### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Engineering Thermodynamics I MAE 301

NCS

GPA 3.62

### View Full Document

## 70

## 0

## Popular in Course

## Popular in Aerospace Engineering (AE)

This 161 page Study Guide was uploaded by Rowan Spinka DVM on Thursday October 15, 2015. The Study Guide belongs to MAE 301 at North Carolina State University taught by Michael Boles in Fall. Since its upload, it has received 70 views. For similar materials see /class/224023/mae-301-north-carolina-state-university in Aerospace Engineering (AE) at North Carolina State University.

## Reviews for Engineering Thermodynamics I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/15/15

Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 1 Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle Energy can cross the boundaries of a closed system in the form of heat or work Energy transfer across a system boundary due solely to the temperature difference between a system and its surroundings is 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 below where z is the elevation of the center of mass above the reference plane and I is the velocity of the center of mass Closed gt I Heat System A Reference Plane Z 0 For the closed system shown above the conservation of energy principle or the rst law of thermodynamics is expressed as Total energy 1 Total energy 1 The change in total entering the system leaving the system energy of the system 0139 Chapter 4 71 Student Study Guide for 5 h edition of Thermod arnics by Y A Qengel amp M A Boles 47 2 E E AE in out system According to classical thermodynamics we consider the energy added to be net heat transfer to the closed system and the energy leaving the closed system to be net work done by the closed system So Qnet W AE net system Where Qnet Qin Qout VVnet VVO VVinother M ut W 392PdV 1 Normally the stored energy or total energy of a system is expressed as the sum of three separate energies The total energy of the system Esystem is given as E Internal energy Kinetic energy Potential energy E U KE PE Recall that U is the sum of the energy contained Within the molecules of the system other than the kinetic and potential energies of the system as a Whole and is called the internal energy The internal energy U is dependent on the state of the system and the mass of the system Chapter 4 72 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 3 For a system moving relative to a reference plane the kinetic energy KE and the potential energy PE are given by I7 a a KEL deV V0 2 PE 2 I270 mg dz mgz The change in stored energy for the system is AEAUAKEAPE Now the conservation of energy principle or the first law of thermodynamics for closed systems is written as QM Wm AUAKE APE If the system does not move with a velocity and has no change in elevation the conservation of energy equation reduces to Qnet VVnet We will find that this is the most commonly used form of the rst law Closed System First Law for a Cycle Chapter 4 73 Studan Stud Guide fury edition of memugmmcs byY A Qangel ampM A Boles 4 4 Since a thermodynamic cycle is composed of processes that cause the Working uid to undergo a series of state changes through a series of processes such that the nal and initial states are identical the change in internal energy of the Working uid is zero for Whole numbers of cycles The rst law for a closed system operating in a thermodynamic cycle becomes 0 QM Wm Mom Qnet VVnet Qnel Wncl Gupta 4 A Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 4 5 Example 41 Complete the table given below for a closed system under going a cycle Process Qnet kJ Wnet kJ U2 U1 M 12 5 5 23 20 10 31 5 Cycle Answer to above problem Row 1 10 Row 2 10 Row 3 10 5 Row 4 15 15 0 Chapter 4 5 Student Study Gmde for 5 edmon of Thermodmam1cs by Y A Qengel Sc M A Boles 4 6 In the next section we will look at boundary work in detail Review the text 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 within the system boundary causes a force the pressure times the surface area to act on the boundary surface and make it move This is what happens when steam the gas in the figure below contained in a pistoncylinder device expands against the piston and forces the piston to move thus boundary work is done by the steam on the piston Boundary work is then calculated from 3911 W f W des JigIds fl dV Chapter 4 76 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 7 Since the work is process dependent the differential of boundary work 8 W1 ESszPdV is called inexact The above equation for W is valid for a quasi equilibrium process and gives the maximum work done during expansion and the minimum work input during compression In an expansion process the boundary work must overcome friction push the atmospheric air out of the way and rotate a crankshaft VVbZVVf W Wrank riction atm c 12 Ffriction P A F atm crank ds To calculate the boundary work the process by which the system changed states must be known Once the process is determined the pressure volume relationship for the process can be obtained and the integral in the boundary work equation can be performed For each process we need to determine PfV So as we work problems we will be asking What is the pressurevolume relationship for the process Remember that this relation is really the forcedisplacement function for the process The boundary work is equal to the area under the process curve plotted on the pressurevolume diagram Chapter 4 77 Student Study Guide for 5 edition of Thermodynamics byY A Qengel amp M A Boles 4r 8 Process alh 1 dAPdV V1 4 clV V2 V Note from the above gure 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 Vdiagram are different the boundary work for each process will be different The next figure shows that each process gives a different value for the boundary work Chapter 4 78 Some Typical Processes Constant volume If the volume is held constant dV 0 and the boundary work equation becomes V PVTdiagram for V constant Wb deV 0 If the working uid is an ideal gas What will happen to the temperature of the gas during this constant volume process Chaptex4 79 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 10 Constant pressure V PrVDlAGRAM FORP CONSTANT If the pressure is held constant the boundary work equation becomes 2 2 W 2de PJdV PV2 Vl l 1 For the constant pressure process shown above is the boundary work positive or negative and why Constant temperature ideal gas If the temperature of an ideal gas system is held constant then the equation of state provides the pressurevolume relation PZmRT V Then the boundary work is Chapter 4 710 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 ll mRTd V W JdeVJf V mRTh1K V 1 Note The above equation is the result of applying the ideal gas assumption for the equation of state For real gases undergoing an isothermal constant temperature process the integral in the boundary work equation would be done numerically The polytropic process The polytropic process is one in which the pressurevolume relation is given as PVnzum mn The eXponent 11 may have any value from minus infinity to plus infinity depending on the process Some of the more common values are given below Process Exponent n Constant pressure 0 Constant volume 00 Isothermal amp ideal gas 1 Adiabatic amp ideal gas k CPCV Here k is the ratio of the specific heat at constant pressure CF to specific heat at constant volume C V The specific heats will be discussed later Chapter 4 711 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 12 The boundary work done during the polytropic process is found by substituting the pressurevolume relation into the boundary work equation The result is Wb fpdV fdV PszPlVl 1 11 7 PVln E nl V 1 For an ideal gas under going a polytropic process the boundary work is 2 2 Consl Wb j Pd V j dV 1 1 V mRltT2 T1gt nil l n mRTln E nl V 1 Chapter 4 712 Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 13 Notice that the results we obtained for an ideal gas undergoing a polytropic process when n l are identical to those for an ideal gas undergoing the isothermal process Example 42 Three kilograms of nitrogen gas at 27 C and 015 MPa are compressed isothermally to 03 MPa in a pistoncylinder device Determine the minimum work of compression in H System Nitrogen contained in a pistoncylinder device Process Constant temperature 2 System P Boundary 1 Nitrogen lVb gas V H PiVDlAGRAM FOR T CONSTANT Property Relation Check the reduced temperature and pressure for nitrogen The critical state properties are found in Table A1 T1 27273K T 238 T R1 Tor 1262K R2 P 015MP PR1 1 a 0044 P 339MPa Chapter 4 713 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 14 Since PRltlt1 and Tgt2Tcr nitrogen is an ideal gas and we use the ideal gas equation of state as the property relation PV 2 mRT Work Calculation W16tlZ VInet0ther12 VVb12 2 2mRT W12 LPdV 1 7w mRTln E 1 For an ideal gas in a closed system mass constant we have Imm2 P1V1 Psz RT1 RT Since the R39s cancel we obtain the combined ideal gas equation Since T 2 Tl R m a m Chapter 4 714 sumsmemmmmmwhyu owed A m P W mRTIn m 1 015MPa 3k 02968 300 1 g kng K quot030Mpa 48451g The networks WWle 0valZ 71845er On a perumt mass basts W k 12 wm12 Equot 7615 39 m kg net work IS negatwe because work IS done an the system dunhg the eompresstoh process Thus e work une an the system IS 184 5 k1 or 184 5 k1 ofwork energy lsrequlred to compress the mtrogeh Example 43 Watens placed m aplstonrcylmder deme mm c 0 11mm Wetghts are placed on the ptstoh to mahtam a constant force on the water as It heatedto 400 c How much work does the water do on the plsf System The wata contamed m the plswnrcyhnder deme Hut Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 16 Property Relation Steam tables Process Constant pressure 5 Steam 10 I I I I I 10quot 103 39339 n E 1 n 102 101 100 I 101 10 3 10 2 io 1 100 101 102 Vm3kg Work Calculation Since there is no Wother mentioned in the problem the net work is 2 2 VVnetlZ VVb12 LPdV PJIdV PV2 Since the mass of the water is unknown we calculate the work per unit mass W P Wb12 1112 Vzm PV2V1 At T1 20 C Psalt 2339 kPa Since P1 gt 2339 kPa state 1 is compressed liquid Thus Chapter 4 716 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 17 v1 vf at 20 C 0001002 1113 kg At P2 P1 01 MPa T2 gt Tsat at 01 MPa 99610C So state 2 is superheated Using the superheated tables at 01 MPa 400 C v2 31027 m3kg wb12 PV2 V1 3 3 01MPa31027 0001002KW kg MPa mkPa 3102 kg The water does work on the piston in the amount of 3102 kJkg Chapter 4 717 Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 18 Example 44 One kilogram of water is contained in a pistoncylinder device at 100 C The piston rests on lower stops such that the volume occupied by the water is 0835 m3 The cylinder is fitted with an upper set of stops When the piston rests against the upper stops the volume enclosed by the piston cylinder device is 0841 m3 A pressure of 200 kPa is required to support the piston Heat is added to the water until the water eXists as a saturated vapor How much work does the water do on the piston System The water contained in the pistoncylinder device P Stops System Boundary Stops Property Relation Steam tables V Process Combination of constant volume and constant pressure processes to be shown on the Pv diagram as the problem is solved Work Calculation The specific volume at state 1 is V1 0835 m3 3 v1 08353 m lkg kg Chapter 4 718 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 19 At T1 100 C 3 3 v 0001044 v 16720 f kg g kg Therefore vflt v1 lt vg and state 1 is in the saturation region so P1 10135 kPa Show this state on the Pv diagram Now let s consider the processes for the water to reach the final state Process 12 The volume stays constant until the pressure increases to 200 kPa Then the piston will move 3 m 2 2 v1 0835 kg Process 23 Piston lifts off the bottom stops while the pressure stays constant Does the piston hit the upper stops before or after reaching the saturated vapor state v Let39s set 3 3 zzmzmm m 1kg kg At P3 P2 200 kPa 3 3 v 0001061 v 088578 f kg g kg Thus vflt v3 lt vg So the piston hits the upper stops before the water reaches the saturated vapor state Now we have to consider a third process Chapter 4 719 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 20 Process 34 With the piston against the upper stops the volume remains constant during the final heating to the saturated vapor state and the pressure increases Because the volume is constant in process 3to4 v4 123 0841 m3kg and v4 is a saturated vapor state lnterpolating in either the saturation pressure table or saturation temperature table at v4 vg gives State 4 P4 2113 kPa I 122 C v4 vg The net work for the heating process is the other work is zero 4 2 3 4 Wm14 W14 JlPdV J1PdV LPdV J3PdV 0mPv3 v20 m3 k 1k 200kPa 0841 0835 lt ggtlt x gt kg mgkpa 12 k Later in Chapter 4 we will apply the conservation of energy or the first law of thermodynamics to this process to determine the amount of heat transfer required Example 45 Air undergoes a constant pressure cooling process in which the temperature decreases by 100 C What is the magnitude and direction of the work for this process Chapter 4 720 Smamsmdmedz xs e mnvabgYACmgdkMAEnks 421 v M menu you commn Prgeerg RelaLiun Ideal gas law PV RT Prucess Constant pressure ka Czlculatinn Neglecting Lhe otherquot work 2 0 W17 LPW PV2 7V1 mm 7 T1 Theworkpaumt mass 15 Rm 4 kl kl 02877 7100 