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					              Solvent Extraction
       Unit Operation in Agro-Industry III

2011          Asst.Prof.Dr.Pakamon Chitprasert
     Process & Unit Operation
2




    Unit I - Basic of engineering calculation, unit
           conversion, mass & energy Balance, thermo, fluid
           mechanics, fluid mixing, particle mechanics
    Unit II - Principles of heat transfer & mass transfer,
           drying involving heat and mass transfer
    Separation Processes
3




    Upstream process      Downstream process
     Unit III - Separation
     process involving heat
     and mass transfer
     Ethanol Production from Corn
4




Ethanol
separation


    Ethanol
    purification
    Importance of Downstream Processes
5


    Downstream processing costs can be as high as
    60 - 70% of the selling price of the product as
    exemplified by the plant Eli Lilly built to produce
    human insulin. Over 90% of the staff are
    involved in the recovery processes. Improvements
    in downstream processing will benefit the overall
    efficiency and cost of processes and will make the
    biotechnology competitive to the conventional
    chemical processes.
Mass & Heat transfer operations –
controlling transport phenomenon
   Mass transfer controlling
    e.g. sedimentation, adsorption, absorption,
    leaching, extraction
   Mass transfer and heat transfer controlling
     e.g. distillation, drying, crystallization,
    evaporation

     Salt field where heat
     and mass transfer
     takes place.
Mass transfer operations – nature of
interface between phases
   Solid-solid contact: sieving
   Gas-liquid contact: absorption, evaporation,
    distillation
   Liquid-liquid contact: extraction
   Liquid-solid contact: crystallisation, adsorption
   Gas-solid contact: adsorption, drying

                A               B
                    Interface:
                    Boundary between any two phases
Solvent Extraction = Liquid-Liquid Extraction,
Liquid Extraction

To separate liquid compound from its mixture using
another liquid compound.
         Solvent           Extract phase (rich in
                           solvent) contains most of
                           desired compounds

                               Raffinate phase (rich
                               in diluent) contains a
                               few of desired
    Mixture                    compounds
Liquid-liquid extraction principles

Feed phase contains a component, i, which is to
be removed. Addition of a second phase
(solvent phase) which is immiscible with feed
phase but component i is soluble in both
phases. Most of component i (solute) is
transferred from the feed phase to the solvent
phase. After extraction the feed and solvent
phases are called the raffinate (R) and extract
(E) phases respectively.
Liquid-liquid extraction principles
Normally one of the two phases is an organic
phase while the other is an aqueous phase.
Under equilibrium conditions the distribution of
solute i over the two phases is determined by the
distribution law. After the extraction the two
phases can be separated because of their
immiscibility. Component i is then separated
from the extract phase by a technique such as
distillation and the solvent is regenerated.
Further extractions may be carried out to remove
more component i.
Examples of Solvent Extraction
Extract sucrose from oil using water




       Adding water in          Shaking
   solution of sucrose in oil
The mixture and the solvent are brought into
intimate contact so that a solute or solutes can
diffuse from one phase to the other phase.
Example of Solvent Extraction
Oil Phase Water Phase Most of sucrose are in water
                       phase at the end of extraction.
                       The amount of sucrose in the
                       water (extract phase) and oil
                       (raffinate phase) depends upon
                        Separation02.ogg




Oil                    the solubilities in one another.
phase
                       Two phases are separated
Water                  according to their immiscibility.
phase
   Separating funnel
Technical Terms
 Mixture = Feed
 Sugar = Solute

 Oil = Diluent

 Water = Solvent, Extractant

 Water layer = Extract layer
                                Separating
 Oil layer = Raffinate Layer
                                  Funnel
When to use solvent extraction?
   When distillation is not appropriate.
        - Boiling point of a desired compound is close
    to those of undesired compounds.
        - Desired compound is heat-sensitive.
        - Need more than one desired compound.
   Chemical properties of desirable compounds are
    much different from those of undesirable
    compounds.
   Cost is lower. e.g. extraction of dilute acetic acid in
    water using ether
Examples of Solvent Extraction
   Extraction of aromatics from paraffin and naptha
    (different chemical structures but about the same
    boiling point range) by liquid sulfur dioxide or
    diethylene glycol
 Extraction of penicillin (heat
  sensitive) from fermentation
  broth by butyl acetate
 Extraction of dilute acetic acid

  from water using ethyl acetate       Fermentor
Choices of separation process

Factors to be considered:
 Feasibility

 Product value

 Cost

 Product quality

 Selectivity
Desirable properties of solvents

 High Kd
 Immiscible with other compounds in mixtures

 Low viscosity

 Inert

 Non-toxic

 Non-flammable

 Low price

 Non-corrosive

 Easy to recover
Partition Coefficient =
Distribution Coefficient (Kd)
Shows the difference in solubility properties
of the desired compound in the diluent and
the solvent.
                           Kd
                Solute            Solute
                       Oil               Water

