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

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 1-1

Chapter 1-1

Study Guide forThermodynamics: an Engineering Approach

By

Michael A. BolesDepartment of Mechanical and Aerospace Engineering

NC State UniversityRaleigh, NC 2795-7910


Chapter 1-2

NomenclatureA area (m2)CP specific heat at constant pressure (kJ/(kg⋅K))CV specific heat at constant volume (kJ/(kg⋅K))COP coefficient of performanced exact differentialE stored energy (kJ)e stored energy per unit mass (kJ/kg)F force (N)g acceleration of gravity ( 9.807 m/s2)H enthalpy (H= U + PV) (kJ)h specific enthalpy (h= u + Pv) (kJ/kg)h convective heat transfer coefficient (W/(m2⋅K)K Kelvin degreesk specific heat ratio, CP/CVk 103

kt thermal conductivity (W/(m-°C))M molecular weight or molar mass (kg/kmol)M 106

m mass (kg) N moles (kmol)n polytropic exponent (isentropic process, ideal gas n = k)η isentropic efficiency for turbines, compressors, nozzlesηth thermal efficiency (net work done/heat added)P pressure (kPa, MPa, psia, psig)Pa Pascal (N/m2)Qnet net heat transfer (∑Qin - ∑Qout) (kJ)qnet Qnet /m, net heat transfer per unit mass (kJ/kg)R particular gas constant (kJ/(kg⋅K))Ru universal gas constant (= 8.314 kJ/(kmol⋅K) )S entropy (kJ/K)s specific entropy (kJ/(kg⋅K))T temperature ( °C, K, °F, R)U internal energy (kJ)


Chapter 1-3

u specific internal energy (kJ/(kg ⋅K))V volume (m3 )V volume flow rate (m3/s) V velocity (m/s)v specific volume (m3/kg)v molar specific volume (m3/kmol) X distance (m)X exergy (kJ)x qualityZ elevation (m)Wnet net work done [(∑Wout - ∑Win)other + Wb] (kJ)

where Wb = PdV1

2z for closed systems and 0 for control volumeswnet Wnet /m, net work done per unit mass (kJ/kg)Wt weight (N)δ inexact differentialε regenerator effectivenessφ relative humidityρ density (kg/m3)ω humidity ratio

Subscripts, superscripts

a actualb boundaryf saturated liquid stateg saturated vapor statefg saturated vapor value minus saturated liquid valuegen generationH high temperatureHP heat pumpL low temperaturenet net heat added to system or net work done by systemother work done by shaft and electrical meansP constant pressure


Chapter 1-4

REF refrigeratorrev reversibles isentropic or constant entropy or reversible, adiabaticsat saturation value v constant volume1 initial state2 finial statei inlet statee exit state⋅ per unit time

REFERENCE

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


Chapter 1-5

Chapter 1: Introduction and Basic Concepts

INTRODUCTION

The study of thermodynamics is concerned with the ways energy is storedwithin a body and how energy transformations, which involve heat andwork, may take place. One of the most fundamental laws of nature is theconservation of energy principle. It simply states that during an energyinteraction, energy can change from one form to another but the totalamount of energy remains constant. That is, energy cannot be created ordestroyed.

This review of thermodynamics is based on the macroscopic approachwhere a large number of particles, called molecules, make up thesubstance in question. The macroscopic approach to thermodynamics doesnot require knowledge of the behavior of individual particles and is calledclassical thermodynamics. It provides a direct and easy way to obtain thesolution of engineering problems without being overly cumbersome. Amore elaborate approach, based on the average behavior of large groups ofindividual particles, is called statistical thermodynamics. This microscopicapproach is rather involved and is not reviewed here and leads to thedefinition of the second law of thermodynamics. We will approach thesecond law of thermodynamics from the classical point of view and willlearn that the second law of thermodynamics asserts that energy has qualityas well as quantity, and actual processes occur in the direction ofdecreasing quality of energy.

Closed, Open, and Isolated Systems

A thermodynamic system, or simply system, is defined as a quantity ofmatter or a region in space chosen for study. The region outside thesystem is called the surroundings. The real or imaginary surface that


Chapter 1-6

separates the system from its surroundings is called the boundary. Theboundary 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 afixed mass or a fixed volume in space is chosen for study.

A closed system consists of a fixed amount of mass and no mass maycross 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 formof heat and work may cross the boundaries of a closed system.


Chapter 1-7

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

An isolated system is a general system of fixed mass where no heat orwork may cross the boundaries. An isolated system is a closed systemwith no energy crossing the boundaries and is normally a collection of amain system and its surroundings that are exchanging mass and energyamong themselves and no other system.

Isolated System Boundary

Mass

System

Surr 2

Surr 3

Surr 4

Mass

Work

Surr 1

Heat = 0Work = 0Mass = 0AcrossIsolatedBoundary Heat


Chapter 1-8

Since some of the thermodynamic relations that are applicable to closedand open systems are different, it is extremely important that we recognizethe 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. Theproperty 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 thesystem.

Some Extensive Propertiesa. massb. volumec. total energyd. mass dependent property

Intensive properties are those that are independent of size.

Some Intensive Properties a. temperature b. pressure c. age d. colore. any mass independent property


Chapter 1-9

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

v Volumemass

Vm

mkg

= =FHGIKJ

3

and density ρ, defined as

ρ = = FHGIKJ

massvolume

mV

kgm3

are intensive properties.

Units

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


Chapter 1-10

solution. Since units present a major hindrance to the correct solution ofthermodynamic problems, we must learn to use units carefully andproperly. The system of units selected for this course is the SI System,also known as the International System (sometimes called the metricsystem). 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 derivedunit from Newton's second law, i.e.,

Force mass accelerationF ma==

( )( )

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

1 1 1 2N kg ms

= ( )( )

This definition of the newton is used as the basis of the conversion factorto convert mass-acceleration units to force units.

The term weight is often misused to express mass. Unlike mass, weightWt is a force. Weight is the gravitational force applied to a body, and itsmagnitude is determined from Newton's second law.

= W mgt

where m is the mass of the body and g is the local gravitationalacceleration (g is 9.807 m/s2 at sea level and 45 °latitude). The weight of aunit volume of a substance is called the specific weight w and isdetermined from w = ρ g, where ρ is density.


Chapter 1-11

Oftentimes, the engineer must work in other systems of units. Comparisonof the United States Customary System (USCS), or English System, andthe slug system of units with the SI system is shown below.

SI USCS Slug mass kilogram (kg) pound-mass (lbm) slug-mass (slug) time second (s) second (s) second (s) length meter (m) foot (ft) foot (ft)force newton (N) pound-force (lbf) pound-force (lbf)

Sometimes we use the mole number in place of the mass. In SI units themole number is in kilogram-moles, or kmol.

Newton’s second law is often written as

F magc

=

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

g maF

kg ms

Nkg mN sc = = =

( )( )1 1

11

2

2

In the USCS 1 pound-force is that force required to accelerate 1 pound-mass 32.176 ft/s2. The gravitational constant in the USCS is


Chapter 1-12

g maF

lbm fts

lbflbm ftlbf sc = = =

( )( . ).

1 32 2

132 2

2

2

In the slug system, the gravitational constant is

g maF

slug fts

lbflbm ftlbf sc = = =

( )( )1 1

11

2

2

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)

W mgt =

( ) .W kg ms

N

kg ms

N

t =FHG

IKJF

HGGG

I

KJJJ

=

400 9 807 1

3922 8

2

2

.


Chapter 1-13

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

(b)

( ) .

W mg

kg ms

N

kg ms

N

t =

= FHG

IKJF

HGGG

I

KJJJ

=

400 9 8076

1

6538

2

2

.

Example 1-2E

An object has a mass of 180 lbm. Find the weight of this object at alocation where the local gravitational acceleration is 30 ft/s2.

( )(30

W mg

lbm fts

lbf

lbm fts

lbf

t =

=

=

180 1

32 2

167 7

2

2

)(.

)

.


Chapter 1-14

State, Equilibrium, Process, and Properties

State

Consider a system that is not undergoing any change. The properties canbe measured or calculated throughout the entire system. This gives us aset of properties that completely describe the condition or state of thesystem. At a given state all of the properties are known; changing oneproperty changes the state.

Equilibrium

A system is said to be in thermodynamic equilibrium if it maintainsthermal (uniform temperature), mechanical (uniform pressure), phase (themass of two phases, e.g., ice and liquid water, in equilibrium) andchemical equilibrium.


Chapter 1-15

Process

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

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

Process Property held constantisobaric pressureisothermal temperatureisochoric volumeisentropic entropy (see Chapter 7)

We can understand the concept of a constant pressure process byconsidering the above figure. The force exerted by the water on the faceof the piston has to equal the force due to the combined weight of thepiston and the bricks. If the combined weight of the piston and bricks isconstant, then F is constant and the pressure is constant even when thewater is heated.

Constant Pressure Process

WaterF

SystemBoundary


Chapter 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 suchas a water heater. The flow is often defined by the terms steady anduniform. 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 thesame conditions are classified as steady-flow devices. The processes forthese devices is called the steady-flow process. The fluid properties canchange from point to point with in the control volume, but at any fixedpoint the properties remain the same during the entire process.

State Postulate

As noted earlier, the state of a system is described by its properties. Butby experience not all properties must be known before the state is


Chapter 1-17

specified. Once a sufficient number of properties are known, the state isspecified and all other properties are known. The number of propertiesrequired to fix the state of a simple, homogeneous system is given by thestate postulate:

The thermodynamic state of a simple compressiblesystem is completely specified by two independent,intensive properties.

Cycle

A process (or a series of connected processes) with identical end states iscalled 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 2back to the initial state 1 along process B. Keep in mind that all otherthermodynamic properties must also change so that the pressure is afunction of volume as described by these two processes.

Pressure

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

ProcessB

ProcessA

1

2P

V


Chapter 1-18

P ForceArea

FA

= =

1 10

1 10 10

32

62

3

kPa Nm

MPa Nm

kPa

=

= =

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

These pressures are related by

P P Pgage abs atm= −

P P Pvac atm abs= −


Chapter 1-19

Or these last two results may be written as

P P Pabs atm gage= ±

Where the +Pgage is used when Pabs > Patm and –Pgage is used for a vacuumgage.

The relation among atmospheric, gage, and vacuum pressures is shownbelow.

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

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


Chapter 1-20

This pressure difference is determined from the manometer fluid displacedheight as

∆P g h kPa= ρ ( )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 1-21

Example 1-3

A vacuum gage connected to a tank reads 30 kPa at a location where theatmospheric pressure is 98 kPa. What is the absolute pressure in the tank?

P P PkPa kPakPa

abs atm gage= −

= −=

98 3068


Chapter 1-22

Example 1-4

A pressure gage connected to a valve stem of a truck tire reads 240 kPa ata location where the atmospheric pressure is 100 kPa. What is the absolutepressure in the tire, in kPa and in psia?

P P PkPa kPakPa

abs atm gage= +

= +=

100 240340

The pressure in psia is

P kPa psiakPa

psiaabs = =340 14 71013

49 3..

.

What is the gage pressure of the air in the tire, in psig?

P P Ppsia psiapsig

gage abs atm= −

= −=

49 3 14 734 6

. .

.

Check the side walls of the tires on your car or truck. What is themaximum allowed pressure? Is this pressure in gage or absolute values?


Chapter 1-23

Example 1-5

Both a gage and a manometer are attached to a gas tank to measure itspressure. If the pressure gage reads 80 kPa, determine the distancebetween the two fluid levels of the manometer if the fluid is mercury,whose density is 13,600 kg/m3.

h Pg

=∆ρ

h kPakgm

ms

N mkPa

Nkg m s

m

=

=

80

13600 9 807

10

1

0 6

3 2

3 2

2.

/

/.

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 measureof the energy content of a mass. When heat energy is transferred to abody, the body's energy content increases and so does its temperature. Infact it is the difference in temperature that causes energy, called heattransfer, to flow from a hot body to a cold body. Two bodies are inthermal equilibrium when they have reached the same temperature. If twobodies are in thermal equilibrium with a third body, they are also inthermal equilibrium with each other. This simple fact is known as thezeroth law of thermodynamics.


Chapter 1-24

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

T F T C° ° + = 95

32

Example 1-6

Water boils at 212 °F at one atmosphere pressure. At what temperaturedoes water boil in °C.

T = ( ) ( ) T F F CF

C° − = − °°°

= °32 59

212 32 59

100

Like pressure, the temperature used in thermodynamic calculations mustbe in absolute units. The absolute scale in the SI system is the Kelvinscale, 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

= F+ 459.67 T R T °

Also, note that

T R T K = 1.8


Chapter 1-25

Below is a comparison of the temperature scales.

This figure shows that that according to the International TemperatureScale of 1990 (ITS-90) the reference state for the thermodynamictemperature scale is the triple point of water, 0.01 °C. The ice point is0°C, but the steam point is 99.975°C at 1 atm and not 100°C as waspreviously established. The magnitude of the kelvin, K, is 1/273.16 of thethermodynamic temperature of the triple point of water.

The magnitudes of each division of 1 K and 1°C are identical, and so arethe magnitudes of each division of 1 R and 1°F. That is,

-273.15 0

32.02 491.69

211.955 671.625

°F R

-273.15 0

0.01 273.16

99.975 373.125

°C K

Absolutezero

Triplepoint ofwater

Boilingpointof waterat 1 atm


Chapter 1-26

∆

∆

∆ ∆

T K T T

T T TT R T

= ( C + 273.15) - ( C + 273.15)

= C - C = CF

2 1

2 1

° °

° ° °

°=


Chapter 2-1

Chapter 2: Energy, Energy Transfer, and General Energy Analysis

We will soon learn how to apply the first law of thermodynamics as theexpression of the conservation of energy principle. But, first we study theways in which energy may be transported across the boundary of a generalthermodynamic system. For closed systems (fixed mass systems) energycan cross the boundaries of a closed system only in the form of heat orwork. For open systems or control volumes energy can cross the controlsurface in the form of heat, work, and energy transported by the massstreams crossing the control surface. We now consider each of thesemodes of energy transport across the boundaries of the generalthermodynamic system.

Energy

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

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

Z

GeneralSystem

CM

Reference Plane, Z=0

V


Chapter 2-2

and potential energy. The internal energy U is that energy associated withthe molecular structure of a system and the degree of the molecularactivity (see Section 2-1 of text for more detail). The kinetic energy KEexists as a result of the system's motion relative to an external referenceframe. When the system moves with velocity V the kinetic energy isexpressed as

KE mV kJ=2

2( )

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

PE mgZ kJ= ( )

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

E U KE PE kJ= + + ( )

or, on a unit mass basis,

e Em

Um

KEm

PEm

kJkg

u V gZ

= = + +

= + +

( )

2

2

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


Chapter 2-3

∆ ∆ ∆ ∆E U KE PE kJ= + + ( )

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

If ∆KE = ∆PE = 0,∆ ∆E U kJ= ( )

Energy Transport by Heat and Work and the Classical SignConvention

Energy may cross the boundary of a closed system only by heat or work.

Energy transfer across a system boundary due solely to the temperaturedifference between a system and its surroundings is called heat.

Energy transferred across a system boundary that can be thought of as theenergy expended to lift a weight is called work.

Heat and work are energy transport mechanisms between a system and itssurroundings. The similarities between heat and work are as follows:

1. Both are recognized at the boundaries of a system as they crossthe boundaries. They are both boundary phenomena.

2. Systems possess energy, but not heat or work.


Chapter 2-4

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 pathfollowed during a process as well as the end states.

Since heat and work are path dependent functions, they have inexactdifferentials designated by the symbol δ. The differentials of heat andwork are expressed as δQ and δW. The integral of the differentials of heatand work over the process path gives the amount of heat or work transferthat occurred at the system boundary during a process.

2

121,

2

121,

(not Q)

(not )

along path

along path

Q Q

W W W

δ

δ

= ∆

= ∆

∫

∫

That is, the total heat transfer or work is obtained by following the processpath and adding the differential amounts of heat (δQ) or work (δW) alongthe way. The integrals of δQ and δW are not Q2 – Q1 and W2 – W1,respectively, which are meaningless since both heat and work are notproperties and systems do not possess heat or work at a state.

The following figure illustrates that properties (P, T, v, u, etc.) are pointfunctions, that is, they depend only on the states. However, heat and workare path functions, that is, their magnitudes depend on the path followed.


Chapter 2-5

0.01 m3 0.03 m3

A sign convention is required for heat and work energy transfers, and theclassical thermodynamic sign convention is selected for these notes.According to the classical sign convention, heat transfer to a system andwork done by a system are positive; heat transfer from a system and worka system are negative. The system shown below has heat supplied to itand work done by it.

In this study guide we will use the concept of net heat and net work.

700 kPa

100 kPa

SystemBoundary


Chapter 2-6

Energy Transport by Heat

Recall that heat is energy in transition across the system boundary solelydue to the temperature difference between the system and its surroundings.The net heat transferred to a system is defined as

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

q Qm

=

Heat transfer has the units of energy measured in joules (we will usekilojoules, kJ) or the units of energy per unit mass, kJ/kg.

Since heat transfer is energy in transition across the system boundary dueto a temperature difference, there are three modes of heat transfer at theboundary that depend on the temperature difference between the boundarysurface and the surroundings. These are conduction, convection, andradiation. However, when solving problems in thermodynamics involvingheat transfer to a system, the heat transfer is usually given or is calculatedby applying the first law, or the conservation of energy, to the system.

An adiabatic process is one in which the system is perfectly insulated andthe heat transfer is zero.


Chapter 2-7

Introduction to the Basic Heat Transfer Mechanisms

For those of us who do not have the opportunity to have a complete coursein heat transfer theory and applications, the following is a shortintroduction to the basic mechanisms of heat transfer. Those of us whohave a complete course in heat transfer theory may elect to omit thismaterial at this time.

Heat transfer is energy in transition due to a temperature difference. Thethree modes of heat transfer are conduction, convection, and radiation.

Conduction through Plane Walls

Conduction heat transfer is a progressive exchange of energy between themolecules of a substance.

Fourier's law of heat conduction is


Chapter 2-8

Q A k dTdxcond t= −

hereQcond = heat flow per unit time (W)kt = thermal conductivity (W/m⋅K)A = area normal to heat flow (m2)dTdx

= temperature gradient in the direction of heat flow (°C/m)

Integrating Fourier's law

cond tTQ k Ax

∆=

∆

Since T2>T1, the heat flows from right to left in the above figure.


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 theleft face temperature of the brick is 20°C. Determine the rate of heatconduction through the wall per unit area of wall.

Q

QA

cond

cond

Convection Heat T

Convection heat trasurface and the adjacombined effects of

Tright = 900°C

Tleft = 20°C

20 cm

Chapter 2-9

. ( ).

k A Tx

k Tx

Wm K

Km

Wm

t

t

=

= =⋅

−FHG

IKJ

=

∆∆

∆∆

0 72 900 200 2

3168 2

ransfer

nsfer is the mode of energy transfer between a solidcent liquid or gas that is in motion, and it involves the conduction and fluid motion.


Chapter 2-10

The rate of heat transfer by convection Qconv is determined from Newton'slaw of cooling, expressed as

( )Q h A T Tconv s f= −here

Qconv = heat transfer rate (W) A = heat transfer area (m2) h = convective heat transfer coefficient (W/m2⋅K)

Ts = surface temperature (K)


Chapter 2-11

Tf = bulk fluid temperature away from the surface (K)

The convective heat transfer coefficient is an experimentally determinedparameter that depends upon the surface geometry, the nature of the fluidmotion, the properties of the fluid, and the bulk fluid velocity. Ranges ofthe convective heat transfer coefficient are given below.

h W/m2⋅K

free convection of gases 2-25free convection of liquids 50-100forced convection of gases 25-250forced convection of liquids 50-20,000convection in boiling and condensation 2500-100,000

Radiative Heat Transfer

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


Chapter 2-12

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

Q A T Trad s surr= −ε σ 4 4c hhere

Qrad = heat transfer per unit time (W)A = surface area for heat transfer (m2) σ = Stefan-Boltzmann constant, 5.67x10-8 W/m2K4

and 0.1713x10-8 BTU/h ft2 R4 ε = emissivity

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 largesurrounding objects. It is desired to know if dew or frost may form on thevehicle top. Assume the following:

• Convection coefficient h from ambient air to vehicle top is 6.0W/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 2-13

Ttop

Tsky = -Tair = 5 C

Qconv Qrad

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

Q Qconv rad=

The energy convected from the ambient air to the vehicle top is

( )Q A h T Tconv top air top= −

The energy radiated from the top to the night sky is

Q A T Trad top top sky= −ε σ 4 4d iSetting these two heat transfers equal gives

A h T T A T T

h T T T T

top air top top top sky

air top top sky

( )

( )

− = −

− = −

ε σ

ε σ

4 4

4 4

d id i


Chapter 2-14

6 0 273

0 84 5 67 10 18 273

2

82 4

4 4 4

. (5 )

( . ) . ( )

Wm K

T K

x Wm K

T K

top

top

+ −

=FHG

IKJ − − +−

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

5 084 676 0

273100

2 554

4− =+F

HGIKJ −

LNMM

OQPPT

Ttop

topd i ( . )(5. ).

( . )

Using the EES software package

Ttop = − °338. C

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

Extra Problem

Explore what happens to Ttop as you vary the convective heat transfercoefficient. On a night when the atmosphere is particularly still and coldand has a clear sky, why do fruit growers use fans to increase the airvelocity in their fruit groves?


Chapter 2-15


Chapter 2-16

Energy Transfer by Work

Electrical Work

The rate of electrical work done by electrons crossing a system boundaryis called electrical power and is given by the product of the voltage drop involts and the current in amps.

(W)eW V I=

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

2

1(kJ)eW V I dt= ∫

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 systemboundary and is done by a system if the sole effect external to theboundaries could have been the raising of a weight.

Mathematically, the differential of work is expressed as


Chapter 2-17

δW F d s Fds= ⋅ = cosΘhere 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 path-dependent function and has an inexact differential. If the angle betweenthe force and the displacement is zero, the work done between two states is

W W Fds12 1

2

1

2= =z zδ

Work has the units of energy and is defined as force times displacement ornewton times meter or joule (we will use kilojoules). Work per unit massof a system is measured in kJ/kg.

Common Types of Mechanical Work Energy (See text for discussionof 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 andboundary work. First, work may cross a system boundary in the form of arotating 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 2-18

with a moving boundary. In thermodynamics electrical energy is normallyconsidered to be work energy rather than heat energy; however, theplacement of the system boundary dictates whether to include electricalenergy as work or heat. Second, the system may do work on itssurroundings because of moving boundaries due to expansion orcompression processes that a fluid may experience in a piston-cylinderdevice.

The net work done by a closed system is defined by

W W W Wnet out in other b= − +∑ ∑d iHere, Wout and Win are the magnitudes of the other work forms crossing theboundary. Wb is the work due to the moving boundary as would occurwhen a gas contained in a piston cylinder device expands and does work tomove the piston. The boundary work will be positive or negativedepending upon the process. Boundary work is discussed in detail inChapter 4.

W W Wnet netother

b= +e jSeveral types of “other” work (shaft work, electrical work, etc.) arediscussed in the text.

Example 2-3

A fluid contained in a piston-cylinder device receives 500 kJ of electricalwork as the gas expands against the piston and does 600 kJ of boundarywork on the piston. What is the net work done by the fluid?


Chapter 2-19

( )( )( )

,

0 500 600100

net net bother

net out in ele bother

net

net

W W W

W W W W

W kJ kJW kJ

= +

= − +

= − +

=

Wele =500 kJ Wb=600 kJ


Chapter 2-20

The First Law of Thermodynamics

The first law of thermodynamics is known as the conservation of energyprinciple. It states that energy can be neither created nor destroyed; it canonly change forms. Joule’s experiments lead to the conclusion: For alladiabatic processes between two specified states of a closed system, thenet work done is the same regardless of the nature of the closed system andthe details of the process. A major consequence of the first law is theexistence 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 ofenergy principle. Energy can cross the boundaries of a closed system inthe form of heat or work. Energy may cross a system boundary (controlsurface) of an open system by heat, work and mass transfer.

A system moving relative to a reference plane is shown below where z isthe elevation of the center of mass above the reference plane and V is thevelocity of the center of mass.

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

CMEnergyin

Energyoutz

System

Reference Plane, z = 0

V


Chapter 2-21

Total energyentering the system

Total energyleaving the system

The change in total energy of the system

FHG

IKJ −FHG

IKJ =FHG

IKJ

or

E E Ein out 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, isgiven as

E Internal energy Kinetic energy Potential energyE U KE PE

= + + = + +

Recall that U is the sum of the energy contained within the molecules ofthe system other than the kinetic and potential energies of the system as awhole and is called the internal energy. The internal energy U isdependent on the state of the system and the mass of the system.

For a system moving relative to a reference plane, the kinetic energy KEand the potential energy PE are given by

2

0

0

2V

V

z

z

mVKE mV dV

PE mg dz mgz

=

=

= =

= =

∫

∫


Chapter 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 ofthermodynamics for closed systems, is written as

in outE E U KE PE− = ∆ + ∆ + ∆

If the system does not move with a velocity and has no change inelevation, it is called a stationary system, and the conservation of energyequation reduces to

in outE E 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 ofa closed (fixed mass) system. Open systems or control volumes haveenergy transfer across the control surfaces by mass flow as well as heatand work.

1. Heat Transfer, Q: Heat is energy transfer caused by a temperaturedifference between the system and its surroundings. When added to asystem heat transfer causes the energy of a system to increase and heat


Chapter 2-23

transfer from a system causes the energy to decrease. Q is zero foradiabatic systems.

2. Work, W: Work is energy transfer at a system boundary could havecaused a weight to be raised. When added to a system, the energy ofthe system increase; and when done by a system, the energy of thesystem decreases. W is zero for systems having no work interactions atits boundaries.

3. Mass flow, m: As mass flows into a system, the energy of the systemincreases by the amount of energy carried with the mass into thesystem. Mass leaving the system carries energy with it, and the energyof the system decreases. Since no mass transfer occurs at the boundaryof a closed system, energy transfer by mass is zero for closed systems.

The energy balance for a general system is

( ) ( )( ), ,

in out in out in out

mass in mass out system

E E Q Q W W

E E E

− = − + −

+ − = ∆

Expressed more compactly, the energy balance is

Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc., energies

( )in out systemE E E kJ− = ∆

or on a rate form, as

E E E kWin out system− =Rate of net energy transfer by heat, work, and mass

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

( )∆


Chapter 2-24

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

, , and ( )Q Q t W W t E E t kJ= ∆ = ∆ ∆ = ∆ ∆

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

( / )in out systeme e e kJ kg− = ∆

and in the differential forms as

( )( / )

in out system

in out system

E E E kJe e e kJ kgδ δ δ

δ δ δ

− =

− =

First Law for a Cycle

A thermodynamic cycle is composed of processes that cause the workingfluid to undergo a series of state changes through a process or a series ofprocesses. These processes occur such that the final and initial states areidentical and the change in internal energy of the working fluid is zero forwhole numbers of cycles. Since thermodynamic cycles can be viewed ashaving heat and work (but not mass) crossing the cycle system boundary,the first law for a closed system operating in a thermodynamic cyclebecomes


Chapter 2-25

net net cycle

net net

Q W E

Q W

− = ∆

=

0


Chapter 2-26

Example 2-4

A system receives 5 kJ of heat transfer and experiences a decrease inenergy in the amount of 5 kJ. Determine the amount of work done by thesystem.

We apply the first law as

( )

5

5

5 5

10

in out system

in in

out out

system

out in system

out

out

E E E

E Q kJE W

E kJ

E E E

W kJ

W kJ

− = ∆

= ==

∆ = −

= − ∆

= − − =

The work done by the system equals the energy input by heat plus thedecrease in the energy of the working fluid.

∆E=-5 kJ∆E = -5 kJQin =5 kJ Wout=?

SystemBoundary


Chapter 2-27

Example 2-5

A steam power plant operates on a thermodynamic cycle in which watercirculates through a boiler, turbine, condenser, pump, and back to theboiler. For each kilogram of steam (water) flowing through the cycle, thecycle receives 2000 kJ of heat in the boiler, rejects 1500 kJ of heat to theenvironment in the condenser, and receives 5 kJ of work in the cyclepump. Determine the work done by the steam in the turbine, in kJ/kg.

