The CRC Handbook of Mechanical Engineering. Chapter 2. Engineering Thermodynamics (776125), страница 12
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For an ideal gas mixture, comparison of Equations 2.63 and 2.72 suggests µi= gi (T, pi); that is, the chemical potential of component i in an ideal gas mixture equals its Gibbsfunction per mole of gas i evaluated at the mixture temperature and the partial pressure of the ith gasof the mixture.The chemical potential is a measure of the escaping tendency of a substance in a multiphase system:a substance tends to move from the phase having the higher chemical potential for that substance to thephase having a lower chemical potential. A necessary condition for phase equilibrium is that the chemicalpotential of each component has the same value in every phase.The Gibbs phase rule gives the number F of independent intensive properties that may be arbitrarilyspecified to fix the intensive state of a system at equilibrium consisting of N nonreacting componentspresent in P phases: F = 2 + N – P.
F is called the degrees of freedom (or the variance). For water asa single component, for example, N = 1 and F = 3 – P.• For a single phase, P = 1 and F = 2: two intensive properties can be varied independently, saytemperature and pressure, while maintaining a single phase.• For two phases, P = 2 and F = 1: only one intensive property can be varied independently if twophases are maintained — for example, temperature or pressure.• For three phases, P = 3 and F = 0: there are no degrees of freedom; each intensive property ofeach phase is fixed. For a system consisting of ice, liquid water, and water vapor at equilibrium,there is a unique temperature: 0.01°C (32.02°F) and a unique pressure: 0.6113 kPa (0.006 atm).The phase rule does not address the relative amounts that may be present in the various phases.With G = H – TS and H = U + pV, Equation 2.72 can be expressed asjU = TS − pV +∑n µi(2.73)ii =1from which can be derivedjdU = TdS − pdV +∑ µ dnii(2.74)i =1When the mixture composition is constant, Equation 2.74 reduces to Equation 2.31a.Ideal SolutionThe Lewis-Randall rule states that the fugacity fi of each component i in an ideal solution is the productof its mole fraction and the fugacity of the pure component, fi, at the same temperature, pressure, andstate of aggregation (gas, liquid, or solid) as the mixture:© 1999 by CRC Press LLC2-57Engineering Thermodynamics(Lewis-Randall rule)fi = yi fi(2.75)The following characteristics are exhibited by an ideal solution: Vi = vi , Ui = ui , Hi = hi .
With these,Equations 2.70a, b, and c show that there is no change in volume, internal energy, or enthalpy on mixingpure components to form an ideal solution. The adiabatic mixing of different pure components wouldresult in an increase in entropy, however, because such a process is irreversible.The volume of an ideal solution isjV=∑jni vi =i =1∑V(ideal solution)i(2.76)i =1where Vi is the volume that pure component i would occupy when at the temperature and pressure ofthe mixture.
Comparing Equations 2.48a and 2.76, the additive volume rule is seen to be exact for idealsolutions. The internal energy and enthalpy of an ideal solution arejU=∑i =1jni ui ,H=∑n hi i(ideal solution)(2.77)i =1where ui and hi denote, respectively, the molar internal energy and enthalpy of pure component i atthe temperature and pressure of the mixture. Many gaseous mixtures at low to moderate pressures areadequately modeled by the Lewis Randall rule. The ideal gas mixtures considered in Section 2.3, IdealGas Model, is an important special case. Some liquid solutions also can be modeled with the LewisRandall rule.© 1999 by CRC Press LLC2-58Section 22.4 CombustionThe thermodynamic analysis of reactive systems is primarily an extension of principles presented inSections 2.1 to 2.3. It is necessary, though, to modify the methods used to evaluate specific enthalpyand entropy.Reaction EquationsIn combustion reactions, rapid oxidation of combustible elements of the fuel results in energy releaseas combustion products are formed.
The three major combustible chemical elements in most commonfuels are carbon, hydrogen, and sulfur. Although sulfur is usually a relatively unimportant contributorto the energy released, it can be a significant cause of pollution and corrosion.The emphasis in this section is on hydrocarbon fuels, which contain hydrogen, carbon, sulfur, andpossibly other chemical substances.
Hydrocarbon fuels may be liquids, gases, or solids such as coal.Liquid hydrocarbon fuels are commonly derived from crude oil through distillation and cracking processes. Examples are gasoline, diesel fuel, kerosene, and other types of fuel oils. The compositions ofliquid fuels are commonly given in terms of mass fractions. For simplicity in combustion calculations,gasoline is often considered to be octane, C8H18, and diesel fuel is considered to be dodecane, C12H26.Gaseous hydrocarbon fuels are obtained from natural gas wells or are produced in certain chemicalprocesses. Natural gas normally consists of several different hydrocarbons, with the major constituentbeing methane, CH4.
