MISS. RAHIMAH BINTI OTHMAN (Email: [email protected]) ERT 313/4 BIOSEPARATION ENGINEERING

MISS. RAHIMAH BINTI OTHMAN(Email: [email protected])

ERT 313/4ERT 313/4BIOSEPARATION BIOSEPARATION ENGINEERINGENGINEERING

LIST OF CHAPTERSLIST OF CHAPTERS

CHAPTER TITLE WEEK

4LIQUID-LIQUID EXTRACTION 5

5SOLID-LIQUID EXTRACTION (LEACHING)

6

6 ADSORPTION 8

LIST OF CHAPTERSLIST OF CHAPTERS

CHAPTER TITLE WEEK

7 CHROMATOGRAPHY 9

8ION EXCHANGE

10CRYSTALLIZATION

COURSE OUTCOMESCOURSE OUTCOMES

COAPPLY principles of extraction. ANALYZE extraction equipments, Batch Extraction, Continuous Extraction and Aqueous Two Phase Extraction. DEVELOP basic design of extractor.

Introduction to extraction.

Equipment for extraction.

Principles of Extraction

Operating modes of extraction (Batch Extraction, Continuous Extraction and Aqueous Two Phase Extraction).

Basic design of extractor.

OUTLINESOUTLINES

INTRODUCTION TO EXTRACTION

Liquid-Liquid extraction is a mass transfer operation in which a liquid solution (the feed) is contacted with an immiscible or nearly immiscible liquid (solvent) that exhibits preferential affinity or selectivity towards one or more of the components in the feed.

Definition of Extraction

To separate closed-boiling point mixture Mixture that cannot withstand high temperature of distillation Example:

- recovery of penicillin from fermentation broth solvent: butyl acetate

- recovery of acetic acid from dilute aqueous solutions solvent: ethyl-acetate

Purpose of Extraction

Selectivity of extraction directly from fermentation broths or from reaction medium in the case of biotransformations wherein whole cells or enzymes are used for conversion of a substrate into a desired product.

Reduction in product loss due to hydrolytic or metabolic/microbial degradation as the product is transferred to a second phase with different physical and chemical properties.

Suitability over a wide range of scales operation.

Advantages of solvent extraction;


1. Compositional complexity of the fermentation broth due to the presence of a variety of dissolved as well as solid substances, which gives rise to phase complexity and influences the extraction of the desired solute (s).

2. The presence of surface active species influences the mass transfer rate.

3. The presence of particulate matter and surface active species affects the phase separation

4. Chemical instability of the desired product due to metabolic or microbial activity and also due to the compositional or pH conditions during extraction affects the overall efficiency in the recovery of the desired product.

5. The rheological properties of the fermentation broths may show time dependence and may be altered affecting the extraction process.

However, solvent extraction of biological products is beset with several problems. These include;



EQUIPMENT FOR EXTRACTION

1. Mixer-settlers2. Packed extraction

towers3. Perforated-plate towers

4. Agitated tower

extractors

5. Pulse columns7. Auxilary equipment[stills, evaporators, heaters

and condenser]

6. Centrifugal extractors

Batchwise or continuous operation

Feed liquid + solvent (put in agitated vessel) = layers (to be settled and separated)

Extract – the layer of solvent + extracted solute

Raffinate – the layer from which solute has been removed

Extract may be lighter or heavier than raffinate.

Continuous flow – more economical for more than one contact process






4. Agitated tower

extractors


and condenser]


MIXER-SETTLERS

For Batchwise Extraction:

→ The mixer and settler may be the same unit.→ A tank containing a turbine or propeller agitator is most common.→ At the end of mixing cycle the agitator is shut off, the layers are

allowed to separate by gravity.→ Extract and raffinate are drawn off to separate receivers through a

bottom drain line carrying a sight glass.→ The mixing and settling times required for a given extraction can

be determined only by experiment.

(e.g: 5 min for mixing and 10 min for settling are typical)

- both shorter and much longer times are common.

MIXER-SETTLERS

Single Stage

Extraction

Feed

Solvent

Raffinate

Extract

Schematic Diagram Representation of a Single Stage Batch Extraction

MIXER-SETTLERS

For Continous Extraction:

→ The mixer and settler are usually separate pieces of equipment.

