Combustion Theory & Combustion Theory & Adiabatic Flame TemperatureAdiabatic Flame Temperature

Brian MooreBrian Moore

Shaun MurphyShaun Murphy

OutlineOutline

Flame TheoryFlame Theory

Combustion Chamber Chemistry Combustion Chamber Chemistry

Adiabatic Flame TemperatureAdiabatic Flame Temperature

Example ProblemExample Problem

Types of FlamesTypes of Flames

Two basic categoriesTwo basic categories– Pre-mixedPre-mixed– DiffusionDiffusion

Both characterized as Both characterized as Laminar or TurbulentLaminar or Turbulent

PremixedPremixed

Results from gaseous Results from gaseous reactants that are mixed reactants that are mixed prior to combustionprior to combustion

Flame propogates at Flame propogates at velocities slightly less velocities slightly less than a few m/sthan a few m/s

Considered constant Considered constant pressure combustionpressure combustion

Reacts quite rapidlyReacts quite rapidly

Example: Spark Ignition Engine

DiffusionDiffusion

Gaseous reactants Gaseous reactants are introduced are introduced separately and mix separately and mix during combustionduring combustion

Energy release rate Energy release rate limited by mixing limited by mixing processprocess

Reaction zone Reaction zone between oxidizer and between oxidizer and fuel zonefuel zone

                                                                                                                 

             

Example: Diesel Engine

LaminarLaminar

PremixedPremixed– Ex. Bunsen BurnerEx. Bunsen Burner– Flame moves at fairly low velocityFlame moves at fairly low velocity– Mechanically create laminar Mechanically create laminar

conditionsconditions

DiffusionDiffusion– Ex. Candle FlameEx. Candle Flame– Fuel: Wax, Oxidizer: AirFuel: Wax, Oxidizer: Air– Reaction zone between wax Reaction zone between wax

vapors and airvapors and air

TurbulentTurbulent

PremixedPremixed– Heat release occurs much fasterHeat release occurs much faster– Increased flame propagationIncreased flame propagation– No definite theories to predict No definite theories to predict

behavior behavior

DiffusionDiffusion– Can obtain high rates of Can obtain high rates of

combustion energy release per combustion energy release per unit volumeunit volume

– Ex. Diesel EngineEx. Diesel Engine– Modeling is very complex, no Modeling is very complex, no

well established approachwell established approach

Flame PropagationFlame Propagation

Initial spark causes Initial spark causes pressure wave formationpressure wave formation

Flame propagation Flame propagation considered constant considered constant pressurepressure

Burned and Unburned Burned and Unburned regionsregions

Unburned portion may Unburned portion may undergo autoignition, undergo autoignition, known as known as “Knock”“Knock”

Chemistry BasicsChemistry Basics

ReactantsReactants– Fuel: Hydro-CarbonFuel: Hydro-Carbon

Octane (COctane (C88HH1818))

– Oxidizer: Dry Air (D.A)Oxidizer: Dry Air (D.A)

21% O21% O22

79% N79% N22

1 mol O1 mol O22 → 3.76 mol N → 3.76 mol N22

ProductsProducts – COCO22

– HH22OO– NN22

Example Using ButaneExample Using Butane

CC44HH1010 + O + O22 → CO→ CO22 + H + H22OO

Balancing the EquationBalancing the EquationConservation of Mass:Conservation of Mass:

CC44HH1010 + 6.5O + 6.5O22 → 4CO→ 4CO22 + 5H + 5H22OO

Ideal Chemical Equation:Ideal Chemical Equation:

Example Cont.Example Cont.

