Chapter 8 - Extraction and Leaching

ChE 221 / ChE 516 Mass Transfer Operations

Main references: McCabe, 7th Ed, Ch 23; Seader, 3rd Ed, Ch 8 & 16

Professor San Ping Jiang

Part I: Liquid-Liquid Extraction

Based on equilateral triangular diagram

Liquid-liquid extraction The method to remove one constituent from a solid or liquid

by means of a liquid solvent Liquid extraction – to recover a valuable product from a

multicomponent solution by contact with an immiscible solvent that has a high affinity for the product

When separation by distillation is ineffective or very difficult, liquid-liquid extraction (or solvent extraction) is one of the main alternatives to consider.

Close-boiling mixtures or substances that cannot withstand the temperature of distillation, even under a vacuum, may often be separated from impurities by extraction, which utilises chemical differences instead of vapour pressure differences.

3

An industrial example

4

Acetic acid: b.p.=118.1oC Water: b.p.=100oC Ethyl-acetate: b.p.=77.1oC

The feed of 30260 1b/h of 22wt% acetic acid in water is in contact with 71100 1b/h of ethyl-acetate solvent in a single-section extraction column. The low-density, solvent rich extract exits from top of the extractor with 99.8% acetic acid in the feed. The high density, carrier-rich raffinate exiting from bottom contains only 0.05wt% acetic acid.

4 major steps in an extraction process Bringing the feed and the solvent into intimate contact by

dispersing one phase into the other as droplets Separation of the extract and the raffinate phases that have

different densities The extract is the layer of solvent plus extracted solute The raffinate is the layer from which solute has been removed.

Removal and recovery of the solute from the extract phase in a relatively pure form (by distillation, evaporation, crystallization, etc)

Removal and recovery of the solvent from each phase, usually by distillation

5

Extraction preferred over distillation for Dissolved or complexed inorganic substances in organic or

aqueous solutions Removal of a contaminant present in small concentration A high-boiling component present in relatively small quantities

in an aqueous waste steam Recovery of heat-sensitive materials, where extraction may be

less expensive than vacuum distillation Separation of mixtures according to chemical type rather than

relative volatility Separation of close-melting or close-boiling liquids, where

solubility differences can be exploited Separation of mixtures that form azeotropes

6

Industrial liquid-liquid leaching process Solute Carrier Solvent

Acetic acid Water Ethyl acetate

Ammonia Butenes Water

Aromatics Paraffins Furfural

Aromatics Kerosene Sulfur dioxide

Benzoic acid Water Benzene

Fatty acids Oil Propane

Acetic acid Water Methyl isobutyl ketone

Formic acid Water Tetrahydrofuran

Penicillin Broth Butyl acetate

Vanilla Oxidized liquors Toluene

Vitamin A Fish-liver oil Propane

7

Principles of Extraction Since most continuous extraction methods use

countercurrent contacts between two phases, many of the fundamentals of countercurrent gas absorption and of distillation carry over into the study of liquid extraction.

Thus questions about ideal stages, stage efficiency, minimum ratio between the two streams, and size of equipment have the same importance in extraction as in distillation.

8

Design Considerations Some factors influencing extraction: Types of stage configuration Min solvent flow rate and actual flow as multiple of

min flow rate (or reflux for more than one stage system)

Operating conditions Emulsification and scum formation tendencies Phase density difference Interfacial tension Types of extractor

9

Desirable properties of a solvent Some characteristics of an ideal solvent: High selectivity for solute / low for carrier High capacity for dissolving solute Volatility difference with solute - recoverability Large density difference with carrier Lack tendency to form scum layer at interface Low viscosity – promote phase separation Stable – minimise make-up solvent Availability and low cost

10

Types of extractor – Mixer-settlers For batchwise extraction the mixer

and settler may be the same unit. For continuous flow the mixer and

settler are usually separate pieces of equipment. The mixer may be a small agitated tank

provided with inlets and drawoff line and baffles to prevent short-circuiting, or it may be other flow mixer;

The settler is often a simple continuous gravity decanter

The extract may be lighter or heavier than the raffinate, and so the extract may come from the top or the bottom of the equipment

11

Perforated-plate towers

The perforations are typically 1½ to 4½ mm in diameter and plate spacings range from 150 to 600 mm;

Usually the light liquid is the dispersed phase, and downcomers carry the heavy continuous phase from one plate to the next;

Depending on gravity flow both for mixing and for separation.