             Kd= Conc. of solute in water layer
                  Conc. of solute in oil layer
                  @ equilibrium
             Need solvent with high Kd.
Extraction Conditions
 Types and amounts of solvents
 Residence time

 Temperature

 pH

 Numbers of extraction stages

  These can be varied to obtain good
extraction efficiency.
Equilibrium Data
Extract acetone from water using methyl
isobutyl ketone (MIK) – 3 component system.
             Acetone
                        Acetone = Solute
              in MIK
            Acetone     Water = Diluent
            in water    MIK     = Solvent

  Complete miscibility of solvent and solute
  Partial miscibility of solvent and diluent
Phases and Components
At equilibrium, the number of phases of the mixture
possibly turns to be:
- One phase

           Components: Acetone,Water, MIK

-Two phases
  MIK     Components: Acetone,Water,MIK
 Water    Components: Acetone,Water,MIK
            Why cann’t it be 3 phases?
Equilateral Triangular diagram
Show the quantity of each component (solute, diluent,
and solvent) in each phase (one or two phases) at
equilibrium.
  One phase region
  Two phase region

  Tie line

  Extract layer

  Raffinate layer

  Plait point
Equilateral triangular   Pure acetone
phase diagram                           Shows mass fraction
of acetone-water-                       of solute-diluent-
MIK system                               solvent
@ 25C                                     Acetone (a) = Solute
                                           Water (b) = Diluent
                                           MIK (c) = Solvent




 (c)
Pure MIK                                           Pure H2O
     Concepts of Triangular Phase Diagram
24

      At the 3 corners of the diagram, the component
       (a, b and c) is pure (mass fraction = 1).
      On the axis of the diagram,

       the mixtures compose of 2
       components: (a,b), (a, c)
       and (b, c).
      Anywhere in the diagram,

       the mixtures compose of 3
       components: a, b and c
       appearing as one phase or 2 phases.
One phase region
                       No phase separation
 Acetone +             in the yellow-shaded
  Water +
    MIK            O   area


                   P




(c)
 Read the mass
 fraction values @ O       xa+xb+xc = 1.00
 xa = 0.70
 xb = 0.15
                       o
 xc = 0.15
                       P




(c)
Two phase region
                       The mixture in the
                       yellow-shaded area
                       appears as 2 phases.


                   P




  Tie line   C

                                D



(c)
 Components of extract and raffinate
 phases
At the beginning of extraction, mass fraction of
 mixture @ M,
      Acetone (xa) = 0.2
      Water (xb) = 0.3      xa + xb + xc = 1
      MIK (xc) = 0.5
At the beginning of extraction, mass fraction of
 extract layer @ C,
      ya = 0.232, yb = 0.043, yc = 0.725 Notice the
                                           change of
 raffinate layer @ D,                      mass fraction.
      xa = 0.132, xb = 0.845, xc = 0.023
 Plait point
Position where 2
phases have the
same components
and the same
mass fractions.
                   P




   Tie line   C

                       D



 (c)
     Concept of Equilateral Phase
30
     Diagram
        Liquid C dissolves completely in A or in B.
        Liquid A is only slightly soluble in B and B slightly
         soluble in A.
        The two-phase region is included inside below the
         curved envelope.
        An original mixture of composition M will separate
         into two phases a and b which are on the
         equilibrium tie line through point M.
        The two phases are identical at point P, the Plait
         point.
     Example of using equilateral triangular diagram
     to calculate compositions @ equilibrium
31


     An original mixture weighing 100 kg and containing
     20 kg of acetone (A), 30 kg of water (B), and 50 kg
     MIK (C) is equilibrated and the equilibrium phases
     separated. What are the compositions of the two
     equilibrium phases?

