Extraction

- separation of a liquid mixture by adding a liquid solvent forming a second phase enriched by

some component(s)

Notation:

A … extracted component

B … added solvent

C … original solvent

F … feed

R … rafinate

S … solvent

E … extract

Two cases:

1) immiscible solvents B and C (only A is transported)

2) partially miscible solvents (A is transported, but B and C also a little)

Example: extraction of acetic acid from water solution by adding an ether; after extraction, ether is

regenerated by distillation; this is cheaper than distilling water solution of acetic acid directly

Extractors

1) staged

a. cascade of mixers and separators (settlers)

b. column with plates (vibration is used to facilitate separation)

2) continuous

a. packed or spray columns (slow)

b. centrifuge (fast)

General extractor is described as a general equilibrium stage (see earlier chapter), only enthalpy balance

is often neglected (no heat effects). For correction to non-equilibrium → efficiency

Special case – immiscible solvents

- only A crosses the interface

- F and R contain only A+C (phase x)

- S and E contain only A+B (phase y)

It is convenient to use relative mass fractions:

𝑋𝐴 =𝑚𝐴𝑚𝐶

=𝑥𝐴

1 − 𝑥𝐴; 𝑥𝐴 =

𝑋𝐴1 + 𝑋𝐴

;𝑚𝐴 = 𝑥𝐴(𝑚𝐴 +𝑚𝐶)

𝑌𝐴 =𝑚𝐴𝑚𝐵

=𝑦𝐴

1 − 𝑦𝐴; 𝑦𝐴 =

𝑌𝐴1 + 𝑌𝐴

; 𝑚𝐴 = 𝑦𝐴(𝑚𝐴 +𝑚𝐵)

𝑚𝐵 and 𝑚𝐶 are constant.

Single stage

For mass balances we use finite time intervals → no flows only masses

- mass balance of A at steady state:

𝑚𝐶𝑋𝐴𝐹 +𝑚𝐵𝑌𝐴𝑆 = 𝑚𝐶𝑋𝐴 +𝑚𝐵𝑌𝐴

or

𝑚𝐶(𝑋𝐴𝐹 − 𝑋𝐴) = 𝑚𝐵(𝑌𝐴 − 𝑌𝐴𝑆)

- equilibrium relation for A:

𝑦𝐴 = 𝜓𝐴𝑥𝐴 or 𝑌𝐴 = 𝜑𝐴𝑋𝐴

By using the definition of relative fractions we obtain:

𝑌𝐴 =𝜓𝐴

1 − (𝜓𝐴 − 1)𝑋𝐴𝑋𝐴

Remark: for low concentrations 𝑥𝐴 =̇ 𝑋𝐴 and 𝑦𝐴 =̇ 𝑌𝐴 and thus 𝜓𝐴 =̇ 𝜑𝐴 = 𝑐𝑜𝑛𝑠𝑡

To find 𝑋𝐴, 𝑌𝐴 we solve the balance and equilibrium equations simultaneously either graphically or

algebraically.

Graphical solution:

We rewrite the balance of A:

𝑌𝐴 = −𝑚𝐶𝑚𝐵

(𝑋𝐴 − 𝑋𝐴𝐹) + 𝑌𝐴𝑆

which is an operational line with slope equal to −𝑚𝐶

𝑚𝐵 passing through point [𝑋𝐴𝐹 , 𝑌𝐴𝑆].

Algebraic solution:

By combining balance of A and equilibrium relation

𝑚𝐶𝑋𝐴𝐹 +𝑚𝐵𝑌𝐴𝑆 = 𝑋𝐴(𝑚𝐶 + 𝜓𝐴𝑚𝐵) /1

𝑚𝐶

𝑋𝐴𝐹 +𝑚𝐵𝑚𝐶

𝑌𝐴𝑆 = 𝑋𝐴 (1 +𝝍𝑨𝒎𝑩

𝒎𝑪)

we introduce an extraction factor

𝜁𝑥 = 𝜓𝐴𝑚𝐵𝑚𝐶

𝑎𝑛𝑑 𝜁𝑦 =𝑚𝐶𝜓𝐴𝑚𝐵

=1

𝜁𝑥

Thus:

𝑋𝐴 = (1 + 𝜁𝑥)−1 (𝑋𝐴𝐹 + 𝑌𝐴𝑆

𝑚𝐵𝑚𝐶)

In general, solution of this equation is by iterations, because 𝜁𝑥 may depend 𝑋𝐴 on through 𝜓𝐴 = 𝜓𝐴(𝑋𝐴).

