AME 513 Principles of Combustion

AME 513

Principles of Combustion

Lecture 3Chemical thermodynamics I – 2nd Law

2AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Outline Why do we need to invoke chemical equilibrium? Degrees Of Reaction Freedom (DORFs) Conservation of atoms Second Law of Thermodynamics for reactive systems Entropy of an ideal gas mixture Equilibrium constants Application of chemical equilibrium to hydrocarbon-air

combustion Application of chemical equilibrium to

compression/expansion

3AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Why do we need chemical equilibrium?

(From lecture 2) What if we assume more products, e.g.1CH4 + 2(O2 + 3.77N2) ? CO2 + ? H2O + ? N2 + ? COIn this case how do we know the amount of CO vs. CO2?

And if if we assume only 3 products, how do we know that the “preferred” products are CO2, H2O and N2 rather than (for example) CO, H2O2 and N2O?

Need chemical equilibrium to decide - use 2nd Law to determine the worst possible end state of the mixture

4AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Degrees of reaction freedom (DoRFs) If we have a reacting “soup” of CO, O2, CO2, H2O, H2 and OH,

can we specify changes in the amount of each molecule independently? No, we must conserve each type of atom (3 in this case)

Conservation of C atoms: nCO + nCO2 = constant O atoms: nCO + 2nCO2 + 2nO2 + nH2O + nOH = constantH atoms: 2nH2O + 2nH2 + nOH = constant

3 equations, 6 unknown ni’s 3 degrees of reaction freedom (DoRFs)

# of DoRFs = # of different molecules (n) - # of different elements

Each DoRF will result in the requirement for one equilibrium constraint

5AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Conservation of atoms Typically we apply conservation of atoms by requiring that

the ratios of atoms are constant, i.e. the same in the reactants as the productsC atoms: nCO + nCO2 = constant O atoms: nCO + 2nCO2 + 2nO2 + nH2O + nOH = constantH atoms: 2nH2O + 2nH2 + nOH = constant

Specifying nO/nH also would be redundant, so the number of atom ratio constraints = # of atoms - 1

What are these “constants?” Depends on initial mixture, e.g. for stoichiometric CH4 in O2, nC/nO = 1/4, nC/nH = 1/4

6AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

2nd law of thermo for reacting systems Constraints for reacting system

First law: dE = Q - W = Q - PdV Second law: dS ≥ Q/T Combine: TdS - dU - PdV ≥ 0 for any allowable change in the

state of the system For a system at fixed T and P (e.g. material in a piston-cylinder

with fixed weight on top of piston, all placed in isothermal bath: d(TS-U-PV) ≥ 0, or per unit mass d(Ts-u-Pv) ≥ 0

Define “Gibbs function” g h - Ts = u + Pv - Ts Thus for system at fixed T and P: d(-g) ≥ 0 or dg ≤ 0 Thus at equilibrium, dg = 0 or dG = 0 (g or G is minimum) and for

each species i

µi = chemical potential of species i

Similarly, for system at fixed U and V (e.g. insulated chamber of fixed volume): ds ≥ 0, at equilibrium ds = 0 or dS = 0 (s is maximum)

7AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Entropy of an ideal gas mixture Depends on P AND T (unlike h and u, which depend ONLY on T)

(1) (2) (3) (1) Start with gases A, B and C at temperature T and Po = 1 atm

(2) Raise/lower each gas to pressure P ≠ 1 atm (same P for all)

AP = Pref

BP = Pref

CP = Pref

AP ≠ Pref

BP ≠ Pref

CP ≠ Pref

A, B, CP ≠ Pref

8AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

(3) Now remove dividers, VA VA + VB + VC (V = volume)

Can also combine Xi and P/Pref terms using partial pressures Pi; for an ideal gas mixture Xi = Pi/P, where (as above) P without subscripts is the total pressure = SPi

Entropy of an ideal gas mixture

9AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Most useful form - entropy per unit mass (m), since mass is constant:

ln(P/Pref) - entropy associated with pressure different from 1 atm - P > Pref leads to decrease in s

Xiln(Xi) - entropy associated with mixing (Xi < 1 means more than 1 specie is present - always leads to increase in entropy

since -Xiln(Xi) > 0) Denominator is just the average molecular weight Units of s are J/kgK

Entropy of an ideal gas mixture

10AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

How to use this result? dG = 0 means there are n equations of the form ∂G/∂ni = 0 for n unknown species concentrations!