      xAM,xBM,                      E = ? y ,y ,y
                                           A B C
                    M
        xCM                         R = ? xA,xB,xC
     M = amount of mixture E = amount of extract layer
                           R = amount of raffinate layer
     Steps to solve
32
     1) Spot the mixture point
                 1.00 Acetone

                                30 kg water




 20 kg acetone    M

     1.00 MIK                         1.00 Water
     2) Spot the extract and raffinate
33
     layer
     by following the tie line that M is located

                1.00 Acetone

     @E: Acetone = 0.25
                                     @R: Acetone = 0.13
         Water = 0.05
                                        Water = 0.85
         MIK = 0.70
                                        MIK = 0.02
                    E     M
                                       R
     1.00 MIK                                  1.00 Water
     3) Do mass balance
34


     to determine the amount of extract and raffinate phase

      xAM,xBM,                       E = ? yA,yB,yC
                     M
        xCM                          R = ? xA,xB,xC
       “Conservation of Mass”
        Overall balance:        E+R=M
              A balance: EyA + RxA = MxAM
              B balance: EyB + RxB = MxBM
              C balance: EyC + RxC = MxCM
      3) Do mass balance
35

     Overall balance:     E+R=M
                          E + R = 20 + 30 + 50 (kg)   (1)
     A balance:       EyA + RxA = MxAM
              E(0.25) + R(0.13) = 100(0.2)            (2)
     B balance:       EyB + RxB = MxBM
              E(0.05) + R(0.85) = 100(0.3)            (3)
     C balance:       EyC + RxC = MxCM
              E(0.70) + R(0.02) = 100(0.5)            (4)
     Solve any 2 of these equations to get:
            E = 58.33 kg and R = 41.67 kg
     Continuous multistage countercurrent
     extraction
Extract                                                       Solvent
E1        E2        E3    En         En+1 EN-1          EN        EN+1
      1         2                n               N-1          N
R0         R1        R2   Rn-1       Rn   RN-2         RN-1       RN
Feed                                                     Raffinate
    Overall balance: R0+EN+1=RN+E1=M
     A balance: R0xA0+EN+1yAN+1=RNxAN+E1yA1=MxAM
     C balance: R0xC0+EN+1yCN+1=RNxCN+E1yC1=MxCM

     A = solute, B = diluent, C = solvent
     Case Study: Improved Ethanol-Water
     Separation Using Fatty Acids
37




      Pros & Cons
     High cost and
     no reuse of
     fermentation
     broth, but high
     yield (95.6%
     v/v)
     Liquid-Liquid Extraction as a
     preprocess of distillation
38


     Solvent normally used in extraction of ethanol
       from fermentation broth
      Good points: good separation (selectivity)

      Bad points: toxicity e.g. chloroform and cresol



                   Any solvent that can give both
                   good separation and nontoxic???
                   Also reduce the cost of
                   separation process???
     Fatty acids: Good solvents for
39
     extraction of ethanol from water?
      Renewable and natural       Solubility in water at 20C
                                   (in grams acid per liter)
       organic solvents
                                        Carbon Solubility
      Various solubility values        number
                                           2    infinite
                                           4    infinite
                                           6      9.7
                                           8      0.7
                                          10     0.15
                                          12     0.055
                                          14     0.02
                                          16     0.007
                                          18     0.003
     Solubility of Fatty acids
40

     C number Chloroform Benzene Cyclohexane Acetone
        10       3260     3980      3420      4070
        12       830       936       680       605
        14       325       292       215       159
        16       151        73        65      53.8
        18        60       24.6       24      15.4

     C number   Ethanol 95% Acetic acid Methanol Acetonitrile
        10          4400      5670       5100       660
        12           912       818       1200        76
        14           189       102        173        18
        16          49.3      21.4         37         4
        18          11.3       1.2         1         <1
     Objectives of Study
41


        Study the relative selectivity and distribution
         coefficients for removing ethanol from
         fermentation media using various kinds of
         fatty acids.

