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AME 436 Energy and Propulsion

AME 436 Energy and Propulsion. Lecture 2 Fuels, chemical thermodynamics (thru 1st Law; 2nd Law next lecture). Outline. Fuels - hydrocarbons, alternatives Balancing chemical reactions Stoichiometry Lean & rich mixtures Mass and mole fractions Chemical thermodynamics Why?

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AME 436 Energy and Propulsion

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  1. AME 436Energy and Propulsion Lecture 2 Fuels, chemical thermodynamics (thru 1st Law; 2nd Law next lecture)

  2. Outline • Fuels - hydrocarbons, alternatives • Balancing chemical reactions • Stoichiometry • Lean & rich mixtures • Mass and mole fractions • Chemical thermodynamics • Why? • 1st Law of Thermodynamics applied to a chemically reacting system • Heating value of fuels • Flame temperature AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  3. Fuels • Usually we employ hydrocarbon fuels, alcohols or hydrogen burning in air, though other possibilities include CO, NH3, CS2, H2S, etc. • For rocket fuels that do not burn air, many possible oxidizers exist - ASTE 470 discusses these - AME 436 focuses on airbreathing devices • Why hydrocarbons? • Many are liquids - high density, easy to transport and store (compared to gases, e.g. CH4), easy to feed into engine (compared to solids) • Lots of it in the earth (often in the wrong places, e.g. Iraq) • Relatively non-toxic fuel and combustion products • Relatively low explosion hazards AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  4. Air • Why air? • Because it’s free, of course (well, not really when you think of all the money we’ve spent to clean up air) • Air ≈ 0.21 O2 + 0.79 N2 (1 mole of air) or 1 O2 + 3.77 N2 (4.77 moles of air) • Note for air, the average molecular weight is 0.21*32 + 0.79*28 = 28.9 g/mole thus the gas constant = (universal gas constant / mole. wt.) = (8.314 J/moleK) / (0.0289 kg/mole) = 287 J/kgK • Also ≈ 1% argon, up to a few % water vapor depending on the relative humidity, trace amounts of other gases, but we’ll usually assume just O2 and N2 AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  5. Hydrocarbons • Alkanes - single bonds between carbons - CnH2n+2, e.g. CH4, C2H6 • Olefins or alkenes - one or more double bonds between carbons • Alkynes - one or more triple bonds between carbons - very reactive, higher heating value than alkanes or alkenes AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  6. Hydrocarbons • Aromatics - one or more ring structures • Alcohols - contain one or more OH groups AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  7. Biofuels • Alcohols - produced by fermentation of food crops (sugars or starches) or cellulose (much more difficult, not an industrial process yet) • Biodiesel - convert vegetable oil or animal fat (which have very high viscosity) into alkyl esters (lower viscosity) through “transesterification” with alcohol Methyl linoleate Generic ester structure (R = any organic radical, e.g. C2H5) Ethyl stearate Methanol + triglyceride Glycerol+ alkyl ester Transesterification process AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  8. Practical fuels • All practical fuels are BLENDS of hydrocarbons and sometimes other compounds • What distinguishes one fuel from another? • Flash point - temperature above which fuel vapor pressure is flammable when mixed with air • Distillation curve - temp. range over which molecules evaporate • Relative amounts of paraffins vs. olefins vs. aromatics vs. alcohols • Amount of impurities, e.g. sulfur • Structure of molecules - affects octane number (Lecture 10) http://static.howstuffworks.com/flash/oil-refining.swf AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  9. Gasoline - typical composition Benzene Toluene J. Burriet al., Fuel,Vol. 83, pp. 187 - 193 (2004) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  10. Practical fuels - properties • Values NOT unique because • Real fuels are a mixture of many molecules, composition varies • Different testing methods & definitions AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  11. Practical fuels - properties http://www.afdc.energy.gov/pdfs/fueltable.pdf AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  12. Practical fuels • What doesn’t distinguish one fuel from another? • Energy content (except for fuels containing alcohols, which are lower) • Examples • Gasoline - low-T distillation point, easy to vaporize, need high octane number; reformulated gasoline contains alcohols • Diesel - high-T distillation point, hard to vaporize, need LOW octane number for easy ignition once fuel is inject • Jet fuel - medium-T distillation point; need low freezing T since it will be used at high alititude / low T AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  13. Stoichiometry • Balancing of chemical reactions with “known” (assumed) products • Example: methane (CH4) in air (O2 + 3.77N2) CH4 + ?