Thermodynamics, Heat Transfer, And Fluid Flow. V.1. Thermodynamics, страница 14

The heat rejected (QC) is graphicallyrepresented as the area under line 1-4. The difference between the heat added and the heatrejected is the net work (sum of all work processes), which is represented as the area of rectangle1-2-3-4.Figure 21Carnot Cycle RepresentationThe efficiency (η) of the cycle is the ratio of the net work of the cycle to the heat input to thecycle. This ratio can be expressed by the following equation.ηwhere:Rev. 0=(QH - QC)/QH = (TH - TC)/TH=1 - (TC/TH)η=cycle efficiencyTC=designates the low-temperature reservoir (°R)TH=designates the high-temperature reservoir (°R)(1-23)Page 73HT-01SECOND LAW OF THERMODYNAMICSThermodynamicsEquation 1-23 shows that the maximum possible efficiency exists when TH is at its largestpossible value or when TC is at its smallest value.

Since all practical systems and processes arereally irreversible, the above efficiency represents an upper limit of efficiency for any givensystem operating between the same two temperatures. The system’s maximum possibleefficiency would be that of a Carnot efficiency, but because Carnot efficiencies representreversible processes, the actual system will not reach this efficiency value. Thus, the Carnotefficiency serves as an unattainable upper limit for any real system’s efficiency. The followingexample demonstrates the above principles.Example 1: Carnot EfficiencyAn inventor claims to have an engine that receives 100 Btu of heat and produces 25 Btuof useful work when operating between a source at 140°F and a receiver at 0°F.

Is theclaim a valid claim?Solution:TH= 140oF + 460 = 600°RTC= 0oF + 460 = 460°Rη= (600-460)/600 x 100 = 23.3%Claimed efficiency = 25/100 = 25%Therefore, the claim is invalid.The most important aspect of the second law for our practical purposes is the determination ofmaximum possible efficiencies obtained from a power system. Actual efficiencies will alwaysbe less than this maximum. The losses (friction, for example) in the system and the fact thatsystems are not truly reversible preclude us from obtaining the maximum possible efficiency.An illustration of the difference that may exist between the ideal and actual efficiency ispresented in Figure 22 and the following example.Example 2: Actual vs.

Ideal EfficiencyThe actual efficiency of a steam cycle is 18.0%. The facility operates from a steamsource at 340°F and rejects heat to atmosphere at 60°F. Compare the Carnot efficiencyto the actual efficiency.HT-01Page 74Rev. 0ThermodynamicsSECOND LAW OF THERMODYNAMICSFigure 22Real Process Cycle Compared to Carnot CycleSolution:η=1 - (Tc / Th)η=1 - (60 + 460)/(340 + 460)=1 - 520/800=35%as compared to 18.0% actual efficiency.An open system analysis was performed using the First Law of Thermodynamics in the previouschapter. The second law problems are treated in much the same manner; that is, an isolated,closed, or open system is used in the analysis depending upon the types of energy that cross theboundary.

As with the first law, the open system analysis using the second law equations is themore general case, with the closed and isolated systems being "special" cases of the open system.The solution to second law problems is very similar to the approach used in the first law analysis.Figure 23 illustrates the control volume from the viewpoint of the second law. In this diagram,the fluid moves through the control volume from section in to section out while work is deliveredexternal to the control volume. We assume that the boundary of the control volume is at someenvironmental temperature and that all of the heat transfer (Q) occurs at this boundary.

We havealready noted that entropy is a property, so it may be transported with the flow of the fluid intoand out of the control volume, just like enthalpy or internal energy. The entropy flow into thecontrol volume resulting from mass transport is, therefore, ṁ insin, and the entropy flow out of thecontrol volume is ṁ outsout, assuming that the properties are uniform atRev. 0Page 75HT-01SECOND LAW OF THERMODYNAMICSThermodynamicssections in and out.

Entropy may also be added to the control volume because of heat transferat the boundary of the control volume.Figure 23 Control Volume for Second Law AnalysisA simple demonstration of the use of this form of system in second law analysis will give thestudent a better understanding of its use.Example 3: Open System Second LawSteam enters the nozzle of a steam turbine with a velocity of 10 ft/sec at a pressure of100 psia and temperature of 500°F at the nozzle discharge. The pressure and temperatureare 1 atm at 300°F.

