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# Thermodynamics I EGN 3343

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This 58 page Class Notes was uploaded by April Prosacco on Wednesday September 23, 2015. The Class Notes belongs to EGN 3343 at University of South Florida taught by Roger Crane in Fall. Since its upload, it has received 8 views. For similar materials see /class/212652/egn-3343-university-of-south-florida in General Engineering at University of South Florida.

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Date Created: 09/23/15

Lecture 14 Today we will continue looking at the First Law of Thermodynamics for Open Systems by going over several examples taken from the problems in the text Turbines amp Compressors Compressors pumps and fans are devices used to increase the pressure of a uid They are often driven by electric motors or steam turbines so that Wout lt 0 These devices are not generally hooked up to provide cooling Q 0 and elevational changes are normally small APE 0 The author states that changes in kinetic energy are also small while this is often the case I would hesitate to make a strong statement in this regard When no information is available on uid velocities at the system inlet and outlet assuming that AKE 0 is certainly reasonable A turbine may be viewed as a compressor working in reverse A high pressure uid is passed through the turbine forcing the blades to rotate a shaft The rotating shaft may be used to turn an electric generator or other device 481 Steam enters an adiabatic turbine at 10 MPa and 400 C and leaves at 20 kPa with a quality of 90 Neglecting the changes in kinetic and potential energies determine the mass ow rate required for a power output of 5 MW Assume Quasistatic process Steady ow process System is adiabatic Negligible change in PE Negligible change in KE Thermo I EGN 3343 Summer 2002 Given State 1 State 2 P1 10 MPa P2 20 kPa T1 400 C X2 90 From the Steam Property Tables h1 30965 kJ h2 hf X2 hfg 2514 M 0923583 M 23739 kJ Thermo I EGN 3343 Sumner 2002 From Conservation of Mass d mm mom 139 5y Steady Process Then m m From the First Law hm ml hm WSW Simplify and rearrange Wm 5000kW k k h1 h2 30965 Ag 23739 Ag kg m 6919 sec quot 1 Throttling Valves Any ow restricting device which produces the primary effect of significantly reducing the pressure of the owing uid No shafts are provided in such devices to introduce work and no mechanisms are present for significant heat transfer Example 498 A well insulated valve is used to throttle steam from 8 MPa and 500 C to 6 MPa Determine the final temperature of the steam Thermo I C C EGN 3343 Summer 2002 Assume Steady ow process System is adiabatic No Work Negligible change in potential energy Negligible change in kinetic energy Given State 1 State 2 P18MPa P26MPa T1 500 C From Conservation of Mass d dr mm maul 5y Steady Process Then mm m From the First Law Simplify and rearrange h1 h2 From Steam Property Tables P1 8 MPaamp T1 500 C gt h1 3398 kJkg Then from First Law of Thermodynamics Thermo I EGN 3343 Summer 2002 h2 3389 kJkg and h2 3389 kJkg amp P2 6 MPa gt T2 4901 C T2 4901 C Mixing Chambers A mixing chamber is a system in which two uids mix but no work is done nor is significant heat transferred The text gives an example as the T in a water system where cold and hot water are mixed prior to owing out of the faucet Assume Steady flow process System is adiabatic No Work Negligible change in potential energy Negligible change in kinetic energy Example Problem 4104 A hot water stream at 80 C enters a mixing chamber with a mass flow rate of 05 kgs where it Q C is mixed with a stream of cold water at 20 C If it is desired that the mixture leave the chamber at 42 C determine the mass flow rate of the cold water stream Assume all the streams are at a pressure of 250 kPa Thermo I EGN 3343 Summer 2002 Given State 1 State 2 State 3 T180 C T220 C T3 42 C P 250kPa P 250kPa P 250kPa m1 05kgs From the Steam Property Tables we find the enthalpy of each stream Since the uid is compressed water in each case properties are taken as saturated liquid at the respective temperatures h1 33491 kJkg h2 8396 kJkg h3 17592 kJkg From Conservation of Mass 61 mm mom I 5y Steady Then 7521 m2 7523 From the First Law ffmmmhml77 mamhm7474f a Simplify m1h1 m2 122 Rearrange mzh3 h2h1 h3m1 Substitute Values mi 1759 8396C1kg 05kgsec m2 0864 kgsec Thermo I EGN 3343 Summer 2002 Lecture 22 o Steady Flow Work We have previously discussed work from pistoncylinder devices We recognize that the system under analysis is a closed system and that the work obtained from the uid can be evaluated as Q szpdV However we have shown other devices that do not lend Q themselves to this type of study Consider a typical open system as represented by a turbine or a compressor Motor or Generator We follow the standard template in evaluating this system Turbine or A IN Compressor Conservation of Mass dAT lm mouljmmz loulzw I 1nin 1noyut Conservation of Energy We will assume steady state no energy storage inside control volume and neglect any heat transfer Note that such devices do Thermo I EGN 3343 Fall 2000 2 not normally have cooling jackets or other significant heat removal systems Wymem mm 39 hm kem pem mum 39 hum keum peuu VVsha Then Wshay t hin how pein peout kein 60111 Second Law Analysis If we do not know how much irreversibility is introduced in the system we may be interested in finding the maximum possible work 5 QS T generated S2 S1 I Since we assume an adiabatic process then 5Q 0 Further if we assume that the process is reversible then Sgen 0 so that S2 S1 0 If we then recall the 2nd TdS relationship TAS dh vdP So that dh vdP Substitute this result back into the lSt Law result Thermo I EGN 3343 Fall 2000 tha Bzw kein keout peout V39 Pout Bu m39kein keout peout V AP m kem keom pew 19600 Note that if we can neglect changes in kinetic and potential energy this equation reduces to W shaft VAP Be careful this is not the same as CW P 39 AV which we obtained for the closed system Example Problem 682 Liquid water enters a 10 kW pump at 100 kPa pressure at a rate of 5 kgs Determine the highest pressure the liquid water can have at the eXit of the pump Neglect the kinetic and potential energy ch3anges of water and take the speci c volume of water to be 0001 m kg Conservation of Mass 