ame 436 energy and propulsion lecture 2 fuels, chemical thermodynamics (thru 1st law; 2nd law next...
TRANSCRIPT
AME 436
Energy and Propulsion
Lecture 2Fuels, chemical thermodynamics
(thru 1st Law; 2nd Law next lecture)
2AME 436 - Spring 2015 - 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
3AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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, e.g. Iraq) Relatively non-toxic fuel and combustion products Relatively low explosion hazards
4AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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/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
5AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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 - very reactive, higher heating value than alkanes or alkenes
6AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Hydrocarbons Aromatics - one or more ring structures
Alcohols - contain one or more OH groups
7AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
8AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
9AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Gasoline - typical composition
BenzeneToluene
J. Burri et al., Fuel, Vol. 83, pp. 187 - 193 (2004)
10AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
11AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Practical fuels - properties
http://www.afdc.energy.gov/pdfs/fueltable.pdf
12AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Practical fuels
What doesn't distinguish one fuel from another?Energy content (except for fuels containing alcohols, which are lower)
ExamplesGasoline - low-T distillation point, easy to vaporize, need high octane
number; reformulated gasoline contains alcoholsDiesel - high-T distillation point, hard to vaporize, need LOW octane
number for easy ignition once fuel is injectJet fuel - medium-T distillation point; need low freezing T since it will
be used at high alititude / low T
13AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
14AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
15AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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)
16AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
17AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
339,000,000 x 106 2H : 3H = 1 : 1
18AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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 + 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
19AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Chemical thermodynamics - intro
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)
20AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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? 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
21AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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)
22AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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)
23AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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!
24AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Chemical thermodynamics - 1st Law Enthalpy of a mixture (sum of thermal and chemical terms)
25AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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)
26AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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)
27AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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:
Chemical thermodynamics - 1st Law
28AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Chemical thermodynamics - 1st Law
Final pressure (for constant volume combustion)
29AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
30AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
where is the constant-pressure specific heat averaged (somehow) over all species and averaged between the product and reactant temperatures
31AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Chemical thermo - heating value Term on right-hand side of boxed equation on page 28 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
32AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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 =
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)
33AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
34AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
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
35AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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 somewhat but not drastically different)
36AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Example of heating value
Iso-octane/air mixture:
C8H18 + 12.5(O2 + 3.77N2) 8 CO2 + 9 H2O + 12.5*3.77 N2
37AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
38AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
39AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
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
40AME 436 - Spring 2015 - Lecture 2 - Chemical Thermodynamics 1
Summary - Lecture 2
Many fuels, e.g. hydrocarbons, when chemically reacted with oxygen or other oxidizers, release large amounts 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
Key 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