crystallization crystallization process of producing crystals from a homogeneous phase which is...

CrystallizationCrystallization

CrystallizationCrystallization• Process of producing crystals from a

homogeneous phase which is obtained from a solution

• Capable of producing bioproducts at very high purity and considered to be a polishing step and a purification step

• Two different application of crystallization:

i. Crystallization for polishing and purification

ii. Crystallization for crystallography

Crystallization from Supersaturated Solutions of Sodium Acetate

• Description: A supersaturated solution of sodium acetate is crystallized by pouring it onto a seed crystal, forming a stalagmite-like solid. Heat is radiated from the solid.

• Source: Shakhashiri, B.Z. Chemical Demonstrations: A Handbook for Teachers of Chemistry

Crystallization PrinciplesCrystallization PrinciplesCrystals• Can appear as polyhedrons, or solids formed by plane faces when allowed to form freely• Angles made by the corresponding faces of the same material do not vary – can be classified

by this characteristics• Relative sizes of the faces of a crystal in a particular system can vary considerably – resulting in

a variety of crystal shapes. This variation is called a habit.• Crystal habit is influence by the conditions of crystallization, particularly by the impurities

present and by the particular solvent or solvents used.• Impurities can stunt the growth of a crystal in certain directions

ACICULARLong and needle-like, thinner than prismatic but thicker than fibrous. Natrolite crystals can be good examples of acicular crystals.

TABULARBook-like (tablets) that are thicker than platy but not as elongated as bladed. Wulfenite forms crystals that are a good example of tabular crystals.

PRISMATICOne of the most common of crystal habits. Prismatic crystals are "pencil-like", elongated crystals that are thicker than needles (see acicular). Indicolite (a variety of elbaite) forms good examples of prismatic crystals.

http://mineral.galleries.com/minerals/silicate/natrolit/natrolit.htm

http://mineral.galleries.com/minerals/sulfates/wulfenit/wulfenit.htm

http://mineral.galleries.com/minerals/gemstone/indicoli/indicoli.htm

http://mineral.galleries.com/minerals/silicate/elbaite/elbaite.htm

Crystals grow in 2 steps:

1. Nucleation - first aggregation.

2. Growth.

• Thermodynamically distinct

• Want a few nuclei to grow big

• Use thermodynamics to understand the required conditions

Nucleation• The generation of ultramicroscopic particles in the process of nucleation is the sum of

contributions by primary nucleation and second nucleation• Primary nucleation : occurs in the absence of crystals, secondary nucleation: attributed to the

influence of existing crystals• Primary nucleation can be either homogeneous (no foreign particles are present) or

heterogeneous (foreign particles present during heterogeneous nucleation)• Rate of primary nucleation has been modeled by the following power law expression:

B: number of nuclei formed per unit volume per unit time; N: number of nuclei per

unit volume; kn : rate constant; c: instantaneous solute concentration; c*:

solute concentration at saturation. (c-c*) term : supersaturation, the exponent of n

can range up to 10 but typically is in the range of 3 to 4• Two types of secondary nucleation : shear nucleation (occurs as a result of fluid shear on

growing crystal faces), contact nucleation ( happens because of crystals colliding with each other and with the impeller and other vessel internal surfaces.

• Rate of secondary nucleation in crystallization is the following:

Crystallization PrinciplesCrystallization Principles

(1)

(2)k1 : rate constant; MT : suspension density, b : can range up to 5 but has a most probable value of 2; j: ranges up to 1.5 with 1 being the most probable value

Crystallization PrinciplesCrystallization PrinciplesFigure 1 Typical phase diagram. The components in solution consist of the product (ordinate) and the precipitating reagent (abscissa). The lines with arrows out line one possible way of performing the crystallization.

-The supersaturation must be above the a certain value before nucleation will begin-Metastable region : the supersaturation is low that nucleation will not start-Once the supersaturation has been raised enough to be in the labile region, nucleation can begin-At this point, crystals begin to grow, and the supersaturation decreases-If the supersaturation becomes too high, the nucleation rate wii be too great, and amorphous precipitate will result

SupersaturationSupersaturation

Phase diagramsPhase diagrams

Precipitatant concentration (salt, PEG etc.)

