Energy and Propulsion - University of Southern California

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AME 436

Energy and Propulsion

Lecture 2Fuels, chemical thermodynamics

(thru 1st Law; 2nd Law next lecture)

2AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1

OutlineØ Fuels - hydrocarbons, alternativesØ Balancing chemical reactions

Ø StoichiometryØ Lean & rich mixturesØ Mass and mole fractions

Ø Chemical thermodynamicsØ Why?Ø 1st Law of Thermodynamics applied to a chemically reacting systemØ Heating value of fuelsØ Flame temperature

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FuelsØ Usually we employ hydrocarbon fuels, alcohols or hydrogen

burning in air, though other possibilities include CO, NH3, CS2, H2S, etc.

Ø For rocket fuels that do not burn air, many possible oxidizers exist - ASTE 470 discusses these - AME 436 focuses on airbreathingdevices

Ø Why hydrocarbons?Ø Many are liquids - high density, easy to transport and store

(compared to gases, e.g. CH4), easy to feed into engine (compared to solids)

Ø Lots of it in the earth (often in the wrong places)Ø Relatively non-toxic fuel and combustion productsØ Relatively low explosion hazards


AirØ Why air?

Ø Because it's free, of course (well, not really when you think of all the

money we’ve spent to clean up air)

Ø Air ≈ 0.21 O2 + 0.79 N2 (1 mole of air) or 1 O2 + 3.77 N2 (4.77 moles of air)

Ø Note for air, the average molecular mass is

thus the gas constant = (universal gas constant / mole. wt.)

= (8.314 J/moleK) / (0.0289 kg/mole) = 287 J/kgK

Ø 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

0.21 Mole O2Mole total

32g O2Mole O2

+0.79 Mole N2Mole total

28g N2Mole N2

=28.9g

Mole total

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HydrocarbonsØ Alkanes - single bonds between carbons - CnH2n+2, e.g. CH4, C2H6

Ø Olefins or alkenes - one or more double bonds between carbons

Ø Alkynes - one or more triple bonds between carbons - very reactive,

higher heating value than alkanes or alkenes

H

H

H

H C

methane

H

H

H

H C

ethane

H

H

C H

H

H

H C

propane

H

H

C

H

H

C

HH

H

H C

propene orpropylene

H

C

H

C

H

H

C

ethene orethylene

H

H

CH

H

C

1, 3 butadiene

H

C

H

C

H

H

C

H C

ethyne oracetylene

HC


HydrocarbonsØ Aromatics - one or more ring structures

Ø Alcohols - contain one or more OH groups

OH

H

H

H C

methanol

OH

H

H

H C

ethanol

H

H

C

H

H

C

benzene

H

CH C

HC

CC

H

H

H

C

toluene

H

CH C

HC

CC

C H

H

H

H

C

napthalene

H

CH C

HC

CC

C

C

C

CH

H H

H

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BiofuelsØ Alcohols - produced by fermentation of food crops (sugars or

starches) or cellulose (much more difficult, not an industrial process yet)

Ø Biodiesel - convert vegetable oil or animal fat (which have very high viscosity) into alkyl esters (lower viscosity) through "transesterification" with alcohol

Methyl linoleate

Ethyl stearateGeneric ester structure (R = any organic radical, e.g. C2H5)

Methanol + triglyceride Glycerol+ alkyl esterTransesterification process


Practical fuelsØ All practical fuels are

BLENDS of hydrocarbons and other compounds

Ø What distinguishes one fuel from another?Ø Flash point - temperature

above which fuel vapor pressure is flammable when mixed with air

Ø Distillation curve - temp. range over which molecules evaporate

Ø Relative amounts of alkanes vs. alkenes vs. aromatics vs. alcohols

Ø Amount of impurities, e.g. sulfur

Ø Structure of molecules -affects octane number (Lecture 10)

http://static.howstuffworks.com/flash/oil-refining.swf

http://static.howstuffworks.com/flash/oil-refining.swf

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Gasoline - typical composition