Practical Chemical Equation:Practical Chemical Equation:

Air used as oxidizer, not pure oxygenAir used as oxidizer, not pure oxygen

CC44HH1010 + 6.5(O + 6.5(O22+3.76N+3.76N22) → 4CO) → 4CO22 + 5H + 5H22O+24.44NO+24.44N22

CC44HH1010 + 31.03D.A. → 4CO + 31.03D.A. → 4CO22 + 5H + 5H22O+ 31.03D.A -6.5OO+ 31.03D.A -6.5O22

Balancing Made EasyBalancing Made Easy

CCHH + a(O + a(O22+3.76N+3.76N22) → bCO) → bCO22 + cH + cH22O + dNO + dN22

a = a = +(+(/4) b = /4) b = c = c = d = 3.76a d = 3.76a

Combustion EnergyCombustion Energy

U = Q - WU = Q - W

Q = Q = U + WU + W

W = PW = PVV

Q = Q = U + PU + PV = V = HH

Q = HQ = Hprod prod -- H Hreactreact

EnthalpyEnthalpyEnthalpy of Formation (Enthalpy of Formation (hhff))– Energy required to form the compoundEnergy required to form the compound

Change in Enthalpy (Change in Enthalpy (h)h)– Difference in enthalpy between Product Temp. and Reference Difference in enthalpy between Product Temp. and Reference

Temp. Temp.

h = h(Th = h(Tprodprod) - h(T) - h(Trefref))

Total Enthalpy (h)Total Enthalpy (h)

h = h = hhff + + hh

H = H = nniihhii) )

Adiabatic AssumptionsAdiabatic Assumptions

No heat transfer through cylinder wallsNo heat transfer through cylinder walls

All energy transferred to engine work & All energy transferred to engine work & exhaust productsexhaust products

Allows Allows Adiabatic Flame Temperature Adiabatic Flame Temperature ((AFTAFT)) to be calculated to be calculated

Q = 0 HQ = 0 Hreact react = H= Hprodprod

Adiabatic Flame TemperatureAdiabatic Flame Temperature

Highest possible temperature that can be achieved Highest possible temperature that can be achieved during combustionduring combustion

Never achieved in practiceNever achieved in practice– No realistic combustion chamber is adiabaticNo realistic combustion chamber is adiabatic– Dissociation lowers temperatureDissociation lowers temperature– Analagous to Carnot cycle for Heat EnginesAnalagous to Carnot cycle for Heat Engines

Useful design parameterUseful design parameter– Upper limit of exhaust temp. is knownUpper limit of exhaust temp. is known

Calculation is an iterative processCalculation is an iterative process

AFT – Example CalculationAFT – Example Calculation

Problem Statement: Liquid Methane (CH4) is burned at a constant pressure. The air and fuel are supplied at 298 K and 1 atm. Determine the adiabatic flame temperature for these conditions assuming complete combustion.

1) Balance Chemical Equation

CHCH44 + 2(O + 2(O22+3.76N+3.76N22) → CO) → CO22 + 2H + 2H22O+7.52NO+7.52N22

2)2) Energy Balance and Adiabatic AssumptionsEnergy Balance and Adiabatic Assumptions

Q = 0 = HQ = 0 = Hprod prod –– HHreactreact Therefore, H Therefore, Hreact react = H= Hprod prod

Calculations ContCalculations Cont..

3)3) Determine Enthalpy of ReactantsDetermine Enthalpy of Reactants

hhf, CH4 f, CH4 = -74.81 kJ/mol (from chart) = -74.81 kJ/mol (from chart)

hhf, O2 f, O2 = = hhf, N2 f, N2 = 0= 0

HHreact react = = nniihhii) (n = # of moles)) (n = # of moles)

HHreact react = 1mol * (-74.81 kJ/mol)= 1mol * (-74.81 kJ/mol)

HHreact react = -74.81 kJ= -74.81 kJ

Calculations Cont.Calculations Cont.