The first law requires for a thermodynamic cycle

( )

Let and

2000 1500 5

505

net net cycle

net net

in out out in

out in out in

out in out in

out

out

Q W E

Q WQ Q W W

W Q Q WW Qw qm m

w q q wkJwkg

kJwkg

− = ∆

=− = −

= − −

= =

= − +

= − +

=

0


Chapter 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 loses100 kW of energy by heat transfer to the surroundings. If the systemexperiences no energy change as the air flows through it, how muchenergy does the air carry as it leaves the system, in kW?

System sketch:

Conservation of Energy:

( )

, ,

, ,

,

0

300 600 100 800

in out system

mass in in mass out out system

mass out mass in in out

mass out

E E E

E W E Q E

E E W Q

E kW kW

− = ∆

+ − − = ∆ =

= + −

= + − =

Open System,mass inE ,mass outE

inW

outQ


Chapter 2-29

Energy Conversion Efficiencies

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

Desired ResultRequired Input

η =

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 fluidundergoes a series of processes and then returns to its original state, so thatthe properties of the system at the end of the cycle are the same as at itsbeginning.

Thus, for whole numbers of cycles

P P T T u u v v etcf i f i f i f i= = = =, , , , .

Heat Engine

A heat engine is a thermodynamic system operating in a thermodynamiccycle to which net heat is transferred and from which net work isdelivered.

The system, or working fluid, undergoes a series of processes thatconstitute the heat engine cycle.


Chapter 2-30

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

Thermal Efficiency, ηth

The thermal efficiency is the index of performance of a work-producingdevice 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 thedesired result).

η th =Desired ResultRequired Input

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


Chapter 2-31

η thnet out

in

WQ

= ,

whereW W W

Q Qnet out out in

in net

, = −

≠

Here, the use of the in and out subscripts means to use the magnitude (takethe positive value) of either the work or heat transfer and let the minus signin 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/kgof pump work, and produced 505 kJ/kg of turbine work. Determine thethermal efficiency for this cycle.

We can write the thermal efficiency on a per unit mass basis as:


( )

,

505 5

2000

0.25 or 25%

net outth

in

out in

in

wq

kJw w kg

kJqkg

η =

−−

= =

=

Combustion Efficiency

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

FuCcatoaiplprco

CombustionChamber

CO2H2ON

FuelCnHm

Chapter 2-32

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

Air2

Qout = HVReactantsTR, PR

ProductsPP, TP


Chapter 2-33

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

2out vaporLHV Q with H O in products=

The lower heating value is often used as the measure of energy per kg offuel supplied to the gas turbine engine because the exhaust gases have sucha high temperature that the water formed is a vapor as it leaves the enginewith other products of combustion.

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

2out liquidHHV Q with H O in products=

The higher heating value is often used as the measure of energy per kg offuel supplied to the steam power cycle because there are heat transferprocesses within the cycle that absorb enough energy from the products ofcombustion that some of the water vapor formed during combustion willcondense.

Combustion efficiency is the ratio of the actual heat transfer from thecombustion process to the heating value of the fuel.

outcombustion

QHV

η =


Chapter 2-34

Example 2-8

A steam power plant receives 2000 kJ of heat per unit mass of steamflowing 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 hasa higher heating value of 40,000 kJ/kg of fuel and the combustionefficiency is 85%, determine the required fuel flow rate, in kg/s.

( )

100 2000

0.85 40000

5.88

steam out to steamoutcombustion

fuel

steam out to steamfuel

combustion

steam

steamfuel

fuel

fuelfuel

m qQHV m HHV

m qm

HHV

kg kJs kg

mkJ

kg

kgm

s

η

η

= =

=

=

=


Chapter 2-35

Generator Efficiency:

electrical outputgenerator

mechanical input

WW

η =

Power Plant Overall Efficiency:

, , ,

, ,

,

in cycle net cycle net electrical outputoverall

fuel fuel in cycle net cycle

overall combustion thermal generator

net electrical outputoverall

fuel fuel

Q W Wm HHV Q W

Wm HHV

η

η η η η

η

=

=

=

Motor Efficiency:

mechanical outputmotor

electrical input

WW

η =


Chapter 2-36

Lighting Efficacy:

Amount of Light in LumensWatts of Electricity Consumed

Lighting Efficacy =

Type of lighting Efficacy, lumens/WOrdinary Incandescent 6 - 20Ordinary Fluorescent 40 - 60

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

Useful Energy Transferred to FoodEnergy Consumed by Appliance

Cooking Efficacy =

Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 3- 1

Chapter 3-1

Chapter 3: Properties of Pure Substances

We now turn our attention to the concept of pure substances and thepresentation of their data.

Simple System

A simple system is one in which the effects of motion, viscosity, fluidshear, capillarity, anisotropic stress, and external force fields are absent.

Homogeneous Substance

A substance that has uniform thermodynamic properties throughout is saidto be homogeneous.

Pure Substance

A pure substance has a homogeneous and invariable chemical compositionand may exist in more than one phase.

Examples: 1. Water (solid, liquid, and vapor phases) 2. Mixture of liquid water and water vapor 3. Carbon dioxide, CO24. Nitrogen, N2 5. 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 theequilibrium state can be determined by specifying any two independentintensive properties.


Chapter 3-2

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 3-3

Real substances that readily change phase from solid to liquid to gas suchas water, refrigerant-134a, and ammonia cannot be treated as ideal gases ingeneral. The pressure, volume, temperature relation, or equation of statefor these substances is generally very complicated, and the thermodynamicproperties are given in table form. The properties of these substances maybe illustrated by the functional relation F(P,v,T)=0, called an equation ofstate. The above two figures illustrate the function for a substance thatcontracts on freezing and a substance that expands on freezing. Constantpressure curves on a temperature-volume diagram are shown in Figure 3-11. These figures show three regions where a substance like water mayexist as a solid, liquid or gas (or vapor). Also these figures show that asubstance 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 saturatedliquid water and saturated water vapor are in equilibrium (called thesaturation region), and the superheated vapor region (the solid or iceregion is not shown).

Let's consider the results of heating liquid water from 20°C, 1 atm whilekeeping the pressure constant. We will follow the constant pressureprocess shown in Figure 3-11. First place liquid water in a piston-cylinderdevice where a fixed weight is placed on the piston to keep the pressure ofthe water constant at all times. As liquid water is heated while thepressure is held constant, the following events occur.

Process 1-2:


P

AcPvc9kpp

The temperature and specific volume willincrease from the compressed liquid, orsubcooled liquid, state 1, to the saturated liquidstate 2. In the compressed liquid region, theproperties of the liquid are approximately equalto the properties of the saturated liquid state atthe temperature.

Chapter 3-4

rocess 2-3:

t state 2 the liquid has reached the temperature at which it begins to boil,alled the saturation temperature, and is said to exist as a saturated liquid.roperties at the saturated liquid state are noted by the subscript f and v2 =f. During the phase change both the temperature and pressure remainonstant (according to the International Temperature Scale of 1990, ITS-0, water boils at 99.975°C ≅ 100°C when the pressure is 1 atm or 101.325Pa). At state 3 the liquid and vapor phase are in equilibrium and anyoint on the line between states 2 and 3 has the same temperature andressure.


Chapter 3-5

Process 3-4:

At state 4 a saturated vapor exists and vaporization is complete. Thesubscript g will always denote a saturated vapor state. Note v4 = vg.

Thermodynamic properties at the saturated liquid state and saturated vaporstate are given in Table A-4 as the saturated temperature table and TableA-5 as the saturated pressure table. These tables contain the sameinformation. In Table A-4 the saturation temperature is the independentproperty, and in Table A-5 the saturation pressure is the independentproperty. The saturation pressure is the pressure at which phase changewill occur for a given temperature. In the saturation region thetemperature and pressure are dependent properties; if one is known, thenthe other is automatically known.

Process 4-5:

If the constant pressure heating is continued, the temperature will begin toincrease above the saturation temperature, 100 °C in this example, and thevolume also increases. State 5 is called a superheated state because T5 isgreater than the saturation temperature for the pressure and the vapor is notabout to condense. Thermodynamic properties for water in thesuperheated region are found in the superheated steam tables, Table A-6.


Chapter 3-6

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

Figure 3-11

Consider repeating this process for other constant pressure lines as shownbelow.

99.975 ≅


Chapter 3-7

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


Chapter 3-8

99.61oC

179.88oC

P2 = 1000 kPa

P1 = 100 kPa


Chapter 3-9

The region to the left of the saturated liquid line and below the criticaltemperature is called the compressed liquid region. The region to the rightof the saturated vapor line and above the critical temperature is called thesuperheated region. See Table A-1 for the critical point data for selectedsubstances.

Review the P-v diagrams for substances that contract on freezing and thosethat expand on freezing given in Figure 3-21 and Figure 3-22.

At temperatures and pressures above the critical point, the phase transitionfrom liquid to vapor is no longer discrete.

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


Chapter 3-10

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 3-11

Plot the following processes on the P-T diagram for water (expands onfreezing) and give examples of these processes from your personalexperiences.

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 3-12

Property Tables

In addition to the temperature, pressure, and volume data, Tables A-4through A-8 contain the data for the specific internal energy u the specificenthalpy h and the specific entropy s. The enthalpy is a convenientgrouping 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 ofmass streams flowing into and out of control volumes. The enthalpy isalso useful in the energy balance during a constant pressure process for asubstance contained in a closed piston-cylinder device. The enthalpy hasunits of energy per unit mass, kJ/kg. The entropy s is a property definedby the second law of thermodynamics and is related to the heat transfer toa system divided by the system temperature; thus, the entropy has units ofenergy divided by temperature. The concept of entropy is explained inChapters 6 and 7.

Saturated Water Tables

Since temperature and pressure are dependent properties using the phasechange, two tables are given for the saturation region. Table A-4 hastemperature as the independent property; Table A-5 has pressure as theindependent property. These two tables contain the same information andoften only one table is given.


Chapter 3-13

For the complete Table A-4, the last entry is the critical point at 373.95 oC.

TABLE A-4Saturated water-Temperature table

Specific volume,m3/kg

Internal energy,kJ/kg

Enthalpy,kJ/kg

Entropy,kJ/kg⋅KTemp.,

T °CSat.Press.,Psat kPa

Sat.liquid, vf

Sat. vapor, vg

Sat.liquid,uf

Evap., ufg

Sat.vapor,ug

Sat.liquid, hf

Evap.,hfg

Sat.vapor,hg

Sat.liquid,sf

Evap.,sfg

Sat.vapor, sg

0.01 0.6117 0.001000 206.00 0.00 2374.9 2374.9 0.00 2500.9 2500.9 0.0000 9.1556 9.15565 0.8725 0.001000 147.03 21.02 2360.8 2381.8 21.02 2489.1 2510.1 0.0763 8.9487 9.024910 1.228 0.001000 106.32 42.02 2346.6 2388.7 42.02 2477.2 2519.2 0.1511 8.7488 8.899915 1.706 0.001001 77.885 62.98 2332.5 2395.5 62.98 2465.4 2528.3 0.2245 8.5559 8.780320 2.339 0.001002 57.762 83.91 2318.4 2402.3 83.91 2453.5 2537.4 0.2965 8.3696 8.666125 3.170 0.001003 43.340 104.83 2304.3 2409.1 104.83 2441.7 2546.5 0.3672 8.1895 8.556730 4.247 0.001004 32.879 125.73 2290.2 2415.9 125.74 2429.8 2555.6 0.4368 8.0152 8.452035 5.629 0.001006 25.205 146.63 2276.0 2422.7 146.64 2417.9 2564.6 0.5051 7.8466 8.351740 7.385 0.001008 19.515 167.53 2261.9 2429.4 167.53 2406.0 2573.5 0.5724 7.6832 8.255645 9.595 0.001010 15.251 188.43 2247.7 2436.1 188.44 2394.0 2582.4 0.6386 7.5247 8.163350 12.35 0.001012 12.026 209.33 2233.4 2442.7 209.34 2382.0 2591.3 0.7038 7.3710 8.074855 15.76 0.001015 9.5639 230.24 2219.1 2449.3 230.26 2369.8 2600.1 0.7680 7.2218 7.989860 19.95 0.001017 7.6670 251.16 2204.7 2455.9 251.18 2357.7 2608.8 0.8313 7.0769 7.908265 25.04 0.001020 6.1935 272.09 2190.3 2462.4 272.12 2345.4 2617.5 0.8937 6.9360 7.829670 31.20 0.001023 5.0396 293.04 2175.8 2468.9 293.07 2333.0 2626.1 0.9551 6.7989 7.754075 38.60 0.001026 4.1291 313.99 2161.3 2475.3 314.03 2320.6 2634.6 1.0158 6.6655 7.681280 47.42 0.001029 3.4053 334.97 2146.6 2481.6 335.02 2308.0 2643.0 1.0756 6.5355 7.611185 57.87 0.001032 2.8261 355.96 2131.9 2487.8 356.02 2295.3 2651.4 1.1346 6.4089 7.543590 70.18 0.001036 2.3593 376.97 2117.0 2494.0 377.04 2282.5 2659.6 1.1929 6.2853 7.478295 84.61 0.001040 1.9808 398.00 2102.0 2500.1 398.09 2269.6 2667.6 1.2504 6.1647 7.4151100 101.42 0.001043 1.6720 419.06 2087.0 2506.0 419.17 2256.4 2675.6 1.3072 6.0470 7.3542

٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠

360 18666 0.001895 0.006950 1726.16 625.7 2351.9 1761.53 720.1 2481.6 3.9165 1.1373 5.0537365 19822 0.002015 0.006009 1777.22 526.4 2303.6 1817.16 605.5 2422.7 4.0004 0.9489 4.9493370 21044 0.002217 0.004953 1844.53 385.6 2230.1 1891.19 443.1 2334.3 4.1119 0.6890 4.8009373.95 22064 0.003106 0.003106 2015.8 0 2015.8 2084.3 0 2084.3 4.4070 0 4.4070


Chapter 3-14

TABLE A-5Saturated water-Pressure table

Specific volume,m3/kg

Internal energy,kJ/kg

Enthalpy,kJ/kg

Entropy,kJ/kg⋅KPress.

P kPa

Sat.Temp.,Tsat °C Sat.

liquid, vf

Sat. vapor, vg

Sat. liquid, uf

Evap., ufg

Sat. vapor, ug

Sat. liquid, hf

Evap., hfg

Sat. vapor, hg

Sat.liquid, sf

Evap., sfg

Sat. vapor, sg

0.6117 0.01 0.001000 206.00 0.00 2374.9 2374.9 0.00 2500.9 2500.9 0.0000 9.1556 9.15561.0 6.97 0.001000 129.19 29.30 2355.2 2384.5 29.30 2484.4 2513.7 0.1059 8.8690 8.97491.5 13.02 0.001001 87.964 54.69 2338.1 2392.8 54.69 2470.1 2524.7 0.1956 8.6314 8.82702.0 17.50 0.001001 66.990 73.43 2325.5 2398.9 73.43 2459.5 2532.9 0.2606 8.4621 8.72272.5 21.08 0.001002 54.242 88.42 2315.4 2403.8 88.42 2451.0 2539.4 0.3118 8.3302 8.64213.0 24.08 0.001003 45.654 100.98 2306.9 2407.9 100.98 2443.9 2544.8 0.3543 8.2222 8.57654.0 28.96 0.001004 34.791 121.39 2293.1 2414.5 121.39 2432.3 2553.7 0.4224 8.0510 8.47345.0 32.87 0.001005 28.185 137.75 2282.1 2419.8 137.75 2423.0 2560.7 0.4762 7.9176 8.39387.5 40.29 0.001008 19.233 168.74 2261.1 2429.8 168.75 2405.3 2574.0 0.5763 7.6738 8.250110 45.81 0.001010 14.670 191.79 2245.4 2437.2 191.81 2392.1 2583.9 0.6492 7.4996 8.148815 53.97 0.001014 10.020 225.93 2222.1 2448.0 225.94 2372.3 2598.3 0.7549 7.2522 8.007120 60.06 0.001017 7.6481 251.40 2204.6 2456.0 251.42 2357.5 2608.9 0.8320 7.0752 7.907325 64.96 0.001020 6.2034 271.93 2190.4 2462.4 271.96 2345.5 2617.5 0.8932 6.9370 7.830230 69.09 0.001022 5.2287 289.24 2178.5 2467.7 289.27 2335.3 2624.6 0.9441 6.8234 7.767540 75.86 0.001026 3.9933 317.58 2158.8 2476.3 317.62 2318.4 2636.1 1.0261 6.6430 7.669150 81.32 0.001030 3.2403 340.49 2142.7 2483.2 340.54 2304.7 2645.2 1.0912 6.5019 7.593175 91.76 0.001037 2.2172 384.36 2111.8 2496.1 384.44 2278.0 2662.4 1.2132 6.2426 7.4558100 99.61 0.001043 1.6941 417.40 2088.2 2505.6 417.51 2257.5 2675.0 1.3028 6.0562 7.3589125 105.97 0.001048 1.3750 444.23 2068.8 2513.0 444.36 2240.6 2684.9 1.3741 5.9100 7.2841

٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠ ٠

20,000 365.75 0.002038 0.005862 1785.84 509.0 2294.8 1826.59 585.5 2412.1 4.0146 0.9164 4.931021,000 369.83 0.002207 0.004994 1841.62 391.9 2233.5 1887.97 450.4 2338.4 4.1071 0.7005 4.807622,000 373.71 0.002703 0.003644 1951.65 140.8 2092.4 2011.12 161.5 2172.6 4.2942 0.2496 4.543922,064 373.95 0.003106 0.003106 2015.8 0 2015.8 2084.3 0 2084.3 4.4070 0 4.4070

For the complete Table A-5, the last entry is the critical point at 22.064MPa.

Saturation pressure is the pressure at which the liquid and vapor phasesare in equilibrium at a given temperature.

Saturation temperature is the temperature at which the liquid and vaporphases are in equilibrium at a given pressure.

In Figure 3-11, states 2, 3, and 4 are saturation states.


Chapter 3-15

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

u u uh h hs s s

fg g f

fg g f

fg g f

= −

= −

= −

The quantity hfg is called the enthalpy of vaporization (or latent heat ofvaporization). It represents the amount of energy needed to vaporize a unitof mass of saturated liquid at a given temperature or pressure. It decreasesas the temperature or pressure increases, and becomes zero at the criticalpoint.

Quality and Saturated Liquid-Vapor Mixture

Now, let’s review the constant pressure heat addition process for watershown in Figure 3-11. Since state 3 is a mixture of saturated liquid andsaturated vapor, how do we locate it on the T-v diagram? To establish thelocation of state 3 a new parameter called the quality x is defined as

xmass

massm

m msaturated vapor

total

g

f g

= =+

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 ofthe quality as follows. Consider a mixture of saturated liquid and saturatedvapor. The liquid has a mass mf and occupies a volume Vf. The vapor hasa mass mg and occupies a volume Vg.


Chapter 3-16

We noteV V Vm m mV mv V m v V m v

f g

f g

f f f g g g

= +

= +

= = =, ,

mv m v m v

vm v

mm v

m

f f g g

f f g g

= +

= +

Recall the definition of quality x

xmm

mm m

g g

f g

= =+

Then

mm

m mm

xf g=−

= −1


Chapter 3-17

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

v x v xvf g= − +( )1

The form that we use most often is

v v x v vf g f= + −( )

It is noted that the value of any extensive property per unit mass in thesaturation region is calculated from an equation having a form similar tothat of the above equation. Let Y be any extensive property and let y bethe corresponding intensive property, Y/m, then

y Ym

y x y y

y x ywhere y y y

f g f

f fg

fg g f

= = + −

= +

= −

( )

The term yfg is the difference between the saturated vapor and the saturatedliquid values of the property y; y may be replaced by any of the variablesv, u, h, or s.

We often use the above equation to determine the quality x of a saturatedliquid-vapor state.

The following application is called the Lever Rule:

xy y

yf

fg

=−


Chapter 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 greaterthan 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 independentproperties. The value of temperature to the right of the pressure is thesaturation temperature for the pressure. The first entry in the table is thesaturated vapor state at the pressure.


Chapter 3-19

Compressed Liquid Water Table

A substance is said to be a compressed liquid when the pressure is greaterthan the saturation pressure for the temperature.

It is now noted that state 1 in Figure 3-11 is called a compressed liquidstate because the saturation pressure for the temperature T1 is less than P1.

Data for water compressed liquid states are found in the compressed liquidtables, Table A-7. Table A-7 is arranged like Table A-6, except thesaturation states are the saturated liquid states. Note that the data in TableA-7 begins at 5 MPa or 50 times atmospheric pressure.


Chapter 3-20

At pressures below 5 MPa for water, the data are approximately equal tothe saturated liquid data at the given temperature. We approximateintensive 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 highpressures the enthalpy can be approximated by

h h v P Pf T f sat≅ + −@ ( )For our work, the compressed liquid enthalpy may be approximated by

h hf T≅ @


Chapter 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 definethe quality as the ratio of the mass that is vapor to the total mass of solidand vapor in the saturated solid-vapor mixture. The process of changingdirectly from the solid phase to the vapor phase is called sublimation. Datafor saturated ice and water vapor are given in Table A-8. In Table A-8, theterm Subl. refers to the difference between the saturated vapor value andthe saturated solid value.

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

y y yy y x yig g i

i ig

= −

= +


Chapter 3-22

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

xm

m mg

i g

=+

How to Choose the Right Table

The correct table to use to find the thermodynamic properties of a realsubstance can always be determined by comparing the known stateproperties to the properties in the saturation region. Given the temperatureor pressure and one other property from the group v, u, h, and s, thefollowing procedure is used. For example if the pressure and specificvolume are specified, three questions are asked: For the given pressure,

Is ?Is ?Is ?

v vv v vv v

f

f g

g

<

< <

<

The answer to one of these questions must be yes. If the answer to the firstquestion is yes, the state is in the compressed liquid region, and thecompressed liquid tables are used to find the properties of the state. If theanswer to the second question is yes, the state is in the saturation region,and either the saturation temperature table or the saturation pressure tableis used to find the properties. Then the quality is calculated and is used tocalculate the other properties, u, h, and s. If the answer to the third


Chapter 3-23

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

u h Pv = −


Chapter 3-24

Example 2-1

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

10-4 10-3 10-2 10-1 100 101 102 1031030

100

200

300

400

500

600

700700

v [m3/kg]

T [C

]

7000 kPa

Steam

10-4 10-3 10-2 10-1 100 101 102102100

101

102

103

104

105

v [m3/kg]

P [k

Pa] 374.1 C

285.9 C

Steam


Chapter 3-25

1. P = 7 MPa, dry saturated or saturated vapor

Using Table A-5,

2581.0gkJu ukg

= =

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,

1258.0fkJu ukg

= =


TriplePoint

P

T, °C

7MPa

0.01 285.8 373.95

CPSteam


Chapter 3-26

3. Moisture = 5%, P = 7 MPa

let moisture be y, defined as

ymm

f= = 0 05.

then, the quality isx y= − = − =1 1 0 05 0 95. .

and using Table A-5,

( )

1258.0 0.95(2581.0 1257.6)

2514.4

f g fu u x u u

kJkg

= + −

= + −

=

Notice that we could have used

u u x uf fg= +


4. P = 7 MPa, T = 600°C

For P = 7 MPa, Table A-5 gives Tsat = 285.83°C. Since 600°C > Tsat forthis pressure, the state is superheated. Use Table A-6.

3261.0 kJukg

=


Chapter 3-27


5. P = 7 MPa, T = 100°C

Using Table A-4, At T = 100°C, Psat = 0.10142 MPa. Since P > Psat, thestate is compressed liquid.

Approximate solution:

@ 100 419.06f T CkJu ukg=≅ =

Solution using Table A-7:

We do linear interpolation to get the value at 100 °C. (We willdemonstrate how to do linear interpolation with this problem even thoughone could accurately estimate the answer.)

P MPa u kJ/kg 5 417.65 7 u = ?10 416.23

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


Chapter 3-28

5 7 417.655 10 417.65 416.23

417.08

u

kJukg

− −=

− −

=


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 u kJ/kg 450 2979.0460 u = ?500 3074.3

Using the above table, form the following ratios.

460 450 2979.0500 450 3074.3 2979.0

2998.1

u

kJukg

− −=

− −

=



Chapter 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 stateof a simple substance. Pressure P is one intensive property and specificvolume is another. Therefore, we calculate the specific volume.

v Volumemass

mkg

mkg

= = =1215

083 3.

..

Using Table A-5 at P = 200 kPa,

vf = 0.001061 m3/kg , vg = 0.8858 m3/kg

Now,

Is ? NoIs ? YesIs ? No

v vv v vv v

f

f g

g

<

< <

<Locate this state on a T-v diagram.

T

v


Chapter 3-30

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

v v x v vf g f= + −( )

0.8 0.0010610.8858 0.0010610.903 (What does this mean?)

f

g f

v vx

v v−

=−

−=

−=

Then,

504.7 (0.903)(2201.6)

2492.7

f fgh h x h

kJkg

= +

= +

=

Example 2-3

Determine the internal energy of refrigerant-134a at a temperature of 0°Cand a quality of 60%.

Using Table A-11, for T = 0°C,

uf = 51.63 kJ/kg ug =230.16 kJ/kg then,


Chapter 3-31

( )

51.63 (0.6)(230.16 51.63)

158.75

f g fu u x u u

kJkg

= + −

= + −

=

Example 2-4

Consider the closed, rigid container of water shown below. The pressureis 700 kPa, the mass of the saturated liquid is 1.78 kg, and the mass of thesaturated vapor is 0.22 kg. Heat is added to the water until the pressureincreases to 8 MPa. Find the final temperature, enthalpy, and internalenergy of the water. Does the liquid level rise or fall? Plot this process ona P-v diagram with respect to the saturation lines and the critical point.

Let’s introduce a solution procedure that we will follow throughout thecourse. 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)

mg, Vg

Sat. Vapor

mf, VfSat. Liquid

P

v


Chapter 3-32

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

v Vm

= = constant

orv v2 1=

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

xm

m mkg

kgg

f g1

1

1 1

0 22178 0 22

011=+

=+

=.

( . . ).

Then, at P = 700 kPa

1 1 1 1 1

3

( )

0.001108 (0.11)(0.2728 0.001108)

0.031

f g fv v x v v

mkg

= + −

= + −

=

State 2 is specified by:

P2 = 8 MPa, v2 = 0.031 m3/kg

At 8 MPa = 8000 kPa,

vf = 0.001384 m3/kg vg = 0.02352 m3/kg at 8 MPa, v2 = 0.031 m3/kg.


Chapter 3-33

Is ? NoIs ? NoIs ? Yes

v vv v vv v

f

f g

g

2

2

2

<

< <

<

Therefore, State 2 is superheated.

Interpolating in the superheated tables at 8 MPa, v = 0.031 m3/kg gives,

T2 = 361 °Ch2 = 3024 kJ/kgu2 = 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 specificvolume had been 0.001 m3/kg and the pressure was 8 MPa?


Chapter 3-34

Equations of State

The relationship among the state variables, temperature, pressure, andspecific volume is called the equation of state. We now consider theequation of state for the vapor or gaseous phase of simple compressiblesubstances.

F P T v, ,b g ≡ 0

Ideal Gas

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

P R Tv

= FHGIKJ

where R is the constant of proportionality and is called the gas constantand takes on a different value for each gas. If a gas obeys this relation, it iscalled an ideal gas. We often write this equation as

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

R RM

u=


Chapter 3-35

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

M ggmol

kgkmol

lbmlbmolair = = =28 97 28 97 28 97. . .

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

m N M=

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

Pv RTVP RTm

PV mRT

=

=

=

PV mM

MR T

PV NR T

P VN

R T

Pv R T

u

u

u

=

=

=

=

b g


Chapter 3-36

Here P = absolute pressure in MPa, or kPa v = molar specific volume in m3/kmol T = absolute temperature in K Ru = 8.314 kJ/(kmol⋅K)

Some values of the universal gas constant are

Universal Gas Constant, Ru8.314 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 oneassumes

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 3-37

R RM

R

kJkmol K

kgkmol

kJkg K

u

air

=

= − =−

8 314

28 970 287

.

