燃烧室和燃烧器课件：Chapter 3 Thermodynamics and Chemical Kinetics of Combustion

1、Chapter 3: Thermodynamics and Chemical Kinetics of Combustion (4h) 3.1 Thermodynamics of CombustionFirst law of thermodynamicsChemical energyProperties of mixturesCombustion stoichiometry（燃烧化学恰当性）First law combustion calculationsChemical equilibriumSecond law analysisStatic Reaction 3.2 Chemical Kin

2、etics of CombustionElementary reactionsGlobal reactionsChain and chain-branching reactionsArrhenius lawCollision reaction rate theorySome important chemical mechanisms 3.1 Thermodynamics of Combustion( 3.1.1 Review of the first law) Q = E + WWhere Q is the heat added to the system E is the Change of

3、 the total system energy W is the work done by system on surroundings 3.1.2 Chemical EnergyStandard enthalpy of formation（标准生成焓）Absolute enthalpy（绝对焓/总焓）Enthalpy of combustion (or reaction)Heating value （热值）Standard enthalpy of formation, hf0hf0 is the heat of reaction per mole of product formed iso

4、thermally from elements in their most stable forms at the standard reference state, i.e., P=1 atm, and T=25 . 标准生成焓：标准状态下，每摩尔产物相对于其组成元素的焓值（与化学键有关）。Enthalpies of formation are zero for the elements in their naturally occurring state (or the most stable form) at the reference state. 所有元素以其自然状态存在时的标准生成

5、焓为0 。For example, hf0,O2=0.Total (Absolute) enthalpy, h h = hf0 + hsWhere hs is the sensible enthalpy（显焓） corresponding to temperature change from Tref to T.About JANAF TableJANAF is the acronym for “Joint Army, Navy and Air Force”JANAF is the most commonly used source of thermo-chemical information

6、, especially for military and their contractors. Almost all students in USA are asked to be familiar with JANAFThe information in JANAF originates either from measurement or from calculations attributed to statistical mechanicsA summary is provided by Glassman, or Borman, in their AppendixEnthalpy o

7、f (combustion) reaction, hr hr = hf,prod - hf,reac = Mihf,i - Mjhf,j Where Mi is the mole number of each product, and Mj is that of each reactantHeating value, hc hc = - hr,HHV: Higher heating Value（高位热值）, calculated with the assumption that all of the water in the products has condensed to liquid.L

8、HV: Lower Heating Value（低位热值）, supposing none of the water is condensed.Example 3.1 Determine the total enthalpy of nitric oxide (NO) at 298K and 1 atm.Solution: since T=298k and P=1 atm, so the total enthalpy is the standard heat of formation. The chemical reaction for NO formation is 0.5N2+0.5O2NO

9、 according to the JANAF table above h0f=h0f,prod-h0f,react=90.29-0-0=90.29 MJ/Kgmol= 90.29/4.1868kcal/mol=21.57 kcal/mol The respective energy levels of N2, O2 and No can be sketched belowcomments: It takes energy to create NO from N2 and O2. More generally, h0f 0 means energy is required to form th

10、e substance; while h0f 0 means energy is released to form itExample 3.2 Determine the total enthalpy of nitric oxide (NO) at 1000K and 1 atm.Solution: Total enthalpy = standard heat of formation + sensible enthalpy = 90.29+22.23 MJ/kgmol = 112.52/4.1868 kcal/mol = 26.88 kcal/mol The energy level（能级）

11、 of NO at 1000K can be sketched as 3.1.3 Properties of mixtures Partial pressure（分压） Partial volume （分容积） Mass fraction （质量百分比） Mole fraction （摩尔百分比） Volume fraction （体积百分比） Mixture molecular weight （混合物分子量） Mixture gas constant （混合物气体常数）Please refer to Chapter 1 of Engineering Thermodynamics（工程热力学）

12、 edited by the 3 universities（三校合编） 3.1.4 Combustion stoichiometryStoichiometric air: the specific theoretical amount (G0 by mass or V0 by volume) of air required to burn the fuel completely without dissociation（理论空气量G0 ：完全燃烧每公斤燃料所需要的空气量）Generally dry air as a mixture of 79%(vol) N2 and 21%(vol) O2