12

Perforations in horizontal plates Cascade weir tray with

mixing and settling zones

Agitated tower extractors Mechanical energy is

provided by internal turbines or other agitators, mounted on a central rotating shaft.

13

Rotating-disk unit York-Scheibel extractor

Types of configuration

14

(a) Single-section cascade; (b) two-section cascade; (c) dual solvent with two-section cascade.

15

The equilateral triangular diagram

16

Ternary phase diagram

The basis of the technique of presentation of equilibrium data on an equilateral triangular graph paper is that the sum of the distances of a point within an equilateral triangle from the three sides is equal to the height of the triangle;

On the equilateral triangle ABC, each apex is a pure component of the mixture, A, B and C, respectively;

Each edge is a mixture of the two pure components at the terminal apexes of the side, A-B, B-C and C-A, respectively.

Consider the point M, for example: MN1+MN2+MN3 = CT; M represents mixture of 16%A, 24%B and 60%C [100-(16+24)=60%].

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The equilateral triangular diagram

18

Exercise: 73.27%A, 6.54%B, 20.19%C (Raffinate phase, x, rich in A) 5.2%A, 91.8%B, 3.0%C (Extract phase, y, rich in B) A: Carrier B: Solvent C: Solute

A B

C

20.19% C

10%

0%

100%

0% 10%

100%

73.27% A

(Print this slide in full size.)

Construct ternary phase diagram

Point Furfural Glycol Water

A 95.0 0.0 5.0

B 90.3 5.2 4.5

C 86.1 10.0 3.9

D 75.1 20.0 4.9

E 66.7 27.5 5.8

F 49.0 41.5 9.5

G 34.3 50.5 15.2

H 27.5 52.5 20.0

I 13.9 47.5 39.6

J 11.0 40.0 49.0

K 9.7 30.0 60.3

L 8.4 15.0 76.6

M 7.7 0.0 92.3

Point Glycol in water layer, wt%

Glycol in furfural layer, wt%

P 41.5 41.5

1 52.5 27.5

2 51.5 20.0

3 47.5 15.0

4 40.0 10.0

5 30.0 7.5

6 20.0 6.2

7 7.3 2.5

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Equilibrium miscibility data Mutual equilibrium (tie-line) data

Water-ethylene glycol-furfural system at 25oC, 101 kPa.

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Ethylene Glycol (B)

Furfural (C)

0.1 0.9

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.4

0.5

0.6

0.7

0.8

Print this slide in full size for Class Problem - 1

0.2

0.3

0.1

0.9 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.1 Water (A)

Water-EG-Furfural phase diagram Water-EG and furfural-

EG are completely miscible pairs

Furfural-water is a partially miscible pair

At plait point, P, the two liquid phases have identical compositions

Miscibility limit for the furfural-water binary system are D and G

Thus, furfural can be used as solvent to remove EG from water Furfural-rich phase is

the extract and water-rich phase is raffinate 21

A single-stage extraction What is composition of

mixture represented by point M?

What is the composition of extract phase and raffinate phase?

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solvent carrier

solute

• Mixture, M: 20% EG; 19% Water; 61% Furfural.

H2O = 0.19

EG = 0.2

•Extract, E: 10% EG; 4% Water; 86% Furfural.

•Raffinate, R: 40% EG; 49% Water; 11% Furfural.

