AME 436 Energy and Propulsion

AME 436 Energy and Propulsion
Lecture 10 Unsteady-flow (reciprocating) engines 5: Combustion in engines

AME 436 - Spring 2015 - Lecture 10 - Combustion in Engines
Outline Combustion in engines Knock Misfire / flammability limits Incomplete combustion Pollutant formation AME Spring Lecture 10 - Combustion in Engines

Knock - what is it? Homogeneous reaction occurring in unburned “end gas” ahead of expanding flame front Occurs when the combination of piston compression + “flame compression” increases T & P of the end gas until the homogeneous reaction rate is fast enough that a very rapid explosion occurs before the usual turbulent flame front reaches & consumes the end gas - “horse race” between flame propagation and homogeneous reaction in end gas What does “rapid” mean? Faster than acoustic waves can relax the pressure gradients - prevents even “push” on the piston How fast is that? Typical engine: Bore = 10 cm = 0.1 m, sound speed = 500 m/s, time scale ≈ 0.1m / 500 m/s = 2 x 10-4 s = 200 µs at most AME Spring Lecture 10 - Combustion in Engines

Knock - movies No knock Knock Videos courtesy Prof. Yuji Ikeda, Kobe University AME Spring Lecture 10 - Combustion in Engines

Knock - why is it bad? Pressures generated by knock are not substantially higher than those of ordinary (deflagration) combustion but the pressure gradients are huge, causing enormous local stresses on the piston; also causes large torque on piston, thus tilting & rubbing against cylinder wall As the shocks propagate into the narrow region between the piston and cylinder wall (the “crevice volume”), the shock strength increases, causing locally even more severe damage - usually the crevice volume is where the most knocking damage occurs AME Spring Lecture 10 - Combustion in Engines

Knock Shock formation causes “ringing” of pressure waves back & forth across cylinder - sounds like you're hitting piston with a hammer, which isn't too far from the truth Töpfer et al., SAE Paper (2000) Post-knock ringing Normal combustion (no knocking) Start of knocking AME Spring Lecture 10 - Combustion in Engines

Knock - how do different fuels compare?
Different fuels are compared by means of an octane number Determine knock-limited compression ratio for your test fuel, using a variable compression ratio engine called a CFR engine (used only for research, not in any vehicle) at specified test conditions (RPM, spark timing, intake T & P, etc.) Define n-heptane as having octane number = zero, 2-2-4 trimethylpentane = 100 Blend these two reference fuels to get the same knock limited compression ratio as the fuel under test The % 2,2,4 trimethylpentane in this blend is the knock rating Two slightly different sets of test conditions used, called “Research” and “Motor” methods; what's displayed at the gas pump is the average of the two methods (R+M)/2 n-heptane 2,2,4 trimethyl pentane (iso-octane) CFR engine AME Spring Lecture 10 - Combustion in Engines

Diesel fuel characterization
For non-premixed-charge engines (i.e. Diesels), you want rapid ignition once the fuel is injected, so fuel requirements are opposite premixed-charge engines The relevant scale is called the cetane scale, with cetane (16 carbon atoms in a straight line) = 100, heptamethylnonane (9 carbon atoms in a row, with the inner 7 each having a methyl group attached) = 15 (not zero!) n-cetane 2, 2, 4, 4, 6, 8, 8 heptamethyl nonane (iso-cetane) AME Spring Lecture 10 - Combustion in Engines

What does octane number measure?
Heating value? Nope - all hydrocarbon fuels have very similar heating values as previous discussed (lecture 2) Laminar burning velocity (SL)? Nope; SL is practically the same for all stoichiometric hydrocarbon-air mixtures (≈ 40 cm/s) since As discussed in Lecture 4, SL ~ (w)1/2 All such mixtures are mostly air, so thermal diffusivity () ≈ same for all Overall reaction rate (w) is controlled by its value at adiabatic flame temperatures (Tad) All such mixtures have nearly equal Tad At these temperatures, controlling reactions are H + O2  OH + O (breaking of O=O double bond, accelerates reaction) H + O2 + M  HO2 + M (competing for H atoms, decelerates reaction) CO + OH  CO2 + H (only way to oxidize CO; regenerates H) and none of these reactions depend on the fuel molecule; fuel decomposition is very fast by comparison at these temperatures Tendency for mixture to exhibit homogeneous reaction at the typical T & P of end gases (typically 30 atm, 900K)? YES!! To understand this, need to look at what reactions occur at these T's and P's … AME Spring Lecture 10 - Combustion in Engines

