AME 513
Principles of Combustion
Lecture 2
Chemical thermodynamics I – 1st Law
Outline
 Fuels - hydrocarbons, alternatives
 Balancing chemical reactions
 Stoichiometry
 Lean & rich mixtures
 Mass and mole fractions
 Chemical thermodynamics
 Why?
 1st Law of Thermodynamics applied to a chemically reacting
system
 Heating value of fuels
 Flame temperature
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Fuels & air
 Usually we employ hydrocarbon fuels, alcohols or coal
burning in air, though other possibilities include H2, CO, NH3,
CS2, H2S, etc.
 For rocket fuels that do not burn air, many possible
oxidizers exist - ASTE 470, 570 & 572 discuss these
 Why air?
 Because it’s free, of course (well, not really when you think of
all the money we’ve spent to clean up air)
 Air ≈ 0.21 O2 + 0.79 N2 (1 mole of air) or 1 O2 + 3.77 N2 (4.77
moles of air)
 Note for air, the average molecular weight is
0.21*32 + 0.79*28 = 28.9 g/mole
thus the gas constant = (universal gas constant / mole. wt.)
= (8.314 J/moleK) / (0.0289 kg/mole) = 287 J/kgK
 Also ≈ 1% argon, up to a few % water vapor depending on the
relative humidity, trace amounts of other gases, but we’ll
usually assume just O2 and N2
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Hydrocarbons
 Alkanes - single bonds between carbons - CnH2n+2, e.g. CH4, C2H6
H
C
H
H
H
H
methane
H
H
C
C
H
H
H
H
H
H
H
C
C
C
H
H
H
H
propane
ethane
 Olefins or alkenes - one or more double bonds between carbons
H
H
C
C
H
H
H
H
ethene or
ethylene
H
H
H
C
C
C
C
C
C
C
H
H
H
H
H
H
H
propene or
propylene
1, 3 butadiene
 Alkynes - one or more triple bonds between carbons - higher
heating value than alkanes or alkenes due to strained
(endothermic) bonds than alkanes or alkenes, also very reactive
H
C
C
H
ethyne or
acetylene
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Hydrocarbons
 Aromatics - one or more ring structures
H
H
C
H
H
H
C
C
C
H
H
H
C
C
H
C
H
H
H
C
C
C
C
H
C
C
C
H
C
H
C
C
C
H
C
C
C
H
C
H
C
H
H
toluene
benzene
H
H
napthalene
 Alcohols - contain one or more OH groups
H
H
C
OH
H
methanol
H
H
H
C
C
H
H
OH
ethanol
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Biofuels
 Alcohols - produced by fermentation of food crops (sugars or
starches) or cellulose (much more difficult, not an industrial
process yet)
 Biodiesel - convert vegetable oil or animal fat (which have very
high viscosity) into alkyl esters (lower viscosity) through
“transesterification” with alcohol
Methyl linoleate
Generic ester structure (R =
any organic radical, e.g. C2H5)
Ethyl stearate
Methanol + triglyceride
Glycerol+ alkyl ester
Transesterification process
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Practical fuels
 All practical fuels are
BLENDS of hydrocarbons and
sometimes other compounds
 What distinguishes fuels?
 Flash point - temperature
above which fuel vapor
pressure is flammable when
mixed with air
 Distillation curve - temp.
range over which molecules
evaporate
 Relative amounts of paraffins
vs. olefins vs. aromatics vs.
alcohols
 Amount of impurities, e.g.
sulfur
 Structure of molecules affects octane number
(gasoline) or cetane number
(Diesel)
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Gasoline - typical composition
Benzene
Toluene
J. Burri et al., Fuel, Vol. 83, pp. 187 - 193 (2004)
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Practical fuels - properties
 Values NOT unique because
 Real fuels are a mixture of many molecules, composition varies
 Different testing methods & definitions
Property
Jet-A
Diesel
Gasoline
Heating value (MJ/kg)
43
43
43
Flash point (˚C) (T at which vapor
makes flammable mixture in air)
38
70
-43
Vapor pressure (at 100˚F) (psi)
0.03
0.02
8
Freezing point (˚C)
−40
-38
-40
Autoignition temperature (˚C) (T at
which fuel-air mixture will ignite
spontaneously without spark or flame)
210
240
260
Density (at 15˚C) (kg/m3)
810
850
720
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Practical fuels - properties
http://www.afdc.energy.gov/afdc/pdfs/fueltable.pdf
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Practical fuels
What doesn’t distinguish one fuel from another?