72877 lt Kx K kg The work dune an the am 28 7 kJkg mym 4x Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 22 Example 46 Find the required heat transfer to the water in Example 44 Review the solution procedure of Example 44 and then apply the first law to the process Conservation of Energy Em Eout AE Qnetl4 Wetl4 AU14 71 In Example 44 we found that WM 12 k n The heat transfer is obtained from the first law as Qnetl4 Wnetl4 AU14 Where AU14 2 U4 U1 2 mu4 1 At state 1T1100 Cv1 0835 m3kg and vflt v1 lt vg at T1 The quality at state 1 is Chapter 4 722 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 23 V1 2 Vf xlvfg V V 0835 0 001043 1 x f o 499 1 vfg 16720 000104339 ul 2 uf xlufg 41906 O49920870 214605 k 8 Because state 4 is a saturated vapor state and v4 0841 m3kg interpolating in either the saturation pressure table or saturation temperature table at v4 vg gives u4 253148 g kg So AU14 mu4 u1 1 kg253148 14605 kg g 10710k The heat transfer is Chapter 4 723 Student Study Guide for 5 h edition of Thetmod arnics by Y A Qengel amp M A Boles 47 24 Qnet14 Wnet14 AU14 12k10710k 10722 k Heat in the amount of 107242 M is added to the water Chapter 4 724 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 25 Specific Heats and Changes in Internal Energy and Enthalpy for Ideal Gases Before the first law of thermodynamics can be applied to systems ways to calculate the change in internal energy of the substance enclosed by the system boundary must be determined For real substances like water the property tables are used to find the internal energy change For ideal gases the internal energy is found by knowing the specific heats Physics defines the amount of energy needed to raise the temperature of a unit of mass of a substance one degree as the specific heat at constant volume CV for a constantvolume process and the specific heat at constant pressure Cp for a constantpressure process Recall that enthalpy h is the sum of the internal energy u and the pressurevolume product Pv huPv In thermodynamics the specific heats are defined as CV e and CF e 6T V 6T P Simple Substance The thermodynamic state of a simple homogeneous substance is specified by giving any two independent intensive properties Let39s consider the internal energy to be a function of T and v and the enthalpy to be a function of T and P as follows uuTv and hhTP Chapter 4 725 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles The total differential of u is du dT dv 6T V 9V T 0 du CvdTj dv 9V T The total differential of h is cm j m dP M P aP T 0 dh C dT j dP aP T Using thermodynamic relation theory we could evaluate the remaining partial derivatives of u and h in terms of functions of Pv and T These functions depend upon the equation of state for the substance Given the specific heat data and the equation of state for the substance we can develop the property tables such as the steam tables Chapter 4 726 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 27 Ideal Gases For ideal gases we use the thermodynamic function theory of Chapter 12 and the equation of state Pv R7 to show that u h C V and Cp are functions of temperature alone For example when total differential for u uTv is written as above the function theory of Chapter 12 shows that V duszdT P dv 6T V Let s evaluate the following partial derivative for an ideal gas 6 Jar P P 9V T 6T V du 2 CV dT dv T For ideal gases RT P V 11 aTv v T P2P P20 9V T v Chapter 4 727 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 28 This result helps to show that the internal energy of an ideal gas does not depend upon specific volume To completely show that internal energy of an ideal gas is independent of specific volume we need to show that the specific heats of ideal gases are functions of temperature only We will do this later in Chapter 12 A similar result that applies to the enthalpy function for ideal gases can be reviewed in Chapter 12 Then for ideal gases CVCVT and 0 9V T 0 6h C C T and p p apl The ideal gas specific heats are written in terms of ordinary differentials as a CV j dT ideal gas a dT ideal gas Chapter 4 728 Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 29 Using the simple dumbbell model for diatomic ideal gases statistical thermodynamics predicts the molar specific heat at constant pressure as a function of temperature to look like the following k p kmolK Vibration mode Ru Rotation mode Ru Dumbbell model Ru Translation mode The following figure shows how the molar specific heats vary with temperature for selected ideal gases Chapter 4 729 Student Study Guide for 5 11 edition of Thermodynamics by Y A Cengel amp M A Boles 47 30 5 p0 klquotkm I K A13 He Ne Kr Xe Rn l l I 1000 2000 3000 39I cmpemlure K The differential changes in internal energy and enthalpy for ideal gases become duzCVdT dh CPdT The change in internal energy and enthalpy of ideal gases can be expressed as 2 Au uz u1J1CVltTgtdT CVWltT2 7 2 Ahhz 471 J1CPltTgtdT CPltT2 Tgt where C Km and C Rave are average or constant values of the specific heats over the temperature range We will drop the ave subscript shortly Chapter 4 730 Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 31 PrVDlAGRAM FOR SEVERAL PROCESSES FOR AN IDEAL GAS In the above figure an ideal gas undergoes three different process between the same two temperatures Process l2a Constant volume Process l2b P a bV a linear relationship Process l2c Constant pressure These ideal gas processes have the same change in internal energy and enthalpy because the processes occur between the same temperature limits Aua Aub Auc fCdeT Aha Ahb Aha fCPde Chapter 4 731 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 32 To find Au and Ah we often use average or constant values of the specific heats Some ways to determine these values are as follows 1 The best average value the one that gives the exact results See Table A2c for variable specific data 2 2 LCV TdT LC TdT vave and Pave T2 T1 T2 T1 2 Good average values are C qxnwraan C annwrgm vave 2 Have 2 an Cvave CV Tave CPave CP Tave where T Zzz ave 2 3 Sometimes adequate and most often used values are the ones evaluated at 300 K and are given in Table A2a C CV300K and C vave Have 2 CP300K Let39s take a second look at the definition of Au and All for ideal gases Just consider the enthalpy for now Np h UMT Chapter 4 732 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 33 Let39s perform the integral relative to a reference state where h href at T Tref Ah TrefC T dT39 TZC T dT39 hz hi T P T P 1 ref or Ah h2 4 CPT39dT CPT39dT 05 href A href At any temperature we can calculate the enthalpy relative to the reference state as T I I h hm Lyef CPT dT OI hh T I I ref ij CPT dT A similar result is found for the change in internal energy T u um j CVT39dT39 Tref These last two relations form the basis of the air tables Table Al7 on a mass basis and the other ideal gas tables Tables A18 through A25 on a mole basis When you review Table Al7 you will find h and u as functions of T in K Since the parameters Pr v and so also found in Table Al7 apply to air only in a particular process call isentropic you should ignore these parameters until we study Chapter 7 The reference state for these tables is defined as Chapter 4 733 rt CnlM PM 4734 urzf 0 at Trzf OK hwf 0 at 22 OK A partial listing of data similar to that found in Table A l 7 is shown in the following figure 39 1 M 2 In the analysis to follow the ave notation is dropped In most applications for ideal gases the values of the specific heats at 300 K given in Table A2 are adequate constants Chapte l 734 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 35 Exercise Determine the average specific heat for air at 305 K C ave Answer 1005 kJkgK approximate the derivative of h with respect to T as differences Relation between CP and CV for Ideal Gases Using the definition of enthalpy h u Pv and writing the differential of enthalpy the relationship between the specific heats for ideal gases is h u Pv dh du dRT CPdT CVdT RdT CP 2 CV R where R is the particular gas constant The specific heat ratio k uids texts often use y instead of k is defined as k 2 g CV Chapter 4 735 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 36 Extra Problem Show that R CPZ and CVz Example 29 Two kilograms of air are heated from 300 to 500 K Find the change in enthalpy by assuming a Empirical specific heat data from Table A2c b Air tables from Table Al7 c Specific heat at the average temperature from Table A2c 1 Use the 300 K value for the specific heat from Table A2a a Table A2c gives the molar specific heat at constant pressure for air as 5P 281l01967x10392T 04802x10395 T2 1966x10399 T3 L kmolK The enthalpy change per unit mole is Chapter 4 736 Student Study Guide for 5 h edition of Thetmod arnics byY A Qengel amp M A Boles 47 37 2 Ah hz 411 j1 CPTdT 500K 2 5 2 J300K2811 01967x1039 T 04802x1039 T 1966x10399 T3 dT 2 5 2811an2 MT3 2 3 1966x10399 T4 500K T 300K 590949 I kmol N7 590949L k AhZ 2039 M 2897 g kg kmol AH mAh 2 kg2039 kg 40798 k g b Using the air tables Table A17 at T1 300 K h1 30019 kJkg and at T2 500 K h2 50302 kJkg AH mAh 2 kg50302 30019 kg 40566 k g Chapter 4 737 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 38 The results of parts a and b would be identical if Table Al7 had been based on the same specific heat function listed in Table A2c c Let s use a constant speci c heat at the average temperature Tave 300 500qu 400 K At Tave Table A2 gives CI 1013 kJkgK For Cp constant Ah 72 hl CPave7g 7l 2 1013 i500 300 K kgK 2026 E kg AH mAh 2 kg2026 kg 4052 k g d Using the 300 K value from Table A2a Cp 1005 kJkg K For C P constant N1hrh CPTzTl 21005 L500 300 K 2010 kgK kg AH mAh 2 kg2010 kg 4020 k g Chapter 4 738 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles Extra Problem Find the change in internal energy for air between 300 K and 500 K in kJkg Chapter 4 739 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles The Systematic Thermodynamics Solution Procedure When we apply a methodical solution procedure thermodynamics problems are relatively easy to solve Each thermodynamics problem is approached the same way as shown in the following which is a modification of the procedure given in the text 1 J U LI 0 Thermodynamics Solution Method Sketch the system and show energy interactions across the boundaries Determine the property relation Is the working substance an ideal gas or a real substance Begin to set up and fill in a property table Determine the process and sketch the process diagram Continue to fill in the property table Apply conservation of mass and conservation of energy principles Bring in other information from the problem statement called physical constraints such as the volume doubles or the pressure is halved during the process Develop enough equations for the unknowns and solve Chapter 4 740 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 41 Example 47 A tank contains nitrogen at 27 C The temperature rises to 127 C by heat transfer to the system Find the heat transfer and the ratio of the final pressure to the initial pressure System Nitrogen in the tank System p boundary T2127 C PVdiagram for a constant volume process Property Relation Nitrogen is an ideal gas The ideal gas property relations apply Let s assume constant specific heats You are encouraged to rework this problem using variable specific heat data Process Tanks are rigid vessels therefore the process is constant volume Conservation of Mass m2 1711 Using the combined ideal gas equation of state Chapter 4 741 Student Study Guide for 5 h edition of Thermod arnics by Y A Qengel amp M A Boles 47 42 13sz PM 7 71 Since R is the particular gas constant and the process is constant volume V2 V1 127273K 21333 P1 7 27 273K Conservation of Energy The first law closed system is For nitrogen undergoing a constant volume process dV 0 the net work is Wother 2 Wet 0 WW LPdV 0 n Using the ideal gas relations with Wnet 0 the first law becomes constant specific heats Chapter 4 742 Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 43 QM 0AUmeVdTmCVltT2 71gt The heat transfer per unit mass is Qne qnet ZCV77l m 0743 L027 27K kgK 743 g kg Example 48 Air is expanded isothermally at 100 C from 04 MPa to 01 MPa Find the ratio of the final to the initial volume the heat transfer and work System Air contained in a pistoncylinder device a closed system Process Constant temperature 1 System P boundary P Vdiagram for T constant Chapter 4 743 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 44 Property Relation Assume air is an ideal gas and use the ideal gas property relations with constant specific heats PV 2 mRT A 2 CV Conservation of Energy Ein Eout AE Qnet VVnet 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 VVnetJZ Maximum VVsz Wm deV fng mRT In E Chapter 4 744 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 45 Conservation of Mass For an ideal gas in a closed system mass constant we have m 2 m2 BKB R7 RT2 Since the R39s cancel and T2 T1 naMMm V1 P2 01MPa Then the work expression per unit mass becomes W wb 12 I2 RTln E 1 Wm 0287 100 273K 1114 kg K kJ 21484 kg The net work per unit mass is wnet12 2 0 wb12 1484 kg Chapter 4 745 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 46 Now to continue with the conservation of energy to find the heat transfer Since T 2 T1 constant AU12 mAuu mCV7g 0 So the heat transfer per unit mass is Qnet qnet m qnet wnet Au 2 0 qnet wnet 1484 g kg The heat transferred to