Combustion basics... We are discussing gaseous combustion in a mixture of perfect gases containing N species indexed with k=1 to N:

Transcription

1 Combustion basics... T. Poinsot Only the things you should know to understand the following courses Mainly elements of laminar flame theory for premixed and diffusion flames 1 Copyright Dr T. Poinsot 2013 We are discussing gaseous combustion in a mixture of perfect gases containing N species indexed with k=1 to N: Ch. 1 2

2 STOECHIOMETRY AND EQUIVALENCE RATIO: Ch. 1 Stoechiometric ratio Equivalence ratio 4

3 FUEL / AIR COMBUSTION COMPOSITION AT EQUIVALENCE RATIO φ With air, combustible mixtures contain typically more than 90 % of... air! This will be useful for theory: a combustible mixture will essentially have the same properties (density, viscosity, conductivity) as pure air. 5 In a combustion code: In a non reacting compressible flow (aerodynamics), we have 5 variables: 3 velocities pressure or temperature or enthalpy or entropy density In combustion, we have 5 + (N-1) unknown 3D fields to determine: 3 velocities pressure or temperature or enthalpy or entropy N species -> actually N-1 since their sum is unity density 6 We need 5 + (N-1) equations: Momentum (3) Energy, or enthalpy or entropy or temperature Continuity for each species (or for N-1 species + total mass)

4 Continuity equations (N): for k = 1 to N Summing them from 1 to N MUST give the continuity equation: So that we need: =0 7 Diffusion velocities: careful... (Williams 1985 or Ern and Giovangigli 1994) The Vk s are obtained by solving this system: Ch. 1 Sec This is the diffusion matrix which must be inverted at each point at each instant... but almost no one does it. We use simplified forms V k = D k Y k Y k FICK s law V k = D k X k X k HIRSCHFELDER s law 8

5 What are the D k s? The diffusion coefficients of species k in the MIXTURE Not the binary diffusion coefficients Dij of kinetic gas theory Obtained by: All Dk s are different in general... Dk s vary significantly but their ratios to thermal diffusivity Dth vary much less. They are measured by the Lewis number of each species: 9 Lewis numbers are not intrinsic parameters of a species. They depend on the mixture 10 Lewis numbers of main species in a stoechiometric CH4/air premixed flame

6 The need for correction velocities if Lewis numbers are different: The Dk s are not equal if Lek differ ρ t + ρu i x i = x i ρ N V k = D k X k X k Summing the species equations with Fick s law: D k W k W X k x i k=1 which is not zero if all Lewis numbers are not equal. 11 A correction velocity Vc is required: Replace: By: Adding all species equations gives: 12

7 ρ t + ρu i x i = x i ρ N k=1 D k W k W X k x i V c can be chosen to ensure mass conservation: ρv c i and the proper expression of the diffusion velocity is: Conclusion: careful of codes using simplified expressions for diffusion velocities with non-equal Lewis numbers. 13 Momentum equations (3) Ch. 1 Sec Same form in non-reacting flows. Except that: - density here has very strong gradients - viscosity changes rapidly with temperature 14

8 Last but not least: energy Ch. 1, Sec Each code uses a different form for energy, enthalpy or temperature All forms of course are equivalent (should be...) The trick is not to confuse them and beware of simplified forms... Compressible vs incompressible is one source of confusion Total versus other forms of energy is another one 15 Full summary in PV2011 Chapter 1 Energy: 4 forms. Enthalpy: 4 forms + one (T) Ch. 1, Sec Cp depends both on temperature (through the Cpk s) and on the composition (the Yk s)

9 7 C m pk/r CO2 Scaled molar heat capacity Perfect gas limit 500 H2O 1000 N Temperature (K) C O 2000 H2 Ideal diatomic gas 2500 Molar heat capacity of each species change with temperature Good starting point: total energy et careful with the heat flux. It is not: but is also includes the flux transported by species diffusion: 18

10 Pick up the right one... The heat release and the reaction rates of species There are N reaction rates (one for each species) ω k in the species equations But only one heat release term: ω T = N k=1 ω k H 0 fk 20

11 Many authors also like to work with the temperature equation. If pressure is constant: Careful with this new heat release 21 Two reaction rate terms: same name in the literature but NOT equal λ T x i x i ρe + (ρu i E)= ω T + t x i N ω T = ω k Hfk 0 k=1 + x j (σ ij u i 22 N N ω T = h k ω k = h sk ω k k=1 k=1 N Hfk 0 ω k k=1

