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Download e-book for kindle: A numerical primer for the chemical engineer by Edwin Zondervan

By Edwin Zondervan

ISBN-10: 1482229447

ISBN-13: 9781482229448

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

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Extra info for A numerical primer for the chemical engineer

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5) where M is the matrix and x and b are vectors. Sometimes a linear system is represented as linear combinations of basis vectors:         1 1 1 4 x 2  + y 1  + z 3  =  7 . 4 The inverse of a matrix If we want to solve a linear system M x = b, we need, in fact, the inverse of the matrix M , provided that the matrix is square. 7) where I is the identity matrix. 5 with M −1 we get: M −1 M x = M −1 b. 9) x = M −1 b. 10) or The question now is, of course, how to determine be found by  C11 C12 1  C21 C22 M −1 = det(M ) C31 C32 the inverse.

Repeat steps 5 and 6. 8. Write down L, U , and P . Let’s do an example: 1. Write down a permutation matrix (initially the identity matrix:   1 0 0 P =  0 1 0 . 24) 0 0 1 2. Write down the matrix you would like to decompose, for example:   0 1 1 M =  2 1 1 . 25) 3. Promote the largest value in the diagonal, so, starting with column 1, row swap to promote the largest value in the column to the diagonal. Do exactly the same row swap with your identity matrix P :     2 1 1 0 1 0 M =  0 1 1 ,P =  1 0 0 .

We can calculate det(M ) by multiplying each element on a row by its co-factor and adding the result:   1 1 1 1 3 2 3 2 1 det  2 1 3  = + det − det + det = −1. 16) 18 A Numerical Primer for the Chemical Engineer Or you can do  1 1 det  2 1 3 1 the same thing for columns:  1 2 1 3  = + det − det 3 1 6 1 3 Now, we have everything to solve our problem:      x 3 −5 2 4 1 1  y =  −3 3 −1   7  = −1 −1 z −1 −2 −1 5 1 1 + det 1 1 2 1 = −1. 17) −13 45   13 =  −4  . 18) −5 For large matrices, computation of determinants and inverses in this way is too difficult (slow), so we need other methods to calculate the inverse of a large matrix.

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A numerical primer for the chemical engineer by Edwin Zondervan


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