0 Conservation of Energy CV min mom gt min mom m Assume no energy storage no heat transfer Neglect changes in potential amp kinetic energy EZySlem mm hm m1 mum 39 hour qu feumll Wsha Thermo I EGN 3343 Fall 2000 Then Wm m 39 hm hour Second Law Analysis S T generated Assume an adiabatic process 5Q 0 and a reversible process Sgen 0 82 81 0 If we then recall the 2nd TdS relationship TAS dh VdP So that dh VdP Substitute this result back into the 1St Law result W sha VAP 10 kW 5 kgs0001m3kgP2 100 kPa P2 2100 kPa Thermo I EGN 3343 Fall 2000 5 Suggestion Perhaps it is too much work to redo this development each time we look at an open system reversible adiabatic device Remember that under these circumstances Wm V39 AP There is a second important insight that the engineer should have from this relationship As the volume of a uid increases the work required to compress this uid in an open system compression will increase proportionately Consider the problem of compressing air to put into a scuba tank There are some special extra highpressure tanks but a normal tank is rated at 3000 psia Assume that an air compressor is situated indoors so that the air source will be near STP 70 F 147 psia Assume that the air compressor operates on a reversible adiabatic process Last period we developed the 1 2nd and 3rd isentropic relations I don t usually try to remember all of these only the 3rd isentropic relationship P Vk Constant Since this is an ideal gas we write Pvv k391 Constant RTv m Now we evaluate the work for an open control volume with an adiabatic process w lvdP lConstPMlt dP Constmltkk1PZGHW WNW Thermo l EGN 3343 Fall 2000 W Piv mikk in 150 FloIn kklP1V1PZp1k1 lt 1 W kltk1gtRT1PZADJW 1 If we compress the gas between 147 and 3000 psia the work per kilogram of air is w 1414 1 02870 kJkg K295 K300014713943911394 1 w 660 kJkg If we compress the gas in two stages first to 210 psia and then recool the gas at constant pressure to 70 F and then compress again to 3000 psia the required work is W 1414102870 kJkg K295 K21014713943911394 1 210 kJkg w2 1414 1 02870 kJkg K295 K300021013943911394 1 210 kJkg w w1 w2 210 210kJkg 420 kmgsee a 36 reduction in the work of compress10n with intercooling Homework 4th Edition Chapter 6 64 69 70 82 89 Thermo I EGN 3343 Fall 2000 Lecture 7 o Compressibility Factor It was stated previously that the ideal gas law is valid for low pressure The question is what is low One way to approach this question is through the compressibility factor defined as follows Pv Z RT For an ideal gas Z 1 As the compressibility factor deviates from 1 the gas may be considered to be increasingly nonideal In attempting to make a general characterization of many gases it has proven useful to try to put the phase diagrams for many gases together In order to normalize the graphs we introduce a normalized pressure and temperature P i I i Putnam Znttcal The critical temperature and pressure of several gases are tabulated in tables A1 and AlE The ratio of the vapour pressure to the critical pressure for the same gas and the temperature to the critical temperature for the same gas is termed the reduced pressure and reduced temperature respectively Phase mam m In figure 251 of the text we see a plot of the compressibility factor Z vs the reduced pressure PR for various reduced temperatures TR Thermo I EGN 3343 Dr R A Crane Spring 2003 The worst correlation for the ideal gas law is near the critical point ie PR l and TR 1 These charts show the conditions for which Z l and the gas behaves as an ideal gas 1 PRlt10andTRgt2 2 PRltlt1 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 In my view the fact that gases behave as ideal gases when the reduced temperature is greater than 2 TR gt 2 is the most commonly encountered Consider the boiling point of several gases taken from Tables A1 and AlE shown below 151K 272R 53K 95R 333K 599 R 1262K 227 R 1548K 2786 R We think of each of these substances as gases because the maximum temperatures at which they are liquids are very low far below temperatures normally encountered in nature For that reason it is common in nature that the TR gt 2 for these substances and they can be treated as ideal gases Thenno I EGN 3343 Dr R A Crane Spring 2003 The Tables for uid properties provided in the appendix of this text will always provide the most accurate means of determining uid properties However for uids for which tables are not available it may prove sufficiently accurate to use the ideal gas law as corrected by the compressibility factor It is the general applicability of the compressibility factor to all gases at all temperatures and pressures that makes the concept such a powerful tool Example Consider air and water first near room temperature amp pressure and again at 15400F 5000 psia PC TR PR TR PR9 0R psia S3S R 147 psi 2000 R 5000 psi Air 2385 547 224 00269 828 914 lWaterl 1165 852 0459 00173 172 567 Thenno I EGN 3343 Dr R A Crane Spring 2003 Legend I Impumnc a nlleplane N ogen Imrhnn dm ud u u Butane 0 Water mama messu 393 Review Example 213 to see how the compressibility factor is used and to get a sense ofthe accuracy Note Working forms of this chart are given in Appendix I as Figures A30a A30b and A30c Figure A30b is similar to the chart shown above and represents a reduced pressure range between 0 and 7 Figure A30a zooms in to better show the pressure range between 0 and 1 Figure A30c zooms out to show an expanded reduced pressure range between 0 and 40 1 Figure from lecture notes for Thermod amics An Engineering Approach 3ml Ed by Cengel and Boles Thermo I EGN 3343 Dr R A Crane Spring 2003 o Other Equations of State A simple accurate equation of state has long been desired in engineering practice The wide spread use of computer programs in engineering has increased the need for such an equation Obviously virtually any algebraic equation either implicit or eXplicit would be easier to program and solve than trying to work with something so cumbersome as the vapour tables When the program must deal with a variety of uids the impetus is increased even more 0 Van Der Waals Thermo l EGN 3343 Van der Waals examined the ideal gas equation and concluded that it failed near the critical point because it failed to fully account for the attraction forces between molecules and molecular volume Molecular repulsive forces in gases