What reactions occur during knock?
Start with fuel molecule RH, where R is an “organic radical”, e.g. propane without an H Remove 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 knocking reactions will be sensitive to rates of H atom removal from fuel molecule RH AME Spring Lecture 10 - Combustion in Engines

How does fuel structure affect knock?
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 Spring Lecture 10 - Combustion in Engines

Does this small difference in bond strength matter? YES because activation energy is high If we use bond strength to estimate activation energy thus reaction rate w (dangerous in general, but ok here…) then at a typical 900K wmethane : wprimary : wsecondary : wtertiary 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 Comparison of fuels Methane (RON=120, MON=120) - highest of any common fuel Larger n-alkanes lower, e.g. propane RON=112, MON=97 Alkenes (C=C double bonds, e.g. ethylene, C2H4) lower Benzene - very high knock resistance kcal/mole C-H bond strength - MON=115 Alcohols good too - methanol: RON=106, MON=92; ethanol RON=106, MON=89 See chart on page 472 of Heywood AME Spring Lecture 10 - Combustion in Engines

Ghosh et al., Ind. Eng. Chem. Res., Vol. 45, p. 337 (2006) AME Spring Lecture 10 - Combustion in Engines

Anti-knock additives Any successful additive must inhibit knock reactions without inhibiting flame propagation reactions! Midgley, Boyd, Kettering (GM research laboratories, Dec. 9, 1921) after trying 1000's of different compounds stumbled onto tetraethyl lead - (C2H5)4Pb Introduced into gasoline in immediately doubled knock-limited compression ratio from 4 to 8, power & efficiency increased (by 40%, from 20.8% to 29.2% according to AirCycles4recips.xls) Last nail in the coffin for steam and electric vehicles in early 1900's Mechanism not understood until 1970's - production of fine PbO particle “fog” that acts as a “scavenger” of HOO and HOOH - takes them out of system before they lead to chain branching … but PbO poisons catalytic converters for exhaust emissions introduced in needed to switch to unleaded gasolines (also Pb not exactly environmentally friendly, so not a bad idea to stop using it anyway) - trend was to increase concentration of aromatics (benzene-like molecules) but leads to more soot formation, also carcinogenic AME Spring Lecture 10 - Combustion in Engines

Other anti-knock additives
Methylcyclopentadienyl manganese tricarbonyl (C9H7MnO3, MMT) Methyl tert-butyl ether (MTBE) (looked good for a while, but now prohibited - water soluble, can be smelled and tasted in water at the parts per billion level Now: ethanol from corn - very desirable to powerful senators from farm states MTBE Ethanol AME Spring Lecture 10 - Combustion in Engines

How do operating conditions affect knock?
Key points in all this knock stuff Whether knock occurs is a horse race between flame propagation and homogeneous reaction of the end gas The chemical reactions that affect flame propagation are different from those that affect knock Homogeneous reaction rates depend primarily on the initial T & P, whereas flame propagation depends primarily on the final T & P Thus, operating conditions that affect knock and flame propagation are different!!! Since knock occurs in end gas, need to see how operating conditions affect end gas T & P AME Spring Lecture 10 - Combustion in Engines

Simple estimate of maximum T & P of end gas in terms of intake conditions T2, P2 and compression ratio r - assume adiabatic compression, constant-v combustion, reversible adiabatic compression of end gas to (maximum) pressure P4 AME Spring Lecture 10 - Combustion in Engines