the air during an isothermal expansion process equals the work done Examples Using Variable Specific Heats Review the solutions in Chapter 4 to the ideal gas examples where the variable specific heat data are used to determine the changes in internal energy and enthalpy Extra Problem for You to Try An ideal gas contained in a pistoncylinder device undergoes a polytropic process in which the polytropic exponent n is equal to k the ratio of specific heats Show that this process is adiabatic When we get to Chapter 7 you will find that this is an important ideal gas process Chapter 4 746 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 47 Internal Energy and Enthalpy Changes of Solids and Liquids We treat solids and liquids as incompressible substances That is we assume that the density or specific volume of the substance is essentially constant during a process We can show that the specific heats of incompressible substances see Chapter 12 are identical CPZCVZC L kgK The speci c heats of incompressible substances depend only on temperature therefore we write the differential change in internal energy as du CV dT CdT and assuming constant specific heats the change in internal energy is Au CAT CT 7 Recall that enthalpy is defined as huPv The differential of enthalpy is dh du Pdv vdP For incompressible substances the differential enthalpy becomes Chapter 4 747 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 48 dv O dh du de 0 vdP dh du vdP Integrating assuming constant specific heats Ah 2 AuvAP CATVAP For solids the specific volume is approximately zero therefore Ahsolid A d VOAP Ahsolid Ausolid 5 CAT soli For liquids two special cases are encountered 1 Constantpressure processes as in heaters AP 0 Ahliquid AI lll39quid E 2 Constanttemperature processes as in pumps AT 0 0 Nahquid Auqu vAP 2 CA T vAP Ahliquid VAP We will derive this last eXpression for All again once we have discussed the first law for the open system in Chapter 5 and the second law of thermodynamics in Chapter 7 The specific heats of selected liquids and solids are given in Table A3 Chapter 4 748 Student Study Guide for 5 h edition of Thermodgmarnics byY A Qengel amp M A Boles 47 49 Example 48 Incompressible Liquid A twoliter bottle of your favorite beverage has just been removed from the trunk of your car The temperature of the beverage is 35 C and you always 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 twoliter bottles in onehalf hour What rate of heat removal in kW is required Assuming that your refrigerator can accomplish this and that electricity costs 85 cents per kWhr how much will it cost to cool these 10 bottles I ystem The liquid in the constant volume closed system container SJrtem boundary l Qout The heat removed Property Relation Incompressible liquid relations let s assume that the beverage is mostly water and takes on the properties of liquid water The specific volume is 0001 m3kg C 418 kJkgK Process Constant volume V2Vl Chapter 4 749 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 50 Conservation of Mass m2 2 m1 m 3 V 2 L m quot72 2 3 lOOOL 22kg V 0001 kg Conservation of Energy The first law closed system is Ein Eout AE Since the container is constant volume and there is no other work done on the container during the cooling process we have VVnet net other Wb0 The only energy crossing the boundary is the heat transfer leaving the container Assuming the container to be stationary the conservation of energy becomes Chapter 4 750 Student Study Guide for 5 h edition of Thermodgmarnics by Y A Qengel amp M A Boles 47 51 E0ut AE Q0m 2 AU 2 mCAT kJ Q0m 2 kg4 Big 1800 35K Q0m 2092 kJ Q0 2092 Id The heat transfer rate to cool the 10 bottles in onehalf hour is k Q 10bottles2092 bottle 1hr k W 05m 3600s E S 1162 kW Cost 1162 kW05 hr kW hr 005 Chapter 4 751 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 61 Chapter 6 The Second Law of Thermodynamics The second law of thermodynamics states that processes occur in a certain direction not in just any direction Physical processes in nature can proceed toward equilibrium 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 will not reverse itself spontaneously Some external inputs energy must be expended to reverse the process As it falls down the waterfall water can be collected in a water wheel cause a shaft to rotate coil a rope onto the shaft and lift a weight So the energy of the falling water is captured as potential energy increase in the weight and the first law of thermodynamics is satisfied However there are losses associated with this process friction Allowing the weight to fall causing the shaft to rotate in the opposite direction will not pump all of the water back up the waterfall Spontaneous processes can proceed only in a particular direction The first law of thermodynamics gives no information about direction it states only that when one form of energy is converted into another identical quantities of energy are involved regardless of the feasibility of the process We know by experience that heat flows spontaneously from a high temperature to a low temperature But heat owing from a low temperature to a higher temperature with no expenditure of energy to cause the process to take place would not violate the first law The first law is concerned with the conversion of energy from one form to another Joule39s experiments showed that energy in the form of heat could not be completely converted into work however work energy can be completely converted into heat energy Evidently heat and work are not Chapter 6l Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 62 completely interchangeable forms of energy Furthermore when energy is transferred from one form to another there is often a degradation of the supplied energy into a less useful form We shall see that it is the second law of thermodynamics that controls the direction processes may take and how much heat is converted into work A process will not occur unless 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 following definitions Heat thermal reservoir A heat reservoir is a sufficiently large system in stable equilibrium to which and from which finite amounts of heat can be transferred without any change in its temperature A high temperature heat reservoir from which heat is transferred is sometimes called a heat source A low temperature heat reservoir to which heat is transferred is sometimes called a heat sink Work reservoir A work reservoir is a sufficiently large system in stable equilibrium to which and from which finite amounts of work can be transferred adiabatically without any change in its pressure Chapter 62 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 63 Thermodynamic cycle A system has completed a thermodynamic cycle when the system undergoes a series of processes and then returns to its original state so that the properties of the system at the end of the cycle are the same as at its beginning Thus for whole numbers of cycles P fZP 1 Tf 7 uf 2 L1 vf vi etc Heat Engine A heat engine is a thermodynamic system operating in a thermodynamic cycle to which net heat is transferred and from which net work is delivered The system or working uid undergoes a series of processes that constitute the heat engine cycle The following gure illustrates a steam power plant as a heat engine operating in a thermodynamic cycle Energy source such as a hlmace Qm swam boundary lf39 om Energy sink such a 1 ammsplum Chapter 63 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 64 Thermal Efficiency 11 h The thermal efficiency is the index of performance of a workproducing device 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 Desired Result nth Required Input For a heat engine the desired result is the net work done and the input is the heat supplied to make the cycle operate The thermal efficiency is always less than 1 or less than 100 percent VVneI out nth Q where VVnet out VVout Wm Qin i Qnet Here the use of the in and out subscripts means to use the magnitude take the positive value of either the work or heat transfer and let the minus sign in the net expression take care of the direction Chapter 64 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles Now apply the first law to the cyclic heat engine 0 Cyclic Qnet in W M net out Whenout Qnetin What out Qin Qout The cycle thermal ef ciency may be written as Qm Q Q Qm 1 Qm nth Cyclic devices such as heat engines refrigerators and heat pumps often operate between a hightemperature reservoir at temperature T H and a low temperature reservoir at temperature T L Chapter 65 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 66 IIigilmnpcmturc reservoir m T Qquot EN age 3 9 p39 Lowtemperature reservoir 11 TL The thermal efficiency of the above device becomes g Tl 1 QH Example 6 1 A steam power plant produces 50 MW of net work while burning fuel to produce 150 MW of heat energy at the high temperature Determine the cycle thermal efficiency and the heat rejected by the cycle to the surroundings W net out QH M 0333 or 333 150 MW nth Chapter 66 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 67 WWW QH QL QL QH Menard 150 MW 50 MW 100 MW Heat Pump A heat pump is a thermodynamic system operating in a thermodynamic cycle that removes heat from a lowtemperature body and delivers heat to a hightemperature body To accomplish this energy transfer the heat pump receives external energy in the form of work or heat from the surroundings While the name heat pump is the thermodynamic term used to describe a cyclic device that allows the transfer of heat energy from a low temperature to a higher temperature we use the terms refrigerator and heat pump to apply to particular devices Here a refrigerator is a device that operates on a thermodynamic cycle and extracts heat from alow temperature medium The heat pump also operates on a thermodynamic cycle but rejects heat to the hightemperature medium The following figure illustrates a refrigerator as a heat pump operating in a thermodynamic cycle Chapter 67 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 68 Surrounding incdiuml such as the kitchen air Qh39 CONDENSER 800 kPal 800 kPal 30 C 60 C W net in 1 20 kPa L 720 C Refrigerated space Coef cient of Performance COP The index of performance of a refrigerator or heat pump is expressed in terms of the coef cient of performance COP the ratio of desired result to input This measure of performance may be larger than 1 and we want the COP to be as large as possible Desired Result Required Input COP For the heat pump acting like a refrigerator or an air conditioner the primary function of the deVice is the transfer of heat from the low temperature system Chapter 68 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles Wann envimmnem at TH gt TL Required input net in quot Q Cold refrigerated a T For the refrigerator the desired result is the heat supplied at the low temperature and the input is the net work into the device to make the cycle operate QL COPR net in Now apply the rst law to the cyclic refrigerator QL QH Aljcycle 0 VVin Wreath QH QL and the coef cient of performance becomes Chapter 69 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 610 COPR L QH QL For the device acting like a heat pump the primary function of the device is the transfer of heat to the hightemperature system The coefficient of performance for a heat pump is COPHP QH QH VVnetin QH QL Note under the same operating conditions the COPHP and COPR are related by COPHP COPR 1 Heat Pump and Air Conditioner Ratings Heat pumps and air conditioners are rated using the SEER system SEER is the seasonal adjusted energy efficiency bad term for HP and AC devices rating The SEER rating is the amount of heating cooling on a seasonal basis in Btuhr 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 12000 Btuhr 211 kJmin Chapter 610 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 61 1 Second Law Statements The following two statements of the second law of thermodynamics are based on the de nitions 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 heat from a single reservoir and produce a net amount of work The KelvinPlanck statement of the second law of thermodynamics states that no heat engine can produce a net amount of work while exchanging heat with a single reservoir only In other words the maximum possible ef ciency is less than 100 percent nmlt 100 39l39hcn39nal energy reservoir HEAT HNGIN H 920 Q 100 kw l Via 0i i 100 kW l Heat engine that violates the KelvinPlanck statement of the second law Chapter 611 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 612 Clausius statement of the second law The Clausius statement of the second law states that it is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lowertemperature body to a higher temperature body Warm environment Q 7 5m 3mm 039 Q 7 5k Cold refrigerated space 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 be expended to force heat to ow from a lowtemperature medium to a high temperature medium Thus the COP of a refrigerator or heat pump must be less than in nity COPltoo A Violation of either the KelvinPlanck or Clausius statements of the second law implies a Violation of the other Assume that the heat engine Chapter 612 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 613 shown below is violating the KelvinPlanck statement by absorbing heat from a single reservoir and producing an equal amount of work W The