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

Transcription

1 AME 436 Energy and Propulsion Lecture 2 Fuels, chemical thermodynamics (thru 1st Law; 2nd Law next lecture Outline Ø Fuels - hydrocarbons, alternatives Ø Balancing chemical reactions Ø Stoichiometry Ø Lean & rich mixtures Ø Mass and mole fractions Ø hemical thermodynamics Ø Ø Ø Ø Why? 1st Law of Thermodynamics applied to a chemically reacting system eating value of fuels Flame temperature 2 1

2 Fuels Ø Usually we employ hydrocarbon fuels, alcohols or hydrogen burning in air, though other possibilities include O, 3, S 2, 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. 4, easy to feed into engine (compared to solids Ø Lots of it in the earth (often the wrong places Ø Relatively non-toxic fuel and combustion products Ø Relatively low explosion hazards 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 (1 mole of air or 1 O (4.77 moles of air Ø ote for air, the average molecular mass is 0.21 Mole O 2 Mole total 32g O Mole 2 Mole O 2 Mole total 28g 2 Mole g Mole total thus the gas constant (universal gas constant / mole. wt. (8.314 J/moleK / ( kg/mole 287 J/kgK 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 2 4 2

3 ydrocarbons Ø Alkanes - single bonds between carbons - n 2n+2, e.g. 4, 2 6 methane ethane Ø Olefins or alkenes - one or more double bonds between carbons ethene or ethylene propene or propylene Ø Alkynes - one or more triple bonds between carbons - very reactive, higher heating value than alkanes or alkenes propane 1, 3 butadiene ethyne or acetylene 5 ydrocarbons Ø Aromatics - one or more ring structures benzene toluene napthalene Ø Alcohols - contain one or more O groups O O methanol ethanol 6 3

4 Biofuels Ø Alcohols - produced by fermentation of food crops (sugars or starches or cellulose (much more difficult, not andustrial 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. 2 5 Ethyl stearate Methanol + triglyceride Glycerol+ alkyl ester Transesterification process 7 Practical fuels Ø All practical fuels are BLEDS 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

5 Gasoline - typical composition Paraffins alkanes Benzene Toluene J. Burri et al., Fuel, Vol. 83, pp ( Practical fuels - properties Ø Values OT unique because Ø Real fuels are a mixture of many molecules, composition varies Ø Different testing methods & definitions Property Jet-A Diesel Gasoline Ethanol atural gas eating value (MJ/kg Flash point ( (T at which vapor makes flammable mixture in air Vapor pressure (at 100 F (psi Freezing point ( Autoignition temperature ( (T at which fuel-air mixture will ignite spontaneously without spark or flame Density (at 15 (kg/m More info:

6 Stoichiometry Ø Balancing of chemical reactions with "known" (assumed products Ø Example: methane ( 4 in air (O a(o b O 2 + c 2 O + d 2 (how do we know this know this set of products is reasonable? From 2nd Law, to be discussed in Lecture 3 onservation of atoms: atoms: n 4 (1 + n O2 (0 + n 2 (0 n O2 (b + n 2O (0 + n 2 (0 atoms: n 4 (4 + n O2 (0 + n 2 (0 n O2 (0 + n 2O (2c + n 2 (0 O atoms: n 4 (0 + n O2 (2a + n 2 (0 n O2 (2b + n 2O (c + n 2 (0 atoms: n 4 (0 + n O2 (0 + n 2 (3.77*2a n O2 (0 + n 2O (0 + n 2 (2d Solve: a 2, b 1, c 2, d (O O O or in general x y + (x + y/4(o x O 2 + (y/2 2 O (x + y/ 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 ; for lower or higher % O 2 in the atmosphere, the numbers would change accordingly 12 6

