The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 17
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This is expressed by the cost rate balance for the turbine-generator:C˙ e + C˙ 2 = C˙1 + Z˙(2.92a)where Ċe is the cost rate associated with the electricity, Ċ1 and Ċ2 are the cost rates associated withthe entering steam and exiting steam, respectively, and Ż accounts for the cost rate associated withowning and operating the system, each annualized in $ per year.FIGURE 2.16 Steam turbine/electric generator used to discuss exergy costing.© 1999 by CRC Press LLC2-77Engineering ThermodynamicsWith exergy costing, the cost rates C˙1 , C˙ 2 , and Ċe are evaluated in terms of the associated rate ofexergy transfer and a unit cost. Equation 2.92a then appears asce W˙ e + c2 E˙ 2 = c1 E˙1 + Z˙(2.92b)The coefficients c1, c2, and ce in Equation 2.92b denote the average cost per unit of exergy for theassociated exergy rate.
The unit cost c1 of the entering steam would be obtained from exergy costingapplied to the components upstream of the turbine. Assigning the same unit cost to the exiting steam:c2 = c1 on the basis that the purpose of the turbine-generator is to generate electricity and thus all costsassociated with owning and operating the system should be charged to the power, Equation 2.92b becomes()ce W˙ e = c1 E˙1 − E˙ 2 + Z˙(2.92c)The first term on the right side accounts for the cost of the net exergy used and the second term accountsfor cost of the system itself. Introducing the exergetic efficiency from Table 2.13, the unit cost of theelectricity isce =c1Z˙+ ˙ε We(2.93)This equation shows, for example, that the unit cost of electricity would increase if the exergetic efficiencywere to decrease owing to a deterioration of the turbine with use.Example 15A turbine-generator with an exergetic efficiency of 90% develops 7 × 107 kW · hr of electricity annually.The annual cost of owning and operating the system is $2.5 × 105.
If the average unit cost of the steamentering the system is $0.0165 per kW · hr of exergy, evaluate the unit cost of the electricity.Solution. Substituting values into Equation 2.93,ce =$2.5 × 10 5 year$0.0165 kW ⋅ h+0.97 × 10 7 kW ⋅ h year= 0.0183 + 0.0036 = $0.0219 kW ⋅ h© 1999 by CRC Press LLC2-78Section 22.6 Vapor and Gas Power CyclesVapor and gas power systems develop electrical or mechanical power from energy sources of chemical,solar, or nuclear origin.
In vapor power systems the working fluid, normally water, undergoes a phasechange from liquid to vapor, and conversely. In gas power systems, the working fluid remains a gasthroughout, although the composition normally varies owing to the introduction of a fuel and subsequentcombustion. The present section introduces vapor and gas power systems. Further discussion is providedin Chapter 8.
Refrigeration systems are considered in Chapter 9.The processes taking place in power systems are sufficiently complicated that idealizations aretypically employed to develop tractable thermodynamic models. The air standard analysis of gas powersystems considered later in the present section is a noteworthy example. Depending on the degree ofidealization, such models may provide only qualitative information about the performance of the corresponding real-world systems.
Yet such information is frequently useful in gauging how changes in majoroperating parameters might affect actual performance. Elementary thermodynamic models can alsoprovide simple settings to assess, at least approximately, the advantages and disadvantages of featuresproposed to improve thermodynamic performance.Rankine and Brayton CyclesIn their simplest embodiments vapor power and gas turbine power plants are represented conventionallyin terms of four components in series, forming, respectively, the Rankine cycle and the Brayton cycleshown schematically in Table 2.14. The thermodynamically ideal counterparts of these cycles arecomposed of four internally reversible processes in series: two isentropic processes alternated with twoconstant pressure processes. Table 2.14 provides property diagrams of the actual and corresponding idealcycles. Each actual cycle is denoted 1-2-3-4-1; the ideal cycle is 1-2s-3-4s-1.
For simplicity, pressuredrops through the boiler, condenser, and heat exchangers are not shown. Invoking Equation 2.29 for theideal cycles, the heat added per unit of mass flowing is represented by the area under the isobar fromstate 2s to state 3: area a-2s-3-b-a. The heat rejected is the area under the isobar from state 4s to state1: area a-1-4s-b-a. Enclosed area 1-2s-3-4s-1 represents the net heat added per unit of mass flowing.For any power cycle, the net heat added equals the net work done.Expressions for the principal energy transfers shown on the schematics of Table 2.14 are provided byEquations 1 to 4 of the table. They are obtained by reducing Equation 2.27a with the assumptions ofnegligible heat loss and negligible changes in kinetic and potential energy from the inlet to the outletof each component.