..

R

kJkmol K

kgkmol

kJkg Khydrogen =

− =−

8 314

2 0164 124

.

..

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

Compressibility Factor

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


Chapter 3-38

Pv Z R Tu=or

Z PvR Tu

=

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

T TT

P PPR

crR

cr

= = and

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 3-51gives a comparison of Z factors for various gases and supports theprinciple of corresponding states.


Chapter 3-39

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

v vR TP

Ractual

cr

cr

=

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


Chapter 3-40

These charts show the conditions for which Z = 1 and the gas behaves asan ideal gas:

1. PR < 10 and TR > 2 or P < 10Pcr and T > 2Tcr 2. PR << 1 or P << Pcr

Note: When PR is small, we must make sure that the state is not in thecompressed liquid region for the given temperature. A compressed liquidstate is certainly not an ideal gas state.

For instance the critical pressure and temperature for oxygen are 5.08 MPaand 154.8 K, respectively. For temperatures greater than 300 K andpressures less than 50 MPa (1 atmosphere pressure is 0.10135 MPa)oxygen is considered to be an ideal gas.

Example 2-6

Calculate the specific volume of nitrogen at 300 K and 8.0 MPa andcompare the result with the value given in a nitrogen table as v = 0.011133m3/kg.

From Table A.1 for nitrogen

Tcr = 126.2 K, P cr = 3.39 MPa R = 0.2968 kJ/kg-K

T TT

KK

P PP

MPaMPa

Rcr

Rcr

= = =

= = =

300126 2

2 38

8 03 39

2 36

..

..

.

Since T > 2T cr and P < 10P cr, we use the ideal gas equation of state


Chapter 3-41

Pv RT

v RTP

kJkg K

K

MPam MPa

kJmkg

=

= = −

=

0 2968 300

8 0 10

0 01113

3

3

3

. ( )

.

.

Nitrogen is clearly an ideal gas at this state.

If the system pressure is low enough and the temperature high enough (Pand T are compared to the critical values), gases will behave as ideal gases.Consider the T-v diagram for water. The figure below shows thepercentage of error for the volume ([|vtable – videal|/vtable]x100) for assumingwater (superheated steam) to be an ideal gas.


Chapter 3-42

We see that the region for which water behaves as an ideal gas is in thesuperheated region and depends on both T and P. We must be cautionedthat in this course, when water is the working fluid, the ideal gasassumption may not be used to solve problems. We must use the real gasrelations, i.e., the property tables.

Useful Ideal Gas Relation: The Combined Gas Law

By writing the ideal gas equation twice for a fixed mass and simplifying,the properties of an ideal gas at two different states are related by

m m1 2= or

PVR T

PVR T

1 1

1

2 2

2

=

But, the gas constant is (fill in the blank), so

PVT

PVT

1 1

1

2 2

2

=


Chapter 3-43

Example 2-7

An ideal gas having an initial temperature of 25 °C under goes the twoprocesses described below. Determine the final temperature of the gas.

Process 1-2: The volume is held constant while the pressuredoubles.

Process 2-3: The pressure is held constant while the volume isreduced to one-third of the original volume.

Process 1-3: m m1 3=

orPVT

PVT

1 1

1

3 3

3

=

but V3 = V1/3 and P3 = P2 = 2P1

Therefore,

IdealGas

3T2

1T1

V

T3

P

2


Chapter 3-44

3 33 1

1 1

1 13 1 1

1 1

3

2 / 3 23

2 (25 273) 198.7 74.33

P VT TP VP VT T T

P V

T K K C

=

= =

= + = = − °

Other Equations of State

Many attempts have been made to keep the simplicity of the ideal gasequation of state but yet account for the intermolecular forces and volumeoccupied by the particles. Three of these are

van der Waals:

( )( )P av

v b R T+ − =2

where

a R TP

b RTP

cr

cr

cr

cr

= =2764 8

2 2

and

Extra Assignment

When plotted on the P-v diagram, the critical isotherm has a point ofinflection at the critical point. Use this information to verify the equationsfor van der Waals’ constants a and b.


Chapter 3-45

Beattie-Bridgeman:

P R Tv

cvT

v B Av

u= −FHGIKJ + −2 3 21 ( )

where

A A av

B B bvo o= −FHG

IKJ = −FHG

IKJ1 1 and

The constants a, b, c, Ao, Bo for various substances are found in Table 3-4.

Benedict-Webb-Rubin:

P R Tv

B R T A CT v

bR T av

av

cv T v

e

uo u o

o u

v

= + − −FHG

IKJ +

−

+ + +FHGIKJ

−

2 2 3

6 3 2 2

1

12α γ γ /

The constants for various substances appearing in the Benedict-Webb-Rubin equation are given in Table 3-4.


Chapter 3-46

Example 2-8

Compare the results from the ideal gas equation, the Beattie-Bridgemanequation, and the EES software for nitrogen at 1000 kPa. The following isan EES solution to that problem.

Notice that the results from the Beattie-Bridgeman equation compare wellwith the actual nitrogen data provided by EES in the gaseous orsuperheated region. However, neither the Beattie-Bridgeman equation northe ideal gas equation provides adequate results in the two-phase region,where the gas (ideal or otherwise) assumption fails.

10-3 10-2 10-110-170

80

90

100

110

120

130

140

150

160

v [m3/kg]

T [K

]

1000 kPa

Nitrogen, T vs v for P=1000 kPa

EES Table ValueEES Table Value

Beattie-BridgemanBeattie-BridgemanIdeal GasIdeal Gas


Chapter 4 -1

Chapter 4: Energy Analysis of Closed Systems

The first law of thermodynamics is an expression of the conservation ofenergy principle. Energy can cross the boundaries of a closed system inthe form of heat or work. Energy transfer across a system boundary duesolely to the temperature difference between a system and its surroundingsis called heat.

Work energy can be thought of as the energy expended to lift a weight.

Closed System First Law

A closed system moving relative to a reference plane is shown belowwhere z is the elevation of the center of mass above the reference planeand V is the velocity of the center of mass.

For the closed system shown above, the conservation of energy principleor the first law of thermodynamics is expressed as

Total energyentering the system

Total energyleaving the system

The change in total energy of the system

FHG

IKJ −FHG

IKJ =FHG

IKJ

or

CMHeat

Workz

ClosedSystem

Reference Plane, z = 0

V


Chapter 4 -2

E E Ein out system− = ∆

According to classical thermodynamics, we consider the energy added tobe net heat transfer to the closed system and the energy leaving the closedsystem to be net work done by the closed system. So

Q W Enet net system− = ∆

Where

2

1

( )net in out

net out in other b

b

Q Q QW W W W

W PdV

= −= − +

= ∫

Normally the stored energy, or total energy, of a system is expressed as thesum of three separate energies. The total energy of the system, Esystem, isgiven as

E Internal energy Kinetic energy Potential energyE U KE PE

= + + = + +

Recall that U is the sum of the energy contained within the molecules ofthe system other than the kinetic and potential energies of the system as awhole and is called the internal energy. The internal energy U isdependent on the state of the system and the mass of the system.


Chapter 4 -3

For a system moving relative to a reference plane, the kinetic energy KEand the potential energy PE are given by

2

0

0

2V

V

z

z

mVKE mV dV

PE mg dz mgz

=

=

= =

= =

∫

∫

The change in stored energy for the system is

∆ ∆ ∆ ∆E U KE PE= + +

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

Q W U KE PEnet net− = + +∆ ∆ ∆

If the system does not move with a velocity and has no change inelevation, the conservation of energy equation reduces to

Q W Unet net− = ∆

We will find that this is the most commonly used form of the first law.

Closed System First Law for a Cycle


Chapter 4 -4

Since a thermodynamic cycle is composed of processes that cause theworking fluid to undergo a series of state changes through a series ofprocesses such that the final and initial states are identical, the change ininternal energy of the working fluid is zero for whole numbers of cycles.The first law for a closed system operating in a thermodynamic cyclebecomes

Q W UQ W

net net cycle

net net

− =

=

∆0


Chapter 4 -5

Example 4-1

Complete the table given below for a closed system under going a cycle.

Process Qnet kJ Wnet kJ U2 – U1 kJ 1-2 +5 -5 2-3 +20 +10 3-1 -5Cycle

(Answer to above problem) Row 1: +10, Row 2: +10, Row 3: -10, -5Row 4: +15, +15, 0


Chapter 4 -6

In the next section we will look at boundary work in detail. Review thetext material on other types of work such as shaft work, spring work,electrical work.

Boundary Work

Work is energy expended when a force acts through a displacement.Boundary work occurs because the mass of the substance contained withinthe system boundary causes a force, the pressure times the surface area, toact on the boundary surface and make it move. This is what happens whensteam, the “gas” in the figure below, contained in a piston-cylinder deviceexpands against the piston and forces the piston to move; thus, boundarywork is done by the steam on the piston. Boundary work is then calculatedfrom

W W Fds FA

Ads

P dV

b b= = =

=

z z zzδ

1

2

1

2

1

2

1

2


Chapter 4 -7

Since the work is process dependent, the differential of boundary workδWb

δW PdVb =

is called inexact. The above equation for Wb is valid for a quasi-equilibrium process and gives the maximum work done during expansionand the minimum work input during compression. In an expansionprocess the boundary work must overcome friction, push the atmosphericair out of the way, and rotate a crankshaft.

b friction atm crank2

friction atm crank1( )

W W W W

F P A F ds

= + +

= + +∫

To calculate the boundary work, the process by which the system changedstates must be known. Once the process is determined, the pressure-volume relationship for the process can be obtained and the integral in theboundary work equation can be performed. For each process we need todetermine

P f V= ( )

So as we work problems, we will be asking, “What is the pressure-volumerelationship for the process?” Remember that this relation is really theforce-displacement function for the process.

The boundary work is equal to the area under the process curve plotted onthe pressure-volume diagram.


Chapter 4 -8

Note from the above figure:

P is the absolute pressure and is always positive.When dV is positive, Wb is positive.When dV is negative, Wb is negative.

Since the areas under different process curves on a P-V diagram aredifferent, the boundary work for each process will be different. The nextfigure shows that each process gives a different value for the boundarywork.


Chapter 4 -9

Some Typical Processes

Constant volume

If the volume is held constant, dV = 0, and the boundary work equationbecomes

W PdVb = =z12 0If the working fluid is an ideal gas, what will happen to the temperature ofthe gas during this constant volume process?

P

V

1

2

P-V diagram for V = constant


Chapter 4 -10

Constant pressure

If the pressure is held constant, the boundary work equation becomes

W PdV P dV P V Vb = = = −z z1

2

1

2

2 1b gFor the constant pressure process shown above, is the boundary workpositive or negative and why?

Constant temperature, ideal gas

If the temperature of an ideal gas system is held constant, then the equationof state provides the pressure-volume relation

P mRTV

=

Then, the boundary work is

P

V

2 1

P-V DIAGRAM FOR P = CONSTANT


Chapter 4 -11

W PdV mRTV

dV

mRT VV

b = =

=FHGIKJ

z z1

2

1

2

2

1

ln

Note: The above equation is the result of applying the ideal gas assumptionfor the equation of state. For real gases undergoing an isothermal(constant temperature) process, the integral in the boundary work equationwould be done numerically.

The polytropic process

The polytropic process is one in which the pressure-volume relation isgiven as

PV n = constantThe exponent n may have any value from minus infinity to plus infinitydepending on the process. Some of the more common values are givenbelow.

Process Exponent nConstant pressure 0Constant volume ∞Isothermal & ideal gas 1 Adiabatic & ideal gas k = CP/CV

Here, k is the ratio of the specific heat at constant pressure CP to specificheat at constant volume CV. The specific heats will be discussed later.


Chapter 4 -12

The boundary work done during the polytropic process is found bysubstituting the pressure-volume relation into the boundary work equation.The result is

W PdV ConstV

dV

PV PVn

PV VV

b n= =

=−−

≠

FHGIKJ

z z1

2

1

2

2 2 1 1

2

1

1,

ln ,

n 1

= n = 1

For an ideal gas under going a polytropic process, the boundary work is

W PdV ConstV

dV

mR T Tn

mRT VV

b n= =

=−

−≠

FHGIKJ

z z1

2

1

2

2 1

2

1

1( ) ,

ln ,

n 1

= n = 1


Chapter 4 -13

Notice that the results we obtained for an ideal gas undergoing a polytropicprocess when n = 1 are identical to those for an ideal gas undergoing theisothermal process.

Example 4-2

Three kilograms of nitrogen gas at 27°C and 0.15 MPa are compressedisothermally to 0.3 MPa in a piston-cylinder device. Determine theminimum work of compression, in kJ.

System: Nitrogen contained in a piston-cylinder device.

Process: Constant temperature

Property Relation: Check the reduced temperature and pressure fornitrogen. The critical state properties are found in Table A-1.

T TT

KK

T

P PP

MPaMPa

P P

Rcr

R

Rcr

R R

11

2

11

2 1

27 273126 2

2 38

0153 39

0 044

2 0 088

= =+

= =

= = =

= =

( ).

.

..

.

.

P-V DIAGRAM FOR T = CONSTANT

P

V

2

1

WbNitrogengas

SystemBoundary


Chapter 4 -14

Since PR<<1 and T>2Tcr, nitrogen is an ideal gas, and we use the ideal gasequation of state as the property relation.

PV mRT=

Work Calculation:

W W Wnet net other b, , ,12 12 12= +b gW PdV mRT

VdV

mRT VV

b ,

ln

12 1

2

1

2

2

1

= =

FHGIKJ

z z=

For an ideal gas in a closed system (mass = constant), we have

m mPVRT

PVRT

1 2

1 1

1

2 2

2

=

=

Since the R's cancel, we obtain the combined ideal gas equation. Since T2= T1,

VV

PP

2

1

1

2

=


Chapter 4 -15

W mRT PP

kg kJkg K

K MPaMPa

kJ

b , ln

( ) . ( ) ln ..

.

121

2

3 0 2968 300 0150 30

184 5

=FHGIKJ

=−

FHG

IKJ

FHG

IKJ

= −The net work is

W W kJnet b, , .12 120 184 5= + = −On a per unit mass basis

wW

mkJkgnet

net,

, .1212 615= = −

The net work is negative because work is done on the system during thecompression process. Thus, the work done on the system is 184.5 kJ, or184.5 kJ of work energy is required to compress the nitrogen.

Example 4-3

Water is placed in a piston-cylinder device at 20 °C, 0.1 MPa. Weights areplaced on the piston to maintain a constant force on the water as it isheated to 400 °C. How much work does the water do on the piston?

System: The water contained in the piston-cylinder device

SystemBoundaryfor water

Wb

Heat


Chapter 4 -16

Property Relation: Steam tables

Process: Constant pressure

10-4 10-3 10-2 10-1 100 101 102102100

101

102

103

104

105

v [m3/kg]

P [k

Pa]

400 C

20 C

Steam

1 2

Work Calculation:

Since there is no Wother mentioned in the problem, the net work is

W W PdV P dV P V Vnet b, ,12 12 1

2

1

2

2 1= = = = −z z b gSince the mass of the water is unknown, we calculate the work per unitmass.

wW

mP V V

mP v vb

b,

,12

12 2 12 1= =

−= −

b g b gAt T1 = 20°C, Psat = 2.339 kPa. Since P1 > 2.339 kPa, state 1 iscompressed liquid. Thus,


Chapter 4 -17

v1 ≅ vf at 20 °C = 0.001002 m3/ kg

At P2 = P1 = 0.1 MPa, T2 > Tsat at 0.1 MPa = 99.61°C.

So, state 2 is superheated. Using the superheated tables at 0.1 MPa, 400°C

v2 = 3.1027 m3/kg

( ),12 2 1

3 3

3

100.1 (3.1027 0.001002)

310.2

bw P v v

m kPa kJMPakg MPa m kPa

kJkg

= −

= −

=

The water does work on the piston in the amount of 310.2 kJ/kg.


Chapter 4 -18

Example 4-4

One kilogram of water is contained in a piston-cylinder device at 100 °C.The piston rests on lower stops such that the volume occupied by the wateris 0.835 m3. The cylinder is fitted with an upper set of stops. When thepiston rests against the upper stops, the volume enclosed by the piston-cylinder device is 0.841 m3. A pressure of 200 kPa is required to supportthe piston. Heat is added to the water until the water exists as a saturatedvapor. How much work does the water do on the piston?

System: The water contained in the piston-cylinder device


Process: Combination of constant volume and constant pressure processesto be shown on the P-v diagram as the problem is solved.

Work Calculation:

The specific volume at state 1 is

v Vm1

1 = = 0.835 m1 kg

= 0.835 mkg

3 3

P

v

WaterWb

SystemBoundary

Stops

Stops


Chapter 4 -19

At T1 = 100°C,3 3

=0.001044 =1.6720f gm mv vkg kg

Therefore, vf < v1 < vg and state 1 is in the saturation region; so P1 = 101.35 kPa. Show this state on the P-v diagram.

Now let’s consider the processes for the water to reach the final state.

Process 1-2: The volume stays constant until the pressure increasesto 200 kPa. Then the piston will move.

v v mkg2 1

3

0 835= = .

Process 2-3: Piston lifts off the bottom stops while the pressure staysconstant. Does the piston hit the upper stops before orafter reaching the saturated vapor state?

Let's set

v Vm

mkg

mkg3

33 30841

10841 = = . = .

At P3 = P2 = 200 kPa

3 3

=0.001061 =0.88578f gm mv vkg kg

Thus, vf < v3 < vg. So, the piston hits the upper stops before the waterreaches the saturated vapor state. Now we have to consider a third process.


Chapter 4 -20

Process 3-4: With the piston against the upper stops, the volumeremains constant during the final heating to the saturatedvapor state and the pressure increases.

Because the volume is constant in process 3-to-4, v4 = v3 = 0.841 m3/kgand v4 is a saturated vapor state. Interpolating in either the saturationpressure table or saturation temperature table at v4 = vg gives

StateP kPav v

T Cg

42113

1224

44

:.=

=UVW

= °

The net work for the heating process is (the “other” work is zero)

W W PdV PdV PdV PdV

mP v v

kg kPa mkg

kJm kPa

kJ

net b, ,

( )

( )( )( . . )

.

14 14 1

4

1

2

2

3

3

4

3 23

3

0 0

1 200 0 841 0 835

12

= = = + +

= + − +

= −

=

z z z z

Later in Chapter 4, we will apply the conservation of energy, or the firstlaw of thermodynamics, to this process to determine the amount of heattransfer required.

Example 4-5

Air undergoes a constant pressure cooling process in which thetemperature decreases by 100°C. What is the magnitude and direction ofthe work for this process?


Chapter 4 -21

T2= T1-100°C

System:

Property Relation: Ideal gas law, Pv = RT

Process: Constant pressure

Work Calculation: Neglecting the “other” work

W W PdV P V V

mR T Tnet b, , ( )

( )12 12 1

2

2 1

2 1

0= + = = −

= −z

The work per unit mass is

,12,12 2 1( )

(0.287 )( 100 ) 28.7

netnet

Ww R T T

mkJ kJK

kg K kg

= = −

= − = −⋅

The work done on the air is 28.7 kJ/kg.

AirWb

SystemBoundary

T1

P-V DIAGRAM FOR T = CONSTANT

P

V

21


Chapter 4 -22

Example 4-6

Find the required heat transfer to the water in Example 4-4.

Review the solution procedure of Example 4-4 and then apply the first lawto the process.


,14 ,14 14

in out

net net

E E EQ W U

− = ∆− = ∆

In Example 4-4 we found that

W kJnet , .14 12=

The heat transfer is obtained from the first law as

,14 ,14 14net netQ W U= + ∆

where ∆U U U m u u14 4 1 4 1= − = −( )

At state 1, T1 = 100°C, v1 = 0.835 m3/kg and vf < v1 < vg at T1. Thequality at state 1 is


Chapter 4 -23

1 1

11

0.835 0.001043 0.4991.6720 0.001043

f fg

f

fg

v v x v

v vx

v

= +

− −= = =

−

1 1

419.06 (0.499)(2087.0)

1460.5

f fgu u x u

kJkg

= +

= +

=

Because state 4 is a saturated vapor state and v4 = 0.841 m3/kg,interpolating in either the saturation pressure table or saturationtemperature table at v4 = vg gives

u kJkg4 253148= .

So

14 4 1( )

(1 )(2531.48 1460.5)

1071.0

U m u ukJkgkg

kJ

∆ = −

= −

=

The heat transfer is


Chapter 4 -24

,14 ,14 14

1.2 1071.01072.2

net netQ W UkJ kJ

kJ

= + ∆

= +=

Heat in the amount of 1072.42 kJ is added to the water.


Chapter 4 -25

Specific Heats and Changes in Internal Energy and Enthalpy for IdealGases

Before the first law of thermodynamics can be applied to systems, ways tocalculate the change in internal energy of the substance enclosed by thesystem boundary must be determined. For real substances like water, theproperty tables are used to find the internal energy change. For ideal gasesthe internal energy is found by knowing the specific heats. Physics definesthe amount of energy needed to raise the temperature of a unit of mass of asubstance one degree as the specific heat at constant volume CV for aconstant-volume process, and the specific heat at constant pressure CP fora constant-pressure process. Recall that enthalpy h is the sum of theinternal energy u and the pressure-volume product Pv.

h u Pv= +

In thermodynamics, the specific heats are defined as

C uT

C hTV

vP

P

=∂∂FHGIKJ =

∂∂FHGIKJ and

Simple Substance

The thermodynamic state of a simple, homogeneous substance is specifiedby giving any two independent, intensive properties. Let's consider theinternal energy to be a function of T and v and the enthalpy to be afunction of T and P as follows:

( , ) and ( , )u u T v h h T P= =


Chapter 4 -26

The total differential of u is

du uT

dT uv

dv

or

du C dT uv

dv

v T

vT

=∂∂FHGIKJ +

∂∂FHGIKJ

= +∂∂FHGIKJ

The total differential of h is

dh hT

dT hP

dP

or

dh C dT hP

dP

P T

PT

=∂∂FHGIKJ +

∂∂FHGIKJ

= +∂∂FHGIKJ

Using thermodynamic relation theory, we could evaluate the remainingpartial derivatives of u and h in terms of functions of P,v, and T. Thesefunctions depend upon the equation of state for the substance. Given thespecific heat data and the equation of state for the substance, we candevelop the property tables such as the steam tables.


Chapter 4 -27

Ideal Gases

For ideal gases, we use the thermodynamic function theory of Chapter 12and the equation of state (Pv = RT) to show that u, h, CV, and CP arefunctions of temperature alone.

For example when total differential for u = u(T,v) is written as above, thefunction theory of Chapter 12 shows that

vT

vv

udu C dT dvv

Pdu C dT T P dvT

∂ = + ∂

∂ = + − ∂

Let’s evaluate the following partial derivative for an ideal gas.

∂∂FHGIKJ =

∂∂FHGIKJ −

uv

T PT

PT v

For ideal gases

P RTv

PT

Rv

uv

T Rv

P P P

v

T

=

∂∂FHGIKJ =

∂∂FHGIKJ = − = − = 0


Chapter 4 -28

This result helps to show that the internal energy of an ideal gas does notdepend upon specific volume. To completely show that internal energy ofan ideal gas is independent of specific volume, we need to show that thespecific heats of ideal gases are functions of temperature only. We will dothis later in Chapter 12. A similar result that applies to the enthalpyfunction for ideal gases can be reviewed in Chapter 12.

Then for ideal gases,

C C T uv

C C T hP

V VT

P PT

=∂∂FHGIKJ ≡

=∂∂FHGIKJ ≡

( )

( )

and

and

0

0

The ideal gas specific heats are written in terms of ordinary differentials as

C dudT

C dhdT

Videal gas

Pideal gas

= FHGIKJ

= FHGIKJ


Chapter 4 -29

Using the simple “dumbbell model” for diatomic ideal gases, statisticalthermodynamics predicts the molar specific heat at constant pressure as afunction of temperature to look like the following

The following figure shows how the molar specific heats vary withtemperature for selected ideal gases.

C kJkmol Kp ⋅

52

Ru

72

Ru

92

Ru

“Dumbbell model”

T

Translation mode

Vibration mode

Rotation mode


Chapter 4 -30

The differential changes in internal energy and enthalpy for ideal gasesbecome

du C dTdh C dT

V

P

==

The change in internal energy and enthalpy of ideal gases can be expressedas

∆

∆

u u u C T dT C T T

h h h C T dT C T T

V V ave

P P ave

= − = = −

= − = = −

zz

2 1 1

2

2 1

2 1 1

2

2 1

( ) ( )

( ) ( )

,

,

where CV,ave and CP,ave are average or constant values of the specific heatsover the temperature range. We will drop the ave subscript shortly.


C

In the above figure an ideal gas uthe same two temperatures.

Process 1-2a: CProcess 1-2b: PProcess 1-2c: C

These ideal gas processes have thenthalpy because the processes o

∆ ∆

∆ ∆

u u

h h

a b

a b

= =

= =

2a

T1

1

P-V DIAGRAM FOR SEVERALIDEAL GAS.

P
2b
hapter 4 -31

ndergoes three different process between

onstant volume = a + bV, a linear relationshiponstant pressure

e same change in internal energy andccur between the same temperature limits.

∆

∆

u C T dT

h C T dT

c V

c P

=

=

zz

( )

( )

1

2

1

2

T2

2c

PROCESSES FOR AN

V


Chapter 4 -32

To find ∆u and ∆h we often use average, or constant, values of the specificheats. Some ways to determine these values are as follows:

1. The best average value (the one that gives the exact results)

See Table A-2(c) for variable specific data.

CC T dT

T TC

C T dT

T Tv aveV

P aveP

, ,

( ) ( )=

−=

−z z1

2

2 1

1

2

2 1

and

2. Good average values are

2 1 2 1, ,

( ) ( ) ( ) ( ) 2 2

V V P Pv ave P ave

C T C T C T C TC C+ += = and

, ,

2 1

( ) ( ) where

2

v ave V ave P ave P ave

ave

C C T C C T

T TT

= =

+=

3. Sometimes adequate (and most often used) values are the onesevaluated at 300 K and are given in Table A-2(a).

C C K C C Kv ave V P ave P, ,( ) ( )= =300 300 and

Let's take a second look at the definition of ∆u and ∆h for ideal gases. Justconsider the enthalpy for now.

∆h h h C T dTP= − = z2 1 1

2( )


Chapter 4 -33

Let's perform the integral relative to a reference state where h = href at T = Tref.

∆h h h C T dT C T dTPT

T

PT

Tref

ref

= − = ′ ′ + ′ ′z z2 11

2( ) ( )

or

∆h h h C T dT C T dT

h h h h

PT

T

PT

T

ref ref

ref ref

= − = ′ ′ − ′ ′

= − − −

z z2 1

2 1

2 1( ) ( )

( ) ( )

At any temperature, we can calculate the enthalpy relative to the referencestate as

h h C T dT

or

h h C T dT

ref PT

T

ref PT

T

ref

ref

− = ′ ′

= + ′ ′

z

z

( )

( )

A similar result is found for the change in internal energy.

( )ref

T

ref vTu u C T dT′ ′= + ∫

These last two relations form the basis of the air tables (Table A-17 on amass basis) and the other ideal gas tables (Tables A-18 through A-25 on amole basis). When you review Table A-17, you will find h and u asfunctions of T in K. Since the parameters Pr, vr, and so, also found in TableA=17, apply to air only in a particular process, call isentropic, you shouldignore these parameters until we study Chapter 7. The reference state forthese tables is defined as


Chapter 4 -34

u T Kh T K

ref ref

ref ref

= =

= =

0 00 0

at at

A partial listing of data similar to that found in Table A.17 is shown in thefollowing figure.

In the analysis to follow, the “ave” notation is dropped. In mostapplications for ideal gases, the values of the specific heats at 300 K givenin Table A-2 are adequate constants.


Chapter 4 -35

Exercise

Determine the average specific heat for air at 305 K.

