1a crystallization
DESCRIPTION
CrystallizationTRANSCRIPT
1
CRYSTALLIZATION
Dr Nurul Ekmi Binti Rabat
06 Oct. 2015
CDB 2063
SEPARATION PROCESS II
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
describe crystal growth
discuss about crystallizing equipment
3
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
4
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
5
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).
6
OBJECTIVE OF CRYSTALLIZATION
Important objectives in crystallization
good yield
high purity
size uniformity
minimize caking
ease of pouring
ease of washing &
filtering
uniform behavior
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
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
9
CRYSTAL GEOMETRY Crystal classification based on the interfacial angle & length of axes
Seven Crystallographic systems
10
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?
11
12
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
13
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
14
SOLUBILITY CHART
generally, the
solubilities of most
salts increase with
increasing temperature
line = saturated
above line = supersaturated
below line = undersaturated
but can be otherwise
15
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.
16
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)
17
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
18
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)]
19
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
20
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
21
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.
22
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?
23
24
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
25
MATERIAL BALANCE
COOLER &
CRYSTALLIZER
L kg solution(solute + solvent)
W kg H2O
S kg solution
xi,S
C kg crystals
xi,C
26
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
27
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.
28
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
29
MATERIAL BALANCE
O10HCONa MW
OH MW
232
2
Cwaterx ,
322
2
CONa OH
OH
kgkg
kgx Swater
,
322
32
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 ,
30
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 ,,,,
31
MATERIAL BALANCE
2. Solving the two equation simultaneously,
C = 6370 kg of Na2CO3•10H2O crystals
S = 3630 kg solution
32
MATERIAL BALANCE
Assume that 6% of the total weight of the
solution is LOST by evaporation of water in
cooling, recalculate C and S ????
33
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
34
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
35
Heat transferred, q = (H2 + HV) – H1
= –78300 + 0 – 350000
= – 428300 btu
Since q is –ve, heat is removed (exothermic)
Example
QUESTION?
36
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.
37
38
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
39
PRIMARY NUCLEATION
40
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.
41
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
32
250
3
16exp10
sRT
NVB aaM
27.10,[McCabe ])
42
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])
43
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.
44
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
45
SPURIOUS NUCLEATION
Nucleation that occurs at large
supersaturations or accompanies poor
(slurry/suspension) magma circulation
Produces irregular crystals
QUESTION?
46