AME 513. " Lecture 2 Chemical thermodynamics I 1 st Law

Transcription

1 AME 513 Priciples of ombustio " Lecture 2 hemical thermodyamics I 1 st Law Outlie" Fuels - hydrocarbos, alteratives Balacig chemical reactios Stoichiometry Lea & rich mixtures Mass ad mole fractios hemical thermodyamics Why? 1st Law of Thermodyamics applied to a chemically reactig system eatig value of fuels Flame temperature 2 1

2 Fuels & air" Usually we employ hydrocarbo fuels, alcohols or coal burig i air, though other possibilities iclude 2, O, N 3, S 2, 2 S, etc. For rocket fuels that do ot bur air, may possible oxidizers exist - ASTE 470, 570 & 572 discuss these Why air? Because it s free, of course (well, ot really whe you thik of all the moey we ve spet to clea up air Air 0.21 O N 2 (1 mole of air or 1 O N 2 (4.77 moles of air Note for air, the average molecular weight is 0.21* * g/mole thus the gas costat (uiversal gas costat / mole. wt. (8.314 J/moleK / ( kg/mole 287 J/kgK Also 1% argo, up to a few % water vapor depedig o the relative humidity, trace amouts of other gases, but we ll usually assume just O 2 ad N 2 3 ydrocarbos" Alkaes - sigle bods betwee carbos - 2+2, e.g. 4, 2 6 methae ethae propae Olefis or alkees - oe or more double bods betwee carbos ethee or ethylee propee or propylee 1, 3 butadiee Alkyes - oe or more triple bods betwee carbos - higher heatig value tha alkaes or alkees due to straied (edothermic bods tha alkaes or alkees, also very reactive ethye or acetylee 4 2

3 ydrocarbos" Aromatics - oe or more rig structures bezee toluee apthalee Alcohols - cotai oe or more O groups O O methaol ethaol 5 Biofuels" Alcohols - produced by fermetatio of food crops (sugars or starches or cellulose (much more difficult, ot adustrial process yet! Biodiesel - covert vegetable oil or aimal fat (which have very high viscosity ito alkyl esters (lower viscosity through trasesterificatio with alcohol Methyl lioleate Geeric ester structure (R ay orgaic radical, e.g. 2 5 Ethyl stearate Methaol + triglyceride Glycerol+ alkyl ester Trasesterificatio process 6 3

4 Practical fuels" All practical fuels are BLENDS of hydrocarbos ad sometimes other compouds What distiguishes fuels? Flash poit - temperature above which fuel vapor pressure is flammable whe mixed with air Distillatio curve - temp. rage over which molecules evaporate Relative amouts of paraffis vs. olefis vs. aromatics vs. alcohols Amout of impurities, e.g. sulfur Structure of molecules - affects octae umber (gasolie or cetae umber (Diesel 7 Gasolie - typical compositio" Bezee Toluee J. Burri et al., Fuel, Vol. 83, pp (

5 Practical fuels - properties" Values NOT uique because Real fuels are a mixture of may molecules, compositio varies Differet testig methods & defiitios Property Jet-A Diesel Gasolie eatig value (MJ/kg Flash poit ( (T at which vapor makes flammable mixture i air Vapor pressure (at 100 F (psi Freezig poit ( Autoigitio temperature ( (T at which fuel-air mixture will igite spotaeously without spark or flame Desity (at 15 (kg/m 3 # Practical fuels - properties"

6 Practical fuels" What does t distiguish oe fuel from aother? Eergy cotet (except for fuels cotaiig alcohols, which are lower Examples Gasolie - low-t distillatio poit, easy to vaporize, eed high octae umber; reformulated gasolie cotais alcohols Diesel - high-t distillatio poit, hard to vaporize, eed LOW octae umber for easy igitio oce fuel is iject Jet fuel - medium-t distillatio poit; eed low freezig T sice it will be used at high altitude / low T 11 Stoichiometry" Balacig of chemical reactios with kow (assumed products Example: methae ( 4 i air (O N a(o N 2 b O 2 + c 2 O + d N 2 (how do we kow this kow this set is reasoable? From 2d Law, to be discussed later oservatio of,, O, N atoms: 4 (1 + O2 (0 + N2 (0 O2 (b + 2O (0 + N2 (0 4 (4 + O2 (0 + N2 (0 O2 (0 + 2O (2c + N2 (0 4 (0 + O2 (2a + N2 (0 O2 (2b + 2O (c + N2 (0 4 (0 + O2 (0 + N2 (3.77*2a O2 (0 + 2O (0 + N2 (2d Solve: a 2, b 1, c 2, d (O N 2 1 O O N 2 or i geeral x y +(x + y 4(O N 2 " xo 2 +(y 2 2 O (x + y 4N

