turbine, pump, compressor, or other steady-flow control volume, work-producing device. The general first law for the steady-flow control volume is

E in = E out

G2 G2 V V Q net + ∑ m i (hi + i + gzi ) = Wnet + ∑ m e (he + e + gze ) 2 2 inlets exits For a one-entrance, one-exit device undergoing an internally reversible process, this general equation of the conservation of energy reduces to, on a unit of mass basis

δ wrev = δ qrev − dh − dke − dpe But δ qrev = T ds δ wrev = T ds − dh − dke − dp Using the Gibb’s second equation, this becomes

dh = T ds + v dP δ wrev = −v dP − dke − dpe Integrating over the process, this becomes

Chapter 7 - 54

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

z

2

wrev = − v dP − ∆ke − ∆pe 1

7-55

FG kJ IJ H kg K

Neglecting changes in kinetic and potential energies, reversible work becomes

z

2

wrev = − v dP 1

FG kJ IJ H kg K

Based on the classical sign convention, this is the work done by the control volume. When work is done on the control volume such as compressors or pumps, the reversible work going into the control volume is

z

2

wrev , in = v dP + ∆ke + ∆pe 1

FG kJ IJ H kg K

Turbine Since the fluid pressure drops as the fluid flows through the turbine, dP < 0, and the specific volume is always greater than zero, wrev, turbine > 0. To perform the integral, the pressure-volume relation must be known for the process. Compressor and Pump Since the fluid pressure rises as the fluid flows through the compressor or pump, dP > 0, and the specific volume is always greater than zero, wrev, in > 0, or work is supplied to the compressor or pump. To perform the integral, the pressure-volume relation must be known for the process. The term compressor is usually applied to the compression of a gas. The term pump is usually applied when increasing the pressure of a liquid. Chapter 7 - 55

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Pumping an incompressible liquid For an incompressible liquid, the specific volume is approximately constant. Taking v approximately equal to v1, the specific volume of the liquid entering the pump, the work can be expressed as

z

2

wrev , in = v dP + ∆ke + ∆pe 1

FG kJ IJ H kg K

= v∆P + ∆ke + ∆pe For the steady-flow of an incompressible fluid through a device that involves no work interactions (such as nozzles or a pipe section), the work term is zero, and the equation above can be expressed as the well-know Bernoulli equation in fluid mechanics.

v ( P2 − P1 ) + ∆ke + ∆pe = 0 Extra Assignment Using the above discussion, find the turbine and compressor work per unit mass flow for an ideal gas undergoing an isentropic process, where the pressure-volume relation is Pvk = constant, between two temperatures, T1 and T2. Compare your results with the first law analysis of Chapter 5 for control volumes.

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Example 7-13 Saturated liquid water at 10 kPa leaves the condenser of a steam power plant and is pumped to the boiler pressure of 5 MPa. Calculate the work for an isentropic pumping process. a. From the above analysis, the work for the reversible process can be applied to the isentropic process (it is left for the student to show this is true) as

 1 ( P2 − P1 ) WC = mv Here at 10 kPa, v1 = vf = 0.001010 m3/kg. The work per unit mass flow is

WC = v1 ( P2 − P1 ) wC = m m3 kJ = 0.001010 (5000 − 10) kPa 3 m kPa kg kJ = 5.04 kg

b. Using the steam table data for the isentropic process, we have

−Wnet = m (h2 − h1 ) −(0 − WC ) = m (h2 − h1 )

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From the saturation pressure table,

kJ  191.81 h = 1 P1 = 10 kPa   kg  kJ Sat. Liquid   s1 = 0.6492  kg ⋅ K Since the process is isentropic, s2 = s1. Interpolation in the compressed liquid tables gives

P2 = 5 MPa

 kJ  kJ  h2 = 197.42 s2 = s1 = 0.6492 kg kg ⋅ K  The work per unit mass flow is

WC = (h2 − h1 ) wC = m = (197.42 − 191.81) = 5.61

kJ kg

kJ kg

The first method for finding the pump work is adequate for this case.