Ternary systems comprising a carrier A, a solvent B and a solute C. Three binary mixtures can be formed out of these components, A-B, B-C, and C-A. The mutual miscibility behaviour of the components in each of these binaries determines the nature of the equilibrium diagram for the ternary system.

Here, the solute C is miscible with A and B in all proportions, but the carrier and the solvent are only partially miscible. The curve RPS is the equilibrium diagram in the equilateral triangular coordinate system.

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Liquid-liquid equilibria (LLE)

Solute C

Carrier A Solvent B

Equilibrium or binodal curve

•

Plait point, P

Separation range The larger the two-phase

region on line R-S, the greater the immiscibility of carrier and solvent.

The closer the top of the two-phase region is to apex A, the greater the range of feed composition, along the line R-C, that can be separated with solvent S.

Limiting cases.

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Solute C

Carrier A Solvent B

F1

F2

M

D •

Feed composition range to be separated

LLE…continue The point R on the line AB represents the solubility of the

solvent B in the carrier A; the point S represents that of the carrier A in the solvent B.

The pair of points G1 and H1 on the curve represents two liquid phases in equilibrium: G1 is rich in the carrier (raffinate phase) and H1 is rich in the solvent (extract phase). The line G1H1 is called tie line. Usually they are not parallel. The point P, that demarcates the raffinate and the extract sides of the equilibrium curve is called the plait point. The type of equilibrium RPS is called binodal because it has two arms RP and PS, representing the raffinate and the extract sides.

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Phase splitting of ternary mixtures

26

Solvent S Component B

Feed F Component A, C

Extract E Component B, C

Raffinate R Component A, C

(a) Components A and B mutually insoluble.

Solvent S Component B

Feed F Component A, C

Extract E Component A, B, C

Raffinate R Component A, B, C

(b) Components A and B partially soluble.

Reverse-level-arm rule Level rule or mixing rule for ternary

liquid systems asserts that the composition that results when two liquid solutions are mixed lies on a straight line connecting their compositions. Thus, if R kg of a mixture represented by point R is combined with E kg of a solution located at point E, the resulting composition M will lie on a straight line connecting points R and E. Furthermore, the location of M will be such that line segments it defines stand in the ratio of the weights of the parent solutions:

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•

C

A B

•

• R

The mixing rule in a triangular diagram

E M

S L P

F N K

• • •

• • •

RMEM

ER= (8.1)

Validation of reverse-level-arm rule First we have: (8.2) M is mass of mixture, and for the component mass balance

Combine eq. (8.2) and (8.3) yields

Re-arrange eq.(8.4)

Because

Thus Because triangle MPR is similar to triangle ESM, consequently

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CMCECR MxExRx =+

CMCECR xERExRx )( +=+

CMCE

CRCM

CECM

CMCR

xxxx

xxxx

RE

−−

=−−

=

RFxEKxMNx CRCECM === ;;

ESSL

MNEKRFMN

RE

=−−

=

EMMR

ESSL

RE

==

MER =+

(8.3)

(8.4)

(8.5)

(8.6)

(8.1)

Type I - Acetone-water-MIK system The solute and solvent are

miscible in all compositions. Partial miscibility of the

solvent (methyl isobutyl ketone, MIK) and the carrier (water): one immiscible pair

The tie lines slope up to the left and the extract is richer in acetone than the raffinate phase.

The most common type.

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solvent

solute

carrier

Type II - Aniline-n-heptane-MCH system The solvent (aniline) is only

partially miscible with both the solute (methylcyclohexane, MCH) and carrier n-heptane: two immiscible pairs.

The tie lines slope up to the right → more solvent would be required since the final extract would not be as rich in the desired component.

In Type II system, there is no plait point.

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solvent carrier

solute

Single stage extraction Locate the feed and solvent

compositions at points F and S, respectively.

Define mixing point M as M = F + S = E + R.

Apply the inverse-level-arm rule. Let wi(E)

be the species I in the extract, wi(R) be

the species I in the raffinate, and wi(M) be

the species I in the feed+solvent phase. For solvent C

Obtain R and E phase compositions. The inverse-level-arm rule applies to

obtain weight fractions of R and E phases.

