Numerical Methods for Problems with Moving Fronts Orthogonal Collocation on Finite Elements

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1 Electronic Text Provided with the Book Numerical Methods for Problems with Moving Fronts by Bruce A. Finlayson Ravenna Park Publishing, Inc., nd Ave. N. E., Seattle, WA ; 6.5. Orthogonal Collocation on Finite Elements The method of orthogonal collocation on finite elements provides a collocation method that is high order. It is applied here to the convective diffusion equation to illustrate its success. When the Peclet number is large, the orthognal collocation method will not provide good results since the number N must be so large to approximate the solution. In that case we are better off using the method of orthogonal collocation on finite elements, as developed by Carey and Finlayson [975]. The domain is divided into elements and the method of orthogonal collocation is applied within each element. The numbering scheme is shown in Figure 6E.. There are three types of nodes: those at which a boundary condition is applied (2 of these), those at which the residual is set to zero (NCOL x NE of these), and those at which the first derivative is made continuous across element boundaries ( NE of these). The total number of points is then NT = (NCOL+) NE +. Within an element we transform the domain using ξ = x x k x k, x k = x k+ x k m2 so that the variable ξ goes from to within an element (rather than - to as we have used for the Galerkin methods). The differential equation can be transformed to give c t + Pe c x k ξ = x 2 k 2 c ξ 2 m2 and the boundary conditions are c(,t) = c x NE ξ (,t) = c(x,) = f(x) m22 Now we are going to use four different numbering systems depending on what we are solving for. The local numbering systems within an element is shown in Figure 6E. and the global number system is shown in Figure 6E.2. The index i represents a global number and the index I represents a local number; in the later case the element must also be identified. Thus we have the correspondence i = (k ) (NCOL + ) + I m23 x i = x (k) + ξ I x k

2 I = = NP = NCOL+2 ξ = Figure 6E.. Local numbering On an element we set the residuals to zero. c t (ξ I, t) + Pe c x k ξ (ξ I, t) = x 2 k 2 c ξ 2(ξ I, t), I =,..., NCOL m24 We also apply the boundary conditions x NE c i= = or c I= () = NCOL+2 A NCOL+2,I c I (NE) = I= m25 c (k) I () = f(x i ) Next we make the first derivatives continuous between the elements by taking Evaluated in the collocation method this is dc (x(k ) NCOL+2 ) = dc (x(k) ) m26 x k NCOL+2 A NCOL+2,I c I (k ) = I= x k NCOL+2 A,I c I (k) I= m27 The method of solution is to use an explicit method to step forward in time. c J (k),n+ c J (k),n t + Pe x k N+2 I= A JI c (k),n I = = N+2 xk 2 I= B JI c (k),n I, J = 2,...,NCOL+,k=,NE m28 If we know the solution at time level n, c i n we can find c I (k),n. Thus we can obtain the value of c i n+ for

3 x k x (k) x (k+) Residual Condition Boundary Conditions Element boundary, continuity x i Global numbering = NT = 2 NE Numbering for solving the ordinary differential equations = NE+ Figure 6E.2. Nodal numbering system each of the points interior to the element, marked by in Figure 6E.2. Implicit methods are also possible, and the linear algebra is discussed by Finlayson [98, p. 6]. Suppose we have made the calculation Eq.(m28) for all of the collocation points interior to each element. We still need the values of c i n+ for each of the points between elements and the two end points. Rearrange the equations Eq.(m25) and Eq.(m27) to have the unknowns on the left-hand side.