BenzeneToluene

J. Burri et al., Fuel, Vol. 83, pp. 187 - 193 (2004)

Paraffins = alkanes


Property Jet-A Diesel Gaso-line

Ethanol Natural gas

Heating value (MJ/kg) 43 43 43 27 47

Flash point (˚C) (T at which vapor makes flammable mixture in air)

38 70 -43 13 -184

Vapor pressure (at 100˚F) (psi) 0.03 0.02 8 2.3 2400

Freezing point (˚C) −40 -38 -40 -114 -182

Autoignition temperature (˚C) (T at which fuel-air mixture will ignite spontaneously without spark or flame)

210 240 260 423 557

Density (at 15˚C) (kg/m3) 810 850 720 792 0.67

Practical fuels - propertiesØ Values NOT unique because

Ø Real fuels are a mixture of many molecules, composition variesØ Different testing methods & definitions

More info: http://www.afdc.energy.gov/pdfs/fueltable.pdf

http://www.afdc.energy.gov/pdfs/fueltable.pdf

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StoichiometryØ Balancing of chemical reactions with "known" (assumed) products Ø 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 in Lecture 3)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

CxHy + (x + y/4)(O2 + 3.77N2) ® x CO2 + (y/2) H2O + 3.77(x + y/4)N2


StoichiometryØ The previous page shows 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

Ø In general, mixtures will have excess air (lean mixture) or excess fuel (rich mixture)

Ø The analysis assumed air = O2 + 3.77 N2; for lower or higher % O2 in the atmosphere, the numbers would change accordingly

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StoichiometryØ Fuel mass fraction (f)

ni = number of moles of species i, Mi = molecular mass 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 fØ For stoichiometric mixtures, f is similar for most hydrocarbons but

depends on the C/H ratio = x/y, e.g.

Ø f = 0.0550 for CH4 (methane) - lowest possible C/H ratio

Ø f = 0.0703 for C6H6 (benzene) or C2H2 (acetylene) - high C/H ratio

Ø Fuel mole fraction Xf

which varies a lot depending on x and y (i.e., much smaller for big

molecules with large x and y)

f = fuel masstotal mass

=nfuelM fuel

n fuelM fuel + nO2MO2

+ nN2MN2

=1⋅ (12x +1y)

1⋅ (12x +1y)+ (x + y 4) ⋅ (32+3.77 ⋅28)

X f =fuel molestotal moles

=nfuel

n fuel + nO2+ nN2

= 11+ 4.76(x + y 4)


StoichiometryØ Fuel-to-air ratio (FAR)

and air-to-fuel ratio (AFR) = 1/(FAR)Ø Note also f = FAR/(1+FAR)Ø Equivalence ratio (f)

f < 1: lean mixture; f > 1: rich mixtureØ What if we assume more products, e.g.

CH4 + ?(O2 + 3.77N2) ® ? CO2 + ? H2O + ? N2 + ? COIn 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 (Lecture 3) to decide how much C and O

are in the form of CO2 vs. CO vs. H2O

€

FAR =fuel massair mass

=fuel mass

total mass - fuel mass=

(fuel mass)/(total mass)1 - (fuel mass)/(total mass)

=f

1 - f

€

φ =FAR (actual mixture)

FAR (stoichiometric mixture)

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Chemical thermodynamics - introØ 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

Ø What is highest temperature flame? H2 + O2 at f = 1? Nope, T = 3079K at 1 atm for reactants at 298K

Ø Probably the highest is diacetylnitrile + ozoneC4N2 + (4/3)O3 ® 4 CO + N2T = 5516K at 1 atm for reactants at 298K

Ø Why should it? The H2 + O2 system has much more energy release per unit mass of reactants, but still a much lower flame temperature


Chemical thermodynamics - introØ The reason is that the product is NOT just H2O, i.e. we don't get