 Energy content (except for fuels containing alcohols, which are
lower)
Examples
 Gasoline - low-T distillation point, easy to vaporize, need high
octane number; reformulated gasoline contains alcohols
 Diesel - high-T distillation point, hard to vaporize, need LOW
octane number for easy ignition once fuel is inject
 Jet fuel - medium-T distillation point; need low freezing T since it
will be used at high altitude / low T
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Stoichiometry
 Balancing of chemical reactions with “known” (assumed) products
 Example: methane (CH4) in air (O2 + 3.77N2)
CH4 + a(O2 + 3.77N2)  b CO2 + c H2O + d N2
(how do we know this know this set is reasonable? From 2nd
Law, to be discussed later)
Conservation of C, H, O, N atoms:
nCH4(1) + nO2(0) + nN2(0) = nCO2(b) + nH2O(0) + nN2(0)
nCH4(4) + nO2(0) + nN2(0) = nCO2(0) + nH2O(2c) + nN2(0)
nCH4(0) + nO2(2a) + nN2(0) = nCO2(2b) + nH2O(c) + nN2(0)
nCH4(0) + nO2(0) + nN2(3.77*2a) = nCO2(0) + nH2O(0) + nN2(2d)
Solve: a = 2, b = 1, c = 2, d = 7.54
CH4 + 2(O2 + 3.77N2)  1 CO2 + 2 H2O + 7.54 N2
or in general
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Stoichiometry
 This is a special case where there is just enough fuel to combine
with all of the air, leaving no excess fuel or O2 unreacted; this is
called a stoichiometric mixture
 In general, mixtures will have excess air (lean mixture) or excess
fuel (rich mixture)
 This development assumed air = O2 + 3.77 N2; for lower or higher %
O2 in the atmosphere, the numbers would change accordingly
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Stoichiometry
 Fuel mass fraction (f)
f=
nfuel M fuel
fuel mass
1×(12x +1y)
=
=
total mass nfuel M fuel + nO2 M O2 + nN2 M N2 1×(12x +1y)+ (x + y 4)×(32+ 3.77× 28)
ni = number of moles of species i, Mi = molecular weight of species i
For the specific case of stoichiometric methane-air (x = 1, y = 4),
f = 0.0550; a lean/rich mixture would have lower/higher f
 For stoichiometric mixtures, f is similar for most
hydrocarbons but depends on the C/H ratio = x/y, e.g.
 f = 0.0550 for CH4 (methane) - lowest possible C/H ratio
 f = 0.0703 for C6H6 (benzene) or C2H2 (acetylene) - high C/H ratio
 Fuel mole fraction Xf
n fuel
fuel moles
1
Xf =
=
=
total moles n fuel + nO2 + n N 2 1+ (x + y 4)
which varies a lot depending on x and y (i.e. much smaller
for big molecules with large x and y)
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Stoichiometry
 Fuel-to-air ratio (FAR)
FAR =
fuel mass
fuel mass
(fuel mass)/(total mass)
f
=
=
=
air mass total mass - fuel mass 1 - (fuel mass)/(total mass) 1 - f
and air-to-fuel ratio (AFR) = 1/(FAR)
 Note also f = FAR/(1+FAR)
 Equivalence ratio ()
f=
FAR (actual mixture)
FAR (stoichiometric mixture)
 < 1: lean mixture;  > 1: rich mixture
 What if we assume more products, e.g.