7 Stoichiometry" This is a special case where there is just eough fuel to combie with all of the air, leavig o excess fuel or O 2 ureacted; this is called a stoichiometric mixture I geeral, mixtures will have excess air (lea mixture or excess fuel (rich mixture This developmet assumed air O N 2 ; for lower or higher % O 2 i the atmosphere, the umbers would chage accordigly 13 Stoichiometry" Fuel mass fractio (f f fuel mass total mass fuelm fuel fuel M fuel + O2 M O2 + N2 M N2 1"(12x +1y 1"(12x +1y+(x + y 4"( " 28 umber of moles of species i, molecular weight of species i For the specific case of stoichiometric methae-air (x 1, y 4, f ; a lea/rich mixture would have lower/higher f For stoichiometric mixtures, f is similar for most hydrocarbos but depeds o the / ratio x/y, e.g. f for 4 (methae - lowest possible / ratio f for 6 6 (bezee or 2 2 (acetylee - high / ratio Fuel mole fractio X f fuel moles X f total moles fuel 1 fuel + O2 + N2 1+ (x + y 4 which varies a lot depedig o x ad y (i.e. much smaller for big molecules with large x ad y 14 7

8 Stoichiometry" 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 ad air-to-fuel ratio (AFR 1/(FAR Note also f FAR/(1+FAR Equivalece ratio (φ " FAR (actual mixture FAR (stoichiometric mixture φ < 1: lea mixture; φ > 1: rich mixture What if we assume more products, e.g. 4 +?(O N 2? O 2 +? 2 O +? N 2 +? O I this case we have 4 atom costraits (1 each for,, O, ad N atoms but 5 ukows (5 questio marks - how to solve? Need chemical equilibrium (discussed later to decide how much ad O are i the form of O 2 vs. O 15 Fuel properties" Fuel eatig value, Q R (J/kg f at stoichiometric Gasolie 43 x Methae 50 x Methaol 20 x Ethaol 27 x oal 34 x Paper 17 x Fruit Loops 16 x 10 6 Probably about the same as paper ydroge 120 x U 235 fissio 83,140,000 x Pu 239 fissio 83,610,000 x fusio 339,00,000 x : 3 1 :

9 hemical thermodyamics - itroductio" Besides eedig to kow how to balace chemical reactios, we eed to determie how much iteral eergy or ethalpy is released by such reactios ad what the fial state (temperature, pressure, mole fractios of each species will be What is highest temperature flame? 2 + O 2 at φ 1? Nope, T 3079K at 1 atm for reactats at K Probably the highest is diacetylitrile + ozoe 4 N 2 + (4/3O 3 4 O + N 2 T 5516K at 1 atm for reactats at K Why should it? The 2 + O 2 system has much more eergy release per uit mass of reactats, but still a much lower flame temperature 17 hemical thermodyamics - itroductio" The problem is that the products are NOT just 2 O, that is, we do t get 2 + (1/2O 2 2 O but rather 2 + (1/2O O O O O i.e. the water dissociates ito the other species Dissociatio does 2 thigs that reduce flame temp. More moles of products to soak up eergy (1.22 vs Eergy is required to break the -O- bods to make the other species igher pressures will reduce dissociatio - Le hatelier s priciple: Whe a system at equilibrium is subjected to a stress, the system shifts toward a ew equilibrium coditio such as way as to reduce the stress (more pressure, less space, system respods by reducig umber of moles of gas to reduce pressure 18 9

