Modelling and Mathematical Methods in Process and Chemical Engineering

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1 FS 07 February 0, 07 Modelling and Mathematical Methods in Process and Chemical Engineering Solution Series. Systems of linear algebraic equations: adjoint, determinant, inverse The adjoint of a (square) matrix A is defined as the transpose of the cofactor matrix of A. The matrix of coefficients is 6 A = 7 () 9 The cofactor matrix is thus Taking the transpose yields the adjoint of A, i.e., 9 8 C A = () 7 adj(a) = () The determinant A is equal to zero and, consequently, the inverse of A is not defined. The rank of A is rank(a) =. Indeed, looking carefully at A, one observes that the first and third row of A are linearly dependent.. System of first-order linear ODEs with constant coefficients We have to consider the system dx dt = x x + () dx dt = x x + with the initial conditions: x (0) = and x (0) =. Rewriting the system in matrix form, we obtain. x x. = + = Ax + b x x x (0) = x (0) () (6)

2 FS 07 February 0, 07 Superimposing the solution of the homogeneous system x h = c j e λjt z j (7) j= (where z j is the jth eigenvector of the coefficient matrix A corresponding to the eigenvalue λ j and c j is a constant yet to be determined) and the particular solution A b (8) results in x = c j e λjt z j A b (9) j= Using the initial conditions, the constant coefficients c j can be uniquely determined: c j = yt j (x 0 + A b) y T j z j (0) where x 0 is the vector of initial conditions x j (0), and yj T Thus, the general solution of the above problem is is the jth eigenrow. x = y T j (x 0 + A b) y T j z e λjt z j A b () j j= Evaluating the determinant D = A λi = λ + λ = 0 () yields the eigenvalues λ = λ = The eigenvectors z j and the eigenrows y T j are the non-zero columns and rows of the adjoint matrix corresponding to λ j, respectively. Here, we obtain (A λ I) = adj(a λ I) = and Analogously, z = (A λ I) = y T = adj(a λ I) =

3 FS 07 February 0, 07 z = y T = Additionally, we have to compute x 0 + A b: adj(a) = A = D adj(a) = x 0 + A b = d = 9 7 The constants c j can easily be calculated to give the following results: c = yt d y T z = c = yt d = y T z Finally, substituting into the general solution for x yields x = e t + e t

4 FS 07 February 0, 07. First-order linear systems in two dimensions () dx dt = x λ, = z = z = () dx dt = x λ, = z = z = () dx dt = 6 x i λ, = i 6 z = i 6 z = + i () dx dt = x + i λ, = i z = i z = + i x 0 0 x () dx dt = x λ, = z = z =

5 FS 07 February 0, 07 Note: The rotation in both problem () and () is counterclockwise, since a > 0.

6 FS 07 February 0, 07. Batch reactor In a batch reactor, two consecutive first-order reactions are taking place: A k B k C The mass balance leads to a linear set of ODEs describing concentrations of species A and B (i.e. C A (t) and C B (t)) dc A (t) dt dc B (t) dt = k C A (t) () = k C A (t) k C B (t) () where it is worth noticing that C i0 is the initial concentration of the species i. Rewriting the above system in matrix form, we obtain. C A (t) k. = 0 C A (t) C B (t) k k C B (t) C C (t) = C A0 + C B0 C A (t) C B (t) () = Ax (6) As indicated on the problem sheet, this system can be solved using either the idea of vector expansion or a more direct way, namely via direct integration of the two simple differential equations. (A) Direct integration Integration of the first ODE, eq, yields C A (t) = C A0 e k t For the second ODE, eq, the solution of the homogeneous problem, i.e. C Bh (t), is found to be C Bh (t) = r e k t where r is a constant coefficient yet to be determined. Assuming that k k, a particular solution of the second ODE could be C Bp (t) = r e k t Substitution in the original, non-homogeneous equation leads to r = k C A0 (k k ). Therefore, the general solution of the second ODE may be written as C B (t) = r e k t + k C A0 (k k ) e k t Using the initial condition C B0 (which will most probably be zero) yields r = C B0 k C A0 (k k ). 6

7 FS 07 February 0, 07 Thus, we have C B (t) = C B0 e k t + k C A0 (k k ) e k t k C A0 (k k ) e k t = C B0 e k t + k C A0 (k k ) (e k t e k t ) (7) and C A (t) = C A0 e k t (8) C B (t) = C B0 e k t + k C A0 (k k ) (e k t e k t ) (9) From these results, it can easily be shown that, for k k, the system approaches pseudo-steadystate conditions for component B; that is, C B = k C A /k. To see this, observe that C B = k C A0 (k k ) ( e k t ) = k C A0 e k t k (0) = k k C A q.e.d. Notice that this can be obtained directly from the equations! (B) Method of eigenvalues and eigenvectors The homogeneous system of first-order linear ordinary differential equations. C A (t) k. = 0 C A (t) = Ax () C B (t) k k C B (t) with the initial conditions C A (0) C = A0 C B (0) C B0 () has the solution C = c j e λjt z j () j= where c j = yt j x 0 y T j z j () Computing the eigenvalues of the coefficient matrix yields λ = k λ = k 7

8 FS 07 February 0, 07 The eigenvectors and eigenrows are z = k k k 0 z = k y T = k 0 y T = k k k Finally, we can write x = C A0 (k k ) k k k e k t + k C A0 + (k k )C B0 k (k k ) 0 k e k t () Simplifying the above results yields exactly the same result as obtained by using method (A). 8