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Feed F Solvent S

Extract E Raffinate R

Solute B

Carrier A C Solvent

F

A

C

B

M

E

R •

S

Sc

Fc

Mc SwFwwSF +=+ )(

MFSM

SF=

• • •

•

Multistage cross-flow extraction Feed goes through stages in series and

to each stage, fresh solvent is added. This process is an extension of single-stage extraction.

For multistage with cross-flow, raffinate from each stage is contacted with following stage with pure solvent. Extracts are withdrawn from each stage.

The concentration of solute (B) in raffinate and extract decreases from stage to stage.

Apply material balance and inverse-level-arm rules, similar to single-stage extraction calculation procedures.

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Feed F

Solvent S1

Extract E1

Raffinate R1

Solute B

Carrier A C Solvent

F

A

C

B M1 E1

R1 •

S

• • •

•

Solvent S2

R2

Extract E2

• • • E2

R2 M2

Right-triangle diagrams Ternary, countercurrent extraction

calculations can also be made on a right-triangle diagram.

A right-triangle diagram can be developed from an equilateral-triangle diagram. Rectangular coordinate are in mass or mole fractions, xA and xC, where A is the solute and C the carrier.

The advantage of the right-triangle diagram is that ordinary, rectangular-coordinate graph paper can be used and either one of the coordinates can be expanded.

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Right-triangle diagrams

34

L L

Point L: Furfural-8.4%; EG-15% (water=100-8.4-15=76.6)

0.15

0.84

Distribution diagram

Plots of solute weight fraction in two phases against each other → two phase behaviour.

Plait point is located on the 45o diagonal → equality of phase concentrations

Compositions that lie above the diagonal are richer in solute content than those of the companion phase, while points below it denote a depletion in solute.

Distribution coefficient is defined as the solute mole-fraction ratio in two phases,:

m = xCB/xCA 35

Part II: Solid-Liquid Extraction

Solid-liquid extraction or leaching Leaching involves the removal of a soluble fraction (the

solute or leachant) of a solid material by a liquid solvent Effluents from a leaching stage are essentially solid-free liquid, called the

overflow, and wet solids, the underflow. To reduce the concentration of solute in the liquid portion of the underflow, leaching is often accompanied by countercurrent-flow washing stages.

The combined process produces a final overflow, called extract, and a final underflow, the extracted or leaching solids.

Ideally, the soluble solids are perfectly separated from the insoluble solids, but solvent is distributed to both products Additional processing of extract and leaching solids is necessary to

recover solvent for recycle.

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Industrial applications of leaching Inorganic and organic materials Metal processing industries e.g. removal as soluble salts

Removal of copper from ore using sulfuric acid Recovery of gold from ore using sodium-cyanide solution

Biological substances Biological and food industries

Extraction of sugar from sugar beets using hot water as solvent Removal of caffeine from green coffee beans using supercritical CO2 Vegetable oils from nuts and seeds using organic solvents (hexane,

acetone, ether)

Pharmaceutical products from plant roots, leaves, stems

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An industrial example

Soybeans contain highest percentage of edible oil

For soybeans, whose oil content is typically less than 0.3 1b per 1b of dry and oil-free solids, leaching is more desirable than expression.

The ideal solvent for commercial leaching of soybeans should have a high oil solubility; a high volatility; nonflammability; low cost; chemical stability; low toxicity.

Nonflammable trichloroethylene is an ideal solvent, but is classified as a hazardous, toxic chemical.

The favoured solvent is the commercial n-hexane, which presents fire hazard but has a low toxicity.