output of the engine drives a heat pump that transfers an amount of heat QL from the lowtemperature thermal reservoir and an amount of heat QH QL to the hightemperature thermal reservoir The combination of the heat engine and refrigerator in the left f1gure acts like a heat pump that transfers heat QL from the lowtemperature reservoir without any external energy input This is a violation of the Clausius statement of the second law I39Iighlcmpcratun reservoir Hightemperature reservoir Qn 311 QL HEAT W r ENG I Rlul R lb 77a 100 OH ikA39I 0R 0L Lowrtempefatui39e reservoir Inwrtempm39ami39e reservoir 11 A i39etngemtor hash is poweredL b The eqluvulem i39emgei39atoi39 by n 100 ailment heat engine Perpetual Motion Machines Any device that violates the rst or second law of thermodynamics is called a perpetualmotion machine If the device violates the rst law it is a perpetualmotion machine of the rst kind If the device violates the second law it is a perpetualmotion machine of the second kind Chapter 613 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 614 Reversible Processes A reversible process is a quasiequilibrium or quasistatic process with a more restrictive requirement Internally reversible process The internally reversible process is a quasiequilibrium process which once having taken place can be reversed and in so doing leave no change in the system This says nothing about what happens to the surroundings about the system Totally or externally reversible process The externally reversible process is a quasiequilibrium process which once having taken place can be reversed and in so doing leave no change in 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 of the following Friction Unrestrained expansion of gases Heat transfer through a finite temperature difference Mixing of two different substances Hysteresis effects 12R losses in electrical circuits Any deviation from a quasistatic process Chapter 614 Student Study Guide for 5 11 edition of Thermodynamics by Y A Qengel amp M A Boles 615 The Carnot Cycle French military engineer Nicolas Sadi Carnot 17691832 was among the rst to study the principles of the second law of thermodynamics Carnot was the rst to introduce the concept of cyclic operation and devised a reversible cycle that is composed of four reversible processes two isothermal and two adiabatic The Carnot Cycle Process 12 Reversible isothermal heat addition at high temperature T H gt T L to the working uid in a piston cylinder device that does some boundary work Process 23 Reversible adiabatic expansion during which the system does work as the working uid temperature decreases from T H to T L Process 34 The system is brought in contact with a heat reservoir at T L lt T H and a reversible isothermal heat exchange takes place while work of compression is done on the system Process 41 A reversible adiabatic compression process increases the working uid temperature from T L to T H Chapter 61 5 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 616 You may have observed that power cycles operate in the clockwise direction when plotted on a process diagram The Carnot cycle may be reversed in which it operates as a refrigerator The refrigeration cycle operates in the counterclockwise direction P Carnot Principles The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the KelvinPlanck and Clausius statements A heat engine cannot operate by exchanging heat with a single heat reservoir and a refrigerator cannot operate without net work input om an external source Consider heat engines operating between two f1xed temperature reservoirs at T H gt T L We draw two conclusions about the thermal efficiency of reversible and irreversible heat engines known as the Carnot principles aThe efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs Tl th lt Tl th Carnot b The efficiencies of all reversible heat engines operating between the same two constanttemperature heat reservoirs have the same efficiency Chapter 616 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 617 As the result of the above Lord Kelvin in 1848 used energy as a thermodynamic property to de ne temperature and deVised a temperature scale that is 39 J J J ofthe ther 39 The following is Lord Kelvin s Carnot heat engine arrangement Thermal energy reservoir W3 Q3 Q1 quotmanual energy Since the thermal ef ciency in general is QL 11m 1 H For the Carnot engine this can be written as nth gTL9TH1fTL TH Chapter 617 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 618 Considering engines A B and C 292 Q3 Q2 Q3 This looks like fT1T3fT1YZfYET3 One way to define the f function is e22eT3 eT3 em 62 em fT1T3 The simplest form of 9 is the absolute temperature itself T T T 3 f 1 3 T1 The Carnot thermal efficiency becomes T Tl Ih rev 2 1 L H This is the maximum possible efficiency of a heat engine operating between two heat reservoirs at temperatures T H and T L Note that the temperatures are absolute temperatures Chapter 61 8 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 619 These statements form the basis for establishing an absolute temperature scale also called the Kelvin scale related to the heat transfers between a reversible device and the high and lowtemperature heat reservoirs by ampi QH TH Then the QHQL ratio can be replaced by T H T L for reversible devices where T H and T L are the absolute temperatures of the high and low temperature heat reservoirs respectively This result is only valid for heat exchange across a heat engine operating between two constant temperature heat reservoirs These results do not apply when the heat exchange is occurring with heat sources and sinks that do not have constant temperature The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows lt n h M irreversible heat engine nth nth rev reversible heat engine gt nth rev impossible heat engine Reversed Carnot Device Coefficient of Performance If the Carnot device is caused to operate in the reversed cycle the reversible heat pump is created The COP of reversible refrigerators and heat pumps are given in a similar manner to that of the Carnot heat engine as Chapter 619 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 620 COPRz QL 2 1 QHQL QH1 L TL 1 THTL TH1 TL QH COPHP QH QL QHQL QH1 QL TH THTL TH1 TL Again these are the maximum possible COPS for a refrigerator or a heat pump operating between the temperature limits of T H and T L The coefficients of performance of actual and reversible such as Carnot refrigerators operating between the same temperature limits compare as follows lt COPR COPR COPR gt C OPR rev impossible refrigerator m 1rrever51ble refrlgerator m rever31ble refrlgerator Chapter 620 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 621 A similar relation can be obtained for heat pumps by replacing all values of COPR by COPHP in the above relation Example 62 A Carnot heat engine receives 500 M of heat per cycle from a high temperature heat reservoir at 652 C and rejects heat to a lowtemperature heat reservoir at 30 C Determine a The thermal efficiency of this Carnot engine b The amount of heat rejected to the lowtemperature heat reservoir a 30 273K 652 273K 0672 or 672 b i 2 i QH TH 30273K 20328 652273K QL 500kJ0328 164kJ Chapter 621 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 622 Example 63 An inventor claims to have invented a heat engine that develops a thermal efficiency of 80 percent when operating between two heat reservoirs at 1000 K and 300 K Evaluate his claim 300K 1000K 2070 or 70 The claim is false since no heat engine may be more efficient than a Carnot engine operating between the heat reservoirs Example 64 An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2 C while operating in a room where the temperature is 25 OC and has a COP of 135 Is there any truth to his claim COPR QL 2 TL QH QL TH TL 2 273K 25 2K 1196 The claim is false since no refrigerator may have a COP larger than the COP for the reversed Carnot device Chapter 622 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 623 Example 65 A heat pump is to be used to heat a building during the Winter The building is to be maintained at 21 C at all times The building is estimated to be losing heat at a rate of 135000 kJh when the outside temperature drops to 5 OC Determine the minimum power required to drive the heat pump unit for this outside temperature The heat lost by the building has to be supplied by the heat pump QH QM 135000 COPHP Q H T H QH QL TH TL 21273K 21 5K 1131 Chapter 623 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 624 Using the basic definition of the COP COPHP QH VIInet in gt COPHP 135000kJh 1h lkW 1131 3600s kJs 3316kW Chapter 624 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 71 Chapter 7 Entropy A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the de nition of a new property called entropy a quantitative measure of microscopic disorder for a system Entropy is a measure of energy that is no longer available to perform useful work within the current environment To obtain the working de nition of entropy and thus the second law let39s derive the Clausius inequality Consider a heat reservoir giving up heat to a reversible heat engine which in turn gives up heat to a pistoncylinder device as shown below Thermal reservoir TR 5 QR I I A I I Reversible 1 cyclic l l O Wrev device I I I I 5 Q l I T I I I System 3 i 5 Wsys I J Combined system system and cyclic device Chapter 7l Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 72 We apply the first law on an incremental basis to the combined system composed of the heat engine and the system Ein Eout AEC 6QR SIKev 6W3 ys dEc where E is the energy of the combined system Let We be the work done by the combined system Then the first law becomes 5W 25W 5Wsys 78V 5QR dEc If we assume that the engine is totally reversible then 332 Now the total work done is found by taking the cyclic integral of the incremental work WT c R lm If the system as well as the heat engine is required to undergo a cycle then Chapter 72 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 73 ME 0 and the total net work becomes 5Q W T1437 If We is positive we have a cyclic device exchanging energy with a single heat reservoir and producing an equivalent amount of work thus the KelvinPlanck statement of the second law is violated But We can be zero no work done or negative work is done on the combined system and not violate the KelvinPlanck statement of the second law Therefore since T R gt 0 absolute temperature we conclude 5Q WC 2 T1437 0 or g 0 T Here Q is the net heat added to the system QM This equation is called the Clausius inequality The equality holds for the reversible process and the inequality holds for the irreversible process Chapter 73 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 74 Example 71 For a particular power plant the heat added and rejected both occur at constant temperature and no other processes experience any heat transfer The heat is added in the amount of 3150 M at 440 C and is rejected in the amount of 1950 M at 20 C Is the Clausius inequality satisfied and is the cycle reversible or irreversible J WT 3150M 4950M lt0 440273K 20273K 4418 6655 0 K 2237E 0 K Calculate the net work cycle efficiency and Carnot efficiency based on T H and T L for this cycle Q Q0m 3150 1950k 1200k Chapter 74 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 75 m EMO381 or 381 Q1 3150k nth Carnal 1 1w0589 0 589 TH 440273K 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 72 For a particular power plant the heat added and rejected both occur at constant temperature no other processes experience any heat transfer The heat is added in the amount of 3150 M at 440 C and is rejected in the amount of 129446 M at 20 C Is the Clausius inequality satis ed and is the cycle reversible or irreversible 3glt0 T ire w we T in T out Q ll Q so Zn 210m 3150M 129446la1 0 440273K 20273K 4418 4418 0 K Chapter 75 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 76 The Clausius inequality is satisfied Since the cyclic integral is equal to zero the cycle is made of reversible processes What cycle can this be Calculate the net work and cycle efficiency for this cycle Wm Qm QM 3150 129446d 18554 M W 185554 kl net n O589 or 589 Q 3150kJ Definition of Entropy Let s take another look at the quantity 6Qnet S 0 T If no irreversibilities occur within the system as well as the reversible cyclic device then the cycle undergone by the combined system will be internally reversible As such it can be reversed In the reversed cycle case all the quantities will have the same magnitude but the opposite sign Therefore the work WC which could not be a positive quantity in the regular case cannot be a negative quantity in the reversed case Then it follows that WeiWW 0 since it cannot be a positive or negative quantity and therefore for internally reversible cycles Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and Chapter 76 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 77 the inequality for the irreversible ones To develop a relation for the de nition of entropy let us examine this last equation more closely Here we have a quantity whose cyclic integral is zero Let us think for a moment what kind of quantities can have this characteristic We know that the cyclic integral of work is not zero It is a good thing that it is not Otherwise heat engines that work on a cycle such as steam power plants would produce zero