10 hemical thermodyamics - itroductio" Actually, evef we somehow avoided dissociatio, the 2 - O 2 flame would be oly 4998K - still ot have as high a flame temp. as the weird 4 N 2 flame Why? 2 O is a triatomic molecule - more degrees of freedom (DOFs (i.e. vibratio, rotatio tha diatomic gases; each DOF adds to the molecule s ability to store eergy So why is the 4 N 2 - O 3 flame so hot? O ad N 2 are diatomic gases - fewer DOFs O ad N 2 are very stable eve at 5500K - almost o dissociatio O 3 decomposes exothermically to (3/2O 2 19 hemical thermodyamics - goals" Give aitial state of a mixture (temperature, pressure, compositio, ad a assumed process (costat pressure, volume, or etropy, usually, fid the fial state of the mixture Three commo processes i egie aalysis ompressio» Usually costat etropy (isetropic» Low P / high V to high P / low V» Usually P or V ratio prescribed» Usually compositio assumed froze - if it reacted before compressio, you would t get ay work out! ombustio» Usually costat P or v assumed» ompositio MUST chage (obviously Expasio» Opposite of compressio» May assume froze (o chage durig expasio or equilibrium compositio (mixture shifts to ew compositio after expasio 20 10

11 hemical thermo - assumptios" Ideal gases - ote may flavors of the ideal gas law PV RT PV mrt Pv RT P ρrt P pressure (N/m 2 ; V volume (m 3 ; umber of moles of gas; R uiversal gas costat (8.314 J/moleK; T temperature (K m mass of gas (kg; R mass-specific gas costat R/M M gas molecular weight (kg/mole; v V/m specific volume (m 3 /kg ρ 1/v desity (kg/m 3 Adiabatic Kietic ad potetial eergy egligible Mass is coserved ombustio process is costat P or V (costat T or s combustios t very iterestig! ompressio/expasios reversible & adiabatic ( isetropic, ds 0 21 hemical thermodyamics - 1st Law" 1st Law of thermodyamics (coservatio of eergy, cotrol mass: de δq - δw E U + PE + KE U U δw PdV ombie: du + PdV 0 ostat pressure: add VdP 0 term du + PdV + VdP 0 d(u+pv 0 d 0 reactats products Recall h /m (m mass, thus h reactats h products ostat volume: PdV 0 du + PdV + VdP 0 d(u 0 U reactats U products, thus u reactats u products h u + Pv, thus (h - Pv reactats (h - Pv products Most property tables report h ot u, so h - Pv form is useful New twist: h or u must iclude BOT thermal ad chemical cotributios! 22 11

12 hemical thermodyamics - 1st Law" Ethalpy of a mixture (sum of thermal ad chemical terms " i h i ; h i ethalpy of i per mole of i [ h (T # h + [ h (T " h ethalpy to raise i from temperature of to T (thermal ethalpy " h o f,i ethalpy of formatio of i at K & 1 atm, i.e. ethalpy chage resultig o h f,i ( (o subscripts umber of species; umber of moles of i from formatio of i from its elemets i their stadard state (chemical ethalpy Note " h o f,i 0 for elemets i their stadard state, e.g. O 2 (gas, (solid m mass of mixture " ; molecular weight of i ombie all these to form h m h o ( f,i [ h (T " h + # 23 hemical thermodyamics - 1st Law" Note we ca also write h as follows h m h m h o ( f,i [ h (T " h + # h o ( f,i X i [ h (T " h + # T [ h (T " h + # h o ( f,i T Moles of i T Total moles of all gases Mole fractio of i X i u U m " PV m X i " mrt m h " RT Use these boxed expressios for h & u with h costat (for costat P combustio or u costat (for costat V combustio 24 12

13 hemical thermodyamics - 1st Law" Examples of tabulated data o h(t - h, Δh f, etc. (double-click table to ope Excel spreadsheet with all data for O, O, O 2,, O 2,, O, 2 O, 2, N 2, NO at 200K K O Molecular weight g/mole!h f o (kj/mole O 2 Molecular weight g/mole!h o f (kj/mole O 2 Molecular weight g/mole!h o f (kj/mole T s h-h_ K J/mole-K kj/mole T s h-h_ K J/mole-K kj/mole T s h-h_ K J/mole-K kj/mole hemical thermodyamics - 1st Law" Example: what are h ad u for a O 2 -O 2 -O at 10 atm, 2500K with X O , X O , X O ? h Pressure does t affect h or u but T does; from the tables: M O ; M O ; M O kg/mole " h o f,o h o ( f,i X i [ h (T " h + # h "3784 kj kg X i J "3.784 %106 kg ( " ( ( " kj /mole ( ( ( kg/mole R & M ; M X i ( ( ( ' R 8.314J molek # ; " h o f,o2 u h " RT "3.784 %10 6 0; " h o f,o2 # kj /mole [ h (2500 # h ] O ; [ h (2500 # h ] O ; [ h (2500 # h ] O kj /mole kg 209.2J mole kgk J kg " 209.2J kgk (2500K "4.307 J %106 kg kg mole 26 13

