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AME 513 Principles of Combustion

AME 513 Principles of Combustion. Lecture 6 Chemical kinetics III – real fuels. Outline. Experimental methods Types of reaction CO-O 2 Hydrocarbons NO x Flame inhibition. Experimental methods. Constant volume vessel

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AME 513 Principles of Combustion

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  1. AME 513Principles of Combustion Lecture 6 Chemical kinetics III – real fuels

  2. Outline • Experimental methods • Types of reaction • CO-O2 • Hydrocarbons • NOx • Flame inhibition AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  3. Experimental methods • Constant volume vessel • Heat evacuated vessel to some temperature, inject reactants quickly (shorter than reaction time scale), see if it reacts explosively or just slowly • Ideally constant T & P • Not well-defined limit, depends on injection / mixing time ~ d2/D (d = vessel dimension, D = diffusivity) being much faster that reaction, only useful for conditions with “slow” reaction • For gases, both mass D and thermal D ~ 1/P, so characteristic reaction time required for explosion varies with P! • Wall effects crucial (radical sink) • Only yields global properties (e.g. overall reaction rates) • Well-stirred reactor • Well defined limits • Must be certain that mixing of fresh reactants with products already in reactor is much shorter than residence time in reactor • Typically need very large mass flows to study flame-like conditions since reaction time scale ~ 10-3 s in flame • Only yields global properties AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  4. Experimental methods • Plug-flow reactors • Flow reactants down a preheated tube (usually constant T), measure species at varying distances along tube (~ time) • Can obtain data on observing evolution of individual species, not just global properties, thus infer elementary reaction rates • Laminar or turbulent flow • Only useful for “slow” reactions (low temperature, not flame-like conditions) (tens of milliseconds) • Need to address issue of axial dispersion of reactants • Shock tube • Pass shock wave through mixture, watch evolution of species • Step-like change in T and P, well defined, nearly homogeneous • Can obtain data on individual reactions • Amenable to quantitative laser diagnostics • Only useful for “fast” reactions (few ms) AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  5. Experimental methods • Laminar flame • “Simple” setup – measure burning velocity SL • Obviously applicable to “real” flame chemistry • Flames are very thin (< 1mm), hard to probe inside to measure species evolution • Interaction with transport via convection and diffusion – compare results with computations using detailed flame models • Not a sensitive instrument – overall reaction rate ~ SL1/2 AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  6. Types of reactions • Global reaction • Example: CH4 + 2O2 CO2+ 2 H2O • NOT an actual reaction that occurs • No relation between order of reaction (3 in this example) and actual pressure effect on reaction rate • Chain initiation • Example: H2 + M  H + H + M • Break stable molecule into radical(s) • High Ea – endothermic, must break strong bond • High Z (i.e. not very orientation sensitive) • Not needed in flames, where radical source (products) exists • Chain branching • Example: H + O2 OH + O • Use radical to create more radical • High Ea - endothermic, must break strong bond, but also make a bond, so not as high as chain initiation • Moderate Z (somewhat orientation sensitive) AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  7. Types of reactions • Chain propagation or “shuffle” reaction • Example: OH + H2 H2O + H • Use radical + stable molecule to create another pair • Moderate Ea – may be nearly thermo-neutral • Moderate Z (somewhat orientation sensitive) • Chain termination • Example: H + OH + M  H2O+ M • Recombine radicals into stable