12 N N N N ω T = h k ω k = h sk ω k Hfk 0 ω k = h sk ω k + ω T k=1 k=1 k=1 k=1 If the Cp,k s are equal: C p,k = C p h sk = C pk dt = C p dt = h k so that Σ N k=1h sk ω k = h k Σ N k=1 ω k =0 ω T = ω T 23 And if the Cp,k s are equal: N C p,k Y k V k,i = C p N k=1 k=1 so that the temperature equation: Y k V k,i =0 C p,k = C p becomes: 24

13 SUMMARY: We have the basic equations We can solve them numerically for real flames Before we do that, we anticipate that we are going to have difficulties with: Kinetics Turbulence Very large domains We are going to need models. Better: we are going to need understanding...otherwise the work is impossible To do this we need to study a few academic (canonical) cases even if it means simplifying things a little bit. 25 So... canonical cases are needed to understand the basic flame features: The premixed laminar flame Ch. 2 The laminar diffusion flame Ch. 3 26

14 PREMIXED LAMINAR FLAME Ch. 2 Under certain assumptions, an explicit resolution of the previous equations is possible This is the case for the premixed laminar flame 27 A one-dimensional steady problem in the reference frame of the flame Ch. 2, Sec 2.3 No momentum equation needed: we need it only to find pressure changes This equation can be solved for a large number of species numerically (CHEMKIN, COSILAB, CANTERA, FLAME MASTER) 28

15 29 But doing so does not tell us much. Let us keep simplifying Additional assumptions Ch. 2 Sec 2.4 Single reaction limited by fuel (lean). Instead of N mass fractions, we can track one only: Yf. We assume that Yo=ct All Lewis numbers equal to unity Can solve this by hand! -> famous field for mathematicians/ combustion experts (Williams, Linan, Matalon) 30

16 Simplest solutions Ch 2 used in Ch. 4, 5, 6 If Lewis = 1 temperature and fuel mass fractions are directly linked if properly scaled: Then: θ + Y =1 We are left with only one variable to solve for: the reduced temperature (also called the progress variable) 31 Master equation for laminar premixed flames: ω T This is where asymptotics enter the field. Not the topic of present lectures. Let us summarize the results. 32

17 In this simplified world, a flame is: dominated by a strongly non linear reaction rate: ω T 33 decomposed in two zones: a preheating region and a reaction zone ω T Preheating Reaction 34

18 has a thickness given by δ = λ 1 ρ 1 C p s L = D1 th s L Heat diffusivity in the fresh gas or δs L D 1 th =1 The Reynolds number of the flame is unity: 35 Re flame = δs0 L ν =1 has a flame speed which depends on the form assumed for reaction rate and molecular diffusion: You can find the results in TNC 2011 and the derivations in asymptotic papers but let us focus on an important result: 36

19 In all premixed flame speed expressions: Ch. 2 Sec Heat diffusivity Kinetics constant Multiply the kinetics constant by 4 : it will increase the flame speed by two Without diffusivity... no premixed flame Flame propagation is the combination of both diffusion and reaction 37 Flame propagation : T(to) T(t1) DIFFUSION T(t1) COMBUSTION 38

20 A useful feature of premixed flames: Ch. 5 Sec Heat diffusivity Kinetics constant Divide the kinetics constant by a factor F>1 and multiply the diffusivity by F. You will get the same flame speed. But the thickness will be larger by a factor F: 39 δ = D1 th s L δ = FD1 th s L A useful property in codes: thicken the flame (O Rourke and Bracco JCP CERFACS 90s) d / dt(!yk) = " (!D "Yk "xi "xi ) + A!Y 1Y 2...Yn exp(# Ta T ) δ = D1 th s L d / dt(!yk) = " "xi (!DF "Yk ) + A "xi F!Y1Y 2... Yn exp(# Ta T ) δ = FD1 th s L Same flame speeds with both equations! 40

21 In a numerical world where we fight with resolution... this can help! 2000 Temperature Non thickened Thickened 20 times Temperature! Abscissa x10-3 Thickened and unthickened flames for CH4/air. F=20 41 For heat release: it is much easier when the flame is thick Reaction rate Non thickened Thickened 20 times Heat release! Abscissa x

22 DIFFUSION FLAME STRUCTURE Ch. 3 The other case where theory can be developed and is heavily used. Most approaches rely on passive scalars and on the mixture fraction z. 43 WHAT IS A PASSIVE SCALAR? A scalar quantity which does NOT have a source term (ie not a chemical species or a temperature) ρz t + ρu iz x i = x i (ρd Z x i ) Ch. 2 Sec