→ The mixer; small agitated tank provided with a drawoff line and

baffles to prevent short-circuiting, or it may be motionless mixer

or other flow mixer.

→ The settler; is often a simple continuous gravity decanter.

→ In common used; several contact stages are required, a train of

mixer-settlers is operated with countercurrent flow.

MIXER-SETTLERS

Note: The raffinate from each settler becomes a feed to the next mixer, where it meets intermediate extract or fresh solvent.




4. Agitated tower

extractors


and condenser]


→ Tower extractors give differential contacts, not stage contacts, and mixing and settling proceed continuously sand simultaneously.

→ Extraction; can be carried out in an open tower, with drops of heavy liquid falling through the rising light liquid or vice versa.

→ The tower is filled with packings such as rings or saddles, which causes the drops to coalesce and reform, and tends to limit axial dispersion.

→ In an extraction tower there is continuous transfer of material between phases, and the composition of each phase changes as it flows through the tower.

→ The design procedure ; is similar to packed absorption towers.

PACKED EXTRACTION TOWERS

PACKED EXTRACTION TOWERS

Tower packings; (a) Raschig rings, (b) metal Pall ring,

(c) plastic Pall ring, (d) Berl saddle, (e) ceramic Intalox saddle, (f) plastic

Super Intalox saddle, (g) metal Intalox saddle




4. Agitated tower

extractors


and condenser]


→ The axial mixing characteristic of an open tower can also be limited by using transverse perforated plate like those in the sieve-plate distillation towers.

→ The perforations are typically 1 , to 4 mm ( to in.) in diameter.

→ Plate spacing range from 150 to 600 mm (6 to 24 in.)→ Usually, the light liquid is the dispersed phase, and downcomers

carry the heavy liquid above.→ Extraction takes place in the mixing zone above the

perforations, with the light liquid (oil) rising and collecting in a space below the next-higher plate, then following transversely over a weir to the next set of perforations.

PERFORATED-PLATE TOWERS

2

1

2

1

16

1

16

3

PERFORATED-PLATE TOWERS




4. Agitated tower

extractors


and condenser]


→ It depends on gravity flow for mixing and for separation.

→ Mechanical energy is provided by internal turbines or other agitators, mounted on a central rotating shaft.

→ Fig (a), flat disks disperse the liquids and impel them outward toward the tower wall, where stator rings create quite zones in which the two phases can separate.

→ In other designs, set of impellers are separated by calming sections to give, in effect, a stack of mixer-settlers one above the other.

AGITATED TOWER EXTRACTORS

AGITATED TOWER EXTRACTORS

→ In the York-Scheibel extractor (Fig. b), the region surrounding the agitators are packed with wire mesh to encounter coalescence and separation of the phases.

→ Most of the extraction takes place in the mixing sections, but some also occurs in the calming sections.

→ The efficiency of each mixer-settler unit is sometimes greater than 100 percent.




4. Agitated tower

extractors


and condenser]


Extraction factor is defined as:

Where:

E = extraction factor

KD = distribution coefficient

V = volume of solvent

L = volume of aqueous

PRINCIPLES OF EXTRACTION→ Most continuous extraction methods use countercurrent contacts

between two phases, one a light liquid and the other a heavier one. → Importance measurements; ideal stage, stage efficiency,

minimum ratio between the two streams, and size of equipment. (same as distillation column)

Extraction of Dilute Solution

For a single-stage extraction with pure solvent;

- the fraction of solute remaining is

- the fraction recovered is

E1

1

E

E

1

PRINCIPLES OF EXTRACTION

BATCH EXTRACTION

EXAMPLE 23.2.

Penicillin F is recovered from a dilute aqueous fermentation broth by extraction with amyl acetate, using 6 volumes of solvent per 100 volumes of the aqueous phase. At pH 3.2 the distribution coefficient KD is 80.

(a) What fraction of the penicillin would be recovered in a single ideal stage?

(b) What would be the recovery with two-stage extraction using fresh solvent in both stages?