That is, we estimate 𝑋𝐴(0)

, then calculate 𝑋𝐴(1)

from the equation, and then use 𝑋𝐴(1)

to calculate 𝑋𝐴(2)

, etc.

until |𝑋𝐴(𝑛+1)

− 𝑋𝐴(𝑛)| < 𝜀 where 𝜀 is small.

For a real stage we have to combine mass balance, equilibrium relation and efficiency. We denote

equilibrium composition by stars:

𝑌𝐴∗ = 𝜑𝐴𝑋𝐴

∗

And efficiency is:

𝐸𝐴 =𝑌𝐴 − 𝑌𝐴𝑆𝑌𝐴∗ − 𝑌𝐴𝑆

=𝑋𝐴𝐹 − 𝑋𝐴𝑋𝐴𝐹 − 𝑋𝐴

∗

Graphical solution

Algebraic solution

1) solution for equilibrium case:

𝑋𝐴∗ = (1 + 𝜁𝑥)

−1 (𝑋𝐴𝐹 + 𝑌𝐴𝑆𝑚𝐵𝑚𝐶)

2) from efficiency we have:

𝑋𝐴 = 𝑋𝐴𝐹 − 𝐸𝐴(𝑋𝐴𝐹 − 𝑋𝐴∗)

Repeated extraction (or cross-flow extraction)

- mass balance of A in the entire cascade:

𝑚𝐶𝑋𝐴𝐹 +∑𝑚𝐵,𝑛𝑌𝐴𝑆 = 𝑚𝐶𝑋𝐴 +∑𝑚𝐵,𝑛⟨𝑌𝐴⟩

or

𝑚𝐶(𝑋𝐴𝐹 − 𝑋𝐴) =∑𝑚𝐵,𝑛(⟨𝑌𝐴⟩ − 𝑌𝐴𝑆) = 𝑚𝐴

where 𝑚𝐴 is the mass of A being transferred from phase x to phase y. This equation serves for

determining 𝑚𝐴 or ⟨𝑌𝐴⟩

- mass balance of A in the stage n

𝑚𝐶𝑋𝐴,𝑛−1 +𝑚𝐵𝑛𝑌𝐴𝑆 = 𝑚𝐶𝑋𝐴,𝑛 +𝑚𝐵,𝑛𝑌𝐴,𝑛; 𝑛 = 1…𝑁

or

𝑌𝐴,𝑛 = −𝑚𝐶𝑚𝐵,𝑛

(𝑋𝐴,𝑛 − 𝑋𝐴,𝑛−1) + 𝑌𝐴𝑆

which are operational lines with slopes −𝑚𝐶

𝑚𝐵,𝑛 passing through points

[𝑋𝐴,𝑛−1, 𝑌𝐴𝑆]; 𝑛 = 1…𝑁

Equilibrium condition (in equilibrium stages):

𝑌𝐴,𝑛 = 𝜑𝐴𝑋𝐴,𝑛; 𝑛 = 1…𝑁

Solution of mass balance and equilibrium conditions leads to all N unknown pairs 𝑋𝐴,𝑛, 𝑌𝐴,𝑛.