What is the increase in entropy for the system if the mass flow rateis 10,000 lbm/hr?HT-01Page 76Rev. 0ThermodynamicsSECOND LAW OF THERMODYNAMICSSolution:ṁsinṗṁsout where ṗ = entropy added to the systemṗṁ (soutsin)sin=1.7088 Btu/lbm -°R (from steam tables)sout=1.8158 Btu/lbm°R (from steam tables)ṗ/ṁ=soutṗ/ṁ=0.107 Btu/lbm -°Rṗ=10,000 (0.107)ṗ=1070 Btu/lbm -°R.

= entropy added to the systemsin1.81581.7088 Btu/lbm-oRIt should always be kept in mind that the Second Law of Thermodynamics gives an upper limit(which is never reached in physical systems) to how efficiently a thermodynamic system canperform. A determination of that efficiency is as simple as knowing the inlet and exittemperatures of the overall system (one that works in a cycle) and applying Carnot’s efficiencyequation using these temperatures in absolute degrees.Diagrams of Ideal and Real ProcessesAny ideal thermodynamic process can be drawn as a path on a property diagram, such as a T-sor an h-s diagram.

A real process that approximates the ideal process can also be representedon the same diagrams (usually with the use of dashed lines).In an ideal process involving either a reversible expansion or a reversible compression, theentropy will be constant. These isentropic processes will be represented by vertical lines oneither T-s or h-s diagrams, since entropy is on the horizontal axis and its value does not change.A real expansion or compression process operating between the same pressures as the idealprocess will look much the same, but the dashed lines representing the real process will slantslightly towards the right since the entropy will increase from the start to the end of the process.Figures 24 and 25 show ideal and real expansion and compression processes on T-s and h-sdiagrams.Rev.

0Page 77HT-01SECOND LAW OF THERMODYNAMICSThermodynamicsFigure 24 Expansion and Compression Processeson T-s DiagramFigure 25 Expansion and Compression Processeson h-s DiagramPower Plant ComponentsIn order to analyze a complete power plant steam power cycle, it is first necessary to analyze theelements which make up such cycles. (See Figure 26) Although specific designs differ, there arethree basic types of elements in power cycles, (1) turbines, (2) pumps and (3) heat exchangers.Associated with each of these three types of elements is a characteristic change in the propertiesof the working fluid.Previously we have calculatedsystem efficiency by knowing thetemperature of the heat source andthe heat sink.

It is also possible tocalculate the efficiencies of eachindividual component.The efficiency of each type ofcomponent can be calculated bycomparing the actual workproduced by the component to thework that would have beenproduced by an ideal componentoperating isentropically betweenthe same inlet and outletconditions.HT-01Figure 26Page 78Steam CycleRev. 0ThermodynamicsSECOND LAW OF THERMODYNAMICSA steam turbine is designed to extract energy from the working fluid (steam) and use it to dowork in the form of rotating the turbine shaft. The working fluid does work as it expandsthrough the turbine.

The shaft work is then converted to electrical energy by the generator. Inthe application of the first law, general energy equation to a simple turbine under steady flowconditions, it is found that the decrease in the enthalpy of the working fluid Hin - Hout equals thework done by the working fluid in the turbine (Wt).Hinṁ (hinwhere:Houthout)(1-24)Wtẇt(1-25)Hin = enthalpy of the working fluid entering the turbine (Btu)Hout = enthalpy of the working fluid leaving the turbine (Btu)Wt = work done by the turbine (ft-lbf)ṁ= mass flow rate of the working fluid (lbm/hr)hin = specific enthalpy of the working fluid entering the turbine (Btu/lbm)hout = specific enthalpy of the working fluid leaving the turbine (Btu/lbm)ẇt = power of turbine (Btu/hr)These relationships apply when the kinetic and potential energy changes and the heat losses ofthe working fluid while in the turbine are negligible.

For most practical applications, these arevalid assumptions. However, to apply these relationships, one additional definition is necessary.The steady flow performance of a turbine is idealized by assuming that in an ideal case theworking fluid does work reversibly by expanding at a constant entropy. This defines the socalled ideal turbine. In an ideal turbine, the entropy of the working fluid entering the turbine Sinequals the entropy of the working fluid leaving the turbine.Sin = Soutsin = soutwhere:SinSoutsinsout====entropy of the working fluid entering the turbine (Btu/oR)entropy of the working fluid leaving the turbine (Btu/oR)specific entropy of the working fluid entering the turbine (Btu/lbm -oR)specific entropy of the working fluid leaving the turbine (Btu/lbm -oR)The reason for defining an ideal turbine is to provide a basis for analyzing the performance ofturbines.