CP ave, =

(Answer: 1.005 kJ/kg⋅K, approximate the derivative of h with respect to Tas differences)

Relation between CP and CV for Ideal Gases

Using the definition of enthalpy (h = u + Pv) and writing the differentialof enthalpy, the relationship between the specific heats for ideal gases is

h u Pvdh du d RTC dT C dT RdTC C R

P V

P V

= += +

= += +

( )

where R is the particular gas constant. The specific heat ratio k (fluidstexts often use γ instead of k) is defined as

k CC

P

V

=


Chapter 4 -36

Extra Problem

Show that

C kRk

C RkP V=

−=

−1 1 and

Example 2-9

Two kilograms of air are heated from 300 to 500 K. Find the change inenthalpy by assuming

a. Empirical specific heat data from Table A-2(c). b. Air tables from Table A-17.c. Specific heat at the average temperature from Table A-2(c).d. Use the 300 K value for the specific heat from Table A-2(a).

a. Table A-2(c) gives the molar specific heat at constant pressure for air as

C x T x T x T kJkmol KP = . + . + . - .

--2 -5 -92811 01967 10 0 4802 10 1966 102 3

The enthalpy change per unit mole is


Chapter 4 -37

∆h h h C T dT

x T x T

x T dT

T x T x T

x T

kJkmol

P

K

K

= − =

=

=

zz

2 1 1

2

2

300

500

3

2 3

4

2811 01967 10 0 4802 10

1966 10

2811 01967 102

0 4802 103

1966 104

5909 49

( )

(

(

.

= . + . + .

- . )

. + . + .

- . )

-2 -5

-9

-2 -5

-9

300K500K

∆∆h hM

kJkmolkg

kmol

kJkg

= = =5909 49

28 972039

.

..

∆ ∆H m h kg kJkg

kJ= = =( )( . ) .2 2039 407 98

b. Using the air tables, Table A-17, at T1 = 300 K, h1 = 300.19 kJ/kg and atT2 = 500 K, h2 = 503.02 kJ/kg


kJ= = − =( )(503. . ) .2 02 30019 40566


Chapter 4 -38

The results of parts a and b would be identical if Table A-17 had beenbased on the same specific heat function listed in Table A-2(c).

c. Let’s use a constant specific heat at the average temperature.

Tave = (300 + 500)K/2 = 400 K. At Tave , Table A-2 gives CP = 1.013 kJ/(kg⋅K). For CP = constant,

∆h h h C T TkJ

kg KK

kJkg

P ave= − = −

=⋅

−

=

2 1 2 1

1013 300

202 6

, ( )

. (500 )

.


kJ= = =( )( . ) .2 202 6 4052

d. Using the 300 K value from Table A-2(a), CP = 1.005 kJ/kg- K.

For CP = constant,

2 1 2 1( )

1.005 (500 300) 201.0

Ph h h C T TkJ kJK

kg K kg

∆ = − = −

= − =⋅


kJ= = =( )( . ) .2 2010 402 0


Chapter 4 -39

Extra Problem

Find the change in internal energy for air between 300 K and 500 K, inkJ/kg.


Chapter 4 -40

The Systematic Thermodynamics Solution Procedure

When we apply a methodical solution procedure, thermodynamicsproblems are relatively easy to solve. Each thermodynamics problem isapproached the same way as shown in the following, which is amodification of the procedure given in the text:

Thermodynamics Solution Method1. Sketch the system and show energy interactions across the

boundaries.

2. Determine the property relation. Is the working substancean ideal gas or a real substance? Begin to set up and fill in aproperty table.

3. Determine the process and sketch the process diagram.Continue to fill in the property table.

4. Apply conservation of mass and conservation of energyprinciples.

5. Bring in other information from the problem statement,called physical constraints, such as the volume doubles orthe pressure is halved during the process.

6. Develop enough equations for the unknowns and solve.


Chapter 4 -41

Example 4-7

A tank contains nitrogen at 27°C. The temperature rises to 127°C by heattransfer to the system. Find the heat transfer and the ratio of the finalpressure to the initial pressure.

System: Nitrogen in the tank.

Property Relation: Nitrogen is an ideal gas. The ideal gas propertyrelations apply. Let’s assume constant specific heats.(You are encouraged to rework this problem usingvariable specific heat data.)

Process: Tanks are rigid vessels; therefore, the process is constant volume.

Conservation of Mass:

m m2 1=

Using the combined ideal gas equation of state,

NITROGENGAS

Systemboundary

P-V diagram for a constantvolume process

2

T2=127°CT1=27°C

P

V

1


Chapter 4 -42

PVT

PVT

2 2

2

1 1

1

=

Since R is the particular gas constant, and the process is constant volume,

V VPP

TT

KK

2 1

2

1

2

1

127 27327 273

1333

=

= =++

=( )( )

.


The first law closed system is

in out

net net

E E EQ W U

− = ∆− = ∆

For nitrogen undergoing a constant volume process (dV = 0), the net workis (Wother = 0)

W W PdVnet b, ,12 12 1

20 0= + = =z

Using the ideal gas relations with Wnet = 0, the first law becomes (constantspecific heats)


Chapter 4 -43

Q U m C dT mC T Tnet V V− = = = −z01

2

2 1∆ ( )

The heat transfer per unit mass is

q Qm

C T T

kJkg K

K

kJkg

netnet

V= = −

=⋅

−

=

( )

.

.

2 1

0 743 127 27

74 3

b g

Example 4-8

Air is expanded isothermally at 100°C from 0.4 MPa to 0.1 MPa. Find theratio of the final to the initial volume, the heat transfer, and work.

System: Air contained in a piston-cylinder device, a closed system

Process: Constant temperature

P-V diagram for T= constant

P

V

1

2

AIR WB

T = const.

Systemboundary


Chapter 4 -44

Property Relation: Assume air is an ideal gas and use the ideal gasproperty relations with constant specific heats.

PV mRTu C T TV

== −∆ ( )2 1


E E EQ W U

in out

net net

− =− =

∆∆

The system mass is constant but is not given and cannot be calculated;therefore, let’s find the work and heat transfer per unit mass.

Work Calculation:

W W Wnet net other b, , ,12 12 12= +b g

2 2

,12 1 1

2

1

= ln

bmRTW P dV dV

VVmRTV

= =

∫ ∫


Chapter 4 -45

Conservation of Mass: For an ideal gas in a closed system (mass =constant), we have

m mPVRT

PVRT

1 2

1 1

1

2 2

2

=

=

Since the R's cancel and T2 = T1

VV

PP

MPaMPa

2

1

1

2

0 401

4= = =..

Then the work expression per unit mass becomes

w

Wm

RT VVb

b,

, ln1212 2

1

= =FHGIKJ

w kJkg K

K

kJkg

b , . ( ) ln

.

12 0 287 100 273 4

148 4

=−

FHG

IKJ +

=

b g

The net work per unit mass is

w w kJkgnet b, , .12 120 148 4= + =


Chapter 4 -46

Now to continue with the conservation of energy to find the heat transfer.Since T2 = T1 = constant,

∆ ∆U m u mC T TV12 12 2 1 0= = − =( )

So the heat transfer per unit mass is

q Qm

q w uq w

kJkg

netnet

net net

net net

=

− = ==

=

∆ 0

148 4.

The heat transferred to the air during an isothermal expansion processequals the work done.

Examples Using Variable Specific Heats

Review the solutions in Chapter 4 to the ideal gas examples where thevariable specific heat data are used to determine the changes in internalenergy and enthalpy.

Extra Problem for You to Try:

An ideal gas, contained in a piston-cylinder device, undergoes a polytropicprocess in which the polytropic exponent n is equal to k, the ratio ofspecific heats. Show that this process is adiabatic. When we get toChapter 7 you will find that this is an important ideal gas process.


Chapter 4 -47

Internal Energy and Enthalpy Changes of Solids and Liquids

We treat solids and liquids as incompressible substances. That is, weassume that the density or specific volume of the substance is essentiallyconstant during a process. We can show that the specific heats ofincompressible substances (see Chapter 12) are identical.

C C C kJkg KP V= =

⋅FHGIKJ

The specific heats of incompressible substances depend only ontemperature; therefore, we write the differential change in internal energyas

du C dT CdTV= =

and assuming constant specific heats, the change in internal energy is

∆ ∆u C T C T T= = −( )2 1

Recall that enthalpy is defined as

h u Pv= +The differential of enthalpy is

dh du Pdv vdP= + +

For incompressible substances, the differential enthalpy becomes


Chapter 4 -48

dvdh du Pdv vdPdh du vdP

=

= + / += +

00

Integrating, assuming constant specific heats

∆ ∆ ∆ ∆ ∆h u v P C T v P= + = +

For solids the specific volume is approximately zero; therefore,

∆ ∆ ∆∆ ∆ ∆

h u v Ph u C T

solid solid

solid solid

= + /= ≅

0

For liquids, two special cases are encountered:

1. Constant-pressure processes, as in heaters (∆P = 0)

∆ ∆ ∆h u C Tliquid liquid= ≅2. Constant-temperature processes, as in pumps (∆T = 0)

∆ ∆ ∆ ∆ ∆

∆ ∆

h u v P C T v Ph v P

liquid liquid

liquid

= + ≅ / +

=

0

We will derive this last expression for ∆h again once we have discussedthe first law for the open system in Chapter 5 and the second law ofthermodynamics in Chapter 7.

The specific heats of selected liquids and solids are given in Table A-3.


Chapter 4 -49

Example 4-8 Incompressible Liquid

A two-liter bottle of your favorite beverage has just been removed fromthe trunk of your car. The temperature of the beverage is 35°C, and youalways drink your beverage at 10°C. a. How much heat energy must be removed from your two liters of

beverage?b. You are having a party and need to cool 10 of these two-liter bottles in

one-half hour. What rate of heat removal, in kW, is required?Assuming that your refrigerator can accomplish this and that electricitycosts 8.5 cents per kW-hr, how much will it cost to cool these 10bottles?

System: The liquid in the constant volume, closed system container

Property Relation: Incompressible liquid relations, let’s assume that thebeverage is mostly water and takes on the properties of liquid water. Thespecific volume is 0.001 m3/kg, C = 4.18 kJ/kg⋅K.

Process: Constant volume

V V2 1=

QoutThe heatremoved

System

boundary

Mybeverage


Chapter 4 -50

Conservation of Mass:

m m m

m Vv

Lmkg

mL

kg

2 1

3

32

0 001 10002

= =

= =FHG

IKJ =.


The first law closed system is

E E Ein out− = ∆

Since the container is constant volume and there is no “other” work doneon the container during the cooling process, we have

W W Wnet net other b= + =b g 0

The only energy crossing the boundary is the heat transfer leaving thecontainer. Assuming the container to be stationary, the conservation ofenergy becomes


Chapter 4 -51

− =− = =

E EQ U mC T

out

out

∆∆ ∆

(2 )(4.18 )(10 35)

209.2209.2

out

out

out

kJQ kg Kkg K

Q kJQ kJ

− = −⋅

− = −=

The heat transfer rate to cool the 10 bottles in one-half hour is

( )( . )

.

.

Qbottles kJ

bottlehr

hrs

kWkJs

kW

out =FHGIKJF

HGGG

I

KJJJ

=

10 209 2

051

3600

1162

Cost kW hrkW hr

=−

=

( . )( . ) $0.

$0.

1162 0 5 085

05


Chapter 5: Mass and Energy Analysis of Control Volumes

Conservation of Energy for Control volumes

The conservation of mass and the conservation of energy principles foropen systems or control volumes apply to systems having mass crossingthe system boundary or control surface. In addition to the heat transfer andwork crossing the system boundaries, mass carries energy with it as itcrosses the system boundaries. Thus, the mass and energy content of theopen system may change when mass enters or leaves the control volume.

Thermodyntwo groupssteady-flowexperiencin

Let’s reviewOne shouldpresented iimportant t

Reference

VCM

Control surface

,m Ve e

,m Vi i

Zi

Wnet

Typical c

amic processes: steady-flow pr process, the flug no change wi

the concepts o study the develn the text. Hereo successful pro

plane

QnetZ

Chapter 5 -

ontrol volum

involving coocesses and id flows thr

th time at a f

f mass flowopment of th we present blem solving

CM

1

e or open system

ntrol volumes can be considered inunsteady-flow processes. During aough the control volume steadily,ixed position.

rate and energy transport by mass.e general conservation of mass

an overview of the concepts techniques.

Ze


Chapter 5 -2

Mass Flow Rate

Mass flow through a cross-sectional area per unit time is called the massflow rate m . Note the dot over the mass symbol indicates a time rate ofchange. It is expressed as

m V dAnA

= z ρwhere Vn is the velocity normal to the cross-sectional flow area.

If the fluid density and velocity are constant over the flow cross-sectionalarea, the mass flow rate is

aveave

V Am V Av

ρ= =

where ρ is the density, kg/m3 ( = 1/v), A is the cross-sectional area, m2;and aveV is the average fluid velocity normal to the area, m/s.

Example 5-1

Refrigerant-134a at 200 kPa, 40% quality, flows through a 1.1-cm insidediameter, d, tube with a velocity of 50 m/s. Find the mass flow rate of therefrigerant-134a.

At P = 200 kPa, x = 0.4 we determine the specific volume from


Chapter 5 -3

3

0.0007533 0.4(0.0999 0.0007533)

0.0404

f fgv v xv

mkg

= +

= + −

=

2

2

3

450 / (0.011 )

0.0404 / 4

0.117

ave aveV A V dmv v

m s mm kg

kgs

π

π

= =

=

=

The fluid volume flowing through a cross-section per unit time is calledthe volume flow rate V . The volume flow rate is given by integrating theproduct of the velocity normal to the flow area and the differential flowarea over the flow area. If the velocity over the flow area is a constant, thevolume flow rate is given by (note we are dropping the “ave” subscript onthe velocity)

( / )V VA m s= 3

The mass and volume flow rate are related by

( / )m V Vv

kg s= =ρ


Chapter 5 -4

Example 5-2

Air at 100 kPa, 50oC, flows through a pipe with a volume flow rate of 40m3/min. Find the mass flow rate through the pipe, in kg/s.

Assume air to be an ideal gas, so

v RTP

kJkg K

KkPa

m kPakJ

mkg

= =⋅

+

=

0 287 273100

0 9270

3

3

. (50 )

.

/ min. /

min

.

m Vv

mm kg s

kgs

= =

=

400 9270

160

0 719

3

3

Conservation of Mass for General Control Volume

The conservation of mass principle for the open system or control volumeis expressed as

Sumof rateof massflowinginto control volume

Sumof rateof massflowingfromcontrol volume

Time ratechangeof massinsidecontrol volume

L

NMMM

O

QPPP−L

NMMM

O

QPPP=L

NMMM

O

QPPP

or

( / )m m m kg sin out system∑ ∑− = ∆


Chapter 5 -5

Steady-State, Steady-Flow Processes

Most energy conversion devices operate steadily over long periods of time.The rates of heat transfer and work crossing the control surface areconstant with time. The states of the mass streams crossing the controlsurface or boundary are constant with time. Under these conditions themass and energy content of the control volume are constant with time.

0CVCV

dm mdt

= ∆ =

Steady-state, Steady-Flow Conservation of Mass:

Since the mass of the control volume is constant with time during thesteady-state, steady-flow process, the conservation of mass principlebecomes

Sumof rateof massflowinginto control volume

Sumof rateof massflowingfromcontrol volume

L

NMMM

O

QPPP=L

NMMM

O

QPPP

or

( / )m m kg sin out∑ ∑=


Chapter 5 -6

Special Case: Steady Flow of an Incompressible Fluid

The mass flow rate is related to volume flow rate and fluid density by

m Vρ=For one entrance, one exit steady flow control volume, the mass flow ratesare related by

in out

in in out out

in out

in out

in in out out

incompressible assumption

(kg/s)

m m

V V

V V

V A V A

ρ ρρ ρ

=

==

=

=

Word of caution: This result applies only to incompressible fluids. Mostthermodynamic systems deal with processes involving compressible fluidssuch as ideal gases, steam, and the refrigerants for which the aboverelation will not apply.

Example 5-3 Geometry Effects on Fluid Flow

An incompressible liquid flows through the pipe shown in the figure. Thevelocity at location 2 is

A) 114

V B) 112

V C) 12V D) 14V

21

Incompressible Liquid

D2D


Chapter 5 -7

Solution:

Answer: D

1 1 2 2

21 1

2 1 122 2

2 21

2 1 12

2 1

/ 4/ 4

2

4

A V A V

A DV V VA D

D DV V VD D

V V

ππ

=

= =

= =

=

1 2

1 2

in outInlets Outlets

m Vm m

V V

V V

ρ

ρ ρ

=

=

=

=

∑ ∑


Chapter 5 -8

Flow work and the energy of a flowing fluid

Energy flows into and from the control volume with the mass. The energyrequired to push the mass into or out of the control volume is known as theflow work or flow energy.

The fluid up steam of the control surface acts as a piston to push a unit ofmass into or out of the control volume. Consider the unit of mass enteringthe control volume shown below.

As the fluid upstream pushes mass across the control surface, work doneon that unit of mass is

flow

flowflow

AW F L F L PV PmvA

Ww Pv

m

= = = =

= =

The term Pv is called the flow work done on the unit of mass as it crossesthe control surface.


Chapter 5 -9

The total energy of flowing fluid

The total energy carried by a unit of mass as it crosses the control surfaceis the sum of the internal energy, flow work, potential energy, and kineticenergy.

θ = + + +

= + +

u Pv V gz

h V gz

2

2

2

2

Here we have used the definition of enthalpy, h = u + Pv.

Energy transport by mass

Amount of energy transport across a control surface:2

(kJ)2mass

VE m m h gzθ

= = + +

Rate of energy transport across a control surface:2

( )2mass

VE m m h gz kWθ

= = + +


Chapter 5 -10

Conservation of Energy for General Control Volume

The conservation of energy principle for the control volume or opensystem has the same word definition as the first law for the closed system.Expressing the energy transfers on a rate basis, the control volume firstlaw is

Sumof rateof energy flowinginto control volume

Sumof rateof energy flowingfromcontrol volume

Time ratechangeof energy insidecontrol volume

L

NMMM

O

QPPP−L

NMMM

O

QPPP=L

NMMM

O

QPPP

or



( )∆

Considering that energy flows into and from the control volume with themass, energy enters because net heat is transferred to the control volume,and energy leaves because the control volume does net work on itssurroundings, the open system, or control volume, the first law becomes

( )Q m W m dEdt

kWnet i i net e eCV+ − − =∑ ∑θ θ

for each inlet for each exit4 :

where θ is the energy per unit mass flowing into or from the controlvolume. The energy per unit mass, θ, flowing across the control surfacethat defines the control volume is composed of four terms: the internalenergy, the kinetic energy, the potential energy, and the flow work.

The total energy carried by a unit of mass as it crosses the control surfaceis


Chapter 5 -11

θ = + + +

= + +

u Pv V gz

h V gz

2

2

2

2

E E E

Q m h V gz W m h V gz E

in out CV

net i ii

i net e ee

e CV

− =

+ + +FHG

IKJ − − + +

FHG

IKJ =∑ ∑

∆

∆2 2

2 2for each inlet for each exit

Where the time rate change of the energy of the control volume has beenwritten as ∆ECV .

Steady-State, Steady-Flow Processes

Most energy conversion devices operate steadily over long periods of time.The rates of heat transfer and work crossing the control surface areconstant with time. The states of the mass streams crossing the controlsurface or boundary are constant with time. Under these conditions themass and energy content of the control volume are constant with time.

dmdt

m

dEdt

E

CVCV

CVCV

= =

= =

∆

∆

0

0


Chapter 5 -12

Steady-state, Steady-Flow Conservation of Mass:

( / )m m kg sin out∑ ∑=Steady-state, steady-flow conservation of energy

Since the energy of the control volume is constant with time during thesteady-state, steady-flow process, the conservation of energy principlebecomes

Sumof rateof energy flowinginto control volume

Sumof rateof energy flowingfromcontrol volume

L

NMMM

O

QPPP=L

NMMM

O

QPPP

or



( )∆

or

E Ein outRate of net energy transfer by heat, work, and mass into the system

Rate of energy transfer by heat, work, and mass from the system

1 :=

Considering that energy flows into and from the control volume with themass, energy enters because heat is transferred to the control volume, andenergy leaves because the control volume does work on its surroundings,the steady-state, steady-flow first law becomes

0


Chapter 5 -13

Q W m h V gz Q W m h V gzin in i ii

i out out e ee

e+ + + +FHG

IKJ = + + + +

FHG

IKJ∑ ∑

2 2


Often this result is written as

Q W m h V gz m h V gznet net e ee

e i ii

i− = + +FHG

IKJ − + +

FHG

IKJ∑ ∑

2 2

2 2for each exit for each inlet

whereQ Q Q

W W Wnet in out

net out in

= −

= −

∑ ∑∑ ∑

Steady-state, steady-flow for one entrance and one exit

A number of thermodynamic devices such as pumps, fans, compressors,turbines, nozzles, diffusers, and heaters operate with one entrance and oneexit. The steady-state, steady-flow conservation of mass and first law ofthermodynamics for these systems reduce to

( / )

( ) ( )

m m kg s

vV A

vV A

Q W m h h V V g z z kW

1 2

11 1

22 2

2 122

12

2 1

1 1

2

=

=

− = − +−

+ −LNM

OQP

where the entrance to the control volume is state 1 and the exit is state 2and m is the mass flow rate through the device.

When can we neglect the kinetic and potential energy terms in the firstlaw?


Chapter 5 -14

Consider the kinetic and potential energies per unit mass.

ke V=

2

2For V = 45 m

s

V = 140 ms

ke m s kJ kgm s

kJkg

ke m s kJ kgm s

kJkg

= =

= =

( / ) //

( / ) //

452

11000

1

1402

11000

10

2

2 2

2

2 2

pe gz=

For z m pe ms

m kJ kgm s

kJkg

z m pe ms

m kJ kgm s

kJkg

= = =

= = =

100 9 8 100 11000

0 98

1000 9 8 1000 11000

9 8

2 2 2

2 2 2

. //

.

. //

.

When compared to the enthalpy of steam (h ≅ 2000 to 3000 kJ/kg) and theenthalpy of air (h ≅ 200 to 6000 kJ/kg), the kinetic and potential energiesare often neglected.

When the kinetic and potential energies can be neglected, the conservationof energy equation becomes

( ) ( )Q W m h h kW− = −2 1

We often write this last result per unit mass flow as

q w h h kJ kg− = −( ) ( / )2 1

where q Qm

= and w Wm

= .


Chapter 5 -15

Some Steady-Flow Engineering Devices

Below are some engineering devices that operate essentially as steady-state, steady-flow control volumes.


Chapter 5 -16

Nozzles and Diffusers

For flow through nozzles, the heat transfer, wnormally neglected, and nozzles have one enconservation of energy becomes

m mm m m

E E

Q m h V gz W

m h V m h

in out

in out

net i ii

i net

== =

=

+ + +FHG

IKJ = +

+FHG

IKJ =FHG

∑

1 2

2

112

2

2

2

for each inlet

Solving for V2

V h h2 1 22= −( )

V1

V1

V V>>
2 1
V V<<

ork, and potential energy aretrance and one exit. The

m h V gz

V

e ee

e+ +FHG

IKJ

+IKJ

∑2

22

2

2

for each exit

V12+

2 1


Chapter 5 -17

Example 5-4

Steam at 0.4 MPa, 300oC, enters an adiabatic nozzle with a low velocityand leaves at 0.2 MPa with a quality of 90%. Find the exit velocity, inm/s.

Control Volume: The nozzle


Process: Assume adiabatic, steady-flow

Conservation Principles:

Conservation of mass:

For one entrance, one exit, the conservation of mass becomes

m m

m m min out∑ ∑== =1 2

Conservation of energy:

According to the sketched control volume, mass crosses the controlsurface, but no work or heat transfer crosses the control surface.Neglecting the potential energies, we have

E E

m h V m h Vin out=

+FHG

IKJ = +FHG

IKJ1

12

222

2 2


Chapter 5 -18

Neglecting the inlet kinetic energy, the exit velocity is

V h h2 1 22= −( )

Now, we need to find the enthalpies from the steam tables.

1 1 2 2

1 2

Superheated Saturated Mix.300 3067.1 0.20.4 0.90

o kJT C h P MPa hkg

P MPa x

= = = = =

At 0.2 MPa hf = 504.7 kJ/kg and hfg = 2201.6 kJ/kg.

2 2= +

= 504.7 + (0.90)(2201.6) = 2486.1

f fgh h x hkJkg2 2

21000 /2(3067.1 2486.1)

/

1078.0

kJ m sVkg kJ kg

ms

= −

=


Chapter 5 -19

Turbines

Turbine control volume

If we neglect the changes in kinetic and potential energies as fluid flowsthrough an adiabatic turbine having one entrance and one exit, theconservation of mass and the steady-state, steady-flow first law becomes

( )

m mm m m

E E

Q m h V gz W m h V gz

m h m h WW m h h

in out

in out

net i ii

i net e ee

e

out

out

== =

=

+ + +FHG

IKJ = + + +

FHG

IKJ

= +

= −

∑ ∑

1 2

2 2

1 1 2 2

1 2


Wout

mout 2

min 1

Controlsurface


Chapter 5 -20

Example 5-5

High pressure air at 1300 K flows into an aircraft gas turbine andundergoes a steady-state, steady-flow, adiabatic process to the turbine exitat 660 K. Calculate the work done per unit mass of air flowing through theturbine when

(a) Temperature-dependent data are used. (b) Cp,ave at the average temperature is used. (c) Cp at 300 K is used.

Control Volume: The turbine.

Property Relation: Assume air is an ideal gas and use ideal gas relations.

Process: Steady-state, steady-flow, adiabatic process


Conservation of mass:m mm m m

in out∑ ∑== =1 2


Q m h V gz W m h V gzin i ii

i out e ee

e+ + +FHG

IKJ = + + +

FHG

IKJ∑ ∑

2 2


According to the sketched control volume, mass and work cross thecontrol surface. Neglecting kinetic and potential energies and noting theprocess is adiabatic, we have


Chapter 5 -21

0 1 1 2 2

1 2

+ = +

= −( )m h W m hW m h h

out

out

The work done by the air per unit mass flow is

w Wm

h houtout= = −1 2

Notice that the work done by a fluid flowing through a turbine is equal tothe enthalpy decrease of the fluid.

(a) Using the air tables, Table A-17 at T1 = 1300 K, h1 = 1395.97 kJ/kgat T2 = 660 K, h2 = 670.47 kJ/kg

w h hkJkg

kJkg

out = −

= −

=

1 2

139597 670 47

7255

( . . )

.

(b) Using Table A-2(c) at Tave = 980 K, Cp, ave = 1.138 kJ/kg⋅K w h h C T T

kJkg K

K

kJkg

out p ave= − = −

=⋅

−

=

1 2 1 2

1138 1300 660

728 3

, ( )

. ( )

.


Chapter 5 -22

c. Using Table A-2(a) at T = 300 K, Cp = 1.005 kJ/kg ⋅K

w h h C T TkJ

kg KK

kJkg

out p= − = −

=⋅

−

=

1 2 1 2

1005 1300 660

643 2

( )

. ( )

.

Compressors and fans

Compressors and fans are essentially the same devices. However,compressors operate over larger pressure ratios than fans. If we neglectthe changes in kinetic and potential energies as fluid flows through anadiabatic compressor having one entrance and one exit, the steady-state,steady-flow first law or the conservation of energy equation becomes

Win

mout 2

min 1

Steady-Flow Compressor

Comp


Chapter 5 -23

( )( ) ( )

( )


W m h hW m h hW m h h

net i ii

i net e ee

e

net

in

in

+ + +FHG

IKJ = + + +

FHG

IKJ

− = −

− − = −

= −

∑ ∑2 2

2 1

2 1

2 1


Example 5-6 Nitrogen gas is compressed in a steady-state, steady-flow, adiabaticprocess from 0.1 MPa, 25oC. During the compression process thetemperature becomes 125oC. If the mass flow rate is 0.2 kg/s, determinethe work done on the nitrogen, in kW.

