By Edwin Zondervan

ISBN-10: 1482229447

ISBN-13: 9781482229448

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

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**Extra info for A numerical primer for the chemical engineer**

**Example text**

We can index a node with k = i + N x(j − 1) such that Ti,j = Tk . Now we can use finite differences to approximate our two-dimensional Laplace equation. Thus, we need some kind of estimate of the second derivative at node k. 1 Domain definition with 4 boundaries We could assume a piece-wise linear profile, where we approximate at each grid interval, a temperature difference by a linear function. 4) which can be written as T −T T −T i+1,j i,j − i,j ∆xi−1,j ∂2T Ti+1,j + 2Ti,j − Ti−1,j ∆x ≈ = . 6) or, in terms of node indices, as Tk+1 − 2Tk + Tk−1 Tk+N x − 2Tk + Tk−N x + = 0.

15) If the inverse of a matrix does not exist, there are either no solutions or infinitely many solutions. The determinant determines the existence of an inverse. If the determinant is zero, an inverse does not exist, and the matrix is called singular. 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.

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. 18) So, how do we decompose A as given before? When we eliminate A21 we can keep multiplying by a matrix that undoes this operation, such that the product remains equal to A: A11 A12 A13 1 0 0 A11 A12 A13 A21 A22 A23 = d21 1 0 0 A22 − d21 A12 A23 − d21 A12 .

### A numerical primer for the chemical engineer by Edwin Zondervan

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