Review -1 School of Aerospace Engineering Copyright © 2004-2005 by Jerry M. Seitzman. All rights...
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Transcript of Review -1 School of Aerospace Engineering Copyright © 2004-2005 by Jerry M. Seitzman. All rights...
AE/ME 6766 CombustionReview -1
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
AE/ME 6766 CombustionAE/ME 6766 Combustion: : Review of Chemical Review of Chemical Thermodynamics Thermodynamics
Methane Flame
0
0.05
0.1
0.15
0.2
0 0.1 0.2 0.3Distance (cm)
Mo
le F
ract
ion
0
500
1000
1500
2000
2500
Tem
per
atu
re (
K)
CH4
H2O
HCO x 1000
Temperature
AE/ME 6766 CombustionReview -2
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Chemical ThermodynamicsChemical Thermodynamics• Calculation of final state based on
some information about initial state– assuming thermodynamic
equilibrium achieved
• Final state properties– temperature (adiabatic flame temperature)
• conservation of energy, 1st law
– equilibrium composition• mass (atom) conservation
• entropy, 2nd law
Reactants Products
AE/ME 6766 CombustionReview -3
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Why Study Thermodynamic Equilibrium?Why Study Thermodynamic Equilibrium?• Equilibrium considers where
reaction is heading to if given enough time– i.e., what levels are species
concentrations, temperatures being “pulled” toward
• Kinetics (later) considers chemical rates– i.e., how fast a reaction occurs
as it tends toward equilibrium
x or t
Teq
Yeq
chem
Y, T
Init
ial
AE/ME 6766 CombustionReview -4
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
RefresherRefresher• Specific heats:
– u=internal energy (kJ/kg)– H=enthalpy (kJ/kg)
• Enthalpy:
• Perfect gas mixtures:– Yi=mass fraction=mi/m– i=mole fraction=Ni/N
( )p p
hC T
T
( )v v
uC T
T
( ) / ( ) ( )ref
Tof ref p
T
h T u p h T C d
i ii
h hY
i ii
h h
Enthalpy of formation (energy in chemical bonds)
Sensible enthalpy
AE/ME 6766 CombustionReview -5
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Specific Heat CharacteristicsSpecific Heat Characteristics
• Specific heats of non-monotonic gases increase with temperature due to excitation of internal energy modes
10’s K Few 1000K
3/2
5/2
7/2Cv/R
Monatomics (He, Ar,..)
Diatomics (N2,O2) Dissociation, ionization
AE/ME 6766 CombustionReview -6
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Specific Heat Characteristics (2)Specific Heat Characteristics (2)
Reproduced from Turns, An Introduction to Combustion, 2000
AE/ME 6766 CombustionReview -7
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
StoichiometryStoichiometry• Stoichiometric quantity of oxidizer is amount
required to completely burn a quantity of fuel• Fuel-oxidizer ratio, f
– sometimes mass fuel/mass oxidizer– or moles fuel/moles oxidizer
– Equivalence ratio (or ) = factual/fstoichiometric
= 1; stoichiometric– just enough oxidizer to
completely consume fuel
< 1; fuel lean (excess ox.) > 1; fuel rich (excess fuel)
AE/ME 6766 CombustionReview -8
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Adiabatic Combustion TemperatureAdiabatic Combustion Temperature• Equilibrium temperature that would be
achieved if reactants wereconverted to equilibrium products without heat addition or loss– energy conservation (1st Law)
provides one equation– need 2nd condition to fix state 2
• Adiabatic Flame Temperature (Tad)
– p constant
• Constant Volume
Reactants Products
Reactants(1)
Products(2)
outin WEQE 21
21 HH
0
21 EE
AE/ME 6766 CombustionReview -9
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
CalculationCalculation• First law:
• Define:
• Final Result:
• If product and reactant Cp’s are equal:
1 2
, , , ,( ( ) ) ( ( ) )ref ref
T To o
i f i p i j f j p ji jT T
Y h C d Y h C d
tani i j j
i reac ts j products
hY h Y
, ,o o
i f i j f ji j
h Yh Y h 2
,
,2
( )ref
T
j p jj T
p aveprodref
Y C d
CT T
1
,
,1
( )ref
T
i p ii T
p avereacref
Y C d
CT T
, ,2 1
, , ,
(1 )p avereac p avereacref
p aveprod p aveprod p aveprod
C C hT T T
C C C
2 1,p aveprod
hT T
C
AE/ME 6766 CombustionReview -10
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Heating Value (HV)Heating Value (HV)
• Heating value represents the amount of chemical to thermal energy conversion from burning one unit of fuel (in defined oxidizer)
– HHV – water in products is liquid
– LHV – water in products is gas
– Typical LHV in air• CH4 = 50,016 kJ/kg
• C3H8=50,368 kJ/kg
• H2 = 120,923 kJ/kg
• CO = 10,107 kJ/kg
/r fuelh h Y
AE/ME 6766 CombustionReview -11
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Typical Stoichiometric TTypical Stoichiometric Tadad
• Methane = 2226 K
• Acetylene = 2539 K
• Propane = 2267 K
• Hydrogen = 2390 K
• Carbon Monoxide = 2275 K
AE/ME 6766 CombustionReview -12
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
dependence of Tdependence of Tadad
• Heat release/(mass oxidizer +mass fuel) peaks near =1
– No extra fuel or air mass to heat up
– Tad usually peaks slightly rich
• CH4: =1.04
• H2: =1.07
Methane/Air (Tin=300K)
1500
1600
1700
1800
1900
2000
2100
2200
2300
0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9
Equivalence Ratio
Adi
abat
ic F
lam
e T
empe
ratu
re
(K)
320
330
340
350
360
370
380
390
400
Cp
(ca
l/kg
/K)
Hydrogen/Air (Tin=300K)
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Equivalence Ratio
Adi
abat
ic F
lam
e T
emp
era
ture
(K
)
300
350
400
450
500
550
600
Cp
(ca
l/kg
/K)
AE/ME 6766 CombustionReview -13
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
EquilibriumEquilibrium• To know Tad, you need to know product species
– to this point, we’ve assumed that they are known– However, determining these is actually a part of the solution
• Typical products: CO2, H2O, H2, CO, O, H, OH• Problem is coupled
– Temperature affects species– Species affects temperature
• As we’ll discuss, reactants never completely go to products• A+B C• Equilibrium is reached when the forward and backward
rates balance
AE/ME 6766 CombustionReview -14
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
How to Determine Equilibrium?How to Determine Equilibrium?