Graphical solution

- intersection of operational lines and equilibrium curve

- either repeat N times (if N is given) or do until a given is reached

Algebraic solution

- the use of combined balance and equilibrium equations with extraction factor 𝜁𝑥,𝑛

- in analogy to one extractor:

𝑋𝐴,𝑛 = (1 + 𝜁𝑥,𝑛)−1(𝑋𝐴,𝑛−1 + 𝑌𝐴𝑆

𝑚𝐵,𝑛𝑚𝐶

)

where

𝜁𝑥,𝑛 = 𝜁𝐴,𝑛𝑚𝐵,𝑛𝑚𝐶

▪ simulation (or control) calculation

o N is given; 𝑋𝐴,𝑛 is calculated consecutively, 𝑛 = 1…𝑁

▪ design calculation

o 𝑋𝐴,𝑛 is given, N is calculated either for 𝑛 = 1,2… until 𝑋𝐴,𝑁 is reached or for 𝑛 = 𝑁,𝑁 −

1… until 𝑋𝐴,0 is reached (more convenient if 𝜑𝐴,𝑛 depends on 𝑋𝐴,𝑛)

For a real cascade efficiency must be used in addition to mass balance and equilibrium condition.

- for equilibrium we use asterisks:

𝑌𝐴,𝑛∗ = 𝜑𝐴,𝑛𝑋𝐴,𝑛

∗

- the efficiency is

𝐸𝐴 =𝑌𝐴,𝑛 − 𝑌𝐴𝑆𝑌𝐴,𝑛∗ − 𝑌𝐴𝑆

=𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛

∗

Graphical solution (analogous to one stage):

Algebraic solution

1) solution for the equilibrium stage n:

𝑋𝐴,𝑛∗ = (1 + 𝜁𝑥,𝑛)

−1(𝑋𝐴,𝑛−1 + 𝑌𝐴𝑆

𝑚𝐵,𝑛𝑚𝐶

)

2) the use of efficiency:

𝑋𝐴,𝑛 = 𝑋𝐴,𝑛−1 − 𝐸𝐴,𝑛(𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛∗ )

We can introduce overall efficiency:

𝐸𝐶 =𝑁

𝑁𝑟𝑒𝑎𝑙=𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑠𝑡𝑎𝑔𝑒𝑠

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑠𝑡𝑎𝑔𝑒𝑠

and use it to calculate 𝑁𝑟𝑒𝑎𝑙.

Special case:

a) 𝑚𝐵,𝑛 = 𝑐𝑜𝑛𝑠𝑡 = 𝑚𝐵

b) 𝜑𝐴,𝑛 = 𝑐𝑜𝑛𝑠𝑡 = 𝜑𝐴 → linear equilibrium line

then also 𝜁𝑥,𝑛 = 𝑐𝑜𝑛𝑠𝑡 = 𝜁𝑥 = 𝜑𝐴𝑚𝐵

𝑚𝐶

Combined mass balance and equilibrium equation reads:

𝑋𝐴,𝑛 = (1 + 𝜁𝑥)−1 (𝑋𝐴,𝑛−1 + 𝑌𝐴𝑆

𝑚𝐵𝑚𝐶)

let us assume for simplicity that the solvent B is pure → 𝑌𝐴𝑆 = 0. Then

𝑋𝐴,𝑛𝑋𝐴,𝑛−1

= (1 + 𝜁𝑥)−1 or

𝑋𝐴,𝑛−1𝑋𝐴,𝑛

= 1 + 𝜁𝑥

For all stages together:

𝑋𝐴,0𝑋𝐴,1

𝑋𝐴,1𝑋𝐴,2

…𝑋𝐴,𝑁−1𝑋𝐴,𝑁⏟

𝑋𝐴,0𝑋𝐴,𝑁

= (1 + 𝜁𝑥)𝑁

A more general form with 𝑌𝐴𝑆 > 0 is:

𝑋𝐴,0 −𝑌𝐴𝑆

𝜑𝐴⁄

𝑋𝐴,𝑁 −𝑌𝐴𝑆

𝜑𝐴⁄= (1 + 𝜁𝑥)

𝑁

Remark: A similar formula with efficiency for real stages can be derived.

