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CRYSTALLISATION

Nucleation Theories

Equipment for Crystallization

OBJECTIVES :

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STUDENTS SHOULD BE ABLE TO:

1. Describe the nucleation theories

2. Describe types of crystallizers

Supersaturation Formation of crystals, follow 2 steps :

1. Birth of new particle- nucleation2. 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.

Very small crystals can be formed by attrition.

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Supersaturation

Question : How to achieve to supersaturated condition?

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Solute solubility increase strongly with increase temperature

Solubility is independent of temperature

Solubility is very high If nearly complete precipitation required

SupersaturationSolute solubility increase

strongly with increase temperature

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COOLING

Solubility is independent of temperature

EVAPORATION OF THE SOLVENT

Solubility is very high

(NEITHER cooling & evaporation is desirable)

Add third agent : SALTING

Supersaturation

• Supersaturation = concentration difference between that of the supersaturated solution in which the crystal is growing and that of a solution in equilibrium with the crystal.

y = y –ys (molar fraction)

c = c –cs (molar concentration)

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Refer FIGURE 1

Supersaturation• When the solubility increases appreciably with

temperature, the supersaturation can be expressed as an equivalent temperature difference instead of concentration difference.

– Tc ----- refer FIGURE 1

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Supersaturation

8Supersaturation Potential

Saturated solution at

Tc

Supersaturated solution at T

FIGURE 1 : Solubility Curve (Ref.2)

Unsaturated

Supersaturated

Nucleation

• Three level of nucleation may occur :

– Primary Nucleation

– Secondary Nucleation

– Spurious Nucleation

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1. Primary Nucleation

• Homogeneous nucleation = Formation of new particles within a phase uninfluenced by other factors (solids, agitation etc.)

• Heterogeneous nucleation = Nucleation occur under the presence of solid particles of foreign substance to increase the nucleation rate.

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2. Homogeneous solution

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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 (up to a few hundred of particles required to form nucleus)

Sequence of stages for crystal evolution

* Important if the solution has high supersaturation and no agitation present

Rate of crystal growth and growth coefficient

• Rate of growth of a crystal is the distance moved per unit time in a direction that is perpendicular to the face.

• Crystal growth is a layer-by-layer process• Growth can occur only at the outer face of the

crystal, hence the solute must be transported to that face from the bulk of solution

• The solute reach the face by diffusion through the liquid phase.

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Rate of crystal growth and growth coefficient

• Overall process consists of two resistances in series

• The solution must be supersaturated for the diffusion and interfacial steps to proceed

• All the formula related please refer to Geankoplis page 824-825

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The ∆L law of crystal

• All crystals that are geometrically similar and of the same material in the same solution grow at the same rate

• Growth is measured as the increase in length ∆L, in mm, in linear dimension of one crystal

• This increase is independent of the initial size of the initial crystals, provided that all the crystals are subject to the same environmental conditions.

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The ∆L law of crystal

• The total growth ∆L is the same for all crystals• The ∆L law fails in cases where the crystals are

given any different treatment based on size.• L/t = G 12.12-4• L = D2 – D1 = G (t2-t1) 12.12-5• G= growth rate mm/h• L= growth measured as the increase in

length, mm15

Particle-size distribution of crystals• An important factor in the design of

crystallization equipment. • The screen/ sieve used is Tyler Standard

screen (App A-5-3)• A common parameter used to characterized

the size distribution is the Coefficient of Variation (CV) as a percent – Eq 12.12-6)

• CV= 100 {(PD16% - PD84%) / 2PD50%)}• PD16% = particle diameter at 16% retained

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Calculation in MSMPR• Crystal population density, n

– n = N/L– N = total number of crystal – n = (slurry density * weight fraction) / ( crystal

density * vp * L)

