1a crystallization

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CRYSTALLIZATION

Dr Nurul Ekmi Binti Rabat

06 Oct. 2015

CDB 2063

SEPARATION PROCESS II

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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

describe crystal growth

discuss about crystallizing equipment

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TODAY’S 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

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CRYSTALLISATION THEORY

Crystallization is a particle formation process by which

solute molecules in a solution are transformed into a

solid phase of regular lattice structure

occurs by precipitation process where particles

form by decreasing solute solubility (i.e. increasing

supersaturation) by cooling, evaporation, anti-

solvent addition, etc.

mass transfer of a solute from liquid solution to

form pure solid crystalline phase

Key point: solid-liquid separation process>>>driving

force: supersaturation

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APPLICATION

One of the oldest and most important unit operation with

enormous economic importance.

- Widely used in fine chemical and pharmaceutical

industries for purification, separation, production step(s).

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OBJECTIVE OF CRYSTALLIZATION

Important objectives in crystallization

good yield

high purity

size uniformity

minimize caking

ease of pouring

ease of washing &

filtering

uniform behavior

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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

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CRYSTAL GEOMETRY

Crystal structure maintain lattice structure

A point lattice is a set of points arranged so that each point has

identical 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

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CRYSTAL GEOMETRY Crystal classification based on the interfacial angle & length of axes

Seven Crystallographic systems

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TYPES OF CRYSTALLINE SOLID

Crystalline solids can be classification based on type of bond to

hold the particles in place in crystal lattice

i.Ionic crystals - charged ions held in place in the lattice by

electrostatic forces (e.g. sodium chloride).

ii.Covalent crystals - constituent atoms do not carry effective

charges; connected by a framework of covalent bonds, the atoms

sharing their outer electrons (e.g. diamond).

iii.Molecular crystals - discrete molecules held together by weak

attractive forces (e.g. VDW force or H bonds) (e.g. organic

compounds, sugar).

iv.Metallic crystals - ordered arrays of identical cations held by

sharing of outer electrons between constituent atoms (e.g.

copper).

QUESTION?

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SOLUBILITY IN CRYSTALLIZATION

Solubility - maximum

amount of solute that can be

dissolved in a given solvent at

a given temperature

EQUILIBRIUM in crystallization is attained

when the solution is SATURATED

Represented by a SOLUBILITY CURVE

Solubility is dependent mainly on

TEMPERATURE

SOLUBILITY IN CRYSTALLIZATION

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Solubility measurements

Polythermal methods — heating solutions

initially 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) solution

Mass (or moles) solute/volume solution

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SOLUBILITY CHART

generally, the

solubilities of most

salts increase with

increasing temperature

line = saturated

above line = supersaturated

below line = undersaturated

but can be otherwise

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SUPERSATURATION

Supersaturated solution

– Solution containing more dissolved solute than that given by

the equilibrium saturation value.

Degree of supersaturation (conc. driving force) is given by: ∆c

= c – cs (molar concentration); or y = y – ys (molar fraction)

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

saturation conc. at a given T, respectively.

Saturated solution

– Solution that is in

thermodynamic

equilibrium with the

solid phase of its solute

at a given temperature.

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GENERATION OF SUPERSATURATION

If solute solubility increase strongly with

increase temperature, supersaturation

generated by temperature reduction

COOLING

If solubility is independent of temperature,

supersaturation generated by evaporating a

portion of the solvent

SOLVENT

EVAPORATION

If solubility is very high (NEITHER cooling &

evaporation is desirable), supersaturation is

generated by addition of common ion salt to

decrease solubility. (e.g. adding ammonium sulphate to protein

solution)

SALTING

Techniques to generate supersaturation

PRECIPITATION

If a nearly complete precipitaion is required,

supersaturation generated by chemical reaction

by adding third component. (e.g. hydrolysis of sodium

benzoate with HCl to crystallize benzoic acid)

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FORMATION OF CRYSTALS

Prerequisites for the formation of crystal is

supersaturation

formation of crystals - 2 steps :