H2 + (1/2)O2 ® H2Obut rather

H2 + (1/2)O2 ® 0.706 H2O + 0.062 O2 + 0.184 H2

+ 0.094 H + 0.129 OH + 0.040 Oi.e. the water dissociates into the other species (how do we know how much of the other species? Wait for Lecture 3 …)

Ø Dissociation does 2 things that reduce flame temp.Ø More moles of products to soak up energy (1.22 vs. 1.00)Ø Energy required to break H-O-H bonds to make the other species

Ø Higher pressures reduce dissociation - Le Chatelier's principle:

When a system at equilibrium is subjected to a stress, the system shifts toward a new equilibrium condition so 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 - introØ 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

Ø 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

Ø So why is the C4N2 - O3 flame so hot?Ø O3 decomposes exothermically to (3/2)O2Ø The products CO and N2 are diatomic gases - fewer DOFsØ CO and N2 are very stable even at 5500K - almost no dissociation


Chemical thermodynamics - goalsØ 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

Ø Three common processes in engine analysis

Ø Compression

» Usually constant entropy S (isentropic)

• Actually, reversible and adiabatic; since dS ≥ dQ/T with = sign applying for

reversible and dQ = 0 for adiabatic, dS = 0

» 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 output!

Ø Combustion

» Usually constant P or V assumed

» Composition MUST change (obviously…)

Ø 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 thermodynamics - assumptionsØ Ideal gases - note many "flavors" of the ideal gas law

• PV = nÂT• PV = mRT• Pv = RT• P = rRT – most useful form in this course; more useful to work with mass

than moles, because moles are not conserved in chemical reactions!

P = pressure (N/m2); V = volume (m3); n = number of moles of gasÂ = universal gas constant (8.314 J/moleK); T = temperature (K)m = mass of gas (kg); R = mass-specific gas constant = Â/MM = gas molecular mass (kg/mole); v = V/m = specific volume (m3/kg)r = 1/v = density (kg/m3)

Ø AdiabaticØ Kinetic and potential energy negligible (we’ll revisit this assumption for

hypersonic propulsion)Ø Mass is conservedØ Combustion process is constant P or V (constant T or s combustion

isn't very interesting!)Ø Compression/expansion is reversible & adiabatic

(Þ isentropic, dS = 0)


Chemical thermodynamics - 1st LawØ 1st Law of thermodynamics (conservation of energy), control

mass: dE = dQ - dWØ E = U + PE + KE = U + 0 + 0 = U Ø dW = PdV for a simple compressible substance; also assume

adiabatic (dQ = 0)Ø Combine: dU + PdV = 0Ø Constant pressure: add VdP = 0 term

Ø dU + PdV + VdP = 0 Þ d(U+PV) = 0 Þ dH = 0Ø Hreactants = HproductsØ Recall h º H/m (m = mass), thus hreactants = hproducts

Ø Constant volume: PdV = 0 Ø dU + PdV = 0 Þ d(U) = 0 Ø Ureactants = Uproducts, thus ureactants = uproductsØ h = u + Pv, thus (h - Pv)reactants = (h - Pv)productsØ Most property tables report h not u, so h - Pv form is useful

Ø h or u must include BOTH thermal and chemical contributions!

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Chemical thermodynamics - 1st LawØ Enthalpy of a mixture (sum of thermal and chemical terms)

(1) H = ni !hii=1

N

∑(2) !hi = enthalpy of i per mole of i = [ !h(T )− !h298]i + Δ !hf ,i

o

(N = number of species; ni = number of moles of i)

[ !h(T )− !h298]i = enthalpy per mole of i to raise i from 298K to T (thermal enthalpy)

Δ !hf ,io = enthalpy of formation per mole of i at 298K & 1 atm, i.e. enthalpy change

from formation of i from its elements in their standard state (chemical enthalpy) Note Δ !hf ,i

o = 0 for elements in their standard state, e.g. O2(gas), C(solid)

