combustion presentation 2005
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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
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