Counter-current cascade (for flow systems)

Mass balance for A in the entire cascade

�̇�𝐶𝑋𝐴𝐹 + �̇�𝐵𝑌𝐴𝑆 = �̇�𝐶𝑋𝐴,𝑁 + �̇�𝐵𝑌𝐴,1

or

�̇�𝐶(𝑋𝐴𝐹 − 𝑋𝐴,𝑁) = �̇�𝐵(𝑌𝐴,1 − 𝑌𝐴𝑆) = �̇�𝐴

where �̇�𝐴 is flow of A from phase x to phase y.

Mass balance of A in stage n:

�̇�𝐵(𝑌𝐴,𝑛 − 𝑌𝐴,𝑛+1) = �̇�𝐶(𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛); 𝑛 = 1…𝑁

or

(𝑌𝐴,𝑛 − 𝑌𝐴,𝑛+1)

(𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛)=�̇�𝐶�̇�𝐵

operational line with slope �̇�𝐶

�̇�𝐵 passing through the points [𝑋𝐴,𝑛−1, 𝑌𝐴,𝑛]; 𝑛 = 1…𝑁

Equilibrium condition:

𝑌𝐴,𝑛 = 𝜑𝐴,𝑛𝑋𝐴,𝑛; 𝑛 = 1…𝑁

Graphical solution

- slope is constant, all points [𝑋𝐴,𝑛−1, 𝑌𝐴,𝑛] lie on one line

- stages are found as rectangular steps

Minimal consumption of the solvent (B)

- two cases

o no tangent

o tangent

Minimal consumption can be calculated from:

�̇�𝐶�̇�𝐵,𝑚𝑖𝑛

=(𝑌𝐴,𝑚𝑎𝑥 − 𝑌𝐴𝑆)

(𝑋𝐴𝐹 − 𝑋𝐴,𝑁) → �̇�𝐵,𝑚𝑖𝑛 = �̇�𝐶

(𝑋𝐴𝐹 − 𝑋𝐴,𝑁)

(𝑌𝐴,𝑚𝑎𝑥 − 𝑌𝐴𝑆)

Remark: if �̇�𝐵 = �̇�𝐵,𝑚𝑖𝑛 then 𝑁 → ∞; in practice �̇�𝐵 = 𝛾�̇�𝐵,𝑚𝑖𝑛 where 𝛾 = 1.2 − 1.5

Algebraic solution

The mass balance in stage n can be rewritten as

�̇�𝐶𝑋𝐴,𝑛−1 + �̇�𝐵𝑌𝐴,𝑛 = �̇�𝐶𝑋𝐴,𝑛 + �̇�𝐵𝑌𝐴,𝑛+1 = ∆�̇�𝐴

where ∆�̇�𝐴 is constant for all stages and can be determined from balance of all stages.

By using equilibrium condition 𝑌𝐴,𝑛 = 𝜑𝐴,𝑛𝑋𝐴,𝑛 we get:

𝑋𝐴,𝑛−1 =𝜑𝐴,𝑛�̇�𝐵�̇�𝐶⏟ 𝜁𝑥,𝑛

+∆�̇�𝐴�̇�𝐶

This formula can be used repeatedly to calculate N if is 𝑋𝐴,𝑁 given or 𝑋𝐴,𝑁 if N is given.

For real stages a new efficiency must be defined → Murphree efficiency

𝐸𝑋,𝑛 =𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛𝑋𝐴,𝑛−1 − 𝑋𝐴,𝑛

∗

where 𝑋𝐴,𝑛∗ =

𝑌𝐴,𝑛𝜑𝐴,𝑛⁄ (i.e. 𝑌𝐴,𝑛 is for real stage, not 𝑌𝐴,𝑛

∗ for equilibrium stage).

𝐸𝑌,𝑛 =𝑌𝐴,𝑛 − 𝑌𝐴,𝑛+1𝑌𝐴,𝑛∗ − 𝑌𝐴,𝑛+1

where 𝑌𝐴,𝑛∗ = 𝜑𝐴,𝑛𝑋𝐴,𝑛 (i.e. 𝑋𝐴,𝑛 is for real stage, not 𝑋𝐴,𝑛

∗ for equilibrium stage).

