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Maximizing Combustion Efficiency in Thermodynamics Analysis

School
Drexel University**We aren't endorsed by this school
Course
MEM 310
Subject
Mechanical Engineering
Date
Dec 10, 2024
Pages
28
Uploaded by SuperHumanSnowButterfly23
Lec15: Combustion (cont)THERMODYNAMICSAn Engineering ApproachCombustion (Chapter 15)Department of Mechanical Engineering & MechanicsDrexel University1
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Announcements•Qz08: 10=??; 5=??; 3=??; 0=??; Null=??•Project notes:–NAME: “MEM310_PrjctRprt_TeamN”–Please let me know if you have “team challenges” –I can step in now (much later and it is TOO LATE)•NOTE: team evaluation will be completed as an assignment on Monday of 10thweek•Final Exam: 2024 December 09 (0800-1000); We will run the exam that somehow overlaps this time –more information as we approach the exam 2
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Project Thoughts•Remember this is a steam power plant and most devices are analyzed as the SSSF devices in the Rankine Cycle were analyzed•Your goal is to identify the pressure of the flash evaporator (PFE) that provides maximum efficiency AND satisfies the design constraints•NOTE: –Boiler effluent enters the flash evaporator through a valve–Specified turbine exit quality is a minimum not necessarily the design value•I suggest:–Start your analysis by creating the T-s diagram for the power plant–Do your trials to find best PFE based on 1 kg/s total flow and once you find the PFE that maximizes efficiency proceed to determine mass flow rate(s) to produce required power and limit cooling water flow3
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Combustion•Balance a chemical reaction equation•Introduce jargon of reactive mixtures: AF; FA; stoichiometry; excess Air; equivalence ratio•Calculate reactant composition from exhaust measurements•Energy Analysis•Limiting Case A: Maximum Temperature4
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Combustion•Energy Analysis26
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What happens in a chemical rxn –227The composition of the working fluid changes either as time progresses or between the inlet and the exit)Fuel+ Oxygen+ NitrogenCO2, H2O,Nitrogen,?CO, ?HC, ?H, . . . timelocationJargon alert: each molecular identity is called a species
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Tools of Energy Analysis•Need new tools to account for this•Tool I: concept of standard reference state -•II: concept of enthalpy of formation:–When a compound is formed from its constituent elements (e.g., methane, CH4, from C (graphite) and H2), energy is either required or released. –This energy is called the “enthalpy of formation”•Released, the reaction is called exothermic;•Required, the reaction is called endothermic. 2825°C and 1 atm (indicated by superscript (°), i.e. h°, u°)The enthalpy of the elements or their naturally occurring stable compounds are assigned a value of zeroat the standard reference state. Applies to molecules such as H2, O2, N2, C etc.
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Example of enthalpy of formation29•The overall reaction equation isC+ 2H2→CH4•The conservation of energy for a steady-flow combustion processEin=EoutQnet+ HReactants= HProductsQnet=HProducts−HReactants
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Example (cont)30Areference state for all reacting components is:The enthalpy of the elements or their stable compounds is defined to beZERO at 25oC (298 K) and 1 atm (or 0.101 MPa).This heat transfer is called theenthalpy of formationfor methane,The superscript (o) implies the 1 atm pressure value and the subscript (f) implies25oCho fdata,is given in Table A-26.During the formation of methane from the elements at 298 K, 0.101 MPa, energy isreleased (an exothermic reaction) such thatho f