13、or 3.764 moles of N2 per mole O2 is used in practical applications unless specified. The molecular weight of dry air is considered to be 29.0 Several definitions concerning combustion stoichiometry: Fuel/air ratio , or air/fuel ratio AFR（油/气比，或气/液比） Ratio of excess air, （余气系数） Equivalence ratio, （当量

14、比）fuel/air ratio (f): the ratio of fuel mass flow rate to air mass flow rate supplied into the combustion systemair/fuel ratio: reciprocal of ratio of excess air( ): the amount of air actually used for 1 kg fuel divided by the stoichiometric air , i.e., =ma/masequivalence ratio(): the actual fuel/ai

15、r mass ratio divided by the stoichiometric fuel/air mass ratio = / sThe relationships between the different definitions AFR=1/ G0 =1 = G0 =1 = 1/(G0)= /G0Example 3.3 For a stoichiometric hydrogen-air reaction at 1 atm pressure, find (a) the fuel-to-air mass ratio and the stoichiometric air G0, (b) t

16、he mass of fuel per mass of reactants, and (c) the partial pressure of water vapor in the products.Solution: React 1mole H2 with enough air to form the complete products H2O and N2: H2 + a (O2 + 3.76 N2) b H2O + 3.76 a N2 From an H balance: 2=2b, so b=1; from an O balance: 2a=b, so a=1/2; from an N

17、balance, 3.76/2=1.88. Hence the stoichiometric equation is: H2 + 1/2 (O2 + 3.76 N2) H2O + 1.88 N2(a) On a mass basis (1 kgmol H2) (2 kg/kgmol) =2 kg H2 reacts with (4.76/2 kgmol air) (29 kg/kgmol) =69.02 kg air, then =mf/ma=2/69.02=0.029, G0=ma/mf=1/=34.5(b) the mass of H2 per unit mass of reactant

18、mixture is mf/(mf+ma) =/(1+)=0.029/1.029=0.0282(c) the partial pressure of the water vapor in the products is obtained from the mole fraction: H2O=moles water/moles products =1/2.88=0.347 So the partial pressure is PH2O= H2O P=0.347 atmH2 + 1/2 (O2 + 3.76 N2) H2O + 1.88 N2Example 3.4 Determine the p

19、ower extracted from the following hypothetical reactor (1 atm, complete combustion), suppose the volume flow rate of methane is 2 liter/sec (measured at 500K, 1 atm)Solution: First, an energy level analysis is needed. The major products of the stoichiometric combustion of methane are given byCH4 + 2

20、(O2 + 3.76 N2) 2 H2O + CO2 + 7.52 N2 Enthalpies from JANAF are tabulated as followsh0f(MJ/kgmol)T(K)Sensible enthalpy(MJ/kgmol)Total enthalpy(MJ/kgmol)Pro-ductsH2O-241.83150048.1-193.73CO2-393.52150061.71-331.81N20150038.438.4Reac-tantsCH4-74.875008.2-66.67O205006.096.09N205005.915.91 So the heat of

21、 reaction corresponding to the combustion of 1 mole of methane in the reactor is hr = Mihf,i - Mjhf,j =2(-193.73) + (-331.81) + 7.52 38.4 - (-66.67) - 2 6.09 - 7.52 5.91 = - 420.4552 MJ/kgmol CH4 That means, the reactants have a higher energy than the products. As a result, they release energy in th

22、e combustion process.A reaction with hr 0 is called an “endothermic（吸热）” reaction, which needs a supply of energy to proceed. The energy levels of the reaction can be sketched belowThe power extracted from the reactor can be calculated as Power = - n (mole/sec) hr (J/mole) = - PCH4 V/(RT) hr (equati

23、on of state) = - NCH4/(NCH4+NO2+NN2) Ptotal V/(RT) hr = 1/(1+2+7.52) 1. 013 105 2 10-3/ (8.314500) 420.4552 103 = 1948 WThe stoichiometric air values for typical fuels are Methane 17 Kerosine / Gasoline 15 Hydrogen 35Tips （小贴士）1（甲）meth2（乙）eth3（丙）prop4（丁）but5（戊）pent6（己）hex7（庚）hep8（辛）oct9（壬）non10（癸）de

24、c11（十一）undec12（十二）dodecTips Molecular formula Family Name Other designations（分子式） （烃类） （其它名称）CnH2n+2 Alkanes（烷烃） ParaffinsCnH2n Alkenes（烯烃） OlefinsCnH2n-2 Alkynes（炔烃） AcetylenesCnH2n-6 Aromatics（芳烃） Benzene 3.1.5 First law combustion calculationsAdiabatic flame temperature (no dissociation，无离解)Adiab