Chapter 7 - 58

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Turbine, Compressor (Pump), and Nozzle Efficiencies Most steady-flow devices operate under adiabatic conditions, and the ideal process for these devices is the isentropic process. The parameter that describes how a device approximates a corresponding isentropic device is called the isentropic or adiabatic efficiency. It is defined for turbines, compressors, and nozzles as follows: Turbine:

The isentropic work is the maximum possible work output that the adiabatic turbine can produce; therefore, the actual work is less than the isentropic work. Since efficiencies are defined to be less than 1, the turbine isentropic efficiency is defined as

ηT =

Actual turbine work w = a Isentropic turbine work ws

ηT ≅

h1 − h2 a h1 − h2 s

Well-designed large turbines may have isentropic efficiencies above 90 percent. Small turbines may have isentropic efficiencies below 70 percent. Chapter 7 - 59

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7-60

Compressor and Pump: The isentropic work is the minimum possible work that the adiabatic compressor requires; therefore, the actual work is greater than the isentropic work. Since efficiencies are defined to be less than 1, the compressor isentropic efficiency is defined as

T1 P1

WC Compressor or pump

T2 P2

Isentropic compressor work ws = Actual compressor work wa h −h ηC ≅ 2s 1 h2 a − h1

ηC =

Well-designed compressors have isentropic efficiencies in the range from 75 to 85 percent. Review the efficiency of a pump and an isothermal compressor on your own.

Chapter 7 - 60

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Nozzle: The isentropic kinetic energy at the nozzle exit is the maximum possible kinetic energy at the nozzle exit; therefore, the actual kinetic energy at the nozzle exit is less than the isentropic value. Since efficiencies are defined to be less than 1, the nozzle isentropic efficiency is defined as

T1 PG1 V1

T2 PG2 V2

G V22a 2

Nozzle

G V22s 2

G Actual KE at nozzle exit V22a / 2 ηN = = G Isentropic KE at nozzle exit V22s / 2 For steady-flow, no work, neglecting potential energies, and neglecting the inlet kinetic energy, the conservation of energy for the nozzle is

G2 V h1 = h2 a + 2 a 2

Chapter 7 - 61

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The nozzle efficiency is written as

ηN ≅

h1 − h2 a h1 − h2 s

Nozzle efficiencies are typically above 90 percent, and nozzle efficiencies above 95 percent are not uncommon.

Example 7-14 The isentropic work of the turbine in Example 7-6 is 1152.2 kJ/kg. If the isentropic efficiency of the turbine is 90 percent, calculate the actual work. Find the actual turbine exit temperature or quality of the steam.

ηT =

w Actual turbine work = a Isentropic turbine work ws

wa = ηT ws = (0.9)(1153.0

kJ kJ ) = 1037.7 kg kg

h1 − h2 a ηT ≅ h1 − h2 s Now to find the actual exit state for the steam. From Example 7-6, steam enters the turbine at 1 MPa, 600oC, and expands to 0.01 MPa.

Chapter 7 - 62

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

From the steam tables at state 1

kJ  h 3698.6 = P1 = 1 MPa   1 kg  kJ T1 = 600o C   s1 = 8.0311  kg ⋅ K At the end of the isentropic expansion process, see Example 7-6.

P2 = 0.01 MPa

kJ  h = 2545.6   2s kg kJ   s2 s = s1 = 8.0311 kg ⋅ K   x2 s = 0.984 The actual turbine work per unit mass flow is (see Example 7-6)

wa = h1 − h2 a h2 a = h1 − wa kJ = (3698.6 − 1037.7) kg kJ = 2660.9 kg For the actual turbine exit state 2a, the computer software gives

U|RT = 86.85 C kJ VS = 2660.9 |TSuperheated kg W

P2 = 0.01 MPa

o

2a

h2 a

Chapter 7 - 63

7-63

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7-64

A second method for finding the actual state 2 comes directly from the expression for the turbine isentropic efficiency. Solve for h2a.

h2 a = h1 − ηT ( h1 − h2 s ) kJ kJ = 3698.6 − (0.9)(3698.6 − 2545.6) kg kg kJ = 2660.9 kg Then P2 and h2a give T2a = 86.85oC.