CH4 + ?(O2 + 3.77N2)  ? CO2 + ? H2O + ? N2 + ? CO
In this case we have 4 atom constraints (1 each for C, H, O, and N
atoms) but 5 unknowns (5 question marks) - how to solve?
Need chemical equilibrium (discussed later) to decide how much
C and O are in the form of CO2 vs. CO
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Fuel properties
Heating value, QR (J/kg)
f at stoichiometric
Gasoline
43 x 106
0.0642
Methane
50 x 106
0.0550
Methanol
20 x 106
0.104
Ethanol
27 x 106
0.0915
Coal
34 x 106
0.0802
Paper
17 x 106
0.122
Fruit Loops
16 x 106
Probably about the same as paper
Hydrogen
120 x 106
0.0283
U235 fission
83,140,000 x 106
1
Pu239 fission
83,610,000 x 106
1
2H
339,00,000 x 106
Fuel
+ 3H fusion
2H
: 3H = 1 : 1
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Chemical thermodynamics - introduction
 Besides needing to know how to balance chemical
reactions, we need to determine how much internal energy
or enthalpy is released by such reactions and what the final
state (temperature, pressure, mole fractions of each species)
will be
 What is highest temperature flame? H2 + O2 at  = 1? Nope,
T = 3079K at 1 atm for reactants at 298K
 Probably the highest is diacetylnitrile + ozone
C4N2 + (4/3)O3  4 CO + N2
T = 5516K at 1 atm for reactants at 298K
 Why should it? The H2 + O2 system has much more energy
release per unit mass of reactants, but still a much lower
flame temperature
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Chemical thermodynamics - introduction
 The problem is that the products are NOT just H2O, that is, we
don’t get
H2 + (1/2)O2  H2O
but rather
H2 + (1/2)O2  0.706 H2O + 0.062 O2 + 0.184 H2
+ 0.094 H + 0.129 OH + 0.040 O
i.e. the water dissociates into the other species
 Dissociation does 2 things that reduce flame temp.
 More moles of products to soak up energy (1.22 vs. 1.00)
 Energy is required to break the H-O-H bonds to make the other species
 Higher pressures will reduce dissociation - Le Chatelier’s principle:
When a system at equilibrium is subjected to a stress, the system shifts
toward a new equilibrium condition in such as way as to reduce the
stress
(more pressure, less space, system responds by reducing number
of moles of gas to reduce pressure)
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Chemical thermodynamics - introduction
 Actually, even if we somehow avoided dissociation, the H2 O2 flame would be only 4998K - still not have as high a flame
temp. as the weird C4N2 flame
 Why? H2O is a triatomic molecule - more degrees of freedom
(DOFs) (i.e. vibration, rotation) than diatomic gases; each
DOF adds to the molecule’s ability to store energy
 So why is the C4N2 - O3 flame so hot?
 CO and N2 are diatomic gases - fewer DOFs
 CO and N2 are very stable even at 5500K - almost no dissociation
 O3 decomposes exothermically to (3/2)O2
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Chemical thermodynamics - goals
 Given an initial state of a mixture (temperature, pressure,
composition), and an assumed process (constant pressure,
volume, or entropy, usually), find the final state of the
mixture
 Three common processes in engine analysis
 Compression
»
»
»
»
Usually constant entropy (isentropic)
Low P / high V to high P / low V
Usually P or V ratio prescribed
Usually composition assumed “frozen” - if it reacted before
compression, you wouldn’t get any work out!