Pro

tein

conce

ntr

ati

on

Under-saturation(protein remains soluble; crystals dissolve)

Nucleation zoneNucleation zone

Precipitation zonePrecipitation zone

Solubility curve

Metastable zoneMetastable zoneCrystals grow, butCrystals grow, but

Nuclei form only Nuclei form only infinitely slowlyinfinitely slowly

Course of Crystallization ExperimentCourse of Crystallization Experiment

[Precipitatant]

Pro

tein

conce

ntr

ati

on

NucleationNucleation

PrecipitationPrecipitationMetastable Metastable

Start w/ soluble protein (undersaturated or metastable)

NucleatNucleates herees here

Incr

ease

[pro

tein

], [p

recipi

tant

]

Crystal growsCrystal growsSequesters Sequesters

proteinprotein[protein] drops[protein] drops

Crystal stops Crystal stops growing @ solubility growing @ solubility

curve curve

Expt incr. [protein], [precipitant]

Xtl grows again, until hits curve

Repeats as follows solubility curve

Crystal Growth• Post nucleation process in which molecules in solution are added to the surface of existing

crystals• The rate of mass deposition R during crystal growth is:

W: mass of crystals per volume of solvent; A : the surface area of crystals per volume

of solvent; kG : overall mass transfer coefficient (depends on temperature, crystal

size, hydrodynamic conditions, the presence of impurities); g : usually 0 and 2.5• Overall linear growth rate can also be expressed as:

• L : characteristics single dimension of the crystal, such as length• Crystal growth is a process that consists of two steps in series – diffusion and surface

integration

• When the exponents are unity, combining Equation 3, 5, 6 gives


(3)

(4)

(5)ci : concentration at the interface between the liquid and solid phase; kd and kr : mass transfer coefficients

(6)

(7)Thus, if surface integration is very fast compared with bulk diffusion, then kr >> kd, and kG kd.

• Crystallization Kinetics From Batch Experiments• Estimates of nucleation and growth rates during crystallization can be obtained from batch

experiments by a method known as moments analysis. • By making a mass balance on the number of crystalline particles within a given size range

between L and L + ΔL, the population balance equation for size- independent growth in a perfectly mixed, constant volume batch crystallizer with negligible attrition and agglomeration becomes:

• The kth moment of the population density distribution in L about the origin, obtained by moment transformation of Equation (8) with respect to size, is defined as

• If the moments of the experimental population density function can be determined at two times that differ by a small time interval Δt over which the linearity of the model can be assumed, then the average values, of the nucleation and growth rates can be expressed as

Δ represents the differences in values of quantity at two different times and the bar an

arithmetic average of the quantities


(8)

(9)

and(10) (11)

• the average size L at each time can be obtained as follows from the moments:

• The meaning of each kth moment in the crystal population is given in Table 2.


(12)

TABLE 2 Moment Analysis of Crystal Size Distribution

Yields and Heat and Material Yields and Heat and Material Balances in CrystallizationBalances in Crystallization

• Yields and material balance in crystallization• The solution (mother liquor) and the solid crystals are in contact for enough time to

reach equilibrium. Hence, the mother liquor is saturated at the final temperature at the final temperature of the process, and the final process, and the final concentration of the solute in the solution can be obtained from the solubility curve

• The yield can be calculated knowing the initial concentration of solute, the final temperature, and the solubility at this temperature.

• In making the material balances, the calculations are straightforward when the solute crystals are anhydrous. Simple water and solute material balances are made. When the crystallizations are hydrated, some of the water in solution is removed with the crystals as a hydrate

Example 1Example 1Yield of a Crystallization ProcessA salt solution weighing 10000 kg with 30 wt % Na2CO3 is cooled to 293 K (20 °C).

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? Do this

for the following cases:

(a) Assume that no water is evaporated.

(b) Assume that 3% of the total weight of the solution is lost by evaporation of water

in cooling.

Solution The molecular weights are 106.0 for Na2CO3,

180.2 for 10H20, and 286.2 for Na2CO3 •10H2O.

The process flow diagram is shown in Fig. 3,

with W being kg H2O evaporated, S kg

solution (mother liquor), and C kg crystals of

Na2CO3 •10H2O.Making a material balance

around the dashed line box for water for part (a), where W = 0.(13)

FIGURE3. Process flow for crystallization

Example 1Example 1where (180.2)/(286.2) is wt fraction of water in the crystals. Making a balance for

Na2CO3,

Solving the two equations simultaneously, C = 6370 kg of Na2CO3 •10H2O crystals and

S = 3630 kg solution.

For part (b), W = 0.03(10000) = 300 kg H2O. Equation (13) becomes

Equation (14) does not change, since no salt is in the W stream. Solving Eqs. (14) and

(15)simultaneously, C = 6630 kg of Na2CO3 •10H2O crystals and S = 3070 kg solution.

(14)

(15)

Heat effects and heat balances in crystallization• When a compound whose solubility increases as temperature increases dissolves,

there is an absorption of heat, called the heat of solution – occurs when the solubility decreases as the temperature increases

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

• The enthalpy H1 of the entering solution at the initial temperature is read off the chart, where H1 is kJ for the total feed. The enthalpy H2 of the final mixture of crystals and mother liquor at the final temperature is also read off. If some evaporation occurs, the enthalpy Hv of the water vapor is obtained from the steam tables. Then the total heat absorbed q in kJ is

• If q is positive, heat must be added to the system. If it is negative, heat is evolved or given off.