molecules (usually products) • Low or zero Ea – exothermic, no activation barrier • Need 3rd body to absorb enthalpy and conserve momentum • Moderate Z (may be orientation sensitive) • Schematic multi-step mechanism (e.g. Hautman et al, 1981) C3H8 1.5 C2H4 + H2 (Initial breakdown of fuel) C2H4 + O22 CO + 2 H2 (C2H4 = surrogate for radicals) 2H2 + O2 2 H2O (oxidation of H2) 2CO + O2 2 CO2(oxidation of CO) Reaction rates crazy (see Turns, p. 158) – sometimes order of reaction is negative – causes problems as concentration  0 AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  8. H2– O2 • Simulate explosion limits using online chemistry calculator • Use time to 50% H2 consumption as measure of explosion limit • Diffusion time ~ P so scale accordingly, e.g. at 10 atm, allow 10x more time than at 1 atm • Results for t at 1 atm = 25 s close to experiments (Fig. 5.1) • Second limit independent of time (vessel size & walls material) • Second limit where branching vs. recombination rates ≈ same H + O2 OH + O = H + O2+ M  HO2+ M when P = 103.4T0.2e-17500/RT AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  9. CO – O2 • H2-O2 already discussed – 3 explosion limits • CO MUCH different because no chain branching and no “shuffle” reaction to create product CO + O2 CO2 + O Very slow, but needed as source of O O2 + M  O+ O Very high activation energy, even slower CO + M  C + O + M No way! C=O is strongest chemical bond! CO + O + M  CO2 + M Creates product but removes radicals • As a result, pure CO – O2 oxidation is extremely slow! • Early experiments showed widely varying results because of contamination with water; with any hydrogen source CO + OH  CO2 + H Shuffle reaction to create CO2 H+ O2 OH + O Regenerate OH plus another O Neither are fast, but better than alternatives! • Stoich. CO + O2, 1 atm, 1500K, time to consume 50% of CO: • No H2O: 1.03 s; 1 ppm H2O: 0.27 s; 10 ppm: 0.038 s; • 100 ppm 0.0043 s; 1,000 ppm 0.00080 s; 10,000 ppm 0.00023 s AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  10. CO – O2 • Case shown: const. T = 1000K, P = 1 atm, CO:H2:O2 = 1:1:10 • To a good approximation, H2-O2 acts as infinitely fast (i.e. steady-state) radical source for CO • CO does not affect H2-O2 process, CO reactionjust too slow • 50% H2 consumption in ≈ 400 µs, ≈ same with or without CO AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  11. CO – O2 • Explosion limit experiments show 2nd-limit behavior, even with “dry” CO – not really dry • CSU homogeneous kinetics model (next page) does not! Dickens et al., 1964 Gordon and Knipe, 1955 AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  12. CO – O2 • Unlike classical experiments, CSU homogenous kinetics calculator predicts that dry CO has only a single limit – but CSU site does not include O3 which may be important • With H2Oaddition, behavior is very similar to H2-O2 with offset to higher T - CO is a “parasite” on the OH source AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  13. Hydrocarbon – O2 • Hydrocarbons inhibit their own oxidation, because they react with radicals more readily than O2 reacts with radicals • Once an H atom is removed from fuel, e.g. C3H8 + O2 C3H7• + HO2• then the fuel radical can lose another H to form an alkene, e.g. C3H7• + O2 C3H6+ HO2• thus alkenes are a key intermediate in alkane oxidation • Nearly all of fuel must be consumed before radical pool needed to consume CO can build up • CO oxidation is the last step • b-scission – fuel molecule breaks apart 1 C-C bond away from C missing an H atom (avoids having to move an H atom to an adjacent C atom) • This stuff only matters at “low” temperatures (<1500K) where H + O2 branching is inhibited by the hydrocarbons; at higher temperatures, H + O2 branching is sufficiently rapid AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  14. Hydrocarbon – O2 AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  15. Hydrocarbon – O2 • Hydrocarbons have no 1st or 2nd explosion limit; instead of H + O2 + M  HO2 + M being