are generally quite small Since these forces drop rapidly as the molecules move apart it is only at very high pressure when molecules are unusually close together that such forces are significant Similarly gaseous molecules are often widely spaced and do not physically occupy a large fraction of the space in which they eXist At very high pressures these molecules are forced together and their volume may become significant Van Der Waals proposed the model Pjv bRT where the constant a accounts for the repulsion between molecules and increases the forces between them The constant b accounts for the volume physically occupied Dr R A Crane Spring 2003 O O O Thermo l EGN 3343 by molecules and decreases the effective open volume Numerical values of a amp b can be calculated as follows R2 Zita a b 64 Pcnttca l R T cum I 8 Putnam BeattieBridgeman Equation of State The Beattie Bridgemen Equation of State represents another attempt patterned after the approach used by van der Waals to improve the accuracy of the ideal gas law by accounting for repulsion forces and the volume of the gaseous molecules While the algebraic for is also slightly different the major difference is that the constants corresponding to a amp b are obtained experimentally for the particular gas While accuracy is improved the model can only be applied to gases for which data is available BenedictWebbRubin Equation of State This represents a more recent attempt to improve on the van der Waals equation by introducing a more complicated model with additional eXperimental constants Again it is successful in improving accuracy but continues to be limited by available data Virial Equation of State This represents an alternate means of improving the van der Waals equation of state this time using an infinite series to represent the dependence of temperature and specific volume on pressure No data is available in the text so that this remains a theoretical approach for us Dr R A Crane Spring 2003 Lecture 24 Introduction We have seen from the 2nd Law that any heat engine operating in a cycle must receive energy from a source at high a high energy potential and reject energy to a low potential sink It follows that no heat engine can achieve a 100 efficiency The question then is what is the maximum efficiency that one might expect from a given power cycle AvailabilityExergy The term availability was made popular in these United States by the MIT School of Engineering in the I 940 s Today an equivalent term exergy introduced in Europe in the 1950 s has found global acceptance partly because it is shorter it rhymes with energy and entropy and it can be adapted without requiring translation Cengelamp Boles Frankly I am accustomed the term availability and will continue to use this term here Candidly I suspect that the real reason that Cengel prefers the European term is that he is accustomed to that usage I find that his reasons for usage of the European terminology here are super uous It is true that exergy is shorter but only by a few letters and l have never found that the somewhat longer word had presented any real difficulty I find myself taking strong exception to the rhyming argument Personally I avoid using terms that sound similar particularly with students new to a topic concerned that it may prove confusing If it were possible I would find an alternative term for either enthalpy or entropy for the same reason I m certainly not going to add to the confusion here by introducing yet another similarly sounding term Thermo l EGN 3343 Spring 2001 Availability Availability the maximum energy useful work that could be obtained from the system at a given state in a speci ed environment 0 Dead State a state in which a system is at equilibrium with its surroundings When a system is at the same temperature pressure elevation and velocity as its surroundings there is no opportunity to construct a heat engine to extract work to operate a piston to raise a weight or turn a turbine No potential differences exist which would allow the engineer to extract additional useful work Unless specified in a problem students may assume that the thermodynamic dead state corresponds to 25 C770F 1 atm sea level and zero velocity A system will deliver the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment that is the dead state This quantity of work is termed the availability of the system Note this is not the work that will be extracted from a system but rather the maximum work that could ever be extracted no matter how much the design of the heat engine is re ned and improved Note that the availability of a system also depends on the state of the surroundings ie the dead state Because the temperature of the surroundings are normally lower in say Scandinavia than in say the Caribbean the availability and efficiency of systems operating in the latter locations is generally smaller Thermo l EGN 3343 Spring 2001 3 Example 72 Consider a large furnace that can supply heat at a temperature of 2000 R at a steady rate of 8000 BTUsec Determine the availability of this energy Assume an environmental temperature of 77 F ncamot 1 TLTH 1 5372460 0732 A neamotQ 07328000BTUsec 5856 BTUsec Note that in the text the author has miscopied the 8000 BTUsec as 3000 BTU sec The author will use the symbol x for specific extropy as has been common in American texts we will use the symbol a for specific availability In my judgment this is confusing and the symbol x will be reserved for quality Reversible Work Reversible work The maximum amount of useful work that can be produced as a system undergoes a process from a specified initial state to a final state If the final state is the dead state then the reversible work is the same as the availability Irreversibility The difference between the reversible work done by a system and the useful work I Wrevput 39 Wusefulput Second Law Ef ciency Cengel amp Boles have provided an excellent example for the need for a 2nd Law efficiency They compare two heat engines of equal lSt Law efficiency but receiving their energy from sources at different temperatures Thermo l EGN 3343 Spring 2001 Heat Engine Heat Engine A B n 30 n 30 Thermodynamically we see that engine A operating between temperatures of 1000K and 300K could if reversible develop efficiencies approaching nA amot l 3001000 70 For heat