• Volume of a particle, vp

– vp = aLav3

– a = crystal shape factor– Lav = average mesh portion

– e.g. for 14-20 mesh portion: Lav = (1.168+0.833)/217

Calculation in MSMPR• Crystal population density when L=0,n0

• Total retention or hold up time, • Average size, La

– La =3.67G• Predominant particle size, Ld

– Ld =3.00G• Nucleation rate, B0

– B0=Gn0

• Growth rate, G

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Calculation in MSMPR

1. Plot ln n vs L2. ln n = (-1/G) L + ln n0

3. (-1/G) = slope4. ln n0 = intercept

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Mesh L Mesh L Lav L Wt% vp = aLav3 n = (slurry density * weight

fraction) / ( crystal density * vp * L)

ln n

INTRODUCTION

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Classification of Crystallizer :

Batch or Continuous

Method used to create supersaturation

Method of suspending the growing product crystals

SUPERSATURATION

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SUPERSATURATION produced by:

Cooling with negligible evaporation tank & batch type crystallizer

Evaporation of solvent with little or no cooling evaporator-crystallizer & crystallizing evaporators

Combined cooling and evaporation in an adiabatic evaporator vacum crystallizer

Types of Industrial Crystallizer

•Draft tube-baffle crystallizer•Vacuum crystallizer

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Draft tube-baffle crystallizer

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Vacuum crystallizer

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OBJECTIVES

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STUDENTS SHOULD BE ABLE TO:

1. Understand the concept of crystallization

2. Perform material & heat balance for crystallization process

CRYSTALLISATION THEORY

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Process where solid particles are formed from homogeneous phase.

Occur in the: freezing of water to form ice,

formation of solid particles from liquid melts,

formation of snow particles from vapour or

formation of solid crystals from a liquid solution (MOST IMPORTANT & COMMERCIALIZED)

MECHANISMS

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The solution is concentrated and usually cooled until the solute concentration becomes greater than its solubility at that temperature.

Then, the solute comes out of the solution, forming crystals of approximately pure solute.

OBJECTIVE OF CRYSTALLIZATION

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Important objectives in crystallization

Good yieldHigh puritySize, shapesUniformity

OBJECTIVE OF CRYSTALLIZATION

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Why uniformity is important???a)Minimize caking in the package

b)For ease of pouringc)For ease of washing and filtering

d)Uniform behavior when used

CRYSTAL GEOMETRY

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Crystallographic systems

CRYSTAL

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A Solid composed of atoms, ions or molecules which are arranged ORDERLY and REPETITIVE MANNER.

Appear as polyhedrons, having flat faces & sharp corners.

The ANGLE between the corresponding faces of all crystals of THE SAME MATERIAL are EQUAL…CHARACTERISTIC OF THE PARTICULAR MATERIAL.

CRYSTAL

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Development of different types of faces of crystal may differ depending on the solute crystallizing.

Eg. NaCl crystallizes from aqueous solution with cubic faces

NaCl crystallizes from aqueous solution, impurity present, will have octahedral faces,

BUT BOTH are in the CUBIC system.

SNOWFLAKES

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A hexagonal prism includes two hexagonal "basal" faces and six rectangular "prism" faces. Hexagonal prism can be plate-like or columnar, depending on which facet surfaces grow most quickly.

When water freezes into ice, the water molecules stack together to form a regular crystalline lattice, and the ice lattice has six-fold symmetry

SNOWFLAKES

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When snow crystals are very small, they are mostly in the form of simple hexagonal prisms.  But as they grow, branches sprout from the corners to make more complex shapes.

SNOWFLAKES CRYSTALS

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EQUILIBRIUM SOLUBILITY IN CRYSTALLIZATION

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ATTAIN EQUILIBRIUM WHEN THE SOLUTION/MOTHER LIQUOR IS

SATURATED.

Represented by a SOLUBILITY CURVE.

Solubility is dependent mainly on TEMPERATURE.

SOLUBILITY CHART

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Generally, the solubilities of most salts increase with increasing temperature.That is why normally we cooled down the solution to get the crystal so that the concentration exceed the solubility at that temperature

SOLUBILITY CHART

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anywhere on it's solubility curve it is saturated above the solubility curve, then it's supersaturated below the solubility curve it's an unsaturated

solution.

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Some salts, when crystallised from aqueous solution, incorporate water molecules into the structure. This is known as 'water of crystallisation', and the 'hydrated' form of the compound.

e.g. magnesium sulphate MgSO4.7H2O.