• Given reaction:
• Equilibrium concentrations given by:
/ ( ...) ( ..)
...
...( ) ( )u
e fG R T e f a bE F
p a bA B o
pK T e
p
... ...aA bB eE fF
, , , ,( ...) ( ...)f E f F f A f BG eg fg ag bg
AE/ME 6766 CombustionReview -15
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Observations on Observations on GGoo
• For endothermic reaction: Go > 0
• For exothermic reaction: Go < 0– Typically strongly negative, indicating that
reaction tends toward products– e.g. CO oxidation at atmospheric pressure,
T2=2000 K
• conclusions: – Mostly CO2 at equilibrium
– However, impossible to get 100% combustion efficiency, even if reaction has infinite time!
2 21/ 2 ... .CO O CO
68,100 / / .G KJ Kmol K / 63uG R Te
2 2
1/ 263CO CO O
AE/ME 6766 CombustionReview -16
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Equilibrium Hydrocarbon/Air Equilibrium Hydrocarbon/Air Combustion ProductsCombustion Products
• Major products:– Lean: CO2, H2O, O2
– Rich: CO2, CO, H2O, H2, O2
AE/ME 6766 CombustionReview -17
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Equilibrium Hydrocarbon/Air Equilibrium Hydrocarbon/Air Combustion Products (2)Combustion Products (2)
• Minor Products: – NO, OH, O, H, H2 (<1),
CO (<1)
AE/ME 6766 CombustionReview -18
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Temperature Effects on EquilibriumTemperature Effects on Equilibrium• Returning to CO oxidation example, note variation in Kp with
temperature:– Kp= 235 (T=1500 K)
= 63 (T=2000 K) = 26 (T=2500 K)
– i.e., shift toward reactants with increasing temperature• This is a general rule for exothermic reactions
– Reaction shifts toward products (reactants) with decreases (increases) in temperature
• Thus, increasing the preheat temperature will cause equilibrium to shift so that the product temperature doesn’t increase by the same amount
• Opposite result is true for endothermic reactions– e.g. O2 dissociation: increasing temperature favors O2 dissociation
AE/ME 6766 CombustionReview -19
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Pressure Effects on EquilibriumPressure Effects on Equilibrium
• Note from Kp equation that the pressure term is raised to the power:
# product moles - # reactant moles
• Equilibrium shifts to less (more) moles as pressure increases (decreases)– Nprod=Nreac: pressure independent– Nprod>Nreac
• p increases (decreases) cause shift toward reactants (products)
– Nprod<Nreac
• p increases (decreases) cause shift toward products (reactants)
AE/ME 6766 CombustionReview -20
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Pressure Effects on DissociationPressure Effects on Dissociation
• Nprod>Nreac
–
– p increases reduce dissociation
– T increases increase dissociation
2... 2 .O O
1E-12
1E-10
1E-08
1E-06
0.0001
0.01
1
1 10 100
Pressure (atm)
O M
ole
Fra
ctio
n
T=1000 K
T=2000 K
T=4000 K
AE/ME 6766 CombustionReview -21
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Pressure Effects on EquilibriumPressure Effects on EquilibriumMethane ExampleMethane Example
• Nprod=Nreac
–
– Thus, above rxn is pressure independent
– Result shows that flame temperature increases slightly with pressure, why?
4 2 2 2 2 22( 3.76 ) 2 7.52CH O N CO H O N Methane/Air (Tin=300K)
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9
Equivalence Ratio
Adi
abat
ic F
lam
e T
empe
ratu
re (
K)
p= 1 atm
p= 30 atm
AE/ME 6766 CombustionReview -22
School of Aerospace Engineering
Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved.
Le Chatelier’s PrincipleLe Chatelier’s Principle
• Provides convenient way to remember effects of pressure, temperature on equilibrium– “Given a change in conditions, reaction will shift in
such a way as to minimize the effect of this change”