(3) m = mass of mixture = niMii=1

N

∑ ; Mi = molecular mass of i

Combine (1)− (3) to obtain

h = Hm

=ni [ !h(T )− !h298]i + Δ !hf ,i

o( )i=1

N

∑

niMii=1

N

∑


Chemical thermodynamics - 1st LawØ Note we can also write h as follows

Ø Use boxed expressions for h & u with h = constant (for constant P combustion) or u = constant (for constant V combustion)

h = Hm

=ni [ !h(T )− !h298]i + Δ !hf ,i

o( )i=1

N

∑

niMii=1

N

∑=

ninT

[ !h(T )− !h298]i + Δ !hf ,io( )

i=1

N

∑ninTMi

i=1

N

∑

ninT

= Moles of iTotal moles of all gases

= Mole fraction of i = Xi

h = Hm

=Xi [ !h(T )− !h298]i + Δ !hf ,i

o( )i=1

N

∑

XiMii=1

N

∑

u = Um

= H − PVm

= H −mRTm

= h− RT

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Chemical thermodynamics - 1st LawØ Examples of tabulated data on h(T) - h298, Dhf, 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)

O2

0.000

T s h-h_298K J/mole-K kJ/mole

200 193.376 -2.866298 205.033 0.000300 205.213 0.054400 213.765 3.029500 220.589 6.088600 226.346 9.247700 231.363 12.502800 235.814 15.841900 239.827 19.2461000 243.475 22.7071100 246.818 26.2171200 249.906 29.7651300 252.776 33.3511400 255.454 36.9661500 257.969 40.6101600 260.337 44.2791700 262.575 47.9701800 264.701 51.6891900 266.726 55.4342000 268.655 59.1992100 270.504 62.9862200 272.278 66.8022300 273.981 70.6342400 275.625 74.4922500 277.207 78.375

Molecular weight = 31.99879 g/moleΔhf

o (kJ/mole)

COMolecular weight = 28.01054 g/moleΔhf

o (kJ/mole) -110.541


200 185.916 -2.858298 197.543 0.000300 197.723 0.054400 206.125 2.975500 212.719 5.929600 218.204 8.941700 222.953 12.021800 227.162 15.175900 230.957 18.3971000 234.421 21.6861100 237.609 25.0331200 240.563 28.4261300 243.316 31.8651400 245.889 35.3381500 248.312 38.8481600 250.592 42.3841700 252.751 45.9401800 254.797 49.5221900 256.743 53.1242000 258.600 56.7392100 260.370 60.3752200 262.065 64.0192300 263.692 67.6762400 265.253 71.3462500 266.755 75.023

CO2

-393.522


200 199.865 -3.414298 213.685 0.000300 213.915 0.067400 225.225 4.008500 234.814 8.314600 243.199 12.916700 250.663 17.761800 257.408 22.815900 263.559 28.0411000 269.215 33.4051100 274.445 38.8941200 279.307 44.4841300 283.847 50.1581400 288.106 55.9071500 292.114 61.7141600 295.901 67.5801700 299.482 73.4921800 302.884 79.4421900 306.122 85.4292000 309.210 91.4502100 312.160 97.5002200 314.988 103.5752300 317.695 109.6712400 320.302 115.7882500 322.808 121.926

Molecular weight = 44.00995 g/moleΔhf

o (kJ/mole)

24

Ø Example: what are h and u for a CO-O2-CO2 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:

h =Xi [ !h(T )− !h298]i + Δ !hf ,i

o( )i=1

N

∑

XiMii=1

N

∑= 0.0129(75.023−110.541)+ 0.3376(78.375+ 0)+ 0.6495(121.926− 393.522)