the dominant recombination reaction, it’s CH4+ H  CH3 + H2, which has same pressure dependence as the branching reaction H + O2 OH + O • C-H bonds stronger in CH4 than C3H8, thus higher explosion T AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  16. Hydrocarbon oxidation • Start with fuel molecule RH, where R is an “organic radical”, e.g. propane without an H • Abstract an H atom from RH RH + O2 R + HOO • Add an O2 to R R + O2 ROO • Produce peroxides with O-O single bond (half as strong as O=O double bond (120 kcal/mole vs. 60 kcal/mole), much easier to break) ROO + RH  R + ROOH or HOO + RH  R + HOOH • Break O-O single bond, create “chain branching” process ROOH + M RO + OH or HOOH + M HO + OH • Newly created radicals generate more organic radicals RH + OH R + HOH or RH + RO R + ROH • Note that rate of reaction will be sensitive to rates of H atom removal from fuel molecule RH AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  17. Hydrocarbon oxidation • Rate of H atom removal depends on strength of C-H bond, which in turn depends on how many other carbons are bonded to that C - stronger bond, slower reaction, less knock • Examples: n-heptane: 6 primary, 12 secondary C-H bonds 2, 2, 4 trimethy pentane: 15 primary, 2 secondary, 1 tertiary AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  18. Hydrocarbon oxidation • Does this small difference in bond strength matter? YES because activation energy is high • If we use bond strength as a measure of activation energy (dangerous in general, but ok here…) then at a typical 900K methane : primary : secondary : tertiary exp(-Emethane/T) : exp(-Eprimary/T) : exp(-Esecondary/T) : exp(-Etertiary/T) ≈ exp(-105,000 cal/mole/(1.987 cal/mole-K)(900K)) : exp(-98000/1.987*900) : exp(-95000/1.987*900) : exp(-93000/1.987*900) ≈ 1 : 50 : 268 : 820 • As a result, fuels with mostly primary C-H bonds will decompose much more slowly than isomers with more secondary & tertiary C-H bonds – higher octane number in gasoline-type fuels AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  19. Hydrocarbon oxidation • How to make CO? RO• formed by cleavage of O-O bond in peroxide then enter aldehyde route RO• + M  R’HCO + H + M, e.g. C2H5O• + M  CH3HCO + H + M • Aldehydes have weakest C-H bond (≈87 kcal/mole) thus R’HCO + M  R’CO• + H• + M, e.g. CH3HCO + M  CH3CO + H• + M R’HCO + O2 R’CO• + HO2•, e.g. CH3HCO + O2 CH3CO• + HO2• • Aldehydes also have weak C-C bond thus R’CO•+ M  R’ + CO + M, e.g. CH3CO• + M  CH3• + CO + M • Somewhat roundabout but easiest way to make CO, still takes a “long” time, see flow reactor result – first aldehydes CH2O and C2H4O form, then CO rises as aldehydes decompose AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  20. Hydrocarbon oxidation • Larger hydrocarbons also have negative temperature coefficient (NTC) behavior at low T (below H + O2 branching) • NTC especially prevalent in rich mixtures – reaction rate decreases with increasing T • R + O2 ROO is very reversible due to weak R-O bond • Equilibrium favors dissociation (ROO R + O2) at higher T, so ROO won’t stick around long enough to make ROOH • At higher temperatures HOO + RH  R + HOOH forms peroxides, lessening the need for ROO • (HOOH reaction has higher Ea than ROO + RH  R + ROOHbecause the former is more exothermic; C-O bond strength 86 vs 111 kcal/mole for H-O, thus HOOH reaction more dominant at higher temperature) • Also forms “cool flames” – exothermic propagating waves that don’t consume all reactants because of NTC shut-down – no longer “homogeneous” reaction • Negative temperature coefficient behavior not seen in calculations because no ROOH chemistry in CSU model AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  21. Hydrocarbon – O2 Hewitt and Thornes, 1937 C3H8-O2, f = 5, 0.6 liter vessel (Note T and P axes are flipped) AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  22. Chemical fire suppressants • Key to suppression is removal of H atoms H + HBr H2 + Br H + Br2 HBr + Br Br + Br + M  Br2 + M -------------------------------- H + H  H2 • Why Br and not Cl or F? HCl and HF too stable, 1st reaction too slow • HBr is a corrosive liquid, not convenient - use CF3Br (Halon 1301) - Br easily removed, remaining CF3 very stable, high CP to soak up thermal enthalpy • Problem - CF3Br very powerful ozone depleter- banned! • Alternatives not very good; best ozone-friendly chemical alternative is probably CF3CH2CF3 or CF3H • Other alternatives (e.g. water mist) AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  23. Chemical fire suppressants AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  24. Zeldovichmechanism for NO formation • Extremely high activation energy due to enormous strength of NN bond (≈ 220 kcal/mole) O + N2 NO + N N + O2 NO + O N + OH  NO + H • Reaction (1) is limiting; Z1exp(-E1/T) < Z2exp(-E2/T) except for T > 3394K • 1 NO molecule formed from (1) yields 2 NO molecules • Assuming steady-state for N, partial equilibrium for O, OH, H, with O2, Heywood (1988) shows: • T = 2200K, P = 1 atm: NO = 0.59 second • By comparison, time scale for chemical reactions in flame front ≈ 0.001 second for stoichiometric hydrocarbon-air • Thus, Zeldovich NO occurs in the burned gases downstream of the flame front, not in the flame front itself AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  25. Summary • Real fuels have very complex chemistry, not just 1 or 2 steps • Need chain branching for fast reaction • Hydrogen, CO, hydrocarbons dominated by • Initiation – H2 + M  H + H + M, RH+ M  R + H + M • Branching • Typically H+ O2 OH + O at high T / low P • Peroxide path (with HOOH or ROOH) at low T / high P where H atoms are lost due to recombination • Recombination - H + O2+ M  HO2+ M • Radical termination at walls in explosion vessel • CO oxidation • Requires CO + OH  CO2+ H • Parasitic on H2– O2 mechanism since CO + OH relatively slow • Hydrocarbons • Inhibit their own oxidation due to RH + H  R + H2 • Decomposition rate depends on C-H bond strength • First fuel decomposes, generates radical pool, generates CO then oxidizes it AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  26. Midterm exam • October 19, 9:00 – 10:30 am • Covering lectures 1 - 6 • Open books / notes / calculators • Laptop computers may be used ONLY to view .pdf versions of lecture notes – NOT .pptx versions • Note .pdf compilation of all lectures: http://ronney.usc.edu/AME513F12/AME513-F12-AllLectures.pdf • GASEQ, Excel spreadsheets, CSU website, etc. NOT ALLOWED • Followed by lecture 10:45 am – 11:50 am AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  27. Midterm exam – topics covered • Chemical thermodynamics • Stoichiometry • Heating value • Flame temperatures • Equilibrium • Degrees of reaction freedom • Equilibrium constraints • Compression/expansion • Chemical kinetics • Law of mass action, collision theory • Arrhenius form of reaction rate expression • Coupling with thermodynamics • Adiabatic constant-volume reaction • Constant-pressure Well-Stirred Reactor • Multistep reactions • Single-step irreversible & reversible • Steady-state & partial equilibrium approximations • Kinetics of real fuels • H2 – O2 • CO – O2 • Hydrocarbons – O2 AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

  28. Midterm exam – types of problems • Chemical thermodynamics • Property values will be given if needed • Stoichiometry • Heating value • Flame temperature • Equilibrium • Chemical kinetics • Exact solution (for very simple chemistry!) • Steady state or partial equilibrium • How would an explosion limit plot change if • Wall conditions changed • New species affecting certain reactions added • “Most likely” reaction steps (similar to Turns 5.7) • General - how would QR, Tad, reaction rates, homogenous explosion time, WSR blowout, etc. be affected by • Ronney Fuels, Inc. – new fuel or additive • Planet X – different atmosphere (pressure, temperature, etc.) AME 513 - Fall 2012 - Lecture 6 - Chemical Kinetics III

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