Crystallization 1may2014

8/12/2019 Crystallization 1may2014

1/35


2/35

2

TOPIC OUTCOMES

By end of topic, students should be able to

understand the concept of crystallization

understand equilibrium solubility of materials

perform material & heat balance for crystallization process

understand Nucleation Theories

discuss basis of equipment

describe types of crystallizers


3/35

3

TODAYS LESSON OUTCOMES

By end of lesson, students should be able to

understand the concept of crystallization

understand equilibrium solubility of materials

perform material balance for crystallization process

understand Nucleation Theories


4/35

4

CRYSTALLISATION THEORY

Crystallization isa particle formation process by whichsolute molecules in a solution are transformed into asolid phase of regular lattice structure

occurs by precipitation process where particles

form by decreasing solute solubility (i.e.increasingsupersaturation ) by cooling, evaporation, anti-solvent addition, etc.mass transfer of a solute from liquid solution toform pure solid crystalline phase

Key point: solid-liquid separation process>>>drivingforce: supersaturation


5/35

5

APPLICATION

One of the oldest and most important unit operation withenormous economic importance.- Widely used in fine chemical and pharmaceuticalindustries for purification, separation, production step(s).


6/35

6

OBJECTIVE OF CRYSTALLIZATION

Important objectives in crystallization

good yield

high purity

size uniformityminimize caking

ease of pouring

ease of washing &filtering

uniform behavior


7/35

7

CRYSTAL GEOMETRY

Crystal solid composed of atoms, ions or molecules

which are arranged in organized, orderly and repetitive manner

appear as polyhedrons

flat faces and sharp corners All crystals of same material possess

equal angle between the corresponding faces(particular shape)

relative sizes of faces can be different

same particular characteristics

Geometry important to recognize crystal characteristics

Different size and similar shape


8/35

8

CRYSTAL GEOMETRY

Crystal structure maintain lattice structure

A point lattice is a set of points arranged so that each point hasidentical surroundings.

A unit cell is a single cell constructed employing the same

parameters ( e.g. bond angles ) as those of lattice.

Point lattice Unit cell


9/35

9

CRYSTAL GEOMETRYCrystal classification based on the interfacial angle & length of axes

Seven Crystallographic systems


10/35

10

TYPES OF CRYSTALLINE SOLID

Crystalline solids can be classification based on type of bond to

hold the particles in place in crystal latticei.Ionic crystals - charged ions held in place in the lattice byelectrostatic forces (e.g. sodium chloride).ii.Covalent crystals - constituent atoms do not carry effectivecharges; connected by a framework of covalent bonds, the atomssharing their outer electrons (e.g. diamond).iii.Molecular crystals - discrete molecules held together by weakattractive forces (e.g. VDW force or H bonds) (e.g. organiccompounds, sugar).

iv.Metallic crystals - ordered arrays of identical cations held bysharing of outer electrons between constituent atoms (e.g.copper).


11/35

QUESTION?

11


12/35

12

SOLUBILITY IN CRYSTALLIZATION

Solubility - maximum

amount of solute that can bedissolved in a given solvent ata given temperature

EQUILIBRIUM in crystallization is attained when the solution is SATURATED

Represented by a SOLUBILITY CURVE

Solubility is dependent mainly on TEMPERATURE


13/35

SOLUBILITY IN CRYSTALLIZATION

13

Solubility measurementsPolythermal methods heating solutionsinitially containing excess solutes.Isothermal methods adding solvents at

constant temperature.

Magnitude of solubility depends on unit used.Mass (or moles) solute/mass (or moles) solvent

Mass (or moles) solute/mass (or moles) solutionMass (or moles) solute/volume solution


14/35

14

SOLUBILITY CHART

generally, thesolubilities of mostsalts increase withincreasing temperature

line = saturatedabove line = supersaturatedbelow line = undersaturated

but can be otherwise


15/35

15

SUPERSATURATION

Supersaturated solution

Solution containing more dissolved solute than that given bythe equilibrium saturation value.

Degree of supersaturation (conc. driving force) is given by: c= c cs (molar concentration); or y = y y s (molar fraction)

where c and cs are the solution conc., and equilibrium

saturation conc. at a given T, respectively.

Saturated solution

Solution that is inthermodynamicequilibrium with thesolid phase of its soluteat a given temperature.


16/35

16

GENERATION OF SUPERSATURATION

If solute solubility increase strongly withincrease temperature, supersaturationgenerated by temperature reduction

COOLING

If solubility is independent of temperature,supersaturation generated by evaporating aportion of the solvent

SOLVENTEVAPORATION

If solubility is very high (NEITHER cooling &evaporation is desirable), supersaturation isgenerated by addition of common ion salt todecrease solubility. (e.g. adding ammonium sulphate to proteinsolution)

SALTING

Techniques to generate supersaturation

PRECIPITATIONIf a nearly complete precipitaion is required,supersaturation generated by chemical reaction

by adding third component. (e.g. hydrolysis of sodiumbenzoate with HCl to crystallize benzoic acid)


17/35


18/35

18

FORMATION OF CRYSTALS

Formation of solid crystals from homogeneoussolution

C on c e n

t r a t i on

of

s ol u

t e , C

Temperature, T

Solubility curve[saturationconcentration, C*(T)]


19/35

19



C on c e n

t r a t i on

of

s ol u

t e , C

Temperature, T


A

Undersaturated


20/35

20



C on c e n

t r a t i on

of

s ol u

t e , C

Temperature, T


AB

Supersaturated


21/35

21



Nucleation

C on c e n

t r a t i on

of

s ol u

t e , C

Temperature, T


Metastablelimit

Metastablezone

CA

B

Metastable limit is influenced by saturation temperature, rate of supersaturationgeneration, impurity level, mixing

For nucleation in metastable zone, seeding (adding small crystal particles) isrequired.