Example 7-15 Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27oC, to a final state of 0.5 MPa. Find the work done on the air for a compressor isentropic efficiency of 80 percent. System: The compressor control volume T1 P1

WC Compressor or pump

T2 P2

Property Relation: Ideal gas equations, assume constant properties. Chapter 7 - 64

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

T

2a

2s

7-65

P2

P1

1

s

Process and Process Diagram: First, assume isentropic, steady-flow and then apply the compressor isentropic efficiency to find the actual work. Conservation Principles: For the isentropic case, Qnet = 0. Assuming steady-state, steady-flow, and neglecting changes in kinetic and potential energies for one entrance, one exit, the first law is

E in = E out m 1h1 + WCs = m 2 h2 s The conservation of mass gives

m 1 = m 2 = m The conservation of energy reduces to

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WCs = m (h2 s − h1 ) WCs wCs = = (h2 s − h1 ) m Using the ideal gas assumption with constant specific heats, the isentropic work per unit mass flow is

wCs = C p (T2 s − T1 ) The isentropic temperature at state 2 is found from the isentropic relation

T2 s

F PI =TG J HPK 1

( k −1)/ k

2

1

F 0.5 MPa IJ = (27 + 273) K G H 01. MPa K

(1.4 −1)/1.4

= 475.4 K The conservation of energy becomes

wCs = C p (T2 s − T1 ) kJ (475.4 − 300) K kg ⋅ K kJ = 176.0 kg . = 1005

Chapter 7 - 66

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The compressor isentropic efficiency is defined as

ηC = wCa =

ws wa wcs

ηC

kJ kJ kg = 220 kg 0.8

176 = Example 7-16

Nitrogen expands in a nozzle from a temperature of 500 K while its pressure decreases by factor of two. What is the exit velocity of the nitrogen when the nozzle isentropic efficiency is 95 percent? System: The nozzle control volume.

T T1 PG1 V1

T2 PG2 V2

P1

1

2s

2a

P2

Nozzle

s

Property Relation: The ideal gas equations, assuming constant specific heats

Chapter 7 - 67

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7-68

Process and Process Diagram: First assume an isentropic process and then apply the nozzle isentropic efficiency to find the actual exit velocity. Conservation Principles: For the isentropic case, Qnet = 0. Assume steady-state, steady-flow, no work is done. Neglect the inlet kinetic energy and changes in potential energies. Then for one entrance, one exit, the first law reduces to

E in = E out

G2 V m 1h1 = m 2 (h2 s + 2 s ) 2 The conservation of mass gives

m 1 = m 2 = m The conservation of energy reduces to

G V2 s = 2(h1 − h2 s ) Using the ideal gas assumption with constant specific heats, the isentropic exit velocity is

G V2 s = 2C p (T1 − T2 s ) The isentropic temperature at state 2 is found from the isentropic relation

Chapter 7 - 68

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

T2 s

F PI =TG J HPK 1

7-69

( k −1)/ k

2

1

= (500) K

FG 0.5P IJ H PK

(1.4 −1)/1.4

1

1

= 410.0 K G V2 s = 2C p (T1 − T2 s ) =

F . kJ IJ (500 − 410.0) K 10 m / s 2G 1005 kJ / kg H kg ⋅ K K

= 442.8

3

2

2

m s

The nozzle exit velocity is obtained from the nozzle isentropic efficiency as

G2 V2 a / 2 ηN = G 2 V2 s / 2 G G m m V2 a = V2 s η N = 442.8 0.95 = 421.8 s s

Chapter 7 - 69

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

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Entropy Balance The principle of increase of entropy for any system is expressed as an entropy balance given by

Ein Sin

System ∆Esystem ∆Ssystem ∆Sgen≥0

Eout Sout

F Total I F Total I F Total I F Change in theI GG entropy JJ − GG entropyJJ + GG entropy JJ = GG total entropy JJ H enteringK H leaving K H generated K H of the system K or

Sin − Sout + S gen = ∆S system The entropy balance relation can be stated as: the entropy change of a system during a process is equal to the net entropy transfer through the system boundary and the entropy generated within the system as a result of irreversibilities. Entropy change of a system The entropy change of a system is the result of the process occurring within the system.