engine B the maximum efficiency would be TlBcarnot 1 Clearly engine A is operating much further below its real potential To account for this we define the 2nd Law ef ciency H E T thTl carnot Thermo I EGN 3343 Spring 2001 Then 1lirA 030070 043 mm 030050 060 This gives a much better indication of the opportunity for efficiency improvement in the two engines than the simple thermal lSt Law efficiency It follows that for any engine T H E WusefulWcamot And for any refrigeration system T H E COPCOPcamot Example 722 A heat engine receives heat from a source at 1500K at a rate of 700 kJs and it rejects the waste heat to a medium at 320 K The measured power output of the heat engine is 320kW and the lowest naturally occurring environment temperature is 25 C Determine a the reversible power b the rate of irreversibility and c the second law efficiency of this heat engine a Reversible Power ncamot1 3201500 0787 Wrev Qncamot 700kW 0787 5509kW b Irreversibility l Wrev Wuseful 5509320kW 2309kW c Second Law Efficiency 1111 nth110mm WoutQm memot 3207000787 0581 Thermo I EGN 3343 Spring 2001 Lecture 21 In the last lecture we introduced the concept of entropy d5 and presented two problems In these problems we found that our template for the solution of thermodynamic problems could be eXpanded to cover 2nd Law considerations as well You will recall that in studying the lSt Law aspects of finding material properties were significant considerations The same consideration will be true in this new material and this will be the primary focus of today s lecture First amp Second Tds Relationships Also known as Gibbs equations From the lSt Law for a closed system containing a simple compressible substance dExystem SIKm Neglecting kinetic and potential energy this reduces to dUsnyem 3sz From the definition of entropy S we have 3Qm T dS For a simple compressible substance we can write a reversible work term as SW PdV After substitution and rearranging TdS dU PdV 1st TdS Relationship Thenno l EGN 3343 Fall 2000 The alternate form of this relationship is obtained using enthalpy H U PV So that after differentiation dH dU PdV VdP Replacing the last two terms in the lSt T39dS relationship TdS dH VdP 2quot 1 TdS Relationship The I and 2nd T a S relationships form the basis for evaluation of entropy as a thermodynamic property When we introduced enthalpy H as a property we saw that we would be dealing primarily with changes in entropy so that it was possible to define zero entropy at an arbitrary point Note that we did the same thing with the common temperature scales Because of this arbitrary selection in thermodynamic scales it is also possible to have negative entropies S as well as positive Entropy for Liquids amp Solids Normally we eXpect volume changes to be small as liquids or solids undergo a change in thermodynamic state Further Cp z CV z C so that T39dsduP392K zC 39dT Then ds CdTT or 52 S1 Cavg 111T2Tl Themio l EGN 3343 Fall 2000 o Entropy for Ideal Gases Recall the lSt and 2nd T39ds relationships We substitute the ideal gas law for the last terms T39ds du P39dv CV dT R T dVV and T39ds dh vdP Cp39dT RTdPP Divide by T and integrate 3952 s1 CVavg lnT2T1 R39lnV2V1 and ls2 s1 13an lnT2T1 RlnP2P1 Note that the above expressions may be placed on a molar basis by dividing by the molecular weight of the particular gas s2 s1 Cmg lnT2T1 mumv1 and s2 s1 13an lnT2lquot1 91nP2P1 Comment There are a lot of equations here to remember I nd it easier to remember only the Iquot T ds relationship and the assumptions necessary to get the rest Comment In working homework problems I would suggest that students use the above relationships It is much simpler for me to supply speci c heats on exams than to provide complete property tables See page 34634 7 and example 69 for use of property tables Thermo l EGN 3343 Fall 2000 o Isentropic Processes AS 0 of Ideal Gases Isentropic refers to a process which occurs at constant entropy s From the de nition of entropy dS IdQT I m we see that for a reversible adiabatic process AS 0 A set of equations which we introduced at the time we discussed the concept of specific heats The equations had no obvious use at the time and may not have left an impression We reintroduce these relationships now Cp CV R Equation 354 k CpCV Equation 35 5 From the lSt TdS relationship we developed the following relationship for an ideal gas s2 s1 Cv vg lnT2T1 R lIlV2V1 For an isentropic process As 0 so that after simplifying and rearranging 111T2T1 R Cvavg 111V1V2 From equation 354 we obtain CpCV l RCV Substitute from 355 and rearrange k l RCV Substitute this result into our simplified TdS relationship lnT2T1 kl lIlV1V2 Raise both sides to the e power 1st Isentmpic Relatiom TZT1V1V2k391 ship for an Ideal Gas Themio l EGN 3343 Fall 2000 For a particular ideal gas P1V1T12 R PZVgTz so that V2V1 P2T1 P1T2 Substitute into lSt isentropic relationship 2nd Isentmpic Relati0n TZTl PZP1k1 lt ship for an Idea Gas P2V2P1V1 V1V2k So that P2V2k P1V1k which is usually as k 3 ll Isentropic Relation PV ConStant ship for an Idea Gas Example Problem 658 A 05 m3 insulated rigid tank contains 09 kg of carbon dioxide at 100 kPa Now paddle wheel work is done on the system until the pressure in the tank rises to 120 kPa Determine the entropy change of carbon dioxide during this process in kJK Assume constant specific heats V 05 m3 Assume Ideal gas m 09 kg Neglect kinetic amp potential energy Adiabatic P1 100 kPa P2 120 kPa Properties V 05 m3 R 01880 kJkg K Table A2 Cp 0846 kJkg K Table A2 CV 0657 kJkg K Table A2 m 09 kg P1 100 kPa v1 Vm 0555 m3kg T1 PlvlR 2955 K ideal gas law Themio I EGN 3343 Fall 2000 P2 120 kPa v2 v1 0555 m3kg T2 P2v2R 3546 K ideal gas law Conservation of Mass AMCM 0 lSt Law of Therm dynami s AU l ALE l AgE in39Wout Wm mCVT2 T1 09 kg0657 kJkg K3546 2955K 3495 kW 2nd Law Analysis 52 Si 2 Cvavg 111T2T1 R391 VZVl 0657 kJkg Kln35462955 012 kJkg K 82 1 ms2 s1 09 kg 012 kJkg K 0108 kJK Sg 81 Frankly I have been of two minds as to whether to include the concept of relative pressure I have been accustomed to using the assumption of constant speci c heats resulting in the 1 2nd and 3rd isentropic relationships to describe isentropic processes The use of relative pressure avoids the limiting assumption of treating specific heat as constant during the process and is therefore more