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

2 AME Fall Lecture 6 - Chemical Kinetics III Outline  Experimental methods  Types of reaction  CO-O 2  Hydrocarbons  NO x  Flame inhibition

3 AME Fall Lecture 6 - Chemical Kinetics III 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 ~ d 2 /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 ~ s in flame  Only yields global properties

4 AME Fall Lecture 6 - Chemical Kinetics III 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)

5 AME Fall Lecture 6 - Chemical Kinetics III Experimental methods  Laminar flame  “Simple” setup – measure burning velocity S L  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 ~ S L 1/2

6 AME Fall Lecture 6 - Chemical Kinetics III Types of reactions  Global reaction  Example: CH 4 + 2O 2  CO H 2 O  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: H 2 + M  H + H + M  Break stable molecule into radical(s)  High E a – 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 + O 2  OH + O  Use radical to create more radical  High E a - endothermic, must break strong bond, but also make a bond, so not as high as chain initiation  Moderate Z (somewhat orientation sensitive)

7 AME Fall Lecture 6 - Chemical Kinetics III Types of reactions  Chain propagation or “shuffle” reaction  Example: OH + H 2  H 2 O + H  Use radical + stable molecule to create another pair  Moderate E a – may be nearly thermo-neutral  Moderate Z (somewhat orientation sensitive)  Chain termination  Example: H + OH + M  H 2 O + M  Recombine radicals into stable molecules (usually products)  Low or zero E a – exothermic, no activation barrier  Need 3 rd body to absorb enthalpy and conserve momentum  Moderate Z (may be orientation sensitive)  Schematic multi-step mechanism (e.g. Hautman et al, 1981) C 3 H 8  1.5 C 2 H 4 + H 2 (Initial breakdown of fuel) C 2 H 4 + O 2  2 CO + 2 H 2 (C 2 H 4 = surrogate for radicals) 2H 2 + O 2  2 H 2 O (oxidation of H 2 ) 2CO + O 2  2 CO 2 (oxidation of CO) Reaction rates crazy (see Turns, p. 158) – sometimes order of reaction is negative – causes problems as concentration  0

8 AME Fall Lecture 6 - Chemical Kinetics III H 2 – O 2  Simulate explosion limits using online chemistry calculator  Use time to 50% H 2 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 + O 2  OH + O = H + O 2 + M  HO 2 + M when P = T 0.2 e /RT

9 AME Fall Lecture 6 - Chemical Kinetics III CO – O 2  H 2 -O 2 already discussed – 3 explosion limits  CO MUCH different because no chain branching and no “shuffle” reaction to create product CO + O 2  CO 2 + O Very slow, but needed as source of O O 2 + 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  CO 2 + MCreates product but removes radicals  As a result, pure CO – O 2 oxidation is extremely slow!  Early experiments showed widely varying results because of contamination with water; with any hydrogen source CO + OH  CO 2 + H Shuffle reaction to create CO 2 H + O 2  OH + O Regenerate OH plus another O Neither are fast, but better than alternatives!  Stoich. CO + O 2, 1 atm, 1500K, time to consume 50% of CO:  No H 2 O: 1.03 s; 1 ppm H 2 O: 0.27 s; 10 ppm: s;  100 ppm s; 1,000 ppm s; 10,000 ppm s

10 AME Fall Lecture 6 - Chemical Kinetics III CO – O 2  Case shown: const. T = 1000K, P = 1 atm, CO:H 2 :O 2 = 1:1:10  To a good approximation, H 2 -O 2 acts as infinitely fast (i.e. steady-state) radical source for CO  CO does not affect H 2 -O 2 process, CO reaction just too slow  50% H 2 consumption in ≈ 400 µs, ≈ same with or without CO

11 AME Fall Lecture 6 - Chemical Kinetics III CO – O 2  Explosion limit experiments show 2 nd -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

12 AME Fall Lecture 6 - Chemical Kinetics III CO – O 2  Unlike classical experiments, CSU homogenous kinetics calculator predicts that dry CO has only a single limit – but CSU site does not include O 3 which may be important  With H 2 O addition, behavior is very similar to H 2 -O 2 with offset to higher T - CO is a “parasite” on the OH source

13 AME Fall Lecture 6 - Chemical Kinetics III Hydrocarbon – O 2  Hydrocarbons inhibit their own oxidation, because they react with radicals more readily than O 2 reacts with radicals  Once an H atom is removed from fuel, e.g. C 3 H 8 + O 2  C 3 H 7 + HO 2 C 3 H 8 + O 2  C 3 H 7 + HO 2 then the fuel radical can lose another H to form an alkene, e.g. C 3 H 7 + O 2  C 3 H 6 + HO 2 C 3 H 7 + O 2  C 3 H 6 + HO 2 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   -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 + O 2 branching is inhibited by the hydrocarbons; at higher temperatures, H + O 2 branching is sufficiently rapid