14 hemical thermodyamics - 1st Law" Fial pressure (for costat volume combustio PV mrt, R " ; " uiversal gas costat J/moleK M M (for mixture Total mass Total moles " " ostat volume combustio : V costat, m costat ombie : P products P reactats (products " T products T reactats i " 27 hemical thermo - heatig value" ostat-pressure eergy coservatio equatio (o heat trasfer, o work trasfer other tha PdV work h reactats [ h (T " h + # h o ( f,i h products (products [ h (T " h + # h o ( f,i (products Deomiator m costat, separate chemical ad thermal terms: # ( " # ([ h (T " h [ h (T " h # (products Term o left-had side is the egative of the total thermal ethalpy chage per uit mass of mixture; term o the right-had side is the chemical ethalpy chage per uit mass of mixture (products # h o f,i " # (products # o h f,i 28 14

15 hemical thermo - heatig value" By defiitio, P ( h/ T P For adeal gas, h h(t oly, thus P dh/dt or dh P dt If P is costat, the for the thermal ethalpy h 2 - h 1 P (T 2 - T 1 m P (T 2 - T 1 /m For a combustio process i which all of the ethalpy release by chemical reactio goes ito thermal ethalpy (i.e. temperature icrease i the gas, the term o the left-had side of the boxed equatio o page 27 ca be writte as # ( " # ([ h (T " h [ h (T " h P # (products m P (T reactats " T products m where is the costat-pressure specific heat averaged (somehow over all species ad averaged betwee the product ad reactat temperatures 29 hemical thermo - heatig value" Term o right-had side of boxed equatio o page 27 ca be rewritte as (products fuel # " h o f,i # (products o " h f,i # " h o f,i # o " h f,i fuel f M fuel M fuel fuel M fuel i " Last term is the chemical ethalpy chage per uit mass of fuel; defie this as -Q R, where Q R is the fuel s heatig value (products % h o f,i # o % h f,i Q R " # fuel M fuel For our stereotypical hydrocarbos, assumig O 2, 2 O ad N 2 as the oly combustio products, this ca be writte as Q R " x # h f,o 2 o o + (y 4 # h f,2 O o "1# h f, fuel 1# M fuel o " (x + y 4h f,o

16 hemical thermo - flame temperature" Now write the boxed equatio o page 27 (coservatio of eergy for combustio at costat pressure oce agai: # ( " # ([ h (T " h [ h (T " h # (products m We ve show that the left-had side P (T reactats " T products m ad the right-had side -fq R ; combiig these we obtai T products T reactats + fq R / P This is our simplest estimate of the adiabatic flame temperature (T products, usually we write this as based o aitial temperature (T reactats, usually writte as T thus (products # h o f,i " # # o h f,i T " + fq R / P (costat pressure combustio, T-averaged P 31 hemical thermo - flame temperature" This aalysis has assumed that there is eough O 2 to bur all the fuel, which is true for lea mixtures oly; i geeral we ca write T " + f burable Q R P where for lea mixtures, f burable is just f (fuel mass fractio whereas for rich mixtures, with some algebra it ca be show that # 1" f & f burable f stoichiometric % ( 1" f stoichiometric ' thus i geeral we ca 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 32 16