4 x k A NP, c (k ),n+ I= + = NP x k I=2 x k A NP,NP c (k ),n+ NP A Np,I c (k ),n+ I + NP x k I=2 x k A, c (k),n+ I= A J c (k),n+ I x k A,NP c (k),n+ NP = m29 c (),n+ I= = x NE A NP, c (NE),n+ + x NE A NP,NP c (NE),n+ NP = NP x NE I=2 A NP,I c (NE),n+ m3 We make the function continuous by taking c (k ),n+ NP = c (k),n+ m3 This system of equations can be represented as a tri-diagonal form (see Eq. 2.5). The values of x i in Eq. 2.5 are the values of c i n+ at the two end points and the points between the elements, using the third numbering system shown in Figure 6E.2. This tri-diagonal system can be solved and the results rearranged to provide the solution for each and every collocation point shown in Figure 6E.2, first numbering system, at the n+-th time level. Results are given for the method, but using the implicit method (which was readily available). For the convective diffusion equation the collocation results for 25 elements and 2 interior points are shown in Figure 6.33a. There are small oscillations. Jensen and Finlayson [98] have shown that oscillations do not occur in the steady state problem provided Pe x m35 This is the number that is comparable with Pe x/2 in the Galerkin method. For this calcualtion the quantity in Eq. (m35) takes the value.5. The Galerkin method with Pe x/2 = gives the results shown in Figure 5.8a; the Taylor-Galerkin method with Pe x/2 = gives the results shown in Figure 5.8e. There are small oscilllations in those solutions, too. When more elements are used, 5 elements and 2 interior collocation points, the method of orthogonal collocation on finite elements gives the results shown in Figure 6.33b. In this case there are no oscillations. All of these methods provide good solutions. The collocation method can also be applied with higher degree polynomials. A solution with 25 elements and 5 interior collocation points (a sixth degree polynomial within each element) gives a solution with oscillations that are very small. Here the key value is Pe x 2 5 m36 and it takes the value 4. The method of orthogonal collocation on finite elements can also be applied to non-linear prob-

5 lems as well. If Burger s equation is solved in the form u t + u u x = ν 2 u x 2 m37 we just replace Eq.(m24) with u t (ξ I, t) + u(ξ I, t) x k u ξ (ξ I, t) = ν xk 2 2 u ξ 2(ξ I, t), I =,..., NCOL m38 An explicit method can be used with this set, as well. Solutions for such a method are not shown here.

6 7.4. Special Petrov-Galerkin Weighting Functions Because S is an even function and satisfies Eq. (7.44) we have ds ds ds = h Since S() = this also says S() =. The integral s ds ds ds h2 can be evaluated by regarding s as the derivative of s 2 /2 and then integrating by parts. s ds ds ds = d ds (s S ) ds S ds = S ds h3 Since S is an even function we have and hence S ds = 2 s ds ds ds = 2 S ds = 2 h4 h5 The following relation can be obtained by integrating by parts the left-hand side. s 2 ds ds ds = 2 s S ds h6 With similar manipulations, (and using the fact that T is an odd function), we obtain dt ds ds =, s dt ds ds = T ds, s 2 dt ds ds = 2 s T ds h7 In the Petrov-Galerkin method the weighted residual is set to zero, Eq. (7.4). Integrate this by parts to obtain ν dw j du dw j V(u) = h8

7 The first term is ν dw j du = ν du + ν du h9 Consider each part in turn. du = NT i= u i dn i h2 But the derivative of the trial function N j is constant within an element so that we get dn j = h ds ds dn j = h ds ds + dn j+ = h ds ds h ds ds h2 Combining all results gives du = u j h ds ds + u j h ds ds ds ds + u j + h ds ds h22 The terms are evaluated using Eq. (7.42) and (7.44). du = h ( u j + 2 u j u j+ ) = u j u j u j+ u j = m h h j m j+ h23 Similarily for the second part of Eq. (h9). du = u j h ds ds + u j h ds ds ds ds + u j + h ds ds h24 By Eq. (7.42) we have

8 ds ds = ds ds h25 and by Eq. (7.44) both terms are zero. Thus there is not contribution due to the T j function. The first term of the weighted residual is then ν dw j du = ν (m j m j+ ) h26 The second term of the weighted residual is dw j V(u) = V(u) + V(u) = ds V(u) ds + ds V(u) ds + ds V(u) ds + ds V(u) ds h27 In the left integral (s = - to ) we use u = u j + ( u j u j ) ( s + ) = u j + m j h s h28 In the right integral (s = to ) we use u = u j + ( u j+ u j ) s = u j + m j+ h s h29 Examine each of the terms in Eq. (h27). V(u) = ds V(u j + m j h s) ds + ds V(u j + m j+ h s) ds h3 Since /ds is an odd function ds V(u j + m j h s) ds = ξ= ξ= dξ V( u j m j h ξ ) dξ = ds V(u j m j h s) ds h3 and we get