Control Volume: The compressor (see the compressor sketched above)

Property Relation: Assume nitrogen is an ideal gas and use ideal gasrelations

Process: Adiabatic, steady-flow



in out∑ ∑== =1 2


Chapter 5 -24


Q m h V gz W m h V gznet i ii

i net e ee

e+ + +FHG

IKJ = + + +

FHG

IKJ∑ ∑

2 2


According to the sketched control volume, mass and work cross thecontrol surface. Neglecting kinetic and potential energies and noting theprocess is adiabatic, we have for one entrance and one exit

0 0 0 0 01 1 2 2

2 1

+ + + = − + + +

= −

( ) ( ) ( )( )

m h W m hW m h h

in

in

The work done on the nitrogen is related to the enthalpy rise of thenitrogen as it flows through the compressor. The work done on thenitrogen per unit mass flow is

w Wm

h hinin= = −2 1

Assuming constant specific heats at 300 K from Table A-2(a), we write thework as


Chapter 5 -25

w C T TkJ

kg KK

kJkg

in p= −

=⋅

−

=

( )

. ( )

.

2 1

1039 125 25

103 9

. .

. .

W m w kgs

kJkg

kJs

kW

in in= =FHG

IKJ

= =

0 2 1039

20 78 20 78

Throttling devices

Consider fluid flowing through a one-entrance, one-exit porous plug. Thefluid experiences a pressure drop as it flows through the plug. No network is done by the fluid. Assume the process is adiabatic and that thekinetic and potential energies are neglected; then the conservation of massand energy equations become


Chapter 5 -26

m m


m h m h

i e

net i ii

i net e ee

e

i i e e

=

+ + +FHG

IKJ = + + +

FHG

IKJ

=

∑ ∑2 2


h hi e=

This process is called a throttling process. What happens when an idealgas is throttled?

h hh h

C T dT

orT T

i e

e i

pi

e

e i

=− =

=

=

z0

0( )

When throttling an ideal gas, the temperature does not change. We willsee later in Chapter 11 that the throttling process is an important process inthe refrigeration cycle.

A throttling device may be used to determine the enthalpy of saturatedsteam. The steam is throttled from the pressure in the pipe to ambientpressure in the calorimeter. The pressure drop is sufficient to superheatthe steam in the calorimeter. Thus, the temperature and pressure in thecalorimeter will specify the enthalpy of the steam in the pipe.


Chapter 5 -27

Example 5-7

One way to determine the quality of saturated steam is to throttle the steamto a low enough pressure that it exists as a superheated vapor. Saturatedsteam at 0.4 MPa is throttled to 0.1 MPa, 100oC. Determine the quality ofthe steam at 0.4 MPa.

Control Volume: The throttle

Property Relation: The steam tables

Process: Steady-state, steady-flow, no work, no heat transfer, neglectkinetic and potential energies, one entrance, one exit



in out∑ ∑== =1 2


E E


in out

net i ii

i net e ee

e

=

+ + +FHG

IKJ = + + +

FHG

IKJ∑ ∑

2 2


1 2

Throttling orifice

ControlSurface


Chapter 5 -28

According to the sketched control volume, mass crosses the controlsurface. Neglecting kinetic and potential energies and noting the processis adiabatic with no work, we have for one entrance and one exit

0 0 0 0 0 01 1 2 2

1 1 2 2

1 2

+ + + = + + +==

( ) ( )m h m hm h m h

h h

22

2

1002675.8

0.1

oT C kJhkgP MPa

==

=

Therefore,

( )1

1 2

1 @ 0.4

2675.8

f fg P MPa

kJh hkg

h x h=

= =

= +

11

2675.8 604.662133.4

0.971

f

fg

h hx

h−

=

−=

=


Chapter 5 -29

Mixing chambers

The mixing of two fluids occurs frequently in engineering applications.The section where the mixing process takes place is called a mixingchamber. The ordinary shower is an example of a mixing chamber.

Example 5-8

Steam at 0.2 MPa, 300oC, enters a mixing chamber and is mixed with coldwater at 20oC, 0.2 MPa, to produce 20 kg/s of saturated liquid water at 0.2MPa. What are the required steam and cold water flow rates?

Control Volume: The mixing chamber


Steam 1

Cold water 2

Saturated water 3

Controlsurface

Mixingchamber


Chapter 5 -30

Process: Assume steady-flow, adiabatic mixing, with no work


Conservation of mass:m m

m m mm m m

in out∑ ∑=+ =

= −1 2 3

2 3 1


E E


in out

net i ii

i net e ee

e

=

+ + +FHG

IKJ = + + +

FHG

IKJ∑ ∑

2 2


According to the sketched control volume, mass crosses the controlsurface. Neglecting kinetic and potential energies and noting the processis adiabatic with no work, we have for two entrances and one exit

( )( ) ( )

m h m h m hm h m m h m h

m h h m h h

1 1 2 2 3 3

1 1 3 1 2 3 3

1 1 2 3 3 2

+ =+ − =

− = −( )( )

m m h hh h1 3

3 2

1 2

=−−

Now, we use the steam tables to find the enthalpies:


Chapter 5 -31

11

1

3003072.1

0.2

oT C kJhkgP MPa

==

=

22 @ 20

2

2083.91

0.2o

o

f C

T C kJh hkgP MPa

=≈ =

= P MPaSat liquid

h h kJkgf MPa

33 0 2

0 2504 7

= UVW = =.

..@ .

3 21 3

1 2

( )( )

(504.7 83.91) /20(3072.1 83.91) /

2.82

h hm mh h

kg kJ kgs kJ kgkgs

−=

−−

=−

=

( . )

.

m m mkgs

kgs

2 3 1

20 2 82

1718

= −

= −

=


Chapter 5 -32

Heat exchangers

Heat exchangers are normally well-insulated devices that allow energyexchange between hot and cold fluids without mixing the fluids. Thepumps, fans, and blowers causing the fluids to flow across the controlsurface are normally located outside the control surface.


Chapter 5 -33

Example 5-9

Air is heated in a heat exchanger by hot water. The water enters the heatexchanger at 45oC and experiences a 20oC drop in temperature. As the airpasses through the heat exchanger, its temperature is increased by 25oC.Determine the ratio of mass flow rate of the air to mass flow rate of thewater.

Control Volume: The heat exchanger

Property Relation: Air: ideal gas relations Water: steam tables or incompressible liquid results

Process: Assume adiabatic, steady-flow



( / )m m m kg sin out system− = ∆

For two entrances, two exits, the conservation of mass becomes

0(steady)

1Air inlet

2Water exit

2Air exit

1Water inlet

Controlsurface


Chapter 5 -34

, , , ,

m mm m m m

in out

air w air w

=+ = +1 1 2 2

For two fluid streams that exchange energy but do not mix, it is better toconserve the mass for the fluid streams separately.

, ,

, ,

m m m

m m mair air air

w w w

1 2

1 2

= =

= =


According to the sketched control volume, mass crosses the controlsurface, but no work or heat transfer crosses the control surface.Neglecting the kinetic and potential energies, we have for steady-flow



( )∆

( ) ( ), , , , , , , ,

, , , ,

E Em h m h m h m h

m h h m h h

in out

air air w w air air w w

air air air w w w

=+ = +

− = −1 1 1 1 2 2 2 2

1 2 2 1

( )( )

, ,

, ,

mm

h hh h

air

w

w w

air air

=−

−2 1

1 2

0(steady)


Chapter 5 -35

We assume that the air has constant specific heats at 300 K, Table A-2(a)(we don't know the actual temperatures, just the temperature difference).Because we know the initial and final temperatures for the water, we canuse either the incompressible fluid result or the steam tables for itsproperties.

Using the incompressible fluid approach for the water, Table A-3, Cp, w = 4.18 kJ/kg⋅K.

( )

( )

, ,2 ,1

, ,1 ,2

( )( )

4.18 20

1.005 25

/3.33/

p w w wair

w p air air air

w

air

air

w

C T Tmm C T T

kJ Kkg K

kJ Kkg K

kg skg s

−=

−

−⋅

=−

⋅

=

A second solution to this problem is obtained by determining the heattransfer rate from the hot water and noting that this is the heat transfer rateto the air. Considering each fluid separately for steady-flow, one entrance,and one exit, and neglecting the kinetic and potential energies, the firstlaw, or conservation of energy, equations become


Chapter 5 -36

,1 ,1 , ,2 ,2

,1 ,1 , ,2 ,2

, ,

:

:

in out

air air in air air air

w w out w w w

in air out w

E E

air m h Q m h

water m h Q m h

Q Q

=

+ =

= +

=

Pipe and duct flow

The flow of fluids through pipes and ducts is often a steady-state, steady-flow process. We normally neglect the kinetic and potential energies;however, depending on the flow situation, the work and heat transfer mayor may not be zero.

Example 5-10

In a simple steam power plant, steam leaves a boiler at 3 MPa, 600oC, andenters a turbine at 2 MPa, 500oC. Determine the in-line heat transfer fromthe steam per kilogram mass flowing in the pipe between the boiler and theturbine.

Control Volume: Pipe section in which the heat loss occurs.


Controlsurface

1Steam fromboiler

Steam toturbine2

Qout


Chapter 5 -37

Process: Steady-flow



( / )in out systemm m m kg s− = ∆

For one entrance, one exit, the conservation of mass becomes

m mm m m

in out== =1 2


According to the sketched control volume, heat transfer and mass cross thecontrol surface, but no work crosses the control surface. Neglecting thekinetic and potential energies, we have for steady-flow

Rate of net energy transfer Rate change in internal, kinetic, by heat, work, and mass potential, etc., energies

( )in out systemE E E kW− = ∆

We determine the heat transfer rate per unit mass of flowing steam as

( )

m h m h Q

Q m h h

q Qm

h h

out

out

outout

1 1 2 2

1 2

1 2

= +

= −

= = −

0(steady)

0(steady)


Chapter 5 -38

We use the steam tables to determine the enthalpies at the two states as

11

1

6003682.8

3

oT C kJhkgP MPa

==

=

22

2

5003468.3

2

oT C kJhkgP MPa

==

=

1 2

(3682.8 3468.3)

214.5

outq h hkJkg

kJkg

= −

= −

=

Example 5-11

Air at 100oC, 0.15 MPa, 40 m/s, flows through a converging duct with amass flow rate of 0.2 kg/s. The air leaves the duct at 0.1 MPa, 113.6 m/s.The exit-to-inlet duct area ratio is 0.5. Find the required rate of heattransfer to the air when no work is done by the air.

Controlsurface

1Air inlet

Air exit2

Qin


Chapter 5 -39

Control Volume: The converging duct

Property Relation: Assume air is an ideal gas and use ideal gas relations

Process: Steady-flow



( / )m m m kg sin out system− = ∆

For one entrance, one exit, the conservation of mass becomesm mm m m

in out== =1 2


According to the sketched control volume, heat transfer and mass cross thecontrol surface, but no work crosses the control surface. Here keep thekinetic energy and still neglect the potential energies, we have for steady-state, steady-flow process

Rate of net energy transfer Rate change in internal, kinetic, by heat, work, and mass potential, etc., energies

( )in out systemE E E kW− = ∆

0(steady)

0(steady)


Chapter 5 -40

( )

m h V Q m h V

Q m h h V V

in

in

1 112

2 222

2 122

12

2 2

2

+FHG

IKJ + = +

FHG

IKJ

= − +−F

HGIKJ

In the first law equation, the following are known: P1, T1 (and h1), 1V , V2 , m , and A2/A1. The unknowns are Qin , and h2 (or T2). We use the

first law and the conservation of mass equation to solve for the twounknowns.

1 2

1 1 2 21 2

1 21 1 2 2

1 2

( / )1 1

m m kg s

V A V Av v

P PV A V ART RT

=

=

=

Solving for T2

T T PP

AA

VV

K MPaMPa

m sm s

K

2 12

1

2

1

2

1

100 273 01015

05 113640

3531

=

= +FHG

IKJFHG

IKJ

=

( ) ..

. . //

.

b g


Chapter 5 -41

Assuming Cp = constant, h2 - h1 = Cp(T2 - T1)

( )

. ( . ( . )

( . ) / //

)

. .

Q m C T T V V

kgs

kJkg K

K

m s kJ kgm s

kJs

kW

in p= − +−F

HGIKJ

=⋅

−

+−

= − = −

2 122

12

2 2 2 2

2 2

2

0 2 1005 3531 373

1136 402 1000

2 87 2 87

Looks like we made the wrong assumption for the direction of the heattransfer. The heat is really leaving the flow duct. (What type of device isthis anyway?)

.Q Q kWout in= − = 2 87

Liquid pumps

PumpWin

Fluid inlet 1

Fluid exit 2

∆ z

Liquid flow through a pump


Chapter 5 -42

The work required when pumping an incompressible liquid in an adiabaticsteady-state, steady-flow process is given by

( ) ( )Q W m h h V V g z z kW− = − +−

+ −LNM

OQP2 1

22

12

2 12

The enthalpy difference can be written as

h h u u Pv Pv2 1 2 1 2 1− = − + −( ) ( ) ( )

For incompressible liquids we assume that the density and specific volumeare constant. The pumping process for an incompressible liquid isessentially isothermal, and the internal energy change is approximatelyzero (we will see this more clearly after introducing the second law).Thus, the enthalpy difference reduces to the difference in the pressure-specific volume products. Since v2 = v1 = v the work input to the pumpbecomes

− = − +−

+ −LNM

OQP

( ) ( ) ( )W m v P P V V g z z kW2 122

12

2 12

W is the net work done by the control volume, and it is noted that work isinput to the pump; so, ,W Win pump= − .

If we neglect the changes in kinetic and potential energies, the pump workbecomes

− − = −

= −

( ) ( ) ( )

( ),

,

W m v P P kW

W m v P Pin pump

in pump

2 1

2 1


Chapter 5 -43

We use this result to calculate the work supplied to boiler feedwater pumpsin steam power plants.

If we apply the above energy balance to a pipe section that has no pump( 0W = ), we obtain.

2 22 1

2 1 2 1

2 22 1

2 1 2 1

2 22 2 1 1

2 1

( ) ( ) ( )2

0 ( ) ( )2

1

2 2

V VW m v P P g z z kW

V Vm v P P g z z

v

P V P Vz zg g

ρ

ρ ρ

−− = − + + −

−

= − + + −

=

+ + = + +

This last equation is the famous Bernoulli’s equation for frictionless,incompressible fluid flow through a pipe.

Uniform-State, Uniform-Flow Problems

During unsteady energy transfer to or from open systems or controlvolumes, the system may have a change in the stored energy and mass.Several unsteady thermodynamic problems may be treated as uniform-state, uniform-flow problems. The assumptions for uniform-state,uniform-flow are

• The process takes place over a specified time period. • The state of the mass within the control volume is uniform at any

instant of time but may vary with time.• The state of mass crossing the control surface is uniform and steady.

The mass flow may be different at different control surface locations.


Chapter 5 -44

To find the amount of mass crossing the control surface at a givenlocation, we integrate the mass flow rate over the time period.

Inlets m m dt Exits m m dti i

t

e e

t: := =z z0 0

The change in mass of the control volume in the time period is

( )m m dmdt

dtCV

t

CV2 1 0− = z

The uniform-state, uniform-flow conservation of mass becomes

m m m mi e CV− = −∑∑ ( )2 1

The change in internal energy for the control volume during the timeperiod is

( )m u m u dUdt

dtCV

t

CV2 2 1 1 0

− = zThe heat transferred and work done in the time period are

Q Qdt and W W dtt t

= =z z0 0

The energy crossing the control surface with the mass in the time period is


Chapter 5 -45

m m hV

gz dtj j j jj

j

tθ∑ z= + +

FHG

IKJ

2

0 2

wherej = i, for inlets

e, for exits

The first law for uniform-state, uniform-flow becomes

E E E

Q W m h V gz m h V gz m e m e

in out CV

e ee

e i ii

i CV

− =

− = + +FHG

IKJ − + +

FHG

IKJ + −∑∑

∆2 2

2 2 1 12 2b g

When the kinetic and potential energy changes associated with the controlvolume and the fluid streams are negligible, it simplifies to

Q W m h m h m u m u kJe e i i CV− = − + −∑∑ 2 2 1 1b g ( )


Chapter 5 -46

Example 5-12

Consider an evacuated, insulated, rigid tank connected through a closedvalve to a high-pressure line. The valve is opened and the tank is filledwith the fluid in the line. If the fluid is an ideal gas, determine the finaltemperature in the tank when the tank pressure equals that of the line.

Control Volume: The tank

Property Relation: Ideal gas relations

Process: Assume uniform-state, uniform-flow



m m m mi e CV− = −∑∑ ( )2 1

Or, for one entrance, no exit, and initial mass of zero, this becomes

m mi CV= ( )2


Chapter 5 -47


For an insulated tank Q is zero and for a rigid tank with no shaft work W iszero. For a one-inlet mass stream and no-exit mass stream and neglectingchanges in kinetic and potential energies, the uniform-state, uniform-flowconservation of energy reduces to

Q W m h m h m u m u kJm h m u

e e i i CV

i i CV

− = − + −

= − +∑∑ 2 2 1 1

2 20b g ( )

( )or

m h m uh u

u Pv uu u Pv

C T T Pv

i i CV

i

i i i

i i i

v i i i

==

+ =− =− =

( )

( )

2 2

2

2

2

2

C T T RT

T C RC

TCC

T

kT

v i i

v

vi

p

vi

i

( )2

2

− =

=+

=

=If the fluid is air, k = 1.4 and the absolute temperature in the tank at thefinal state is 40 percent higher than the fluid absolute temperature in thesupply line. The internal energy in the full tank differs from the internalenergy of the supply line by the amount of flow work done to push thefluid from the line into the tank.

Extra AssignmentRework the above problem for a 10 m3 tank initially open to theatmosphere at 25oC and being filled from an air supply line at 90 psig,25oC, until the pressure inside the tank is 70 psig.


Chapter 6-1

Chapter 6: The Second Law of Thermodynamics

The second law of thermodynamics states that processes occur in a certaindirection, not in just any direction. Physical processes in nature canproceed toward equilibrium spontaneously:

Water flows down a waterfall. Gases expand from a high pressure to a low pressure. Heat flows from a high temperature to a low temperature.

Once it has taken place, a spontaneous process can be reversed, but it willnot reverse itself spontaneously. Some external inputs, energy, must beexpended to reverse the process. As it falls down the waterfall, water canbe collected in a water wheel, cause a shaft to rotate, coil a rope onto theshaft, and lift a weight. So the energy of the falling water is captured aspotential energy increase in the weight, and the first law ofthermodynamics is satisfied. However, there are losses associated withthis process (friction). Allowing the weight to fall, causing the shaft torotate in the opposite direction, will not pump all of the water back up thewaterfall. Spontaneous processes can proceed only in a particulardirection. The first law of thermodynamics gives no information aboutdirection; it states only that when one form of energy is converted intoanother, identical quantities of energy are involved regardless of thefeasibility of the process. We know by experience that heat flowsspontaneously from a high temperature to a low temperature. But heatflowing from a low temperature to a higher temperature with noexpenditure of energy to cause the process to take place would not violatethe first law.

The first law is concerned with the conversion of energy from one form toanother. Joule's experiments showed that energy in the form of heat couldnot be completely converted into work; however, work energy can becompletely converted into heat energy. Evidently heat and work are not


Chapter 6-2

completely interchangeable forms of energy. Furthermore, when energy istransferred from one form to another, there is often a degradation of thesupplied energy into a less “useful” form. We shall see that it is thesecond law of thermodynamics that controls the direction processes maytake and how much heat is converted into work. A process will not occurunless it satisfies both the first and the second laws of thermodynamics.

Some Definitions

To express the second law in a workable form, we need the followingdefinitions.

Heat (thermal) reservoir A heat reservoir is a sufficiently large system in stable equilibrium towhich and from which finite amounts of heat can be transferred withoutany change in its temperature.

A high temperature heat reservoir from which heat is transferred issometimes called a heat source. A low temperature heat reservoir to whichheat is transferred is sometimes called a heat sink.

Work reservoir A work reservoir is a sufficiently large system in stable equilibrium towhich and from which finite amounts of work can be transferredadiabatically without any change in its pressure.


Chapter 6-3

Thermodynamic cycle

A system has completed a thermodynamic cycle when the systemundergoes a series of processes and then returns to its original state, so thatthe properties of the system at the end of the cycle are the same as at itsbeginning.

Thus, for whole numbers of cycles

P P T T u u v v etcf i f i f i f i= = = =, , , , .

Heat Engine A heat engine is a thermodynamic system operating in a thermodynamiccycle to which net heat is transferred and from which net work isdelivered.

The system, or working fluid, undergoes a series of processes thatconstitute the heat engine cycle.

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


Chapter 6-4

Thermal Efficiency, η th

The thermal efficiency is the index of performance of a work-producingdevice or a heat engine and is defined by the ratio of the net work output(the desired result) to the heat input (the costs to obtain the desired result).

η th =Desired ResultRequired Input

For a heat engine the desired result is the net work done and the input isthe heat supplied to make the cycle operate. The thermal efficiency isalways less than 1 or less than 100 percent.

η thnet out

in

WQ

= ,

where

W W W

Q Qnet out out in

in net

, = −

≠

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


Chapter 6-5

Now apply the first law to the cyclic heat engine.

Q W U

W Q

W Q Q

net in net out

net out net in

net out in out

, ,

, ,

,

− =

=

= −

∆

The cycle thermal efficiency may be written as

η thnet out

in

in out

in

out

in

WQ

Q QQQQ

=

=−

= −

,

1

Cyclic devices such as heat engines, refrigerators, and heat pumps oftenoperate between a high-temperature reservoir at temperature TH and a low-temperature reservoir at temperature TL.

0 (Cyclic)


Chapter 6-6

The thermal efficiency of the above device becomes

η thL

H

QQ

= −1

Example 6-1

A steam power plant produces 50 MW of net work while burning fuel toproduce 150 MW of heat energy at the high temperature. Determine thecycle thermal efficiency and the heat rejected by the cycle to thesurroundings.

η thnet out

H

WQ

MWMW

=

= =

,

.50150

0 333 or 33.3%


Chapter 6-7

W Q Q

Q Q W

MW MWMW

net out H L

L H net out

,

,

= −

= −

= −=

150 50100

Heat Pump

A heat pump is a thermodynamic system operating in a thermodynamiccycle that removes heat from a low-temperature body and delivers heat toa high-temperature body. To accomplish this energy transfer, the heatpump receives external energy in the form of work or heat from thesurroundings.

While the name “heat pump” is the thermodynamic term used to describe acyclic device that allows the transfer of heat energy from a lowtemperature to a higher temperature, we use the terms “refrigerator” and“heat pump” to apply to particular devices. Here a refrigerator is a devicethat operates on a thermodynamic cycle and extracts heat from alow-temperature medium. The heat pump also operates on a thermodynamiccycle but rejects heat to the high-temperature medium.

The following figure illustrates a refrigerator as a heat pump operating in athermodynamic cycle.


Chapter 6-8

Coefficient of Performance, COP

The index of performance of a refrigerator or heat pump is expressed interms of the coefficient of performance, COP, the ratio of desired result toinput. This measure of performance may be larger than 1, and we want theCOP to be as large as possible.

COP =Desired ResultRequired Input

For the heat pump acting like a refrigerator or an air conditioner, theprimary function of the device is the transfer of heat from the low-temperature system.


Chapter 6-9

For the refrigerator the desired result is the heat supplied at the lowtemperature and the input is the net work into the device to make the cycleoperate.

COP QWR

L

net in

=,

Now apply the first law to the cyclic refrigerator.

( ) ( )

,

Q Q W UW W Q Q

L H in cycle

in net in H L

− − − = =

= = −

0 0∆

and the coefficient of performance becomes


Chapter 6-10

COP QQ QR

L

H L

=−

For the device acting like a “heat pump,” the primary function of thedevice is the transfer of heat to the high-temperature system. Thecoefficient of performance for a heat pump is

COP QW

QQ QHP

H

net in

H

H L

= =−,

Note, under the same operating conditions the COPHP and COPR arerelated by

COP COPHP R= + 1

Heat Pump and Air Conditioner Ratings

Heat pumps and air conditioners are rated using the SEER system. SEERis the seasonal adjusted energy efficiency (bad term for HP and A/Cdevices) rating. The SEER rating is the amount of heating (cooling) on aseasonal basis in Btu/hr per unit rate of power expended in watts, W.

The heat transfer rate is often given in terms of tons of heating or cooling.One ton equals 12,000 Btu/hr = 211 kJ/min.


Chapter 6-11

Second Law Statements

The following two statements of the second law of thermodynamics arebased on the definitions of the heat engines and heat pumps.

Kelvin-Planck statement of the second law

It is impossible for any device that operates on a cycle to receive heatfrom a single reservoir and produce a net amount of work.

The Kelvin-Planck statement of the second law of thermodynamics statesthat no heat engine can produce a net amount of work while exchangingheat with a single reservoir only. In other words, the maximum possibleefficiency is less than 100 percent.

η th < 100%

Heat engine that violates the Kelvin-Planck statement of the second law


Chapter 6-12

Clausius statement of the second law

The Clausius statement of the second law states that it is impossible toconstruct a device that operates in a cycle and produces no effect otherthan the transfer of heat from a lower-temperature body to a higher-temperature body.

Heat pump that violates the Clausius statement of the second law

Or energy from the surroundings in the form of work or heat has to beexpended to force heat to flow from a low-temperature medium to a high-temperature medium.

Thus, the COP of a refrigerator or heat pump must be less than infinity.

COP < ∞

A violation of either the Kelvin-Planck or Clausius statements of thesecond law implies a violation of the other. Assume that the heat engine


Chapter 6-13

shown below is violating the Kelvin-Planck statement by absorbing heatfrom a single reservoir and producing an equal amount of work W. Theoutput of the engine drives a heat pump that transfers an amount of heat QLfrom the low-temperature thermal reservoir and an amount of heat QH +QL to the high-temperature thermal reservoir. The combination of the heatengine and refrigerator in the left figure acts like a heat pump that transfersheat QL from the low-temperature reservoir without any external energyinput. This is a violation of the Clausius statement of the second law.

Perpetual-Motion Machines

Any device that violates the first or second law of thermodynamics iscalled a perpetual-motion machine. If the device violates the first law, it isa perpetual-motion machine of the first kind. If the device violates thesecond law, it is a perpetual-motion machine of the second kind.


Chapter 6-14

Reversible Processes

A reversible process is a quasi-equilibrium, or quasi-static, process with amore restrictive requirement.

Internally reversible process

The internally reversible process is a quasi-equilibrium process, which,once having taken place, can be reversed and in so doing leave no changein the system. This says nothing about what happens to the surroundingsabout the system.

Totally or externally reversible process

The externally reversible process is a quasi-equilibrium process, which,once having taken place, can be reversed and in so doing leave no changein the system or surroundings.

Irreversible Process

An irreversible process is a process that is not reversible.

All real processes are irreversible. Irreversible processes occur because ofthe following:

FrictionUnrestrained expansion of gases

Heat transfer through a finite temperature difference Mixing of two different substances

Hysteresis effects I2R losses in electrical circuits Any deviation from a quasi-static process


Chapter 6-15

The Carnot Cycle

French military engineer Nicolas Sadi Carnot (1769-1832) was among thefirst to study the principles of the second law of thermodynamics. Carnotwas the first to introduce the concept of cyclic operation and devised areversible cycle that is composed of four reversible processes, twoisothermal and two adiabatic.

The Carnot Cycle

Process 1-2 Reversible isothermal heat addition at hightemperature, TH > TL, to the working fluid in a piston-cylinder device that does some boundary work.

Process 2-3 Reversible adiabatic expansion during which thesystem does work as the working fluid temperaturedecreases from TH to TL.

Process 3-4 The system is brought in contact with a heatreservoir at TL < TH and a reversible isothermal heatexchange takes place while work of compression is doneon the system.

Process 4-1 A reversible adiabatic compression processincreases the working fluid temperature from TL to TH

TL =const.


Chapter 6-16

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

Carnot Principles

The second law of thermodynamics puts limits on the operation of cyclicdevices as expressed by the Kelvin-Planck and Clausius statements. Aheat engine cannot operate by exchanging heat with a single heat reservoir,and a refrigerator cannot operate without net work input from an externalsource.