14 AME Fall Lecture 6 - Chemical Kinetics III Hydrocarbon – O 2

15 AME Fall Lecture 6 - Chemical Kinetics III Hydrocarbon – O 2  Hydrocarbons have no 1 st or 2 nd explosion limit; instead of H + O 2 + M  HO 2 + M being the dominant recombination reaction, it’s CH 4 + H  CH 3 + H 2, which has same pressure dependence as the branching reaction H + O 2  OH + O  C-H bonds stronger in CH 4 than C 3 H 8, thus higher explosion T

16 AME Fall Lecture 6 - Chemical Kinetics III 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 + O 2  R  + HOO   Add an O 2 to R  R  + O 2  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

17 AME Fall Lecture 6 - Chemical Kinetics III 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

18 AME Fall Lecture 6 - Chemical Kinetics III 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(-E methane /  T) : exp(-E primary /  T) : exp(-E secondary /  T) : exp(-E tertiary /  T) ≈ exp(-105,000 cal/mole/(1.987 cal/mole-K)(900K)) : exp(-98000/1.987*900) : exp(-95000/1.987*900) : exp( /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

19 AME Fall Lecture 6 - Chemical Kinetics III 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. C 2 H 5 O + M  CH 3 HCO + H + M  Aldehydes have weakest C-H bond (≈87 kcal/mole) thus R’HCO + M  R’CO + H + M, e.g. CH 3 HCO + M  CH 3 CO + H + M R’HCO + O 2  R’CO + HO 2, e.g. CH 3 HCO + O 2  CH 3 CO + HO 2 R’HCO + O 2  R’CO + HO 2, e.g. CH 3 HCO + O 2  CH 3 CO + HO 2  Aldehydes also have weak C-C bond thus R’CO + M  R’ + CO + M, e.g. CH 3 CO + M  CH 3 + CO + M R’CO + M  R’ + CO + M, e.g. CH 3 CO + M  CH 3 + CO + M  Somewhat roundabout but easiest way to make CO, still takes a “long” time, see flow reactor result – first aldehydes CH 2 O and C 2 H 4 O form, then CO rises as aldehydes decompose

20 AME Fall Lecture 6 - Chemical Kinetics III Hydrocarbon oxidation  Larger hydrocarbons also have negative temperature coefficient (NTC) behavior at low T (below H + O 2 branching)  NTC especially prevalent in rich mixtures – reaction rate decreases with increasing T  R  + O 2  ROO  is very reversible due to weak R-O bond  Equilibrium favors dissociation (ROO   R  + O 2 ) 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 E a than ROO  + RH  R  + ROOH because 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

21 AME Fall Lecture 6 - Chemical Kinetics III Hydrocarbon – O 2 Hewitt and Thornes, 1937 C 3 H 8 -O 2,  = 5, 0.6 liter vessel (Note T and P axes are flipped)

22 AME Fall Lecture 6 - Chemical Kinetics III Chemical fire suppressants  Key to suppression is removal of H atoms H + HBr  H 2 + Br H + Br 2  HBr + Br Br + Br + M  Br 2 + M H + H  H 2  Why Br and not Cl or F? HCl and HF too stable, 1st reaction too slow  HBr is a corrosive liquid, not convenient - use CF 3 Br (Halon 1301) - Br easily removed, remaining CF 3 very stable, high C P to soak up thermal enthalpy  Problem - CF 3 Br very powerful ozone depleter - banned!  Alternatives not very good; best ozone-friendly chemical alternative is probably CF 3 CH 2 CF 3 or CF 3 H  Other alternatives (e.g. water mist)

23 AME Fall Lecture 6 - Chemical Kinetics III Chemical fire suppressants

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

25 AME Fall Lecture 6 - Chemical Kinetics III 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 – H 2 + M  H + H + M, RH + M  R + H + M  Branching »Typically H + O 2  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 H atoms are lost due to recombination  Recombination - H + O 2 + M  HO 2 + M  Radical termination at walls in explosion vessel  CO oxidation  Requires CO + OH  CO 2 + H  Parasitic on H 2 – O 2 mechanism since CO + OH relatively slow  Hydrocarbons  Inhibit their own oxidation due to RH + H  R + H 2  Decomposition rate depends on C-H bond strength  First fuel decomposes, generates radical pool, generates CO then oxidizes it

26 AME Fall Lecture 6 - Chemical Kinetics III Midterm exam  October 19, 9:00 – 10:30 am  Covering lectures  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:  GASEQ, Excel spreadsheets, CSU website, etc. NOT ALLOWED  Followed by lecture 10:45 am – 11:50 am

27 AME Fall Lecture 6 - Chemical Kinetics III 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 »H 2 – O 2 »CO – O 2 »Hydrocarbons – O 2

28 AME Fall Lecture 6 - Chemical Kinetics III 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 Q R, T ad, 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.)