AME 436 Energy and Propulsion Lecture 2 Fuels

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

Outline Ø Fuels - hydrocarbons, alternatives Ø Balancing chemical reactions Ø Stoichiometry Ø Lean & rich mixtures Ø Mass and mole fractions Ø Chemical thermodynamics Ø Ø Why? 1 st Law of Thermodynamics applied to a chemically reacting system Heating value of fuels Flame temperature AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 2

Fuels Ø Usually we employ hydrocarbon fuels, alcohols or hydrogen burning in air, though other possibilities include CO, NH 3, CS 2, H 2 S, etc. Ø For rocket fuels that do not burn air, many possible oxidizers exist - ASTE 470 discusses these - AME 436 focuses on airbreathing devices Ø Why hydrocarbons? Ø Many are liquids - high density, easy to transport and store (compared to gases, e. g. CH 4), easy to feed into engine (compared to solids) Ø Lots of it in the earth (often in the wrong places) Ø Relatively non-toxic fuel and combustion products Ø Relatively low explosion hazards AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 3

Air Ø Why air? Ø Because it's free, of course (well, not really when you think of all the money we’ve spent to clean up air) Ø Air ≈ 0. 21 O 2 + 0. 79 N 2 (1 mole of air) or 1 O 2 + 3. 77 N 2 (4. 77 moles of air) Ø Note for air, the average molecular mass is thus the gas constant = (universal gas constant / mole. wt. ) = (8. 314 J/mole. K) / (0. 0289 kg/mole) = 287 J/kg. K Ø 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 O 2 and N 2 AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 4

Hydrocarbons Ø Alkanes - single bonds between carbons - Cn. H 2 n+2, e. g. CH 4, C 2 H 6 Ø Olefins or alkenes - one or more double bonds between carbons Ø Alkynes - one or more triple bonds between carbons - very reactive, higher heating value than alkanes or alkenes AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 5

Hydrocarbons Ø Aromatics - one or more ring structures Ø Alcohols - contain one or more OH groups AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 6

Biofuels Ø Alcohols - produced by fermentation of food crops (sugars or starches) or cellulose (much more difficult, not an industrial process yet) Ø 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. C 2 H 5) Ethyl stearate Methanol + triglyceride Glycerol+ alkyl ester Transesterification process AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 7

Practical fuels Ø All practical fuels are BLENDS of hydrocarbons and other compounds Ø What distinguishes one fuel from another? Ø Flash point - temperature above which fuel vapor pressure is flammable when mixed with air Ø Distillation curve - temp. range over which molecules evaporate Ø Relative amounts of alkanes vs. alkenes vs. aromatics vs. alcohols Ø Amount of impurities, e. g. sulfur Ø Structure of molecules affects octane number (Lecture 10) http: //static. howstuffworks. com/flash/oil-refining. swf AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 8

Gasoline - typical composition Paraffins = alkanes Benzene Toluene J. Burri et al. , Fuel, Vol. 83, pp. 187 - 193 (2004) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 9

Practical fuels - properties Ø Values NOT unique because Ø Real fuels are a mixture of many molecules, composition varies Ø Different testing methods & definitions Property Jet-A Diesel Gasoline Ethanol Natural gas Heating value (MJ/kg) 43 43 43 27 47 Flash point (˚C) (T at which vapor makes flammable mixture in air) 38 70 -43 13 -184 Vapor pressure (at 100˚F) (psi) 0. 03 0. 02 8 2. 3 2400 Freezing point (˚C) − 40 -38 -40 -114 -182 Autoignition temperature (˚C) (T at which fuel-air mixture will ignite spontaneously without spark or flame) 210 240 260 423 557 Density (at 15˚C) (kg/m 3) 810 850 720 792 0. 67 More info: http: //www. afdc. energy. gov/pdfs/fueltable. pdf AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 10