net work Neither is the cyclic integral of heat Now consider the volume occupied by a gas in a pistoncylinder device undergoing a cycle as shown below 3 m it 1 m3 cigar2 Arley r 0 When the piston returns to its initial position at the end of a cycle the volume of the gas also returns to its initial value Thus the net change in volume during a cycle is zero This is also expressed as dV0 Chapter 77 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 78 We see that the cyclic integral of a property is zero A quantity whose cyclic integral is zero depends only on the stale and not on the process path thus it is a property Therefore the quantity aneDimev must be a property Inclass Example Consider the cycle shown below composed of two reversible processes A and B Apply the Clausius inequality for this cycle What do you conclude about these two integrals 5 5 f f along pathA along pathB A cycle composed oftwo reversible processes Apply the Clausius inequality for the cycle made of two internally reversible processes 6Qnet Z 0 T intrev Chapter 78 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 79 2 5QM 2 5Qm You should find ii T 1 iii T 1 111 rev 111 rev alongth A alongth B Since the quantity QneDinlm is independent of the path and must be a property we call this property the entropy S The entropy change occurring during a process is related to the heat transfer and the temperature of the system The entropy is given the symbol S kJK and the specific entropy is s kJkgK The entropy change during a reversible process sometimes called an internally reversible process is defined as dS 5Qnet T int rev 25Qnet S 511 T int rev Consider the cycle lAZBl shown below where process A is arbitrary that is it can be either reversible or irreversible and process B is internally reversible 3911 Jof v LL and in v i w processes A cycle r Chapter 79 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 710 5th KT 3 0 f5Qnerf5Qmj lt0 1 T I 2 T intrev alongl along The integral along the internally reversible path process B is the entropy change 51 752 Therefore 2 5 j S1 52 g o 1 T or 2 5 S2 SI Qnet 1 T In general the entropy change during a process is de ned as d3 2 5Qnet T where holds for the internally reversible process gt holds for the irreversible process Consider the effect of heat transfer on entropy for the internally reversible case Chapter 710 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 7ll dS 5Qnet T Which temperature T is this one If SQM gt 0 then 615 gt 0 SQM 0 then 615 0 SQM lt 0 then 615 lt 0 This last result shows why we have kept the subscript net on the heat transfer Q It is important for you to recognize that Q has a sign depending on the direction of heat transfer The net subscript is to remind us that Q is positive when added to a system and negative when leaving a system Thus the entropy change of the system will have the same sign as the heat transfer in a reversible process From the above we see that for a reversible adiabatic process d3 0 S2 2 S1 The reversible adiabatic process is called an t39sentropt39c process Entropy change is caused by heat transfer and irreversibilities Heat transfer to a system increases the entropy heat transfer from a system decreases it The effect of irreversibilities is always to increase the entropy In fact a process in which the heat transfer is out of the system may be so irreversible that the actual entropy change is positive Friction is one source of irreversibilities in a system Chapter 71 1 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 712 The entropy change during a process is obtained by integrating the dS equation over the process 2 SQM kJ ASsys S S1 2J1 T Here the inequality is to remind us that the entropy change of a system during an irreversible process is always greater than fSQT called the entropy transfer That is some entropy is generated or created during an irreversible process and this generation is due entirely to the presence of irreversibilities The entropy generated during a process is called entropy generation and is denoted as Sgen We can remove the inequality by noting the following 2 5Qne k ASsys S S1 fl 5g f Sgen is always a positive quantity or zero Its value depends upon the process and thus it is not a property Sgen is zero for an internally reversible process The integral fSQ T is performed by applying the first law to the process to obtain the heat transfer as a function of the temperature The integration is not easy to perform in general Chapter 712 Student Study Guide for 5 h edition of Thermod gmi cs by Y A Cengel amp M A Boles 713 De nition of Second Law of Thermodynamics Now consider an isolated system composed of several subsystems exchanging energy among themselves Since the isolated system has no energy transfer across its system boundary the heat transfer across the system boundary is zero Isolated Subsystem 1 N AStotal 2 ASi gt 0 Subsystem F1 2 Subsystem ff Subsystem 3 l N Applying the definition of entropy t0 the isolated system 0 adiabatic gt 2 6QMI isolated T l The total entropy change for the isolated system is AS Isolated Z O This equation is the working definition of the second law of thermodynamics The second law known as the principle of increase of entropy is stated as The total entropy change of an isolated system during a process always increases or in the limiting case of a reversible process remains constant Chapter 713 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 714 Now consider a general system exchanging mass as well as energy with its surroundings Isolated syslcm m tdary occ Surroundings SAS gen total Assys Z AS314 2 0 where holds for the totally reversible process gt holds for the irreversible process Thus the entropy generated or the total entropy change sometimes called the entropy change of the universe or net entropy change due to the process of this isolated system is positive for actual processes or zero for reversible processes The total entropy change for a process is the amount of entropy generated during that process Sgen and it is equal to the sum of the entropy changes of the system and the surroundings The entropy changes of the important system closed system or control volume and its surroundings do not both have to be positive The entropy for a given system important or surroundings may decrease during a process but the Chapter 714 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 715 sum of the entropy changes of the system and its surroundings for an isolated system can never decrease Entropy change is caused by heat transfer and irreversibilities Heat transfer to a system increases the entropy and heat transfer from a system decreases it The effect of irreversibilities is always to increase the entropy The increase in entropy principle can be summarized as follows gt 0 Irreversible processes S AST0ml 0 Rever31ble processes gen lt 0 Impossible processes Some Remarks about Entropy i Processes can occur in a certain direction only not in just any direction such that Sgen 2 0 J Entropy is a nonconserved property and there is no such thing as the conservation of entropy principle The entropy of the universe is continuously increasing 3 The performance of engineering systems is degraded by the presence of irreversibilities and entropy generation is a measure of the magnitudes of the irreversibilities present during that process Chapter 715 Student Study Guide for 5Lh edition of Thermod amics by Y A Qengel amp M A Boles 716 Heat Transfer as the Area under a TS Curve For the reversible process the equation for dS implies that dS 5Qnet T SQM T dS or the incremental heat transfer in a process is the product of the temperature and the differential of the entropy the differential area under the process curve plotted on the TS diagram Q deS Internally reversible process aATdS 5Q In the above figure the heat transfer in an internally reversible process is shown as the area under the process curve plotted on the TS diagram Chapter 716 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 7l7 Isothermal Reversible Process For an isothermal reversible process the temperature is constant and the integral to find the entropy change is readily performed If the system has a constant temperature T 0 the entropy change becomes 28Qnet Qnet ASZS2S1J1 77 For a process occurring over a varying temperature the entropy change must be found by integration over the process Adiabatic Reversible Isentropic Process For an adiabatic process one in which there is no heat transfer the entropy change is 0 adiabatic 25f Aszg azi 7f ASamp amp20 If the process is adiabatic and reversible the equality holds and the entropy change is Meg azo ampamp or on a per unit mass basis S s m S2 2 S1 Chapter 71 7 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 718 The adiabatic reversible process is a constant entropy process and is called isentropic As will be shown later for an ideal gas the adiabatic reversible process is the same as the polytropic process Where the polytropic exponent n k CICV The principle of increase of entropy for a closed system exchanging heat with its surroundings at a constant temperature T sun is found by using the equation for the entropy generated for an isolated system Surroundings A general closed system a cup of coffee exchanging heat with its surroundings s ASm AssysZAs 20 gen surr AS39sys S2 Slsys AS Qnet surr surr SAS gen total Qnet surr ms2 s1sys ZO surr Qnet surr Qnet sys Qout sys Qout sys Where Chapter 71 8 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 719 Effect of Heat Transfer on Entropy Let39s apply the second law to the following situation Consider the transfer of heat from a heat reservoir at temperature T to a heat reservoir at temperature T AT gt 0 where AT gt 0 as shown below HR Two heat reservoirs at exchanging heat over a T nite temperature difference T Q HR at T AT The second law for the isolated system composed of the two heat reservoirs is Sgen A Sltotal Assys ZASsurr Z O Sgen Astotal HRT ASHRT AT In general if the heat reservoirs are internally reversible AS Zf5Qm 5 1 T Q ASHRT T Q ASHRT AT m Chapter 719 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 720 Q Q ng ASTotal Tm Q AT Sgen A slTotal T T AT Now as AT gt 0 Sgen gt 0 and the process becomes totally reversible Therefore for reversible heat transfer AT must be small As AT gets large Sgen increases and the process becomes irreversible Example 73 Find the total entropy change or entropy generation for the transfer of 1000 M of heat energy from a heat reservoir at 1000 K to a heat reservoir at 500 K HR at T Areas T1000 K 1000 Id 1000 K l Q1000 kJ HR 500 K V at T AT 500K l 2 S kJK The second law for the isolated system is Chapter 720 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 721 Q Q ng ASTotal 7 m lOOOkJ lOOOJ 1000K 500K kJ l2f zl K What happens when the lowtemperature reservoir is at 750 K The effect of decreasing the AT for heat transfer is to reduce the entropy generation or total entropy change of the universe due to the isolated system and the irreversibilities associated with the heat transfer process Third Law of Thermodynamics The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero This law provides an absolute reference point for the determination of entropy The entropy determined relative to this point is called absolute entropy Entropy as a Property Entropy is a property and it can be expressed in terms of more familiar properties P v 7 through the T ds relations These relations come from the analysis of a reversible closed system that does boundary work and has heat added Writing the first law for the closed system in differential form on a per unit mass basis Chapter 721 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 722 5Qim m 6 VVint rev out System used to nd expressions for ds 5Qint rev 5 VVint rev out 5Qint rev 2 W PdV 1nt rev out TdS PdedU On a unit mass basis we obtain the first T ds equation or Gibbs equation as TdsduPdv Recall that the enthalpy is related to the internal energy by h u Pv Using this relation in the above equation the second T ds equation is Tdszdh vdP These last two relations have many uses in thermodynamics and serve as the starting point in developing entropychange relations for processes The successful use of T ds relations depends on the availability of property relations Such relations do not eXist in an easily used form for a general pure substance but are available for incompressible substances liquids Chapter 722 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 723 solids and ideal gases So for the general pure substance such as water and the refrigerants we must resort to property tables to nd values of entropy and entropy changes The temperatureentropy and enthalpyentropy diagrams for water are shown below Pl N T1 Sl SfT1 T3 P3 S3 Superheated vapor Compressed liquid D Saturated liquidivapor mixture T2 x2 32 7 37 xzsfg Chapter 723 400 h kJkg 3200 3000 2800 2600 2400 Satllrmed hquldlme V Salumlcd vapor lme y kJkg K 7 220 U 5 6 7 Shown above are the temperatureentropy and enthalpyentropy diagrams for water The h s diagram called the Mollier diagram is a useful aid in solving steam power plant problems Chapter 724 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles Example 74 725 Find the entropy andor temperature of steam at the following states P T Region s kJkg K 5 MPa 1200 C 1 MPa 50 C 18 MPa 4000 C 40 kPa Quality x 09 40 kPa 71794 