7 Stoichiometry Ø Fuel mass fraction (f f fuel mass total mass n M fuel fuel 1 (12x +1y n fuel M fuel + n O2 M O2 + n 2 M 2 1 (12x +1y + (x + y 4 ( number of moles of species i, molecular mass of species i For the specific case of stoichiometric methane-air (x 1, y 4, f ; a lean/rich mixture would have lower/higher f Ø For stoichiometric mixtures, f is similar for most hydrocarbons but depends on the / ratio x/y, e.g. Ø f for 4 (methane - lowest possible / ratio Ø f for 6 6 (benzene or 2 2 (acetylene - high / ratio Ø Fuel mole fraction X f fuel moles X f total moles n fuel 1 n fuel + n O2 + n (x + y 4 which varies a lot depending on x and y (i.e., much smaller for big molecules with large x and y 13 Stoichiometry Ø Fuel-to-air ratio (FAR fuel mass FAR air mass fuel mass total mass - fuel mass (fuel mass/(total mass 1 - (fuel mass/(total mass f 1 - f and air-to-fuel ratio (AFR 1/(FAR Ø ote also f FAR/(1+FAR Ø Equivalence ratio (f φ FAR (actual mixture FAR (stoichiometric mixture f < 1: lean mixture; f > 1: rich mixture Ø What if we assume more products, e.g. 4 +?(O ? O 2 +? 2 O +? 2 +? O In this case we have 4 atom constraints (1 each for,, O, and atoms but 5 unknowns (5 question marks - how to solve? eed chemical equilibrium (Lecture 3 to decide how much and O are in the form of O 2 vs. O vs. 2 O 14 7

8 hemical 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? 2 + O 2 at f 1? ope, T 3079K at 1 atm for reactants at 298K Ø Probably the highest is diacetylnitrile + ozone (4/3O 3 4 O + 2 T 5516K at 1 atm for reactants at 298K Ø Why should it? The 2 + O 2 system has much more energy release per unit mass of reactants, but still a much lower flame temperature 15 hemical thermodynamics - intro Ø The reasos that the product is OT just 2 O, i.e. we don't get 2 + (1/2O 2 2 O but rather 2 + (1/2O O O O 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 Ø Energy required to break -O- bonds to make the other species Ø igher pressures reduce dissociation - Le hatelier'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 16 8

9 hemical thermodynamics - intro Ø Actually, evef we somehow avoided dissociation, the 2 - O 2 flame would be only 4998K - still not have as high a flame temp. as the weird 4 2 flame Ø Why? 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 O 3 flame so hot? Ø O 3 decomposes exothermically to (3/2O 2 Ø The products O and 2 are diatomic gases - fewer DOFs Ø O and 2 are very stable even at 5500K - almost no dissociation 17 hemical thermodynamics - goals Ø Given anitial 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 ompression» Usually constant entropy S (isentropic Actually, reversible and adiabatic; since ds dq/t with sign applying for reversible and dq 0 for adiabatic, ds 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 ombustion» Usually constant P or V assumed» omposition 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 18 9

10 hemical thermodynamics - assumptions Ø Ideal gases - note many "flavors" of the ideal gas law PV nât PV mrt Pv RT P rrt most useful form in this course; more useful to work with mass than moles, because moles are not conserved in chemical reactions! P pressure (/m 2 ; V volume (m 3 ; 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 Â/M M gas molecular mass (kg/mole; v V/m specific volume (m 3 /kg r 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 ombustion process is constant P or V (constant T or s combustion isn't very interesting! Ø ompression/expansios reversible & adiabatic (Þ isentropic, ds 0 19 hemical thermodynamics - 1st Law Ø 1st Law of thermodynamics (conservation of energy, control mass: de dq - dw Ø E U + PE + KE U U Ø dw PdV for a simple compressible substance; also assume adiabatic (dq 0 Ø ombine: du + PdV 0 Ø onstant pressure: add VdP 0 term Ø du + PdV + VdP 0 Þ d(u+pv 0 Þ d 0 Ø reactants products Ø Recall h º /m (m mass, thus h reactants h products Ø onstant volume: PdV 0 Ø du + PdV 0 Þ d(u 0 Ø U reactants U products, thus u reactants u products Ø h u + Pv, thus (h - Pv reactants (h - Pv products Ø Most property tables report h not u, so h - Pv form is useful Ø h or u must include BOT thermal and chemical contributions! 20 10

11 hemical thermodynamics - 1st Law Ø Enthalpy of a mixture (sum of thermal and chemical terms (1! hi i1 (2 h! i enthalpy of i per mole of i + Δ h! o ( number of species; number of moles of i [! h(t! h 298 enthalpy per mole of i to raise i from 298K to T (thermal enthalpy Δ h! o enthalpy of formation per mole of i at 298K & 1 atm, i.e. enthalpy change from formation of i from its elements in their standard state (chemical enthalpy ote Δ h! o 0 for elements in their standard state, e.g. O 2 (gas, (solid (3 m mass of mixture ; molecular mass of i ombine (1 (3 to obtain i1 o ( h + Δ h! m i1 i1 21 hemical thermodynamics - 1st Law Ø ote we can also write h as follows o ( h + Δ h! m i1 i1 [ h(t! h! n Δ h! o T ( i1 n i M i1 T Moles of i n T Total moles of all gases Mole fraction of i X i o ( h X i + Δ h! m i1 u U m PV m i1 X i mrt m h RT Ø Use boxed expressions for h & u with h constant (for constant P combustion or u constant (for constant V combustion 22 11