Consider heat engines operating between two fixed temperature reservoirsat TH > TL. We draw two conclusions about the thermal efficiency ofreversible and irreversible heat engines, known as the Carnot principles.

(a) The efficiency of an irreversible heat engine is always less than theefficiency of a reversible one operating between the same tworeservoirs.

η ηth th Carnot< ,

(b) The efficiencies of all reversible heat engines operating between thesame two constant-temperature heat reservoirs have the sameefficiency.

TL =const.


Chapter 6-17

As the result of the above, Lord Kelvin in 1848 used energy as athermodynamic property to define temperature and devised a temperaturescale 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 thL

H

QQ

= −1

For the Carnot engine, this can be written as

η th L H L Hg T T f T T= = −( , ) ( , )1


Chapter 6-18

Considering engines A, B, and C

QQ

QQ

QQ

1

3

1

2

2

3

=

This looks like

f T T f T T f T T( , ) ( , ) ( , )1 3 1 2 2 3=

One way to define the f function is

f T T TT

TT

TT

( , ) ( )( )

( )( )

( )( )1 3

2

1

3

2

3

1

= =θθ

θθ

θθ

The simplest form of θ is the absolute temperature itself.

f T T TT

( , )1 33

1

=

The Carnot thermal efficiency becomes

η th revL

H

TT, = −1

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


Chapter 6-19

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

QQ

TT

L

H

L

H

=

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 low-temperature heat reservoirs, respectively. This result is only valid for heatexchange across a heat engine operating between two constant temperatureheat reservoirs. These results do not apply when the heat exchange isoccurring with heat sources and sinks that do not have constanttemperature.

The thermal efficiencies of actual and reversible heat engines operatingbetween the same temperature limits compare as follows:

η

η

η

ηth

th rev

th rev

th rev

< irreversible heat engine

reversible heat engine

impossible heat engine

,

,

,

=

>

RS|

T|

Reversed Carnot Device Coefficient of Performance

If the Carnot device is caused to operate in the reversed cycle, thereversible heat pump is created. The COP of reversible refrigerators andheat pumps are given in a similar manner to that of the Carnot heat engineas


Chapter 6-20

COP QQ Q Q

QT

T T TT

RL

H L H

L

L

H L H

L

=−

=−

=−

=−

1

1

1

1

COP QQ Q

QQ

QQ

TT T

TT

TT

HPH

H L

H

L

H

L

H

H L

H

L

H

L

=−

=−

=−

=−

1

1

Again, these are the maximum possible COPs for a refrigerator or a heatpump 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 asfollows:

COP

COP

COP

COPR

R rev

R rev

R rev

< irreversible refrigerator

reversible refrigerator

impossible refrigerator

,

,

,

=

>

RS|

T|


Chapter 6-21

A similar relation can be obtained for heat pumps by replacing all valuesof COPR by COPHP in the above relation.

Example 6-2

A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature heat reservoir at 652oC and rejects heat to a low-temperatureheat 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.

η th revL

H

TT

KK

or

,

( )( )

. .

= −

= −++

=

1

1 30 273652 273

0 672 67 2%

b.QQ

TT

KK

Q kJkJ

L

H

L

H

L

=

=++

=

==

( )( )

.

( . )

30 273652 273

0 328

500 0 328164

QL

WOUT

QH

TH = 652oC

TL = 30oC

HE


Chapter 6-22

Example 6-3

An inventor claims to have invented a heat engine that develops a thermalefficiency of 80 percent when operating between two heat reservoirs at1000 K and 300 K. Evaluate his claim.

η th revL

H

TT

KK

or

,

.

= −

= −

=

1

1 3001000

0 70 70%

The claim is false since no heat engine may be more efficient than aCarnot engine operating between the heat reservoirs.

Example 6-4

An inventor claims to have developed a refrigerator that maintains therefrigerated space at 2oC while operating in a room where the temperatureis 25 oC and has a COP of 13.5. Is there any truth to his claim?

COP QQ Q

TT T

KK

RL

H L

L

H L

=−

=−

=+−

=

( )( )

.

2 27325 2

1196

The claim is false since no refrigerator may have a COP larger than theCOP for the reversed Carnot device.

QL

WOUT

QH

TH = 1000 K

TL = 300 K

HE

QL

Win

QH

TH = 25oC

TL = 2oC

R


Chapter 6-23

Example 6-5

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

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

Q Q kJhH Lost= = 135000

COP QQ Q

TT T

KK

HPH

H L

H

H L

=−

=−

=+− −

=

( )( ( ))

.

21 27321 5

1131

QH

21 oC

HP

Win

QL

QLost

-5 oC


Chapter 6-24

Using the basic definition of the COP

COP QW

W QCOP

kJ h hs

kWkJ s

kW

HPH

net in

net inH

HP

=

=

=

=

, /. /

.

,

,

135 0001131

13600

1

3316


Chapter 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 newproperty called entropy, a quantitative measure of microscopic disorder fora system. Entropy is a measure of energy that is no longer available toperform useful work within the current environment. To obtain theworking definition of entropy and, thus, the second law, let's derive theClausius inequality.

Consider a heat reservoir giving up heat to a reversible heat engine, whichin turn gives up heat to a piston-cylinder device as shown below.


Chapter 7-2

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

E E EQ W W dE

in out c

R rev sys c

− =− + =

∆δ δ δ( )

where Ec is the energy of the combined system. Let Wc be the work doneby the combined system. Then the first law becomes

δ δ δ

δ δ

W W WQ W dE

c rev sys

R c c

= +

− =

If we assume that the engine is totally reversible, then

δ δ

δδ

QT

QT

Q T QT

R

R

R R

=

=

The total net work done by the combined system becomes

δδW T QT

dEc R c= −

Now the total work done is found by taking the cyclic integral of the incremental work.

W T QT

dEc R c= −z zδ

If the system, as well as the heat engine, is required to undergo a cycle,then


Chapter 7-3

dEcz = 0

and the total net work becomes

W T QTc R= z δ

If Wc is positive, we have a cyclic device exchanging energy with a singleheat reservoir and producing an equivalent amount of work; thus, theKelvin-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) andnot violate the Kelvin-Planck statement of the second law. Therefore,since TR > 0 (absolute temperature), we conclude

W T QTc R= ≤z δ 0

or

δQTz ≤ 0

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

δQT

netz ≤ 0

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


Chapter 7-4

Example 7-1

For a particular power plant, the heat added and rejected both occur atconstant 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 theamount of 1950 kJ at 20oC. Is the Clausius inequality satisfied and is thecycle reversible or irreversible?

δ

δ δ

QT

QT

QT

QT

QT

QT

QT

kJK

kJKkJKkJK

net

net

in

net

out

net

in

net

out

in

in

out

out

zz z

≤

FHGIKJ + FHG

IKJ ≤

FHGIKJ + FHG

IKJ ≤

FHGIKJ +

−FHGIKJ ≤

+FHG

IKJ +

−+

FHG

IKJ ≤

− ≤

− ≤

0

0

0

0

3150440 273

195020 273

0

4 418 6 655 0

2 237 0

( ) ( )

( . . )

.

Calculate the net work, cycle efficiency, and Carnot efficiency based on THand TL for this cycle.

(3150 1950) 1200net in outW Q Q kJ kJ= − = − =


Chapter 7-5

,

1200 0.381 38.1%3150

(20 273)1 1 0.589 58.9%(440 273)

netth

in

Lth Carnot

H

W kJ orQ kJ

T K orT K

η

η

= = =

+= − = − =

+

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 atconstant temperature; no other processes experience any heat transfer. Theheat is added in the amount of 3150 kJ at 440oC and is rejected in theamount of 1294.46 kJ at 20oC. Is the Clausius inequality satisfied and isthe cycle reversible or irreversible?

δ

δ δ

QT

QT

QT

QT

QT

kJK

kJKkJK

net

net

in

net

out

in

in

out

out

zz z

≤

FHGIKJ + FHG

IKJ ≤

FHGIKJ +

−FHGIKJ ≤

+FHG

IKJ +

−+

FHG

IKJ ≤

− =

0

0

0

3150440 273

1294 4620 273

0

4 418 4 418 0

( ).

( )

( . . )


Chapter 7-6

The Clausius inequality is satisfied. Since the cyclic integral is equal tozero, the cycle is made of reversible processes. What cycle can this be?

Calculate the net work and cycle efficiency for this cycle.

W Q Q kJ kJWQ

kJkJ

or

net in out

thnet

in

= − = − =

= = =

( . ) .. . .

3150 1294 46 18554185554

31500589 58 9%η

Definition of Entropy

Let’s take another look at the quantity

δQT

netz ≤ 0

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

δQT

netFHGIKJ =z

int rev

0

for internally reversible cycles. Thus we conclude that the equality in theClausius inequality holds for totally or just internally reversible cycles and


Chapter 7-7

the inequality for the irreversible ones.

To develop a relation for the definition of entropy, let us examine this lastequation more closely. Here we have a quantity whose cyclic integral iszero. Let us think for a moment what kind of quantities can have thischaracteristic. We know that the cyclic integral of work is not zero. (It is agood thing that it is not. Otherwise, heat engines that work on a cycle suchas steam power plants would produce zero net work.) Neither is the cyclicintegral of heat.

Now consider the volume occupied by a gas in a piston-cylinder deviceundergoing a cycle, as shown below.

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

dVb gz = 0


Chapter 7-8

We see that the cyclic integral of a property is zero. A quantity whosecyclic integral is zero depends only on the state and not on the processpath; thus it is a property. Therefore the quantity (δQnet/T)int rev must be aproperty.

In-class Example

Consider the cycle shown below composed of two reversible processes Aand B. Apply the Clausius inequality for this cycle. What do youconclude about these two integrals?

δ δQT

and QT

net

along path A

net

along path B

FHGIKJ

FHGIKJz z1

2

1

2

int rev int rev

Apply the Clausius inequality for the cycle made of two internallyreversible processes:

δQT

netFHGIKJ =z

int rev

0

B

A

2

V

P

1

A cycle composed of two reversible processes.


Chapter 7-9

You should find: δ δQ

TQT

net

along path A

net

along path B

FHGIKJ = FHG

IKJz z1

2

1

2

int rev int rev

Since the quantity (δQnet/T)int rev is independent of the path and must be aproperty, we call this property the entropy S. The entropy change occurring during a process is related to the heattransfer and the temperature of the system. The entropy is given thesymbol S (kJ/K), and the specific entropy is s (kJ/kg⋅K).

The entropy change during a reversible process, sometimes called aninternally reversible process, is defined as

dS QT

S S QT

net

net

=

− = z

δ

δint rev

int rev2 1 1

2

Consider the cycle 1-A-2-B-1, shown below, where process A is arbitrarythat is, it can be either reversible or irreversible, and process B is internallyreversible.

B

A

2

V

P

1

A cycle composed of reversible and irreversible processes.


Chapter 7-10

δ

δ δ

QT

QT

QT

net

net

along A

net

along B

FHGIKJ ≤

+ FHGIKJ ≤

zz z

int rev

int rev2

1

0

01

2

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

δQT

S Snet1

2

1 2 0z + − ≤

or

S S QT

net2 1 1

2− ≥ z δ

In general the entropy change during a process is defined as

dS QT

net≥δ

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

Consider the effect of heat transfer on entropy for the internally reversiblecase.


Chapter 7-11

dS QT

net=δ

Which temperature T is this one? If

δδδ

Q then dSQ then dSQ then dS

net

net

net

> >= =< <

0 00 00 0

,,,

This last result shows why we have kept the subscript net on the heattransfer Q. It is important for you to recognize that Q has a sign dependingon the direction of heat transfer. The net subscript is to remind us that Q ispositive 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 heattransfer in a reversible process.

From the above, we see that for a reversible, adiabatic process

dSS S

==

0

2 1

The reversible, adiabatic process is called an isentropic process.

Entropy change is caused by heat transfer and irreversibilities. Heattransfer to a system increases the entropy; heat transfer from a systemdecreases it. The effect of irreversibilities is always to increase theentropy. In fact, a process in which the heat transfer is out of the systemmay be so irreversible that the actual entropy change is positive. Frictionis one source of irreversibilities in a system.


Chapter 7-12

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

∆S S S QT

kJKsys

net= − ≥ FHGIKJz2 1 1

2 δ

Here, the inequality is to remind us that the entropy change of a systemduring an irreversible process is always greater than δQ T/

1

2z , called theentropy transfer. That is, some entropy is generated or created during anirreversible process, and this generation is due entirely to the presence ofirreversibilities. The entropy generated during a process is called entropygeneration and is denoted as Sgen.

We can remove the inequality by noting the following

∆S S S QT

S kJKsys

netgen= − = + FHGIKJz2 1 1

2 δ

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

The integral δQ T/1

2z is performed by applying the first law to the process toobtain the heat transfer as a function of the temperature. The integration isnot easy to perform, in general.


Chapter 7-13

Definition of Second Law of Thermodynamics

Now consider an isolated system composed of several subsystemsexchanging energy among themselves. Since the isolated system has noenergy transfer across its system boundary, the heat transfer across thesystem boundary is zero.

Applying the definition of entropy to the isolated system

∆S QTisolated

net≥ z δ1

2

The total entropy change for the isolated system is

0isolatedS∆ ≥

This equation is the working definition of the second law ofthermodynamics. The second law, known as the principle of increase ofentropy, is stated as

The total entropy change of an isolated system during a processalways increases or, in the limiting case of a reversible process,remains constant.

0, adiabatic


Chapter 7-14

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

S S S Sgen total sys surr= = + ≥∑∆ ∆ ∆ 0

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

Thus, the entropy generated or the total entropy change (sometimes calledthe entropy change of the universe or net entropy change) due to theprocess of this isolated system is positive (for actual processes) or zero (forreversible processes). The total entropy change for a process is the amountof entropy generated during that process (Sgen), and it is equal to the sum ofthe entropy changes of the system and the surroundings. The entropychanges of the important system (closed system or control volume) and itssurroundings do not both have to be positive. The entropy for a givensystem (important or surroundings) may decrease during a process, but the


Chapter 7-15

sum of the entropy changes of the system and its surroundings for anisolated system can never decrease.

Entropy change is caused by heat transfer and irreversibilities. Heattransfer to a system increases the entropy, and heat transfer from a systemdecreases it. The effect of irreversibilities is always to increase theentropy.

The increase in entropy principle can be summarized as follows:

S S sgen Total=>=<

RS|T|

∆000

Irreversible processesReversible processeImpossible 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 theconservation of entropy principle. The entropy of the universe iscontinuously increasing.

3. The performance of engineering systems is degraded by the presence ofirreversibilities, and entropy generation is a measure of the magnitudesof the irreversibilities present during that process.


Chapter 7-16

Heat Transfer as the Area under a T-S Curve

For the reversible process, the equation for dS implies that

dS QT

Q TdS

net

net

=

=

δ

δ

or the incremental heat transfer in a process is the product of thetemperature and the differential of the entropy, the differential area underthe process curve plotted on the T-S diagram.

Q TdSnet = z12

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


Chapter 7-17

Isothermal, Reversible Process

For an isothermal, reversible process, the temperature is constant and theintegral to find the entropy change is readily performed. If the system hasa constant temperature, T0, the entropy change becomes

∆S S S QT

QT

net net

o

= − = =z2 1 1

2 δ

For a process occurring over a varying temperature, the entropy changemust be found by integration over the process.

Adiabatic, Reversible (Isentropic) Process

For an adiabatic process, one in which there is no heat transfer, the entropychange is

∆

∆

S S S QT

S S S

net= − ≥

= − ≥

z2 1 1

2

2 1 0

δ

If the process is adiabatic and reversible, the equality holds and theentropy change is

∆S S SS S

= − ==

2 1

2 1

0

or on a per unit mass basis

s Sm

s s

=

=2 1

0, adiabatic


Chapter 7-18

The adiabatic, reversible process is a constant entropy process and iscalled isentropic. As will be shown later for an ideal gas, the adiabatic,reversible process is the same as the polytropic process where thepolytropic exponent n = k = Cp/Cv.

The principle of increase of entropy for a closed system exchanging heatwith its surroundings at a constant temperature Tsurr is found by using theequation for the entropy generated for an isolated system.

S S S SS S S

SQ

T

gen total sys surr

sys sys

surrnet surr

surr

= = + ≥

= −

=

∑

∑

∆ ∆ ∆

∆

∆

0

2 1( )

,

S S m s sQ

Tgen total sysnet surr

surr

= = − + ≥∆ ( ) ,2 1 0

where

Q Q Q Qnet surr net sys out sys out sys, , , ,( )= − = − − =0

Qout, sys

A general closed system (acup of coffee) exchangingheat with its surroundings

SurroundingsTsurr

SystemBoundary


Chapter 7-19

Effect of Heat Transfer on Entropy

Let's apply the second law to the following situation. Consider the transferof heat from a heat reservoir at temperature T to a heat reservoir attemperature T - ∆T > 0 where ∆T > 0, as shown below.

The second law for the isolated system composed of the two heatreservoirs is

S S S SS S S S

gen total sys surr

gen total HR T HR T T

= = + ≥

= = +∑

−

∆ ∆ ∆

∆ ∆ ∆ ∆

0

@ @

In general, if the heat reservoirs are internally reversible

∆

∆

∆∆∆

S QT

SQT

SQ

T T

sysnet

HR T

HR T T

=

=−

=+

−

z

−

δ1

2

@

@

Q

HRatT

HRatT-∆T

Two heat reservoirsexchanging heat over afinite temperaturedifference


Chapter 7-20

S SQT

QT Tgen Total= =

−+

+

−∆

∆

S SQT

TT Tgen Total= =−

LNM

OQP∆

∆∆( )

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 of1000 kJ of heat energy from a heat reservoir at 1000 K to a heat reservoirat 500 K.

The second law for the isolated system is

Q=1000 kJ

HRatT=1000 K

HRatT-∆T = 500K 0 1 2 S, kJ/K

1000 K

500 K

T Areas= 1000 kJ


Chapter 7-21

S SQT

QT T

kJK

kJK

kJK

kJK

gen Total= =−

++

−

=−

+

= − +

=

∆∆

10001000

1000500

1 2

1

( )

What happens when the low-temperature reservoir is at 750 K?

The effect of decreasing the ∆T for heat transfer is to reduce the entropygeneration or total entropy change of the universe due to the isolatedsystem and the irreversibilities associated with the heat transfer process.Third Law of Thermodynamics

The third law of thermodynamics states that the entropy of a purecrystalline substance at absolute zero temperature is zero. This lawprovides an absolute reference point for the determination of entropy. Theentropy 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 familiarproperties (P,v,T) through the Tds relations. These relations come fromthe analysis of a reversible closed system that does boundary work and hasheat added. Writing the first law for the closed system in differential formon a per unit mass basis


Chapter 7-22

δ δ

δ

δ

Q W dU

Q T dS

W P dV

TdS P dV dU

int rev int rev, out

int rev

int rev, out

=−

=

=

− =

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 asthe starting point in developing entropy-change relations for processes.The successful use of Tds relations depends on the availability of propertyrelations. Such relations do not exist in an easily used form for a generalpure substance but are available for incompressible substances (liquids,

δWint rev, outδQint rev

System used to find expressions for ds


Chapter 7-23

solids) and ideal gases. So, for the general pure substance, such as waterand the refrigerants, we must resort to property tables to find values ofentropy and entropy changes.

The temperature-entropy and enthalpy-entropy diagrams for water areshown below.


Chapter 7-24

Shown above are the temperature-entropy and enthalpy-entropy diagramsfor water. The h-s diagram, called the Mollier diagram, is a useful aid insolving steam power plant problems.


Chapter 7-25

Example 7-4

Find the entropy and/or temperature of steam at the following states:

P T Region s kJ/(kg K)5 MPa

120o

C

1 MPa 50oC

1.8MPa 400o

C

40 kPa 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 itchanges phase from saturated liquid to saturated vapor when the pressureis 0.1 MPa and constant. Why is the entropy change positive for thisprocess?

System: The water contained in the system (a piston-cylinder device)

Steam

s

T


Chapter 7-26


Process and Process Diagram: Constant pressure (sketch the processrelative to the saturation lines)


Using the definition of entropy change, the entropy change of the waterper mass is

2 1

6.0562

g f fgs s s s s s

kJkg 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 processis isentropic, find the final temperature, the final enthalpy of the steam,and the turbine work.

System: The control volume formed by the turbine

Control surface

1

2

Wout

s

T


Chapter 7-27


Process and Process Diagram: Isentropic (sketch the process relative tothe saturation lines on the T-s diagram)

Conservation Principles: Assume: steady-state, steady-flow, one entrance, one exit, neglect KE and

PE


m m m1 2= =

First Law or conservation of energy:

The process is isentropic and thus adiabatic and reversible; therefore Q =0. The conservation of energy becomes

1 1 2 2

in out

out

E E

m h m h W

=

= +

Since the mass flow rates in and out are equal, solve for the work done perunit mass


Chapter 7-28

( )

E EW m h m h

m h h

w Wm

h h

in out

out

out

=

= −= −

= = −

1 1 2 2

1 2

1 2

Now, let’s go to the steam tables to find the h’s.

11

11

3698.61

600 8.0311o

kJhP MPa kg

kJT C skg 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 thestate.


Chapter 7-29

2 2

22

2 2

8.0311 0.6492 0.9847.4996

191.8 (0.984)(2392.1)

2545.6

f fg

f

fg

f fg

s s x ss s

xs

h h x h

kJkg

= +

−=

−= =

= +

= +

=

Since state 2 is in the two-phase region, T2 = Tsat at P2 = 45.81oC.

1 2

(3698.6 2545.6)

1153

w h hkJkg

kJkg

= −

= −

=

Entropy Change and Isentropic Processes

The entropy-change and isentropic relations for a process can be sum-marized as follows:

1. Pure substances: Any process: ∆s s s= −2 1 (kJ/kg⋅K)


Chapter 7-30

Isentropic process: s s2 1=

2. Incompressible substances (Liquids and Solids):

ds duT

PT

dv= +

The change in internal energy and volume for an incompressible substanceis

du C dTdv

=≅ 0

The entropy change now becomes

ds C dTT

s C T dTT

= +

= z0

1

2∆

( )

If the specific heat for the incompressible substance is constant, then theentropy change is

Any process: s s C TTav2 1

2

1

− = ln (kJ/kg⋅K)

Isentropic process: T T2 1=

3. Ideal gases:


Chapter 7-31

a. Constant specific heats (approximate treatment):

Any process: (can you fill in the steps?)

s s C TT

R vvv av2 1

2

1

2

1

− = +, ln ln (kJ/kg⋅K)

and (can you fill in the steps?)

2 22 1 ,

1 1

ln lnp avT Ps s C RT P

− = − (kJ/kg⋅K)

Or, on a unit-mole basis,

s s C TT

R vvv av u2 1

2

1

2

1

− = +, ln ln (kJ/kmol⋅K)

and2 2

2 1 ,1 1

ln lnp av uT Ps s C RT P

− = − (kJ/kmol⋅K)


Chapter 7-32

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

TT

vvs const

k

2

1

1

2

1FHGIKJ =

FHGIKJ=

−

.

TT

PPs const

k k

2

1

2

1

1FHGIKJ =

FHGIKJ=

−

.

( ) /

2 1

1 2.

k

s const

P vP v

=

=

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

b. Variable specific heats (exact treatment):

From Tds = dh - vdP, we obtain

2 21

1

( )lnpC T Ps dT R

T P∆ = −∫

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

C TT

dTC T

TdT

C TT

dT

C TT

dTC T

TdT

p p

T

T p

T

T

p

T

T p

T

T

ref

ref

ref ref

( ) ( ) ( )

( ) ( )1

2

1

2

2 1

z z zz z

= +

= −


Chapter 7-33

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

sC T

TdT

sC T

TdT

o p

T

T

o p

T

T

ref

ref

1

2

1

2

=

=

zz

( )

( )

Therefore, for any process:

s s s s R PP

o o2 1 2 1

2

1

− = − − ln (kJ/kg⋅K)

or

s s s s R PP

o ou2 1 2 1

2

1

− = − − ln (kJ/kmol⋅K)

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

Isentropic process: ∆s = 0


Chapter 7-34

s s R PP

o o2 1

2

1

= + ln (kJ/kg⋅K)

If we are given T1, P1, and P2, we find so1 at T1, calculate so

2, and thendetermine from the tables T2, u2, and h2.

When air undergoes an isentropic process when variable specific heat dataare required, there is another approach to finding the properties at the endof the isentropic process. Consider the entropy change written as

2 21

1

( )lnpC T Ps dT R

T P∆ = −∫

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

PP

EXPR

C TT

dTref s const

p

T

T

ref

FHGIKJ =

FHG

IKJ

=

z1 ( ' )'

'

We define the relative pressure Pr as the above pressure ratio. Pr is thepressure ratio necessary to have an isentropic process between thereference temperature and the actual temperature and is a function of theactual temperature. This parameter is a function of temperature only andis found in the air tables, Table A-17. The relative pressure is notavailable for other gases in this text.

P EXPR

C TT

dTr s constp

T

T

refb g = =

FHG

IKJz1 ( ' )

''


Chapter 7-35

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

22 2

1 1 1. .

//

ref r

ref rs const s const

P PP PP P P P

= =

= =

There is a second approach to finding data at the end of an ideal gasisentropic process when variable specific heat data are required. Considerthe following entropy change equation set equal to zero.

From Tds = du + Pdv, we obtain for ideal gases

∆s C TT

dT R vv

v= +z ( ) ln1

22

1

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

vv

EXPR

C TT

dTref s const

vT

T

ref

FHGIKJ = −FHG

IKJ

=

z1 ( ' )'

'

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

v EXPR

C TT

dTr s constv

T

T

refb g = = −FHG

IKJz1 ( ' )

''


Chapter 7-36

vv

v vv v

vvs const

ref

ref s const

r

r

2

1

2

1

2

1

FHGIKJ =

FHG

IKJ =

= =. .

//

Extra Assignment

For an ideal gas having constant specific heats and undergoing apolytropic process in a closed system, Pvn = constant, with n = k, find theheat transfer by applying the first law. Based on the above discussion ofisentropic processes, explain your answer. Compare your results to thisproblem 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 ofwater at a temperature of 30oC. If the mass of the aluminum is 0.5 kg, findthe final equilibrium temperature of the aluminum and water, the entropychange of the aluminum and the water, and the total entropy change of theuniverse because of this process. Before we work the problem, what doyou think the answers ought to be? Are entropy changes going to bepositive or negative? What about the entropy generated as the processtakes place?

System: Closed system including the aluminum and water.

Water

AL

Tank insulatedboundary


Chapter 7-37

Property Relation: ?

Process: Constant volume, adiabatic, no work energy exchange betweenthe aluminum and water.


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

Q W UU U

system

water AL

− =

− = +

∆

∆ ∆0 0

Using the solid and incompressible liquid relations, we have

m C T T m C T Twater water water AL AL AL( ) ( )2 1 2 1 0− + − =

But at equilibrium, T2,AL = T2,water = T2

T m C T m C Tm C m C

kg kJ kg K K kg kJ kg K Kkg kJ kg K kg kJ kg K

K

water water water AL AL AL

water water AL AL

water water AL AL

water water AL AL

21 1

10 4 18 303 0 5 0 941 37310 418 0 5 0 941

3038

=++

=⋅ + ⋅

⋅ + ⋅=

( ) ( )

( . / )( ) . ( . / )( )( . / ) . ( . / )

.

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


Chapter 7-38

S S S Sgen total water AL= = + ≥∆ ∆ ∆ 0

Using the entropy change equation for solids and liquids,

∆S m C TT

kg kJkg K

KK

kJK

AL AL ALAL

=

=⋅ +FHG

IKJ

= −

ln

. ( . ) ln .( )

.

,

2

1

0 5 0 941 3038100 273

0 0966

∆S m C TT

kg kJkg K

KK

kJK

water water waterwater

=

=⋅ +FHG

IKJ

= +

ln

( . ) ln .( )

.

,

2

1

10 4 177 303830 273

01101

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

kJK

kJK

gen total water AL= = +

= −

= +

∆ ∆ ∆

( . . )

.

01101 0 0966

0 0135

Why is Sgen or ∆STotal positive?