4)4) Determine Enthalpies of ProductsDetermine Enthalpies of ProductsGuess value for temperature required: try 1000KGuess value for temperature required: try 1000K

hCO2 = hhf, CO2 f, CO2 + (h+ (hCO2CO2(T(Tprodprod) – h) – hCO2CO2(T(Trefref))))

hH2O = hhf, H2O f, H2O + (h+ (hH2OH2O(T(Tprodprod) – h) – hH2OH2O(T(Trefref))))

hN2 = hhf, N2 f, N2 + (h+ (hN2N2(T(Tprodprod) – h) – hN2N2(T(Trefref))))

– Use tables provided to find hUse tables provided to find h ff and and hh

Calculations Cont.Calculations Cont.

Enthalpy of Formation values:

hf,CO2 = -393.5 kJ/mol

hf,H2O = -241.8 kJ/mol

hf,N2 = 0 kJ/mol

h values:h values:

hCO2(Tprod) – hCO2(Tref) = 33.41 kJ/mol

hH2O(Tprod) – hH2O(Tref) = 25.98 kJ/mol

hN2(Tprod) – hN2(Tref) = 21.46 kJ/mol

Calculations Cont.Calculations Cont.

5)5) Total Enthalpy of each molecule: h = Total Enthalpy of each molecule: h = hhff + + hh

hhCO2 CO2 = = -393.5 kJ/mol + 33.41 kJ/mol = -360.09 kJ/mol

hhH2O H2O = = -241.8 kJ/mol + 25.98 kJ/mol = -215.82 kJ/mol

hhN2 N2 = = 0 kJ/mol + 21.46 kJ/mol = 21.46 kJ/mol

Total Enthalpy of Products:

Hprod = nniihhii))

Hprod = (1) -360.09 + (2) -215.82 + (7.5) 21.46

Hprod = -630.78 kJ

Calculations Cont.Calculations Cont.

6)6) HHprod prod << H<< Hreact react Try Higher Temperature (2300 K)Try Higher Temperature (2300 K)

hhCO2 CO2 = = -393.5 kJ/mol + 109.67 kJ/mol = -283.83 kJ/mol

hhH2O H2O = = -241.8 kJ/mol + 88.29 kJ/mol = -153.51 kJ/mol

hhN2 N2 = = 0 kJ/mol + 67.01 kJ/mol = 67.01 kJ/mol

Hprod = nniihhii))

Hprod = 1( –283.83) + 2( –153.51) + 7.5( 67.01)

Hprod = -88.28 kJ

Calculations Cont.Calculations Cont.

7)7) HHprod prod < H< Hreact react Try Higher Temperature (2400 K)Try Higher Temperature (2400 K)

hhCO2 CO2 = = -393.5 kJ/mol + 115.79 kJ/mol = -277.71 kJ/mol

hhH2O H2O = = -241.8 kJ/mol + 93.60 kJ/mol = -148.20 kJ/mol

hhN2 N2 = = 0 kJ/mol + 70.65 kJ/mol = 70.62 kJ/mol

Hprod = nniihhii))

Hprod = (1) –302.05 + (2) -169.11 + (7.5) 56.14

Hprod = -44.46 kJ

Calculations Cont.Calculations Cont.

8)8) Interpolate to find proper valueInterpolate to find proper valueTprod 2300K

74.81 kJ 88.28 kJ( )2400K 2300K

44.46 kJ 88.28 kJ( )

Find Tprod 2331K

SummarySummary

Premixed and Diffusion FlamesPremixed and Diffusion Flames– LaminarLaminar– TurbulentTurbulent

Combustion ChemistryCombustion Chemistry– Balancing Chemical equationsBalancing Chemical equations– First Law Energy BalanceFirst Law Energy Balance

Adiabatic Flame TemperatureAdiabatic Flame Temperature– AssumptionsAssumptions– DeterminationDetermination– IterationIteration

Homework ProblemHomework Problem

Problem Statement: Liquid Octane (C8H18) is burned at a constant pressure. The air and fuel are supplied at 298 K and 1 atm. Determine the adiabatic flame temperature for these conditions assuming complete combustion.