23 A USEFUL PROPERTY OF PASSIVE SCALARS: IF TWO PASSIVE SCALARS Z1 and Z2 HAVE THE SAME BOUNDARY CONDITIONS: IF THEY ARE EQUAL AT t=0, THEY WILL ALWAYS REMAIN EQUAL IF THEY DIFFER AT t=0, THEY WILL CONVERGE TO THE SAME VALUE AFTER A FEW CHARACTERISTIC TIMES DEMO: by considering the norm of (Z2-Z1) OR by intuition 45 INTUITIVE DEMO: ρz t + ρu iz x i = x i (ρd Z x i ) can also be written: ρ DZ Dt = x i (ρd z x i ) where D/Dt is the total derivative (the variations along the trajectories) 46

24 Consider two passive scalars Z1 and Z2. Along trajectories Z=Z2-Z1 changes only because of diffusion ρ DZ Dt = (ρd z ) x i x i with boundary condition: Z=0 If the initial condition is Z=0, all gradients are zero and Z will remain 0 If the initial condition is not Z=0, any non zero fluid element will go to Z=0 rapidly. 47 WHAT IS MIXTURE FRACTION? Ch. 3 Sec A PASSIVE SCALAR WHICH GOES FROM 1 IN THE FUEL TO 0 IN OXIDIZER STREAM ρz t + ρu iz x i = x i (ρd z x i ) z=1 z=0 ANY PASSIVE SCALAR GOING FROM 1 IN THE FUEL TO 0 IN THE OXIDIZER IS EQUAL TO THE MIXTURE FRACTION 48

25 WHAT DOES IT TAKE TO BE ABLE TO CONSTRUCT A MIXTURE FRACTION? H1: Constant pressure and equal Cp s H2: All Lewis numbers equal to unity H3: Global single step reaction Ch. 2 Sec So that we can track only T, YF and YO 49 IN THE ABSENCE OF COMBUSTION : PURE MIXING THREE PASSIVE SCALARS WITH THE SAME EQUATION. IF WE CAN NORMALIZE THEM TO HAVE THE SAME BOUNDARY CONDITIONS, THEY ARE EQUAL TO z 50

26 NORMALIZE THE PASSIVE SCALARS: /Y 0 F /Y 0 0 /(T 0 F T 0 O) 51 THREE SCALED PASSIVE SCALARS: Z 1 = Y F /Y 0 F Z 2 =1 Y O /Y 0 O Z 3 =(T T 0 O)/(T 0 F T 0 O) Z 1 Z 2 Z 3 FUEL OXI THREE PASSIVE SCALARS WITH THE SAME EQUATION AND SAME BC: THEY ARE EQUAL. WE CALL THEM THE MIXTURE FRACTION z Z 1 = Z 2 = Z 3 = z 52

27 THE MIXING LINES (NO COMBUSTION) Ch. 2 Sec INTERPRETATION OF z: A MASS WEIGHTED MEASUREMENT OF MIXING z=1 z=0 At a given point in the chamber: mass coming from stream 1 = m1 mass of fuel = m1 YF 0 mass coming from stream 2 = m2 total mass= m1+m2 fuel mass fraction YF= m1 YF 0 / (m1+m2) 54 mixture fraction z = YF/YF 0 = m1/(m1+m2)

28 INTERPRETATION OF z: z=1 z=0 If the mixture fraction at this point at time t is z, it means that at this point, in 1 kg of mixture: z kg come from stream 1 (1-z) kg come from stream 2 Careful: this works only if all Lewis numbers are equal 55 With combustion, z can also be introduced: Combining them two by two to eliminate source terms and normalizing to have the same boundary conditions leads to three passive scalars with identical bcs: they are all equal to the mixture fraction z 56

29 The mixture fraction z: In these expressions, only YF, YF and T are variable. All other quantities are fixed: - s is the stoichiometric ratio - YF 0 and YF 0 are the fuel and oxidizer inlet mass fractions in stream 1 and 2 respectively - Q is the heat of reaction per unit mass 57 Another property of the mixture fraction z: it does not change in a reacting zone State 1: Y 1 F,Y 1 O z 1 = sy 1 F Y 1 O + Y 0 O sy 0 F + Y 0 O State 2: Y 1 F Y 1 F,Y 1 O s Y 1 F z 2 = s(y 1 F Y 1 f ) (Y 1 O s Y 1 F )+Y 0 O sy 0 F + Y 0 O = z 1 58

30 How can we use the mixture fraction z?: Nice to have a quantity without source term But not sufficient. We can compute z in multiple flows but from z we cannot obtain T, YF and YO : in the above expressions, we have relations for couples of variables but not for individual variables. We need additional assumptions: the simplest one is the infinitely fast assumption 59 The infinitely fast chemistry assumption Suppose that kinetics are faster than all flow processes: fuel and oxidizer cannot co exist. There will be a fuel side (YF =0) and an oxidizer side (YO =0) That is enough to close the problem. For example, on the fuel side: 60