CONTINUOUS SINGLE STAGE EXTRACTION

EXAMPLE

An inlet water solution of 100 kg/h containing 0.010 wt fraction nicotine in water is stripped with a kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a single stage extraction unit. It is desired to reduce the concentration of the exit water to 0.0010 wt fraction nicotine. Calculate the flow rate of the nicotine in both of the exit streams.


SOLUTION

1. Nicotine in the feed solution = 100 (0.01) = 1 kg/h nicotine

Water in feed = 100 (1 - 0.01) = 99 kg/h water

2. Nicotine in solvent = 200 (0.0005) = 0.1 kg/h nicotine

Kerosene = 200 (1 – 0.0005) = 199.9 kg/h kerosene

3. Exit stream of aqueous phase, L1

Water = 99 kg/h = (1 – 0.0010) L1

L1 = 99.099 kg/h (nicotine + water)

Nicotine = 99.099 – 99 = 0.099 kg/h nicotine in exit stream

4. Exit stream of solvent phase, V1

Solvent = 199.9 kg/h

Nicotine in solvent = 0.1 + (1 – 0.099) = 1.001 kg/h in exit stream

Solvent + Nicotine = 199.9 + 1.001 = 200.9 kg/h

Equilibrium relationship are more complicated – 3 or more components present in each phase.

Equilibrium data are often presented on a triangular diagram such as Fig 23.7 and 23.8.


Extraction of Concentrated Solution

PRINCIPLES OF EXTRACTION Consider Fig 23.7

Line ACE shows extract phase

Line BDE shows raffinate phase

Point E is the plait point – the composition of extract & raffinate phases approach each other

Tie line – a straight line joining the composition of extract & raffinate phases.

Tie line in Fig 23.7 slope up to the left – extract phase is richer in acetone than the raffinate phase.

This suggest that most of the acetone could be extract from water phase using moderate amount of solvent.

• Consider Fig 23.8• Line AD shows extract phase • Line BC shows raffinate phase • Tie line in Fig 23.8 slope up to the right – extraction

would still be possible• But more solvent would have to use.• The final extract would not be as rich in desired

component (MCH)


How to obtain the phase composition using the triangular diagram?

- Point M: 0.2 Acetone, 0.3 water, 0.5 MIK

- Draw a new tie line

- Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK

- Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK

- Ratio of acetone to water in the product = 0.232/0.043 = 5.4

- Ratio of acetone to water in the raffinate = 0.132/0.845 =0.156

Let’s compare with Fig 23.8. Which system is more effective?


COORDINATES SCALE Refer to Treybal, Mass Transfer Operation, 3rd ed., McGraw

Hill

The book use different triangular system

The location of solvent (B) is on the right of the triangular diagram (McCabe use on the left)

Coordinate scales of equilateral triangles can be plotted as y versus x as shown in Fig 10.9

Y axis = wt fraction of component C (acetic acid)

X axis = wt fraction of solvent B (ethyl acetate)


COORDINATES SCALE

SINGLE-STAGE EXTRACTION

The triangular diagram in Fig 10.12 (Treybal) is a bit different as compared to Fig. 23.7 (McCabe)

Extract phase – on the left

Raffinate phase - on the right

Fig 10.12 shows that we want to extract component C from A by using solvent B.

Total material balance:

Material balance on C:

SINGLE-STAGE EXTRACTION

Amount of solvent to provide a given location for M1 on the line FS:

The quantities of extract and raffinate:

Minimum amount of solvent is found by locating M1 at D

Maximum amount of solvent is found by locating M1 at K

MULTISTAGE CROSSCURRENT EXTRACTION

Continuous or batch processes

Refer to Fig 10.14

Raffinate from the previous stage will be the feed for the next stage

The raffinate is contacted with fresh solvent

The extract can be combined to provide the composited extract

The total balance for any stage n:

Material balance on C:



EXAMPLE

If 100 kg of a solution of acetic acid (C) and water (A) containing 30% acid is to be extracted three times with isopropyl ether (B) at 20°C, using 40 kg of solvent in each stage, determine the quantities and compositions of the various streams. How much solvent would be required if the same final raffinate concentration were to be obtained with one stage?

The equilibrium data at 20°C are listed below [Trans. AIChE, 36, 628 (1940), with permission].