All quantities are positive in the directions of the arrows on the figure. Using theseexpressions, the thermal efficiency isη=(h3− h4 ) − (h2 − h1 )h3 − h2(2.94)To obtain the thermal efficiency of the ideal cycle, h2s replaces h2 and h4s replaces h4 in Equation 2.94.Decisions concerning cycle operating conditions normally recognize that the thermal efficiency tendsto increase as the average temperature of heat addition increases and/or the temperature of heat rejectiondecreases.
In the Rankine cycle, a high average temperature of heat addition can be achieved bysuperheating the vapor prior to entering the turbine, and/or by operating at an elevated steam-generatorpressure. In the Brayton cycle an increase in the compressor pressure ratio p2/p1 tends to increase theaverage temperature of heat addition. Owing to materials limitations at elevated temperatures andpressures, the state of the working fluid at the turbine inlet must observe practical limits, however.
Theturbine inlet temperature of the Brayton cycle, for example, is controlled by providing air far in excessof what is required for combustion. In a Rankine cycle using water as the working fluid, a low temperatureof heat rejection is typically achieved by operating the condenser at a pressure below 1 atm. To reduce© 1999 by CRC Press LLC2-79Engineering ThermodynamicsTABLE 2.14 Rankine and Brayton CyclesRankine CycleBrayton CycleW˙ p = m˙ (h2 − h1 )W˙ c Q˙ = m˙ (h − h )(> 0 )(1)(> 0 )(2)W˙ t = m˙ (h3 − h4 )(> 0 )(3)in32Q˙ out = m˙ (h1 − h4 )(> 0 )(4)erosion and wear by liquid droplets on the blades of the Rankine cycle steam turbine, at least 90%quality should be maintained at the turbine exit: x4 > 0.9.The back work ratio, bwr, is the ratio of the work required by the pump or compressor to the workdeveloped by the turbine:bwr =© 1999 by CRC Press LLCh2 − h1h3 − h4(2.95)2-80Section 2As a relatively high specific volume vapor expands through the turbine of the Rankine cycle and a muchlower specific volume liquid is pumped, the back work ratio is characteristically quite low in vaporpower plants — in many cases on the order of 1 to 2%.
In the Brayton cycle, however, both the turbineand compressor handle a relatively high specific volume gas, and the back ratio is much larger, typically40% or more.The effect of friction and other irreversibilities for flow-through turbines, compressors, and pumps iscommonly accounted for by an appropriate isentropic efficiency.
The isentropic turbine efficiency isηt =h3 − h4h3 − h4 s(2.95a)ηc =h2 s − h1h2 − h1(2.95b)The isentropic compressor efficiency isIn the isentropic pump efficiency, ηp, which takes the same form as Equation 2.95b, the numerator isfrequently approximated via Equation 2.30c as h2s – h1 ≈ v1∆p, where ∆p is the pressure rise across thepump.Simple gas turbine power plants differ from the Brayton cycle model in significant respects. In actualoperation, excess air is continuously drawn into the compressor, where it is compressed to a higherpressure; then fuel is introduced and combustion occurs; finally the mixture of combustion products andair expands through the turbine and is subsequently discharged to the surroundings.
Accordingly, thelow-temperature heat exchanger shown by a dashed line in the Brayton cycle schematic of Table 2.14is not an actual component, but included only to account formally for the cooling in the surroundingsof the hot gas discharged from the turbine.Another frequently employed idealization used with gas turbine power plants is that of an air-standardanalysis. An air-standard analysis involves two major assumptions: (1) as shown by the Brayton cycleschematic of Table 2.14, the temperature rise that would be brought about by combustion is effectedinstead by a heat transfer from an external source; (2) the working fluid throughout the cycle is air,which behaves as an ideal gas.
In a cold air-standard analysis the specific heat ratio k for air is taken asconstant. Equations 1 to 6 of Table 2.7 together with data from Table A.8 apply generally to air-standardanalyses. Equations 1′ to 6′ of Table 2.7 apply to cold air-standard analyses, as does the followingexpression for the turbine power obtained from Table 2.1 (Equation 27c″):[kRT3( k −1) kW˙ t = m˙1 − ( p4 p3 )k −1](2.96)(Equation 2.96 also corresponds to Equation 5′ of Table 2.8 when n = k.) An expression similar in formcan be written for the power required by the compressor.For the simple Rankine and Brayton cycles of Table 2.14 the results of sample calculations are providedin Table 2.15. The Brayton cycle calculations are on an air-standard analysis basis.Otto, Diesel, and Dual CyclesAlthough most gas turbines are also internal combustion engines, the name is usually reserved toreciprocating internal combustion engines of the type commonly used in automobiles, trucks, and buses.Two principal types of reciprocating internal combustion engines are the spark-ignition engine and thecompression-ignition engine.
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