The two efficiencies 𝐸𝑋,𝑛 and 𝐸𝑌,𝑛 are not equal.

Graphical representation

Graphical solution of the cascade

Special case

- 𝜑𝐴,𝑛 = 𝑐𝑜𝑛𝑠𝑡 = 𝜑𝐴 → linear equilibrium

then 𝜁𝑋 =𝜑𝐴�̇�𝐵

�̇�𝐶= 𝑐𝑜𝑛𝑠𝑡

we introduce effectiveness factors

η𝑋 =𝑋𝐴𝐹 − 𝑋𝐴𝑁

𝑋𝐴𝐹 −𝑌𝐴𝑆𝜑𝐴

; η𝑌 =𝑌𝐴1 − 𝑌𝐴𝑆𝜑𝐴𝑋𝐴𝐹 − 𝑌𝐴𝑆

Graphical representation

For η𝑋 = 1 or η𝑌 = 1 the cascade would have 𝑁 → ∞.

It can be shown that

1 − η𝑋1 − η𝑌

= (1

𝜁𝑋)𝑁

= 𝜁𝑌𝑁 and η𝑋 =

η𝑌𝜁𝑌; η𝑌 =

η𝑋𝜁𝑋

then

𝑁 =

ln(1 −

η𝑍𝜁𝑍⁄

1 − η𝑍)

ln 𝜁𝑍; where 𝑍 is either 𝑋 or 𝑌

for 𝜁𝑍 = 1:

𝑁 =η𝑍

1 − η𝑍

General case – partially miscible solvents

In addition to A, also B and C cross the interface. Therefore, all three mass balances must be used now →

relative mass fractions are not useful → ordinary mass fractions should be used together with masses of

the streams, not just.

Equilibrium:

- formally 𝑦𝑘 = 𝜓𝑘𝑥𝑘; 𝑘 = A,B,C

- 𝜓𝑘 depends on composition and temperature

- graphical representation in a triangle diagram is more convenient

𝑥𝐶 = 1 − 𝑥𝐴−𝑥𝐵; 𝑦𝐶 = 1 − 𝑦𝐴−𝑦𝐵

- data for the equilibrium curve are typically given in tables

Mass balance of A, B, C is graphically represented in the triangle diagram by the lever rule.

Extractor is thought of as a combination of a mixer and settler.

- overall mass balance: 𝑚𝐹 +𝑚𝑆 = 𝑚𝑀 = 𝑚𝑅 +𝑚𝐸

- component k: 𝑚𝐹𝑥𝑘𝐹 +𝑚𝑆𝑦𝑘𝑆 = 𝑚𝑀𝑧𝑘𝑀 = 𝑚𝑅𝑥𝑘 +𝑚𝐸𝑦𝑘

Combination yields: 𝑚𝐹(𝑥𝑘𝐹 − 𝑦𝑘𝑆) = 𝑚𝑀(𝑧𝑘𝑀 − 𝑦𝑘𝑆)

or

𝑚𝐹𝑚𝑀

=𝑦𝑘𝑆 − 𝑧𝑘𝑀𝑦𝑘𝑆 − 𝑥𝑘𝐹

=𝑆𝑀

𝑆𝐹 lever rule for mixer

in analogy the lever rule for the settler is

𝑚𝑅𝑚𝑀

=𝑀𝐸

𝑅𝐸

Other variants of the lever rule (less convenient for calculations)

- mixer: 𝑚𝐹

𝑚𝑆=𝑀𝑆

𝐹𝑀

- settler: 𝑚𝑅

𝑚𝐸=𝑀𝐸

𝑅𝑀

Repeated extraction (cross-flow)

mass balance of stage n (assume constant mass of solvent)