AirCycles4recips.xls end gas calculation Assumes a small portion of the total mixture has no heat added to it but undergoes reversible compression/expansion according due to the pressure increase/decrease of the rest of the gas in the cylinder that DOES have heat input due to combustion End gas undergoes heat loss according to the usual formula The following plot shows effects of compression ratio, spark advance, mixture strength, Tintake and Pintake on the P-T history of end gas AME Spring Lecture 10 - Combustion in Engines

P-T trajectory of end gas (combustion process only)
For lean mixture case: f = 0.051, BurnDuration = 0.3, BurnStart = (best efficiency timing) Baseline: r = 8,  = 1.3, f = 0.068, QR = 4.5 x 107 J/kg, Tin = 300K, Pin = Pexh = 1 atm BurnStart = 0.045, BurnDuration= 0.105, BurnRateProfile = 0, h = 0.01, Twall = 400K comp = exp = 0.9 AME Spring Lecture 10 - Combustion in Engines

Compression ratio As r , Tend gas  and Pend gas , thus knock tendency  Recall d[fuel]/dt ~ Pnexp(-E/T), (n = order of reaction, E = activation energy) thus both P and T affect knock P increases more than T, but reaction rate is more sensitive to T than P Tintake - increasing Tintake increases Tend gas, thus knock tendency  (that's why your car knocks more on a hot day) (though Pend gas isn't affected much by increasing Tintake) Pintake - increasing Pintake increases Pend gas, thus knock tendency  (that's why your car knocks when you put your foot to the floor) (though Tend gas isn't affected) AME Spring Lecture 10 - Combustion in Engines

Spark timing As mentioned in Lecture 9, page 24, as spark timing is advanced, there is more “burn then compress” compared to “compress then burn” which leads to higher T This also leads to higher P, thus higher Tend gas Also, igniting earlier means more time for knock to occur Fundamental tradeoff between increasing spark advance to obtain best thermal efficiency, and decreasing spark advance to minimize knock and NOx Spark advance up to ≈ 0.05 increases thermal efficiency (see next page), but there's a penalty to pay in terms of knock (Tend gas increasing) and NOx (Tmax increasing) Tmax and Tend gas don't increase much as advance increases, but both knock and NOx are high activation energy processes so a little increase in T matters Note knock and NOx are controlled by very different reactions AME Spring Lecture 10 - Combustion in Engines

Spark advance AME Spring Lecture 10 - Combustion in Engines

Mixture strength Many experiments show less knock with leaner mixtures Leaner mixtures burn slower, so need more spark advance, but even when spark timing adjusted for best efficiency, still less knock that stoichiometric mixtures Less heat release, thus lower peak P, thus lower Tend gas Ronney, et al., J. Auto. Eng., (Proc. Instit. Mech. Eng., Part D), Vol. 208, pp (1994). AME Spring Lecture 10 - Combustion in Engines

Engine RPM - generally most important effect is less time for knock to occur - at higher RPM, more turbulence, ST increases, time needed for flame propagation to occur decreases, but turbulence has no effect on time for homogeneous reaction of end gas AME Spring Lecture 10 - Combustion in Engines

HCCI engines Actually, burning rapidly at minimum volume yields the best possible thermal efficiency, but damage due to knocking means we want to burn “fast but not too fast” HCCI - Homogeneous Charge Compression Ignition engines take advantages of this - “controlled knocking” Video courtesy Prof. Yuji Ikeda, Kobe University AME Spring Lecture 10 - Combustion in Engines

HCCI engines By using homogeneous reaction instead of flame propagation, conventional flammability/misfire limits don't exist - recall for homogeneous reaction, burn time is nearly independent of equivalence ratio () As a result, one can burn very lean mixtures, low Tad, low peak temperature, low NOx formation Lean mixtures - can obtain part-load operation without throttling and its losses Since we're asking for knock, use high compression ratios, thus high th Much more difficult to control the rate and timing of homogeneous reaction than a propagating spark-ignited flame; various control schemes being studied Variable intake temperature Variable exhaust gas recirculation Variable compression ratio and valve timing Cycle-to-cycle control probably needed As a result, not in commercial use yet AME Spring Lecture 10 - Combustion in Engines