Chapter 7 - 70

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Entropy change = Entropy at final state – Entropy at initial state

∆Ssystem = S final − Sinitial = S2 − S1 Mechanisms of Entropy Transfer, Sin and Sout Entropy can be transferred to or from a system by two mechanisms: heat transfer and mass flow. Entropy transfer occurs at the system boundary as it crosses the boundary, and it represents the entropy gained or lost by a system during the process. The only form of entropy interaction associated with a closed system is heat transfer, and thus the entropy transfer for an adiabatic closed system is zero. Heat transfer The ratio of the heat transfer Q at a location to the absolute temperature T at that location is called the entropy flow or entropy transfer and is given as

Entropy transfer by heat transfer:

Sheat =

Q (T = constant ) T

Q/T represents the entropy transfer accompanied by heat transfer, and the direction of entropy transfer is the same as the direction of heat transfer since the absolute temperature T is always a positive quantity. When the temperature is not constant, the entropy transfer for process 1-2 can be determined by integration (or by summation if appropriate) as

Sheat =

z

2

1

δQ Qk ≅∑ T Tk

Chapter 7 - 71

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Work Work is entropy-free, and no entropy is transferred by work. Energy is transferred by both work and heat, whereas entropy is transferred only by heat and mass.

Entropy transfer by work :

S work = 0

Mass flow Mass contains entropy as well as energy, and the entropy and energy contents of a system are proportional to the mass. When a mass in the amount m enters or leaves a system, entropy in the amount of ms enters or leaves, where s is the specific entropy of the mass.

Entropy transfer by mass:

Smass = ms

Entropy Generation, Sgen Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, non-quasiequilibrium expansion, or compression always cause the entropy of a system to increase, and entropy generation is a measure of the entropy created by such effects during a process.

Chapter 7 - 72

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For a reversible process, the entropy generation is zero and the entropy change of a system is equal to the entropy transfer. The entropy transfer by heat is zero for an adiabatic system and the entropy transfer by mass is zero for a closed system. The entropy balance for any system undergoing any process can be expressed in the general form as

3

Sin − Sout + S gen = ∆Ssystem   



Net entropy transfer by heat and mass

Entropy generation

( kJ / K )

Change in entropy

The entropy balance for any system undergoing any process can be expressed in the general rate form, as

Sin − Sout   

Rate of net entropy transfer by heat and mass

+

Sgen

3

Rate of entropy generation

= ∆Ssystem 

( kW / K )

Rate of change of entropy

where the rates of entropy transfer by heat transferred at a rate of  . mass flowing at a rate of m are Sheat = Q / T and Smass = ms

Q

and

The entropy balance can also be expressed on a unit-mass basis as

( sin − sout ) + sgen = ∆ssystem

( kJ / kg ⋅ K )

The term Sgen is the entropy generation within the system boundary only, and not the entropy generation that may occur outside the system boundary during the process as a result of external irreversibilities. Sgen = 0 for the internally reversible process, but not necessarily zero for the totally reversible process. The total entropy generated during any process is Chapter 7 - 73

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7-74

obtained by applying the entropy balance to an Isolated System that contains the system itself and its immediate surroundings.

∆

Closed Systems Taking the positive direction of heat transfer to the system to be positive, the general entropy balance for the closed system is

Qk ∑ T + S gen = ∆Ssystem = S2 − S1 k

( kJ / K )

For an adiabatic process (Q = 0), this reduces to

Adiabatic closed system: S gen = ∆Sadiabatic system Chapter 7 - 74

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-75

A general closed system and its surroundings (an isolated system) can be treated as an adiabatic system, and the entropy balance becomes

System + surroundings: S gen = ∑ ∆S = ∆S system + ∆S surroundings

Control Volumes The entropy balance for control volumes differs from that for closed systems in that the entropy exchange due to mass flow must be included.

Qk

∑ T + ∑m s − ∑m s i i

e e

+ S gen = ( S2 − S1 ) CV ( kJ / K )

k

In the rate form we have

Q k ∑ T + ∑ m i si − ∑ m e se + Sgen = ∆SCV ( kW / K ) k This entropy balance relation is stated as: the rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities.

Chapter 7 - 75

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For a general steady-flow process, by setting ∆SCV = 0 the entropy balance simplifies to

S gen

Q k = ∑ m e se − ∑ m i si − ∑ Tk

For a single-stream (one inlet and one exit), steady-flow device, the entropy balance becomes

Q k  S gen = m ( se − si ) − ∑ Tk

For an adiabatic single-stream device, the entropy balance becomes

S gen = m ( se − si )

This states that the specific entropy of the fluid must increase as it flows

through an adiabatic device since Sgen ≥ 0 . If the flow through the device is reversible and adiabatic, then the entropy will remain constant regardless of the changes in other properties.