1. birth of new particle - nucleation

2. its growth to macroscopic size

neither crystal growth nor formation of nuclei

from the solution can occur in a saturated or

unsaturated solution

driving potential for both rates is supersaturation

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FORMATION OF CRYSTALS

• Formation of solid crystals from homogeneous

solution

Concen

tration o

f solu

te, C

Temperature, T

Solubility curve

[saturation

concentration, C*(T)]

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FORMATION OF CRYSTALS

• Formation of solid crystals from homogeneous

solution

Concen

tration o

f solu

te, C

Temperature, T

Solubility curve

[saturation

concentration, C*(T)]

A

Undersaturated

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FORMATION OF CRYSTALS

• Formation of solid crystals from homogeneous

solution

Concen

tration o

f solu

te, C

Temperature, T

Solubility curve

[saturation

concentration, C*(T)]

AB

Supersaturated

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FORMATION OF CRYSTALS

• Formation of solid crystals from homogeneous

solution

Nucleation

Concen

tration o

f solu

te, C

Temperature, T

Solubility curve

[saturation

concentration, C*(T)]

Metastable

limit

Metastable

zone

CA

B

• Metastable limit is influenced by saturation temperature, rate of supersaturation

generation, impurity level, mixing

• For nucleation in metastable zone, seeding (adding small crystal particles) is

required.

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FORMATION OF CRYSTALS

• Formation of solid crystals from homogeneous

solution

Growth

Concen

tration o

f solu

te, C

Temperature, T

Solubility curve

[saturation

concentration, C*(T)]

Metastable

limit

D

Metastable

zone

CA

B

QUESTION?

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YIELD & MATERIAL BALANCE

material balance is

straightforward if

solutes are anhydrous

in crystallization

some water is removed as water

some water in the solution is

removed with the crystals as

hydrate

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MATERIAL BALANCE

COOLER &

CRYSTALLIZER

L kg solution(solute + solvent)

W kg H2O

S kg solution

xi,S

C kg crystals

xi,C

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MATERIAL BALANCE

COOLER &

CRYSTALLIZER

L kg solution

xi,L

W kg H2O

= 0 (no evap)

xi,W

S kg solution

xi,S

C kg crystals

xi,C

solutewater,

i

xCxWxSxL CiWiSiLi ,,,,

MATERIAL BALANCE

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Example:

A salt solution weighing 10 000 kg with 30%

Na2CO3 is cooled to 293 K (20C). The salt

crystallizes as the decahydrate. What will be the

yield of Na2CO3•10H2O crystals if the solubility

is 21.5 kg anhydrous Na2CO3 per 100 kg of total

water? Assume that no water is evaporated.

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MATERIAL BALANCE

COOLER &

CRYSTALLIZER

10,000 kg

solution

30% Na2CO3

W kg H2O

=0, no evap.

S kg soln

21.5 kg Na2CO3/

100 kg H2O

C kg crystals,

Na2CO3•10H2O

Molecular Weight:

10H2O = 180.2

Na2CO3 = 106

Na2CO3 • 10H2O = 286.2

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MATERIAL BALANCE

O10HCONa MW

OH MW

232

2

Cwaterx ,

322

2

CONa OH

OH

kgkg

kgx Swater

,

322

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CONa OH

CONa

kgkg

kgx SCONa

,32

,

1. Perform material balance for water and Na2CO3

Feed = Solution stream + Crystals stream + Vapor stream

Solution stream

Given: 21.5 kg Na2CO3 per 100 kg H2O in Solution stream

Vapor stream

W = 0 as no evaporation

Feed stream: given

Crystal stream contains Na2CO3•10H2O

O10HCONa MW

CONa MW ,

232

32CONa 32

Cx ,

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MATERIAL BALANCE

Feed = Solution stream + Crystals stream + Vapor stream

Water: 0)(2.286

2.180)(

5.21100

100)10000(7.0

CS

Na2CO3: 0)(2.286

106)(

5.21100

5.21)10000(3.0

CS

solutewater,

i

xCxWxSxL CiWiSiLi ,,,,

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MATERIAL BALANCE

2. Solving the two equation simultaneously,

C = 6370 kg of Na2CO3•10H2O crystals

S = 3630 kg solution

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MATERIAL BALANCE

Assume that 6% of the total weight of the

solution is LOST by evaporation of water in

cooling, recalculate C and S ????