        Calculate the thermal energy requirements
         for purification of ethanol using liquid-liquid
         extraction compared to distillation.
     Variety of Fatty Acids
42



     Pentanoic acid   C5H10O2

     Hexanoic acid    C6H12O2

     Octanoic acid    C8H16O2

     Nonanoic acid    C9H18O2

     Oleic acid       C18H34O2
     Experimental Procedures
43


                    2-25 ml of
                    each fatty acid
                    2 ml of aqueous
                    phase: ethanol
                    7.81 g/dl

                   25 C,
                   200 rpm,
                   72 h
                   (equilibrium)
      Extraction                      Phase partition
     Separation of Ethanol from
44
     Fermentation Broth

                           Nonanoic
              EtOH + H2O




         Nonanoic +
          EtOH + H2O         Nonanoic +
         (Water phase)        EtOH + H2O
                             (Fatty acid phase)
     What are we looking for?
45


              Mixture of fatty acid, ethanol
              and water
              Ethanol
                                  Distillation
               Water
              Mixture of fatty acid, ethanol
              and water
              Ethanol
                                  Fermentation
               Fatty
                acid
     Factors determining separation
46
     efficiency
     1. Concentration of ethanol in each
       phase
           : distribution coefficient
       should be
     2. Amount of water in fatty acid
       phase
           : water absorption ability of
       fatty acid should be
     3. Amount of fatty acid in water
       phase
           : solubility of fatty acid in
       water should be
     Equilibrium Distribution of Ethanol
     between Water and Fatty Acids @ 25C
47


                Shortest
                 fatty acid




                                 Longest fatty acid




          1. Concentration of EtOH in each phase
     Effect of MW on D (1) and Water
48
     Absorption (2) in Fatty acids @ 25 C
                        Distribution coeff.
                        Water absorption

                        Pentanoic acid


                            Octanoic acid




     D = 3.88 x 105 x MW-2.75   g/dL in fatty acid phase
                                  g/dL in water phase
     Distribution Coefficients & Water
49
     Absorption
                                         g/dL in fatty acid phase
     D = 3.88 x    105 x   MW-2.75
                                           g/dL in water phase
                                            Abs.    Solubility
     Fatty acids      MW             D
                                           (g/dL)    (g/dL)
     Valeric         102.1       1.10        12       4.97
     Hexanoic        116.2      0.818         6        1.11
     Octanoic        144.2      0.494         2        0.68
     Nonanoic        158.2      0.354        1.5       0.03
     Oleic           282.5      0.069        0.8    insoluble
     What is the suitable fatty acid?
50

        According to D value, it is……………..
        According to water absorption value, it is…………
                          Distillation
                       To save energy:
                       High concentration of ethanol
                       Low concentration of water
                        Oleic acid is not our choice.
                          Fermentation
                       To be safe for yeast:
                       Low concentration of fatty acid
                       valeric and hexanoic acid aren’t
                       our choices.
     Octanoic acids Vs. Nonanoic acids
51




     Nonanoic acid is a better solvent because it gives
     1) higher ethanol recovery
     2) higher ethanol purity with negligible amount of
        nonanoic acid
     Energy Requirement
52

                 Nonanoic acid extraction combined
                 with the flash process requires 38%
                 less thermal energy to separate an
                 equivalent amount of ethanol from
                 fermentation broth compared to the
                 traditional distillation process.
     Solvent Extractors
•High degree of turbulence
•Good contact between phases
•High mass transfer rate
   Mixer-settlers
   Static column e.g. Spray column,
       Sieve tray column
 Agitated column e.g. Scheibel
     column
All can run batch-wise or continuously.
Mixer-Settlers
Solvent                                             Feed




      Small droplets            3
                                       Separation by gravity
   Turbine, propeller

                       Mixer Settler
                   Raffinate              Extract
Coalesce
of light           Light     Spray Columns
liquid             liquid
  Heavy            outlet
                            The drops of light liquid
  liquid
                            rise through the heavy
                            liquid. The drops are
                            coalesced at the top and
                            the heavy liquid leaves
                            the bottom.
 Light
 liquid            Heavy
      Nozzle       liquid
     distributor   outlet
           Perforated Plate
           (Sieve Tray) Column
Heavy
Liquid   The dispersed droplets coalesce
         to the thin layer below each
         tray and jet into the thick layer
         of liquid above by passing
         through the perforations.




Light
Liquid
Scheibel
Extractor
A series of agitators
mounts on a central
rotating shaft. Each
agitator is separated
by baffle to encourage
coalescence of the
droplets and phase
separation.
 Characteristics of Extractors

                  Mixer-        Static     Agitated
  Properties
                  Settlers     columns     columns
No. of stages   low          moderate    high
Flow rate      high      moderate        moderate
Residence time very high moderate        moderate
Viscosity       low-high low-moderate low-high
Density         low-high low-moderate low-high
difference

				
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