Answers are on the last page of Chapter 7 Example 75 Determine the entropy change of water contained in a closed system as it changes phase from saturated liquid to saturated vapor when the pressure is 01 MPa and constant Why is the entropy change positive for this process System The water contained in the system a pistoncylinder device T Steam Chapter 725 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 726 Property Relation Steam tables Process and Process Diagram Constant pressure sketch the process relative to the saturation lines Conservation Principles Using the de nition of entropy change the entropy change of the water per mass is Ass2 s1sg sf ng 60562L kgK The entropy change is positive because Heat is added to the water Example 76 Steam at 1 MPa 600 C expands in a turbine to 001 MPa If the process is isentropic find the nal temperature the final enthalpy of the steam and the turbine work System The control volume formed by the turbine 1 T Control surface Chapter 726 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 727 Property Relation Steam tables Process and Process Diagram Isentropic sketch the process relative to the saturation lines on the T s diagram Conservation Principles Assume steadystate steady ow one entrance one eXit neglect KB and PE Conservation of mass m 2 m2 2 m First Law or conservation of energy The process is isentropic and thus adiabatic and reversible therefore Q 0 The conservation of energy becomes Ein Eout VVout Since the mass ow rates in and out are equal solve for the work done per unit mass Chapter 727 Student Study Guide for 53911 edition of Thermod namics by Y A Qengel amp M A Boles 728 Ein Eout VVout mlhl mm gt Ei h e w i m Now let s go to the steam tables to find the h s a 236986 kg k RlePa s180311 kgK T1 6000C The process is isentropic therefore s2 s1 80311 kJkg K At P2 001 MPa sf 06492 kJkgK and sg 81488 kJkg K thus sflt S lt sg State 2 is in the saturation region and the quality is needed to specify the state Chapter 728 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 729 s2 sf xzsfg s s 62 2 2 f sfg 80311 06492 74996 h2 hf xzhfg 1918 09842392 1 0984 25456 kg Since state 2 is in the twophase region T2 TsatatPZ 45810C W h h2 36986 25456 kg 21153 kg Entropy Change and Isentropic Processes The entropychange and isemropz39c relations for a process can be sum marized as follows 1 Pure substances Any process AS 52 51 kJkgK Chapter 729 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 730 Isentropic process 52 51 2 Incompressible substances Liquids and Solids du ds dv T T The change in internal energy and volume for an incompressible substance is duszT dVEO The entropy change now becomes dsCdTO T AS 2CTdT 1 T If the specific heat for the incompressible substance is constant then the entropy change is T2 Any process S2 51 2 CW 111 kJkgK l Isentropic process TZ 7 3 Ideal gases Chapter 730 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 731 a Constant speci c heats approximate treatment Any process can you fill in the steps T 52 s1 Cv v 111439 1 1 and can you ll in the steps T P S2 S1 Cp v 11 12 Rll 12 1 1 Or on a unitmole basis T v S2 S1 Cum111 Ru lnv 2 1 1 and T P S2 S1 Cpav 111 Ru 111 1 1 Chapter 731 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 732 lsentropic process Can you fill in the steps here kil I 2 5 const ijacDk Ti 5 const 1 k 1 s const V2 For an isentropic process this last result looks like ka constant which is the polytropic process equation Pvquot constant with n k CpCv b Variable speci c heats exact treatment From T ds dh vdP we obtain C T As 2Lair Ring 1 T p1 The first term can be integrated relative to a reference state at temperature Trefn Tmf CPT T2 CPT T 2CPT L707 dT d T1 T Tref T 2 C T 1 C T 2f 1mm T plt gtT Tref T Tref T Chapter 732 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 733 The integrals on the righthand side of the above equation are called the standard state entropies so at state 1 T1 and state 2 T 2 s0 is a function of temperature only 1C T s1 ij 07 Tref T T C T s j 2 P dT Tref T Therefore for any process 0 0 P2 s2 s1 2 s2 s1 Rln kJkgK l or 0 0 P2 S2 S1 2 S2 S1 Ru 111 kJkmolK l The standard state entropies are found in Tables Al7 for air on a mass basis and Tables A18 through A25 for other gases on a mole basis When using this variable specific heat approach to finding the entropy change for an ideal gas remember to include the pressure term along with the standard state entropy tennsthe tables don t warn you to do this lsentropic process As 0 Chapter 733 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 734 P 0 0 2 s2 s1 R1n P kJkgK 1 If we are given T1 P1 and P2 we find s01 at T1 calculate s02 and then determine from the tables T 2 L12 and hz When air undergoes an isentropic process when variable speci c heat data are required there is another approach to finding the properties at the end of the isentropic process Consider the entropy change written as C T As 2LdT Rlni 1 T p1 Letting T1 Tm P1 Pref latm T 2 T P2 P and setting the entropy change equal to zero yield C T39 i ZEXP if Lair P R Tref 7quot39 ref s00nst We define the relative pressure P as the above pressure ratio P is the pressure ratio necessary to have an isentropic process between the reference temperature and the actual temperature and is a function of the actual temperature This parameter is a function of temperature only and is found in the air tables Table Al7 The relative pressure is not available for other gases in this text 13 sconst C T39 EXP if Lair R Tref T39 Chapter 734 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 735 The ratio of pressures in an isentropic process is related to the ratio of relative pressures g Z we 82 1 s00nSl RBef sconsl r1 There is a second approach to nding data at the end of an ideal gas isentropic process when variable speci c heat data are required Consider the following entropy change equation set equal to zero From T ds du Pdv we obtain for ideal gases 2 C T v As dT R 111 2 1 T v1 Letting T1 Tm v1 vref T 2 T v2 v and setting the entropy change equal to zero yield L EXPiJT air R Tref T39 V ref s00nst We define the relative volume vr as the above volume ratio vr is the volume ratio necessary to have an isentropic process between the reference temperature and the actual temperature and is a function of the actual temperature This parameter is a function of temperature only and is found in the air tables Table Al7 The relative volume is not available for other gases in this text if CVT39 W v 7 I 503 SiC UnS R TV Chapter 735 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 736 V2 V v2 vref v V1 vref vrl 1 s const 3 comp Extra Assignment For an ideal gas having constant specific heats and undergoing a polytropic process in a closed system Pvquot constant with n k find the heat transfer by applying the first law Based on the above discussion of isentropic processes explain your answer Compare your results to this problem to a similar extra assignment problem in Chapter 4 Example 77 Aluminum at 100 C is placed in a large insulated tank having 10 kg of water at a temperature of 30 C If the mass of the aluminum is 05 kg find the final equilibrium temperature of the aluminum and water the entropy change of the aluminum and the water and the total entropy change of the universe because of this process Before we work the problem what do you think the answers ought to be Are entropy changes going to be positive or negative What about the entropy generated as the process takes place System Closed system including the aluminum and water Tank insulated boundary Chapter 736 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 737 Property Relation Process Constant volume adiabatic no work energy exchange between the aluminum and water Conservation Principles Apply the first law closed system to the aluminumwater system Q W2AU O OAU water system AU Using the solid and incompressible liquid relations we have m CwaterT2 mALCALT2 T1AL 0 water water But at equilibrium TLAL T2Water 2 I2 mwater Cwater water mAL CAL AL mwater Cwater mAL CAL mgmyuwngmfmammmg mmmngg mum lOkwa 418k kgwm K 05kgAL 09411 kgAL K 3038K 1 The second law gives the entropy production or total entropy change of the universe as Chapter 737 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 738 Sgen Aston AS water ASAL20 Using the entropy change equation for solids and liquids T ASAL mALCAL1n 2 LAL 05kg0941 1 ln kgK 100273K 00966 K Tz T1 AS m C 1n water water water water 10kg4177i1n 3038K kg K 30 273K 01101 K Why is ASAL negative Why is ASwmer positive Sgen total AS water AS AL 01101 00966 2 00135 K Why is Sgen or ASTotal positive Chapter 738 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 739 Example 78 Carbon dioxide initially at 50 kPa 400 K undergoes a process in a closed system until its pressure and temperature are 2 MPa and 800 K respectively Assuming ideal gas behavior find the entropy change of the carbon dioxide by first assuming constant specific heats and then assuming variable specific heats Compare your results with the real gas data obtained from the EES software a Assume the Table A2a data at 300 K are adequate then C 0846 kJkgK and R 01889 kJkgK s2 s1 CWV lng Rlni 1 Pl In 2 0846 ln800K 01889 k lnwj kg K 400K kg K 50kPa 01 104i kg K b For variable specific heat data use the carbon dioxide data from Table A20 E0 0 s2 s1 Rln CO2 Pl 257408 225225kkmolK 01889 k 1112000kPa 44kg kmol kg K 50kPa 00346i kgK Chapter 739 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 740 c Using EES for carbon dioxide as a real gas Deltas ENTROPYCarbonDioxideT800P2000 ENTROPYCarbonDioxideT400P50 003452 kJkgK 1 Repeat the constant specific heat calculation assuming Cp is a constant at the average of the specific heats for the temperatures Then C p 1054 kJkgK see Table A2b T P s2 s1 CMv Inf Rm2 l l 1054 k 1n800K 01889 k 111Mj kgK 400K kgK 50kPa 00337i kgK It looks like the 300 K data give completely incorrect results here If the compression process is adiabatic Why is As positive for this process Chapter 740 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 1 Chapter 3 Properties of Pure Substances We now turn our attention to the concept of pure substances and the presentation of their data Simple System A simple system is one in which the effects of motion viscosity uid shear capillarity anisotropic stress and external force fields are absent Homogeneous Substance A substance that has uniform thermodynamic properties throughout is said to be homogeneous Pure Substance A pure substance has a homogeneous and invariable chemical composition and may exist in more than one phase Examples Water solid liquid and vapor phases Mixture of liquid water and water vapor Carbon dioxide C02 Nitrogen N2 Mixtures of gases such as air as long as there is no change of phase LIIPWNH State Postulate Again the state postulate for a simple pure substance states that the equilibrium state can be determined by specifying any two independent intensive properties Chapter 31 Student Study Guide for 5 edition ofThermodmamics by y A gengel ampM A Boles The PVT Surface for a Real Substance 9 PVT Surface for a Substance that contracts upon freezing Pressure Pressure Chapter 3 2 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 3 3 Real substances that readily change phase from solid to liquid to gas such as water refrigerant134a and ammonia cannot be treated as ideal gases in general The pressure volume temperature relation or equation of state for these substances is generally very complicated and the thermodynamic properties are given in table form The properties of these substances may be illustrated by the functional relation F PvT0 called an equation of state The above two figures illustrate the function for a substance that contracts on freezing and a substance that expands on freezing Constant pressure curves on a temperaturevolume diagram are shown in Figure 3 11 These figures show three regions where a substance like water may exist as a solid liquid or gas or vapor Also these figures show that a substance may exist as a mixture of two phases during phase change solidvapor solidliquid and liquidvapor Water may exist in the compressed liquid region a region where saturated liquid water and saturated water vapor are in equilibrium called the saturation region and the superheated vapor region the solid or ice region is not shown Let39s consider the results of heating liquid water from 20 C 1 atrn while keeping the pressure constant We will follow the constant pressure process shown in Figure 311 First place liquid water in a pistoncylinder device where a fixed weight is placed on the piston to keep the pressure of the water constant at all times As liquid water is heated while the pressure is held constant the following events occur Process 12 Chapter 33 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 3 4 The temperature and specific volume will STATE 1 increase from the compressed liquid or subcooled liquid state 1 to the saturated liquid state 2 1n the compressed liquid region the properties of the liquid are 1 J 39 equal r i to the properties of the saturated liquid state at 1 the temperature Heat A Process 23 At state 2 the liquid has reached the temperature at which it begins to boil called the saturation temperature and is said to exist as a saturated liquid Properties at the saturated liquid state are noted by the subscript f and v2 Vf During the