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HEAT BALANCES IN CRYSTALLIZATION

q = (H2 + HV) – H1

H1 = enthalpy of the entering solution (feed) at the

initial temperature

H2 = enthalpy of the final mixture of crystals and

mother liquor at the final temperature

HV = 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,

H1

Hv , Water vapor

Two phase mixture

(crystal + saturated

solution), H2

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Example

A feed of 10000 lbm solution is flowed into the

system at 130F. The concentrated solution is

flowed out at 80F. The yield of crystals

FeSO4.7H2O is 2750 lbm. The average heat

capacity of the feed is 0.70 btu/lbmF. The heat of

solution at 80F is -28.47 btu/lbm FeSO4.7H2O.

Heat of feed, H1 = 10000(0.70)(130-80) = 350000 btu

Heat of crystallization, H2

= – 28.47 2750 lbm FeSO4.7H2O

= – 78300 btu

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Heat transferred, q = (H2 + HV) – H1

= –78300 + 0 – 350000

= – 428300 btu

Since q is –ve, heat is removed (exothermic)

Example

QUESTION?

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NUCLEATION THEORIES

The first formed ‘embryos’ due to clustering or

aggregation of ions or molecules in a supersaturated

solution.

Minute solid particles, seeds, small crystals- alternative

form of nucleation

Act as centers for crystal growth.

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NUCLEATION

Cluster – Several particles accumulate to form

loose aggregate

Embryo – Enough particles to form a new and

separate phase

Nucleus – Smallest group of particles, not

redissolve and grow to form crystal

Sequence of stages for crystal evolution

NUCLEATION

Types of nucleation

–Primary

• Homogeneous (spontaneous)

• Heterogeneous (induced by foreign

particles)

–Secondary (induced by crystals)

–Spurious Nucleation

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PRIMARY NUCLEATION

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Primary nucleation - nucleation takes place in the

absence of crystalline solid phase of the solute.

Occurs at high levels of supersaturation via

homogeneous or heterogeneous means.

– When the solution is absolutely clear, nucleation is

referred to as homogeneous/spontaneous

nucleation.

– While in the presence of a foreign solid phase (e.g.

dust, colloidal particles), nucleation is referred to as

heterogeneous nucleation.

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NUCLEATION RATE

Number of new particles formed per unit time

per unit volume of magma (crystal+mother

liquor) or solids-free mother liquor.

Unit = number/cm3 s

Rate of nucleation must be estimated to

determine the volume or residence time of

magma

223

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250

3

16exp10

sRT

NVB aaM

27.10,[McCabe ])

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NUCLEATION RATE

VM = Molar volume of crystals, cm3/g mol

Na = Avogadro constant, 6.022 x 1023 molecules/g mol

a = Average interfacial tension between solid and

liquid, ergs/cm2

R = Gas constant, 8.3143 x 107 ergs/g mol K

T = Temperature, K

= Number of ions per molecule of solute

s = Fractional supersaturation = ln

= Ratio of concentrations of supersaturated and

saturated solutions = c/cs

RTL

VM

4ln , L= crystal size . (27.8, [McCabe])

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NUCLEATION RATE

the rate of homogeneous nucleation of

potassium chloride is consistent with an

apparent interfacial tension of 2.8 ergs/cm2 at

300K, and the density of crystals is 1.97 g/cm3

Question : Determine the nucleation rate as a

function of s at a temperature of 300K.

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SECONDARY NUCLEATION

Formation of new particle influence by the

existing macroscopic crystal in the magma

1. Induced from fragments arising

from the suspended solute

crystals or from seeding

2. Induced from fluid shear &

collisions between existing

crystals with one another or with

wall of the crystallizer/rotary

impeller

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SPURIOUS NUCLEATION

Nucleation that occurs at large

supersaturations or accompanies poor

(slurry/suspension) magma circulation

Produces irregular crystals

QUESTION?

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