Review -1 School of Aerospace Engineering Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion AE/ME 6766 Combustion:

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




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




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




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




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




Specific Heat Characteristics (2)Specific Heat Characteristics (2)

Reproduced from Turns, An Introduction to Combustion, 2000




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)




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




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




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




Typical Stoichiometric TTypical Stoichiometric Tadad

• Methane = 2226 K

• Acetylene = 2539 K

• Propane = 2267 K

• Hydrogen = 2390 K

• Carbon Monoxide = 2275 K




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)




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




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




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




Equilibrium Hydrocarbon/Air Equilibrium Hydrocarbon/Air Combustion ProductsCombustion Products

• Major products:– Lean: CO2, H2O, O2

– Rich: CO2, CO, H2O, H2, O2




Equilibrium Hydrocarbon/Air Equilibrium Hydrocarbon/Air Combustion Products (2)Combustion Products (2)

• Minor Products: – NO, OH, O, H, H2 (<1),

CO (<1)




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




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)




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




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




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”

Documents

Review -1 School of Aerospace Engineering Copyright © 2004-2005 by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion AE/ME 6766 Combustion: