Numerical Methods Applied to Chemical Engineering Homework #3. Nonlinear algebraic equations and matrix eigenvalue problems SOLUTION

Transcription

1 0.34. Numerical Methods pplied to Chemical Engineering Homework #3. Nonlinear algebraic equations and matrix eigenalue problems SOLUTION Problem. Modeling chemical reaction with diffusion in a catalyst pellet In HW #, we considered the use of the finite difference method to conert a boundary alue problem into a set of linear algebraic equations. In HW #, we considered the use of Newton s method to sole sets of nonlinear algebraic equations. In this problem, we combine these two approaches to sole a boundary alue problem that contains a nonlinear term inoling chemical reaction. This problem is inspired by the discussion of nonisothermal catalyst particles in section 3.9 of G. F. Froment and K. B. Bischoff, Chemical Reactor nalysis and Design, nd ed., Wiley, 990. We consider the case of a spherical catalyst pellet of radius R (see figure below). We want to write a MTLB program that will employ a finite difference method to sole for the dimensionless concentration profile and the effectieness factor. The goerning equation in dimensionless form is d dψ γβ( Ψ) ξ ξ Φ exp Ψ = 0 (EQ 56) dξ dξ + β( Ψ) subect to the boundary conditions dψ Ψ () = = 0 (EQ 57) d ξ ξ= 0 where Ψ is the dimensionless concentration inside the catalyst pellet and ξ is the dimensionless radius of the particle. The program should take three input parameters: Φ, γ, and β, which are defined in EQ (50), (5), and (54) of the assignment sheet, respectiely.

2 We are going to sole this problem using finite differences. Let s number the grid points in the following way: ξ = 0 ξ = 0 N- N N+ Discretizing the second deriatie in (EQ 56) using central finite differences d d ξ Ψ Ψ ξ d dξ dξ ξ+ / ξ / where + / / d dξ dξ Ψ ξ + / ξ / ξ = = ξ = ξ Ψ Ψ Ψ Ψ ξ ξ ξ, +,,, + / / ξ+ ξ ξξ + / / * =, Ψ, +, Ψ, +, + Ψ, + (EQ 7 ),,, + ξ / = ( ξ ξ )( ξ+ / ξ /) ξ / ξ + / = + ξ+ / ξ / ξ+ ξ ξ ξ ξ + / = ( ξ+ ξ)( ξ+ /ξ/) The nonlinear algebraic equation for an interior grid point, [, N], takes the form γβ( Ψ, ) *, Ψ, +, Ψ, +, + Ψ, + ξ Φ exp Ψ, = 0 (EQ 8 ) + β( Ψ, ) s described on page 5 of the assignment sheet, some modification of the aboe equation is required for the first and last grid points. t the last grid point, = N, the discretized form of the differential equation is γβ( Ψ, ) N * N, N Ψ N, + N, NΨ N, + N, N+ ΨN, + ξnφ exp Ψ N, = 0 (EQ 30 ) + β( ΨN, ) where Ψ, = Ψ = from the boundary condition. N+ ξ=

3 The modification for grid point = is slightly more inoled and is described in detail on pp. 6-7 of the assignment sheet. Using the second boundary condition, we arried at the relation 4Ψ Ψ =,0 Ψ (EQ 4 ),, * 3 and the algebraic equation therefore became 4 γβ( Ψ,) *, +,0 Ψ, +,,0 Ψ, ξ Φ exp Ψ, = 0 (EQ 43 ) β( Ψ,) s was mentioned in the hints for this assignment, the fsole() routine is executed much faster if the Jacobian matrix is calculated analytically and supplied to the subroutine. The general form of the Jacobian is df dx df dx df dx N df dx df dx df dx N Jac = dfn dx dfn dx dfn dx N In our case, [f.f N ] will be expressions in (EQ 8 * ), (EQ 30 * ), and (EQ 43 * ) and [x.x N ] are [Ψ,.Ψ,N ]. Gien that these three equations are at most functions of three different Ψ,, the Jacobian matrix will hae at most three non-zero entries in each row. For completeness, let s find the entries of the Jacobian for an interior point. df Jac(, ) = = dψ, df Jac(, + ) = = dψ, +,, + df d γβ( Ψ ) Jac(, ) = = ξ Φ exp Ψ dψ Ψ + ( Ψ ) dθ =, ξ Φ Θ+ Ψ, dψ, where Θ =,,,, d, β, γβ( ) Ψ + β( Ψ, ), exp and therefore 3