Stoichiometry Ø Balancing of chemical reactions with "known" (assumed) products Ø Example: methane (CH 4) in air (O 2 + 3. 77 N 2) CH 4 + a(O 2 + 3. 77 N 2) b CO 2 + c H 2 O + d N 2 (how do we know this set of products is reasonable? From 2 nd Law, to be discussed in Lecture 3) Conservation of atoms: C atoms: n. CH 4(1) + n. O 2(0) + n. N 2(0) = n. CO 2(b) + n. H 2 O(0) + n. N 2(0) H atoms: n. CH 4(4) + n. O 2(0) + n. N 2(0) = n. CO 2(0) + n. H 2 O(2 c) + n. N 2(0) O atoms: n. CH 4(0) + n. O 2(2 a) + n. N 2(0) = n. CO 2(2 b) + n. H 2 O(c) + n. N 2(0) N atoms: n. CH 4(0) + n. O 2(0) + n. N 2(3. 77*2 a) = n. CO 2(0) + n. H 2 O(0) + n. N 2(2 d) Solve: a = 2, b = 1, c = 2, d = 7. 54 CH 4 + 2(O 2 + 3. 77 N 2) 1 CO 2 + 2 H 2 O + 7. 54 N 2 or in general Cx. Hy + (x + y/4)(O 2 + 3. 77 N 2) x CO 2 + (y/2) H 2 O + 3. 77(x + y/4)N 2 AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 11

Stoichiometry Ø The previous page shows a special case where there is just enough fuel to combine with all of the air, leaving no excess fuel or O 2 unreacted; this is called a stoichiometric mixture Ø In general, mixtures will have excess air (lean mixture) or excess fuel (rich mixture) Ø The analysis assumed air = O 2 + 3. 77 N 2; for lower or higher % O 2 in the atmosphere, the numbers would change accordingly AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 12

Stoichiometry Ø Fuel mass fraction (f) ni = number of moles of species i, Mi = molecular mass 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 Ø For stoichiometric mixtures, f is similar for most hydrocarbons but depends on the C/H ratio = x/y, e. g. Ø f = 0. 0550 for CH 4 (methane) - lowest possible C/H ratio Ø f = 0. 0703 for C 6 H 6 (benzene) or C 2 H 2 (acetylene) - high C/H ratio Ø 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 2019 - Lecture 2 - Chemical Thermodynamics 1 13

Stoichiometry Ø Fuel-to-air ratio (FAR) and air-to-fuel ratio (AFR) = 1/(FAR) Ø Note also f = FAR/(1+FAR) Ø Equivalence ratio ( ) < 1: lean mixture; > 1: rich mixture Ø What if we assume more products, e. g. CH 4 + ? (O 2 + 3. 77 N 2) ? CO 2 + ? H 2 O + ? N 2 + ? 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 (Lecture 3) to decide how much C and O are in the form of CO 2 vs. CO vs. H 2 O AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 14

Chemical thermodynamics - intro Ø 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 Ø What is highest temperature flame? H 2 + O 2 at = 1? Nope, T = 3079 K at 1 atm for reactants at 298 K Ø Probably the highest is diacetylnitrile + ozone C 4 N 2 + (4/3)O 3 4 CO + N 2 T = 5516 K at 1 atm for reactants at 298 K Ø Why should it? The H 2 + O 2 system has much more energy release per unit mass of reactants, but still a much lower flame temperature AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 15

Chemical thermodynamics - intro Ø The reason is that the product is NOT just H 2 O, i. e. we don't get H 2 + (1/2)O 2 H 2 O but rather H 2 + (1/2)O 2 0. 706 H 2 O + 0. 062 O 2 + 0. 184 H 2 + 0. 094 H + 0. 129 OH + 0. 040 O i. e. the water dissociates into the other species (how do we know how much of the other species? Wait for Lecture 3 …) Ø Dissociation does 2 things that reduce flame temp. Ø More moles of products to soak up energy (1. 22 vs. 1. 00) Ø Energy required to break H-O-H bonds to make the other species Ø Higher pressures reduce dissociation - Le Chatelier's principle: When a system at equilibrium is subjected to a stress, the system shifts toward a new equilibrium condition so as to reduce the stress (more pressure, less space, system responds by reducing number of moles of gas to reduce pressure) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 16

Chemical thermodynamics - intro Ø Actually, even if we somehow avoided dissociation, the H 2 - O 2 flame would be only 4998 K - still not have as high a flame temp. as the weird C 4 N 2 flame Ø Why? H 2 O 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 Ø So why is the C 4 N 2 - O 3 flame so hot? Ø O 3 decomposes exothermically to (3/2)O 2 Ø The products CO and N 2 are diatomic gases - fewer DOFs Ø CO and N 2 are very stable even at 5500 K - almost no dissociation AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 17