Chapter 7 - 76

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Therefore, for steady-flow, single-stream, adiabatic and reversible process:

se = si Example 7-17 An inventor claims to have developed a water mixing device in which 10 kg/s of water at 25oC and 0.1 MPa and 0.5 kg/s of water at 100oC, 0.1 MPa, are mixed to produce 10.5 kg/s of water as a saturated liquid at 0.1 MPa. If the surroundings to this device are at 20oC, is this process possible? If not, what temperature must the surroundings have for the process to be possible? System: The mixing chamber control volume. 10 kg/s P1 = 0.1 MPa T1 = 25oC 0.5 kg/s P1 = 0.1 MPa T1 = 100oC

Mixing chamber

10.5 kg/s P3 = 0.1 MPa Saturated liquid

Qsys ? Surroundings To = 20oC

Property Relation: The steam tables Process and Process Diagram: Assume steady-flow

Chapter 7 - 77

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7-78

Conservation Principles: First let’s determine if there is a heat transfer from the surroundings to the mixing chamber. Assume there is no work done during the mixing process, and neglect kinetic and potential energy changes. Then for two entrances and one exit, the first law becomes

FG H

IJ K

FG H

G2 G2 V V Q net + ∑ m i hi + i + gzi = Wnet + ∑ m e he + e + gze 2 2 inlets exits Q net + m 1h1 + m 2 h2 = m 3h3 kJ  ≅ = h h 104.83 f @ T1 P1 = 0.1 MPa   1 kg  kJ T1 = 25o C   s1 ≅ s f @ T1 = 0.3672  kg ⋅ K kJ  = h 2675.8 P2 = 0.1 MPa   2 kg  kJ T2 = 100o C   s2 = 7.3611  kg ⋅ K kJ  = h 417.51 3 P3 = 0.1 MPa   kg  kJ Sat. liquid   s3 = 1.3028  kg ⋅ K

Chapter 7 - 78

IJ K

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles

7-79

Q net = m 3 h3 − m 1h1 − m 2 h2 kg kJ kg kJ kg kJ 417.51 − 10 104.83 − 0.5 2675.8 s kg s kg s kg kJ = +1997.7 s = 10.5

So, 1996.33 kJ/s of heat energy must be transferred from the surroundings to this mixing process, or Q net , surr = −Q net , CV . For the process to be possible, the second law must be satisfied. Write the second law for the isolated system,

Q k ∑ T + ∑ m i si − ∑ m e se + Sgen = ∆SCV k For steady-flow

∆SCV = 0 .

Solving for entropy generation, we have

Q k  S gen = ∑ m e se − ∑ m i si − ∑ Tk Q cv = m 3 s3 − m 1s1 − m 2 s2 − Tsurr kg kJ kg kJ − 10 0.3672 1.3028 s kg ⋅ K s kg ⋅ K kg kJ 1997.7 kJ / s − 0.5 7.3611 − s kg ⋅ K (20 + 273) K kJ = −0.491 K ⋅s

= 10.5

Chapter 7 - 79

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7-80

Since Sgen must be ≥ 0 to satisfy the second law, this process is impossible, and the inventor's claim is false. To find the minimum value of the surrounding temperature to make this mixing process possible, set Sgen = 0 and solve for Tsurr.

Q k    =0 S gen = ∑ me se − ∑ mi si − ∑ Tk Q cv Tsurr = m 3 s3 − m 1s1 − m 2 s2 1997.7 kJ / s kg kJ kg kJ kg kJ − 10 0.3672 − 0.5 7.3611 10.5 1.3028 s kg ⋅ K s kg ⋅ K s kg ⋅ K = 315.75K =

One way to think about this process is as follows: Heat is transferred from the surroundings at 315.75 K (42.75oC) in the amount of 1997.7 kJ/s to increase the water temperature to approximately 42.75oC before the water is mixed with the superheated steam. Recall that the surroundings must be at a temperature greater than the water for the heat transfer to take place from the surroundings to the water.

Chapter 7 - 80

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Answer to Example 7-4 Find the entropy and/or temperature of steam at the following states: P 5 MPa

T 120oC

1 MPa

50oC

1.8 400oC MPa 40 kPa T=Tsat =75.87o C 40 kPa T=Tsat =75.87o C

Region Compressed Liquid and in the table Compressed liquid but not in the table Superheated

s kJ/kg⋅K 1.5233

Quality, x = 0.9 Saturated mixture

s = sf + x sfg = 7.0056

sf 85 percent).