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Soybean flakes (10.67%moisture, 0.2675g oil/dry flake): 6.375 1b/h

Leached solid contains 0.0151 g oil/dry oil, free flake

Extract of 7.313 1b/h contains 15.35% oil after 11 h operation

Solvent flow: 10.844 1b/h

Kennedy extractor

Espresso machine A batch-leaching machine

In the machine, 7-9 grams of coffee beans are ground to a powder of particle size of 250-750 µm

Water is pumped to a pressure of 9-15 atm and heated to 88-92oC. The high pressure is required for pressure infusion of hot water through the bed of coffee powder, so extraction can proceed rapidly

During a period of 20-30 s, hot water is percolated through the coffee powder to produce a 45 mL shot.

Maximise extraction of the flavour-and-aroma chemicals and minimise extraction of the chemicals associated with bitterness such as quinine and caffeine.

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Moving-bed leaching: Bollman extractor Buckets are loaded with flaky seeds and

beans such as soybeans and are sprayed with intermediate solvent, half miscella.

As solids and solvent flow concurrently down the right-hand side of the machine, the solvent extracts more oil from the beans.

As the partially extracted beans rise through the left side of the machine, a stream of pure solvent percolates countercurrently through them.

Fully extracted beans are dumped at the top of the elevator. Each basket contains ~350kg of solids and

~200,000 kg of solids can be extracted per day

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Moving-bed leaching: Rotocel extractor A horizontal basket is divided into

walled compartments with floor that is permeable to the liquid.

Solids are admitted to each compartment at the feed point; the compartments then successively pass a number of solvent sprays, a drainage section, and a discharge point at which the floor of the compartment opens to discharge the extracted solids.

To give countercurrent extraction, fresh solvent is fed only to the last compartment before discharge point.

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Rotocel extractors are typically 3.4-11.3 m in diameter, 6.4-7.3m in height, and with bed depths of 1.8-3.0m.

Continuous, perforated-belt extractor Used to process sugar cane, sugar

beets, oil seeds and apples (for apply juice)

The feed solids are fed from a hopper to a slow-moving, continuous and nonpartitioned perforated belt. Below the belt are compartments for collecting solvent

Fresh solvent is sprayed over solids and above the compartments in a countercurrent fashion, starting from the discharge end of the belt.

Units from 7 to 37m long with belts from 0.5 to 9.5m wide and 0.8 to 2.6m deep can process as much as 7,000,000 kg/day of sugar cane or sugar beets. 43

Factors influencing leaching processes Stages involved in leaching: (1) dissolution of solute in the

solvent; (2) diffusion of the solute through the solvent; (3) transfer of the solute from the solution in contact with the particles to the bulk of the solution.

Particle size, nature of the solvent used, temperature and agitation are the important factors that influence extraction rates.

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Countercurrent leaching cascade

S = mass flow rate of inert solid and is constant from stage to stage

V = mass flow rate of entering solvent or overflow liquid (solvent + solute), which varies from stage to stage.

L = mass flow rate of underflow liquid (solvent + solute) which varies from stage to stage.

y = mass fraction of solute in the overflow liquid

x = mass fraction of solute in the underflow liquid Alternatively, V and L can refer to mass flow rate of solvent on solute-free basis and

symbols Y and X can be used as mass ratios of solute to solvent in overflow and underflow liquids, respectively.

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overflow

underflow

yN+1 yL

xN xL

Equilibrium-stage model Overflow contains no solid; solvent is not vaporised, adsorbed,

or crystallised Any entering solid solute is completely dissolved into liquid in

the stage; composition of liquid in the stage is uniform Solute is not adsorbed on the surface of inert solid; inert solid

leaving in underflow from each stage are wet with liquid, such that mass ratio of solvent in that liquid to inert solid is constant from stage to stage

Concentration of solute in the overflow is equal to that in the liquid portion of solute in the underflow. This is equivalent to an equilibrium assumption

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Equilibrium-stage model For the continuous, countercurrent system of ideal

leaching stages, solute and total-liquid material balances can be used to solve problems: Determination of ideal stages required to achieve a specified

degree of washing Determination of the effect of washing for a specified degree

of washing with a certain number of ideal stages

For most problems, it is best to consider the leaching stage separately from the washing stage

Depending on the problem, either an algebraic or a graphic method (e.g., McCabe-Thiele method) can be used.

