OpenSpecEngineering

Finds the number of equilibrium stages a liquid–liquid extraction needs to recover a target fraction of a solute, or the recovery that a set number of stages gives. It covers countercurrent cascades (columns, mixer-settler trains) and cross-current cascades, where fresh solvent goes to every stage. Process engineers use it to screen solvent rates and stage counts.

Equations

E = m·S/F

Countercurrent: x_r = (E − 1)/(E^N+1 − 1)

Cross-current: x_r = 1/(1 + E/N)^N

F is the flow of the feed carrier (the raffinate phase) and S the flow of solvent (the extract phase), in kg/h. m is the distribution coefficient y = m·x, where y and x are the solute concentrations in the extract and raffinate phases, on the same basis as F and S (kg of solute per kg of solvent and per kg of carrier). E is the extraction factor, N the number of equilibrium stages, and x_r the fraction of the solute left in the raffinate. The recovery is 1 − x_r.

The countercurrent form is the Kremser equation [1], [2]; at E = 1 it becomes x_r = 1/(N + 1). The cross-current form splits the solvent equally among the N stages [3]. For a target recovery, the page solves either equation for N, which comes out as a fraction; round it up.

Limits and common mistakes

• The method assumes a dilute solute, immiscible solvent and carrier, and constant m. For partly miscible systems, use a triangular diagram.
• The solvent is assumed to enter free of solute. Recycled solvent with some solute left in it lowers the recovery.
• With E < 1, countercurrent recovery can never exceed E. Cross-current recovery can never exceed 1 − e^−E, however many stages are used. The page says so when a target is out of reach.
• A coefficient in mol/L, or one defined the other way round (raffinate over extract), gives a wrong E.
• N counts ideal stages. Allow for stage efficiency.

Worked example

A solute must be extracted from 1,000 kg/h of an aqueous stream with 1,500 kg/h of solvent, recovering 99 % of it. Lab tests give m = 1.6. How many countercurrent stages are needed, and what would a cross-current train with the same solvent give?

1. Leave Configuration on Countercurrent and Solve For on Stages (given recovery).
2. Enter Feed (raffinate-phase) Flow 1000 kg/h, Solvent (extract-phase) Flow 1500 kg/h and Distribution Coefficient 1.6.
3. Enter Target Solute Recovery 0.99 and press Solve: Extraction factor E = m·S/F: 2.4
   Fraction of solute left in raffinate: 0.01
   Required stages N: 4.6527
4. Round up to 5 stages. Choose Solve For Recovery (given stages), enter Number of Stages 5 and press Solve: Fraction left in raffinate: 0.00736443
   Solute recovery into extract: 99.2636 %
5. Choose Configuration Cross-current (equal solvent per stage) and press Solve again: Fraction left in raffinate: 0.140829
   Solute recovery into extract: 85.9171 %

Check by hand: E = 1.6 × 1500/1000 = 2.4. For five countercurrent stages, 2.4⁶ = 191.1, so x_r = 1.4/190.1 = 0.00736. For cross-current, each stage gets E/5 = 0.48, so x_r = 1/1.48⁵ = 0.1408. Countercurrent flow makes far better use of the same solvent.