•

Increase boiler pressure (for fixed maximum temperature) Availability of steam is higher at higher pressures. Moisture is increased at turbine exit.

•

Lower condenser pressure Less energy is lost to surroundings. Moisture is increased at turbine exit.

Extra Assignment For the above example, find the heat rejected by the cycle and evaluate the thermal efficiency from

η th =

wnet q = 1 − out qin qin 14

Reheat Cycle As the boiler pressure is increased in the simple Rankine cycle, not only does the thermal efficiency increase, but also the turbine exit moisture increases. The reheat cycle allows the use of higher boiler pressures and provides a means to keep the turbine exit moisture (x > 0.85 to 0.90) at an acceptable level.

Let’s sketch the T-s diagram for the reheat cycle.

T

15 s

Rankine Cycle with Reheat Component Process First Law Result Boiler Const. P qin = (h3 - h2) + (h5 - h4) Turbine Isentropic wout = (h3 - h4) + (h5 - h6) Condenser Const. P qout = (h6 - h1) Pump Isentropic win = (h2 - h1) = v1(P2 - P1) The thermal efficiency is given by

η th = =

wnet qin (h3 - h4 ) + (h5 - h6 ) - (h2 - h1 ) (h3 - h2 ) + ( h5 - h4 )

= 1−

h6 − h1 (h3 - h2 ) + (h5 - h4 )

16

Example 10-2 Compare the thermal efficiency and turbine-exit quality at the condenser pressure for a simple Rankine cycle and the reheat cycle when the boiler pressure is 4 MPa, the boiler exit temperature is 400oC, and the condenser pressure is 10 kPa. The reheat takes place at 0.4 MPa and the steam leaves the reheater at 400oC.

No Reheat With Reheat

ηth 35.3% 35.9%

xturb exit 0.8159 0.9664

17

Regenerative Cycle To improve the cycle thermal efficiency, the average temperature at which heat is added must be increased. One way to do this is to allow the steam leaving the boiler to expand the steam in the turbine to an intermediate pressure. A portion of the steam is extracted from the turbine and sent to a regenerative heater to preheat the condensate before entering the boiler. This approach increases the average temperature at which heat is added in the boiler. However, this reduces the mass of steam expanding in the lowerpressure stages of the turbine, and, thus, the total work done by the turbine. The work that is done is done more efficiently. The preheating of the condensate is done in a combination of open and closed heaters. In the open feedwater heater, the extracted steam and the condensate are physically mixed. In the closed feedwater heater, the extracted steam and the condensate are not mixed.

18

Cycle with an open feedwater heater

19

Rankine Steam Power Cycle with an Open Feedwater Heater 600

3000 kPa 500

5 500 kPa

T [C]

400 300

2

100

10 kPa

6

4

200

3

7

1

0 0

2

4

6

8

10

12

s [kJ/kg-K]

Cycle with a closed feedwater heater with steam trap to condenser

20

Let’s sketch the T-s diagram for this closed feedwater heater cycle. T

s

21

Cycle with a closed feedwater heater with pump to boiler pressure

22

Let’s sketch the T-s diagram for this closed feedwater heater cycle. T

s

Consider the regenerative cycle with the open feedwater heater. To find the fraction of mass to be extracted from the turbine, apply the first law to the feedwater heater and assume, in the ideal case, that the water leaves the feedwater heater as a saturated liquid. (In the case of the ideal closed feedwater heater, the feedwater leaves the heater at a temperature equal to the saturation temperature at the extraction pressure.) Conservation of mass for the open feedwater heater: 23

& 6 / m& 5 be the fraction of mass extracted from the turbine for the feedwater Let y = m heater. m& in = m& out m& 6 + m& 2 = m& 3 = m& 5 m& 2 = m& 5 − m& 6 = m& 5 (1 − y ) Conservation of energy for the open feedwater heater:

E& in = E& out m& 6h6 + m& 2 h2 = m& 3h3 ym& 5h6 + (1 − y )m& 5h2 = m& 5h3 y=

h3 − h2 h6 − h2

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Example 10-3 An ideal regenerative steam power cycle operates so that steam enters the turbine at 3 MPa, 500oC, and exhausts at 10 kPa. A single open feedwater heater is used and operates at 0.5 MPa. Compute the cycle thermal efficiency. The important properties of water for this cycle are shown below. States with selected properties State

