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7/30/2019 HunterNash

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Lecture 17: Hunter Nash 1

Hunter Nash Liquid-Liquid Extraction

The following slides discuss:

Liquid-Liquid extraction

Specifications for liquid-liquid extraction cascades

Product points

Operating lines and operating points

The Hunter Nash ProcedureHunter Nash graphical construction

Stepping off stages

Minimum solvent


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Liquid-Liquid Extraction

Liquid-liquid Extraction of ternary systems involves contacting two equilibrium liquids where

the solvent liquid is immiscible or nearly immiscible with one of the components of the feed

liquid and miscible with one or more of the other components.

Liquid-liquid extraction is also referred to as:

Extraction

Solvent extraction

Liquid extraction

In Section 4.5 we used the Ternary Phase Diagram to analyze Liquid-liquid Extraction for a single equilibrium stage.

Ternary Phase Diagram

Solvent: TCE

Solute: Acetone

Carrier: Water


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Ternary Single-Stage Flash Separation

Plait Point

P

Tie-lines

Extract

Raffinate

Solvent C Carrier

Solute

Mixing point

E

R

Feed

S

F


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Liquid-Liquid Extraction Cascades

What if we have a countercurrent cascade of Liquid-Liquid Contacting Stages?

Can we use a similar analysis to the one we used for countercurrent leaching where

we had two condensed phases and complete immiscibility of the carrier in the solvent?

Yes, but also different since in the liquid-liquid case we assume that we can

disengage the phases and we dont have complete solubility.

Can we use a similar analysis to the one we used for countercurrent absorption or stripping?

Yes, but also different since equilibrium here is given by a liquid-liquid ternary diagram rather

than a vapor-liquid equilibrium.

S

F

Extract1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C

R1 Rn-1 Rn RN-2 RN-1 RNR2

E2 En En+1EN-1 ENE3E1

Raffinate


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Liquid-Liquid Extraction: Specifications

Specifications: F, (xi)F, (yi)S, T and one of:

1) S and (xi)RN2) S and(yi)E13) (xi)RN and(yi)E1

4) N and (xi)RN5) N and (yi)E16) S and N

S

F

Extract

1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C



Raffinate


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Product Points:

Step 1) Find the mixing point M=F+S

Step 2) Determine mixing point compositions from

component material balances or inverse lever rule

Step 3) Since we know RN lies on the equilibriumcurve and we know (xA)RN we can determine (xB)RN and (xC)RNStep 4) Since we know RN, M and E1 lie on a mixing

line we can locate E1 by extending a line from RN through

M to the equilibrium curve where it intersects E1.

Hunter Nash Solution for Liquid-Liquid Extraction

Plait Point

P

Tie-lines

Extract

Raffinate

Solvent C Carrier

Solute

E1

R1

Feed

S

F

Extract1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C

R1 Rn-1 Rn

RN-2

RN-1 RN

R2


Raffinate

RN

M


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S

F

Extract1 2 n N1 N

Carrier A (FA)Solute B (FB)

Solvent C

R1 Rn-1 Rn RN-2RN-1 RNR2

E2 En En+1 EN-1 ENE3E1

Raffinate

Liquid-Liquid Extraction: Operating Lines

S

F

Extract1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C



Raffinate

Operating Points and Lines

Mass Balance around entire cascade:

We define the operating point P as thedifference between passing streams:

F S RN E

1

F E1 RN S P

F En1

Rn E

1

F E1 R

n E

n1 P

Mass Balance around the first n stages:

We rearrange this equation to find that

all passing streams are related by the same

operating point P.



Operating Point


Mass Balance around an internal stage:

We can rearrange the above expression to

find that Rn is just a mixing point between

P and En+1

.