 Combustion
» Usually constant P or v assumed
» Composition MUST change (obviously…)
 Expansion
» Opposite of compression
» May assume frozen (no change during expansion) or equilibrium
composition (mixture shifts to new composition after expansion)
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Chemical thermo - assumptions
 Ideal gases - note many “flavors” of the ideal gas law
•
•
•
•
PV = nT
PV = mRT
Pv = RT
P = RT
P = pressure (N/m2); V = volume (m3); n = number of moles of gas;
 = universal gas constant (8.314 J/moleK); T = temperature (K)
m = mass of gas (kg); R = mass-specific gas constant = /M
M = gas molecular weight (kg/mole); v = V/m = specific volume (m3/kg)
 = 1/v = density (kg/m3)




Adiabatic
Kinetic and potential energy negligible
Mass is conserved
Combustion process is constant P or V (constant T or s
combustion isn’t very interesting!)
 Compression/expansion is reversible & adiabatic
( isentropic, dS = 0)
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Chemical thermodynamics - 1st Law
 1st Law of thermodynamics (conservation of energy),
control mass: dE = Q - W
 E = U + PE + KE = U + 0 + 0 = U
 W = PdV
 Combine: dU + PdV = 0
 Constant pressure: add VdP = 0 term
 dU + PdV + VdP = 0  d(U+PV) = 0  dH = 0
 Hreactants = Hproducts
 Recall h  H/m (m = mass), thus hreactants = hproducts
 Constant volume: PdV = 0




dU + PdV + VdP = 0  d(U) = 0
Ureactants = Uproducts, thus ureactants = uproducts
h = u + Pv, thus (h - Pv)reactants = (h - Pv)products
Most property tables report h not u, so h - Pv form is useful
 New twist: h or u must include BOTH thermal and chemical
contributions!
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Chemical thermodynamics - 1st Law
 Enthalpy of a mixture (sum of thermal and chemical terms)
(n (no subscripts) = number of species; ni = number of moles of i)
• [ h˜ (T) - h˜298 ]i = enthalpy to raise i from temperature of 298 to T (thermal enthalpy)
n
• m = mass of mixture = å n i M i ; M i = molecular weight of i
i=1
Combine all these to form
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Chemical thermodynamics - 1st Law
 Note we can also write h as follows
ni
Moles of i
=
= Mole fraction of i = X i
nT Total moles of all gases
 Use these boxed expressions for h & u with h = constant (for constant P
combustion) or u = constant (for constant V combustion)
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Chemical thermodynamics - 1st Law
 Examples of tabulated data on h(T) - h298, hf, etc.
(double-click table to open Excel spreadsheet with all data for CO,
O, CO2, C, O2, H, OH, H2O, H2, N2, NO at 200K - 6000K)
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Chemical thermodynamics - 1st Law
 Example: what are h and u for a CO2-O2-CO at 10 atm, 2500K
with XCO = 0.0129, XO2 = 0.3376, XCO2 = 0.6495?
Pressure doesn’t affect h or u but T does; from the tables:
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Chemical thermodynamics - 1st Law
 Final pressure (for constant volume combustion)
Â
PV = mRT, R = ; Â = universal gas constant = 8.314 J/moleK
M
n
ni M i
å
Total mass i=1
M (for mixture) =
= n
Total moles
åni
i=1
Constant volume combustion : V = constant, m = constant
n (products)
Pproducts
Combine :
=
Preactants
ån
i
i=1
n (reactants)
ån
Tproducts
Treactants
i
i=1
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Chemical thermo - heating value
 Constant-pressure energy conservation equation (no heat
transfer, no work transfer other than PdV work)
Denominator = m = constant, separate chemical and thermal terms:
 Term on left-hand side is the negative of the total thermal enthalpy
change per unit mass of mixture; term on the right-hand side is
the chemical enthalpy change per unit mass of mixture
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Chemical thermo - heating value
 By definition, CP  (∂h/∂T)P
 For an ideal gas, h = h(T) only, thus CP = dh/dT or dh = CPdT
 If CP is constant, then for the thermal enthalpy
h2 - h1 = CP(T2 - T1) = mCP(T2 - T1) /m
 For a combustion process in which all of the enthalpy release by
chemical reaction goes into thermal enthalpy (i.e. temperature