P kPa

T °C

h kJ/kg

Selected saturation properties s kJ/kg-K

P kPa

Tsat °C

vf 3/kg m

hf kJ/kg

1

10

10

45.81

0.00101

191.8

2

500

500

151.83

0.00109

640.1

3

500

3000

4

3000

5

3000

500 3457.2

7.2359

6

500

2942.6

7.2359

7

10

2292.7

7.2359

233.85 0.00122

1008.3

25

The work for pump 1 is calculated from

w pump 1 = v1 ( P2 − P1 ) kJ m3 = 0.00101 (500 − 10) kPa 3 m kPa kg kJ = 0.5 kg Now, h2 is found from

h2 = w pump 1 + h1 kJ kJ + 1918 . kg kg kJ = 192.3 kg = 0.5

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The fraction of mass extracted from the turbine for the open feedwater heater is obtained from the energy balance on the open feedwater heater, as shown above.

kJ h −h kg y= 3 2 = = 0.163 h6 − h2 (2942.6 − 192.3) kJ kg (640.1 − 192.3)

This means that for each kg of steam entering the turbine, 0.163 kg is extracted for the feedwater heater. The work for pump 2 is calculated from

w pump 2 = v3 ( P4 − P3 ) = 0.00109 = 2.7

m3 kJ (3000 − 500) kPa 3 kg m kPa

kJ kg

27

Now, h4 is found from the energy balance for pump 2 for a unit of mass flowing through the pump.

Eout = Ein h4 = wpump 2 + h3 kJ kJ + 640.1 kg kg kJ = 642.8 kg = 2.7

Apply the steady-flow conservation of energy to the isentropic turbine.

E& in = E& out m& 5 h5 = W&turb + m& 6 h6 + m& 7 h7 W&turb = m& 5 [h5 − yh6 − (1 − y )h7 ] W& wturb = turb = h5 − yh6 − (1 − y )h7 m& 5 = [3457.2 − (0.163)(2942.1) − (1 − 0.163)(2292.7)] = 1058.6

kJ kg

kJ kg 28

The net work done by the cycle is

W&net = W&turb − W& pump 1 − W& pump 2 m& 5 wnet = m& 5 wturb − m& 1wpump 1 − m& 3 wpump 2 m& 5 wnet = m& 5 wturb − m& 5 (1 − y ) wpump 1 − m& 5 wpump 2 wnet = wturb − (1 − y ) wpump 1 − wpump 2 = [1058.6 − (1 − 0.163)(0.5) − 2.7] = 1055.5

kJ kg

kJ kg

Apply the steady-flow conservation of mass and energy to the boiler.

m& 4 = m& 5 m& 4 h4 + Q& in = m& 5h5 Q& in = m& 5 (h5 − h4 ) Q& qin = in = h5 − h4 m& 5 29

The heat transfer per unit mass entering the turbine at the high pressure, state 5, is

qin = h5 − h4 = (3457.2 − 642.8)

kJ kJ = 2814.4 kg kg

The thermal efficiency is

kJ w kg ηth = net = qin 2814.4 kJ kg = 0.375 or 37.5% 1055.5

If these data were used for a Rankine cycle with no regeneration, then ηth = 35.6 percent. Thus, the one open feedwater heater operating at 0.5 MPa increased the thermal efficiency by 5.3 percent. However, note that the mass flowing through the lower-pressure turbine stages has been reduced by the amount extracted for the feedwater and the net work output for the regenerative cycle is about 10 percent lower than the standard Rankine cycle based on a unit of mass entering the turbine at the highest pressure.