accurate After having reviewed the material I have decided that the assumption of constant specific heats is sufficient for our needs in this course I will however include an example of the relative pressure treatment for completeness I do not plan to test on this material Students may solve all homework problems specifying relative pressure using the constant specific heat assumption Themio l EGN 3343 Fall 2000 Relative Pressure Problem 665 is a good example of the use of the concept In the first part of the problem we assume constant specific heats and in the second assume variable specific heat to get a somewhat more accurate answer This is the first text that I ve seen that lists relative pressure and allows for computation with variable specific heat Relative Pressure From the 2nd T39dS relationship we have T39ds dh V39dP For an ideal gas dh Cp39dT and v R39TP so that T39ds Cp39dT RTPdP Rearrange and integrate Ids I opT dT R I1PdP Define a new function so so a If C dT So that s sf 2 Jffcpur Then 52 1 02 01 R111P2P1 For a reversible adiabatic process s2 s1 0 so that 502 SO R 111P2P1 Or 502R SolR 2111P2P1 Thenno l EGN 3343 Fall 2000 Take the exponential of each side eXPSozRCXPSoiR PZPl Define the relative pressure Pr E exps R So that PrZPrl PzP 1 Relative speci c volume Pl39v M v 12132 11132 T T 3 v T1PlT1Blv 1 2 1 rl Both Pr and vr are tabulated for various gases in the Appendix These relationships are to be used Whenever we Wish to have integrated values of specific heat used in evaluating changes involving reversible adiabatic process for ideal gases Thenno l EGN 3343 Fall 2000 Lecture 19 o Carnot Cycle The Carnot cycle was rst proposed in 1824 so it is hardly a new development The interest in the cycle is largely theoretical as no practical Carnot cycle engine has yet been built Nevertheless it can be shown to be the most efficient cycle possible so that considerable attention has been given at discovering ways of making the more practical cycles look as much as possible like the Carnot Process 12 Process 23 Process 34 Process 41 0 Process 12 A constant T heat addition 0 Process 23 An adiabatic expansion 0 Process 3 4 A constant T heat rejection 0 Process 41 An adiabatic compression Thermo l EGN 3343 Fall 2000 The Carnot Cycle Carnot Cycle Ef ciency Like other heat engines the Carnot cycle efficiency can be attained from the relationship 8 W0utnetQin Wout 39 Win Qin Qin QoutQin 1 QoutQin Since the heat comes into the system from a high temperature reservoir and is discharged to a low temperature reservoir 8 1 QLQH Carnot cycle principles The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs The efficiencies of all reversible heat engines operating between the same two reservoirs are the same Thermo I EGN 3343 Fall 2000 o Thermodynamic Temperature Consider the Carnot cycle engines shown Since a control volume could be drawn about engines A amp B together they may be considered as a single reversible engine The ef ciency of Engine AampB must be the same as that of Engine c since both are reversible If the heat input GA is equal then 0 must also be equal for equal ef ciency Also WA WB Wc Since energy reservoirs are characterized by their temperatures the thermal efficiency of reversible heat engines is a function of the reservoir temperatures only That is g T T QH mm Applying this idea to the three engines separately QAQc Z TA Tc QAQB Z TA TB QBQc Z TB Tc Thermo I EGN 3343 Fall 2000 For the efficiency of AampB to be equal to that of C TA Tc 2 TA TB39fTB3 Tc A careful examination of this equation reveals that the left hand side is a function of T A and Tc and therefore the right hand side must also be a function of T A and Tc only and not TB This condition will be satisfied only if the function has the following form CI3TA CI3Tc CDTA CI3TB39CDTB CI3Tc From this relationship Kelvin proposed a temperature scale in which CDT T such that Szl TLTH Example Problem 551 An air conditioner removes heat steadily from a house at a rate of 750 kJmin while drawing electric power at a rate of 6 kW Determine a the COP of this air conditioner and b the rate of heat transfer to the outside air Assume Steady State Operating Conditio Analysis a Determine the COP r 4A0 COPRef QLWin 6kw COP 750 kJmin60 smin6 kW J COP 208 House Thermo I EGN 3343 Fall 2000 b Determine the rate of heat transfer to the outlet From the 1St Law of Thermodynamics Qin 39 Wout 0 750 kJmin60 secmin Qout 6 kW Qout 185 kW Example Problem 581 A Carnot heat engine receives 500 M of heat from a source of unknown temperature and rejects 200 M of it to a sink at 17 C Determine a the temperature of the source and b the thermal ef ciency of the heat engine 7 7 7 Assume Steady State Process E Qin 39 Wout Woutnet Qin 39 Qout Wout net 300 M n WannaQin 300 kJ500 M 60 n60 For the Carnot Cycle LowTemperature Heat Sink T 17 C nzl TLTH 06 1 290 KTH TH 725 K Thermo I EGN 3343 Fall 2000 Three Applications of the Carnot Cycle 1 Carnot Cycle Engine I OH gy Carnot ows fbal a ieEi39 C cle I W NO From the de nition of ef ciency we nd for the Carnot engine EZLIOZ QH 0L 1 g 0H 0H 0H Using the thermodynamic temperature scale T 8 1 L TH Thermo I EGN 3343 Fall 2000 2 Carnot Cycle Refrigerator b39f E eitg39iygle t x From the de nition of Coef cient of Performance we nd for the Carnot refrigeration cycle W oH q 6 1 Q Using the thermodynamic temperature scale COP 1 TH 1 r Thermo I EGN 3343 Fall 2000 2 Carnot Cycle Heat Pump From CbniSteIV39 ai ion ca39fEn ieirgtye ner x From the de nition of Coef cient of Performance we nd for the Carnot refrigeration cycle COP i W 0H 0L 1 0H Using the thermodynamic temperature scale COP 1 1 r rH Thermo I EGN 3343 Fall 2000 Reversibility of the Carnot Cycle We use a Carnot Cycle engine to power a Carnot cycle heat pump The efficiency of the engine and the COP of the heat pump depend upon the temperatures of the respective reservoirs In this case assume a high temperature of 1500 R and a low temperature of 500 R The efficiency of the engine is then 8 1 TLTH 1 5001500 or 6667 If for example the engine receives 3000 kW of heat from the high temperature source it Will produce 2000 kW of power Thermo I EGN 3343 Fall 2000 Now let us put the 2000 of power into the heat pump The COP of the heat pump is 11 TLTH 11 5001500 15 We see then that the heat pump returns exactly 152000 kW 3000 kW to