YIELD, HEAT & MATERIAL BALANCE IN CRYSTALLIZATION

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DURING CRYSTALLIZATION PROCESS, THE SOLUTION (MOTHER LIQUOR) & THE SOLID CRYSTALS ARE IN CONTACT FOR ENOUGH TIME TO REACH EQUILIBRIUM.

At the end of the process, AT FINAL TEMPERATURE T, the solution is saturated, hence the final concentration of the solution can be obtained from SOLUBILITY CURVE.

YIELD, HEAT & MATERIAL BALANCE IN CRYSTALLIZATION

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THE YIELD OF CRYSTALS CAN BE CALCULATED FROM:

INITIAL CONCENTRATION OF SOLUTE

FINAL TEMPERATURE &

ITS SOLUBILITY AT THE TEMPERATURE

YIELD, HEAT & MATERIAL BALANCE IN CRYSTALLIZATION

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Material balance is straightforward , when SOLUTES are anhydrous.

When the crystals are hydrated, some of the water in the solution is removed with the crystals are hydrate.

EXAMPLE 12.11-1

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A salt solution weighing 10,000 kg with 30% Na2CO3 is cooled to 293 K (200 oC). The salt crystallizes as the decahydrate. What will be the yield of Na2CO3 • 10H2O crystals if the solubility is 21.5 kg anhydrous Na2CO3/100 kg of total water? Assume that no water is evaporated.

SOLUTION FOR EXAMPLE 12.11-1

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COOLER & CRYSTALLIZER

10,000 KG SOLUTION

30% Na2CO3

W kg H2O

S kg solution

21.5 kg Na2CO3/100 kg H2O

C kg crystals, Na2CO3 • 10H2O

SOLUTION FOR EXAMPLE 12.11-1

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1. Perform material balance for water and Na2CO3

Feed = Solution stream + Crystals stream + Other stream

Water

0)(2.286

106)(5.21100

5.21)10000(3.0

0)(2.2862.180)(

5.21100100)10000(7.0

CS

CS

Na2CO3

Molecular Weight: 10H2O = 180.2

Na2CO3 = 106

Na2CO3 • 10H2O = 286.2

SOLUTION FOR EXAMPLE 12.11-1

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2. Solving the two equation simultaneously,

C = 6370 kg of Na2CO3 • 10H2O crystals

S = 3630 kg solution

Assume that 3% of the total weight of the solution is LOST by evaporation of water in cooling….RECALCULATE the C& S VALUES.

Exercise 1

A salt weighing 8,000 kg with 25 wt% Na2CO3 is cooled to 293 K. The salt crystallized as dehydrate and solubility is 21.5 kg anhydrous Na2CO3 /100 kg of total water.

Assuming that 3% of the total weight of the solution is lost by evaporation of water in cooling and no water is evaporated, calculate the yield of Na2CO3.10H2O crystals.

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Heat effect and balance

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Heat effects and balance

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When a compound (whose solubility increases as temperature increases) dissolves, there is an absorption of heat; called the heat of solution.

When a compound (whose solubility decreases as temperature increases) dissolves, there is an evolution of heat.

For compounds dissolving whose solubility does not change with temperature, there is no heat of evolution or heat of dissolution.

Heat effects and balance

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In crystallization, the opposite of dissolution occurs.

At equilibrium, the heat of crystallization is equal to the negative of the heat of solution at the same concentration in solution.

HEAT BALANCE IN CRYSTALLIZATION

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q + H1 = H2 +HV H1 = enthalpy of the entering solution 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 absorbed (kJ)

= +ve: heat must be added to the system

= -ve: heat is evolved or given off

REFERENCES

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Websites:

http://www.its.caltech.edu/~atomic/snowcrystals/primer/primer.htm

1. Geankoplis C. J., Transport Processes and Separation Process Principles, 4th Edition, Prentice Hall, 2003.

2. McCabe W. M., Smith J. C. and Harriott P., Unit Operations of Chemical Engineering, 7th Ed., McGraw Hill, 2005.

Books

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THANK YOU.