Chapter 7-39

Example 7-8

Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closedsystem until its pressure and temperature are 2 MPa and 800 K,respectively. Assuming ideal gas behavior, find the entropy change of thecarbon dioxide by first assuming constant specific heats and then assumingvariable specific heats. Compare your results with the real gas dataobtained from the EES software.

a. Assume the Table A-2(a) data at 300 K are adequate; then Cp = 0.846kJ/kg⋅K and R = 0.1889 kJ/kg-K.

s s C TT

R PP

kJkg K

KK

kJkg K

kPakPa

kJkg K

p av2 12

1

2

1

0 846 800400

01889 200050

01104

− = −

=⋅FHGIKJ − ⋅

FHG

IKJ

= −⋅

, ln ln

. ln . ln

.

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

s ss sM

R PP

kJ kmol Kkg kmol

kJkg K

kPakPa

kJkg K

To

To

CO2 1

2

1

2 1

2

257 408 225 22544

01889 200050

0 0346

− =−F

HGIKJ −

=− ⋅F

HGIKJ − ⋅

FHG

IKJ

= +⋅

ln

( . . ) //

. ln

.


Chapter 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 constantat the average of the specific heats for the temperatures. Then Cp =1.054 kJ/kg⋅K (see Table A-2(b)).

s s C TT

R PP

kJkg K

KK

kJkg K

kPakPa

kJkg K

p av2 12

1

2

1

1054 800400

01889 200050

0 0337

− = −

=⋅FHGIKJ − ⋅

FHG

IKJ

= +⋅

, ln ln

. ln . ln

.

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 - 41

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

a. Constant specific heats, isentropic process

TT

PPs const

k k

2

1

2

1

1FHGIKJ =

FHGIKJ=

−

.

( ) /

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

T T PP

K

K C

k k

o

2 12

1

1

1 4 1 1 417 273 8

525 3 252 3

=FHGIKJ

= +

= =

−

−

( ) /

( . ) / .( )

. .b g

b. Variable specific heat method

PP

P PP P

PP

s const

ref

ref s const

r

r

2

1

2

1

2

1

FHGIKJ =

FHG

IKJ =

= =. .

//

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

P P PPr r2 1

2

1

12311 9 8488

=

= =. (8) .Interpolating in the air table at this value of Pr2, givesT2 = 522.4 K = 249.4oC


Chapter 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

s s R PPkJ

kg KkJ

kg KkJ

kg K

o o2 1

2

1

166802 0 287 8

2 26482

= +

=⋅

+⋅

=⋅

ln

. . ln

.

b g

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_1T_2 = TEMPERATURE(Air, P=800, s=s_2)The solution is:

s_1 = 5.668 kJ/kg⋅Ks_2 = 5.668 kJ/kg⋅KT_2 = 249.6oC


Chapter 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.

s s C TT

R PP

kJkg K

KK

kJkg K

MPaMPa

kJkg K

p av2 12

1

2

1

1005 227 27327 273

0 287 0 501

0 0507

− = −

=⋅

++

FHG

IKJ − ⋅

FHG

IKJ

= +⋅

, ln ln

. ln ( )( )

. ln ..

.

b.

s s C TT

R PP

kJkg K

KK

kJkg K

MPaMPa

kJkg K

p av2 12

1

2

1

1005 180 27327 273

0 287 0 501

0 0484

− = −

=⋅

++

FHG

IKJ − ⋅

FHG

IKJ

= −⋅

, ln ln

. ln ( )( )

. ln ..

.


Chapter 7 - 44

c.

T T PP

K MPaMPa

K C

k k

o

2 12

1

1

1 4 1 1 4

27 273 0 501

475 4 202 4

=FHGIKJ

= + FHG

IKJ

= =

−

−

( )/

( . ) / .

( ) ..

. .The T-s plot is

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

cb

a

1

s

T

P1

P22


Chapter 7 - 45

Example 7-11

Nitrogen expands isentropically in a piston cylinder device from atemperature of 500 K while its volume doubles. What is the finaltemperature of the nitrogen, and how much work did the nitrogen doagainst the piston, in kJ/kg?

System: The closed piston-cylinder device

Property Relation: Ideal gas equations, constant properties

Process and Process Diagram: Isentropic expansion


Second law:

s = const.

W

Nitrogensystemboundary 2

1

s

T P1

P2

2

1

v

P

P1

P2


Chapter 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, thetemperatures are related by

T T vv

K

K C

k

o

2 11

2

1

1 4 1

500 12

378 9 105 9

=FHGIKJ

= FHGIKJ

= =

−

−.

. .

Why did the temperature decrease?

First law, closed system:

Note, for the isentropic process (reversible, adiabatic); the heat transfer iszero. The conservation of energy for this closed system becomes

E E EW UW U

in out− =− =

= −

∆∆∆

Using the ideal gas relations, the work per unit mass is


Chapter 7 - 47

W mC T T

w Wm

C T T

kJkg K

K

kJkg

v

v

= − −

= = − −

= −⋅

−

=

( )

( )

. ( . )

.

2 1

2 1

0 743 378 9 500

90 2

Why is the work positive?

Extra Assignment

For the isentropic process Pvk = constant. Use the definition of boundarywork to show that you get the same result as the last example. That is,determine the boundary work and show that you obtain the sameexpression 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 tothe air at 800 K and rejected by the air at 300 K. At the beginning of theheat addition process, the pressure is 0.8 MPa and during heat addition thevolume triples.

(a) Calculate the net cycle work assuming air is an ideal gas with constantspecific heats.

(b) Calculate the amount of work done in the isentropic expansion process.


Chapter 7 - 48

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

Property Relation: Ideal gas equations, constant properties.

Process and Process Diagram: Constant temperature heat addition.


a.Apply the first law, closed system, to the constant temperature heataddition process, 1-2.

W

Q

Airsystemboundary

Carnot Cycle

4 QL

1

s

T2

3

QH

TH

TL


Chapter 7 - 49

Q W UmC T T

Q W

net net

v

net net

, ,

, ,

( )12 12 12

2 1

12 12

0− =

= − ==

∆

So for the ideal gas isothermal process,

W W W

PdV

mRT dVV

mRT VV

kg kJkg K

K

kJ

net other b, , ,

ln

( . )(800 ) ln( )

.

12 12 12

1

2

1

2

2

1

1 0 287 3

252 2

= +

=

=

=FHGIKJ

=⋅

=

zz

But

,12

252.2net H

H

Q QQ kJ

=

=


Chapter 7 - 50

The cycle thermal efficiency is

η thnet cycle

H

WQ

= ,

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

3001 1800

0.625

Lth

H

T KT K

η = − = −

=

The net work done by the cycle is

W Q

kJkJ

net cycle th H,

. ( . ).

=

==

η

0 625 252 2157 6

b. Apply the first law, closed system, to the isentropic expansion process, 2-3.

But the isentropic process is adiabatic, reversible; so, Q23 = 0.


Chapter 7 - 51

E E EW U

W U

in out− =− =

= −

∆∆∆23 23

Using the ideal gas relations, the work per unit mass is

W mC T T

kg kJkg K

K

kJ

v23 3 2

1 0 718 300 800

359 0

= − −

= −⋅

−

=

( )

( )( . )( )

.

This is the work leaving the cycle in process 2-3.

c.Using equation (6-34)

∆s C TT

R PPp34

4

3

4

3

=FHGIKJ −FHGIKJln ln

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 - 52

1 2

1 1 2 2

1 2

12 1

131(800 )3

266.7

m mPV PVT T

VP PV

kPa

kPa

=

=

=

=

=

Consider process 2-3 where s3 = s2.

TT

PP

P P TT

kPa Kk

kPa

k k

k k

3

2

3

2

1

3 23

2

1

1 4 1 4 1

266 7 300800

8 613

=FHGIKJ

=FHGIKJ

= FHGIKJ =

−

−

−

( )/

/( )

. /( . )

. .


Chapter 7 - 53

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

/( 1)

44 1

1

1.4 /(1.4 1)3008000800

25.834

k kTP PT

KkPak

kPa

−

−

=

=

=

Now,

∆S m C TT

R PP

mR PP

kg kJkg K

kPakPa

kJK

p=FHGIKJ −FHGIKJ

LNM

OQP

= −FHGIKJ

=⋅FHG

IKJ

= −

ln ln

ln

( )( . ) ln ..

.

4

3

4

3

4

3

1 0 287 258348 613

0 315

Extra Problem

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


Chapter 7 - 54

Reversible Steady-Flow Work

Isentropic, Steady Flow through Turbines, Pumps, and Compressors

Consider a turbine, pump, compressor, or other steady-flow controlvolume, work-producing device.

The general first law for the steady-flow control volume is

( ) ( )

E E


in out

net iinlets

ii

i net eexits

ee

e

=

+ + + = + + +∑ ∑2 2

2 2

For a one-entrance, one-exit device undergoing an internally reversibleprocess, this general equation of the conservation of energy reduces to, ona unit of mass basis

rev rev

rev

rev

w q dh dke dpeBut q T ds

w T ds dh dke dp

δ δδδ

= − − −== − − −

Using the Gibb’s second equation, this becomes

rev

dh T ds v dPw v dP dke dpeδ

= += − − −

Integrating over the process, this becomes


Chapter 7 - 55

w v dP ke pe kJkgrev = − − −FHGIKJz12 ∆ ∆

Neglecting changes in kinetic and potential energies, reversible workbecomes

w v dP kJkgrev = −FHGIKJz12

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

w v dP ke pe kJkgrev in, = + +FHGIKJz ∆ ∆

1

2

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. Toperform the integral, the pressure-volume relation must be known for theprocess.

Compressor and Pump

Since the fluid pressure rises as the fluid flows through the compressor orpump, 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 termcompressor is usually applied to the compression of a gas. The termpump is usually applied when increasing the pressure of a liquid.


Chapter 7 - 56

Pumping an incompressible liquid

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

w v dP ke pe kJkg

v P ke pe

rev in, = + +FHGIKJ

= + +

z ∆ ∆

∆ ∆ ∆

1

2

For the steady-flow of an incompressible fluid through a device thatinvolves no work interactions (such as nozzles or a pipe section), the workterm is zero, and the equation above can be expressed as the well-knowBernoulli equation in fluid mechanics.

v P P ke pe( )2 1 0− + + =∆ ∆

Extra Assignment

Using the above discussion, find the turbine and compressor work per unitmass flow for an ideal gas undergoing an isentropic process, where thepressure-volume relation is Pvk = constant, between two temperatures, T1and T2. Compare your results with the first law analysis of Chapter 5 forcontrol volumes.


Chapter 7 - 57

Example 7-13

Saturated liquid water at 10 kPa leaves the condenser of a steam powerplant and is pumped to the boiler pressure of 5 MPa. Calculate the workfor an isentropic pumping process.

a. From the above analysis, the work for the reversible process can beapplied to the isentropic process (it is left for the student to show this istrue) as

( )W mv P PC = −1 2 1

Here at 10 kPa, v1 = vf = 0.001010 m3/kg.

The work per unit mass flow is

w Wm

v P P

mkg

kPa kJm kPa

kJkg

CC= = −

= −

=

( )

. (5000 )

.

1 2 1

3

30 001010 10

5 04

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

− = −

− − = −

( )( ) ( )

W m h hW m h h

net

C

2 1

2 10


Chapter 7 - 58

From the saturation pressure table,

11

1

191.8110

. 0.6492

kJhP kPa kg

kJSat Liquid skg K

== = ⋅

Since the process is isentropic, s2 = s1. Interpolation in the compressedliquid tables gives

2

22 1

5197.42

0.6492

P MPakJhkJs s kg

kg K

= == = ⋅

The work per unit mass flow is

2 1( )

(197.42 191.81)

5.61

CC

Ww h hm

kJkg

kJkg

= = −

= −

=

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


Chapter 7 - 59

Turbine, Compressor (Pump), and Nozzle Efficiencies

Most steady-flow devices operate under adiabatic conditions, and theideal process for these devices is the isentropic process. The parameterthat describes how a device approximates a corresponding isentropicdevice is called the isentropic or adiabatic efficiency. It is defined forturbines, compressors, and nozzles as follows:

Turbine:

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

ηTa

s

Actual turbine workIsentropic turbine work

ww

= =

1 2

1 2

aT

s

h hh h

η −≅

−

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


Chapter 7 - 60

Compressor and Pump:

The isentropic work is the minimum possible work that the adiabaticcompressor requires; therefore, the actual work is greater than theisentropic work. Since efficiencies are defined to be less than 1, thecompressor isentropic efficiency is defined as

ηCs

a

Isentropic compressor workActual compressor work

ww

= =

ηCs

a

h hh h

≅−−

2 1

2 1

Well-designed compressors have isentropic efficiencies in the range from75 to 85 percent.

Review the efficiency of a pump and an isothermal compressor on yourown.

T2P2Compressor

or pump

T1P1

WC


Chapter 7 - 61

Nozzle:

The isentropic kinetic energy at the nozzle exit is the maximum possiblekinetic energy at the nozzle exit; therefore, the actual kinetic energy at thenozzle exit is less than the isentropic value. Since efficiencies are definedto be less than 1, the nozzle isentropic efficiency is defined as

ηNActual KE at nozzle ex

Isentropic KE at nozzle e=

For steady-flow, no work, neglecting potential einlet kinetic energy, the conservation of energy

h h Va

a1 2

22

2= +

Nozzle

T1P1

V1

T2P2

V2

V a22

2V 2

a

s

itxit

VV

= 22

22

22

//

nergies, and neglecting thefor the nozzle is

s2

2


Chapter 7 - 62

The nozzle efficiency is written as

ηNa

s

h hh h

≅−−

1 2

1 2

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

Example 7-14

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

(0.9)(1153.0 ) 1037.7

aT

s

a T s

wActual turbine workIsentropic turbine work w

kJ kJw wkg kg

η

η

= =

= = =

ηTa

s

h hh h

≅−−

1 2

1 2

Now to find the actual exit state for the steam.

From Example 7-6, steam enters the turbine at 1 MPa, 600oC, and expandsto 0.01 MPa.


Chapter 7 - 63

From the steam tables at state 1

11

11

3698.61

600 8.0311o

kJhP MPa kg

kJT C skg K

==

= = ⋅

At the end of the isentropic expansion process, see Example 7-6.

22

2 12

0.01 2545.68.0311

0.984

s

ss

kJP MPa hkgkJs s

xkg K

= == = =⋅

The actual turbine work per unit mass flow is (see Example 7-6)

1 2

2 1

(3698.6 1037.7)

2660.9

a a

a a

w h hh h w

kJkg

kJkg

= −= −

= −

=

For the actual turbine exit state 2a, the computer software gives

P MPa

h kJkg

T CSuperheateda

ao2

2

2

0 01

2660 986 85

=

=

UV|W|

=RST.

..


Chapter 7 - 64

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

2 1 1 2( )

3698.6 (0.9)(3698.6 2545.6)

2660.9

a T sh h h hkJ kJkg kgkJkg

η= − −

= − −

=

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 acompressor isentropic efficiency of 80 percent.

System: The compressor control volume

Property Relation: Ideal gas equations, assume constant properties.

Compressoror pump

T1P1

WC

T2P2


Chapter 7 - 65

Process and Process Diagram: First, assume isentropic, steady-flow andthen apply the compressor isentropicefficiency to find the actual work.


For the isentropic case, Qnet = 0. Assuming steady-state, steady-flow, andneglecting changes in kinetic and potential energies for one entrance, oneexit, the first law is

E Em h W m h

in out

Cs s

=

+ =1 1 2 2

The conservation of mass gives

m m m1 2= =

The conservation of energy reduces to

2a2s

1

s

T P2

P1


Chapter 7 - 66

( )

( )

W m h h

w Wm

h h

Cs s

CsCs

s

= −

= = −

2 1

2 1

Using the ideal gas assumption with constant specific heats, the isentropic work per unit mass flow is

w C T TCs p s= −( )2 1

The isentropic temperature at state 2 is found from the isentropic relation

T T PP

K MPaMPa

K

s

k k

2 12

1

1

1 4 1 1 4

27 273 0 501

475 4

=FHGIKJ

= + FHG

IKJ

=

−

−

( )/

( . )/ .

( ) ..

.

The conservation of energy becomes

w C T TkJ

kg KK

kJkg

Cs p s= −

=⋅

−

=

( )

. ( . )

.

2 1

1005 4754 300

176 0


Chapter 7 - 67

The compressor isentropic efficiency is defined as

176220

0.8

sC

a

csCa

C

wwww

kJkJkgkg

η

η

=

=

= =

Example 7-16

Nitrogen expands in a nozzle from a temperature of 500 K while itspressure decreases by factor of two. What is the exit velocity of thenitrogen when the nozzle isentropic efficiency is 95 percent?

System: The nozzle control volume.

Property Relation: The ideal gas equations, assuming constant specificheats

Nozzle

T1P1

V1

T2P2

V2 2a2s

1

s

T P1

P2


Chapter 7 - 68

Process and Process Diagram: First assume an isentropic process andthen apply the nozzle isentropic efficiency to find the actual exit velocity.


For the isentropic case, Qnet = 0. Assume steady-state, steady-flow, nowork is done. Neglect the inlet kinetic energy and changes in potentialenergies. Then for one entrance, one exit, the first law reduces to

( )

E E

m h m h Vin out

ss

=

= +1 1 2 222

2

The conservation of mass gives

m m m1 2= =

The conservation of energy reduces to

V h hs s2 1 22= −( )

Using the ideal gas assumption with constant specific heats, the isentropicexit velocity is

V C T Ts p s2 1 22= −( )

The isentropic temperature at state 2 is found from the isentropic relation


Chapter 7 - 69

T T PP

K PP

K

s

k k

2 12

1

1

1

1

1 4 1 1 40 5

410 0

=FHGIKJ

=FHGIKJ

=

−

−

( )/

( . )/ .

(500) .

.

V C T T

kJkg K

K m skJ kg

ms

s p s2 1 2

3 2 2

2

2 1005 410 0 10

442 8

= −

=⋅

FHG

IKJ −

=

( )

. (500 . ) //

.

The nozzle exit velocity is obtained from the nozzle isentropic efficiencyas

222

2

2 2

/ 2/ 2

442.8 0.95 421.8

aN

s

a s N

VV

m mV Vs s

η

η

=

= = =


Chapter 7 - 70

Entropy Balance

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

Totalentropyentering

Totalentropyleaving

Totalentropygenerated

Change in thetotal entropyof the system

F

HGG

I

KJJ −F

HGG

I

KJJ +F

HGG

I

KJJ =F

HGG

I

KJJ

or

S S S Sin out gen system− + = ∆

The entropy balance relation can be stated as: the entropy change of asystem during a process is equal to the net entropy transfer through thesystem boundary and the entropy generated within the system as a resultof irreversibilities.

Entropy change of a system

The entropy change of a system is the result of the process occurringwithin the system.

System∆Esystem∆Ssystem∆Sgen≥0

Ein Eout

Sin Sout


Chapter 7 - 71

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

Mechanisms of Entropy Transfer, Sin and Sout

Entropy can be transferred to or from a system by two mechanisms: heattransfer and mass flow. Entropy transfer occurs at the system boundary asit crosses the boundary, and it represents the entropy gained or lost by asystem during the process. The only form of entropy interactionassociated with a closed system is heat transfer, and thus the entropytransfer for an adiabatic closed system is zero.

Heat transfer

The ratio of the heat transfer Q at a location to the absolute temperature Tat that location is called the entropy flow or entropy transfer and is givenas

Entropy transfer by heat transfer S QT

Theat: ( )= = constant

Q/T represents the entropy transfer accompanied by heat transfer, and thedirection of entropy transfer is the same as the direction of heat transfersince the absolute temperature T is always a positive quantity.

When the temperature is not constant, the entropy transfer for process 1-2can be determined by integration (or by summation if appropriate) as

S QT

QTheat

k

k

= ≅z ∑δ1

2

∆S S S S Ssystem final initial= − = −2 1


Chapter 7 - 72

Work

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

Entropy transfer by work Swork: = 0

Mass flow

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

Entropy transfer by mass S msmass: =

Entropy Generation, Sgen

Irreversibilities such as friction, mixing, chemical reactions, heat transferthrough a finite temperature difference, unrestrained expansion, non-quasi-equilibrium expansion, or compression always cause the entropy of asystem to increase, and entropy generation is a measure of the entropycreated by such effects during a process.


Chapter 7 - 73

For a reversible process, the entropy generation is zero and the entropychange of a system is equal to the entropy transfer. The entropy transferby heat is zero for an adiabatic system and the entropy transfer by mass iszero for a closed system.

The entropy balance for any system undergoing any process can be ex-pressed in the general form as

S S S S kJ Kin out gen system− + =Net entropy transfer by heat and mass

Entropygeneration

Change in entropy

3 ∆ ( / )

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

( / )S S S S kW Kin out gen system− + =Rate of net entropy transfer by heat and mass

Rate of entropy generation

Rate of change of entropy

3 ∆

where the rates of entropy transfer by heat transferred at a rate of Q andmass flowing at a rate of m are /S Q T S msheat mass= = and .

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

( ) ( / )s s s s kJ kg Kin out gen system− + = ⋅∆

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


Chapter 7 - 74

obtained by applying the entropy balance to an Isolated System thatcontains 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

QT

S S S S kJ Kk

kgen system+ = = −∑ ∆ 2 1 ( / )

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

Adiabatic closed system S Sgen adiabatic system: = ∆

∆


Chapter 7 - 75

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

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

Control Volumes

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

QT

m s m s S S S kJ Kk

ki i e e gen CV∑ ∑ ∑+ − + = −( ) ( / )2 1

In the rate form we have

( / )QT

m s m s S S kW Kk

ki i e e gen CV∑ ∑ ∑+ − + = ∆

This entropy balance relation is stated as: the rate of entropy change withinthe control volume during a process is equal to the sum of the rate ofentropy transfer through the control volume boundary by heat transfer, thenet rate of entropy transfer into the control volume by mass flow, and therate of entropy generation within the boundaries of the control volume as aresult of irreversibilities.


Chapter 7 - 76

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

S m s m s QTgen e e i i

k

k

= − −∑ ∑ ∑

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

( )S m s s QTgen e i

k

k

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

( )S m s sgen e i= −

This states that the specific entropy of the fluid must increase as it flowsthrough an adiabatic device since Sgen ≥ 0 . If the flow through the deviceis reversible and adiabatic, then the entropy will remain constantregardless of the changes in other properties.


Chapter 7 - 77

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

s se i=

Example 7-17

An inventor claims to have developed a water mixing device in which 10kg/s of water at 25oC and 0.1 MPa and 0.5 kg/s of water at 100oC, 0.1MPa, are mixed to produce 10.5 kg/s of water as a saturated liquid at 0.1MPa. If the surroundings to this device are at 20oC, is this processpossible? If not, what temperature must the surroundings have for theprocess to be possible?

System: The mixing chamber control volume.

Property Relation: The steam tables

Process and Process Diagram: Assume steady-flow

SurroundingsTo = 20oC

10.5 kg/sP3 = 0.1 MPaSaturated liquid

10 kg/sP1 = 0.1 MPaT1 = 25oC

0.5 kg/sP1 = 0.1 MPaT1 = 100oC

Mixingchamber

Qsys ?


Chapter 7 - 78


First let’s determine if there is a heat transfer from the surroundings to themixing chamber. Assume there is no work done during the mixingprocess, and neglect kinetic and potential energy changes. Then for twoentrances and one exit, the first law becomes


Q m h m h m h

net iinlets

ii

i net eexits

ee

e

net

+ + +FHG

IKJ = + + +

FHG

IKJ

+ + =

∑ ∑2 2

1 1 2 2 3 3

2 2

1

1

1 @1

11 @

104.830.125 0.3672

f T

o

f T

kJh hP MPa kg

kJT C s skg K

≅ ==

= ≅ = ⋅

22

22

2675.80.1100 7.3611

o

kJhP MPa kg

kJT C skg K

==

= = ⋅

33

3

417.510.1

. 1.3028

kJhP MPa kg

kJSat liquid skg K

== = ⋅


Chapter 7 - 79

3 3 1 1 2 2

10.5 417.51 10 104.83 0.5 2675.8

1997.7

netQ m h m h m hkg kJ kg kJ kg kJs kg s kg s kg

kJs

= − −

= − −

= +

So, 1996.33 kJ/s of heat energy must be transferred from the surroundingsto this mixing process, or , ,Q Qnet surr net CV= − .

For the process to be possible, the second law must be satisfied. Write thesecond law for the isolated system,

QT

m s m s S Sk

ki i e e gen CV+ − + =∑ ∑∑ ∆

For steady-flow ∆SCV = 0 . Solving for entropy generation, we have

3 3 1 1 2 2

10.5 1.3028 10 0.3672

1997.7 /0.5 7.3611(20 273)

0.491

kgen e e i i

k

cv

surr

QS m s m sT

Qm s m s m sT

kg kJ kg kJs kg K s kg Kkg kJ kJ ss kg K K

kJK s

= − −

= − − −

= −⋅ ⋅

− −⋅ +

= −⋅

∑ ∑ ∑


Chapter 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 thismixing process possible, set Sgen = 0 and solve for Tsurr.

3 3 1 1 2 2

0

1997.7 /

10.5 1.3028 10 0.3672 0.5 7.3611

315.75

kgen e e i i

k

cvsurr

QS m s m sT

QTm s m s m s

kJ skg kJ kg kJ kg kJs kg K s kg K s kg K

K

= − − =

=− −

=− −

⋅ ⋅ ⋅=

∑ ∑ ∑

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


Chapter 7 - 81

Answer to Example 7-4

Find the entropy and/or temperature of steam at the following states:

P T Region s kJ/kg⋅K5 MPa 120oC Compressed

Liquid andin the table

1.5233

1 MPa 50oC Compressed liquidbutnot in the table

s = sf at 50oC = 0.7038

1.8MPa

400oC Superheated 7.1794

40 kPa T=Tsat

=75.87o

C

Quality, x = 0.9Saturated mixture

s = sf + x sfg = 7.0056

40 kPa T=Tsat

=75.87o

C

sf<s<sg at PSaturated mixtureX = (s-sf)/sfg = 0.9262

7.1794


Chapter 8-1

Chapter 8: Exergy: A Measure of Work Potential

The energy content of the universe is constant, just as its mass content is.Yet at times of crisis we are bombarded with speeches and articles on howto “conserve” energy. As engineers, we know that energy is alreadyconserved. What is not conserved is exergy, which is the useful workpotential of the energy. Once the exergy is wasted, it can never berecovered. When we use energy (to heat our homes, for example), we arenot destroying any energy; we are merely converting it to a less usefulform, a form of less exergy.

Exergy and the Dead State

The useful work potential of a system is the amount of energy we extractas useful work. The useful work potential of a system at the specified stateis called exergy. Exergy is a property and is associated with the state of thesystem and the environment. A system that is in equilibrium with itssurroundings has zero exergy and is said to be at the dead state. Theexergy of the thermal energy of thermal reservoirs is equivalent to thework output of a Carnot heat engine operating between the reservoir andthe environment.

Exergy Forms

Now let’s determine the exergy of various forms of energy.

Exergy of kinetic energy

Kinetic energy is a form of mechanical energy and can be converteddirectly into work. Kinetic energy itself is the work potential or exergy ofkinetic energy independent of the temperature and pressure of theenvironment.


Chapter 8-2

Exergy of kinetic energy:

2

ke (kJ/kg)2

Vx ke= =

Exergy of potential energy

Potential energy is a form of mechanical energy and can be converteddirectly into work. Potential energy itself is the work potential or exergyof potential energy independent of the temperature and pressure of theenvironment.

Exergy of potential energy: pe (kJ/kg)x pe gz= =

Useful Work

The work done by work producing devices is not always entirely in auseable form. Consider the piston-cylinder device shown in the followingfigure.


Chapter 8-3

The work done by the gas expanding in the piston-cylinder device is theboundary work and can be written as

0 0

b, useful 0

( )W P dV P P dV P dVW P dV

δδ

= = − += +

The actual work done by the gas is

b, useful 0

b, useful 0 2 1( )

W W P dV

W P V V

= +

= + −∫

The word done on the surroundings is

surr 0 0 2 1( )W P dV P V V= = −∫Any useful work delivered by a piston-cylinder device is due to thepressure above the atmospheric level.

u surrW W W= −


Chapter 8-4

Reversible Work

Reversible work Wrev is defined as the maximum amount of useful workthat can be produced (or the minimum work that needs to be supplied) as asystem undergoes a process between the specified initial and final states.This is the useful work output (or input) obtained when the processbetween the initial and final states is executed in a totally reversiblemanner.