the high temperature source Conclusions 0 The Carnot cycle is truly reversible We can recreate exactly the same conditions that we started with We can see that this is the general case in that 3 39 39 1 If an engine could be devised which could produce a greater ef ciency than a Carnot engine then the excess power could be used the power say a car while the Carnot cycle power could be used to power a Heat Pump returning the initial energy to the gas tank If that were possible we could build a car in which the gas tank was initially lled and sealed it would never require gas again 0 If a heat pump could be built which had a greater COP than the Carnot Heat pump we could create high temperature energy analogous to gasoline driving it with a motor powered from the same source That is we could all have our own free gas pump outside our own houses which would never need to be powered from the outside Homework 4th Edition Chapter 5 80 84 E 85 87 101E 120 136 Answers 67 25 313 079 hp 0937 hr 17719 kJhr Thermol EGN 3343 Fall 2000 Thermo I EGN 3343 Fall 2000 Lecture 20 As we have undertaken our study of possible processes we have been discussing the 2quotd Law of Thermodynamics from a conceptual perspective In Chapter 6 we begin to consider these concepts in a more analytical manner In the process we will introduce a new property entropy which is de ned as follows 5 d3 2 e T 19V That is a differential change in entropy corresponds to a differential quantity of heat being transferred divided by the temperature at which is it transferred in a reversible process The development of this concept is rather lengthy but we will present a brief overview here The intention is not that the student should be able to derive the equations being discussed but to present a rationale for the concepts which we will be using Clausius Inequality Recall the KelvinPlank statement of 2quotd 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 Clausius proceeded to consider a device shown on the right which violates the 2quotd Law Work is produced but the device receives heat from a single reservoir Begin our analysis by writing the 1st Law for the control volume AE QM W system out The total work output is the sum of that for the system and the heat engine sz 5wrev swsys For an in nitesimal part of the engine cycle we write dEWstem 5Q 5W7 Thermo I EGN 3343 Spring 2002 Since the cyclic device is a reversible heat engine it follows from the de nition of the Kelvin temperature scale that SQRSQ TR T Combine these results to obtain 51E 6QTR T 5WT Consider the operation of the device through a complete engine cycle For a cycle the energy of the control volume must return to its initial value so that the energy term drops out 5Q 0 T3 T WT Or 5Q mm T As noted previously if the total work is positive then this device would violate the Kelvin Plank statement of the 2quotd Law It follows that the total work must be zero or negative A P 0 V Now let us run the engine in reverse If the path for the cycle is reversed then the sign of the integral is reversed and the integral will become greater than or equal to zero This would likewise result in a violation of the 2quotd Law The only way that the process can be reversed is if the integral gives a value of zero If the integral of 5QT about a cycle is equal to zero then the process is said to be reversible In the special case that the cyclic integral is zero then the quantity under the integral 5Q T returns to its initial value during the cycle It behaves like a thermodynamic property Introduce the entropy S 5157 rev 5Q T Thermo I EGN 3343 Spring 2002 Increase in Entropy Principle Since 5QT is zero for a reversible process and less than zero for an irreversible process 5Q 5Q i7 i7m and 5Q dS T Finally for any process between two states as 2 Q T or 2 SZ S1 2 or Z 5 S2 S1 I2 S T generated A process can only occur if the entropy generated is greater than or equal to zero Now at last we have a quantative statement of the second law Example Problem 625 Air is compressed by a 8 kW compressor from P1 to P2 The air temperature is maintained constant at 25 C during this process as a result of heat transfer to the surrounding medium at 10 C Determine the rate of entropy change of the air State the assumptions made in solving this problem Air Assume Steady State Operation T125 c T225 c Air behaves as an ideal gas F1 1 pa 1 Neglect PE amp KE Consv of Mass w s k w dMAT mm m0m mnmwm CV Consv of Energy E stem mm m m maul 39heuz aut feaut Qm WshqH Recall that for an ideal gas h hT Thus for a constant temperature process hi ham After canceling the enthalpy terms the equation reduces to Thermo I EGN 3343 Spring 2002 Qm Wshajh SkW From the de nition of entropy 4110 S 7 Z af rmed 1 T g S Assume a reversible process 25Q S2 S1 I7Sfmrmed 1 As 8 kW25 273K 0027 kWK ASair 0027 kWK ASH 8 kW10 273K 002826 kWK ASH 002826 kWK Asuniverse AS air ASan 000126 kW K Note that the Entropy of the Universe has increased Example Problem 631 The radiator of a steam heating system has a volume of 20 L and is lled with superheated water vapour at 200 kPa and 200 C At this moment both the inlet and exit valves to the radiator are closed After a while the temperature of the steam drops to 80 C as a result of heat transfer to the room air Determine the entropy change of the steam during this process in kJK Radiator Assume Closed System V 002m3 Transient Process Neglect PE amp KE P1 200 kPa No work input no volume change T1 200 c From Property Tables T2 80 c v1 10803 m3kg 2 M1 Vv1 00185 kg 111 263411 1113115 s1 75066 ldkg K Table A6 Consv of Mass AM 0 3 V2 V1 VZ 10803 m3kg T2 80 0C Thermol X2 VVfVfg 10803 7 00010293406 0317 EGN 3343 13 u 8311 33486 0317211139741 ltJ 1ltg 10156 111 39 Spring 2002 s2 sf mfg 10753 0317653691dkg K 31475 kJkg K AS MsZ 7 s1 00185 kg3147575066 kJkg K 0806 kJK AS 0806 kJK Consv of Energy AECM AU fE Qin 7 Qquot Muz 7111 De nition of Entropy 25Q SS S21T generated Homework 4th EditonChapter 6 28E 29 32 42E 46 Answers 3885 BTU 07 BTUR 0 0456 ldK 0448 ldK 0008 kJK 3883 kJK 3441 kJkg 0422 ldK 3118BTUlb 3rd Edition 907 1d 731 k 4 h Edition 218 k 175 1d Thermo I EGN 3343 Spring 2002 Lecture 17 Introduction to the Second Law of Thermodynamics The First Law of Thermodynamics is a necessary but not a sufficient condition for a process to occur The text provides a good