increase) in the gas, the term on the left-hand side of the boxed
equation on page 27 can be written as
n (reactants)
å
i=1
(
)
n i [ h˜ (T) - h˜298 ]i -
n (products)
å n ([h˜ (T) - h˜
i
i=1
n (reactants)
ån M
i
]
298 i
)
mCP (Treactants - Tproducts )
=
m
i
i=1
where CP is the constant-pressure specific heat averaged
(somehow) over all species and averaged between the product
and reactant temperatures
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Chemical thermo - heating value
 Term on right-hand side of boxed equation on page 27 can be rewritten as
 Last term is the chemical enthalpy change per unit mass of fuel;
define this as -QR, where QR is the fuel’s heating value
 For our stereotypical hydrocarbons, assuming CO2, H2O and N2 as
the only combustion products, this can be written as
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Chemical thermo - flame temperature
 Now write the boxed equation on page 27 (conservation of energy
for combustion at constant pressure) once again:
mCP (Treactants - Tproducts )
 We’ve shown that the left-hand side =
m
and the right-hand side = -fQR; combining these we obtain
Tproducts = Treactants + fQR /CP
 This is our simplest estimate of the adiabatic flame temperature
(Tproducts, usually we write this as Tad) based on an initial
temperature (Treactants, usually written as T∞) thus
Tad = T¥ + fQR /CP
(constant pressure combustion,
T-averaged CP)
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Chemical thermo - flame temperature
 This analysis has assumed that there is enough O2 to burn all the
fuel, which is true for lean mixtures only; in general we can write
Tad = T¥ + f burnable
QR
CP
where for lean mixtures, fburnable is just f (fuel mass fraction)
whereas for rich mixtures, with some algebra it can be shown that
f burnable
æ
ö
1- f
= f stoichiometric ç
÷
1f
è
stoichiometric ø
thus in general we can write
QR
(if f £ f stoichiometric )
CP
æ
ö QR
1- f
Tad = T¥ + f stoichiometric ç
(if f ³ f stoichiometric )
÷
è1- f stoichiometric ø CP
Tad = T¥ + f
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Chemical thermo - flame temperature
 For constant-volume combustion (instead of constant pressure),
everything is the same except u = const, not h = const, thus the term on
the left-hand side of the boxed equation on page 27 must be re-written as
ù
én (reactants)
ù én (products)
˜
˜
˜
˜
ê å n i [ h(T) - h298 ]i - (PV ) reactantsú - ê å n i [h (T) - h298 ]i - (PV ) productsú
ë i=1
û ë i=1
û
(
)
(
)
n (reactants)
ån M
i
i
i=1
The extra PV terms (= mRT for an ideal gas) adds an extra mR(TproductsTreactants) term, thus
which means that (again, Tproducts = Tad; Treactants = T∞)
Tad = T¥ + fQR /Cv (constant volume combustion, T-averaged CP)
which is the same as for constant-pressure combustion except for the Cv
instead of CP
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Chemical thermo - flame temperature
 The constant-volume adiabatic flame (product) temperature on the
previous page is only valid for lean or stoichiometric mixtures; as
with constant-pressure for rich mixtures we need to consider how
much fuel can be burned, leading to
QR
Tad = T¥ + f
(if f < f stoichiometric )
Cv
æ
ö QR
1- f
Tad = T¥ + f stoichiometric ç
(if f > f stoichiometric )
÷
è1 - f stoichiometric ø Cv
 Note that the ratio of adiabatic temperature rise due to
combustion for constant pressure vs. constant volume is
(Tad - T¥ ) constant v CP
=
=g
(Tad - T¥ ) constant P CV
 In practice, one can determine CP by working backwards from a
detailed analysis; for stoichiometric CH4-air, f = 0.055, QR = 50 x
106 J/kg, constant-pressure combustion, Tad = 2226K for T∞ =
300K, thus CP ≈ 1429 J/kg-K (for other stoichiometries or other
fuels CP will be moderately different)
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Example of heating value
 Iso-octane/air mixture:
C8H18 + 12.5(O2 + 3.77N2)  8 CO2 + 9 H2O + 12.5*3.77 N2
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Comments on heating value
 Heating values usually computed assuming that due to
reaction with air all C  CO2, H  H2O, N  N2, S  SO2, etc.