0.0129(0.02801)+ 0.3376(0.03200)+ 0.6495(0.04401)kJ / molekg / mole

h = −3784 kJkg

= −3.784×106 Jkg

R = ℜM

; M = XiMii=1

N

∑ = 0.0129(0.02801)+ 0.3376(0.03200)+ 0.6495(0.04401) = 0.03975 kgmole

⇒ R = 8.314JmoleK

0.03975 kgmole

= 209.2JkgK

u = h− RT = −3.784×106 Jkg

− 209.2JkgK

(2500K ) = −4.307 ×106 Jkg

AME 436 - Spring 2019 - Lecture 2 - Chemical Thermodynamics 1

Chemical thermodynamics - 1st Law

€

MCO= 0.02801; MO2= 0.03200; MCO2

= 0.04401 kg /mole

Δ ˜ h f ,COo = −110.541; Δ ˜ h f ,O2

o = 0; Δ ˜ h f ,CO2

o = −393.522 kJ /mole

[ ˜ h (2500) − ˜ h 298]CO = 75.023; [ ˜ h (2500) − ˜ h 298]O2= 78.375; [ ˜ h (2500) − ˜ h 298]CO2

=121.926 kJ /mole

•13


Chemical thermodynamics - 1st LawØ Final pressure (for constant volume combustion)

€

PV = mRT, R =ℜM

; ℜ = universal gas constant = 8.314 J/moleK

M (for mixture) = Total massTotal moles

=niMi

i=1

N

∑

nii=1

N

∑=

niMii=1

N

∑nTotal

= X iMii=1

N

∑

€

Constant volume combustion : V = constant, m = constant

Combine:PproductsPreactants

=ni

i=1

N (products)

∑

nii=1

N (reactants)

∑

TproductsTreactants


Chemical thermodynamics - heating valueØ Constant-pressure energy conservation equation (no heat transfer, no

work transfer other than PdV work)

Denominator = m = constant, separate chemical and thermal terms:

Ø This scary-looking boxed equation is simply conservation of energy for a chemically reacting mixture at constant pressure

Ø 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

hreactants =ni [ !h(T )− !h298]i + Δ !hf ,i

o( )i=1

N (reactants)

∑

niMii=1

N (reactants)

∑= hproducts =

ni [ !h(T )− !h298]i + Δ !hf ,io( )

i=1

N (products)

∑

niMii=1

N (products)

∑

ni [ !h(T )− !h298]i( )− ni [ !h(T )− !h298]i( )i=1

N (products)

∑i=1

N (reactants)

∑

niMii=1

N (reactants)

∑=

niΔ !hf ,io

i=1

N (products)

∑ − niΔ !hf ,io

i=1

N (reactants)

∑

niMii=1

N (products)

∑

•14


Chemical thermo - heating valueØ By definition, CP º (∂h/∂T)PØ For an ideal gas, h = h(T) only, thus CP = dh/dT or dh = CPdTØ If CP is constant, then for the thermal enthalpy

h2 - h1 = CP(T2 - T1) = mCP(T2 - T1) /mØ 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 26 can be written as

where is the constant-pressure specific heat averaged (somehow) over all species and averaged between the product and reactant temperatures

ni [ !h(T )− !h298]i( )− ni [ !h(T )− !h298]i( )i=1

N (products)

∑i=1

N (reactants)

∑

niMii=1

N (reactants)

∑=mCP(Treactants −Tproducts )

m

€

C P


Chemical thermo - heating valueØ Term on right-hand side of boxed equation on page 26 can be re-

written as

Ø Last term is the chemical enthalpy change per unit mass of fuel; define this as -QR, where QR is the fuel's heating value

Ø For our stereotypical hydrocarbons, assuming CO2, H2O and N2 as the only combustion products, this can be written as

QR = −x ⋅ Δhf ,CO2

o + (y 2) ⋅ Δhf ,H2Oo −1⋅ Δhf , fuel

o − (x + y 4)Δhf ,O2o

1⋅M fuel

nfuel M fuel

niMii=1

N (reactants)

∑

niΔ !hf ,io

i=1

N (products)


i=1

N (reactants)

