MaximaLabs
All equations

Distillation MESH stage balances (tridiagonal)

Separation

A three-stage slice of a distillation column at fixed K-values and constant molal overflow — the component material balances form the classic tridiagonal system the inside-out inner loop solves.

Governing equation

Lxj1+VKxj+1(L+VK)xj+fj=0L\,x_{j-1} + V K\,x_{j+1} - (L + V K)\,x_j + f_j = 0
xjx_j
Liquid mole fraction of the light component on stage j — the unknowns.
LL
Liquid molal overflow (mol s⁻¹), constant.
VV
Vapor molal overflow (mol s⁻¹), constant.
KK
Equilibrium ratio y/x (fixed in the inner loop).
fjf_j
Feed / boundary source on stage j (reflux on top, feed on the middle stage).

Derivation

  1. 1
    Lj1xj1+Vj+1yj+1+fj=Ljxj+VjyjL_{j-1} x_{j-1} + V_{j+1} y_{j+1} + f_j = L_j x_j + V_j y_j

    Component material balance on stage j: liquid down from above plus vapor up from below plus feed equals liquid and vapor leaving.

  2. 2
    yj=Kjxjy_j = K_j x_j

    Equilibrium closes the vapor composition; in the inside-out inner loop the K-values are frozen, making the balances linear in x.

  3. 3
    Lxj1+VKxj+1(L+VK)xj+fj=0L\,x_{j-1} + V K\,x_{j+1} - (L + V K)\,x_j + f_j = 0

    Constant molal overflow makes L and V stage-independent; each row couples only to its neighbours, so the system is tridiagonal (the Thomas algorithm).

Jacobian

The matrix Jij=Fi/xjJ_{ij} = \partial F_i / \partial x_j of the residual vector with respect to the unknowns, at the representative operating point. Every value below is computed by the exact forward-mode autodiff engine the solver uses — click a non-zero cell to see the symbolic derivative, each one verified against that engine in our test suite. This is the block a Newton step factorizes.

∂/∂x1∂/∂x2∂/∂x3
Stage 1 balance
Stage 2 balance (feed)
Stage 3 balance

Shaded cells are non-zero (the sparsity pattern); blanks are structural zeros.

Symbolic derivative
F1x1=\dfrac{\partial F_{1}}{\partial\, \mathrm{x1}} =
(L+VK)-(L + V K)
Evaluated at the operating point: -208

For the residual Stage 1 balance with respect to x1. This exact expression is asserted equal to the autodiff value in the test suite.

Operating point

Representative operating point — a real solution of this instance.

x1 = 0.83398x2 = 0.72655x3 = 0.3493L = 100V = 120K = 0.9x0 = 0.95x4 = 0Fz = 30

References

  • Seader, Henley & Roper, Separation Process Principles, 3rd ed., Ch. 10.
  • Boston, J.F. & Sullivan, S.L. (1974). A new class of solution methods for multicomponent, multistage separation processes. Can. J. Chem. Eng. 52.

Stop fighting legacy software. Build your first flowsheet in 60 seconds.