12 hemical thermodynamics - 1st Law Ø Examples of tabulated data on h(t - h 298, Dh f, etc. (double-click table to open Excel spreadsheet with all data for O, O, O 2,, O 2,, O, 2 O, 2, 2, O at 200K K O Molecular weight g/mole Δh f o (kj/mole O 2 Molecular weight g/mole Δh o f (kj/mole O 2 Molecular weight g/mole Δh o f (kj/mole T s h-h_298 K J/mole-K kj/mole T s h-h_298 K J/mole-K kj/mole T s h-h_298 K J/mole-K kj/mole hemical thermodynamics - 1st Law Ø Example: what are h and u for a O-O 2 -O 2 mixture at 10 atm & 2500K with X O , X O , X O ? h Pressure doesn't affect h or u but T does; from the tables: M O ; M O ; M O kg/mole Δh o f,o i1 ( X i + Δ h! o h 3784 kj kg R R M ; M i1 R 8.314J molek ; Δh o f,o2 X i J kg i1 X i ( ( ( kj / mole ( ( ( kg / mole kg ( ( ( mole kg 209.2J mole kgk 0; Δh o f,o kj /mole [ h (2500 h 298 ] O ; [ h (2500 h 298 ] O ; [ h (2500 h 298 ] O kj /mole J u h RT kg 209.2J kgk (2500K J 106 kg 24 12

13 hemical thermodynamics - 1st Law Ø Final pressure (for constant volume combustion PV mrt, R R ; R universal gas constant J/moleK M M (for mixture Total mass Total moles i1 i1 i1 n Total X i onstant volume combustion : V constant, m constant ombine: P products P reactants (products T i1 products (reactants T n reactants i i1 i1 25 hemical thermodynamics - heating value Ø onstant-pressure energy conservation equation (no heat transfer, no work transfer other than PdV work h reactants (reactants i1 ( + Δ h! o (reactants i1 h products (products + Δ h! o ( Denominator m constant, separate chemical and thermal terms: (reactants i1 [ h(t! h! (products ( 298 (reactants i1 i1 ( Ø This scary-looking boxed equatios 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 26 i1 (products (products i1 Δ h! o Δ h! o i1 (products i1 (reactants i1 13

14 hemical thermo - heating value Ø By definition, P º ( h/ T P Ø For adeal gas, h h(t only, thus P dh/dt or dh P dt Ø If P is constant, then for the thermal enthalpy h 2 - h 1 P (T 2 - T 1 m P (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 (reactants i1 [ h(t! h! (products ( 298 P (reactants i1 i1 ( m P (T reactants T products m where is the constant-pressure specific heat averaged (somehow over all species and averaged between the product and reactant temperatures 27 hemical thermo - heating value Ø Term on right-hand side of boxed equation on page 26 can be rewritten as (products n n fuel Δ! (reactants h o fuel Δ! (products o h Δ! (reactants h o Δ! o h i1 i1 i1 i1 f n M fuel M fuel n fuel M fuel i (reactants i1 Ø Last term is the chemical enthalpy change per unit mass of fuel; define this as -Q R, where Q R is the fuel's heating value (products Δ h! o Δ! o h i1 i1 Q R n fuel M fuel Ø For our stereotypical hydrocarbons, assuming O 2, 2 O and 2 as the only combustion products, this can be written as Q R x Δh f,o 2 o (reactants o + (y 2 Δh f,2 O o 1 Δh f, fuel 1 M fuel o (x + y 4Δh f,o

15 hemical thermo - flame temperature Ø ow write the boxed equation on page 26 (conservation of energy for combustion at constant pressure once again: (reactants i1 [ h(t! h! (products ( 298 (reactants i1 i1 ( m Ø We've shown that the left-hand side P (T reactants T products m and the right-hand side -fq R ; combining these we obtain T products T reactants + fq R / P Ø This is our simplest estimate of the adiabatic flame temperature (T products, usually we write this as based on anitial temperature (T reactants, usually written as T thus (products Δ h! o Δ h! o i1 (reactants i1 (reactants i1 T + fq R / P (constant pressure combustion, T-averaged P 29 hemical 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 T + f burnable Q R P where for lean mixtures, f burnable is just f (fuel mass fraction whereas for rich mixtures, with some algebra it can be shown that # 1 f & f burnable f stoichiometric % ( $ 1 f stoichiometric ' thus in general we can write T + f Q R P (if f f stoichiometric % 1 f ( T + f stoichiometric ' * Q R (if f f stoichiometric & 1 f stoichiometric P 30 15

