Thermodynamics, Heat Transfer, And Fluid Flow. V.1. Thermodynamics (776129), страница 15
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An ideal turbine performs the maximum amount of work theoretically possible.Rev. 0Page 79HT-01SECOND LAW OF THERMODYNAMICSThermodynamicsAn actual turbine does less work because of friction losses in the blades, leakage past the bladesand, to a lesser extent, mechanical friction. Turbine efficiency η t , sometimes called isentropicturbine efficiency because an ideal turbine is defined as one which operates at constant entropy,is defined as the ratio of the actual work done by the turbine Wt, actual to the work that would bedone by the turbine if it were an ideal turbine Wt,ideal.Wt,actualηt(1-26)Wt,idealη(hinhout) actual(hinhout) ideal(1-27)where:ηt= turbine efficiency (no units)Wt,actualWt,ideal(hin - hout)actual(hin - hout)ideal====actual work done by the turbine (ft-lbf)work done by an ideal turbine (ft-lbf)actual enthalpy change of the working fluid (Btu/lbm)actual enthalpy change of the working fluid in an ideal turbine(Btu/lbm)In many cases, the turbineefficiency η t has been determinedindependently.
This permits theactual work done to be calculateddirectly by multiplying the turbineefficiency η t by the work done byan ideal turbine under the sameconditions. For small turbines, theturbine efficiency is generally 60%to 80%; for large turbines, it isgenerally about 90%.The actual and idealizedperformances of a turbine may becompared conveniently using a T-sdiagram. Figure 27 shows such aFigure 27 Comparison of Ideal and Actual Turbine Performancescomparison. The ideal case is aconstant entropy. It is representedby a vertical line on the T-s diagram. The actual turbine involves an increase in entropy.
Thesmaller the increase in entropy, the closer the turbine efficiency η t is to 1.0 or 100%.HT-01Page 80Rev. 0ThermodynamicsSECOND LAW OF THERMODYNAMICSA pump is designed to move the working fluid by doing work on it. In the application of thefirst law general energy equation to a simple pump under steady flow conditions, it is found thatthe increase in the enthalpy of the working fluid Hout - Hin equals the work done by the pump,Wp, on the working fluid.Hout Hin Wp(1-28)ṁ (houthin)ẇp(1-29)where:HoutHinWp= enthalpy of the working fluid leaving the pump (Btu)= enthalpy of the working fluid entering the pump (Btu)= work done by the pump on the working fluid (ft-lbf)ṁ= mass flow rate of the working fluid (lbm/hr)houthin= specific enthalpy of the working fluid leaving the pump (Btu/lbm)= specific enthalpy of the working fluid entering the pump (Btu/lbm)ẇp= power of pump (Btu/hr)These relationships apply when the kinetic and potential energy changes and the heat losses ofthe working fluid while in the pump are negligible.
For most practical applications, these arevalid assumptions. It is also assumed that the working fluid is incompressible. For the idealcase, it can be shown that the work done by the pump Wp is equal to the change in enthalpyacross the ideal pump.Wp ideal = (Hout - Hin)ideal(1-30)ẇp(1-31)ideal= ṁ (hout - hin)idealwhere:WpHoutHin= work done by the pump on the working fluid (ft-lbf)= enthalpy of the working fluid leaving the pump (Btu)= enthalpy of the working fluid entering the pump (Btu)ẇp= power of pump (Btu/hr)ṁ= mass flow rate of the working fluid (lbm/hr)houthin= specific enthalpy of the working fluid leaving the pump (Btu/lbm)= specific enthalpy of the working fluid entering the pump (Btu/lbm)The reason for defining an ideal pump is to provide a basis for analyzing the performance ofactual pumps.
A pump requires more work because of unavoidable losses due to friction andfluid turbulence. The work done by a pump Wp is equal to the change in enthalpy across theactual pump.Rev. 0Page 81HT-01SECOND LAW OF THERMODYNAMICSThermodynamicsWp actual = (Hout - Hin)actual(1-32)ẇp(1-33)actual= ṁ (hout - hin)actualPump efficiency, η p , is defined as the ratio of the work required by the pump if it were an idealpump wp, ideal to the actual work required by the pump wp, actual.Wp, idealηp(1-34)Wp, actualExample:A pump operating at 75% efficiency has an inlet specific enthalpy of 200 Btu/lbm.
Theexit specific enthalpy of the ideal pump is 600 Btu/lbm. What is the exit specificenthalpy of the actual pump?Solution:Using Equation 1-34:ηpwp, idealwp, actualwp, idealwp, actual(houtηphin) actualhout, actualhout, actual(hout(houthin) idealηphin) idealhin, actualηp(600 Btu/lbm200 Btu/lbm).75200 Btu/lbmhout, actual = 533.3 Btu/lbm + 200 Btu/lbmhout, actual = 733.3 Btu/lbmHT-01Page 82Rev. 0ThermodynamicsSECOND LAW OF THERMODYNAMICSPump efficiency, η p , relates the work required by an ideal pump to the actual work required bythe pump; it relates the minimum amount of work theoretically possible to the actual workrequired by the pump.
