An 1

Blocks Aj, Bj, and Cj are (2C + 1) by (2C + 1) submatrices of partial derivatives of the functions on stage j with respect to unknowns on stage j _ 1, j. and j + 1 respectively. The solution to Eq. (13-100) is readily obtained by a matrix-algebra equivalent of the Thomas algorithm for a tridiagonal-matrix equation. Computer storage requirements are minimized by making the following replacements. Starting at top stage 1, using forward-block elimination,

For stages j from 2 to (N - 1), Cj —^ (Bj — AjCj-1) 1Cj,

For final stage N,

This completes the forward steps to give AXN = _IFn. Remaining values of corrections_AXj are obtained by successive backward substitution from AXj = _Fj ^ _(Fj _ CjFj+1). Matrix inversions are best done by LU decomposition. Efficiency is best for a small number of components C.

The Newton iteration is initiated by providing reasonable guesses for all unknowns. However, these can be generated from guesses of just T. Tn, and one interstage value of Fj or Lj. Remaining values of Tj are obtained by linear interpolation. By assuming constant molal over

Was this article helpful?

0 0
Making Your Own Wine

Making Your Own Wine

At one time or another you must have sent away for something. A

Get My Free Ebook

Post a comment