9 V(u) = ds V(u j m j h s) ds + ds V(u j + m j+ h s) ds h32 The same steps applied to the term involving /ds, which is an even function, gives V(u) = ds V(u j m j h s) ds + ds V(u j + m j+ h s) ds h33 Thus the weighted residual becomes ν (m j+ m j ) ds V(u j m j h s) ds + ds V(u j + m j+ h s) ds + + ds V(u j m j h s) ds + dt h34 j ds V(u j + m j+ h s) ds = Now for this application we have V(u j m j h s) = 2 u2 j m j u j h s + 2 m2 j h 2 s 2 V(u j + m j+ h s) = 2 u2 j + m j+ u j h s + 2 m2 j+ h2 s 2 h35 Substitution into Eq. (h34) gives ν + u j h α 2ν ( m j+ m j ) u j 2 ( u j+ u j ) = = h 2 ( mj+ 2 m2 j ) s S(s) ds + h 2 ( mj+ 2 + m2 j ) s T (s) ds h36 In this equation we have let α = 2 T (s) ds h37 The parameter α thus determines the amount of upstream weighting; if both α and u j have the same sign then the inclusion of the function T j increases the coefficient of the diffusion term. The presence of the function T j also decreases the truncation error. This development shows how to get an upstream method using the weighting function (7.39).

10 .3. Combustion in Flames The equations governing the propagation of flames in one dimension are given by Ramos [987]: ρ t + x ( ρ u ) = ρ Y i t + u Y i = R x i x ( ρ Y i V i ) b ρc T p t + u T x = p t N i = h i R i q x N i= ρ Y i V i C pi T x b2 Y N = N i= p = ρ R T N i= h i = h i + T Y i Y i Mˆ i T C pi dt,, C p = N i= q = λ T x Y i C pi b3 These are the continuity equation, conservation of mass, the energy equation, an equation saying the mass fractions add to one, the ideal gas law, the equation for heat capacity of a mixture, the equation for enthalpy of a species in an ideal gas, and q is the heat flux. The variables are ρ the density, u the velocity, t the time, x the distance, Y i the mole fraction of the i-th species, w i dot the reaction rate for the i-th species, V i the diffusion velocity of the i-th species, p the pressure, T the absolute temperature, R squiggle the gas constant, C p the heat capacity of the mixture, h i the enthalpy of the i-th species, and λ the thermal conductivity. In these equations we have assumed that the flame speed is slow enough that the Mach number is low; in that case the pressure is constant in space. We have also neglected the Soret and Dufour effects (or thermal diffusion). One form of the equations is for an infinite premixed combustible fluid mixture. The x-position goes from - < x < and the flame propagates from one boundary (x = - ) to the other. If the mixture is stationary, the fluid velocity ahead of the flame is zero. We transform the equations into a Lagrangian coordinate system in order to eliminate the convection term. This is possible because of the infinite extent of the physical domain. We go from where ( t, x ) ( τ, ψ ) b4

11 τ = t x = ρ t = x ρ t = x (ρ u ) x = ρ u b5 The derivatives are then calculated using ρ t = ρ τ (ρu) x τ t + ρ = (ρu) τ t τ x + (ρu) x b6 Since τ t =, t = ρ u τ x =, x = ρ b7 we obtain ρ t = ρ τ (ρu) x ρu ρ = ρ (ρu) b8 The continuity equation is then ρ τ ρu ρ + ρ (ρu) = b9 which becomes ρ τ + u ρ2 = b Likewise the derivatives of temperature and mass fraction are given by T x = T τ τ x + T x = ρ T b Thus the convective derivatives are

