The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 16
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The chemicalexergy data of Table 2.12 correspond to two alternative standard exergy reference environments, calledhere model I and model II, that have gained acceptance for engineering evaluations.Although the use of standard chemical exergies greatly facilitates the application of exergy principles,the term standard is somewhat misleading since there is no one specification of the environment that© 1999 by CRC Press LLC2-72Section 2suffices for all applications. Still, chemical exergies calculated relative to alternative specifications ofthe environment are generally in good agreement.
For a broad range of engineering applications thesimplicity and ease of use of standard chemical exergies generally outweigh any slight lack of accuracythat might result. In particular, the effect of slight variations in the values of T0 and p0 about the valuesused to determine the standard chemical exergies reported in Table 2.12 can be neglected.The literature of exergy analysis provides several expressions allowing the chemical exergy to beevaluated in particular cases of interest. The molar chemical exergy of a gas mixture, for example, canbe evaluated fromje CH =∑jyi eiCH + RT0i =1∑ y ln yii(2.90)i =1where eiCH is the molar chemical exergy of the ith component.Example 13Ignoring the kinetic and potential exergies, determine the exergy rate, in kJ/kg, associated with each ofthe following streams of matter:(a) Saturated water vapor at 20 bar.(b) Methane at 5 bar, 25°C.Let T0 = 298 K, p0 = 1.013 bar (1 atm).Solution.
Equation 2.88 reduces to reade = (h − h0 ) − T0 (s − s0 ) + e CH(a) From Table A.5, h = 2799.5 kJ/kg, s = 6.3409 kJ/kg · K. At T0 = 298 K (25°C), water would bea liquid; thus with Equations 2.50c and 2.50d, h0 ≈ 104.9 kJ/kg, s0 ≈ 0.3674 kJ/kg · K.
Table 2.12(model I) gives eCH = 45/18.02 = 2.5 kJ/kg. Thene = (2799.5 − 104.9) − 298(6.3409 − 0.3674) + 2.5= 914.5 + 2.5 = 917.0 kJ kgHere the specific exergy is determined predominately by the physical component.(b) Assuming the ideal gas model for methane, h – h0 = 0.
Also, Equation 2.58 reduces to give s –s0 = –Rlnp/p0. Then, Equation 2.88 readse = RT0 ln p p0 + e CHWith eCH = 824,350/16.04 = 51,393.4 kJ/kg from Table 2.12 (model I), 8.314 kJ kJ5e=+ 51, 393.4 (298 K) lnkg1.013 16.04 kg ⋅ K = 246.6 + 51, 393.4= 51,640 kJ kgHere the specific exergy is determined predominately by the chemical component.© 1999 by CRC Press LLC2-73Engineering ThermodynamicsTABLE 2.12 Standard Molar Chemical Exergy, eCH (kJ/kmol),of Various Substances at 298 K and p0SubstanceNitrogenOxygenCarbon dioxideWaterCarbon (graphite)HydrogenSulfurCarbon monoxideSulfur dioxideNitrogen monoxideNitrogen dioxideHydrogen sulfideAmmoniaMethaneEthaneMethanolEthyl alcoholabFormulaModel IaModel IIbN2(g)O2(g)CO2(g)H2O(g)H2O(l)C(s)H2(g)S(s)CO(g)SO2(g)NO(g)NO2(g)H2S(g)NH3(g)CH4(g)C2H6(g)CH3OH(g)CH3OH(l)C2H5OH(g)C2H5OH(l)6403,95014,1758,63545404,590235,250598,160269,410301,94088,85055,565799,890336,685824,3501,482,035715,070710,7451,348,3301,342,0857203,97019,8709,500900410,260236,100609,600275,100313,40088,90055,600812,000337,900831,6501,495,840722,300718,0001,363,9001,357,700Ahrendts, J.
1977. Die Exergie Chemisch Reaktionsfähiger Systeme,VDI-Forschungsheft. VDI-Verlag, Dusseldorf, 579. Also seeReference States, Energy — The International Journal, 5: 667–677,1980. In Model I, p0 = 1.019 atm. This model attempts to impose acriterion that the reference environment be in equilibrium. Thereference substances are determined assuming restricted chemicalequilibrium for nitric acid and nitrates and unrestrictedthermodynamic equilibrium for all other chemical components ofthe atmosphere, the oceans, and a portion of the Earth’s crust. Thechemical composition of the gas phase of this model approximatesthe composition of the natural atmosphere.Szargut, J., Morris, D. R., and Steward, F.
R. 1988. Energy Analysisof Thermal, Chemical, and Metallurgical Processes. Hemisphere,New York. In Model II, p0 = 1.0 atm. In developing this model areference substance is selected for each chemical element fromamong substances that contain the element being considered and thatare abundantly present in the natural environment, even though thesubstances are not in completely mutual stable equilibrium. Anunderlying rationale for this approach is that substances foundabundantly in nature have little economic value. On an overall basis,the chemical composition of the exergy reference environment ofModel II is closer than Model I to the composition of the naturalenvironment, but the equilibrium criterion is not always satisfied.The small difference between p0 = 1.013 bar and the value of p0 for model I has been ignored.Exergetic EfficiencyThe exergetic efficiency (second law efficiency, effectiveness, or rational efficiency) provides a truemeasure of the performance of a system from the thermodynamic viewpoint.