Yields and Heat and Material Yields and Heat and Material Balances in CrystallizationBalances in Crystallization

(16)

Example 2Example 2Heat Balance in CrystallizationA feed solution of 2268 kg at 327.6 K (54.4 °C) containing 48.2 kg MgSO4/100 kg total

water is cooled to 293.2 K (20°C), where MgSO4•7H2O crystals are removed. The

solubility of the salt is 35.5 kg MgSO4/100 kg total water (P1). The average heat capacity

of the feed solution can be assumed as 2.93 kJ/kg• K (H1). The heat of solution at

291.2 K (18 °C) is -13.31 x 103 kJ/kg mol MgSO4•7H2O (P1). Calculate the yield of

crystals and make a heat balance to determine the total heat absorbed, q, assuming that

no water is vaporized.

Solution Making a water balance and a balance for MgSO4 using equations similar to (13) and

(14) in Example 1, C = 616.9 kg MgSO4•7H2O crystals and S = 1651.1 kg solution.

To make a heat balance, a datum of 293.2 K (20°C) will be used. The molecular weight

of MgSO4•7H20 is 246.49. The enthalpy of the feed is H1:

Example 2Example 2The heat of solution is -(13.31x103)/246.49 = -54.0 kJ/kg crystals. Then the heat of

crystallization is -(-54.0) = +54.0 kJ/kg crystals, or 54.0(616.9) = 33312 kJ. This

assumes that the value at 291.2 K is the same as at 293.2 K. The total heat absorbed, q,

is

Since q is negative, heat is given off and must be removed.

Equipment for CrystallizationEquipment for Crystallization• Tank Crystallization• Hot saturated solutions are allowed to cool in open tanks

• After a period of time, the mother liquor is drained and the crystals removed

• Nucleation and the size of crystals are difficult to control

• Labor cots are very high

• Has limited application; used to produce certain fine chemical and pharmaceutical products

Scraped surface crystallizers• One type of scraped surface crystallizer is the Swenson-Walker crystallizer, which

consists of an open trough 0.6 m wide with a semicircular bottom having a cooling jacket inside

• Slow-speed spiral agitator rotates and suspends the growing crystals on turning

• Blades pass close to the wall and break off any deposits of crystals on the cooled wall

• In the double-pipe scraped surface crystallizer, cooling water passes in the annular space

• Internal agitator is fitted with spring-loaded scrapers that wipe the wall and provide good heat-transfer coefficients which called a votator

• Used in crystallizing ice cream and plasticizing margarine

Equipment for CrystallizationEquipment for Crystallization

Circulating-liquid evaporated-crstallizer• Supersaturation is generated by evaporation

• Circulating liquid is drawn by the screw pump down inside the tube side of condensing steam heater

• Heated liquid then flows into the vapor space, where flash evaporation occurs, giving some supersaturation

• The vapor leaving is condensed

• The supersaturated liquid flow down the downflow tube and then up through the bed fluidized and agitated crystals, which are growing in size

• The living saturated liquid then goes back as a recycle stream to the heater, where it is joined by the entering feed

• The larger crystals settle out and a slurry of crystals and mother liquor is withdrawn as product

• Also called Oslo crystallizer


Circulating-magma vacuum crystallizer• The magma or suspension of crystals is circulated out the main body through a

circulating pipe by a screw pump

• Magma flows through a heater, where its temperature is raised 2-6 K.

• The heated liquor then mixes with body slurry and boiling occurs at liquid surface

• This cause supersaturation in the swirling liquid near the surface, which results in deposits on the swirling suspended crystals until they leave again via the circulating pipe

• The vapors leave through the top

• A steam-jet ejector provides the vacuum


Model for Mixed- Suspension-Mixed Model for Mixed- Suspension-Mixed Product Removal CrystallizerProduct Removal Crystallizer

Introduction and model assumptions

• Derived for a mixed suspension— mixed product removal (MSMPR) crystallizer, which is by far the most important type of crystallizer in use in industry today.

• Conditions assumed are as follows: steady state, suspension completely mixed, no product classification, uniform concentration, no crystals in feed, and the ΔL law of crystal growth applies.

• All continuous crystallizers have mixing by an agitator or by pump around.• In ideal mixing, the effluent composition is the same as in the vessel.

Crystal population-density function n• To analyze data from a crystallizer, an overall theory must consider

combining the effects of nucleation rate, growth rate, and material balance.• Randolph and Larson derived such a model - plotted the total cumulative

number of crystals N per unit volume of suspension (usually 1 L) of size L and smaller versus this size L, as in Fig.4.