∑nfuel M fuel

= fniΔ !hf ,i

o

i=1

N (products)


i=1

N (reactants)

∑nfuel M fuel

QR ≡ −niΔ !hf ,i

o

i=1

N (products)


i=1

N (reactants)

∑nfuel M fuel

•15


Chemical thermo - flame temperatureØ Now write the boxed equation on page 26 (conservation of energy for

combustion at constant pressure) once again:

Ø We've shown that the left-hand side =

and the right-hand side = -fQR; combining these we obtain

Ø 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

(constant pressure combustion, T-averaged CP)

ni [ !h(T )− !h298]i( )− ni [ !h(T )− !h298]i( )i=1

N (products)

∑i=1

N (reactants)

∑

niMii=1

N (reactants)

∑=

niΔ !hf ,io

i=1

N (products)


i=1

N (reactants)

∑

niMii=1

N (reactants)

∑

€

mC P (Treactants −Tproducts)m

€

Tproducts = Treactants + fQR /C P

€

Tad = T∞ + fQR /C P


Chemical thermo - flame temperatureØ 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

where for lean mixtures, fburnable is just f (fuel mass fraction) whereas for rich mixtures, with some algebra it can be shown that

thus in general we can write

€

Tad = T∞ + fburnableQR

C P

€

Tad = T∞ + f QR

C P (if f ≤ f stoichiometric )

Tad = T∞ + fstoichiometric1− f

1− f stoichiometric

%

& '

(

) *

QR

C P (if f ≥ fstoichiometric )

€

fburnable = fstoichiometric1− f


#

$ %

&

' (

•16


Chemical thermo - flame temperatureØ For constant-volume combustion (instead of constant P), everything is

the same except u = const, not h = const, thus the term on the left-hand side of the boxed equation on page 29 must be re-written as

The extra PV terms (= mRT for an ideal gas) adds an extra mR(Tproducts-Treactants) term, thus

which means that (again, Tproducts = Tad; Treactants = T∞)(constant volume combustion, T-averaged CP)

which is the same as for constant-pressure combustion except for the Cv instead of CP

ni [ !h(T )− !h298]i( )− (PV )reactantsi=1

N (reactants)

∑⎡

⎣⎢

⎤

⎦⎥ − ni [ !h(T )− !h298]i( )− (PV )products

i=1

N (products)

∑⎡

⎣⎢

⎤

⎦⎥

niMii=1

N (reactants)

∑

€

mC P (Tproducts −Treactants)m

→mC P (Tproducts −Treactants) −mR(Tproducts −Treactants)

m= (C P − R)(Tproducts −Treactants) = C v (Tproducts −Treactants)

€

Tad = T∞ + fQR /C v


Chemical thermo - flame temperatureØ 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

Ø Note that the ratio of adiabatic temperature rise due to combustion for constant pressure vs. constant volume is

Ø One can determine 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 ≈ 1429 J/kgK(for other stoichiometries or other fuels, the effective will be somewhat but not drastically different)

€

Tad −T∞( )constant v

Tad −T∞( )constant P

=C PC V

= γ

€

C P

€

C P

€

C P

€

Tad = T∞ + fQR

C v (if f < f stoichiometric )

Tad = T∞ + fstoichiometric1− f


$

% &

'

( )

QR

C v (if f > fstoichiometric )

•17


Example of heating valueØ Iso-octane/air mixture:

C8H18 + 12.5(O2 + 3.77N2) ® 8 CO2 + 9 H2O + 12.5*3.77 N2

QR ≡ −niΔ !hf ,i

o

i=1

N (products)


i=1

N (reactants)

∑nfuel M fuel

= −8Δ !hf ,CO2

o + 9Δ !hf ,H 2Oo +12.5(3.77)Δ !hf ,N 2

o( )− 1Δ !hf ,C8H18o +12.5Δ !hf ,O2

o +12.5(3.77)Δ !hf ,N 2o( )