17 hemical thermo - flame temperature" For costat-volume combustio (istead of costat pressure, everythig is the same except u cost, ot h cost, thus the term o the left-had side of the boxed equatio o page 27 must be re-writte as ([ h (T " h ' (products & # " (PV reactats " [ h (T " h ' & # ( " (PV products % ( % ( # The extra PV terms ( mrt for adeal gas adds a extra mr(t products - T reactats term, thus m P (T products " T reactats # m P (T products " T reactats " mr(t products " T reactats m m ( P " R(T products " T reactats v (T products " T reactats which meas that (agai, T products ; T reactats T T " + fq R / v (costat volume combustio, T-averaged P which is the same as for costat-pressure combustio except for the v istead of P 33 hemical thermo - flame temperature" The costat-volume adiabatic flame (product temperature o the previous page is oly valid for lea or stoichiometric mixtures; as with costat-pressure for rich mixtures we eed to cosider how much fuel ca be bured, leadig 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 Note that the ratio of adiabatic temperature rise due to combustio for costat pressure vs. costat volume is ( " T # costat v " T # ( costat P P V I practice, oe ca determie by workig backwards from a detailed aalysis; for stoichiometric 4 -air, f 0.055, Q R 50 x 10 6 J/kg, costat-pressure combustio, 2226K for T 300K, thus P 1429 J/kg-K (for other stoichiometries or other fuels P will be moderately differet P 34 17

18 Example of heatig value" Iso-octae/air mixture: (O N 2 8 O O *3.77 N 2 Q R " # # 8 (products % h o f,i # % fuel M fuel o h f,i h o f,o h o f, 2O +12.5(3.77 h o ( # 1 h o f,n 2 f, M moles(#393.5 kj/mole + 9(# (3.77(0 # (1 mole(0.114 kg/mole h o f,o (3.77 h o ( f,n 2 ( # ( 1(# ( (3.77(0 44,500 kj/kg 4.45 x 10 7 J/kg 35 ommets o heatig value" eatig values usually computed assumig that due to reactio with air all O 2, 2 O, N N 2, S SO 2, etc. If oe assumes liquid water, the result is called the higher heatig value; if oe (more realistically assumes gaseous water, the result is called the lower heatig value Most hydrocarbos have similar Q R ( x 10 7 J/kg sice the same - ad - bods are beig broke ad same -O ad -O bods are beig made Foods similar - o a dry weight basis, about same Q R for all Fruit Loops ad Shredded Wheat have same heatig value (110 kcal/oz 1.6 x 10 7 J/kg although Fruit Loops is mostly sugar whereas Shredded Wheat has oe (the above does ot costitute a commercial edorsemet Fats slightly higher tha starches or sugars Foods with (o-digestible fiber lower 36 18

19 ommets o heatig value" Acetylee is higher (4.8 x 10 7 J/kg due to - triple bod Methae is higher (5.0 x 10 7 J/kg due to high / ratio 2 is MU higher (12.0 x 10 7 J/kg due to heavy atoms Alcohols are lower (2.0 x 10 7 J/kg for methaol, 3 O due to useless O atoms - add mass but o ethalpy release 37 Example of adiabatic flame temperature" Leaso-octae/air mixture, equivalece ratio 0.8, iitial temperature 300K, average P 1400 J/kgK, average v 1100 J/kgK: Stoichiometric: (O N 2 8 O O *3.77 N 2 " FAR (actual mixture, " 0.8 FAR (stoichiometric mixture, " 1 f /(1# f " 0.8 " f " 1 /(1# f " 1 f " 1 fuel M fuel (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 N 2 (0.028 kg/mole FAR " 1 f " 1 /(1# f " /(1# " 0.8 : f " 0.8 /(1# f " % f " T & + fq R / P 300K + ( (4.45 '10 7 J /kg /(1400J /kgk 1906K (cost. P T & + fq R / V 300K + ( (4.45 '10 7 J /kg /(1100J /kgk 2345K (cost. V 38 19

20 Summary - Lecture 2" May fuels, e.g. hydrocarbos, whe chemically reacted with oxyge or other oxidizig agets, will release a large amout of ethalpy This chemical eergy or ethalpy is coverted ito thermal eergy or ethalpy, thus i a combustio process the product temperature is much higher tha the reactat temperature Oly 2 priciples are required to compute flame temperatures oservatio of each type of atom oversatio of eergy (sum of chemical + thermal but the resultig equatios required to accout for chages i compositio ad eergy ca look formidable The key properties of a fuel are its heatig value Q R ad its stoichiometric fuel mass fractio f stoichiometric The key property of a fuel/air mixture is its equivalece ratio (φ A simplified aalysis leads to T " + fq R / P (costat pressure T " + fq R / V (costat volume 39 20