(O2 + 3.77N2)  ? CO2 + ? H2O + ? N2 (how do we know this know this set is reasonable? From 2nd Law, to be discussed later) I’ll assume you can balance this type of reaction to obtain CH4 + 2(O2 + 3.77N2)  1 CO2 + 2 H2O + 7.54 N2 or in general • This is a special case where there is just enough fuel to combine with all of the air, leaving no excess fuel or O2 unreacted; this is called a stoichiometric mixture • In general, mixtures will have excess air (lean mixture) or excess fuel (rich mixture) • This development assumed air = O2 + 3.77 N2; for lower or higher % O2 in the atmosphere, the numbers would change accordingly AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  14. Stoichiometry • In any case, the fuel mass fraction (f) is ni = number of moles of species i, Mi = molecular weight of species i For the specific case of stoichiometric methane-air (x = 1, y = 4), f = 0.0550; a lean/rich mixture would have lower/higher f • For stoichiometric mixtures, f is similar for most hydrocarbons but depends on the C/H ratio = x/y, e.g. • f = 0.0550 for CH4 (methane) - lowest possible C/H ratio • f = 0.0703 for C6H6 (benzene) or C2H2 (acetylene) - high C/H ratio • Fuel mole fraction Xf which varies a lot depending on x and y (i.e. much smaller for big molecules with large x and y) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  15. Stoichiometry • Fuel-to-air ratio (FAR) and air-to-fuel ratio (AFR) = 1/(FAR) • Note also f = FAR/(1+FAR) • Equivalence ratio ()  < 1: lean mixture;  > 1: rich mixture • What if we assume more products, e.g. CH4 + ?(O2 + 3.77N2)  ? CO2 + ? H2O + ? N2+ ? CO In this case we have 4 atom constraints (1 each for C, H, O, and N atoms) but 5 unknowns (5 question marks) - how to solve? Need chemical equilibrium (discussed later) to decide how much C and O are in the form of CO2 vs. CO AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  16. Fuel properties AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  17. Chemical thermodynamics - intro • Besides needing to know how to balance chemical reactions, we need to determine how much internal energy or enthalpy is released by such reactions and what the final state (temperature, pressure, mole fractions of each species) will be • What is highest temperature flame? H2 + O2 at  = 1? Nope, T = 3079K at 1 atm for reactants at 298K • Probably the highest is diacetylnitrile + ozone C4N2 + (4/3)O3  4 CO + N2 T = 5516K at 1 atm for reactants at 298K • Why should it? The H2 + O2 system has much more energy release per unit mass of reactants, but still a much lower flame temperature AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  18. Chemical thermodynamics - intro • The problem is that the products are NOT just H2O, that is, we don’t get H2 + (1/2)O2  H2O but rather H2 + (1/2)O2  0.706 H2O + 0.062 O2 + 0.184 H2 + 0.094 H + 0.129 OH + 0.040 O i.e. the water dissociates into the other species • Dissociation does 2 things that reduce flame temp. • More moles of products to soak up energy (1.22 vs. 1.00) • Energy is required to break the H-O-H bonds to make the other species • Higher pressures will reduce dissociation - Le Chatelier’s principle: When a system at equilibrium is subjected to a stress, the system shifts toward a new equilibrium condition in such as way as to reduce the stress (more pressure, less space, system responds by reducing number of moles of gas to reduce pressure) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  19. Chemical thermodynamics - intro • Actually, even if we somehow avoided dissociation, the H2 - O2 flame would be only 4998K - still not have as high a flame temp. as the weird C4N2 flame • Why? H2O is a triatomic molecule - more degrees of freedom (DOFs) (i.e. vibration, rotation) than diatomic gases; each DOF adds to the molecule’s ability to store energy • So why is the C4N2 - O3 flame so hot? • CO and N2 are diatomic gases - fewer DOFs • CO and N2 are very stable even at 5500K - almost no dissociation • O3 decomposes exothermically to (3/2)O2 AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  20. Chemical thermodynamics - goals • Given an initial state of a mixture (temperature, pressure, composition), and an assumed process (constant pressure, volume, or entropy, usually), find the final state of the mixture • Three common processes in engine analysis • Compression • Usually constant entropy (isentropic) • Low P / high V to high P / low V • Usually P or V ratio