Break system into multiple sub-systems, each with 1 DoRF, e.g.. for a chemical reaction of the form

AA + BB CC + DD

e.g. 1 H2 + 1 I2 2 HIA = H2, A = -1, B = I2, B = -1, C = HI, C = 2, D = nothing, D = -0

conservation of atoms requires that

G = H – TS, where

Equilibrium of an ideal gas mixture

11AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Equilibrium of an ideal gas mixture

12AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

It is fairly inconvenient to work with the exp[-g/RT] terms, but we can tabulate for every species its equilibrium constant Ki, i.e. the value of this term for the equilibrium of the species with its forming elements in their standard states, e.g.

So in general for a reaction of the form AA + BB CC + DD

Equilibrium of an ideal gas mixture

13AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Equilibrium constants Examples of tabulated data on K - (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)

Note K = 1 at all T for elements in their standard state (e.g. O2)

14AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Chemical equlibrium - example For a mixture of CO, O2 and CO2 (and nothing else, note 1 DoRF

for this case) at 10 atm and 2500K with C:O = 1:3, what are the mole fractions of CO, O2 and CO2?1 CO2 1 CO + .5 O2

3 equations, 3 unknowns: XCO2 = 0.6495, XO2 = 0.3376, XCO = 0.0129 At 10 atm, 1000K: XCO2 = .6667, XO2 = 0.3333, XCO = 2.19 x 10-11

At 1 atm, 2500K: XCO2 = 0.6160, XO2 = 0.3460, XCO = 0.0380 With N2 addition, C:O:N = 1:3:6, 10 atm, 2500K: XCO2 = 0.2141, XO2 =

0.1143, XCO = 0.0073, XN2 = 0.6643 (XCO2/XCO = 29.3 vs. 50.3 without N2 dilution) (note still 1 DoRF in this case)

Note high T, low P and dilution favor dissociation

15AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Chemical equlibrium How do I know to write the equilibrium as

1 CO2 1 CO + .5 O2

and not (for example)2 CO + 1 O2 2 CO2

For the first form

which is the appropriate expression of equilibrium for the 2nd form - so the answer is, it doesn’t matter as long as you’re consistent

16AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Chemical equlibrium - hydrocarbons Reactants: CxHy + rO2 + sN2 (not necessarily stoichiometric) Assumed products: CO2, CO, O2, O, H2O, OH, H, H2, N2, NO How many equations?

10 species, 4 elements 6 DoRFs 6 equil. constraints 4 types of atoms 3 atom ratio constraints Conservation of energy (hreactants = hproducts) (constant pressure

reaction) or ureactants = uproducts (constant volume reaction)

Pressure = constant or (for const. vol.) 6 + 3 + 1 + 1 + 1 = 12 equations

How many unknowns? 10 species 10 mole fractions (Xi) Temperature Pressure 10 + 1 + 1 = 12 equations

17AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Equilibrium constraints - not a unique set, but for any set Each species appear in at least one constraint Each constraint must have exactly 1 DoRF Note that the initial fuel molecule does not necessarily appear in the set

of products! If the fuel is a large molecule, e.g. C8H18, its entropy is so low compared to other species that the probability of finding it in the equilibrium products is negligible!

Example set (not unique)

Chemical equlibrium - hydrocarbons

18AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Chemical equlibrium - hydrocarbons Atom ratios

Sum of all mole fractions = 1

Conservation of energy (constant P shown)

19AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Chemical equlibrium - hydrocarbons This set of 12 simultaneous nonlinear algebraic equations looks

hopeless, but computer programs (using slightly different methods more amenable to automation) (e.g. GASEQ) exist

Typical result, for stoichiometric CH4-air, 1 atm, constant P

20AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Chemical equlibrium - hydrocarbons Most of products are CO2, H2O and N2 but some dissociation (into

the other 7 species) occurs Product is much lower than reactants - affects estimation of

compression / expansion processes using Pv relations Bad things like NO and CO appear in relatively high concentrations

in products, but will recombine to some extent during expansion By the time the expansion is completed, according to equilibrium

calculations, practically all of the NO and CO should disappear, but in reality they don’t - as T and P decrease during expansion, reaction rates decrease, thus at some point the reaction becomes “frozen”, leaving NO and CO “stuck” at concentrations MUCH higher than equilibrium

21AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Adiabatic flame temp. - hydrocarbons

22AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Adiabatic flame temp - hydrocarbons Adiabatic flame temperature (Tad) peaks slightly rich of

stoichiometric - since O2 is highly diluted with N2, burning slightly rich ensures all of O2 is consumed without adding a lot of extra unburnable molecules

Tad peaks at ≈ 2200K for CH4, slightly higher for C3H8, iso-octane (C8H18) practically indistinguishable from C3H8

H2 has far heating value per unit fuel mass, but only slightly higher per unit total mass (due to “heavy” air), so Tad not that much higher Also - massive dissociation as T increases above ≈ 2400K,

keeps peak temperature down near stoichiometric Also - since stoichiometric is already 29.6% H2 in air (vs. 9.52%

for CH4, 4.03% for C3H8), so going rich does not add as many excess fuel molecules

CH4 - O2 MUCH higher - no N2 to soak up thermal energy without contributing enthalpy release

Constant volume - same trends but higher Tad

23AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Compression / expansion Compression / expansion processes are typically assumed

to occur at constant entropy (not constant h or u) Use sreactants = sproducts constant for compression / expansion

instead of hreactants = hproducts or ureactants = uproducts All other relations (atom ratios, SXi = 1, equilibrium

constraints) still apply

24AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Compression / expansion Three levels of approximation

Frozen composition (no change in Xi’s), corresponds to infinitely slow reaction

Equilibrium composition (Xi’s change to new equilibrium), corresponds to infinitely fast reaction (since once we get to equilibrium, no further change in composition can occur)

Reacting composition (finite reaction rate, not infinitely fast or slow) - more like reality but MUCH more difficult to analyze since rate equations for 100’s or 1000’s of reactions are involved

Which gives most work output? Equilibrium - you’re getting everything the gas has to offer;

recombination (e.g. H + OH H2O) gives extra enthalpy release, thus more push on piston or more kinetic energy of exhaust

Frozen - no recombination, no extra heat release Reacting - somewhere between, most realistic

25AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Compression / expansion Example - expansion of CO2-O2-CO mixture from 10 atm, 2500K to 1

atm in steady-flow control volume (e.g. nozzle) or control mass (e.g piston/cylinder)

Initial state (mixture from lecture 2 where h was calculated):XCO = 0.0129, XO2 = 0.3376, XCO2 = 0.6495, T = 2500 , P = 10 atmh = -3784 kJ/kg, u = -4397 kJ/kg

26AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Compression / expansion Expand at constant entropy to 1 atm, frozen composition:

T = 1738K, XCO = 0.0129, XO2 = 0.3376, XCO2 = 0.6495, , h = -4795 kJ/kg, u = -5159 kJ/kg, s = 7381J/kgK

Work done (control volume, steady flow) = hbefore - hafter = +1011 kJ/kg

Work done (control mass) = ubefore - uafter = +762 kJ/kg Expand at constant entropy to 1 atm, equilibrium composition:

T = 1794K, XCO = 0.00022, XO2 = 0.3334, XCO2 = 0.6664(significant recombination)

h = -4811 kJ/kg, u = -5184 kJ/kg, s = 7382 J/kgK Work done (control volume, steady flow) = +1027 kJ/kg

(1.6% higher) Work done (control mass) = 787 kJ/kg (3.3% higher)

Moral: let your molecules recombine!

27AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Summary - Lecture 3 In order to understand what happens to a chemically reacting

mixture if we wait a very long time, we need to apply 1st Law of Thermodynamics (conservation of energy) - but this

doesn’t tell us what the allowable direction of the reaction is; A B or B A are equally valid according to the 1st Law

2nd Law of Thermodynamics (increasing entropy) - invokes restrictions on the direction of allowable processes (if A B is allowed then B A isn’t)

Equilibrium occurs when the worst possible end state is reached; once this point is reached, no further chemical reaction is possible unless something is changed, e.g. T, P, V, etc.

The statements of the 2nd law are ds ≥ 0 (constant u & v, e.g. a rigid, insulated box) dg = d(h - Ts) ≤ 0 (constant T and P, e.g. isothermal

piston/cylinder)

28AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

Summary - Lecture 3 The application of the 2nd law leads to complicated looking

expressions called equilibrium constraints that involve the concentrations of each type of molecule present in the mixture

These equilibrium constraints are coupled with conservation of energy, conservation of each type of atom, and the ideal gas law (or other equations of state, not discussed in this class) to obtain a complete set of equations that determine the end state of the system, e.g. T, P and composition of the combustion products