4 dθ ( γβ) γβ ( Ψ, ) = Θ + dψ, + β( Ψ, ) + β( Ψ, ) t this point, we are ready to code in the problem in order to calculate the concentration profile. Once that is done, this profile can be used to calculate the effectieness factor from (EQ 6) using the trapz() function. The first of the attached plots is a concentration profile for a particular set of input parameters, and the second plot is the master plot required by Question.. 4

5 % PROBLEM CODE % catalyst_nonisothermal_scan_.m % % This MTLB program plots the effectieness % factor s. the Thiele modulus for a catalyst % pellet with first order reaction at steady % state in nonisothermal operation. % No external mass or heat transfer resistance % is included, and Diriclet boundary conditions % are employed at the particle surface. % % To sole the dimensionless PDE's, finite % differences are used with a non-uniform % grid (except near r=0 for discretized % symmetry boundary condition). % % The dimensionless input parameters are : % Phi_TM = Thiele modulus % = product of particle radius and % the square root of the rate % constant diided by the diffusiity % = measure of importance of diffusion % limitations % gamma = actiation energy / (R*Ts), where % Ts is the known surface temperature % beta = Da*(-DH)*Cs/lambda/Ts, where Da is % the diffusiity, DH is the heat of % reaction, Cs is the known surface % concentration, lambda is the thermal % conductiity, and Ts is the known % surface temperature % = measure of importance of heat generation % by reaction s. thermal conductiity % % Written K. Beers. MIT ChE. 9//00 %. 9//00. modified from catalyst_nonisothermal_scan_.m % to compute the Jacobian analystically, saing % time required to perform finite differences % % COMPRISON : beta = 0, gamma = 0, % phi_tm_min = e-, phi_tm_max = e, % num_tm = 0, dz = 0.0, dz_fine = 0.00 % catalyst_nonisothermal_scan.m: CPU =.56 % catalyst_nonisothermal_scan_.m: CPU =.6 % (BUT HD WRNINGS SINCE USED SPRSE % JCOBIN, STILL WORKS) function iflag_main = catalyst_nonisothermal_scan() cpu_start = cputime; iflag_main = 0; % First, hae the user input the simulation parameters. %disp('enter simulation parameters,...'); 5

6 % problem parameters %beta = input('enter beta (row ector) : '); beta = [-0.;-0.;0;0.05] num_beta = length(beta); gamma = input('enter gamma : '); % Thiele modulus bounds (log scale) phi_tm_min = input('enter min. Thiele modulus : '); phi_tm_max = input('enter max. Thiele modulus : '); num_tm = input('enter number of Thiele modulus points : '); phi_tm = logspace(log0(phi_tm_min),log0(phi_tm_max),num_tm); % number of physical grid points, use simple uniform scale with fine grid near surface dz = input('enter interior grid point spacing : '); dz_fine = input('enter surface (0.9-.0) spacing : '); z_grid = [ [dz : dz : 0.9-dz], [0.9 : dz_fine :.0-dz_fine] ]; num_grid = length(z_grid); % We now scan the alues of the Thiele modulus, starting at the lowest alue. For each alue of % the Thiele modulus, we need to sole the set of nonlinear algebraic equations obtained by finite % difference disrectization of the goerning equations. We start at the lowest alue of the Thiele % modulus, and use as an initial guess a completely uniform concentation profile. In the limit of % the Thiele modulus goes to zero, this initial guess is correct. For each subsequent calculations, % we use the final result of the last calculation as an initial guess. % allocate space to store the effectieness factor eta_eff = zeros(num_beta,num_tm); figure; for k_beta = :num_beta; % form initial guess - uniform concentration psi_guess = ones(size(z_grid)); % set some options flags for fsole() options = optimset('tolfun',e-8, 'Display','off', 'LargeScale','off',... 'Jacobian','on'); % iterate oer each alue of the Thiele modulus do_plot = 0; for i_tm = :num_tm % compute dimensionless profile [psi,fal,exiflag] =... fsole(@catalyst_nonisothermal_calc_f,... psi_guess, options,... phi_tm(i_tm), beta(k_beta), gamma, z_grid); if(do_plot) figure; plot(z_grid,psi); xlabel('\zeta (dimensionless radius)'); ylabel('\psi (dimensionless conc.)'); title('nonisothermal catalyst pellet'); end 6