Comparison of gasoline, diesel & HCCI
AME Spring Lecture 10 - Combustion in Engines

Comparison of gasoline, diesel & HCCI
J. Dec., Proc. Combust. Instit., Vol. 32, p (2009). AME Spring Lecture 10 - Combustion in Engines

HCCI experiments in a single-cylinder engine

HCCI experiments in 6 cyl. engine (1 cyl. HCCI)

HCCI - disadvantages (opportunities?)
Difficult to control timing and rate of combustion If misfire occurs, gas mixture during the next cycle will be too cold for auto-ignition to occur (unless intake air heating is used), the engine will stop Cold starting? Operating window for HCCI operation (load and engine speed) is small - most HCCI concepts use conventional spark-ignited operation at higher loads (less lean mixtures) Additional components for control system - increased cost Relatively high friction losses due to low IMEP, thus FMEP is a higher % of BMEP AME Spring Lecture 10 - Combustion in Engines

Misfire / ignition limits
If an engine operates too lean or too rich, the engine will start to run roughly, misfire and may quit altogether The limit is usually defined by a percentage of misfires (cycles with negative MEP) or a certain percent standard deviation in IMEP With conventional spark ignition premixed-charge engines, the limit is at equivalence ratio () ≈ , whereas in a laboratory experiment (e.g. on a bunsen burner)  ≈ 0.5 will burn Why the difference? In the laboratory, the mixture can take as long as it wants to burn, whereas in the engine, there is only a limited time ~ 1/N available for burning - minimum turbulent burning velocity (ST) requirement To avoid misfire Increase burning velocity Increase time available for burning Decrease time needed for burning AME Spring Lecture 10 - Combustion in Engines

Misfire effects of operating conditions
Inlet temperature As T∞  with  fixed, Tad , SL , thus ST  (slightly) Since SL depends more on Tad than any other property, limit criterion is basically a minimum Tad criterion Recall Tad = T∞ + fQR/Cv - can get a given Tad with a high f (thus high ) and low T∞ or vice versa “Brilliant” experiments show this - limit Tad 2000K K Ronney, et al., J. Auto. Eng., (Proc. Instit. Mech. Eng., Part D), Vol. 208, pp (1994). AME Spring Lecture 10 - Combustion in Engines

Misfire - effects of operating conditions
Intake pressure Recall SL ~ P(n-2)/2, with n typically slightly less than 2, not much effect But as Pintake decreases, more exhaust gas (at 1 atm) relative to fresh mixture (at P < 1 atm), so more dilution of mixture with exhaust gas, thus lower Tad, lower SL Thus more problems with misfire at throttled conditions (for fixed N) Turbulence - more turbulence, higher ST - helps limit problem, but remember too much turbulence will extinguish the flame! Engine RPM (N) Time available for burning ~ 1/N But u' ~ upiston ~ N and ST ~ u' So (time available for burning)/(time needed) ≈ constant But actually ST ~ u'(1-), so you have more problems with misfire at higher N AME Spring Lecture 10 - Combustion in Engines

Misfire - effects of operating conditions
Multiple spark plugs - reduces distance each evolving flame needs to propagate, less chance of misfire, so less misfire problem Knock additives (e.g. (C2H5)4Pb) - no effect (affects knock reactions, not flame propagation reactions) AME Spring Lecture 10 - Combustion in Engines

Lean mixtures - effect on temperatures
Frequently I hear the comment from non-thermodynamicists who know engines that “engine temperature increases as you go leaner - if you go too lean you'll burn up the engine” Thermodynamically this makes no sense - leaner mixtures have lower Tad = T∞ + fQR/Cv (constant volume) BUT - leaner mixtures burn slower - more heat release after some expansion already completed - less expansion, higher T Since chemical reactions have high activation energy, a small decrease in Tad (leaner mixture) causes a large decrease in SL, thus much slower burn, higher exhaust T AME Spring Lecture 10 - Combustion in Engines