1MC8H18

= −8 moles)(−393.5 kJ/mole)+ 9(−241.8)+12.5(3.77)(0)( )− 1(−250.0)+12.5(0)+12.5(3.77)(0)( )

(1 mole)(0.114 kg/mole)

= 44,500 kJ/kg = 4.45 x 107 J/kg

Δ !hf ,C8H18o = −250.0 kJ /mole( )


Fuel properties

Fuel Heating value, QR (J/kg) f at stoichiometric

Gasoline 43 x 106 0.0642Methane 50 x 106 0.0550Methanol 20 x 106 0.104Ethanol 27 x 106 0.0915

Coal 34 x 106 0.0802Paper 17 x 106 0.122Fruit Loops™ 16 x 106 Probably about the same as paperHydrogen 120 x 106 0.0283U235 fission 83,140,000 x 106 1

Pu239 fission 83,610,000 x 106 12H + 3H fusion 339,000,000 x 106 2H : 3H = 1 : 1

•18


Comments on heating valueØ Heating values are usually computed assuming all C ® CO2, H ®

H2O, N ® N2, S ® SO2, etc.

Ø 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

Ø Most hydrocarbons have similar QR (4.0 – 5.0 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

Ø Foods similar - on a dry weight basis, about same QR for all

Ø 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)

Ø Fats slightly higher than starches or sugars

Ø Foods with (non-digestible) fiber lower


Comments on heating value

Ø Acetylene higher (4.8 x 107 J/kg) because of CºC triple bondØ Methane somewhat higher (5.0 x 107 J/kg) because of high H/C

ratioØ H2 MUCH higher (12.0 x 107 J/kg) because no "heavy" C atomsØ Alcohols lower (2.0 x 107 J/kg for methanol, CH3OH) because of

"useless" O atoms - add mass but no enthalpy release

•19


Example of adiabatic flame temperatureØ Lean iso-octane/air mixture, equivalence ratio f = 0.8, initial temperature

T∞ = 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

€

φ =FAR (actual mixture, φ = 0.8)

FAR (stoichiometric mixture, φ =1)=

fφ =0.8 /(1− fφ =0.8)fφ =1 /(1− fφ =1)

= 0.8

fφ =1 =n fuel M fuel

niMii=1

n (reactants)

∑

= 0.06218FARφ =1 = fφ =1 /(1− fφ =1) = 0.06218 /(1− 0.06218) = 0.06630

φ = 0.8 :fφ =0.8 /(1− fφ =0.8)

0.06630= 0.8⇒ fφ =0.8 = 0.0504

Tad = T∞ + fQR /C P = 300K + (0.05054)(4.45 ×107 J /kg) /(1400J /kgK) =1906K (const. P)Tad = T∞ + fQR /C V = 300K + (0.05054)(4.45 ×107 J /kg) /(1100J /kgK) = 2345K (const. V)

€

=(1 mole C8H18)(0.114 kg/mole)

(1 mole C8H18)(0.114 kg/mole) + (12.5 mole O2)(0.032 kg/mole) + (12.5* 3.77 mole N2)(0.028 kg/mole)


Summary - Lecture 2Ø Many fuels, e.g. hydrocarbons, when chemically reacted with an

oxidizer, e.g. O2, release large amounts of energy or enthalpyØ 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

Ø Only 2 principles are required to compute flame temperaturesØ Conservation of each type of atomØ Conversation of energy (sum of chemical + thermal)

… but the resulting equations required to account for changes in composition and energy can look formidable

Ø Key thermodynamic properties of a fuel are its heating value QR

and its stoichiometric fuel mass fraction fstoichiometric

Ø Key property of a fuel/air mixture is its equivalence ratio (f)Ø A simplified analysis leads to

€

Tad = T∞ + fQR /C P (constant pressure)Tad = T∞ + fQR /C V (constant volume)

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Energy and Propulsion - University of Southern California