Rn E

n R

n1 E

n1

S

F

Extract1 2 n N1 N

Solvent C

R1 Rn-1 Rn RN-2 RN-1R

NR2


Raffinate

Rn R

n1 E

n E

n1 P E

n1

n

Rn

En En+1

Rn-1

n

Rn

P

En+1

F E1 R

n1 E

n R

N S PThe N mass balances around the N individual

stages result in:

The following figure illustrates this concept:

The stream Rn is the mixing point between

P and En+1 because P is the net flow into

stage n from passing streams Rn-1, and En.

S

F

Extract

1 2 n N1 N

Solvent C


E2 En En+1 EN-1 ENE3

E1

Raffinate

Replace Rn-1

and Enby P



Operating Lines

Operating Lines:

The raffinate points are mixing points between P and corresponding extract points. This is shown graphically

in the following diagram. Notice that to get the point P we need just F, S, E1 and RN.

S

Carrier

Solute

E1

R1

F

RN

E2

E3

E4

E5

E6

Operating Point

P





Step 1) Locate the Operating Point by finding the intersection

of operating lines for the leftmost and rightmost stages

a) Draw a line through E1

and F

b) Draw a line through S and RN

c) Locate the intersection P. This point

is the operating point P.

Plait Point

Carrier

Solute

Feed

RN

M

Operating Point

P

E1

S

S

F

Extract1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C

R1 Rn-1 Rn RN-2R

N-1

RNR2


Raffinate




Plait Point

Solvent C

Carrier

Solute

E1

R1

Feed

RN

E2

E3

E4

E5

E6

M

Operating Point

P

Operating Lines and Tie Lines: Stepping Off Stages:

Step 1) Locate point R1 from the tie line intersecting E1

Step 2) Draw a line from the operating point P through R1to the extract side of the equilibrium curve.

The intersection locates E2.

Step 3) Locate point R2 from a tie line.

Step 4) Repeat Steps 2 and 3 until RN is obtained.

S

F

Extract1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C


NR2


Raffinate




Operating Lines and Tie Lines: Stepping Off Stages:

Step 1) Locate point R1 from the tie line intersecting E1

Step 2) Draw a line from the operating point P through R1 to the extract side of the equilibrium curve.

The intersection locates E2.

Step 3) Locate point R2 from a tie line.

Step 4) Repeat Steps 2 and 3 until RN

is obtained.


Step 1) Locate the Operating Point by finding the intersection of operating lines for the leftmost

and rightmost stage

1a) Draw a line through E1 and F

1b) Draw a line through S and RN

1c) Locate the intersection P. This point is the operating point P.

Product Points:

Step 1) M=F+S

Step 2) Determine mixing point compositions from component material balances or inverse lever rule

Step 3) Since we know RN lies on a tie line and we know (xA)RN we can determine (xB)RN and (xC)RNStep 4) Since we know RN, M and E1 lie on a mixing line we can locate E1 by extending a line from RN throughM to the equilibrium curve where it intersects E1.

S

F

Extract1 2 n N1 N

Carrier A (FA)

Solute B (FB)

Solvent C


NR2


Raffinate



Liquid-Liquid Extraction: Minimum Solvent


Step 1) Locate the raffinate Operating Line by extending

a line from S through RNStep 2) Extend the tie lines to intersect the operating line

Step 3) The tie line that intersects furthest from RN

gives the minimum operating point Pmin.

Step 4) Extend a line from Pmin through F to the extract side

of the equilibrium curve to find E1.

Step 5) Extend a line from E1 to RN. The intersection with the

line SF gives the minimum mixing point.

Plait Point

Carrier

Solute

F

Pmin

E1

RN

Mmin

Mmax

Note: If the tie lines slope down towards

the solvent side of the diagram, then

the minimum operating point will lie on

the operating line at an intersection with

a tie line nearest S.

S


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Liquid-Liquid Extraction: Minimum Solvent

Plait Point

Carrier

Solute

F

Pmin

E1

RN

Mmin

S

Stepping off Stages for the minimum solvent case

After locating the points Pmin, E1, and Mmin

the stages can be stepped off. If the minimumsolvent is used then the separation will bepinched

offand will require an infinite number of stages.

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