7 % compute effectieness factor eta_eff(k_beta,i_tm) = calc_eta_eff(psi,... phi_tm(i_tm), beta(k_beta), gamma, z_grid) % disp(['eta = ' intstr(eta_eff(k_beta,i_tm))]); end % Now, plot out the results semilogx(phi_tm,eta_eff); %gtext(numstr(beta(k_beta))); hold on; end legend('\beta = -0.','\beta = -0.','\beta = 0','\beta = 0.05') xlabel('\phi (Thiele modulus)'); ylabel('\eta (effectieness factor)'); title(['nonisothermal catalyst: ', '\gamma = ', numstr(gamma), ' \beta aried']); iflag_main = ; cpu_end = cputime; cpu_elapsed = cpu_end - cpu_start; disp(['elapsed CPU time : ', numstr(cpu_elapsed)]); return; % ========================================================= % This routines calculates the function alues for the % set of equations that sole for the dimensionless % concentration profile within the catalyst pellet % for a nonisothermal first order reaction. % K. Beers. MIT ChE. 9//00 function [fal,jac] = catalyst_nonisothermal_calc_f(psi,... phi_tm, beta, gamma, z_grid); num_grid = length(z_grid); % allocate space for function ector % and Jacobian fal = zeros(size(psi)); Jac = spalloc(num_grid,num_grid,3*num_grid); % For each point except the first and last, compute % the alue of the function. for =:(num_grid-) % estimate the local second deriatie in % spherical coordinates z_mid_up = (z_grid(+)+z_grid())/; z_mid_lo = (z_grid()+z_grid(-))/; dz_mid = z_mid_up - z_mid_lo; dz_up = z_grid(+) - z_grid(); 7

8 end dz_lo = z_grid() - z_grid(-); _lo = (z_mid_lo^)/(dz_lo*dz_mid); _mid = (-/dz_mid)*((z_mid_up^)/dz_up + (z_mid_lo^)/dz_lo); _up = (z_mid_up^)/(dz_up*dz_mid); second_deri = _lo*psi(-) + _mid*psi() +_up*psi(+); % compute nonlinear reaction term theta = exp( gamma*beta*(-psi()) / ( + beta*(-psi()))); rxn = (z_grid()^)*(phi_tm^)*theta*psi(); % compute function alue fal() = second_deri - rxn; % compute Jacobian elements Jac(,-) = _lo; Jac(,+) = _up; dtheta_dpsi = theta* (... (-gamma*beta)/(+beta*(-psi())) +... (gamma*beta^*(-psi()))/((+beta*(-psi()))^)); drxn_dpsi = (z_grid()^)*(phi_tm^) * (theta + psi()*dtheta_dpsi); Jac(,) = _mid - drxn_dpsi; % Now, calculate function alue for last point, employing known concentration at the % catalyst surface. z_mid_up = (+z_grid(num_grid))/; z_mid_lo = (z_grid(num_grid) +... z_grid(num_grid-))/; dz_mid = z_mid_up - z_mid_lo; dz_up = - z_grid(num_grid); dz_lo = z_grid(num_grid) - z_grid(num_grid-); _lo = (z_mid_lo^)/(dz_lo*dz_mid); _mid = (-/dz_mid)*((z_mid_up^)/dz_up +... (z_mid_lo^)/dz_lo); _up = (z_mid_up^)/(dz_up*dz_mid); second_deri = _lo*psi(num_grid-) +... _mid*psi(num_grid) + _up; % compute nonlinear reaction term theta = exp( gamma*beta*(-psi(num_grid)) /... ( + beta*(-psi(num_grid)))); rxn = (z_grid(num_grid)^)*(phi_tm^)*theta*psi(num_grid); % compute function alue fal(num_grid) = second_deri - rxn; % compute Jacobian elements = num_grid; Jac(,-) = _lo; dtheta_dpsi = theta* (... (-gamma*beta)/(+beta*(-psi())) +... (gamma*beta^*(-psi()))/((+beta*(-psi()))^)); drxn_dpsi = (z_grid()^)*(phi_tm^) *... (theta + psi()*dtheta_dpsi); Jac(,) = _mid - drxn_dpsi; 8