16 hemical 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 (reactants i1 [ h(t! h! (products ( 298 (PV reactants i1 (reactants ( (PV products The extra PV terms ( mrt for adeal gas adds an extra mr(t products -T reactants term, thus i1 m P (T products T reactants m P (T products T reactants mr(t products T reactants m m ( P R(T products T reactants v (T products T reactants which means that (again, T products ; T reactants T T + fq R / v (constant volume combustion, T-averaged P which is the same as for constant-pressure combustion except for the v instead of P 31 hemical 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 T + f Q R v (if f < f stoichiometric $ 1 f ' T + f stoichiometric & Q R (if f > f % 1 f stoichiometric ( stoichiometric v Ø ote that the ratio of adiabatic temperature rise due to combustion for constant pressure vs. constant volume is ( T constant v P γ ( T constant P V P Ø One can determine by working backwards from a detailed analysis; for stoichiometric 4 -air, f 0.055, Q R 50 x 10 6 J/kg, constantpressure combustion, 2226K for T 300K, thus P 1429 J/kgK (for other stoichiometries or other fuels, the effective P will be somewhat but not drastically different 32 16

17 Example of heating value Ø Iso-octane/air mixture: ( Δ h! o f, kj / mole (O O O * (products (reactants Δ h! o Δ! o h i1 i1 Q R n fuel M fuel 8Δ h! o f,o2 + 9Δ h! o f, 2O +12.5(3.77Δ h! o ( f, 2 1Δ h! o f, Δ h! o f,o (3.77Δ h! o ( f, 2 1M 818 ( ( 1( ( (3.77(0 8 moles( kj/mole + 9( (3.77(0 (1 mole(0.114 kg/mole 44,500 kj/kg 4.45 x 10 7 J/kg 33 Fuel properties Fuel eating value, Q R (J/kg f at stoichiometric Gasoline 43 x Methane 50 x Methanol 20 x Ethanol 27 x oal 34 x Paper 17 x Fruit Loops 16 x 10 6 Probably about the same as paper ydrogen 120 x U 235 fission 83,140,000 x Pu 239 fission 83,610,000 x fusion 339,000,000 x : 3 1 :

18 omments on heating value Ø eating values are usually computed assuming all O 2, 2 O, 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 Q R ( x 10 7 J/kg since the same - and - bonds are being broken and same -O and -O bonds are being made Ø Foods similar - on a dry weight basis, about same Q R for all Ø Fruit Loops and Shredded Wheat have same "heating value" (110 kcal/oz 1.6 x 10 7 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 35 omments on heating value Ø Acetylene higher (4.8 x 10 7 J/kg because of º triple bond Ø Methane somewhat higher (5.0 x 10 7 J/kg because of high / ratio Ø 2 MU higher (12.0 x 10 7 J/kg because no "heavy" atoms Ø Alcohols lower (2.0 x 10 7 J/kg for methanol, 3 O because of "useless" O atoms - add mass but no enthalpy release 36 18

19 Example of adiabatic flame temperature Ø Leaso-octane/air mixture, equivalence ratio f 0.8, initial temperature T 300K, average P 1400 J/kgK, average v 1100 J/kgK: Stoichiometric: (O O O * φ FAR (actual mixture, φ 0.8 FAR (stoichiometric mixture, φ 1 f /(1 f φ 0.8 φ f φ 1 /(1 f φ 1 f φ 1 n fuel M fuel n (reactants i1 (1 mole 8 18 (0.114 kg/mole (1 mole 8 18 (0.114 kg/mole + (12.5 mole O 2 (0.032 kg/mole + (12.5* 3.77 mole 2 (0.028 kg/mole FAR φ 1 f φ 1 /(1 f φ /( φ 0.8 : f /(1 f φ 0.8 φ f φ T + fq R / P 300K + ( ( J /kg /(1400J /kgk 1906K (const. P T + fq R / V 300K + ( ( J /kg /(1100J /kgk 2345K (const. V 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 onservation of each type of atom Ø onversation 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 Q R and its stoichiometric fuel mass fraction f stoichiometric Ø Key property of a fuel/air mixture is its equivalence ratio (f Ø A simplified analysis leads to T + fq R / P (constant pressure T + fq R / V (constant volume 38 19