- overall: 𝑚𝑅,𝑛−1 +𝑚𝑆 = 𝑚𝑀,𝑛 = 𝑚𝑅,𝑛 +𝑚𝐸,𝑛

- components: 𝑚𝑅,𝑛−1𝑥𝑘,𝑛−1 +𝑚𝑆𝑦𝑘𝑆 = 𝑚𝑀,𝑛𝑧𝑘,𝑛 = 𝑚𝑅,𝑛𝑥𝑘,𝑛 +𝑚𝐸,𝑛𝑦𝑘,𝑛

This implies lever rules for points

- mixer (R𝑛−1, M𝑛, S); 𝑛 = 1…𝑁

- settler (R𝑛, M𝑛, 𝐸𝑛); 𝑛 = 1…𝑁

Equilibrium in stage n: 𝑦𝑘,𝑛 = 𝜓𝑘,𝑛𝑥𝑘,𝑛; 𝑛 = 1…𝑁 or via graph

Graphical solution

a) lever ruler for stage 1:

o mixer: 𝑚𝐹

𝑚𝑀1

=𝑆𝑀1

𝑆𝐹 → determine 𝑀1

o then tie line and lever rule for settler: 𝑚𝑅1

𝑚𝐸1

=𝑀1𝐸1

𝑅1𝐸1 → determine 𝑚𝑅1

b) for stage 2

o mixer: 𝑚𝑅1

𝑚𝑅1+𝑚𝑆=𝑆𝑀2

𝑆𝑅1 → determine 𝑀2

o settler: 𝑚𝑅2

𝑚𝑅1+𝑚𝑆=𝑀2𝐸2

𝑅2𝐸2 → determine 𝑚2

c) etc.

Counter current cascade of extractors (or counter-current staged column)

Mass balance of all stages – summing form:

- overall: �̇�𝐹 + �̇�𝑆 = �̇�𝑀 = �̇�𝑅𝑁 + �̇�𝐸1

- components: �̇�𝐹𝑥𝑘𝐹 + �̇�𝑆𝑦𝑘𝑆 = �̇�𝑀𝑧𝑘𝑀 = �̇�𝑅𝑁𝑥𝑘,𝑁 + �̇�𝐸1𝑦𝑘,1

These equations imply lever rules connecting points (F, M, S) and (RN, M, E1)

Mass balance of stages 1 to n – difference form:

- overall: �̇�𝐹 − �̇�𝐸1 = �̇�𝑅𝑛 − �̇�𝐸𝑛+1 = �̇�𝑃 = 𝑐𝑜𝑛𝑠𝑡

- components: �̇�𝐹𝑥𝑘𝐹 − �̇�𝐸1𝑦𝑘,1 = �̇�𝑅𝑛𝑥𝑘,𝑁 − �̇�𝐸𝑛+1𝑦𝑘,𝑛+1 = �̇�𝑃𝑧𝑘𝑃 = 𝑐𝑜𝑛𝑠𝑡

Here the point P with composition 𝑧𝑘𝑃 is hypothetical but useful; lever rule applies on the lines connecting

(F, E1, P) and (Rn, En+1, P) for 𝑛 = 1…𝑁 in particular (RN, EN+1=S, P )

Equilibrium in stage n

- formally: 𝑦𝑘,𝑛 = 𝜓𝑘,𝑛𝑥𝑘,𝑛; 𝑛 = 1…𝑁

- in graph: equilibrium curve + tie lines

Graphical solution

1) Summing form of mass balances serves to find the point M and then the point E1 provided that the

compositions of F, S and RN are given.

Lever rule: �̇�𝐹

�̇�𝐹+�̇�𝑆=𝑆𝑀

𝑆𝐹 → determine M, then connect RN and M to find E1

2) Difference form of mass balances serves first to find the point P (=pole) as intersection of line

(F, E1) and (RN, S). The tie line from E1 determines R1 and connection of P with R1 determines E2

(see the difference form of balances). This is repeated until RN is reached. In the figure below six

step were needed to reach RN → N = 6.