example Consider a weight attached to a pulley which turns a paddlewheel inside a liquid Paddlewheel Allowing the weight to fall will turn the paddlewheel which will do work on the uid system Initially this may increase the kinetic energy of the uid as it churns but in time the circulation will stop When this happens the temperature of the uid will rise and the heat will be transferred to the surroundings From a lSt Law perspective there is no reason why one could not simply supply this same amount of heat to the uid so that the paddlewheel would turn and the weight would be raised However no one has ever built a success ll engine based on this principle It simply wouldn t work The 2quotd Law provides us with a basis to determine why such a process isn t feasible Heat Engines While the 2quotd Law of Thermodynamics is applicable to a wide range to topics much of our understanding of the principle comes from the study of heat engines We will therefore use engines as a basis for much of our study As an aside the author makes a major distinction between internal and external combustion engines The distinction is in my judgment over stated We will be looking at both types of engines and perruruuug Luer J I ofboth With that said we will be looking at a variety of heat engines I ve done some work in the past on a closed loop gas turbine that we will use for our analysis here The components of the system are shown below Theran I EGN 3343 Fall 2000 Qm 2 Workm gt E Compressor turbine I Workuu Cooler 1 In this cycle air or some other uid will circulate through the heat engine to transport energy from one component to the next This uid is often referred to as the working uid Win Work required to compress the working uid Wont Gross usable work produced by the engine This work may be used to drive a truck or car turn a propeller an electric generator or to perform some other useful task Qin Thermal energy supplied to the system This may arise from combustion a solar collector a nuclear reactor or any other available heat source Qom Thermal energy rejected to the environment The cooler could be a radiator similar to that used with an automobile engine Since operation of the engine requires a work input to turn the compressor a portion of the output work may be utilized for this purpose iwoutmeti iwouti iwini This is a thermodynamic cycle By de nition the cycle begins and ends at the same point In the case of an open cycle heat rejection occurs as the exhaust is emitted into the atmosphere Hot exhaust gases are cooled to the temperature of the surroundings before being drawn into the engine at the compressor From the 1St Law of Thermodynamics AESW Q WM For the cycle since the starting and ending points are the same there is no change in internal energy Qin net Woutnet 0 and Woutnet Qin Qout Thermo I EGN 3343 Fall 2000 Cycle Diagram Constant Pressure Heat Addition Adiabatic Expansion P Adiabatic Constant Pressure Compression V Heat Rejection Thermal Ef ciency 11 of the Cycle Conceptually we may think of the ef ciency of the engine as the ratio of the useful work out to the energy that must be input the energ that we pay for T E WnetoutQ in Since we have shown that Womynet Q 7 Qom we may substitute into the above equation and nd T E Qin QoutQin 1 39 QoutQin It probably isn t too surprising that we want to minimize the fraction of rejected heat in the last part of the equation in order to attain the maximum ef ciency Example Problem 522 A steam power plant receives heat from a furnace at a rate of 280 GJhr Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 8 GJhr If the waste heat is transferred to the cooling water at a rate of 145 GJhr determine a net power output and b the thermal ef ciency of this power plant Given Qin 280 GJhr Qlosses to surroundings 8 Qwaste heat 145 Qout Qlosses to surroundings Qwaste heal 8 145 153 Thermo I EGN 3343 Fall 2000 Then Woutynet Qquot 7 Qom 280 7 153GJhr 127 GJhr 127000 MJhr Womm 127000 MJhr 1 hr3600 secl MW secMJ 3527 MW Also 11 1 QomQm 1 7 153 GJhr280 GJhr 04536 Homework 43911 Edition Chapter 5 18 19E 27 Answers 454lbmmin 6109 kW 838ms 321100year Thermo I EGN 3343 Fall 2000 Lecture 18 Heat Rejection from a Heat Engine Case I Brayton Cycle gas turbine engine Qm Workn gt E Compressor turbine I Workuut G 1 Constant Pressure Heat Addition Cycle Diagram Adiabatic Expansion Compression V Constant Pressure Heat Rejection Case 11 Piston Cylinder System EGN 3343 Fall 2000 In the drawing to the left a pistoncylinder contains a gas at 30 C A weight is placed on the piston Heat is then added to the gas from a reservoir at 100 C so that the gas expands As the gas expands it does work to raise the weight and piston The weight now is at a higher potential energy than at state one We slide the weight off of the piston at the higher elevation Assume that the frictional forces in sliding the weight are negligible Now we wish to raise a second weight That is we wish the heat engine to continue to do work The gas inside the cylinder is still heated and will not return to its initial position unless the gas is cooled Only if heat is rejected can we return the system to it s initial position and repeat the process We have now considered two different heat engines In both cases we have seen that it is necessary to reject heat at the end of the work process in order to return the cycle to its initial state and repeat the process We generalize this observation as the 2quotd Law of Thermodynamics 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 K elvinPlank statement of 2quot 1 Law o Refrigeration Cycles Up to this point we have only considered heat engines Refrigeration cycles operate in a similar manner Heat To39Outdoor Env39ironm ent Q Condenser Expansion Compressor Valve A Evaporator Workmg fluld 1 s compressed ralsmg its temperature and pressure Heat From IJTdQOI Enviromnem Thermo I I EGN 3343 Fall 2000 V Vapour Refrigeration Cycle The vapour refrigeration cycle involves four components designed to produce four respective process on the working uid 0 Process 12 The compressor is designed to raise the pressure of the working uid adiabatically As the vapor is compressed the temperature will rise In a typical residential unit temperatures of about 150 F can be produced at the end of the process Process 23 The uid is passed through the condenser