• The slope dN/dL of this line is defined as the crystal population density n:

n: number of crystals/ (L•mm)


(17)

FIGURE 4. Determination of population density n of crystals.• This population density is obtained experimentally by screen analysis of the total crystal content of a given volume, such as 1.0 L of magma suspension. • Each sieve fraction by weight is obtained by collection between two closely spaced and adjacent screens. • Then Lav= (L1 + L2)/2. where L1 and L2 are the openings in mm in the two adjacent screens. • Also, ΔL = (L1 - L2), where L1 is the size opening of the upper screen. • Then the volume of a particle vp is

vp: mm3/particle ; a : constant shape factor

(18)

• Knowing the total weight of the crystals in this fraction, the density ρ in g/mm3, and the weight of each crystal, which is ρvp, the total number of crystals ΔN is obtained for the size range ΔL.

• Then, rewriting Eq. (17) for this ΔL size,

• This method permits the calculation of n for each fraction collected in the screen analysis with an average size of Lav mm.

Population material balance. • To make a population material balance, in Δt time, Δn ΔL crystals are withdrawn.

• Since the effluent composition in the outflow of Q L/h is the same as that in the crystallizer, then the ratio (Δn ΔL )/(n ΔL ) or fraction of particles withdrawn during Δt time, is the same as the volume ratio Q t withdrawn divided by the total volume V of ne crystallizer. Hence,

• During this time period Δt, the growth ΔL of a crystal is


(19)

(20)

(21) G : growth rate in mm/h

• Combining Eqs. 20 and 21,

• Letting ΔL→0, Δn→0, and integrating,

n° : population of nuclei when L = 0; n : population when the size is L. and V/Q

is τ the total retention or holdup time in h in the crystallizer.

• A plot of Eq. (244) of In n versus L is a straight line with intercept n° and slope

-1/Gτ.

• If the line is not straight, this could be an indication of the violation of the ΔL law. These calculations are well suited for being done with a computer spreadsheet.


(22)

(23)

(24)

Average particle size and nucleation rate. • Further derivation of the population approach results in the equation for the average

size La in mm of the mass distribution:

• Here, 50% of the mass of the product is smaller or larger in size than this value. Also, the predominant particle size is given as

• This predominant particle size means more of the mass is in this differential size interval than in any other size interval.

• Another relation can be obtained from this model, which relates the nucleation rate B° to the value of the zero-size particle population density n° and the growth rate G.

• For the condition L→0, the limit of dN/dt (nucleation rate) can be written as

• However, when L→0, the slope dL/dt = G, the slope dN/dL = n°, and B° = dN/dt. Hence,


(25)

(26)

(27)

(28) B° = nucleation rate in number of nuclei/h•L

Example 3Example 3Growth and Nucleation Rates in MSMPR CrystallizerCalculate the population density and nucleation growth rates for crystal samples of urea from a screen analysis. The slurry density (g of crystals) is 450 g/liter, the crystal shape factor a is 1.00, the crystal density ρ is 1.335 g/cm3, and the residence time τ is 3.38 h.The screen analysis from reference (B5) is as follows:

Solution The data above are tabulated in Table1 using data from Appendix 1. The value of L is

the screen opening. For the 14-20 mesh portion, Lav = (1.168 + 0.833)/2 = 1.001 mm

and ΔL = 1.168 -0.833 = 0.335 mm. For Lav = 1.001 mm, using Eq. (18), vp = aLav3

=1.00(1 .001)3 = 1.003 mm3/particle. The density ρ = 1.335 g/cm3 = 1.335 x 10-3 g/mm3

and the mass/particle = ρvp = 1.335x10-3 (1.003) = 1.339 X 10-3 g. The total mass of crystals = (450 g/L)(0.044 wt frac).Using Eq. (19) to calculate n for the 14-20 mesh size range,

Then In n = In 4.414 X 104 = 10.695. For the 20-28 mesh size range, Lav = 0.711/mm,

and ΔL = 0.244. Then,

Then In n = 13.224. Other values are calculated in a similar manner using a computer

spreadsheet and are given in Table 1.

The In n is plotted versus L in Fig. 5. The equation of this line is Eq. (24), where the

slope is -9.12 and the intercept = 19.79:

The slope -9.12 = -1/Gτ = -1/(G)(3.38). Hence, G = 0.03244 mm/h. The intercept ln n°

=19.79. Hence, n° = 3.933 X 108. From Eq. (28), the nucleation rate B° is

Example 3Example 3

The average size is, from Eq. (25), La = 3.67Gτ = 3.67(0.03244)(3.38) = 0.402 mm. The

predominant size, from Eq. (26), is Ld = 3.00Gτ = 3.00(0.03244)(3.38) 0.329 mm.

Example 3Example 3

TABLE I. Data and Calculations for Example 3

FIGURE 5. Plot of population density n versus length for Example 3.

Example 3Example 3Tyler Standard Screw Sale

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