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Ideal leaching stage In an ideal leaching stage, all of

the solute is dissolved by the solvent, whereas the solid is not dissolved. Also the composition of the retained liquid phase in the underflow slurry is identical to the composition of the liquid overflow, and that overflow is free of solids.

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Mixer-settler

Overflow

Liquid (B,C)

Liquid (B,C)

Solid (A)

underflow

Solid feed, F

Insoluble A Soluble B

Liquid solvent, S

Leaching and washing

When leaching is very rapid, it is common to countercurrently wash the solids to reduce the solute concentration in the liquid adhering to the solids. This can be accomplished in a series of gravity thickeners or centrifugal thickeners, called hydroclones.

The arrangement ensures that a clear overflow is obtained.

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Strong solution

Operating lines for washing stages Material balances for the portion of the cascade consisting of the N units

are as follows: Total solution: LL + VN+1 = LN + V1 (8.7)

Solute: LLxL + VN+1yN+1 = LNxN + V1y1 (8.8)

Solving for yN+1 gives the operating line equation:

As usual, the operating line passes through the two end points (XL, Y1) and (XN, YN+1), and if the flow rates are constant, the slope is L/V.

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1

11

11

+++

−+=

N

LLN

N

NN V

xLyVxVLy (8.9)

McCabe-Smith algebraic method McCabe and Smith derived an algebraic method based on Kremser equation

for the calculation of number of ideal stages, N

The term log(L/V) can be written in terms of end points

Combining eq(8.10) and (8.11) gives

When V and L are liquid flows of solute-free solvent, x and y in (8.10) and (8.11) are replaced by X and Y solute mass ratios.

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( )VLyy

yx

N LL

NN

/log

log 1

−−

=

+

NL

N

NL

N

xyyy

xxyy

VL

−−

=−−

= ++ 1111

(8.10)

(8.11)

−−

−−

=+

+

NL

N

LL

NN

xyyyyy

yx

N11

1

log

log

(8.12)

Tutorial 9

Study questions When a binary feed is contacted with a solvent to form

two equilibrium liquid phases, which is the extract and which is the raffinate?

What are the characteristics of triangular diagrams? On such a diagram, what are the miscibility boundary, plait point, and tie lines?

What are the conditions for an ideal, equilibrium leaching stage?

What are the similarities and differences between extraction and distillation?

What is inverse-level-arm rule and its application in extraction?

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Class problem - 1 Construction of equilateral triangular diagram for

Furfural-Ethylene Glycol-Water system at 25oC and 101 kPa.

53

Class problem - 2 A mixture containing 40wt% acetone and 60wt% water is

contacted with an equal amount of MIK (a) What fraction of the acetone can be extracted in a single-

stage process? (b) What fraction of the acetone could be extracted if the

fresh solvent were divided into two parts and two successive extractions used?

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Answers: (a) fraction of acetone extracted = 78.9%; (b) Fraction of acetone extracted = 86.3%.

Class problem - 3 4000 1b per day of waxed paper containing 25wt% soluble wax and 75wt%

insoluble pulp are to be dewaxed by leaching with kerosene in the continuous, countercurrent system. The wax is completely dissolved by kerosene in the leaching stage, L. Subsequent washing stages reduce the wax content in the liquid adhering to the pulp leaving the last stage, N, to 0.2 1b wax/100 1b pulp. Kerosene entering the system is recycled from a solvent-recovery system and contains 0.05 1b wax/100 1b kerosene. The final extract is to contain 5 1b wax/100 1b kerosene. Experiments show that underflow from each stage contains 2 1b kerosene/1b insoluble pulp. Determine the required washing stage using McCabe-Thiele graphic method.

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YN+1, XN

Y1, XL

Answers: three washing stages are required.