To define the exergeticefficiency both a product and a fuel for the system being analyzed are identified. The product representsthe desired result of the system (power, steam, some combination of power and steam, etc.). Accordingly,the definition of the product must be consistent with the purpose of purchasing and using the system.© 1999 by CRC Press LLC2-74Section 2The fuel represents the resources expended to generate the product and is not necessarily restricted tobeing an actual fuel such as a natural gas, oil, or coal. Both the product and the fuel are expressed interms of exergy.For a control volume at steady state whose exergy rate balance readsE˙ F = E˙ P + E˙ D + E˙ Lthe exergetic efficiency isE˙E˙ + E˙ε = ˙P = 1 − D ˙ LEFEF(2.91)where the rates at which the fuel is supplied and the product is generated are ĖF and E˙ P , respectively.Ė D and Ė L denote the rates of exergy destruction and exergy loss, respectively.
Exergy is destroyedby irreversibilities within the control volume, and exergy is lost from the control volume via stray heattransfer, material streams vented to the surroundings, and so on. The exergetic efficiency shows thepercentage of the fuel exergy provided to a control volume that is found in the product exergy. Moreover,the difference between 100% and the value of the exergetic efficiency, expressed as a percent, is thepercentage of the fuel exergy wasted in this control volume as exergy destruction and exergy loss.To apply Equation 2.91, decisions are required concerning what are considered as the fuel and theproduct.
Table 2.13 provides illustrations for several common components. Similar considerations areused to write exergetic efficiencies for systems consisting of several such components, as, for example,a power plant.Exergetic efficiencies can be used to assess the thermodynamic performance of a component, plant,or industry relative to the performance of similar components, plants, or industries. By this means theperformance of a gas turbine, for instance, can be gauged relative to the typical present-day performancelevel of gas turbines. A comparison of exergetic efficiencies for dissimilar devices — gas turbines andheat exchangers, for example — is generally not significant, however.The exergetic efficiency is generally more meaningful, objective, and useful than other efficienciesbased on the first or second law of thermodynamics, including the thermal efficiency of a power plant,the isentropic efficiency of a compressor or turbine, and the effectiveness of a heat exchanger.
Thethermal efficiency of a cogeneration system, for instance, is misleading because it treats both work andheat transfer as having equal thermodynamic value. The isentropic turbine efficiency (Equation 2.95a)does not consider that the working fluid at the outlet of the turbine has a higher temperature (andconsequently a higher exergy that may be used in the next component) in the actual process than in theisentropic process. The heat exchanger effectiveness fails, for example, to identify the exergy destructionassociated with the pressure drops of the heat exchanger working fluids.Example 14Evaluate the exergetic efficiency of the turbine in part (a) of Example 1 for T0 = 298 K.Solution. The exergetic efficiency from Table 2.13 isW˙W˙ε= ˙=˙˙E1 − E2 m(e1 − e2 )Using Equations 2.88 and 2.89a, and noting that the chemical exergy at 1 and 2 cancels,© 1999 by CRC Press LLCEngineering ThermodynamicsTABLE 2.13 The Exergetic EfÞciency for Selected Components at Steady StateaTurbine orExpanderExtractionTurbineCompressor,Pump, or FanHeatExchangerbMixing UnitGasiÞer orCombustionChamberEÇPWÇWÇEÇ2 - EÇ1EÇ2 - EÇ1EÇ3EÇ3EÇFEÇ1 - EÇ2EÇ1 - EÇ2 - EÇ3WÇEÇ3 - EÇ4EÇ1 + EÇ2EÇ1 + EÇ2eWÇÇE1 - EÇ2WÇÇE1 - EÇ2 - EÇ3EÇ2 - EÇ1WÇEÇ2 - EÇ1EÇ - EÇEÇ3ÇE1 + EÇ2EÇ3ÇE1 + EÇ2BoilerComponentab34(EÇ - EÇ ) + (EÇ - EÇ )(EÇ + EÇ ) + (EÇ + EÇ )(EÇ - EÇ ) + (EÇ - EÇ )(EÇ + EÇ ) - (EÇ + EÇ )6587123465871234For discussion, see Bejan et al.
(1996).This deÞnition assumes that the purpose of the heat exchanger is to heat the cold stream (T1 ³ T0). If the purpose of the heat exchanger is to provide cooling (T3 ³ T0), then thefollowing relations should be used: EÇP = EÇ4 - EÇ3 and EÇF = EÇ1 - EÇ2 .2-75© 1998 by CRC Press LLC2-76Section 2ε=[W˙]m˙ (h1 − h2 ) − T0 (s1 − s2 )Since W˙ = m˙ (h1 − h2 ),W˙ε= ˙˙ 0 (s2 − s1 )W + mTFinally, using data from Example 1 and s2 = 6.8473 kJ/kg · K,ε==30 MW kJ 1 MW 162, 357 kg 30 MW +(298 K) (6.8473 − 6.6022) 3600 s kg ⋅ K 10 3 kJ sec 30 MW= 0.9(90%)(30 + 3.29) MWExergy CostingSince exergy measures the true thermodynamic values of the work, heat, and other interactions betweenthe system and its surroundings as well as the effect of irreversibilities within the system, exergy is arational basis for assigning costs. This aspect of thermoeconomics is called exergy costing.Referring to Figure 2.16 showing a steam turbine-electric generator at steady state, the total cost toproduce the electricity and exiting steam equals the cost of the entering steam plus the cost of owningand operating the device.
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