31 The z diagram with infinitely fast chemistry and unity Lewis numbers EQUILIBRIUM LINES + MIXING LINES (NO COMBUSTION) 61 There are multiple other z diagrams of growing complexity Ch. 3, Sect 3.5 Strained flamelet 62 Reversible single step Flamelet strained assumption

32 OK we have T, Yk = f(z). So what? What we really want is to have T and all Yk s as functions of time and spatial positions. Does z help us? Yes: the introduction of z represents a significant simplification as described now: 63 Instead of solving: Only solve the z equation: ρz t + x i (ρu i z)= x i ρd z x i Z diagram structure: T(z) YF(z) Yo(z) 64 And read T and Yk in the z diagram. We ll use this in turbulent flames

33 Think of the z diagram as trajectories: MIXING PHASE: First you mix and reach a given z Combustion phase: after mixing, you burn. And you stop burning when you reach the equilibrium lines z 66

34 Possible states: cant be outside the triangle limited by equilibrium and mixing lines POSSIBLE STATES 67 Parenthesis - a useful tool for codes: In 3D codes, one often needs to know the local equivalence ratio φ = sy F /Y O SinceYF and YO change because of combustion, the above expression cannot be used to know the local equivalence ratio except in the unburnt gases Mixing lines allow to do this: wherever you are in a reacting flow, if you know z, you can find the local equivalence ratio 68 φ = sy 0 F Y 0 O z (1 z) ONLY IF LEWIS =1

35 Example: Ch. 2 Sec Through a premixed flame front, z does not change but the local equivalence ratio does change from the fresh gas equivalence ratio to 0 in the burnt gas (for lean flames) φ = sy F /Y O Since z does not change through the flame front, we can compute the equivalence ratio of the mixture before combustion (on the mixing line) with: φ = sy 0 F Y 0 O z (1 z) or φ = sy 0 F Y 0 O sy F Y 0 + Y 0 0 s(y 0 F Y F )+Y O 69 Simple exercice: using Cantera, Cosilab or Chemkin Compute a 1D premixed flame with Lewis not equal for a chosen equivalence ratio φ 1 Compute z =(syf-yo+yo0)/(syf0+yo0) Plot z versus x: it will be constant Compute the equivalence ratio from z: And check it is equal to φ 1 φ = sy 0 F Y 0 O Repeat with real (non equal) values of Lewis numbers... z (1 z) 70

36 Will also work in 3D to know the local equivalence ratio in a code (useful for diffusion, partially premixed or spray flames): BEFORE COMBUSTION. EASY. What is the equivalence ratio of the gases in the fresh gas? φ = sy F /Y O AFTER COMBUSTION: cant use the same formula. Compute z then get the equivalence ratio: 71 φ = sy 0 F Y 0 O z (1 z) The limits of the z diagram The concept of splitting the resolution of diffusion flame structures in two parts (z diagram and mixing problem) is common in many codes, even for turbulent diffusion flames But can we really do this? can we base all our models on z? let us come back to the assumptions required to define a mixture fraction z. 72

37 Equal Lewis numbers?? Never happens. Lewis numbers are not equal and not equal to unity. Lewis numbers measured in a 1D premixed flame 73 Alternative definitions for z? Ch. 3 Sec The literature contains multiple improved definitions of z (based on elements, or on combination of elements). The mixture fraction defined using a single step global reaction is of course limited. For H2+1/2O2 -> H2O, define a mixture fraction z1: We know that this is not a good definition in practice since in the real world, we have more than one reaction Solution: work on elements (C, H or O) because elements are always conserved. Generalized mixture fractions 74

38 Other mixture fractions: For example, build a passive scalar on H atom: and a mixture fraction from it: See Bilger, Barlow, Pope s papers for other extensions. For example in the TNF workshop, Bilger s definition is: zb= 75 But all mixture fractions are... different: Measurements of various mixture fractions (based on C and H) in a piloted diffusion flame vs Bilger s z: 76

39 Alternative definitions for z? In the end, there is no such thing as a real mixture fraction because nothing (species or elements) diffuses at the same speed in a multispecies gas. This leads to multiple problems in practice since the multiple definitions of z lead to z s which are not equal Last comment on Lewis numbers When we will discuss turbulent flames, we will see that many authors use unity Lewis number assumptions. Why? It seems to work better... It may be vaguely justified by the fact that small scale turbulence leads to turbulent transport which dominates transport and is the same for all species, thereby justifying the equal Lewis number assumption 78