The compositions of gaseous fuels are commonly given in terms of mole fractions.Both gaseous and liquid hydrocarbon fuels can be synthesized from coal, oil shale, and tar sands. Thecomposition of coal varies considerably with the location from which it is mined.
For combustioncalculations, the makeup of coal is usually expressed as an ultimate analysis giving the composition ona mass basis in terms of the relative amounts of chemical elements (carbon, sulfur, hydrogen, nitrogen,oxygen) and ash. Coal combustion is considered further in Chapter 8, Energy Conversion.A fuel is said to have burned completely if all of the carbon present in the fuel is burned to carbondioxide, all of the hydrogen is burned to water, and all of the sulfur is burned to sulfur dioxide. Inpractice, these conditions are usually not fulfilled and combustion is incomplete.
The presence of carbonmonoxide (CO) in the products indicates incomplete combustion. The products of combustion of actualcombustion reactions and the relative amounts of the products can be determined with certainty only byexperimental means. Among several devices for the experimental determination of the composition ofproducts of combustion are the Orsat analyzer, gas chromatograph, infrared analyzer, and flame ionization detector. Data from these devices can be used to determine the makeup of the gaseous productsof combustion. Analyses are frequently reported on a “dry” basis: mole fractions are determined for allgaseous products as if no water vapor were present. Some experimental procedures give an analysisincluding the water vapor, however.Since water is formed when hydrocarbon fuels are burned, the mole fraction of water vapor in thegaseous products of combustion can be significant.
If the gaseous products of combustion are cooled atconstant mixture pressure, the dew point temperature (Section 2.3, Ideal Gas Model) is reached whenwater vapor begins to condense. Corrosion of duct work, mufflers, and other metal parts can occur whenwater vapor in the combustion products condenses.Oxygen is required in every combustion reaction. Pure oxygen is used only in special applicationssuch as cutting and welding. In most combustion applications, air provides the needed oxygen.
Idealizations are often used in combustion calculations involving air: (1) all components of air other thanoxygen (O2) are lumped with nitrogen (N2). On a molar basis air is then considered to be 21% oxygenand 79% nitrogen. With this idealization the molar ratio of the nitrogen to the oxygen in combustionair is 3.76; (2) the water vapor present in air may be considered in writing the combustion equation orignored. In the latter case the combustion air is regarded as dry; (3) additional simplicity results byregarding the nitrogen present in the combustion air as inert.
However, if high-enough temperatures areattained, nitrogen can form compounds, often termed NOX, such as nitric oxide and nitrogen dioxide.© 1999 by CRC Press LLC2-59Engineering ThermodynamicsEven trace amounts of oxides of nitrogen appearing in the exhaust of internal combustion engines canbe a source of air pollution.The minimum amount of air that supplies sufficient oxygen for the complete combustion of all thecombustible chemical elements is the theoretical, or stoichiometic, amount of air. In practice, the amountof air actually supplied may be greater than or less than the theoretical amount, depending on theapplication. The amount of air is commonly expressed as the percent of theoretical air or the percentexcess (or percent deficiency) of air.
The air-fuel ratio and its reciprocal the fuel-air ratio, each of whichcan be expressed on a mass or molar basis, are other ways that fuel-air mixtures are described. Anotheris the equivalence ratio: the ratio of the actual fuel-air ratio to the fuel-air ratio for complete combustionwith the theoretical amount of air. The reactants form a lean mixture when the equivalence ratio is lessthan unity and a rich mixture when the ratio is greater than unity.Example 9Methane, CH4, is burned with dry air.
The molar analysis of the products on a dry basis is CO2, 9.7%;CO, 0.5%; O2, 2.95%; and N2, 86.85%. Determine (a) the air-fuel ratio on both a molar and a massbasis, (b) the percent of theoretical air, (c) the equivalence ratio, and (d) the dew point temperature ofthe products, in °F, if the pressure is 1 atm.Solution.(a) The solution is conveniently conducted on the basis of 100 lbmol of dry products. The chemicalequation then readsaCH 4 + b(O 2 + 3.76N 2 ) → 9.7CO 2 + 0.5CO + 2.95O 2 + 86.85N 2 + cH 2 Owhere N2 is regarded as inert. Water is included in the products together with the assumed 100lbmol of dry products. Balancing the carbon, hydrogen, and oxygen, the reaction equation is10.2CH 4 + 23.1(O 2 + 3.76 N 2 ) → 9.7CO 2 + 0.5CO + 2.95O 2 + 86.85N 2 + 20.4H 2 OThe nitrogen also balances, as can be verified.
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