However, the work required by a pump is normally only an intermediateform of energy. Normally a motor or turbine is used to run the pump. Pump efficiency doesnot account for losses in this motor or turbine. An additional efficiency factor, motor efficiencyη m , is defined as the ratio of the actual work required by the pump to the electrical energy inputto the pump motor, when both are expressed in the same units.Wp, actualηmWm, inCwhere:ηmWp, actualWm, inC====motor efficiency (no units)actual work required by the pump (ft-lbf)electrical energy input to the pump motor (kw-hr)conversion factor = 2.655 x 106 ft-lbf/kw-hrLike pump efficiency η p , motor efficiency η m is always less than 1.0 or 100% for an actualpump motor.
The combination of pump efficiency η p and motor efficiency η m relates the idealpump to the electrical energy input to the pump motor.η mη pWp, ideal(1-35)Wm, inCwhere:ηm= motor efficiency (no units)ηp= pump efficiency (no units)Wp, idealWm, inC= ideal work required by the pump (ft-lbf)= electrical energy input to the pump motor (kw-hr)= conversion factor = 2.655 x 106 ft-lbf/kw-hrA heat exchanger is designed to transfer heat between two working fluids. There are several heatexchangers used in power plant steam cycles.
In the steam generator or boiler, the heat source(e.g., reactor coolant) is used to heat and vaporize the feedwater. In the condenser, the steamexhausting from the turbine is condensed before being returned to the steam generator. Inaddition to these two major heat exchangers, numerous smaller heat exchangers are usedthroughout the steam cycle. Two primary factors determine the rate of heat transfer and thetemperature difference between the two fluids passing through the heat exchanger.Rev. 0Page 83HT-01SECOND LAW OF THERMODYNAMICSThermodynamicsIn the application of the first law general energy equation to a simple heat exchanger understeady flow conditions, it is found that the mass flow rates and enthalpies of the two fluids arerelated by the following relationship.ṁ1 (hout, 1hin, 1)ṁ2 (hout, 2hin, 2)(1-36)where:ṁ1= mass flow rate of the working fluid 1 (lbm/hr)ṁ2= mass flow rate of the working fluid 2 (lbm/hr)hout, 1hin, 1hout, 2hin, 2====specific enthalpy of the working fluid 1 leaving the heat exchanger (Btu/lbm)specific enthalpy of the working fluid 1 entering the heat exchanger (Btu/lbm)specific enthalpy of the working fluid 2 leaving the heat exchanger (Btu/lbm)specific enthalpy of the working fluid 2 entering the heat exchanger (Btu/lbm)In the preceding sections we have discussed the Carnot cycle, cycle efficiencies, and componentefficiencies.
In this section we will apply this information to allow us to compare and evaluatevarious ideal and real cycles. This will allow us to determine how modifying a cycle will affectthe cycle’s available energy that can be extracted for work.Since the efficiency of a Carnot cycle is solely dependent on the temperature of the heat sourceand the temperature of the heat sink, it follows that to improve a cycles’ efficiency all we haveto do is increase the temperature of the heat source and decrease the temperature of the heat sink.In the real world the ability to do this is limited by the following constraints.1. For a real cycle the heat sink is limited by the fact that the "earth" is our final heatsink.
And therefore, is fixed at about 60°F (520°R).2. The heat source is limited to the combustion temperatures of the fuel to be burned orthe maximum limits placed on nuclear fuels by their structural components (pellets,cladding etc.). In the case of fossil fuel cycles the upper limit is ~3040°F (3500°R).But even this temperature is not attainable due to the metallurgical restraints of theboilers, and therefore they are limited to about 1500°F (1960°R) for a maximum heatsource temperature.Using these limits to calculate the maximum efficiency attainable by an ideal Carnot cycle givesthe following.ηHT-01TSOURCETSINKTSOURCE1960 oR 520 oR1960 oRPage 8473.5%Rev.
0ThermodynamicsSECOND LAW OF THERMODYNAMICSThis calculation indicates that the Carnot cycle, operating with ideal components under real worldconstraints, should convert almost 3/4 of the input heat into work. But, as will be shown, thisideal efficiency is well beyond the present capabilities of any real systems.Heat RejectionTo understand why an efficiency of 73% is not possible we must analyze the Carnot cycle, thencompare the cycle using real and ideal components. We will do this by looking at the T-sdiagrams of Carnot cycles using both real and ideal components.The energy added to a working fluid during the Carnot isothermal expansion is given by qs. Notall of this energy is available for use by the heat engine since a portion of it (q r) must be rejectedto the environment.
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