Combustion chemistry definition

chemistry combustion analysis and chemistry combustion reaction problems and combustion chemistry and processes
Dr.BenjaminClark Profile Pic
Dr.BenjaminClark,United States,Teacher
Published Date:21-07-2017
Your Website URL(Optional)
Summary of Part 1: Overview, Chem Basics • Fuels are both Boon and Bane – Oil: cheap-to-produce convenient high-density energy carrier – Current system not sustainable, esp. greenhouse – So we need: 1) higher efficiency 2) new energy carrier • Maybe an Alternative Fuel? • If so, need to understand how it behaves… • Fuel Chemistry is Tricky – NOT Arrhenius single step – Details matter to understand why fuels behave differently • Many clever (tricky) methods for experimentally measuring rates…but you cannot measure everything • …Most of the rest of the lectures focus on understanding fuel chemistry theoretically / computationally Combustion Chemistry, Part 2 2 Plan of attack • Start with Thermodynamics/Thermochemistry – Including equilibrium Statistical Mechanics • Next simple Rate Theory – conventional Transition State Theory – high-P-limit (thermalized) • Then Fancier Rate Theory – variational TST – P-dependence of rate coefficients • Then Mechanisms combining many species, reactions 3 Thermodynamics/Thermochemistry All Kinetics is Leading Toward Equilibrium. So good to start by figuring out where we are going (later we can worry about how fast we will get there…) Part of Thermo is about phase behaviour (e.g. volatility, miscibility) “Thermochemistry” is about reactions. st 1 Law gives energy density, final temperature nd 2 Law related to detailed balance (and so reverse rate coefficients), final composition at equilibrium 4 Fuel volatility must match fuel injector, and certain volatility ranges are particularly hazardous to store. Gasoline boils 150 C, jet 200 C, diesel 300 C Large molecules pyrolyze at lower T than b.p., and are solids at room temperature. 5 Volatility, Solvation • Phase behavior of mixtures is complicated subject in itself – How liquid fuel evaporates in engine – Formation of aerosols in atmosphere (particulate pollution) – Distillation is main separation process in refinery – Some liquids are inmiscible, so can have several liquid phases • Don’t want this to occur in fuel tank or fuel injector • Strongly affects mobility in environment • Volatility (Partial Pressure of species in gas phase) depends on solvation of the molecule in the liquid phase – Non-bonded interactions (Enthalpic interactions) – Entropy of molecule in liquid (mostly hard-sphere) • Boiling Point is just when sum of all the partial pressures equals atmospheric pressure. Pure-compound boiling point not directly related to Henry’s Law constant when species is in solution 6 Thermodynamics • We need thermodynamic data to: – Determine the heat release in a combustion process (need enthalpies and heat capacities) – Calculate the equilibrium constant for a reaction – this allows us to relate the rate coefficients for forward and reverse reactions (need enthalpies, entropies (and hence Gibbs energies, and heat capacities). • This lecture considers: – Classical thermodynamics and statistical mechanics – relationships for thermodynamic quantities – Sources of thermodynamic data – Measurement of enthalpies of formation for radicals – Active Thermochemical Tables – Representation of thermodynamic data for combustion models 7 Various thermodynamic relations are needed to determine heat release and the relations between forward and reverse rate coefficients A statement of Hess’s Law n is the i stoichiometric coefficient o For ideal gas or ideal solution, a =P /P or i i o a = C /C . Must use same standard state i i o used when computing DG . Solids have a =1. i For surface sites use fractional occupation. 8 Tabulated thermodynamic quantities. 