prescribed • Usually composition assumed “frozen” - if it reacted before compression, you wouldn’t get any work out! • Combustion • Usually constant P or v assumed • Composition MUST change (obviously…) • Expansion • Opposite of compression • May assume frozen (no change during expansion) or equilibrium composition (mixture shifts to new composition after expansion) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  21. Chemical thermo - assumptions • Ideal gases - note many “flavors” of the ideal gas law • PV = nT • PV = mRT • Pv = RT • P = RT P = pressure (N/m2); V = volume (m3); n = number of moles of gas; • = universal gas constant (8.314 J/moleK); T = temperature (K) m = mass of gas (kg); R = mass-specific gas constant = /M M = gas molecular weight (kg/mole); v = V/m = specific volume (m3/kg)  = 1/v = density (kg/m3) • Adiabatic • Kinetic and potential energy negligible • Mass is conserved • Combustion process is constant P or V (constant T or s combustion isn’t very interesting!) • Compression/expansion is reversible & adiabatic ( isentropic, dS = 0) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  22. Chemical thermodynamics - 1st Law • 1st Law of thermodynamics (conservation of energy), control mass: dE = Q - W • E = U + PE + KE = U + 0 + 0 = U • W = PdV • Combine: dU + PdV = 0 • Constant pressure: add VdP = 0 term • dU + PdV + VdP = 0  d(U+PV) = 0  dH = 0 • Hreactants = Hproducts • Recall h  H/m (m = mass), thus hreactants = hproducts • Constant volume: PdV = 0 • dU + PdV + VdP = 0  d(U) = 0 • Ureactants = Uproducts, thus ureactants = uproducts • h = u + Pv, thus (h - Pv)reactants = (h - Pv)products • Most property tables report h not u, so h - Pv form is useful • New twist: h or u must include BOTH thermal and chemical contributions! AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  23. Chemical thermodynamics - 1st Law • Enthalpy of a mixture (sum of thermal and chemical terms) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  24. Chemical thermodynamics - 1st Law • Note we can also write h as follows • Use these boxed expressions for h & u with h = constant (for constant P combustion) or u = constant (for constant V combustion) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  25. Chemical thermodynamics - 1st Law • Examples of tabulated data on h(T) - h298, hf, etc. (double-click table to open Excel spreadsheet with all data for CO, O, CO2, C, O2, H, OH, H2O, H2, N2, NO at 200K - 6000K) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  26. Chemical thermodynamics - 1st Law • Example: what are h and u for a CO2-O2-CO mixture at 10 atm&2500K with XCO = 0.0129, XO2 = 0.3376, XCO2 = 0.6495? Pressure doesn’t affect h or u but T does; from the tables: AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  27. Chemical thermodynamics - 1st Law • Final pressure (for constant volume combustion) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  28. Chemical thermo - heating value • Constant-pressure energy conservation equation (no heat transfer, no work transfer other than PdV work) Denominator = m = constant, separate chemical and thermal terms: • This scary-looking boxed equation is simply conservation of energy for a chemically reacting mixture at constant pressure • Term on left-hand side is the negative of the total thermal enthalpy change per unit mass of mixture; term on the right-hand side is the chemical enthalpy change per unit mass of mixture AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  29. Chemical thermo - heating value • By definition, CP (∂h/∂T)P • For an ideal gas, h = h(T) only, thus CP= dh/dT or dh = CPdT • If CP is constant, then for the thermal enthalpy h2 - h1 = CP(T2 - T1) = mCP(T2 - T1) /m • For a combustion process in which all of the enthalpy release by chemical reaction goes into thermal enthalpy (i.e. temperature increase) in the gas, the term on the left-hand side of the boxed equation on page 28 can be written as where is the constant-pressure specific heat averaged (somehow) over all species and averaged between the product and reactant temperatures AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  30. Chemical thermo - heating value • Term on right-hand side of boxed equation on page 28 can be re-written as • Last term is the chemical enthalpy change per unit mass of fuel; define this as -QR, where QR is the fuel’s heating value • For our stereotypical hydrocarbons, assuming CO2, H2O and N2 as the only combustion products, this can be written as AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  31. Chemical thermo - flame temperature • Now write the boxed equation on page 28 (conservation of energy for combustion at constant pressure) once again: • We’ve shown that the