Here no work is done on the uid but it is cooled to a saturated liquid This is the part of a residential AC unit which is located outdoors Air is circulated around tubes containing the working uid by a fan Process 34 An expansion valve is simply a device which does no work nor does it transfer heat to or from the working uid It is simply a device to allow the pressure to drop to a lower value A First Law analysis of this process will show that it is a constant enthalpy process It is common for some of the liquid to ash to vapour as the uid expands At the same time the temperature will drop significantly The final temperature will be around 50 F in a typical residential unit Process 41 The uid is passed through the evaporator This device consists of cold tubes often seen located inside the house in a residential unit Here air from inside the house is blown over the cold tubes containing the working uid As the cold uid gains heat no work is done on the uid The working uid is largely vaporized in this process as it returns to state 1 to begin the cycle again The second semester of thermodynamics includes a more detailed study of the Otto gasoline engine cycle the Brayton gas turbine cycle the Rankine Steam Power Plant cycle the Diesel Diesel engine cycle the Refrigeration Air Conditioning amp Heat Pump cycle and other minor cycles The details are more in the area of specialization for mechanical engineers Our purpose here is simply to note the similarity to the heat eng1ne Let us return to our discussion of the Second Law of Thermodynamics At the beginning of this section it was noted that energy has both a magnitude and quality Temperature is one measure of quality Heat will not ow from a low temperature to a high temperature Thermo I EGN 3343 Fall 2000 without the expenditure of work We see the input of work in the basic refrigeration cycle in operating the compressor This principle is accepted as an alternate statement of the Second Law of Thermodynamics 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 highertemperature body Clausius statement of 239 1 Law The two statements of the 2quotd Law can be shown to be equivalent Students are expected to know both since the principle forms one of the two major concepts of the course the lst Law being the other To me neither statement really states the underlying principle explicitly so let me try to tell you what I think the statements are trying to say In the normal course of events when things are not actively maintained they deteriorate Suns burn out mountains crumble cars rust houses fall down etc The reverse process does not happen without intervention You might buy an old rusty car and restore it but it will take work In a like fashion process involving an energy transfer will only occur in such a way so as to decrease the quality of energy We can within a nite control volume transfer energy to the system so as to increase the quality of energy within that system However the work that we input required that we decrease the quality of energy from an area outside the system When we consider the universe as a whole the quality of energy in the universe has still deteriorated These comments are intended to leave students with a qualitative sense of what the 2quotd Law is all about We will put these concepts into a quantative form with the introduction of entropy s in Chapter 6 Coef cient of Performance When describing heat engines we introduced the concept of engine efficiency as the desired output divided by the energy input Similarly we de ne the coefficient of performance COP of a refrigerator COP W QL Re quired Input W QH QL mm The definition is basically the same as that of thermal efficiency The primary reason for the change in terminology is that a COP is often greater than 1 The term is used to avoid speaking of any efficiency greater than 100 Thermo I EGN 3343 Fall 2000 Another term closely related to the COP is the Energy Efficiency Rating or EER If we use an electric motor to operate a refrigerator we pay for the number of kW hours of electricity used to operate the motor What we want is to get a certain amount of cooling normally expressed in BTUs 39 BTUh EER WmtinkW Reversibility Consider a system of pulleys and weights as shown In order for the weight on the right to raise the weight on the left it is essential that W 5W gt W How large must 5W be Well it must be large enough that it compensates for the effects of friction in the pulley the friction associated with the exing of the cable etc If we use very good bearings then the required extra weight on the right may approach zero Can it reach zero W No this would result in static equilibrium and the weights will no move It is essential that 5W be greater than zero it may approach but never reach zero even with perfect bearings W6W We see that in order for the process of raising the left weight to occur the loss in potential energy on the right must be greater than the gain in potential energy on the left As an idealization only we may assume that the weight 5W may be so small that it can be treated as zero In such a case the weight on the right and left are equal and the direction that the pulley moves would be completely reversible Such processes never occur in nature but may be approached in the limit Thermo I EGN 3343 Fall 2000 Carnot Cycle The Carnot cycle was rst proposed in 1824 so it is hardly a new development The interest in the cycle is largely theoretical as no practical Carnot cycle engine has yet been built Nevertheless it can be shown to be the most ef cient cycle possible so that considerable attention has been given at discovering ways of making the more practical cycles look as much as possible like the Carnot l Process 12 Process 23 Process 34 Process 41 0 Process 12 A constant T heat addition 0 Process 23 An adiabatic expansion 0 Process 34 A constant T heat rejection 0 Process 41 An adiabatic compression 1 2 P 3 4 V The Carnot Cycle Thermo I EGN 3343 Fall 2000 Carnot cycle principles The ef ciency of an irreversible heat engine is always less than the ef ciency of a reversible one operating between the same two reservoirs The efficiencies of all reversible heat engines operating between the same two reservoirs are the same Thermo I EGN 3343 Fall 2000

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