12 T t + u T x = T T ρu τ + u ρ T = T τ b2 With these manipulations the equations become Y i τ = ρ R i ( ρ Y i V i ) T τ = ρc p N i = h i R i ρ q N i= ρ 2 Y i V i C T pi b3 The boundary conditions are T T (,τ) = (,τ) = Y i (,τ) = Y i (,τ) = T(,τ) = T un, Y i (,τ) = Yi un b4 and the initial conditions are T(ψ,) = T un (ψ) Y i (ψ,) = Yi un (ψ) b5 where the superscript un represents the value in the unburned region. The Fickian model is obtained by making further assumptions. Assume that the binary diffusion coefficients of all pairs of chemicals are the same, the specific heats of all species are the same and that ρ 2 D = β = constant The diffusion velocity is given by Fick s law b6 V i = D x ( ln Y i ) = D ρ ( ln Y i ) = Dρ Y i ρ Y i V i = D ρ 2 Y i = β Y i Y i b8 The Lewis number is taken as one.

13 λ Le = = ρdc p b45 The equations are then Y i τ = β 2 Y i 2 + R i ρ T τ = β 2 T 2 N i= h i ρ R i ρc p b7 Another simplification of Eq.(b3) applies when a burner-stabilized flame is achieved. The premixed gas enters the burner at the flame holder at x = and flows to the flame zone and on to the burned zone (see Figure E.3). In this case we make the same transformation in Eq. (b.5) except that we replace one equation with: t = x ρ t = x (ρ u ) x = ρ u + ṁ(t) b9 where ṁ (t) = ρ(,t) u(,t) b2 is the mass flux at the flame holder. Eq. (b3) are then T b Y b flame T un Y un flow Figure E.3. Burner-stabilized Flame Geometry

14 Y i τ + ṁ Y i = ρ R i T τ + ṁ T = ρc p N ( ρ Y i V i ) i = h i R i + ρ q N i= ρ 2 Y i V i C T pi b2 We consider only steady-state solutions of these equations ṁ dy i dψ = R i ṁ dt dψ = C p N d dψ ( ρ Y i V i ) i = h i R i dq dψ N i= ρ Y i V i C dt pi dψ b22 together with boundary conditions T() = T un, Y i () = Y un i dt () = dt ( ) = dy i () = dy i ( ) = b23 We must solve these equations for the distribution of mole fractions, Y i, and temperature, T; we must also solve for the mass flux, m dot.

15 .4. Transient Catalytic Muffler The unsteady problem uses the full equations Eq. (.5)-(.5). A finite difference formulation of this equation would give α 5 α dts i 3 dt r hs L 2 s T i + Ti s + Ti s z Nu ( Ts i T i ) + β 5 < r i > = c which is to be solved subject to the steady equations Eq. (.35)-(.36) and Eq. (.38). Young and Finlayson [976] did this using orthogonal collocation on finite elements for the solid temperature and solving the integro-differential equations, Eq. (.5)-(.5). Here we consider several other options. If we evaluate the integral in Eq. Eq. (.48)-(.49) from one grid point to the next we get T i+ = e 2 α Nu z T i + 2 α Nu z i+ If the integral is evaluated using the trapezoid rule, the result is z i e 2 α Nu ( z i+ z') T s (z' ) dz', z i+ = z i + z c2 T i+ = e 2 α Nu z T i + 2 α Nu z T s 2 α Nu z 2 i + T s 2 i+ c3 This equation is of the form T i+ = γ T i + γ 2 T s i+ + γ 3 T s i c4 If the Euler method is applied to Eq. (.35)-(.36) we get T i+ T i z + 2 α Nu T i = 2 α Nu T s i c5 or if a second-order Runge-Kutta method is applied we get instead T i+ T i z + 2 α Nu 2 ( T i + T i+ ) = 2 α Nu 2 ( T s i + T s i+ ) c6 Both Eq. (c5-6) can be rearranged to give or T i+ = T i ( 2 α Nu z ) + 2 α Nu z T s i α Nu z T i+ = T + α Nu z i + z 2 2 α Nu + α Nu z ( Ts i + T s i+ ) c7

