Dr Saad Al-Shahrani

Example: A liquid-liquid ternary phase diagram for isopropyl alcohol (IPA), toluene, and water at 25oC is shown in the next figure. 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 kg/hr. 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


Solution:

In this system, water is the solvent and IPA is the solute that we wish to extract from the toluene-IPA feed mixture.

The raffinate stream leaving the extractor will be mostly toluene.

Firstly, the points F and So are plotted, and a straight line is drawn joining them.

Then the point M is determined as follows:

Next a straight line is drawn from the point RN up through M. The intersection of this straight line with the solubility curve gives us the point S1.

x1M,= 40/ (100+100) = 0.2 , x3

M= 100/(100+100) = 0.5



The composition of the solvent phase leaving the first stage is read off the graph: 27.5 % IPA and 70 % water.

Two straight lines are drown to locate Δ point. The first line pass through F and S1, the second through RN and S0. The intersection of the two lines occurs at Δ.

An LLE tie line is then followed from S1 down to R1 on the solubility curve. This line represent the first stage.

The composition of the raffinate phase leaving the first stage is 21 % IPA and 2 % water.

An operating line is drown from R1 to Δ. This line intersects the solubility curve at S2.



Finally, an LLE tie line is drown from S2 to R2. since the raffinate phase from stage 2 contain 3% IPA, achieved the required separation in two stages.

The composition of the solvent phase leaving the second stage is 12% IPA and 86% water.


S1

S2

S0

R1

= R2

.

.F

RN

F = 100 kg/hr z1= 40

S0= 100 Kg/hr z2= 60

x1RN = 0.03 x3

S0 = 1.0

Δ

M1

2


Minimum SOLVENT RATE

In the normal design situation, the solvent rate must be specified by the designer.

Usually, the smaller the solvent flow rate the lower the energy costs because less solvent must be circulated between the extractor and the stripping column. Hence, separation costs are reduced.

However, the smaller the solvent flow, the higher the capital costs, because more stages are required in the liquid-liquid extractor.

The minimum solvent flow rate Somin occurs when an LLE tie-line is

coincident with an "operating line" drawn to the Δ point.

The minimum solvent rate is usually found by extending the LLE tie-line that passes through the feed point.




The intersection of this straight line with the straight line through the points S0 and RN gives the minimum Δ (Δmin).

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

The point S1min is found on straight line from the feed point to Δmin, and a

straight line is drown from it to the point RN which has been specified.



The point Mmin is then given by the intersection of this line with the straight connecting the point F and S0.

S0min can then be calculated from equations (13), (14) using the known

compositions x1min and x3

min of Mmin:(13) 0SFM

(14) 01110 SxFzMx SM

(15)



DESIGN PROCEDURE

The summarized specific sequence of steps to design a countercurrent liquid-liquid extractor for a ternary system are as follow:

1. Parameters specified.

(a) Feed flow rate and compositions (F, z1, z3).

(b) Composition of solute in the raffinate stream leaving the last stage

(x1RN).

(c) LLE data for the ternary system.

(d) Economic ratio of actual to minimum solvent flow rates (S0/ S0min).

(e) Composition of fresh solvent (x1S0, x3

S0)



2. Calculate the minimum solvent flow rate S0min.

(a) Extend the tie-line through the point F.

(b) Draw a line through the points S0 and RN.

(c) Locate S1min on the tie-line through F.

(d) Draw a line from S1min to RN.

(e) The intersection of this line with a line through F and S0 gives the point Mmin.

(f) Read off the composition (x1min) at Mmin.

(g) Calculate S0min from equation (15).

3. Set the actual fresh solvent flow rate

4. Calculate the point M from equations (13) and (14).

5. Draw a line from RN through M to find S1 on the solubility curve.



6. Locate the Δ point by drawing two lines, one through S0 and RN and one through the other through S1 and F.

7. Step off stages, alternately using the LLE tie-lines and the A point, as follows:

...32LLE

21LLE

1 SRSRS



Example: Design a countercurrent liquid-liquid extractor to separate acetone from methyl isobutyl ketone (MIK) using water as a solvent. The column is to operate at 25°C. The organic feed rate is 10,000 kg/h, and the composition is 45 weight percent acetone and 55 weight percent ketone. The fresh solvent is pure water, and twice the minimum solvent rate is to be used. The concentration of acetone in the organic raffinate leaving the top of the extractor is to be 2.5 weight percent acetone.

Solution:acetone is the solute to be separated from MIK using water as the solvent. Thus, component 1 is acetone and component 3 is water.

The point RN is plotted on the solubility curve at x1RN = 2.5 weight percent acetone.

The point S0 is plotted at the right lower comer of the diagram, since x3So = 100

weight percent water. The point F is plotted on the vertical axis at z1 = 45 weight percent acetone.



The LLE tie-line that passes through F is used to determine the points Δmin and S1

min.

A straight line is drawn from RN to S1min, and another from F to S0.

The intersection of these lines is the point Mmin, giving a composition x1min = 24

weight percent acetone.

The minimum solvent flow rate is calculated from equation (15):



●

●●

S1min

RN

Mmin

●x1min



The actual solvent rate is twice the minimum:

Calculate the point M



Determine S1:

Draw a straight line from RN through M to the solubility curve. The composition of the solvent phase on stage 1 is found from Figure to be x1

S1= 19 weight percent acetone.



The original organic feed rate was 10,000 kg/hr, but the organic raffinate leaving the unit is only 4,400 kg/hr. The recovery of acetone is found as follows:

Thus, 97.6 percent of the acetone fed to the unit is captured by the solvent.

S1

S2

S3

S4

S5

S0

R1

R2

R3

R4

RN

Δ .

M.



F

=R5

1

2

34

5

X1=16.4

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Dr Saad Al-Shahrani