 If one assumes liquid water, the result is called the higher
heating value; if one (more realistically) assumes gaseous
water, the result is called the lower heating value
 Most hydrocarbons have similar QR (4.0 - 4.5 x 107 J/kg)
since the same C-C and C-H bonds are being broken and
same C-O and H-O bonds are being made
 Foods similar - on a dry weight basis, about same QR for all
 Fruit Loops™ and Shredded Wheat™ have same “heating
value” (110 kcal/oz = 1.6 x 107 J/kg) although Fruit Loops™ is
mostly sugar whereas Shredded Wheat™ has none (the above
does not constitute a commercial endorsement)
 Fats slightly higher than starches or sugars
 Foods with (non-digestible) fiber lower
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Comments on heating value




Acetylene is higher (4.8 x 107 J/kg) due to C-C triple bond
Methane is higher (5.0 x 107 J/kg) due to high H/C ratio
H2 is MUCH higher (12.0 x 107 J/kg) due to “heavy” C atoms
Alcohols are lower (2.0 x 107 J/kg for methanol, CH3OH) due
to “useless” O atoms - add mass but no enthalpy release
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Example of adiabatic flame temperature
 Lean iso-octane/air mixture, equivalence ratio 0.8, initial temperature
300K, average CP = 1400 J/kgK, average Cv = 1100 J/kgK:
Stoichiometric: C8H18 + 12.5(O2 + 3.77N2)  8 CO2 + 9 H2O + 12.5*3.77 N2
f=
f
/(1- f f =0.8 )
FAR (actual mixture, f = 0.8)
= f =0.8
= 0.8
FAR (stoichiometric mixture, f = 1)
f f =1 /(1- f f =1 )
ff =1 =
n fuel M fuel
n (reactants)
ån M
i
i
i=1
=
(1 mole C 8 H18 )(0.114 kg/mole)
(1 mole C8 H18 )(0.114 kg/mole) + (12.5 mole O 2 )(0.032 kg/mole) + (12.5 * 3.77 mole N 2 )(0.028 kg/mole)
= 0.06218
FARf =1 = ff =1 /(1- f f =1 ) = 0.06218 /(1- 0.06218) = 0.06630
f = 0.8 :
ff =0.8 /(1- f f =0.8 )
= 0.8 Þ ff =0.8 = 0.0504
0.06630
Tad = T¥ + fQR /CP = 300K + (0.05054)(4.45 ´10 7 J /kg) /(1400J /kgK) = 1906K (const. P)
Tad = T¥ + fQR /CV = 300K + (0.05054)(4.45 ´10 7 J /kg) /(1100J /kgK) = 2345K (const. V)
AME 513 - Fall 2012 - Lecture 2 - Chemical thermodynamics 1
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Summary - Lecture 2
 Many fuels, e.g. hydrocarbons, when chemically reacted with oxygen
or other oxidizing agents, will release a large amount of enthalpy
 This chemical energy or enthalpy is converted into thermal energy or
enthalpy, thus in a combustion process the product temperature is
much higher than the reactant temperature
 Only 2 principles are required to compute flame temperatures
 Conservation of each type of atom
 Conversation of energy (sum of chemical + thermal)
… but the resulting equations required to account for changes in
composition and energy can look formidable
 The key properties of a fuel are its heating value QR and its
stoichiometric fuel mass fraction fstoichiometric
 The key property of a fuel/air mixture is its equivalence ratio ()
 A simplified analysis leads to
Tad = T¥ + fQR /CP (constant pressure)
Tad = T¥ + fQR /CV (constant volume)
AME 513 - Fall 2012 - Lecture 2 - Chemical thermodynamics 1
39