16 Notice that these are appoximations in various degrees to Eq. (c3). They can also be put into the form of Eq. (c4). Thus the temperature equations are Eq. (c) and Eq. (c4). The equations are similar for the mole fraction Y and Y s. The equation (.47) or Eq. (.49) can be written in finite difference form as where the Euler method uses Y i+ = δ Y i + δ 2 Y s i+ + δ 3 Y s i c8 δ = 2 α 2 Sh z, δ 2 =, δ 3 = 2 α 2 Sh z c9 and the Runge-Kutta method uses δ = α 2 Sh z + α 2 Sh z, δ 2 = α 2 Sh z + α 2 Sh z, δ 3 = α 2 Sh z + α 2 Sh z c2 The Eq. (.38) is then α 6 β 4 < r > i + 2 Sh (Ys i Y i ) = c2 Consider the case when the variables T i, Y i, T i s, Y i s are known at some time t. Then Eq. (c) can be used to step forward to determine T i s at time t + t for all i. Eq. (c4) can then be used to step forward to determine T i at time t + t for all i. Next look at the concentration equation for the first node. α 6 β 4 < r > + 2 Sh (Ys Y ) = c22 The value of Y would be known from the boundary condtions at every time t, including time t + t. Thus we must evaluate the reaction rate. Use y = Y in Eq. (.39), then y = Y in Eq. (.42), and then calculate < r > from Eq. (.43). Then Eq. (c22) gives Y s. At this point in the calculations we have the value of T i and T i s at time t + t for all i, and we know Y, Y s at time t + t. We then solve Eq. (c8) and (c2) for Y 2, Y 2 s at time t + t. These are written in terms of the unknowns Y i+ n+, Y i+ s,n+ and the known quantities Y i n+, Y i s,n+. Yi+ n+ δ 2 Yi+ s,n+ = δ Yi n+ + + δ 3 Yi s,n+ c23 α 6 β 4 < r > n+ i+ + Sh (Ys,n+ 2 i+ Yi+ n+ ) = c24 This set of two equations must be solved simultaneously for i=. Then we can increase i and repeat until we reach an i representing the end of the reactor. Then we know T i, Y i, T i s, Y i s at time t + t for all i. We then repeat the process. Of course, all the techniques discussed in the book can be applied. In particular the nodes can be moved as the front moves and concentrated near the reaction zone. Travelling waves could also be evaluated.

17 2.3. Compressibility Method The first method originated with Chorin [967] and is called the artificial compressibility method because the incompressible fluid is allowed to have a small compressibility. The chief problem with the Navier-Stokes equations and the continuity equation is that the Navier-Stokes equations are equations for velocities, and they include the pressure gradient. The continuity equation is a constraint on the velocity but is in essence the equation for pressure. The continuity equation for an incompressible fluid does not have a time derivative in it, so in the artificial compressibility method a time derivative is added. The continuity equation is expanded to be p t + c2 u = u The constant parameter c 2 is chosen to aid in convergence and the transient itself does not have any physical significance. The method is termed artificial compressibility because Eq. (u) is the continuity equation for a compressible fluid with an equation of state p = c 2 ρ u2 The method itself is listed in Table 2.2. Now we have three (or four) parabolic equations in time for the pressure and the two (or three) velocity components. Standard algorithms for parabolic equations can be applied. When the algorithm is applied each variable can be identified with the same node, as shown in Figure 2.E(a), or different grids can be used for different variables, as in Figure 2.E(b). u v p u u p u v p u p v p (a) regular mesh v v (b) staggered mesh Figure 2.E. Finite difference meshes The first grid, Figure 2.E(a) is not a preferred one because the boundary conditions for pressure are inconvenient to derive. We discuss here the equations involving boundary terms for the staggered grid, Figure 2.E(b), which is known as the MAC mesh. Consider the problem shown in Figure 2.E2, an entry flow problem with a solid boundary, a line of symmetry, an entry condition and an exit condition. For side A the u nodes occur on the boundaries, so that we set the u velocity there. The same thing is true for the v velocity on B and C. However, the v velocity nodes are not on the boundary A, nor are the u velocity nodes on the boundary B. On A we force the boundary condition by taking the v-

