AME 513 Principles of Combustion Lecture 2 Chemical thermodynamics I – 1 st Law

AME 513

Principles of Combustion

Lecture 2Chemical thermodynamics I – 1st Law

2AME 513 - Fall 2012 - Lecture 2 - Chemical thermodynamics 1

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


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/molethus 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


Hydrocarbons

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

Olefins or alkenes - one or more double bonds between carbons

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


Hydrocarbons Aromatics - one or more ring structures

Alcohols - contain one or more OH groups


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

Ethyl stearate

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

Methanol + triglyceride Glycerol+ alkyl ester

Transesterification process


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)


Gasoline - typical composition

BenzeneToluene

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


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

Practical fuels - properties

Values NOT unique because Real fuels are a mixture of many molecules, composition varies Different testing methods & definitions


Practical fuels - properties

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

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


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


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


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


Stoichiometry

Fuel mass fraction (f)

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

which varies a lot depending on x and y (i.e. much smaller for big molecules with large x and y)


Stoichiometry

Fuel-to-air ratio (FAR)

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

< 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


Fuel properties

Fuel 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 12H + 3H fusion

339,00,000 x 106 2H : 3H = 1 : 1


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 + ozoneC4N2 + (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



The problem is that the products are NOT just H2O, that is, we don’t getH2 + (1/2)O2 H2O

but ratherH2 + (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

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)



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


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)


Chemical thermo - assumptions

Ideal gases - note many “flavors” of the ideal gas law• PV = nT• PV = mRT• Pv = RT• P = RT

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


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!


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


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

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



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)


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:




Final pressure (for constant volume combustion)


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


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

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


Chemical thermo - heating value Term on right-hand side of boxed equation on page 27 can be re-

written 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


Chemical thermo - flame temperature Now write the boxed equation on page 27 (conservation of energy

for combustion at constant pressure) once again:

We’ve shown that the left-hand side =

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

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)


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

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


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

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


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

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

In practice, 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/kg-K (for other stoichiometries or other fuels will be moderately different)


Example of heating value

Iso-octane/air mixture:

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


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


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


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


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

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AME 513 Principles of Combustion Lecture 2 Chemical thermodynamics I – 1 st Law