Chemical thermodynamics - goals Ø 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 Ø Three common processes in engine analysis Ø Compression » Usually constant entropy S (isentropic) • Actually, reversible and adiabatic; since d. S ≥ d. Q/T with = sign applying for reversible and d. Q = 0 for adiabatic, d. S = 0 » 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 output! Ø Combustion » Usually constant P or V assumed » Composition MUST change (obviously…) Ø Expansion » Opposite of compression » May assume frozen (no change during expansion) or equilibrium composition (mixture shifts to new composition after expansion) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 18

Chemical thermodynamics - assumptions Ø Ideal gases - note many "flavors" of the ideal gas law • • PV = n T PV = m. RT Pv = RT P = RT – most useful form in this course; more useful to work with mass than moles, because moles are not conserved in chemical reactions! P = pressure (N/m 2); V = volume (m 3); n = number of moles of gas = universal gas constant (8. 314 J/mole. K); T = temperature (K) m = mass of gas (kg); R = mass-specific gas constant = /M M = gas molecular mass (kg/mole); v = V/m = specific volume (m 3/kg) = 1/v = density (kg/m 3) Ø Adiabatic Ø Kinetic and potential energy negligible (we’ll revisit this assumption for hypersonic propulsion) Ø Mass is conserved Ø Combustion process is constant P or V (constant T or s combustion isn't very interesting!) Ø Compression/expansion is reversible & adiabatic ( isentropic, d. S = 0) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 19

Chemical thermodynamics - 1 st Law Ø 1 st Law of thermodynamics (conservation of energy), control mass: d. E = Q - W Ø E = U + PE + KE = U + 0 = U Ø W = Pd. V for a simple compressible substance; also assume adiabatic ( Q = 0) Ø Combine: d. U + Pd. V = 0 Ø Constant pressure: add Vd. P = 0 term Ø d. U + Pd. V + Vd. P = 0 d(U+PV) = 0 d. H = 0 Ø Hreactants = Hproducts Ø Recall h H/m (m = mass), thus hreactants = hproducts Ø Constant volume: Pd. V = 0 Ø Ø d. U + Pd. V = 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 Ø h or u must include BOTH thermal and chemical contributions! AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 20

Chemical thermodynamics - 1 st Law Ø Enthalpy of a mixture (sum of thermal and chemical terms) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 21

Chemical thermodynamics - 1 st Law Ø Note we can also write h as follows Ø Use boxed expressions for h & u with h = constant (for constant P combustion) or u = constant (for constant V combustion) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 22

Chemical thermodynamics - 1 st Law Ø Examples of tabulated data on h(T) - h 298, hf, etc. (double-click table to open Excel spreadsheet with all data for CO, O, CO 2, C, O 2, H, OH, H 2 O, H 2, NO at 200 K - 6000 K) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 23

Chemical thermodynamics - 1 st Law Ø Example: what are h and u for a CO-O 2 -CO 2 mixture at 10 atm & 2500 K with XCO = 0. 0129, XO 2 = 0. 3376, XCO 2 = 0. 6495? Pressure doesn't affect h or u but T does; from the tables: AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 24

Chemical thermodynamics - 1 st Law Ø Final pressure (for constant volume combustion) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 25

Chemical thermodynamics - heating value Ø Constant-pressure energy conservation equation (no heat transfer, no work transfer other than Pd. V work) Denominator = m = constant, separate chemical and thermal terms: Ø This scary-looking boxed equation is simply conservation of energy for a chemically reacting mixture at constant pressure Ø 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 2019 - Lecture 2 - Chemical Thermodynamics 1 26

Chemical thermo - heating value Ø By definition, CP (∂h/∂T)P Ø For an ideal gas, h = h(T) only, thus CP = dh/d. T or dh = CPd. T Ø If CP is constant, then for thermal enthalpy h 2 - h 1 = CP(T 2 - T 1) = m. CP(T 2 - T 1) /m Ø 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 26 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 2019 - Lecture 2 - Chemical Thermodynamics 1 27

Chemical thermo - heating value Ø Term on right-hand side of boxed equation on page 26 can be rewritten as Ø Last term is the chemical enthalpy change per unit mass of fuel; define this as -QR, where QR is the fuel's heating value Ø For our stereotypical hydrocarbons, assuming CO 2, H 2 O and N 2 as the only combustion products, this can be written as AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 28