SOLUTION

The horizontal rows give the concentrations in equilibrium solutions. The system is of the type shown in Fig. 10.9, except that the tie lines slope downward toward the B apex. The rectangular coordinates of Fig. l0.9a will be used, but only for acid concentrations up to x = 0.30. These are plotted in Fig. 10.15.


Point M1 is located on line FB. With the help of a distribution curve, the tie line passing through M1 is located as shown, and x1 = 0.258, y1 = 0.117 wt fraction acetic acid. Eq. (10.8):

Stage 3

In a similar manner, B3 = 40, M3 = 130.1, xM3 = 0.1572,

x3 = 0.20, y3 = 0.078, E3 = 45.7, and R3 = 84.4.

The acid acetic content of the final raffinate is

= x3 R3 = 0.20(84.4) = 16.88 kg.

The composited extract is

E1 + E2 + E3 = 43.6 + 46.3 + 45.7 = 135.6 kg,

The acid content in the composited extract:

E1y1 + E2y2 + E3y3 = 13.12 kg.

If an extraction to give the same final raffinate concentration, x = 0.20.

were to be done in one stage, the point M would be at the intersection of tie line R3E3 and line BF of Fig. 10.15,

XM = 0.12.

The solvent required would then be, by Eq. (10.6),

S1 = 100(0.30 - 0.12)/(0.12 - 0) = 150 kg,

Hence, 150 kg of solvent is required for single stage extraction

120 kg of solvent is required in the three-stage extraction.

EXAMPLE 1A countercurrent system is to be used for the water-acetic acid-isoprophyl ether extraction (see Table 1). Feed is 39 wt % acetic acid and 59 wt % water. Feed flow rate is 1000 kg/hr. Solvent added contains 1 wt % acetic acid but no water. Total flow rate of added solvent is 1500 kg/hr. We desire a raffinate that is 7 wt % acetic acid. What are the weight fractions of EN and R1? What are the flow rates of EN and R1? How many stages are required?

EXAMPLE 1Water Layer, wt % Isopropyl Ether, wt %

Acetic acid xA

Water xD

Isopropyl Ether xS

Acetic Acid yA

Water yD

Isopropyl Ether yS

0.69 98.1 1.2 0.18 0.5 99.3

1.41 97.1 1.5 0.37 0.7 98.9

2.89 95.5 1.6 0.79 0.8 98.4

6.42 91.7 1.9 1.93 1.0 97.1

13.30 84.4 2.3 4.82 1.9 93.3

25.50 71.1 3.4 11.40 3.9 84.7

36.70 58.9 4.4 21.60 6.9 71.5

44.30 45.1 10.6 31.10 10.8 58.1

46.40 37.1 16.5 36.20 15.1 48.7

Table 1: Equilibrium data for water-acetic acid-isopropyl ether at 20 oC and 1 atm

STEPS1. Plot equilibrium data and construct conjugate line.2. Plot locations of streams Eo = S, RN+1=F, and R1

3. Find mixing point M on line through points S and F at xA, M value calculated from Equation below;

4. Line R1M gives pint EN.5. Find point as intersection of straight lines EoR1 and ENRN+1

6. Step off stages, using the procedure shown in Fig 4-12. To keep the diagram less crowded, the operating lines, EjRj+1, are not shown. You can use straight edge on Fig 14-14 to check the operating lines.

10

1,10,0,

10

1,10,0,

N

NDNDMD

N

NANAMA

RE

xRyEx

RE

xRyEx

EXAMPLE 2We wish to remove acetic acid from water using isopropyl ether as solvent. The operation is at 20 oC and 1 atm (See Table 1). The feed is 0.45 wt frac acetic acid, and 0.55 wt frac water. Feed flow rate is 2000 kg/hr. A countercurrent system is used. Pure solvent (no acetic acid and no water) is used. We desire an extract stream that is 0.20 wt frac acetic acid and a raffinate that is 0.20 wt frac acetic acid.

(a) How much solvent is required? (b) How many equilibrium stages are needed?

THANK YOU

Documents

MISS. RAHIMAH BINTI OTHMAN (Email: [email protected]) ERT 313/4 BIOSEPARATION ENGINEERING