Slow-burn effect > lower-Tad effect r = 3,  = 1.3, f = or , QR = 4.5 x 107 J/kg, Tin = 300K, Pin = 1 atm, Pexh = 1 atm, BurnStart = , BurnDuration = 0.15, 0.3 or 0.45, ExhRes = FALSE, Const-v comb, ComplExp = FALSE, h = 0.01, Twall = 400K, comp = exp = 0.9 AME Spring Lecture 10 - Combustion in Engines

After expansion After blowdown Const. v After exhaust stroke Const. P AME Spring Lecture 10 - Combustion in Engines

Incomplete combustion (review from Lecture 4)
Most fuel is burned in engines - very little unburned fuel under normal operating conditions unless mixture is too lean, too rich or otherwise “disabled” Flame quenching may occur in the “crevice volumes” where gap < quenching distance - main source of hydrocarbon emissions from engines Before we get to the point that raw fuel is emitted, some fuel reacts only partially - sequence of events: Fuel + O2  CO + H2 CO + H2 CO + H2O CO + H2O  CO2 + H2O so CO emissions are the first sign of trouble! AME Spring Lecture 10 - Combustion in Engines

Example #1 For the Otto cycle example in Lecture 9 with r = 9,  = 1.3, M = kg/mole, f = 0.062, QR = 4.3 x 107 J/kg, T2 = 300K, P2 = Pin = 0.5 atm, P6 = Pex = 1 atm, h = 0, comp = exp = 0.9, determine the end gas temperature and pressure If this were an ideal cycle we could use the formula on page 16, i.e. But since this is not an ideal cycle we compress the end gas isentropically from its pressure (9.165 atm) and temperature (611K) after compression to the pressure after combustion (51.02 atm) , i.e. which is not much different. AME Spring Lecture 10 - Combustion in Engines

Example #2 During WWII, fighter aircraft engines were equipped with systems to inject water into the cylinders during compression. How would water injection affect Knock tendency: The water would cool the intake charge and increase its , thus reducing the end-gas temperature substantially. Also, while the compression ratio could not be changed, the intake pressure could be increased (these were turbocharged engines) without knocking, thus allowing more air to be ingested Misfire tendency: The greater mass of un-burnable water vapor and its higher Cp (compared to air) led to lower Tad, thus more tendency to misfire. This limited the amount of water that could be injected. Power: This was the main reason for water injection. The liquid water requires much less Work/mass to compress than air (Work = ∫PdV thus Work/m = ∫Pd(V/m) = ∫Pd(1/ρ), and ρwater >> ρair), yet when the liquid water vaporizes, its volume increases tremendously, generating more expansion work / compression work (this is why steam cycles are more efficient that gas cycles!) Also as mentioned above, the intake air pressure could be increased, thus allowing more fuel to be added and burned Brake thermal efficiency: The slower burning would decrease efficiency, but the steam cycle has inherently higher efficiency, so it could go either way. Work per unit mass of (fuel + water): This would definitely decrease, since the water provided no additional heat release but much additional mass. In actuality, these aircraft used a methanol/water mixture, not just water, to recover some energy release from the injected liquid (though note methanol has only 40% of QR of gasoline) AME Spring Lecture 10 - Combustion in Engines

Summary Knock A very rapid homogeneous reaction in the “end gas” ahead of the flame front Horse race between knock (bad) and flame propagation (good) Knock tendency depends on reaction rates at initial T, P & composition of reactants (end gas), flame propagation depends on final (burned gas) temperature, so factors affecting knock & flame propagation are very different Knock tendency characterized by octane number - higher ON, more resistant to homogeneous reaction Diesel fuels - opposite characteristics desired Misfire / flammability limits Due to insufficient ignition source or time to burn - roughly corresponds to fixed Tad Depends on required propagation time (~ d/ST) vs. available time (~1/N) Incomplete combustion Due to insufficient time to burn - CO is first sign of trouble AME Spring Lecture 10 - Combustion in Engines

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