1. Standard enthalpy of formation o Standard enthalpy change of formation, D H f The standard enthalpy change when 1 mol of a substance is formed from its elements in their reference states, at a stated temperature (usually 298 K). The reference state is the most stable state at that temperature, and at a pressure of 1 bar. e.g. C(s) + 2H (g)  CH (g) 2 4 o -1 D H = -74.8 kJ mol f The standard enthalpies of formation of C(s) and H (g) are both zero 2 9 Computing K (T), G(T), H(T), S(T) eq 𝑇 𝑜 0 𝑜 ′ 𝐻 𝑇 =𝑈 + =𝐻 (𝑇 ) + 𝐶 𝑇 𝑑𝑇′ 𝑜 𝑝 𝑇 𝑇 0 0 ′ o 𝑆 𝑇 =𝑆 (T ) + (𝐶 𝑇 /𝑇′ )𝑑𝑇′ 𝑜 𝑝 𝑇 0 0 0 𝐺 𝑇 = 𝐻 (𝑇 )−𝑇 𝑆 (𝑇 ) 0 𝐾 𝑇 =exp (−𝐺 𝑇 ) None of these depend on Pressure (they are for standard state) Same K works at all pressures. For non-ideal gases, the eq non-idealities are conventionally hidden in the activities as Fugacities or “activity coefficients”. C (T) is expressed in several different formats causing some confusion: p NIST/Benson tabulated C (T ) or several different polynomial-type expansions: p i Shomate, Wilhoit (beware typos in original paper), two different NASA formats. C (T) can also be expressed by statistical mechanics formulas. p 10 𝑒𝑞 𝑃𝑉11 Now maintained by Elke Goos. 12 LHV and UHV • Fuels are classified by their Heating Value, i.e. their heat of combustion. • Two variants are commonly used: Lower Heating Value and Upper Heating Value. • LHV assumes all the H O formed is in gas phase, this is realistic 2 for engines where the H O leaves in the exhaust. Note in a real 2 engine the H O in the exhaust would be hot, but the LHV 2 calculation usually assumes room temperature steam. • UHV assumes all the H O formed is in liquid phase. This is 2 realistic for bomb calorimetry experiments, where the final temperature is usually pretty low. So UHV is easier to measure. But it can be a big overestimate of the true heat delivered by the fuel in an engine. 13 Adiabatic Flame Temperatures Compute by noting H (T ) = H (T ) out out in in for an adiabatic process o Given H ’s and C (T)’s p and assuming stoichiometry, you can solve for T out At true combustion T, equilibrium concentrations of Species other than CO2 & H2O are significant… 14 At low P, high T, CO+O2 equilibrium important so Adiabatic Flame T varies a bit with P 15 Standard entropy Standard entropy rd Based on the 3 law of Thermodynamics: C /T p The entropy of any perfectly crystalline material at T = 0 is zero  Standard molar entropy, S m 0 T The entropy of 1 mol of a NB – calculation using substance in its standard state statistical mechanics – next rd based on the 3 law slide Sometimes entropies of formation are used, but this makes no difference to entropies of reaction provided consistency is maintained 16 Statistical Mechanics Basics • There is a quantity Q called the “partition function” Q = S g exp(-E /k T) i i B where Ei are the possible energies of the molecule (quantum mechanics only allows certain quantized energies), and g is the i number of quantum states with energy E . i • Q contains enough information to compute all the normal thermochemical quantities. For example Helmholtz Free Energy = U – TS = G - PV = - k T ln Q B 2 U(T,V) = k T ∂(ln Q)/∂T B 17 Quantized Energies & Partition Functions We usually approximate each of the vibrations in a molecule as a harmonic oscillator. (This is not always an accurate approximation, but it really simplifies the math) The quantized energy of a harmonic oscillator with characteristic frequency n are: I recommend you choose the zero of energy to be the lowest state (all the vibrations have n =0), and handle the zero-point-energy i (ZPE) = ½ h S n separately. Then E = h S n n and i vib i i -1 q = P ( 1 – exp (-hn /k T)) vib i B 18 3 Translational Partition Function (Particle in a Cube, V=L ) 19 Rotational Partition Functions For a Linear Molecule, assumed to be Rigid with Moment of Inertia “I”: Each J state has (2J+1) M states, so g = 2J+1 Making a similar approximation as for translation, and considering the effects of symmetry (with symmetry number s), we obtain 2 2 q  8p Ik T/sh rot B For nonlinear molecules there are 3 distinct moments of inertia Ia,Ib,Ic ½ 3/2 q  p q rot_nonlinear rot 3/2 where I is replaced by sqrt(I I I ) a b c 20