25、atic flame temperature (with dissociation，有离解, second law)Variation of the adiabatic flame temperature with equivalence ratio and air preheating（空气预热）Adiabatic flame temperatureFuel TypeFuel NameEquivalence Ratio 0.81.01.2Gaseous FuelsMethane（甲烷）202022502175Ethane （乙烷）204022652200Propane （丙烷）2045227

26、02210Octane （辛烷）215023552345Liquid FuelsOctane （辛烷）205022752215Methanol（甲醇）175519751810Ethanol（乙醇）193521552045No.2 fuel oil208523052260Dry Solid FuelsBituminous（烟煤）199022152120Lignite（褐煤）196021852075Wood（木材）193021452040(25% Moisture)Solid FuelsLignite（褐煤）176019901800Wood148017001480Adiabatic Flame T

27、emperature (K) （绝热火焰温度）of selected fuels with air at 1 atm. and initially at 298 KConstant pressure adiabatic flame temperature: the fuel-air mixture burns adiabatically at constant pressure, such as the situation in gas turbine combustors and furnaces. The absolute enthalpy of the reactants at the

28、initial state equals that of the products at the final state.（定压燃烧）Constant volume adiabatic flame temperature: the fuel-air mixture burns adiabatically at constant volume, such as the situation in internal combustion engines. The absolute internal energy of the reactants at the initial state equals

29、 that of the products at the final state.（定容燃烧）Example 3.5 constant pressure adiabatic flame temperature Estimate the constant pressure adiabatic flame temperature for the combustion of a stoichiometric CH4-air mixture at the initial state of 1 atm and 298 K.Use the following assumptions : 1. Comple

30、te combustion and no dissociation. 2. The product mixture enthalpies are estimated using constant specific heats evaluated at 1200 K (0.5(Ti+Tad), Tad is guessed to be about 2100 K).Solution: mixture composition: CH4 + 2(O2+3.76N2) CO2 + 2H2O + 7.52N2SpeciesEnthalpy of formation at 298K (kJ/kmol)Spe

31、cific heat at 1200K (kJ/kmol-K)CH4-74831CO2-39354656.21H2O-24184543.87N2033.71O20Gas Properties (refer to other textbooks or handbooks) are as follow:According to the first law: Hreact = Hprodwhile Hreact = nihi0 = (1)(-74831)+(2)(0)+(7.52)(0) = -74831 kJCH4 + 2(O2+3.76N2) CO2 + 2H2O + 7.52N2Hprod =

32、 njhj0 + njCpj(Tad-298) = (1)-393546+56.21(Tad-298) + (2)-241845+43.87(Tad-298) + (7.52)0+33.71(Tad-298)then Tad=2318 K CommentsComparing the above results with the experiment (or equilibrium-composition based computation) shown as before (2250 K) shows that the simplified approach overestimates（高估）

33、 Tad by slightly less than 100K. Removing assumption 2 and re-calculating Tad using variable Cp (or find the sensible enthalpy from JANAF tables) Hprod = njhj0 + njCpjdT will yield Tad=2328 K. Then we conclude that the 100K difference is the result of neglecting dissociation.Example 3.6 constant vol

34、ume adiabatic flame temperature Estimate the constant volume adiabatic flame temperature for the combustion of a stoichiometric CH4-air mixture using the same assumptions as in the last example at the initial state of 1 atm and 298 K.Solution: The same composition and properties used in the last exa

35、mple apply here. We note, however, that the Cp,i values should be evaluated at a temperature somewhat greater than 1200 K, since the constant volume Tad will be higher than the constant pressure Tad. Nevertheless, we will use the same values as before. First law: Ureact(Tinit, Pinit) = Uprod(Tad, Pf

36、), i.e., Hreact VPinit = Hprod VPf, i.e., Hreact Hprod Ru(nreactTinit nprodTad) = 0While Hraect = (1)(-74831) + (2)(0) + (7.52)(0) = -74831 kJ Hprod = (1)-393546 + 56.21(Tad-298) + (2)-241845 + 43.87(Tad-298) + (7.52)0 + 33.71(Tad-298) = -877236 + 397.5(Tad-298) kJAnd Ru(nreactTinit nprodTad) = 8.31

