New PDF release: A numerical primer for the chemical engineer

By Edwin Zondervan

ISBN-10: 1482229447

ISBN-13: 9781482229448

"This booklet emphasizes the deriviation and use of various numerical tools for fixing chemical engineering difficulties. The algorithms are used to unravel linear equations, nonlinear equations, traditional differential equations and partial differential equations. additionally it is chapters on linear- and nonlinear regression and ond optimizaiton. MATLAB is followed because the programming surroundings in the course of the book. Read more...

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Additional info for A numerical primer for the chemical engineer

Example text

11) Linear equations 17 where Cij are the co-factors of the matrix M . Calculation of co-factors is best illustrated by an example. Consider our matrix M   1 1 1 M =  2 1 3 . 13) ⋆ 1 6 and calculate C11 = + det 1 3 1 6 The plus sign comes from the following  + −  − + + − = 6 × 1 − 3 × 1 = 3. 14) matrix:  + − . + After calculating the co-factors for each element in the matrix, the result is:  1 1  2 1 3 1 −1   1 3 −5 2 1  −3 3  = 3 −1  . 15) If the inverse of a matrix does not exist, there are either no solutions or infinitely many solutions.

Now we have completed an LU factorization, and we can solve the L and U matrices with forward and back substitution. But what if  A11  A21 A31 we, for example, obtain   A12 A13 1 A22 A23  =  d21 A32 A33 d31 a matrix like this:  0 0 A11 A12 1 0  0 A′22 0 1 0 A′32  A13 A′23  . 22) We would like to exchange rows 2 and 3. That can be done by multiplication with a permutation matrix, resulting in       1 0 0 A11 A12 A13 1 0 0 A11 A12 A13  0 0 1   A21 A22 A23  =  d31 0 1   0 A′32 A′33  .

B1 b2 b3  . 13) 1 2 3     .. .. .. . . . By Gaussian elimination and U , so that  A11 A12  A21 A22 A31 A32 we could factor the matrix A into two matrices, L   A13 1 A23  =  ⋆ A33 ⋆  0 0 ⋆ 1 0  0 ⋆ 1 0  ⋆ ⋆ ⋆ ⋆ . 14) Then we could solve each right-hand side using only forward and back substitution. 15) we could rewrite A in terms of L and U : LU x = b. 16 and solve by forward substitution as: Ly = b. 17) Elimination methods 31 And subsequently we solve by back substitution: U x = y.

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A numerical primer for the chemical engineer by Edwin Zondervan


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