18 value x/2 outside the mesh to equal minus the v-value x/2 inside the mesh. Then the average value is zero, and the average value represents the v-value on A. u = v = u = 2(-y ) v = 2 A B C D u/ x = v/ x = v =, u/ y = Figure 2.E2. Entry-length problem v,j+/2 = v,j+/2 In an analogous fashion we use u i+/2,m+ = u i+/2,m for the u-boundary conditions on surface B. To force the first derivative of u to zero on boundary C we use u i+/2,m+ = u i+/2,m This represents the value of the first derivative centered on y=. On the exit boundary D the v derivative is centered on the exit. v N+,j+/2 = v N,j+/2 The u-velocity is expressed at nodes that are on the boundary. To set the first derivative to zero we take the value x from the exit to equal to the value + x from the exit. u N+3/2,j = u N /2,j Following these rules, and using an explicit method for integrating in time, gives the following algorithm for this case. u n+ i+/2,j un i+/2,j = pn i+,j pn i,j u n u n i+3/2,j un i /2,j t x i+/2,j + 2 x 4 (vn i+,j+/2 + vn i,j+/2 + vn i,j /2 + vn i+,j /2 )un i+/2,j+ un i+/2,j 2 y + + Re u n i + 3/2,j 2un i+/2,j + un i /2,j x 2 i =,..., N; j=,..., M u n i+/2,j+ + i+/2,j + un i+/2,j y 2 u5

19 v n+ i,j+/2 vn i,j+/2 = pn i,j+ pn i,j t y v n i,j+/2 v n i,j+3/2 vn i,j /2 2 y 4 (un i+/2,j + un i+/2,j+ + un i /2,j + un i /2,j+ ) vn i+,j+/2 vn i,j+/2 2 x + + v n i +,j+/2 2vn i,j+/2 + vn i,j+/2 Re x 2 + v n i,j+3/2 vn i,j+/2 + vn i,j /2 y 2 i =,..., N; j=,..., M- p n+ i,j p n i,j t = c 2 u n i + + /2,j u n+ x i /,j + vn+ i,j+/2 v n+ i,j /2 y i =,..., N; j=,..., M surface A: u / 2, j = 2 ( y 2 j ), v,j+/2 = v,j+/2 surface B: u i+/2,m+ = u i+/2,m, v i,m+/2 = surface C: u i+/2, = u i+/2,, v i,/2 = surface D: u N+3/2,j = u N /2,j, v N+,j+/2 = v N,j+/2

20 3.4. u-v Formulation Another way to solve this problem is to tackle the second-order problem, Eq. (3.9), but with the new variable v = u t f3 Then Eq. (3.9) is v t = v E + E 2 u y 2 f4 coupled with u t = v f2 We now have a system of two equations to solve for u(y,t) and v(y,t). Note that the two equations look like a diffusion Eq.(f4) with a source term and an equation with only a source term, Eq. (f2). The convective nature of the problem is not apparent with this formulation. Only a centered finite difference method is suitable here, since upstream derivatives cannot be used on the non-existent convection terms. dv i dt = v i E + u i+ 2 u i + u i E y 2 du i dt = v i i = 2,..., N f5 This set is applied at the points i = 2,..., N-. The two end points must be handled with boundary conditions. u(,t) = or u (t) = u(,t) = or u N (t) = f6 Thus we know the value of u and u N that are needed in the first of Eq. (f5) when i = 2 or i = N-. We do not know the corresponding values of v and v N, but fortunately we do not need them.. Thus we have enough equations for the solution. The initial conditions are u(y,) = or u i () =, i =,...N u t (y,) = or v i () =, i=2,...,n f7

21 We then integrate the 2(N-2) equations, Eq.(f5), together with the boundary conditions, Eq.(f6), beginning with the initial conditions, Eq.(f7). The stability of this set of equations can be derived using the Fourier analysis, which gives t y2 4 f2 or a time step four times smaller than for the other set of equations.