left-hand side = and the right-hand side = -fQR; combining these we obtain • This is our simplest estimate of the adiabatic flame temperature (Tproducts, usually we write this as Tad) based on an initial temperature (Treactants, usually written as T∞) thus (constant pressure combustion, T-averaged CP) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  32. Chemical thermo - flame temperature • This analysis has assumed that there is enough O2 to burn all the fuel, which is true for lean mixtures only; in general we can write where for lean mixtures, fburnable is just f (fuel mass fraction) whereas for rich mixtures, with some algebra it can be shown that thus in general we can write AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  33. Chemical thermo - flame temperature • For constant-volume combustion (instead of constant pressure), everything is the same except u = const, not h = const, thus the term on the left-hand side of the boxed equation on page 28 must be re-written as The extra PV terms (= mRT for an ideal gas) adds an extra mR(Tproducts-Treactants) term, thus which means that (again, Tproducts = Tad; Treactants = T∞) (constant volume combustion, T-averaged CP) which is the same as for constant-pressure combustion except for the Cv instead of CP AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  34. Chemical thermo - flame temperature • The constant-volume adiabatic flame (product) temperature on the previous page is only valid for lean or stoichiometric mixtures; as with constant-pressure for rich mixtures we need to consider how much fuel can be burned, leading to • Note that the ratio of adiabatic temperature rise due to combustion for constant pressure vs. constant volume is • In practice, one can determine by working backwards from a detailed analysis; for stoichiometric CH4-air, f = 0.055, QR = 50 x 106 J/kg, constant-pressure combustion, Tad = 2226K for T∞ = 300K, thus ≈ 1429 J/kg-K (for other stoichiometries or other fuels will be somewhat but not drastically different) AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  35. Example of heating value • Iso-octane/air mixture: C8H18 + 12.5(O2 + 3.77N2)  8 CO2 + 9 H2O + 12.5*3.77 N2 AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  36. Comments on heating value • Heating values are usually computed assuming all C  CO2, H  H2O, N N2, S  SO2, etc. • If one assumes liquid water, the result is called the higher heating value; if one (more realistically, as we have been doing) assumes gaseous water, the result is called the lower heating value • Most hydrocarbonshave similar QR (4.0 - 4.5 x 107 J/kg) since the same C-C and C-H bonds are being broken and same C-O and H-O bonds are being made • Foods similar - on a dry weight basis, about same QR for all • Fruit Loops™ and Shredded Wheat™ have same “heating value” (110 kcal/oz = 1.6 x 107 J/kg) although Fruit Loops™ mostly sugar, Shredded Wheat™ has none (the above does not constitute a commercial endorsement) • Fats slightly higher than starches or sugars • Foods with (non-digestible) fiber lower AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  37. Comments on heating value • Acetylene higher (4.8 x 107 J/kg) because of C-C triple bond • Methane higher (5.0 x 107 J/kg) because of high H/C ratio • H2MUCH higher (12.0 x 107 J/kg) because no “heavy” C atoms • Alcohols lower (2.0 x 107 J/kg for methanol, CH3OH) because of “useless” O atoms - add mass but no enthalpy release AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  38. Example of adiabatic flame temperature • Lean iso-octane/air mixture, equivalence ratio 0.8, initial temperature 300K, average CP = 1400 J/kgK, average Cv = 1100 J/kgK: Stoichiometric: C8H18 + 12.5(O2 + 3.77N2)  8 CO2 + 9 H2O + 12.5*3.77 N2 AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

  39. Summary - Lecture 2 • Many fuels, e.g. hydrocarbons, when chemically reacted with oxygen or other oxidizers, release large amounts of enthalpy • This chemical energy or enthalpy is converted into thermal energy or enthalpy, thus in a combustion process the product temperature is much higher than the reactant temperature • Only 2 principles are required to compute flame temperatures • Conservation of each type of atom • Conversation of energy (sum of chemical + thermal) … but the resulting equations required to account for changes in composition and energy can look formidable • Key properties of a fuel are its heating value QR and its stoichiometric fuel mass fraction fstoichiometric • Key property of a fuel/air mixture is its equivalence ratio () • A simplified analysis leads to AME 436 - Spring 2013 - Lecture 2 - Chemical Thermodynamics 1

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