9 % Then, calculate function alue for % first point, using symmetry boundary % condition. z_mid_up = (z_grid()+z_grid())/; z_mid_lo = (z_grid())/; dz_mid = z_mid_up - z_mid_lo; dz_up = z_grid() - z_grid(); dz_lo = z_grid(); _lo = (z_mid_lo^)/(dz_lo*dz_mid); _mid = (-/dz_mid)*((z_mid_up^)/dz_up +... (z_mid_lo^)/dz_lo); _up = (z_mid_up^)/(dz_up*dz_mid); second_deri = (_mid+4*_lo/3)*psi() +... (_up - _lo/3)*psi(); % compute nonlinear reaction term theta = exp( gamma*beta*(-psi()) /... ( + beta*(-psi()))); rxn = (z_grid()^)*(phi_tm^)*theta*psi(); % compute function alue fal() = second_deri - rxn; % compute Jacobian elements = ; Jac(,+) = _up - _lo/3; dtheta_dpsi = theta* (... (-gamma*beta)/(+beta*(-psi())) +... (gamma*beta^*(-psi()))/((+beta*(-psi()))^)); drxn_dpsi = (z_grid()^)*(phi_tm^) *... (theta + psi()*dtheta_dpsi); Jac(,) = _mid + 4*_lo/3; - drxn_dpsi; return; % ======================================================= % This routine computes the effectieness factor from % the dimensionless concentration profile for a % catalyst particle with a nonisothermal first order % reaction. % K. Beers. MIT ChE. 9//00 function eta_eff = calc_eta_eff(psi,... phi_tm, beta, gamma, z_grid); % compute the augmented z grid including the end % points z_aug = [0, z_grid, ]; psi_aug = [0, psi, ]; psi_aug() = (4*psi()-psi())/3; % Now, compute on this augmented grid the alues % of the integrand. integrand = zeros(size(psi_aug)); for =:length(integrand) ar = gamma*beta*(-psi_aug()) /... ( + beta*(-psi_aug())); integrand() = 3*psi_aug()*(z_aug()^)*exp(ar); 9

10 end % Use the trapezoid rule to estimate the alue of the % integral. eta_eff = trapz(z_aug,integrand) return; 0

11 Problem. Eigenalues of 3x3 matrices In this problem we are gien the following three 3x3 matrices: = 3 0 B= 3 4 C= Question.. Which of these matrices can you tell by inspection, without computation, must hae all eigenalues be real numbers? Since = T and C = C T, matrices and C will hae real eigenalues. Question.B. Using Gershogen s theorem, what are the bounds on the possible alues of the eigenalues of? Is it possible that any of the eigenalues of are negatie? From Eqn. (3..-9) of the class notes, r = + = 3 r = + = 3 r = + = Using Eqn. (3..-0), we can get the bounds on the three eigenalues. λ r λ 0 λ 4 λ r λ 3 λ 4 λ r λ 4 3 λ Based on the limits, we see that all of the eigenalues of will be positie. Question.C. Which of these matrices are normal? real matrix D is said to be normal if the following condition holds : DD T = D T D. s a special case, we see that a real, symmetric matrix has D T = D, so that it is obiously

12 normal. Haing said this, based on our answer to Q..., matrices and C are normal. What about matrix B? T BB = T BB= and Since the condition stated aboe does not hold, matrix B is NOT normal. Question.D. For eery matrix that you know has all real eigenalues, compute the eigenalues and the normalized (unit-length) eigenectors by hand. Show your calculations. Let s look at matrix first. To find the eigenalues, we need to derie a characteristic polynomial in λ s, which is done in the following way. λ det( λi) = 3 λ 0 = λ The characteristic polynomial is then ( λ)(3 λ)(4 λ) (4 λ) + ( )(0 + (3 λ)) = 0 Rearranging, we get 3 λ λ λ =0 Soling for the three eigenalues, we get λ =.06, λ =3.3473, and λ 3 =4.53. Let s put expression = λ, where is the eigenector corresponding to a particular eigenalue, in the equation form. 3 0 = λ

13 + = λ () = λ () + 4 = λ (3) = (4) 3 Equation (4) ensures that the eigenector has unit length. From (), = 3(4 λ) (5) dding () and (), we get = λ( + ) 3 3 4λ 3λ = 3 (6) Substituting (5) and (6) into (4), 4λ [ λ ] (4 ) + + = (7) 3 λ Soling for 3 from (7) / 4 λ (4 λ) (8) 3 = λ Thus, we use Eqns. (8), (5), and (6) to get expressions for the three components of the eigenector for a gien alue of λ. The results are: λ = = λ = = λ = = Now let s look at matrix C. λ 0 0 det( C λi) = 0 λ = 0 0 λ and the characteristic polynomial is 3

14 ( λ )( λ) = 0 λ =, λ =, and λ = = λ We can apply exactly the same methodology here as we did for matrix, but gien the simple structure of matrix C, it might be quicker to find eigenalues by inspection. For λ=, we can take =, and = 3 = 0. nother option for this repeated root is = 0 and = 3 = (/) 0.5 = For λ = -, we can choose = 0 and = (/) 0.5 = and 3 = -. In summary, λ = 0 = 0 λ = = λ = =