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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
discuss basis of equipment
describe types of crystallizers
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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
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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
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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).
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OBJECTIVE OF CRYSTALLIZATION
Important objectives in crystallization
good yield
high purity
size uniformityminimize 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 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
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CRYSTAL GEOMETRYCrystal 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 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).
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QUESTION?
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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
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SOLUBILITY IN CRYSTALLIZATION
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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
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SOLUBILITY CHART
generally, thesolubilities of mostsalts increase withincreasing temperature
line = saturatedabove line = supersaturatedbelow line = undersaturated
but can be otherwise
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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.
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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)
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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)]
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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)]
A
Undersaturated
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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)]
AB
Supersaturated
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FORMATION OF CRYSTALS
Formation of solid crystals from homogeneoussolution
Nucleation
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)]
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.
FORMATION OF CRYSTALS
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FORMATION OF CRYSTALS
Formation of solid crystals from homogeneoussolution
Growth
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)]
Metastablelimit
D
Metastablezone
CA
B
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QUESTION?
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YIELD & MATERIAL BALANC
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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
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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
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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
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MATERIAL BALANCE
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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
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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
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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
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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
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MATERIAL BALANCE
2. Solving the two equation simultaneously,
C = 6370 kg of Na 2CO 310H 2O crystals
S = 3630 kg solution
MATERIAL BALANCE
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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
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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
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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
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Heat transferred , q = (H 2 + H V ) H 1
=
78300 + 0
350000= 428300 btu
Since q is ve, heat is removed (exothermic)
Example