AME 436
Energy and Propulsion
Lecture 2
Fuels, chemical thermodynamics
(thru 1st Law; 2nd Law next lecture)
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
 Usually we employ hydrocarbon fuels, alcohols or hydrogen
burning in air, though other possibilities include CO, NH3, CS2,
H2S, etc.
 For rocket fuels that do not burn air, many possible oxidizers exist
- ASTE 470 discusses these - AME 436 focuses on airbreathing
devices
 Why hydrocarbons?
 Many are liquids - high density, easy to transport and store
(compared to gases, e.g. CH4), easy to feed into engine (compared
to solids)
 Lots of it in the earth (often in the wrong places)
 Relatively non-toxic fuel and combustion products
 Relatively low explosion hazards
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Air
 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 - very reactive,
higher heating value than alkanes or alkenes
H
C
C
H
ethyne or
acetylene
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Hydrocarbons
 Aromatics - one or more ring structures
H
C
H
H
H
C
C
C
H
H
C
C
H
C
C
H
C
H
H
H
C
C
H
C
C
H
H
C
C
C
C
C
H
C
H
C
H
H
toluene
benzene
H
C
H
C
C
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 one fuel
from another?
 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
(Lecture 10)
http://static.howstuffworks.com/flash/oil-refining.swf
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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
AME 436 - Spring 2016 - Lecture 2 - Chemical Thermodynamics 1
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Practical fuels - properties
http://www.afdc.energy.gov/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 alititude / 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 of products is reasonable? From 2nd
Law, to be discussed later)
Conservation of atoms:
C atoms: nCH4(1) + nO2(0) + nN2(0) = nCO2(b) + nH2O(0) + nN2(0)
H atoms: nCH4(4) + nO2(0) + nN2(0) = nCO2(0) + nH2O(2c) + nN2(0)
O atoms: nCH4(0) + nO2(2a) + nN2(0) = nCO2(2b) + nH2O(c) + nN2(0)
N atoms: 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
Fuel
2H
+ 3H fusion
339,000,000 x 106
2H
: 3H = 1 : 1
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Chemical thermodynamics - intro
 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 - intro
 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 - intro
 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?
 O3 decomposes exothermically to (3/2)O2
 CO and N2 are diatomic gases - fewer DOFs
 CO and N2 are very stable even at 5500K - almost no dissociation
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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)
AME 436 - Spring 2016 - Lecture 2 - Chemical Thermodynamics 1
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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 = 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!
AME 436 - Spring 2016 - Lecture 2 - Chemical Thermodynamics 1
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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)
AME 436 - Spring 2016 - Lecture 2 - Chemical Thermodynamics 1
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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 mixture 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:
 This scary-looking boxed equation is simply conservation of energy for
a chemically reacting mixture at constant pressure
 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 28 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 28 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
QR = -
x × Dh of ,CO2 + (y 2)× Dh of ,H 2O -1× Dh of , fuel - (x + y 4)Dh of ,O2
1× M fuel
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Chemical thermo - flame temperature
 Now write the boxed equation on page 28 (conservation of energy for
combustion at constant pressure) once again:
 We've shown that the left-hand side =
mCP (Treactants - Tproducts )
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 28 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, thusCP
≈ 1429 J/kg-K (for other stoichiometries or other fuels CP will be
somewhat but not drastically 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 are usually computed assuming 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, as we have been doing) 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™ mostly sugar,
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 higher (4.8 x 107 J/kg) because of CC triple bond
Methane higher (5.0 x 107 J/kg) because of high H/C ratio
H2 MUCH higher (12.0 x 107 J/kg) because no "heavy" C atoms
Alcohols lower (2.0 x 107 J/kg for methanol, CH3OH) because of
"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 436 - Spring 2016 - Lecture 2 - Chemical Thermodynamics 1
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Summary - Lecture 2
 Many fuels, e.g. hydrocarbons, when chemically reacted with an
oxidizer, e.g. O2, release large amounts of energy or 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
 Key thermodynamic properties of a fuel are its heating value QR
and its stoichiometric fuel mass fraction fstoichiometric
 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 436 - Spring 2016 - Lecture 2 - Chemical Thermodynamics 1
40