dr saad al-shahraniche 334: separation processes liquid-liquid extraction choice of solvent...

Dr Saad Al-ShahraniChE 334: Separation Processes

Liquid-Liquid Extraction

CHOICE OF SOLVENT

Choosing the best solvent is the most critical aspect of developing a

liquid-liquid extraction process. The solvent should have a high

selectivity for the extracted solute. The selectivity of a solvent is similar

to relative volatility and is given by

RR

SS

xx

xx

21

21

/

/

x1S = weight fraction of component 1 (solute) in the solvent phase

x2S = weight fraction of component 2 in the solvent phase

x1R = weight fraction of component 1 (solute) in the raffinate phase

x2R = weight fraction of component 2 in the raffinate phase



SINGLE-STAGE CALCULATIONS

In one-stage liquid-liquid extractor as shown in the next Figure, Component 1 is

the solute, component 2 is the other component in the feed that we are trying to

separate from component 1, and component 3 is the solvent.

Feed

F, z1, z2 , z3

Fresh solvent

So, x1So, x2

So , x3So

solvent phase

S1, x1S1, x2

S1 , x3S1

Raffinate phase

R1, x1R1, x2

R1 , x3R1Stage

In the normal situation, the process feed contains no solvent, so that z3 = 0. We need to specify only two weight fractions in this ternary system.

If the solvent is essentially pure, as is often the case, x3So = 1, x1

So = 0, x2So = 0



The total mass balance at this stage is

A component balance on the jth component yields

Now we define the parameter

Where the point M must lie on the straight line joining F and So. It also must lie on a straight line joining S1 and R1

(4)

(5)

(6)



Since the two liquid phases leaving the system

are in phase equilibrium, the points R1 and S1

must be connected by an LLE tie-line.

1. Calculate M from equation (7).

2. Calculate x1M and x3

M from equations (8) and (9).

3. Locate the point M using x1M and x3

M .

4. Find the LLE tie-line that passes through the

point M.

(7)

(8)

(9)

the procedure for determining the compositions and flow rates of the two liquid streams leaving the system is as follows:

6. Calculate the flow rates S1 and R1 by solving equations (4) and (5)

simultaneously:



5. The points at the two ends of the LLE tie-line give the compositions of the two

phases leaving the system: the raffinate-rich phase with composition x1R1 and x3

R1

and the solvent-rich phase with composition x1S1 and x3

S1

Splitting the mixture of F and So into raffinate-rich phase R1 and solvent-rich phase S1 at ends of a tie-line.



Example: An organic stream, with composition 30 weight percent acetone

and 70 weight percent methyl isobutyl ketone and flow rate 10,000 kg/h,

is mixed with a pure water solvent with flow rate 5,000 kg/h. What are

the compositions and flow rates of the two liquid phases leaving a single-

stage liquid-liquid extractor operating at 25oC.

Solution:

x1So = 0

x2So = 0

x3So= 1.0

z1 = 0.3

z2 =0.7

z3 = 0

Feed= 10000

z1=0.3 z2= 0.7, z3=0

Fresh solvent=5000

x1So,=0 x2

So =0, x3So=1.0

solvent phase

S1, x1S1, x2

S1 , x3S1

Raffinate phase

R1, x1R1, x2

R1 , x3R1Stag

e



M

xSFzx

SM

0101

1

M

xSFzx

SM

0303

3



The point M is thus located at (0.333,0.20), as shown in Figure

Note that M lies on the straight line connecting the points F and So



The LLE conjugate line is used to determine the other end of the LLE tie-line on the solvent phase solubility curve. If the tie-line goes through the point M, the compositions of the water solvent phase and organic raffinate phase have been found. We have:



MULTIPLE STAGES WITH CROSSFLOW OF SOLVENT

If the process liquid stream from the first stages fed into a second extractor and mixed with more fresh solvent, as shown in the Figure, we have what is called cross-flow extraction.

The process can be described the same as for a single stage .it is simply repeated again for each stage, using the raffinate phase from the upstream stage as the feed to each stage.



Mix points for cross-flow extractor.



MULTISTAGE COUNTERCURRENT EXTRACTION

Multistage countercurrent extraction is the most commonly encountered liquid-liquid extraction process. The raffinate and solvent streams travel countercurrent to each other through N stages. The flow rate of the raffinate leaving the last stage (tray N) is R.

Mass and component balances around the entire cascade give

(10) 10 SRSF N

(11) 11101110 SxRxSxFz S

NRS N

(12) 13303310 SxRxSxFz S

NRS N



We next define a pseudo flow rate M and pseudo compositions x1M and x3

M :

(13) 0SFM

(14) 01110 SxFzMx SM

(15) 03330 SxFzMx SM



If the fresh solvent flow rate So

and composition are given

along with the feed flow rate F

and composition, we can

locate the point M on the

straight line connecting the

points F and So.

From equations (10) through

(12), it is clear that M must

also lie on the straight line

connecting S1 and RN, as

shown in the figure.



The point RN will be given, i.e., the concentration of the raffinate phase leaving the final stage will be specified so as to recover the desired amount of the solute from the feed.

In the typical design problem, you will be given:

Note: in any real system we cannot recover all of the solute from the feed

Since the points RN and M are known, a straight line can be drawn to the solubility curve to determine the composition of the x1

S1 and

x3S1 of the S1 stream.

Equations (10) and (11) can be used to solve for the flow rates RN and S1.



Since S1 and R1 are in phase equilibrium, we can use an LLE tie-line to determine the point R1 on the solubility curve.


Component and mass balances around the first stage can then be used to calculate the flow rate S2 and compositions x1

S2 and x3S2 of the solvent entering

stage 1 from stage 2:

(16) 112 SRSF

(17) 1111211112 SxRxSxFz SRS

(18) 1313233112 SxRxSxFz SRS

Once S2 is known, R2 can be found at the other end of the LLE tie-line.

This computational procedure can be repeated from stage to stage until enough stages have been used to produce a process stream that meets or exceeds the specifications on RN (the final raffinate phase leaving the unit).



Graphical procedure

First, define another pseudo flow rate Δ and pseudo compositions x1Δ and x3

Δ

This fictitious Δ stream serves the same purpose as the operating lines did in

binary distillation. We can calculate the compositions of "passing" streams in the

column. It provides a graphical way to solve the three mass balances that

describe this ternary system simultaneously:

(19) 0SRN

(20) 01110 SxRxx S

NRN

(21) 03330 SxRxx S

NRN



The pseudo composition xjΔ is entirely fictitious and, therefore, can be

less than zero or greater than unity.

Using equations (10) and (19), we see that

(22) 10 SFSRN

Therefore, the Δ point must lie on two straight lines, one through the points RN and S0 and the other through the points F and S1.



Δ can lie either to the left or to the right of the phase diagram



Liquid-Liquid Equilibrium

The mass balances for the first stage (equations (16) through (18), the definition of Δ (equations (19) through (21), and equation(22) can be combined to give

(23) 2110 SRSFSRN

The two streams R1 and S2 that pass each other between the first and second stages are related to each other by the Δ point. Hence, if we know R1 we can determine S2 by using the straight line that connects R1 and Δ.

Example: A liquid-liquid ternary phase diagram for isopropyl alcohol (IPA), toluene, and water at 25oC. Feed flow rate is 100 kg/hr, and feed compositions 40 weight percent IPA and 60 weight percent toluene. Fresh solvent is pure water at a flow rate of 100 kghr. Determine the number of equilibrium stages required to produce a raffinate stream that contains 3 weight percent IPA.



dr saad al-shahraniche 334: separation processes liquid-liquid extraction choice of solvent...

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