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Some enthalpy of formation valuesSubstanceFormulaMho kJ/kmolfAir28.970OxygenO2320NitrogenN2280Carbon dioxideCO244-393,520Carbon monoxideCO28-110,530Water (vapor)H2Ovap18-241,820Water (liquid)H2Oliq18-285,830MethaneCH416-74,850AcetyleneC2H226+226,730EthaneC2H630-84,680PropaneC3H844-103,850ButaneC4H1058-126,150Octane (vapor)C8H18114-208,450DodecaneC12H26170-291,01031
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Must bring all molecules (species) to the same base32The enthalpies at a temperature are represented as the combination of the enthalpy of formationevaluated at 25oC or 298 K, 1 atm and the “sensible enthalpy” energy required to raise the temperature of this species to the temperature of the systemh=h+ (h–h)fTooHere the term is the enthalpy of formation of any component at 298 K.The enthalpies at the T and 298K are tabulated (our text A-18 to -25)ho fIf tables are not available, the enthalpy difference due to the temperaturedifference can be calculated fromBased on the sign convention, the net heat transfer to the reacting system is
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Graphical Desc of Energy Analysis33•h500–hoDuplicated from a draft of the book Internal Combustion Engines and Automotive Engineering, R.D. Matthews, SAE International Pressh2000–ho
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Stoichiometric Butane Combustion34What is Qnet?Balanced combustion equation for tcc (stoichiometric) rxn:C4H10+ 6.5 (O2+ 3.76N2)→4CO2+5H2O+ 24.44N2The steady-flow heat transfer is
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Energy of Reactants35Reactants: TR= 298KCompNi kmol/kmol fuelho fkJ/kmolhTkJ/kmolhokJ/kmolN[ho + (h −ho)]ifT ikJ/kmol fuelC4H101-126,150-----126,150O26.508,6828,6820N224.4408,6698,6690HR =[h+ (hT −h)]ikJkmol C4H10∑i f=−126,150NReactantsoo
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36CompNe kmol/kmol fuelho fkJ/kmolhTkJ/kmolhokJ/kmolN [ho + (h −ho)]efTekJ/kmol fuelCO24-393,52042,7699,364-1,440,460H2O5-241,82035,8829,904-1,079,210N224.44030,1298,669+524,482Products:TP= 1000 KHP =NProducts[h+ (hT −h)]ekJkmol C4H10∑efoo=−1,995,188Energy of Products
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Q net= -1,870 MJ/kmol C4H1037Qnet =HP −HR=−1,869,038kJkmol C4H10
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Other Energy Terms•Enthalpy of Reaction, hR: HP-HRwhen both at same temperature•Enthalpy of Combustion, hC: complete combustion with tcc stoichiometry, T of both R and P is 25C, HP-HR; normalized to unit mass of fuel•Higher Heating Value, HHV: absolute value of hCwhen water is in liquid phase •Lower Heating Value, LHV: absolute value of hCwhen water is in gaseous phase–HHV = LHV + NH2O* hfg, H2O 38
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Goals•Limiting Case –Max Temperature39
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Limiting Case•Adiabatic Flame Temperature (AFT) or Adiabatic Combustion Temperature (ACT)–Temperature products have if combustion takes place without any energy loss/gain from heat transfer, i.e., adiabatically- alternatively, HP= HR 40
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Look at QnetWe can regroup:?𝒏?𝒕= σ?𝑵?ഥ𝒉𝑻?+ σ?𝑵?ഥ𝒉?𝟎?− σ?𝑵?ഥ𝒉𝟎?- σ?𝑵?ഥ𝒉?𝟎+ഥ𝒉𝑻− ഥ𝒉𝟎?Once fuel, stoichiometry and products are known: B, C and D are knownMust have an AFT to evaluate AGoal is to have Qnet= 0 41?𝒏?𝒕= σ?𝑵?ഥ𝒉?𝟎+ഥ𝒉𝑻− ഥ𝒉𝟎?- σ?𝑵?ഥ𝒉?𝟎+ഥ𝒉𝑻− ഥ𝒉𝟎?ABCD