40 Stretch and scalar dissipation All previous derivations for premixed like diffusion cases use assumptions which are too simple. In practice, at least one additional notion is required for turbulent flames: stretch Stretch applies to any flame (premixed or diffusion) and measures the rate at which the flame area increases: κ = 1 A da dt Intuition tells us that stretching a flame must have a limit. The flame is affected by stretch (see Williams book or Pr Matalon s course). This effect is different for premixed and diffusion flames 79 Stretch and premixed flames Strain in the tangent plane to the flame Strain due to flame propagation Displacement speed Front curvature S. M. Candel and T. Poinsot. Flame stretch and the balance equation for the flame surface area. Combust. Sci. Tech., 70:1-15, Ch

41 Examples of stretched premixed flames 81 Evaluations of stretch: κ = U 1 + U 2 d This is an average value: stretch is not constant along the flame normal Defining flame stretch at the flame location is difficult 82

42 Flame location in a stagnation point flow: Flame thickness 83 Evaluations of stretch: spherical flames A =4πr 2 84 Stretch is constant on the flame front Knowing r(t) gives directly stretch Stretch changes like 1/r: very large at initial times

43 Variations of flame speeds with stretch Premixed flames are affected by stretch. Theory can be used to derive the variations of speeds: Markstein length Markstein number 85 Measuring Markstein numbers is extremely difficult... But the dependence of premixed flame speeds on stretch is not very strong, compared to diffusion flames Flame speeds versus stretch: depend on the Lewis number of the deficient reactant AND on heat losses: ADIABATIC Lewis=1 86

44 ADIABATIC Lewis < 1 NON ADIABATIC Lewis > 1 87 Stretch and diffusion flames: In diffusion flames, flame strain is often replaced by a more useful quantity called scalar dissipation : Why? - Scalar dissipation does not depend on the flame orientation (as flame stretch does). - For simple cases (stretched diffusion flame), stretch and scalar dissipation are directly linked. Ch

45 The diffusion flame with infinitely fast chemistry: Flame stretch φ = s Y 0 F Y 0 O 89 Scalar dissipation on the flame front We can work with stretch OR with scalar dissipation 90

46 Diffusion flames are VERY sensitive to stretch (or scalar dissipation) Ch Total fuel consumption in a stretched diffusion flame Flame stretch When you stretch a diffusion flame, it burns more: we all know we must blow on fires...: we just bring them more air If you stretch it too much, you can quench it: we also know that you should not blow too much on a candle, otherwise you kill it Stretching a diffusion flame: Total fuel consumption Ch

47 Implications for turbulent flames: In turbulent flames, the unsteady velocity field will induce strain and therefore stretch: Strain in the tangent plane to the flame Strain due to flame propagation 93 In turbulent flames: We will need to establish a link between the velocity field (created by an ensemble of vortices) and the flame stretch: 94

48 Are all flames premixed or non-premixed? No and two other cases must be mentioned: triple flames partially premixed flames 95 TRIPLE FLAMES: THE STRUCTURE WHICH SEPARATES IGNITED FROM NON IGNITED DIFFUSION LAYERS COMBUSTION MIXING Ch. 6 Sec In any flame where you inject pure fuel and pure oxidizer, you can be either on the mixing line or on the combusfon line. How do you go from one to the other? (in other words: how do you ignite a diffusion flame 96?)

49 TRIPLE FLAMES: THE STRUCTURE WHICH SEPARATES IGNITED FROM NON IGNITED DIFFUSION LAYERS Fuel T T Oxidizer MIXING STATE BURNING STATE Fuel Pr Kioni flame Oxidizer 97 TRIPLE FLAME POINTS IN A z DIAGRAM: Mixing lines Infinitely fast chemistry Fuel Premixing zone Rich premixed flame Diffusion flame Oxidizer Lean premixed flame 98

50 1/ TRIPLE FLAMES PROPAGATE. 2/ THEY PROPAGATE FASTER THAN PREMIXED FLAMES A TRIPLE FLAME SPEED SCALES LIKE: ρ1 s Triple = s 0 L ρ 2 Stoechiometric laminar flame speed Density rafo 99 Partially premixed flames Almost noone uses perfectly premixed flames: too dangerous And almost noone uses pure diffusion flames: not efficient Most flames will be produced in devices where we TRY to mix at the last moment to reach premixed conditions but never really make it... Partially premixed flames Mean equivalence ratio Real distribution on flame front PDF Equivalence ratio 1.2

51 101 Example: the PRECCINSTA burner (DLR)