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Unformatted text preview: ChemE 109  Numerical and Mathematical Methods in Chemical and Biological Engineering Fall 2007 Solution to Homework Set 5 1. The set of linear ordinary differential equations which describe the monomolecular kinetics of the chemical reaction: A B * ) C has the form: dC A dt = k 1 C A dC B dt = k 1 C A k 2 C B + k 3 C C dC C dt = k 2 C B k 3 C C where C A ,C B ,C C denote the concentrations of species A , B , C , respectively, k 1 = 1, k 2 = 2, k 3 = 3 and the initial conditions for the three components are C A (0) = 1, C B (0) = 0 and C C (0) = 0. For the above system of ODEs: (a) Compute its analytical solution and the steadystate values of the variables C A , C B and C C . (b) Write a MATLAB code, which uses Euler predictorcorrector, with h = 0 . 1, to compute C A (10), C B (10), C C (10). (c) Write a MATLAB code, which uses fourthorder RungeKutta, with h = 0 . 1, to compute C A (10), C B (10), C C (10). Solution: (a) Compute its analytical solution and the steadystate values of the variables C A , C B and C C . A =  k 1 k 1 k 2 k 3 k 2 k 3 =  1 1 2 3 2 3 1 = 5 2 = 0 3 = 1 ( A 1 I ) 1 = 0 1 = 1 1 ( A 2 I ) 2 = 0 2 = 3 2 ( A 3 I ) 3 = 0 3 = 2 1 1 1 T = [ 1 2 3 ] = 0 0 2 1 3 1 1 2 1 T 1 =  . 1 0 . 4 . 6 . 2 0 . 2 . 2 . 5 e At = T e 5 t 0 1 0 0 e t T 1 = e t . 6 . 1 e 5 t . 5 e t . 6 + 0 . 4 e 5 t . 6 . 6 e 5 t . 4 + 0 . 1 e 5 t . 5 e t . 4 . 4 e 5 t . 4 + 0 . 6 e 5 t x ( t ) = e At x (0) = e t . 6 . 1 e 5 t . 5 e t . 4 + 0 . 1 e 5 t . 5 e t At steady state: dC A dt = dC B dt = dC C dt = 0 So  C A,ss = 0 C A,ss 2 C B,ss + 3 C C,ss = 0 2 C B,ss 3 C C,ss = 0 ( C A,ss = 0 C B,ss = 1 . 5 C C,ss We also have the following equation from the mass conservation law C A + C B + C C = Const. C A,ss + C B,ss + C C,ss = C A (0) + C B (0) + C C (0) = 1 Then the values of steady state can be solved as C A,ss = 0 C B,ss = 0 . 6 C C,ss = 0 . 4 We can also solve get the same steady state from the analytical solution by assigning t = . (b) Write a MATLAB code, which uses Euler predictorcorrector, with h = 0 . 1, to compute C A (10), C B (10), C C (10). The results using Euler predictorcorrector are C A (10) = 0 . 0000 C B (10) = 0 . 6000 C C (10) = 0 . 4000 From the results of the first question, the solutions at t = 10 are approaching the steady state....
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This homework help was uploaded on 04/07/2008 for the course CBE 109 taught by Professor Christofides during the Fall '07 term at UCLA.
 Fall '07
 Christofides

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