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Finding AFT, aka ACT•Iterative solution, A: Qnet=0 means A=C+D-B0. Balance the overall reaction1. Use thermochemical information for reactants and products to calculate C+D-B, save the value2. As the products are primarily N2: set all kmols of products as N2kmols and look at N2tables to suggest TP,ithat has this enthalpy value C+D-B3. Evaluate real Ai using tabular data for real products at somewhere near ~97% of this Tp,i –use a temperature in tables4. If this Ai = C+D-B, stop5. If not, assume another T from tables (Tp,i+1) - go up or down based upon whether Ai> or < B-C-D, 6. Repeat steps 3 –5, until you stop at step 4{For details see example 15-8 (Ed 9)}42
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Finding AFT, aka ACT•Iterative solution, B0. Calculate B, C and D –keep handy1. Assume initial Tp,i is 2000 K (for air); 3000K (for oxygen)2. Evaluate Qnet= Qerror= Ai+ B –C - D; the only quantity to calculate is Ai•Qerror= 0 stop•Qerror< 0 then increase TP,i•Qerror> 0 then decrease TP,i 3. Define new test temperature: TP,i+1 = TP,i –Qerror/{~cpproducts}: round to nearest value in the tables4. If |TP,i+1 –TP,i |< 25 !!!! STOP5. Replace TP,i with TP,i+1 to evaluate Ai+1repeat 2-4, until you stop•~cpproducts: my suggestion is (Ai-C)/(Tp,i –298)43
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AFT Example –Technique B•Propane is burned in a SSSF boiler using 20% excess air•Unfortunately, the reaction still does not go to completion and scaled to 1 kmol of propane the measured products are as follows–CO2= 2.82 kmol -O2= 1.09 kmol–CO = 0.18 kmol -H2O=4.00 kmol•What is the maximum temperature that could be achieved? 44
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Solution: 1/5•Must first balance the tcc equation–C3H8+ NO2tccO2=> aCO2+ bH2O•C-balance: 1*3 = a*1, so a=3•H-balance: 1*8 = b*2, so b=4•O-balance: ntcc*2 = 3*2 + 4*1, so ntcc=5•Invoke excess air specification to determine stoichiometry of reactants–O2: (NO2act–NO2tcc)/NO2tcc= 0.2–Therefore NO2act= 645
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Solution: 2/5•Reactants (per kmol of C3H8)–C3H8+ 6 O2+6*3.76 N2•Measured Products (per kmol of fuel): –2.82 CO2+ 0.18 CO +1.09 O2+ 4 H2O•Plus non-participating N2: (6*3.76) N2•So the overall reaction is:C3H8+6O2+22.56N2=>2.82CO2+0.18CO+1.09O2+4 H2O+ 22.56N246
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Solution: 3/5•Need to calculate D, C, B•Reactants & Products: need enthalpies of formation enthalpy at 298 and enthalpy at 298K•D: σ?𝑵?ഥ𝒉?𝟎+ഥ𝒉𝑻− ഥ𝒉𝟎?=-103850 kJ/(kmol of fuel)•C: σ?𝑵?ഥ𝒉𝟎?= 272618.9 kJ/(kmol of fuel)•B: σ?𝑵?ഥ𝒉?𝟎?= -2111002 kJ/(kmol of fuel)47R#kmolP#kmol h298hf,298C3H810 27784-103850CO22.829364-398,520CO 0.188669-110,530O261.0986820N222.5622.5686690H2O49904-241820
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Solution: 4/5•A: h at first guess of products at 2000 K•A2000= 2162517 kJ/(kmol of fuel)•Qerr= -117254 kJ/(kmol of fuel)•Cp eff = (Ai-C)/(Tp,i –298) = 1110.398 kJ/(kmol of fuel-K)•Delta T = Qerr/Cp eff = 105.6 K48P#kmolh2000C3H80CO22.82100,804CO 0.1865408O21.0967881N222.5664810H2O482593
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Solution: 5/5•Repeat for 2100K (my table only has 100K increments)•A2100= 2286330 kJ/ (kmol of fuel)•Qerr= 6559.32 kJ/(kmol of fuel)•Cp eff = 1117.487 kJ/((kmol of fuel-K)•Delta T = Qerr/Cp eff = -5.8697 K, STOP!!!49P#kmolh2100C3H80CO22.82106864CO 0.1869044O21.0971668N222.5668417H2O487735
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