phase change both the temperature and pressure remain constant according to the International Temperature Scale of 1990 ITS 90 water boils at 99975 C 5 100 C when the pressure is 1 atm 0r 101325 kPa At state 3 the liquid and vapor phase are in equilibrium and any point on the line between states 2 and 3 has the same temperature and pressure STATEZ STATES 1 E i Pl mm P 1 atm T 1003C T 100 C Heat Heat 13 A Chapter 34 Student Study Guide for 5quot edition ofThermOMamics by Y A Qengel amp M A Boles 3 5 Process 34 At state 4 a saturated vapor exists and vaporization is complete The subscript g will always denote a saturated vapor state Note v4 vg STATE 4 I I P 1 atm T 100 C Heat A Thermodynamic properties at the saturated liquid state and saturated vapor state are given in Table A4 as the saturated temperature table and Table A5 as the saturated pressure table These tables contain the same information In Table A4 the saturation temperature is the independent property and in Table A5 the saturation pressure is the independent property The saturation pressure is the pressure at which phase change will occur for a given temperature In the saturation region the temperature and pressure are dependent properties if one is known then the other is automatically known Process 45 If the constant pressure heating is continued the temperature will begin to increase above the saturation temperature 100 C in this example and the volume also increases State 5 is called a superheated state because T5 is greater than the saturation temperature for the pressure and the vapor is not about to condense Thermodynamic properties for water in the superheated region are found in the superheated steam tables Table A6 Chapter 35 Student Study Guide for 5 h edition ofrhermogynamics by Y A cenge1 amp M A Boles STATE 5 P1atm T 300 C Heat A This constant pressure heating process is illustrated in the following gure 300 99 975 100 2 Saluxaled 3 in mm A a Figure 311 Consider repeating this process for other constant pressure lines as shown below Chapter 35 Student Study Guide for 5Lh edition ofThennod amics by Y A Qengel amp M A Boles 3 7 Critical point 7 Saturated Vapor liquid 0003155 u If all of the saturated liquid states are connected the saturated liquid line is established If all of the saturated vapor states are connected the saturated vapor line is established These two lines intersect at the critical point and form what is often called the steam dome The region between the saturated liquid line and the saturated vapor line is called by these terms saturated liquidvapor mixture region wet region ie a mixture of saturated liquid and saturated vapor twophase region and just the saturation region Notice that the trend of the temperature following a constant pressure line is to increase with increasing volume and the trend of the pressure following a constant temperature line is to decrease with increasing volume Chapter 37 Student Study Guide for 5Lh edition of Thermod amics by Y A Qengel amp M A Boles 179880C 99610C T Critical point 33 P2 kPa u 739 95 P1 100 kPa Q SUPERHEATED VAPOR REGION SATURATED 4 LIQUID7VAPOR REGION SUPERHEATED VAPOR REGION l COMPRESSED LIQUID REGI ON Satumed Chapter 38 Student Study Guide for 53911 edition of Thennodmarnics by Y A Cengel amp M A Boles 3 9 The region to the left of the saturated liquid line and below the critical temperature is called the compressed liquid region The region to the right of the saturated vapor line and above the critical temperature is called the superheated region See Table A1 for the critical point data for selected substances Review the Pv diagrams for substances that contract on freezing and those that eXpand on freezing given in Figure 321 and Figure 322 At temperatures and pressures above the critical point the phase transition from liquid to vapor is no longer discrete T Critical point c y Phase Chang e Figure 325 shows the PT diagram often called the phase diagram for pure substances that contract and eXpand upon freezing Chapter 39 Student Study Guide for 5Lh edition of Therrnod amics by Y A Qengel amp M A Boles P 3 10 Substances Substances that expand that contract on freezing on freezing Critical 4 oint 2 P g 20 LIQUID 639 s4 e 639 0 Q 4 SOLID Triple point t 50 6 VAPOR The triple point of water is 0010C 06117 kPa See Table 33 The critical point of water is 373950C 22064 MPa See Table Al Chapter 310 St udent Study Guide for 5Lh edition of Therrnod amics by Y A Qengel amp M A Boles 3 ll Plot the following processes on the PT diagram for water expands on freezing and give examples of these processes from your personal experiences 1 process ab liquid to vapor transition 2 process cd solid to liquid transition 3 process ef solid to vapor transition Substances that expand on freezmg Substances that contract on freezing Critical point SOLID Triple point VAPOR Chapter 3ll Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 12 Property Tables In addition to the temperature pressure and volume data Tables A4 through A8 contain the data for the specific internal energy u the specific enthalpy h and the speci c entropy s The enthalpy is a convenient grouping of the internal energy pressure and volume and is given by HUPV The enthalpy per unit mass is huPv We will find that the enthalpy h is quite useful in calculating the energy of mass streams owing into and out of control volumes The enthalpy is also useful in the energy balance during a constant pressure process for a substance contained in a closed pistoncylinder device The enthalpy has units of energy per unit mass kJkg The entropy s is a property defined by the second law of thermodynamics and is related to the heat transfer to a system divided by the system temperature thus the entropy has units of energy divided by temperature The concept of entropy is explained in Chapters 6 and 7 Saturated Water Tables Since temperature and pressure are dependent properties using the phase change two tables are given for the saturation region Table A4 has temperature as the independent property Table A5 has pressure as the independent property These two tables contain the same information and often only one table is given Chapter 312 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 13 For the complete Table A4 the last entry is the critical point at 37395 0C TABLE A4 Saturated waterTemperature table Temp Sat T C Press PmkPa liquid vapor liquid Evap vapor liquid Evap vapor liquid Evap vapor Chapter 313 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 14 TABLE A5 Saturated waterPressure table Press Temp P kPa Tm c liquid vapor liquid ufg vapor liquid hfg vapor liquid xfg For the complete Table A5 the last entry is the critical point at 22064 MPa Saturation pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature Saturation temperature is the temperature at which the liquid and vapor phases are in equilibrium at a given pressure In Figure 311 states 2 3 and 4 are saturation states Chapter 314 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 15 The subscript fg used in Tables A4 and A5 refers to the difference between the saturated vapor value and the saturated liquid value region That is Zl gZZl uf g The quantity hfg is called the enthalpy of vaporization or latent heat of vaporization It represents the amount of energy needed to vaporize a unit of mass of saturated liquid at a given temperature or pressure It decreases as the temperature or pressure increases and becomes zero at the critical p01nt Quality and Saturated LiquidVapor Mixture Now let s review the constant pressure heat addition process for water shown in Figure 311 Since state 3 is a mixture of saturated liquid and saturated vapor how do we locate it on the Tv diagram To establish the location of state 3 a new parameter called the quality x is defined as m as S saturated vapor mg x m asstotal mfmg The quality is zero for the saturated liquid and one for the saturated vapor 0 s x 1 The average specific volume at any state 3 is given in terms of the quality as follows Consider a mixture of saturated liquid and saturated vapor The liquid has a mass mf and occupies a volume The vapor has a mass mg and occupies a volume Vg Chapter 315 Student Study Guide for 53911 edition ofThermod amics by Y A Qengel amp M A Boles 3 16 PorT Saturated vapor Hg Dav Saturated Hf liquidvapor mlxture Saturated llqllld We note Vnn m mf mg V mv Vf mfvf Vg mng mvszvf 171ng mv mv v f f g g m m Recall the definition of quality x Then Chapter 3 1 6 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 17 Note quantity 1 x is often given the name moisture The specific volume of the saturated mixture becomes v l xvf xvg The form that we use most often is v vf xvg vf It is noted that the value of any extensive property per unit mass in the saturation region is calculated from an equation having a form similar to that of the above equation Let Y be any extensive property and let y be the corresponding intensive property Ym then Y y Z yf xyg yf yfxyfg where yfg yg y The term yfg is the difference between the saturated vapor and the saturated liquid values of the property y y may be replaced by any of the variables v u h or S We often use the above equation to determine the quality x of a saturated liquidvapor state The following application is called the Lever Rule yyf yfg x Chapter 317 Student Study Guide for 53911 edition ofTheImodmamics by Y A Qengel amp M A Boles 3 18 The Lever Rule is illustrated in the following gures PorT Sat vapor ng Sat liquid va Superheated Water Table A substance is said to be superheated if the given temperature is greater than the saturation temperature for the given pressure State 5 in Figure 311 is a superheated state In the superheated water Table A6 T and P are the independent properties The value of temperature to the right of the pressure is the saturation temperature for the pressure The rst entry in the table is the saturated vapor state at the pressure Chapter 318 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 3 l9 Compressed Liquid Water Table A substance is said to be a compressed liquid when the pressure is greater than the saturation pressure for the temperature It is now noted that state 1 in Figure 311 is called a compressed liquid state because the saturation pressure for the temperature T1 is less than P1 Data for water compressed liquid states are found in the compressed liquid tables Table A7 Table A7 is arranged like Table A6 except the saturation states are the saturated liquid states Note that the data in Table A7 begins at 5 MPa or 50 times atmospheric pressure Chapter 319 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles o 8287 3134 I 2 A38 1285 11349 0 00 2755 At pressures below 5 MPa for water the data are approximately equal to the saturated liquid data at the given temperature We approximate intensive parameter y that is v u h and 3 data as yEyfT The enthalpy is more sensitive to variations in pressure therefore at high pressures the enthalpy can be approximated by h E hfT vfP PM For our work the compressed liquid enthalpy may be approximated by h E hfT Chapter 320 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 3 21 Saturated Ice Water Vapor Table When the temperature of a substance is below the triple point temperature the saturated solid and liquid phases exist in equilibrium Here we define the quality as the ratio of the mass that is vapor to the total mass of solid and vapor in the saturated solidvapor mixture The process of changing directly from the solid phase to the vapor phase is called sublimation Data for saturated ice and water vapor are given in Table A8 In Table A8 the term Subl refers to the difference between the saturated vapor value and the saturated solid value TABLE A l Salnmted ice am apm Tmp Sal 7 c Prcss P ice The speci c volume internal energy enthalpy and entropy for a mixture of saturated ice and saturated vapor are calculated similarly to that of saturated liquidvapor mixtures yig yg yi yzyi xyig Chapter 321 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 22 where the quality x of a saturated icevapor state is How to Choose the Right Table The correct table to use to find the thermodynamic properties of a real substance can always be determined by comparing the known state properties to the properties in the saturation region Given the temperature or pressure and one other property from the group v u h and s the following procedure is used For example if the pressure and specific volume are specified three questions are asked For the given pressure ls vltvf ls vfltvltvg ls vgltv The answer to one of these questions must be yes If the answer to the first question is yes the state is in the compressed liquid region and the compressed liquid tables are used to find the properties of the state If the answer to the second question is yes the state is in the saturation region and either the saturation temperature table or the saturation pressure table is used to find the properties Then the quality is calculated and is used to calculate the other properties u h and s If the answer to the third Chapter 322 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 23 question is yes the state is in the superheated region and the superheated tables are used to nd the other properties Some tables may not always give the internal energy When it is not listed the internal energy is calculated from the de nition of the enthalpy