Irreversibility

The difference between the reversible work Wrev and the useful work Wu isdue to the irreversibilities present during the process and is called theirreversibility I. It is equivalent to the exergy destroyed and is expressed as

destroyed 0 gen rev, out u, out u, in rev, inI X T S W W W W= = = − = −

where Sgen is the entropy generated during the process. For a totallyreversible process, the useful and reversible work terms are identical andthus irreversibility is zero.

Irreversibility can be viewed as the wasted work potential or the lostopportunity to do work. It represents the energy that could have beenconverted to work but was not.

Exergy destroyed represents the lost work potential and is also called thewasted work or lost work.


Chapter 8-5

Second-Law Efficiency

The second-law efficiency is a measure of the performance of a devicerelative to the performance under reversible conditions for the same endstates and is given by

u

, rev

thII

th rev

WW

ηηη

= =

for heat engines and other work-producing devices and

rev

uII

rev

WCOPCOP W

η = =

for refrigerators, heat pumps, and other work-consuming devices.

In general, the second-law efficiency is expressed as

η II = = −Exergy recoveredExergy supplied

Exergy destroyedExergy supplied

1

Exergy of change of a system

Consider heat transferred to or from a closed system whenever there is atemperature difference across the system boundary. The exergy for a


Chapter 8-6

system may be determined by considering how much of this heat transferis converted to work entirely. Let’s take a second look at the followingfigure.

Taking the heat transfer to be from the system to its surroundings, theconservation of energy is

in out system

0

E E dE

Q W dUδ δ

δ δ

− =

− − =

The work is the boundary work and can be written as

0 0

b, useful 0

( )W P dV P P dV P dVW P dV

δδ

= = − += +

Any useful work delivered by a piston-cylinder device is due to thepressure above the atmospheric level.

To assure the reversibility of the process, the heat transfer occurs through areversible heat engine.


Chapter 8-7

0HE th 0

net

HE 0

HE 0

(1 )T QW Q Q Q TT T

Q QdST T

W Q T dSQ W T dS

δδ η δ δ δ

δ δ

δ δδ δ

= = − = −

−= =

= += −

( ) ( )HE 0 b, useful 0W T dS W P dV dUδ δ− − − + =

total useful b, useful HE

0 0

W W W

dU P dV T dS

δ δ δ= +

= − − +

Integrating from the given state (no subscript) to the dead state (0subscript), we have

total useful 0 0 0 0 0

0 0 0 0 0

( ) ( ) ( )

( ) ( ) ( )

W U U P V V T S S

U U P V V T S S

= − − − − + −

= − + − − −

This is the total useful work due to a system undergoing a reversibleprocess from a given state to the dead state, which is the definition ofexergy.

Including the kinetic energy and potential energy, the exergy of a closedsystem is


Chapter 8-8

2

0 0 0 0 0( ) ( ) ( )2

VX U U P V V T S S m mgz= − + − − − + +

on a unit mass basis, the closed system (or nonflow) exergy is

2

0 0 0 0 0

0 0 0 0 0

( ) ( ) ( )2

( ) ( ) ( )

Vu u P v v T s s gz

e e P v v T s s

φ = − + − − − + +

= − + − − −

Here, u0, v0, and s0 are the properties of the system evaluated at the deadstate. Note that the exergy of the internal energy of a system is zero at thedead state is zero since u = u0, v = v0, and s = s0 at that state.

The exergy change of a closed system during a process is simply thedifference between the final and initial exergies of the system,

2 1 2 1

0 0 0 0 0

2 22 1

0 0 0 0 0 2 1

( )( ) ( ) ( )

( ) ( ) ( ) ( )2

X X X mE E P V V T S S

V VU U P V V T S S m mg z z

φ φ∆ = − = −= − + − − −

−= − + − − − + + −

On a unit mass basis the exergy change of a closed system is


Chapter 8-9

2 1

0 0 0 0 0

2 22 1

0 0 0 0 0 2 1

( )( ) ( ) ( )

( ) ( ) ( ) ( )2

e e P v v T s s

V Vu u P v v T s s g z z

φ φ φ∆ = −= − + − − −

−= − + − − − + + −

Exergy of flow

The energy needed to force mass to flow into or out of a control volume isthe flow work per unit mass given by (see Chapter 3).

floww Pv=

The exergy of flow work is the excess of flow work done againstatmospheric air at P0 to displace it by volume v. According to the abovefigure, the useful work potential due to flow work is

flow, energy 0w Pv P v= −

Thus, the exergy of flow energy is


Chapter 8-10

flow energy 0 0( )x Pv P v P P v= − = −

Flow Exergy

Since flow energy is the sum of nonflow energy and the flow energy, theexergy of flow is the sum of the exergies of nonflow exergy and flowexergy.

flowing fluid nonflowing fluid flow exergy

2

0 0 0 0 0 0

2

0 0 0 0 0

2

0 0 0

( ) ( ) ( ) ( )2

( ) ( ) ( )2

( ) ( )2

x x x

Vu u P v v T s s gz P P v

Vu Pv u P v T s s gz

Vh h T s s gz

= +

= − + − − − + + + −

= + − + − − + +

= − − − + +

The flow (or stream) exergy is given by

2

0 0 0( ) ( )2

Vh h T s s gzψ = − − − + +

The exergy of flow can be negative if the pressure is lower thanatmospheric pressure.

The exergy change of a fluid stream as it undergoes a process from state 1to state 2 is


Chapter 8-11

2 22 1

2 1 2 1 0 2 1 2 1( ) ( ) ( )2

V Vh h T s s g z zψ ψ ψ −∆ = − = − − − + + −

Exergy Transfer by Heat, Work, and Mass

Exergy can be transferred by heat, work, and mass flow, and exergytransfer accompanied by heat, work, and mass transfer are given by thefollowing.

Exergy transfer by heat transfer

By the second law we know that only a portion of heat transfer at atemperature above the environment temperature can be converted intowork. The maximum useful work is produced from it by passing this heattransfer through a reversible heat engine. The exergy transfer by heat is

Exergy transfer by heat:0

heat 1 TX QT

= −


Chapter 8-12

Note in the above figure that entropy generation is always by exergydestruction and that heat transfer Q at a location at temperature T is alwaysaccompanied by entropy transfer in the amount of Q/T and exergy transferin the amount of (1-T0/T)Q.

Note that exergy transfer by heat is zero for adiabatic systems.Exergy transfer by work

Exergy is the useful work potential, and the exergy transfer by work cansimply be expressed as

Exergy transfer by work:surr

work

(for boundary work) (for other forms of work)

W WX

W−

=

where surr 0 2 1( )W P V V= − , P0 is atmospheric pressure, and V1 and V2 are the initialand final volumes of the system. The exergy transfer for shaft work andelectrical work is equal to the work W itself.


Chapter 8-13

Note that exergy transfer by work is zero for systems that have no work.

Exergy transfer by mass

Mass flow is a mechanism to transport exergy, entropy, and energy into orout of a system. As mass in the amount m enters or leaves a system theexergy transfer is given by

Exergy transfer by mass: X mmass = ψ

Note that exergy transfer by mass is zero for systems that involve no flow.


Chapter 8-14

The Decrease of Exergy Principle and Exergy Destruction

The exergy of an isolated system during a process always decreases or, inthe limiting case of a reversible process, remains constant. This is knownas the decrease of exergy principle and is expressed as

isolated 2 1 isolated( ) 0X X X∆ = − ≤

Exergy Destruction

Irreversibilities such as friction, mixing, chemical reactions, heat transferthrough finite temperature difference, unrestrained expansion, non-quasi-equilibrium compression, or expansion always generate entropy, andanything that generates entropy always destroys exergy. The exergydestroyed is proportional to the entropy generated as expressed as

destroyed 0 genX T S=

The decrease of exergy principle does not imply that the exergy of asystem cannot increase. The exergy change of a system can be positive ornegative during a process, but exergy destroyed cannot be negative. Thedecrease of exergy principle can be summarized as follows:

destroyed

0 Irreversible proces0 Reversible process0 Impossible process

X>=<


Chapter 8-15

Exergy Balances

Exergy balance for any system undergoing any process can be expressedas

Total Total Total Change in theexergy exergy exergy total exergyentering leaving destroyed of the system

− − =

General:

in out destroyed system

Net exergy transfer Exergy Changeby heat, work, and mass destruction in exergy

X X X X− − = ∆

General, rate form:


Rate of net exergy transfer Rate of exergy Rate of change by heat, work, and mass destruction of exergy

X X X X− − = ∆

General, unit-mass basis:

in out destroyed system( )x x x x− − = ∆where

0heat

work useful

mass

system system

1

/

TX QT

X W

X m

X dX dt

ψ

= −

=

=

∆ =


Chapter 8-16

For a reversible process, the exergy destruction term, Xdestroyed, is zero.

Considering the system to be a general control volume and taking thepositive direction of heat transfer to be to the system and the positivedirection of work transfer to be from the system, the general exergybalance relations can be expressed more explicitly as

[ ]0k 0 2 1 destroyed 2 1

k

1 ( ) i i e eT Q W P V V m m X X XT

ψ ψ − − − − + − − = −

∑ ∑ ∑

0 CV CV0 destroyed

k

1 k i i e eT dV dXQ W P m m XT dt dt

ψ ψ − − − + − − =

∑ ∑ ∑

where the subscripts are i = inlet, e = exit, 1 = initial state, and 2 = finalstate of the system. For closed systems, no mass crosses the boundariesand we omit the terms containing the sum over the inlets and exits.

Example 8-1

Oxygen gas is compressed in a piston-cylinder device from an initial stateof 0.8 m3/kg and 25oC to a final state of 0.1 m3/kg and 287oC. Determinethe reversible work input and the increase in the exergy of the oxygenduring this process. Assume the surroundings to be at 25oC and 100 kPa.

We assume that oxygen is an ideal gas with constant specific heats. FromTable A-2, R = 0.2598 kJ/kg⋅K. The specific heat is determined at theaverage temperature

1 2av

(25 287) C 156 C2 2

(156 273)K 429K

T TT + += = =

= + =

Table A-2(b) gives Cv, ave = 0.690 kJ/kg⋅K.


Chapter 8-17

The entropy change of oxygen is

2 22 1 v, ave

1 1

3

3

ln ln

m0.1kJ (287 273) kJ kg0.690 ln 0.2598 ln

mkg K (25 273) kg K 0.8kg

kJ0.105kg K

T vs s C RT v

KK

− = +

+ = + ⋅ + ⋅

= −⋅

We calculate the reversible work input, which represents the minimumwork input Wrev,in in this case, from the exergy balance by setting theexergy destruction equal to zero.


Net exergy transfer Exergy Changeby heat, work, and mass destruction in exergy

X X X X− − = ∆

rev,in 2 1W X X= −

Therefore, the change in exergy and the reversible work are identical inthis case. Substituting the closed system exergy relation, the reversiblework input during this process is determined to be

0


Chapter 8-18

rev,in 2 1

2 1 0 2 1 0 2 1

v,ave 2 1 0 2 1 0 2 1

3

3

( ) ( ) ( )( ) ( ) ( )

kJ m kJ0.690 (287 25)K 100 kPa(0.1 0.8)kg K kg m kPa

kJ(25 273)K( 0.105 )kg K

kJ142.1kg

w

u u P v v T s sC T T P v v T s s

φ φ= −

= − + − − −= − + − − −

= − + −⋅

− + −⋅

=

The increase in exergy of the oxygen is

2 1 2 1 rev, inkJ142.1kg

x x wφ φ− = − = =

Example 8-2

Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leavesat 50 kPa, 100°C, and 140 m/s. The surroundings to the turbine are at25°C. If the power output of the turbine is 5MW, determine(a) the power potential of the steam at its inlet conditions, in MW.(b) the reversible power, in MW.(c) the second law efficiency.

We assume steady-flow and neglect changes in potential energy.


Chapter 8-19

The mass flow rate of the steam is determined from the steady-flow energyequation applied to the actual process,

in out systems

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

2 21 2

1 1 2 2 out( ) ( ) 02 2

E E E

V Vm h m h W

− = ∆

+ − + − =

Conservation of mass for the steady flow gives

in out system

Rate of net mass transfer Rate of change of mass

1 2

1 2

0

m m m

m mm m m

− = ∆

− == =

Steam turbine1V1 = 80 m/sP1 = 6 MPaT1 = 600oC

2V2 = 140 m/sP2 = 50 kPaT2 = 100oC

Wout = 5 MW

0 (steady)


Chapter 8-20

The work done by the turbine and the mass flow rate are

2 21 2

out 1 2

out

1 2

( )2 2

( )

V VW m h h

Wmh h ke

= − + −

=− −∆

where

2 22 1

2 2

2 2

2 2

(140m/s) (80 m/s) 1kJ/kg2 1000m /s

kJ6.6kg

V Vke

∆ = −

−=

=


Chapter 8-21

From the steam tables:

o

o

11

o1

1

22

o2

2

0 f@25 C0

o0

0 f@25 C

kJ3658.86MPa kg

kJ600 7.1693kg K

kJ2682.450kPa kg

kJ100 C 7.6953kg K

kJ104.83100 kPa kg

kJ25 C 0.3672kg K

hP

T C s

hP

T s

h hP

T s s

==

= = ⋅ ==

= =

⋅ ≅ ==

= ≅ =

⋅

out

1 2( )5 MW 1000kJ/s

kJ MW(3658.8 2682.4 6.6)kg

kg5.16s

Wmh h ke

=− −∆

= − −

=

The power potential of the steam at the inlet conditions is equivalent to itsexergy at the inlet state. Recall that we neglect the potential energy of theflow.


Chapter 8-22

21

1 1 1 0 0 1 0 1( ) ( )2

Vm m h h T s s gzψ

Ψ = = − − − + +

1 2

2 2

kJ kJ(3658.8 104.83) (298 )(7.1693 0.3672)kg kg Kkg5.16

s (80m/s) kJ/kg2 1000 m /s

kg kJ MW5.16 1533.3s kg 1000 kJ/s

7.91MW

K − − − ⋅ Ψ = +

=

=

The power output of the turbine if there are no irreversibilities is thereversible power and is determined from the rate form of the exergybalance applied on the turbine and setting the exergy destruction termequal to zero.


Rate of net exergy transfer Rate of exergy Rate of change by heat, work, and mass destruction of exergy

in out

1 rev, out 2

X X X X

X X

m W mψ ψ

− − = ∆

=

= +

[ ]rev, out 2 1

1 2 0 1 2

( )

( ) ( )

W m

m h h T s s ke pe

ψ ψ= −

= − − − −∆ −∆

0

0

0 (steady)0


Chapter 8-23

rev, out

kJ kJ(3658.8 2682.4) (298 )(7.1693 7.6953)kg kg Kkg5.16

kJs 6.6kg

kg kJ MW5.16 1126.5s kg 1000 kJ/s

5.81MW

KW

− − − ⋅ = −

=

=

The second-law efficiency is determined from

5MW 86.1%5.81MWII

rev

WW

η = = =

Chapter 10

Vapor and Combined Power Cycles

Study Guide in PowerPoint

to accompanyto accompany

Thermodynamics: An Engineering Approach, 5th edition

by Yunus A. Çengel and Michael A. Boles

We consider power cycles where the working fluid undergoes a phase change. The

best example of this cycle is the steam power cycle where water (steam) is the

working fluid.

Carnot Vapor Cycle

2

The heat engine may be composed of the following components.

3

The working fluid, steam (water), undergoes a thermodynamic cycle from 1-2-3-4-1.

The cycle is shown on the following T-s diagram.

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

0

100

200

300

400

500

600

700700

s [kJ/kg-K]

T [C]

6000 kPa

100 kPa

Carnot Vapor Cycle Using Steam

1

23

4

The thermal efficiency of this cycle is given as

4

The thermal efficiency of this cycle is given as

η th Carnotnet

in

out

in

L

H

W

Q

Q

Q

T

T

, = = −

= −

1

1

Note the effect of TH and TL on ηth, Carnot.•The larger the TH the larger the ηth, Carnot•The smaller the TL the larger the ηth, Carnot

To increase the thermal efficiency in any power cycle, we try to increase the

maximum temperature at which heat is added.

Reasons why the Carnot cycle is not used:

•Pumping process 1-2 requires the pumping of a mixture of saturated liquid and

saturated vapor at state 1 and the delivery of a saturated liquid at state 2.

•To superheat the steam to take advantage of a higher temperature, elaborate

controls are required to keep TH constant while the steam expands and does work.

To resolve the difficulties associated with the Carnot cycle, the Rankine cycle was

devised.

Rankine Cycle

5

Rankine Cycle

The simple Rankine cycle has the same component layout as the Carnot cycle

shown above. The simple Rankine cycle continues the condensation process 4-1

until the saturated liquid line is reached.

Ideal Rankine Cycle Processes

Process Description

1-2 Isentropic compression in pump

2-3 Constant pressure heat addition in boiler

3-4 Isentropic expansion in turbine

4-1 Constant pressure heat rejection in condenser

The T-s diagram for the Rankine cycle is given below. Locate the processes for heat

transfer and work on the diagram.

0 2 4 6 8 10 1212

0

100

200

300

400

500

T [C]

6000 kPa

10 kPa

Rankine Vapor Power Cycle

1

2

3

4

6

0 2 4 6 8 10 1212

s [kJ/kg-K]

Example 10-1

Compute the thermal efficiency of an ideal Rankine cycle for which steam leaves

the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa.

We use the power system and T-s diagram shown above.

P2 = P3 = 6 MPa = 6000 kPa

T3 = 350oC

P1 = P4 = 10 kPa

Pump

The pump work is obtained from the conservation of mass and energy for steady-flow

but neglecting potential and kinetic energy changes and assuming the pump is

adiabatic and reversible.

& & &

& & &

& & ( )

m m m

m h W m h

W m h h

pump

pump

1 2

1 1 2 2

2 1

= =

+ =

= −

Since the pumping process involves an incompressible liquid, state 2 is in the

compressed liquid region, we use a second method to find the pump work or the ∆h

7

across the pump.

Recall the property relation:

dh = T ds + v dP

Since the ideal pumping process 1-2 is isentropic, ds = 0.

The incompressible liquid assumption allows

v v const

h h v P P

≅ =

− ≅ −1

2 1 1 2 1

.

( )

The pump work is calculated from

& & ( ) & ( )

&

&( )

W m h h mv P P

wW

mv P P

pump

pump

pump

= − ≅ −

= = −

2 1 1 2 1

1 2 1

Using the steam tables

1 191.8110

f

kJh h

kgP kPa

= ==

8

1

3

1

10

.0.00101f

kgP kPa

Sat liquid mv v

kg

= = =

w v P P

m

kgkPa

kJ

m kPa

kJ

kg

pump = −

= −

=

1 2 1

3

30 00101 6000 10

6 05

( )

. ( )

.

Now, h2 is found from

2 1

6.05 191.81

197.86

pumph w h

kJ kJ

kg kg

kJ

kg

= +

= +

=

Boiler

To find the heat supplied in the boiler, we apply the steady-flow conservation of mass

and energy to the boiler. If we neglect the potential and kinetic energies, and note

9

and energy to the boiler. If we neglect the potential and kinetic energies, and note

that no work is done on the steam in the boiler, then

& & &

& & &

& & ( )

m m m

m h Q m h

Q m h h

in

in

2 3

2 2 3 3

3 2

= =

+ =

= −

We find the properties at state 3 from the superheated tables as

3

3

33

3043.96000

3506.3357

o

kJh

P kPa kg

kJT Cs

kg K

==

= = ⋅

The heat transfer per unit mass is

3 2

(3043.9 197.86)

inin

Qq h h

m

kJ

= = −

= −

&

&

10

(3043.9 197.86)

2845.1

kJ

kg

kJ

kg

= −

=

Turbine

The turbine work is obtained from the application of the conservation of mass and

energy for steady flow. We assume the process is adiabatic and reversible and

neglect changes in kinetic and potential energies.

& & &

& & &

& & ( )

m m m

m h W m h

W m h h

turb

turb

3 4

3 3 4 4

3 4

= =

= +

= −

We find the properties at state 4 from the steam tables by noting s4 = s3 = 6.3357

11

4 3

kJ/kg-K and asking three questions.

4

4

4

4

10 : 0.6492 ; 8.1488

?

?

?

f g

f

f g

g

kJ kJat P kPa s s

kg K kg K

is s s

is s s s

is s s

= = =⋅ ⋅

<

< <

<

4 4

4

4

6.3357 0.64920.758

7.4996

f fg

f

fg

s s x s

s sx

s

= +

− −= = =

4 4

191.81 0.758(2392.1)

2005.0

f fgh h x h

kJ kJ

kg kg

kJ

kg

= +

= +

=

The turbine work per unit mass is

12

The turbine work per unit mass is

3 4

(3043.9 2005.0)

1038.9

turbw h h

kJ

kg

kJ

kg

= −

= −

=


(1038.9 6.05)

1032.8

net turb pumpw w w

kJ

kg

kJ

kg

= −

= −

=

The thermal efficiency is

1032.8net

kJ

w kgη = =

13

2845.1

0.363 36.3%

netth

inkJq

kg

or

η = =

=

Ways to improve the simple Rankine cycle efficiency:

• Superheat the vapor

Average temperature is higher during heat addition.

Moisture is reduced at turbine exit (we want x4 in the above example > 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

14

• 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 thnet

in

out

in

w

q

q

q= = −1

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.

15

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

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 thnet

in

w

q

h h h h h h

h h h h

=

=( - ) + ( - ) - ( - )

( - ) + ( - )

3 4 5 6 2 1

16

h h h h

h h

h h h h

=

= −−

( - ) + ( - )

( - ) + ( - )

3 2 5 4

6 1

3 2 5 4

1

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.

ηth xturb exit No Reheat 35.3% 0.8159

With Reheat 35.9% 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 lower-

pressure stages of the turbine, and, thus, the total work done by the turbine. The

work that is done is done more efficiently.

18

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.

Cycle with an open feedwater heater

19

0 2 4 6 8 10 1212

0

100

200

300

400

500

600

s [kJ/kg-K]

T [C]

3000 kPa

500 kPa

10 kPa

Rankine Steam Power Cycle with an Open Feedwater Heater

1

23

4

5

6

7

Cycle with a closed feedwater heater with steam trap to condenser

20

Cycle with a closed feedwater heater with steam trap to condenser

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

T

21

s

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

23

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:

y m m= & / &6 5Let be the fraction of mass extracted from the turbine for the feedwater

heater.

& &

& & & &

& & & & ( )

m m

m m m m

m m m m y

in out=

+ = =

= − = −6 2 3 5

2 5 6 5 1

Conservation of energy for the open feedwater heater:

& &

& & &

& ( ) & &

E E

m h m h m h

ym h y m h m h

in out=

+ =

+ − =

−

6 6 2 2 3 3

5 6 5 2 5 31

24

yh h

h h=

−−

3 2

6 2

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 Selected saturation properties

State P

kPa

T

°Ch

kJ/kg

s

kJ/kg-K

P

kPa

Tsat

°Cvf

m3/kg

hf

kJ/kg

1 10 10 45.81 0.00101 191.8

25

1 10 10 45.81 0.00101 191.8

2 500 500 151.83 0.00109 640.1

3 500 3000 233.85 0.00122 1008.3

4 3000

5 3000 500 3457.2 7.2359

6 500 2942.6 7.2359

7 10 2292.7 7.2359

The work for pump 1 is calculated from

w v P P

m

kgkPa

kJ

m kPa

kJ

kg

pump 1 1 2 1

3

30 00101 10

05

= −

= −

=

( )

. (500 )

.

Now, h2 is found from

h w h

kJ kJ

pump2 1 1

05 1918

= +

= +. .

26

kg kg

kJ

kg

05 1918

192 3

= +

=

. .

.

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.

3 2

6 2

(640.1 192.3)

0.163

(2942.6 192.3)

kJ

h h kgy

kJh h

kg

−−

= = =− −

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

27

w v P P

m

kgkPa

kJ

m kPa

kJ

kg

pump 2 3 4 3

3

30 00109 3000 500

2 7

= −

= −

=

( )

. ( )

.

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

through the pump.

4 2 3

2.7 640.1

642.8

out in

pump

E E

h w h

kJ kJ

kg kg

kJ

kg

=

= +

= +

=

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

28

5 5 6 6 7 7

5 5 6 7

5 6 7

5

[ (1 ) ]

(1 )

[3457.2 (0.163)(2942.1) (1 0.163)(2292.7)]

1058.6

in out

turb

turb

turbturb

E E

m h W m h m h

W m h yh y h

Ww h yh y h

m

kJ

kg

kJ

kg

=

= + +

= − − −

= = − − −

= − − −

=

& &

&& & &

& &

&

&


1 2

5 5 1 1 3 2

5 5 5 1 5 2

1 2

(1 )

(1 )

[1058.6 (1 0.163)(0.5) 2.7]

1055.5

net turb pump pump

net turb pump pump

net turb pump pump

net turb pump pump

W W W W

m w m w m w m w

m w m w m y w m w

w w y w w

kJ

kg

kJ

kg

= − −

= − −

= − − −

= − − −

= − − −

=

& & & &

& & & &

& & & &

29

kg

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

& &

& & &

& & ( )

&

&

m m

m h Q m h

Q m h h

qQ

mh h

in

in

inin

4 5

4 4 5 5

5 5 4

5

5 4

=

+ =

= −

= = −

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

5 4

(3457.2 642.8) 2814.4

inq h h

kJ kJ

kg kg

= −

= − =

The thermal efficiency is

1055.5

2814.4

0.375 37.5%

netth

in

kJ

w kg

kJq

kg

or

η = =

=

30

0.375 37.5%or=

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.

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.

0.364

0.366

0.368

0.370

0.372

0.374

0.3760.376

ηη ηηth

ηth vs OFWH Pressure

31

0 450 900 1350 1800 22500.360

0.362

0.364

Pofwh [kPa]

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?

1100

1150

1200

0.18

0.20

0.23

0.25

wnet and y vs OFWH Pressure

32

0 450 900 1350 1800 2250

900

950

1000

1050

0.03

0.05

0.08

0.10

0.13

0.15

Pofwh [kPa]

wnet kJ/kg

y

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.

∆ ∆T T etccond to FWH boiler to FWH= , .

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.

33

used. Select starting values for the feedwater heater extraction pressures.

0 2 4 6 8 10 1212

0

100

200

300

400

s [kJ/kg-K]

T [C]

3000 kPa 815 kPa

136.2 kPa

10 kPa

Steam

∆ Τ = 62.68 ∆ Τ = 62.68 ∆ Τ = 62.68 ∆ Τ = 62.68

∆ Τ = 62.68 ∆ Τ = 62.68 ∆ Τ = 62.68 ∆ Τ = 62.68

∆ Τ = 62.68 ∆ Τ = 62.68 ∆ Τ = 62.68 ∆ Τ = 62.68

C

C

C

233.9 C

45.85 C45.8

1

Deviation from Actual Cycles

•Piping losses--frictional effects reduce the available energy content of the steam.

•Turbine losses--turbine isentropic (or adiabatic) efficiency.

4a4s

3

s

T P3

P4

η actual aw h h= =

−3 4

34

η turbactual

isentropic

a

s

w

w

h h

h h= =

−−

3 4

3 4

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

component) is

h h h ha turb s4 3 3 4= − −η ( )

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

2a2s

1

s

T P2

P1

η pump

isentropic

actual

s

a

w

w

h h

h h= =

−−

2 1

2 1

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

35

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

component) is

h h h ha

pump

s2 1 2 1

1= + −

η( )

•Condenser losses--relatively small losses that result from cooling the condensate

below the saturation temperature in the condenser.

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

36

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

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

38

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%.

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.

40

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

Documents

Transcript of Student Study Guide for 5. th. edition of Thermodynamics by Y. A. Çengel & M. A. Boles Chapter_ 1...