Chemical thermo - flame temperature Ø Now write the boxed equation on page 26 (conservation of energy for combustion at constant pressure) once again: Ø We've shown that the left-hand side = and the right-hand side = -f. QR; combining these we obtain Ø 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, CP) T-averaged AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 29

Chemical thermo - flame temperature Ø This analysis has assumed that there is enough O 2 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 2019 - Lecture 2 - Chemical Thermodynamics 1 30

Chemical thermo - flame temperature Ø For constant-volume combustion (instead of constant P), everything is the same except u = const, not h = const, thus the term on the left-hand side of the boxed equation on page 29 must be re-written as The extra PV terms (= m. RT for an ideal gas) adds an extra m. R(Tproducts. Treactants) term, thus which means that (again, Tproducts = Tad; Treactants = T∞) (constant volume combustion, T-averaged CP) which is the 436 same as for constant-pressure combustion except for 1 the 31 AME - Spring 2019 - Lecture 2 - Chemical Thermodynamics

Chemical thermo - flame temperature Ø 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 Ø Note that the ratio of adiabatic temperature rise due to combustion for constant pressure vs. constant volume is Ø One can determine by working backwards from a detailed analysis; for stoichiometric CH 4 -air, f = 0. 055, QR = 50 x 106 J/kg, constantpressure combustion, Tad = 2226 K for T∞ = 300 K, thus ≈ 1429 J/kg. K (for other stoichiometries or other fuels, the effective will be somewhat but not drastically different) AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 32

Example of heating value Ø Iso-octane/air mixture: C 8 H 18 + 12. 5(O 2 + 3. 77 N 2) 8 CO 2 + 9 H 2 O + 12. 5*3. 77 N 2 AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 33

Fuel properties Fuel Heating value, QR (J/kg) f at stoichiometric Gasoline 43 x 106 0. 0642 Methane 50 x 106 0. 0550 Methanol 20 x 106 0. 104 Ethanol 27 x 106 0. 0915 Coal 34 x 106 0. 0802 Paper 17 x 106 0. 122 Fruit Loops™ 16 x 106 Probably about the same as paper Hydrogen 120 x 106 0. 0283 U 235 fission 83, 140, 000 x 106 1 Pu 239 fission 83, 610, 000 x 106 1 2 H + 3 H fusion 339, 000 x 106 2 H : 3 H = 1 : 1 AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 34

Comments on heating value Ø Heating values are usually computed assuming all C CO 2, H H 2 O, N N 2, S SO 2, etc. Ø 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 Ø Most hydrocarbons have similar QR (4. 0 – 5. 0 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 Ø Foods similar - on a dry weight basis, about same QR for all Ø 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) Ø Fats slightly higher than starches or sugars Ø Foods with (non-digestible) fiber lower AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 35

Comments on heating value Ø Acetylene higher (4. 8 x 107 J/kg) because of C C triple bond Ø Methane somewhat higher (5. 0 x 107 J/kg) because of high H/C ratio Ø H 2 MUCH higher (12. 0 x 107 J/kg) because no "heavy" C atoms Ø Alcohols lower (2. 0 x 107 J/kg for methanol, CH 3 OH) because of "useless" O atoms - add mass but no enthalpy release AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 36

Example of adiabatic flame temperature Ø Lean iso-octane/air mixture, equivalence ratio f = 0. 8, initial temperature T∞ = 300 K, average CP = 1400 J/kg. K, average Cv = 1100 J/kg. K: Stoichiometric: C 8 H 18 + 12. 5(O 2 + 3. 77 N 2) 8 CO 2 + 9 H 2 O + 12. 5*3. 77 N 2 AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 37

Summary - Lecture 2 Ø Many fuels, e. g. hydrocarbons, when chemically reacted with an oxidizer, e. g. O 2, release large amounts of energy or enthalpy Ø 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 Ø Only 2 principles are required to compute flame temperatures Ø Conservation of each type of atom Ø Conversation of energy (sum of chemical + thermal) … but the resulting equations required to account for changes in composition and energy can look formidable Ø Key thermodynamic properties of a fuel are its heating value QR and its stoichiometric fuel mass fraction fstoichiometric Ø Key property of a fuel/air mixture is its equivalence ratio ( ) Ø A simplified analysis leads to AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1 38