as uzh Pv Chapter 323 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 24 Example 21 Find the internal energy of water at the given states for 7 MPa and plot the states on T v Pv and PT diagrams Steam 700 I I I I I I 600 7000 kPa 500 400 g I 300 200 100 0 10394 103 3 vm kg 5 Steam 10 I I I I I 10quot 2859C 103 u 3741 C n E n 102 101 10 I I I I I 10quot i03 i02 io l 10 i01 i02 vm3lkg Chapter 324 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 25 P Steam CP MPa Triple Point 001 2858 37395 T cc 1 P 7 MPa dry saturated or saturated vapor Using Table A5 u u 2581 OH g Locate state 1 on the T v Pv and PT diagrams 2 P 7 MPa wet saturated or saturated liquid Using Table A5 u uf 212580 kg Locate state 2 on the T v Pv and PT diagrams Chapter 325 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 26 3 Moisture 5 P 7 MPa let moisture be y defined as m y f 005 m then the quality is xl yl 005O95 and using Table A5 u uf xug uf 12580 095258l0 12576 25144 kg Notice that we could have used L I Hf X Mfg Locate state 3 on the T v Pv and PT diagrams 4 P 7 MPa T 600 C For P 7 MPa Table A5 gives TSat 285830C Since 600 C gt TSat for this pressure the state is superheated Use Table A6 u 232610 kg Chapter 326 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 27 Locate state 4 on the T v Pv and PT diagrams 5 P 7 MPa T 100 C Using Table A4 At T 100 C Psat 010142 MPa Since P gt Psat the state is compressed liquid Approximate solution kJ M 241906 kg 2 ufT100C Solution using Table A7 We do linear interpolation to get the value at 100 C We will demonstrate how to do linear interpolation with this problem even though one could accurately estimate the answer P MPa ukJkg 5 41765 7 n 7 10 41623 The interpolation scheme is called the ratio of corresponding differences Using the above table form the following ratios Chapter 327 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 28 5 7 41765 u 5 10 41765 41623 u4l708 kg Locate state 5 on the T v Pv and PT diagrams 6 P 7 MPa T 460 C Since 460 C gt Tsat 385830C at P 7 MPa the state is superheated Using Table A6 we do a linear interpolation to calculate u T C 11 kJkg 450 29790 460 u 500 30743 Using the above table form the following ratios 460 450 21 29790 500 450 30743 29790 u 29981 kg Locate state 6 on the T v Pv and PT diagrams Chapter 328 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 29 Example 22 Determine the enthalpy of 15 kg of water contained in a volume of 12 m3 at 200 kPa Recall we need two independent intensive properties to specify the state of a simple substance Pressure P is one intensive property and speci c volume is another Therefore we calculate the specific volume 3 v Volume 12m3 m mass 15 kg 39 kg Using Table A5 at P 200 kPa vf 0001061 m3kg vg 08858 m3kg ls vltvf No Is vfltvltvg Yes Is vgltv No Locate this state on a T v diagram Chapter 329 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 3 30 We see that the state is in the twophase or saturation region So we must find the quality x rst v vf xvg vf v v xf vg vf 08 0001061 08858 0001061 0903 What does this mean Then h hf thg 5047 O90322016 24927 kg Example 23 Determine the internal energy of refrigerant134a at a temperature of 0 C and a quality of 60 Using Table A1 1 for T 0 C uf 5163 kJkg ug 23016 kJkg then Chapter 330 Student Study Guide for 53911 edition of Thermodynamics by Y A Cengel amp M A Boles 3 31 u uf xug uf 5163 O623016 5163 215875 k Example 24 Consider the closed rigid container of water shown below The pressure is 700 kPa the mass of the saturated liquid is 178 kg and the mass of the saturated vapor is 022 kg Heat is added to the water until the pressure increases to 8 MPa Find the final temperature enthalpy and internal energy of the water Does the liquid level rise or fall Plot this process on a Pv diagram with respect to the saturation lines and the critical point P mg Vg Sat Vapor 111 VI Sat Liquid V Let s introduce a solution procedure that we will follow throughout the course A similar solution technique is discussed in detail in Chapter 1 System A closed system composed of the water enclosed in the tank Property Relation Steam Tables Process Volume is constant rigid container Chapter 331 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 32 For the closed system the total mass is constant and since the process is one in which the volume is constant the average speci c volume of the saturated mixture during the process is given by V v constant m 01 v2 v1 Now to find v1 recall that in the twophase region at state 1 mg1 022 kg mf1 mg1 178 022 kg Then at P 700 kPa V1 Vf1x1vg1vf1 0001108 01102728 0001108 3 m 0031 kg State 2 is specified by P2 8MPa v2 0 031 m3kg At 8 MPa 8000 kPa vf 0001384 m3kg vg 0 02352 m3kg at 8 MPa v2 0031 m3kg Chapter 332 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 33 Is v2 lt vf No Is vf ltv2 ltvg No Is vg lt v2 Yes Therefore State 2 is superheated Interpolatng in the superheated tables at 8 MPa v 0031 m3kg gives T 2 361 C hz 3024 kJkg L12 2776 kJkg Since state 2 is superheated the liquid level falls Extra Problem What would happen to the liquid level in the last example if the specific volume had been 0001 m3kg and the pressure was 8 MPa Chapter 333 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 34 Equations of State The relationship among the state variables temperature pressure and specific volume is called the equation of state We now consider the equation of state for the vapor or gaseous phase of simple compressible substances FPTvE 0 Ideal Gas Based on our experience in chemistry and physics we recall that the combination of Boyle s and Charles laws for gases at low pressure result in the equation of state for the ideal gas as 10sz v where R is the constant of proportionality and is called the gas constant and takes on a different value for each gas If a gas obeys this relation it is called an ideal gas We often write this equation as PszT The gas constant for ideal gases is related to the universal gas constant valid for all substances through the molar mass or molecular weight Let Ru be the universal gas constant Then Chapter 334 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 35 The mass m is related to the moles N of substance through the molecular weight or molar mass M see Table A1 The molar mass is the ratio of mass to moles and has the same value regardless of the system of units kg 22897 NW 0 lbmol MW 2897 g 2897 gmol km Since 1 kmol 1000 gmol or 1000 grammole and 1 kg 1000 g l kmol of air has a mass of 2897 kg or 28970 grams mNM The ideal gas equation of state may be written several ways PV 2 RT V P 2 RT m PVszT PVzllhme M NENET P ng N Pvz f Chapter 335 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles Here P absolute pressure in MPa or kPa v molar specific volume in m3kmol T absolute temperature in K Ru 8314 kJkmolK Some values of the universal gas constant are Universal Gas Constant Ru 8314 kJkmolK 8314 kPam3kmolK 1986 BtulbmolR 1545 ftlbflbmolR 1073 psiaft3lbmolR The ideal gas equation of state can be derived from basic principles if one assumes l Intermolecular forces are small 2 Volume occupied by the particles is small Example 25 Determine the particular gas constant for air and hydrogen Chapter 336 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 37 R R M 8314L k Rm kmokl K 0287 2897 g kg K 01 8314L kmol K k h d 4124 y rogen k K 2016 g kmol The ideal gas equation of state is used when l the pressure is small compared to the critical pressure or 2 when the temperature is twice the critical temperature and the pressure is less than 10 times the critical pressure The critical point is that state where there is an instantaneous change from the liquid phase to the vapor phase for a substance Critical point data are given in Table A1 Compressibility Factor To understand the above criteria and to determine how much the ideal gas equation of state deviates from the actual gas behavior we introduce the compressibility factor Z as follows Chapter 337 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 38 P17ZRuT 0139 P17 RuT For an ideal gas Z l and the deviation of Z from unity measures the deviation of the actual PVT relation from the ideal gas equation of state The compressibility factor is expressed as a function of the reduced pressure and the reduced temperature The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure which are defined as T P T2 1 P R T an R P or or Where Par and T c are the critical pressure and temperature respectively The critical constant data for various substances are given in Table A1 This is known as the principle of corresponding slates Figure 351 gives a comparison of Z factors for various gases and supports the principle of corresponding states Chapter 338 Student Study Guide for 5 edition ofThen nodynamics by Y A Qengel amp M A Boles 3 39 LUgHmI I Impcmm Ethylene a quotNeptune 4 Wm Avemg sun based an dzla an hydrucmbuns a n x s 4 0 Reduced przssurcI When either P or T is unknown Z can be determined from the compressibility chart with the help of the pseudo reduced speci c volume de ned as V actual VR Cr P or Figure A 15 presents the generalized compressibility chart based on data for a large number of gases Chapter 339 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 40 These charts show the conditions for which Z 1 and the gas behaves as an ideal gas 1 PRlt10and TRgt20rPlt10PcrandTgt2Tcr 2 PR ltlt 1 orP ltlt Pcr Note When PR is small we must make sure that the state is not in the compressed liquid region for the given temperature A compressed liquid state is certainly not an ideal gas state For instance the critical pressure and temperature for oxygen are 508 MPa and 1548 K respectively For temperatures greater than 300 K and pressures less than 50 MPa 1 atmosphere pressure is 010135 MPa oxygen is considered to be an ideal gas Example 26 Calculate the specific volume of nitrogen at 300 K and 80 MPa and compare the result with the value given in a nitrogen table as v 0011133 m3kg From Table A1 for nitrogen Tc 1262 K PC 339MPa R 02968 kJkgK R l 3OOK 238 Tr 1262K P 80MPa PR P 339MPa cr Since T gt 2Tcr and P lt 10P or we use the ideal gas equation of state Chapter 340 Student Study Guide for 53911 edition ofThennodynamics by Y A Qengel amp M A Boles 3 41 Pv RT 02968L300K 3 v kg K m MPa P 80MPa 103k 3 001113 quot kg Nitrogen is clearly an ideal gas at this state If the system pressure is low enough and the temperature high enough P and T are compared to the critical values gases will behave as ideal gases Consider the T v diagram for water The gure below shows the percentage of error for the volume lvtable videallvtableklOO for assuming water superheated steam to be an ideal gas Chapter 341 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 42 We see that the region for which water behaves as an ideal gas is in the superheated region and depends on both T and P We must be cautioned that in this course when water is the working uid the ideal gas assumption may not be used to solve problems We must use the real gas relations ie 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 m1 2 m2 or PM 2 1 R I R T But the gas constant is fill in the blank so PM 2 1 I 1 Chapter 342 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 43 Example 27 An ideal gas having an initial temperature of 25 C under goes the two processes described below Determine the final temperature of the gas Process 12 The volume is held constant While the pressure doubles Process 23 The pressure is held constant While the volume is reduced to onethird of the original volume V Process 13 m1 2 m3 or PM P3V3 T T 1 3 but V3V13 andP3P22P1 Therefore Chapter 343 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 44 4155 P1V1 gnarl321 P1 V1 3 T3 25 273K 1987K 743 C Other Equations of State Many attempts have been made to keep the simplicity of the ideal gas equation of state but yet account for the intermolecular forces and volume occupied by the particles Three of these are van der Waals P12v bRT V Where 27R2T2 a and b RT 64 7 8P6quot Extra Assignment When plotted on the Pv diagram the critical isotherm has a point of in ection at the critical point Use this information to verify the equations for van der Waals constants a and 3 Chapter 344 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 45 BeallieBridgeman RT p u 1 2 V c A B VTJO V2 Where A AUG 2 and B 341 V V The constants a b 0 A0 B0 for various substances are found in Table 34 Benedict WebbRubin RT 1 9R T P BORuT AO ng 3a V V V doc C Y A72 i39V 6 i39V3T2 1 Zje y The constants for various substances appearing in the BenedictWebb Rubin equation are given in Table 34 Chapter 345 Student Study Guide for 53911 edition of Thermodynamics by Y A Qengel amp M A Boles 3 46 Example 28 Compare the results from the ideal gas equation the BeattieBridgeman equation and the EES software for nitrogen at 1000 kPa The following is an EES solution to that problem Nitrogen T vs v for P1000 kPa EI dea Gas I e BeattieBridgeman 140 I EES Table Value T K 1000 kPa 0 I i0393 i0392 1039 v makg Notice that the results from the BeattieBridgeman equation compare well with the actual nitrogen data provided by EES in the gaseous or superheated region However neither the BeattieBridgeman equation nor the ideal gas equation provides adequate results in the twophase region where the gas ideal or otherwise assumption fails Chapter 346

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.