37、510.52(298-Tad)where nreact = nprod = 10.52 kmol.Finally, the Tad can be solved to be: Tad = 2889 KComments:For the same initial conditions, constant volume combustion results in much higher temperatures (571 K higher in this example) than for constant pressure combustion. This is a consequence of t

38、he pressure forces doing no work（不作功） when the volume is fixed.The final pressure is well above the initial pressure: Pf = Pinit(Tad/Tinit) = 9.69 atm 3.1.6 Chemical equilibriumWhat is a chemical equilibrium: Chemical equilibrium is achieved for constant temperature and pressure systems when the rat

39、e of change of concentration goes to zero for all species.It is not safe to assume equilibrium for a shock wave（激波）Strictly speaking, chemical equilibrium does not exist in a flame zone because of the steep temperature gradients and the occurrence of many short-lived species.Why is chemical equilibr

40、ium importantThermodynamics alone cant determine what species may be in the products. When the products have reached chemical equilibrium, the composition of the products can be determined.Only when the composition of products is determined, the thermodynamic properties of the mixture, such as u, h,

41、 etc., may be calculatedDetermination of equilibrium constantConsidering a typical reaction in equilibrium: aA + bB cC + dDIn equilibrium, (dG) = Gprod Greact =0The Gibbs function（吉布斯函数） for a mixture of ideal gases can be expressed as: Gmix = ni gi,TFor a mixture of ideal gas, the Gibbs function fo

42、r the ith species is given by（见工程热力学中，热力学一般关系式）: gi,T=g0i,T + RuT ln(Pi/P0)where g0i,T is the Gibbs function of the pure species at the standard-state pressure and can be found in JANAF tables, and Pi is the partial pressure, Ru= 8.314 kJ/(kgmol-k) is the universal gas constant（普适气体常数）Then the chang

43、e of Gibbs function of the system after reaction is: (dG) = Gprod Greact = c gc,T + d gd,T a ga,T b gb,T = c g0c,T + d g0d,T a g0a,T b g0b,T + RuT ln(PccPdd/PaaPbb)+ RuT ln P0a+b-c-dAccording to the definition of equilibrium constant Kp = PccPdd/PaaPbband suppose P0 = 1 atm, when in equilibrium, the

44、 above equation can be reduced to: c g0c,T + d g0d,T a g0a,T b g0b,T + RuT ln Kp = 0Finally, ln Kp=a g0a,T/ RuT+b g0b,T/ RuTc g0c,T/RuTd g0d,T/ RuTgi,T=g0i,T + RuT ln(Pi/P0)Example 3.7 equilibrium constant Equal moles of H2 and O2 react to produce H2O, H2, and O2 at 2500 K and 1 atm. Find the percen

45、t by volume of H2, O2, and H2O due to the equilibrium reaction H2O = H2 + 0.5O2 using the method of equilibrium constants.Solution:Let z moles of H2 react with z moles of O2 to form products. Suppose the mole fraction of each species to be H2O, H2, and O2, then, the reaction can be written asA hydro

46、gen atom balance gives: 2z = 2 H2O + 2 H2Similarly, an oxygen atom balance gives: 2z = H2O + 2 O2Combining the two equations results in: 2 H2O + 2 H2 = H2O + 2 O2 (1)Obviously, the sum of the mole fractions of the products should be 1: H2O + H2 + O2 = 1 (2)According to previous derivation, ln Kp=a g

47、0a,T/ RuT+b g0b,T/ RuT c g0c,T/RuTd g0d,T/ RuT = g0H2O/ RuTg0H2/RuT0.5g0O2/ RuTUsing the data from JANAF tables,H2O = H2 + 0.5O2 ln Kp= (-106416-0 0 )/8.314/2500= -5.12then Kp= 0.005976according to the definition of Kp, Kp = PccPdd/PaaPbb = H2(O2)0.5P0.5/ H2O since P=1 atm, then H2(O2)0.5/ H2O = 0.0

48、05976 (3)solving equations (1), (2), and (3) simultaneously yields: H2O = 0.658 O2 = 0.335 H2 = 0.007Comments:For a given global reaction, if T is given, then Kp could be determined 3.1.7 Second law analysisConsider the following combustion in a fixed volume, adiabatic vessel in which a fixed mass o

