 Residual correction in tie-line sensitivity

The residual correction in tie-line sensitivity accounts for rotation of a fitted liquid-liquid tie line when the measured overall composition does not lie exactly on it. In independent mole fraction coordinates, set $\mathbf{d}=\mathbf{b}-\mathbf{a}$, hold a symmetric positive-definite matrix $W$ fixed, and use the interior unconstrained tie-line phase-fraction fit. With $q={\mathbf{d}}^{T}W\mathbf{d}$, residual $\mathbf{r}=\mathbf{y}-\mathbf{a}-{\lambda }_0\mathbf{d}$, and row vectors $\mathbf{g}={\mathbf{d}}^{T}W/q$, $\mathbf{h}={\mathbf{r}}^{T}W/q$, its differential isThe full derivation is in tie-line phase-fraction sensitivity. The normal equation makes $\mathbf{r}$ orthogonal to $\mathbf{d}$ in the chosen metric; it does not make the residual zero.

For an original example, take $\mathbf{a}=(0.60,0.10)$, $\mathbf{b}=(0.20,0.50)$, $\mathbf{y}=(0.35,0.40)$ and $W=I$. The third mole fraction is fixed by closure. These choices giveIncreasing only the first coordinate of $\mathbf{b}$ by $0.001$ predicts a phase fraction change $0.000859375+0.000078125=0.0009375$. Omitting the correction misses the second contribution even though the fit satisfies its normal equation. A centered finite-difference recalculation as the perturbation shrinks tests this derivative directly.

This is a local sensitivity, not an uncertainty until combined with the joint input covariance matrix.^[ref] The formula assumes distinct endpoints and an interior fit with fixed $W$. A changing metric requires sensitivity of a tie-line fit to its weight matrix; a constrained boundary requires a separate local analysis. A nonzero residual can reflect composition noise or model mismatch and does not independently establish liquid-liquid equilibrium.