22/35

22



Growth

C on c e n

t r a t i on

of

s ol u

t e , C

Temperature, T


Metastablelimit

D

Metastablezone

CA

B


23/35

QUESTION?

23

YIELD & MATERIAL BALANC


24/35

24

YIELD & MATERIAL BALANC

material balance isstraightforward if

solutes are anhydrous

in crystallizationsome water is removed as water

some water in the solution isremoved with the crystals ashydrate

MATERIAL BALANCE


25/35

25

MATERIAL BALANCE

COOLER &

CRYSTALLIZER

L kg solution(solute + solvent)

W kg H 2O

S kg solution xi,S

C kg crystals xi,C

MATERIAL BALANCE


26/35

26

MATERIAL BALANCE

COOLER &

CRYSTALLIZER

L kg solution xi,L

W kg H 2O

= 0 (no evap) xi,W

S kg solution xi,S

C kg crystals xi,C

solutewater,i

xC xW xS x L C iW iS i Li ,,,,

MATERIAL BALANCE


27/35

MATERIAL BALANCE

27

Example:

A salt solution weighing 10 000 kg with 30%

Na2CO3 is cooled to 293 K (20 C). The saltcrystallizes as thedecahydrate . What will be theyield of Na 2CO 310H 2O crystals if the solubilityis 21.5 kg anhydrous Na2CO3 per 100 kg of total

water? Assume that no water is evaporated.

MATERIAL BALANCE


28/35

28

MATERIAL BALANCE

COOLER &CRYSTALLIZER

10,000 kgsolution

30% Na 2CO 3

W kg H 2O

=0, no evap.

S kg soln

21.5 kg Na 2CO 3/100 kg H 2O

C kg crystals, Na 2CO 310H 2O

Molecular Weight:10H 2O = 180.2

Na 2CO 3 = 106

Na 2CO 3 10H 2O = 286.2

MATERIAL BALANCE


29/35

29

MATERIAL BALANCE

O10HCO NaMWOHMW

232

2C water x ,

322

2

CO NaOHOH

kg kg kg

x S water ,322

32

CO NaOHCO Na

kg kg

kg x S CO Na ,32,

1. Perform material balance for water and Na 2CO 3

Feed = Solution stream + Crystals stream + Vapor stream

Solution stream

Given: 21.5 kg Na 2CO 3 per 100 kg H 2O in Solution stream

Vapor stream

W = 0 as no evaporation

Feed stream: given

Crystal stream contains Na2CO

310H

2O

O10HCO NaMWCO NaMW

,232

32CO Na 32 C

x ,

MATERIAL BALANCE


30/35

30

MATERIAL BALANCE

Feed = Solution stream + Crystals stream + Vapor stream

Water: 0)(2.2862.180

)(5.21100

100)10000(7.0 C S

Na 2CO 3: 0)(2.286

106)(

5.211005.21

)10000(3.0 C S

solutewater,i

xC xW xS x L C iW iS i Li ,,,,

MATERIAL BALANCE


31/35

31

MATERIAL BALANCE

2. Solving the two equation simultaneously,

C = 6370 kg of Na 2CO 310H 2O crystals

S = 3630 kg solution

MATERIAL BALANCE


32/35

32

MATERIAL BALANCE

Assume that 6% of the total weight of thesolution is LOST by evaporation of water incooling, recalculate C and S ????

HEAT BALANCES IN CRYSTALLIZATIO


33/35

33

HEAT BALANCES IN CRYSTALLIZATIO

q = (H 2 + H V ) H 1

H 1 = enthalpy of the entering solution (feed) at the initialtemperature

H 2 = enthalpy of the final mixture of crystals and mother

liquor at the final temperature H V = enthalpy of water vapor (if evaporation occurs)

q = total heat transferred (kJ) (+ve: heat must be added(endothermic), -ve: heat must be removed (exothermic))

CRYSTALLIZER

Feed, H 1

H v , Water vapor

Two phase mixture(crystal + saturatedsolution), H 2

Example


34/35

34

Example

A feed of 10000 lbm solution is flowed into thesystem at 130 F. The concentrated solution isflowed out at 80 F. The yield of crystalsFeSO4.7H2O is 2750 lbm. The average heat

capacity of the feed is 0.70 btu/lb m F. The heat ofsolution at 80 F is -28.47 btu/lb m FeSO4.7H2O.

Heat of feed, H 1 = 10000(0.70)(130-80) = 350000 btu

Heat of crystallization, H 2 = 28.47 2750 lb m FeSO4.7H 2O= 78300 btu

Example


35/35

35

Heat transferred , q = (H 2 + H V ) H 1

=

78300 + 0

350000= 428300 btu

Since q is ve, heat is removed (exothermic)

Example

Crystallization 1may2014

Documents