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Below is a plot of cycle thermal efficiency versus the open feedwater heater pressure. The feedwater heater pressure that makes the cycle thermal efficiency a maximum is about 400 kPa. ηth vs OFWH Pressure

0.376 0.374 0.372

ηth

0.370 0.368 0.366 0.364 0.362 0.360 0

450

900

1350 Pofw h [kPa]

1800

2250

31

Below is a plot of cycle net work per unit mass flow at state 5 and the fraction of mass y extracted for the feedwater heater versus the open feedwater heater pressure. Clearly the net cycle work decreases and the fraction of mass extracted increases with increasing extraction pressure. Why does the fraction of mass extracted increase with increasing extraction pressure?

wnet and y vs OFWH Pressure 1200

0.25 0.23

1150 0.20 0.18 0.15 1050 0.13 1000

y

w net kJ/kg

1100

0.10 0.08

950 0.05 900 0

450

900

1350

1800

0.03 2250

Pofw h [kPa] 32

Placement of Feedwater Heaters The extraction pressures for multiple feedwater heaters are chosen to maximize the cycle efficiency. As a rule of thumb, the extraction pressures for the feedwater heaters are chosen such that the saturation temperature difference between each component is about the same.

∆Tcond to FWH = ∆Tboiler to FWH , etc.

Example 10-4 An ideal regenerative steam power cycle operates so that steam enters the turbine at 3 MPa, 500oC, and exhausts at 10 kPa. Two closed feedwater heaters are to be used. Select starting values for the feedwater heater extraction pressures. Steam

T [C]

400

300

3000 kPa

233.9 C

815 kPa

∆ Τ = 62.68 C

200

136.2 kPa

∆ Τ = 62.68 C 100

∆ Τ = 62.68 C 45.85 C 45.8 1 0

10 kPa 0

2

4

6

8

10

12

33

s [kJ/kg-K]

Deviation from Actual Cycles •Piping losses--frictional effects reduce the available energy content of the steam. •Turbine losses--turbine isentropic (or adiabatic) efficiency. P3

T 3

4a

P4

4s

s

η turb =

wactual h −h = 3 4a wisentropic h3 − h4 s

The actual enthalpy at the turbine exit (needed for the energy analysis of the next component) is

h4 a = h3 − η turb (h3 − h4 s )

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•Pump losses--pump isentropic (or adiabatic) efficiency. 2a

T

P2

2s

P1

1

s

η pump =

wisentropic wactual

=

h2 s − h1 h2 a − h1

The actual enthalpy at the pump exit (needed for the energy analysis of the next component) is

h2 a = h1 +

1

η pump

(h2 s − h1 )

•Condenser losses--relatively small losses that result from cooling the condensate below the saturation temperature in the condenser.

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The following examples you should try on your own. Regenerative Feedwater Heater problem Consider an ideal steam regenerative Rankine cycle with two feedwater heaters, one closed and one open. Steam enters the turbine at 10 MPa and 500 C and exhausts to the condenser at 10 kPa. Steam is extracted from the turbine at 0.7 MPa for the closed feedwater heater and 0.3 MPa for the open one. The extracted steam leaves the closed feedwater heater and is subsequently throttled to the open feedwater heater. Show the cycle on a T-s diagram with respect to saturation lines, and using only the data presented in the data tables given below determine a) the fraction of steam leaving the boiler that is extracted at 0.3 MPa z=0.1425 b) the fraction of steam leaving the boiler that is extracted at 0.7 MPa y=0.06213 c) the heat transfer from the condenser per unit mass leaving the boiler q_out=1509 kJ/kg d) the heat transfer to the boiler per unit mass leaving the boiler q_in=2677 kJ/kg e) the mass flow rate of steam through the boiler for a net power output of 250 MW m_dot=214.1 kg/s f) the thermal efficiency of the cycle. Eta_th=0.4363 36

37

Cogeneration Plant A cogeneration plant is to generate power and process heat. Consider an ideal cogeneration steam plant. Steam enters the turbine from the boiler at 7 MPa, 500 C and a mass flow rate of 30 kg/s. One-fourth of the steam is extracted from the turbine at 600-kPa pressure for process heating. The remainder of the steam continues to expand and exhausts to the condenser at 10 kPa. The steam extracted for the process heater is condensed in the heater and mixed with the feedwater at 600 kPa. The mixture is pumped to the boiler pressure of 7 MPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine a) the heat transfer from the process heater per unit mass leaving the boiler Qdot,process = 15,774 kW. b) the net power produced by the cycle. Wdot,net = 32,848 kW. c) the utilization factor of the plant Qdot,in = 92,753 kW, Utilization factor = 52.4%.

38

39

Combined Gas-Steam Power Cycle Example of the Combined Brayton and Rankine Cycles (a) Explain what’s happening in the various process for the hardware shown below.

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