49、f reactants form products. CO + 0.5O2 CO2If the final temperature is high enough, CO2 will dissociate.Supposing the products to consist only of CO2, CO, and O2, then we can write: CO + 0.5O2 (1-) CO2 + CO + 0.5 O2where is the fraction of the CO2 dissociated. We can calculate the adiabatic flame temp

50、erature Tf as a function of, according to the first law. What constraints are imposed by the second law on the reaction ? (dS)U,V,m = 0 （熵增为0）The entropy of the product mixture can be calculated by summing the product species entropies, i.e., Smix = nisi = (1- ) sCO2 + sCO + 0.5 sO2And the individua

51、l species entropy are obtained from si = s0i(Tref) + cp,i/T dT Ruln(Pi/P0) = s0i(T) Ruln(Pi/P0)Where s0i(T) can be found in JANAF tables, and Pi is the partial pressure.Obviously, The mixture entropy is also a function of the dissociation fraction.If we plot Smix , and Tf in the same figure, the cou

52、ld be determined once the point at which Smix has the maximum value is found. Given the initial temperature T0 is 298 K, and initial pressure is 1 atm, then Hreact=1hco+0.5hO2 =-110.53+0.50=-110.53 MJ Hprod=(1- )hCO2+ hco+0.5hO2According to previous deduction, for constant volume combustion Hprod=Hr

53、eact+Ru(njTf-niT0)Suppose Tf=1500K, then Hprod=(1- )(-395.67+61.71)+ (-115.23+38.85)+ 0.5(0+40.61) =267.885 -323.96 MJ Ru(njTf-niT0)=8.31410-3 (1+0.5) 1500-1.5298 =6.2355+8.7546 MJSubstituting to above equation, one obtains 267.885-323.96=-110.53+6.2355+8.7546 CO + 0.5O2 (1-) CO2 + CO + 0.5 O2So, =0

54、.85474Since si = s0i(T) Ruln(Pi/P0)And Pi/P0=(Pi/PT)(PT/P0) PT=P0(Tad/T0)(1+0.5)/1.5=4.7634P0And Smix = (1- ) sCO2 + sCO + 0.5 sO2 Then the entropy of the product mixture can be determined asSmix=(1- )292.11-8.314ln4.7634(1- )/(1+0.5 ) +248.31-8.314ln4.7634/ (1+0.5 ) +0.5 257.97-8.314ln4.76340.5 /(1

55、+0.5 ) =357.25 kJ/KSimilarily, suppose Tf=1800K, one can obtains =0.81383, Smix =362.43 kJ/KCO + 0.5O2 (1-) CO2 + CO + 0.5 O2Tf(K)Smix(kJ/K)Tf(K)Smix(kJ/K)15000.85474357.2538000.49647376.18000.81383362.4339000.47869375.4425000.71126370.8340000.46071375.6534000.56578375.2542000.42419375.1436000.53149

56、375.2946000.3488374.2837000.51408375.6849000.29014372.93For different flame temperature Tf, one can obtain the following table by the same calculations (by means of the JANAF table in Glassmans book)These data could be plotted asComments and summaryFrom the figure we see that a maximum value of the

57、mixture entropy （混合物的熵） is reached at some intermediate value of (around 0.5).Once the maximum entropy is reached, no further change in composition is allowed, since this would require the system entropy to decrease in violation of the second law.In summary, if we fix the internal energy, volume, an

58、d mass of an isolated system, the application of the first law, second law, and equation of state（状态方程） will define the equilibrium temperature, pressure, and chemical composition.本节要点 理论空气量（定义、常用值） 当量比、油/气比、余气系数等 绝热火焰温度（理论燃烧温度）（定义、常用值、定压与定容差异） 焓（标准焓、显焓、绝对焓）与能级 热力学第二定律确定离解度 熟悉JANAF表 3.2 Chemical Kin

59、etics of Combustion3.2.1 Elementary reactions （基元反应） and the law of mass action（质量作用定律）Two problems in chemical kinetics:To find the elementary reactions by which the given reaction proceedsTo determine the reaction rate constant（反应速率常数） for each of the stepsAn elementary reaction occurs at the mole

60、cular level by collision process when two or more molecules or atoms approach one another, such as H + O2 O + OH , O + H2 H + OH , etc.Three major types of elementary reactions:Bimolecular atom exchange reactions: AB + C BC + ATermolecular recombination reactions: A + B + M AB + MBimolecular decompo