CRYSTALLIZATION

1. Basic Concept

2. Crystal Size Distribution

3. Batch Crystallization

4. Recrystallization

5. Conclusion

• 결정화의중요성1. 결정은순물질의일반적인형태이다.2. 일정한형태의결정의생산은마무리작업을용이하게한다.3. 결정화는생성물의외양을개선한다.

1. Saturation

포화 : 용액상에서열역학적으로안정한용질의최대농도상평형의결과

1 BASIC CONCEPT

결정은비생물물질중가장정렬된 3차원배열을하고있다.이러한격자는 constituent 입자로구성되어있으며, 원자·이온·분자일수있다.

특성

SaturationPurityNucleationSingle crystal growth

고체결정상과주변의용액(모액) 사이의 chemical potential의결과

Fig. 1. Solubility versus temperature. Solubility almost always varies withtemperature, but in widely different ways

•다양한농도-온도계수를보여준다.•각각의계수는특정용질과용매의성질

•포화상태에서용액은과포화상태로갈수있다.

•이러한과포화용액은열역학적으로불안정하다.

Fig. 2. Regions of supersaturation. The nature of crystal nucleation and growth is often different in the three regions shown above the solubility.

Metastable region : 용질의평형농도이상함유, 결정이존재하지만새로운결정핵은생성되지않는다.

Intermediate region:결정의성장과생성이이루어진다.

Lable region: 핵이자발적으로 clear solution으로부터생성

Metastability: 작은결정의표면에너지의결과

2. PurityBioseparation에서가장중요

The separation factor ββ = EA / EB (1-3) β가클경우분리는효과적이다.

3. Nucleation

2)-(1 filtratein Bimpurity of weightcrystalsin Bimpurity of weightE

Bimpurity for equation similar a

1)-(1 filtratein A solute of weight

cake crystalin A solute of weightE

E,factor a of useby purity quantify

B

A

=

=

(1) homogeneous nucleation : 단지과포화에의해용액으로부터결정이생성(2) heterogeneous nucleation : 액체내결정의성장에큰영향을미치는비용해성입자로부터결정이생성

* 결정핵생성 mechanism에의한분류

(3) secondary nucleation : 다른결정간의접촉에의해핵이생성

(4) attrition nucleation : 기존의결정이파쇄되어새로운결정으로성장 secondary nucleation과유사

모든결정핵생성의 mechanism은새로운 crystal의형성에기여하지만자세한각반응의 mechanism은밝혀지지않았다. 대신 nucleation rate에대한실험식이사용된다.

4)-(1 B =

)c-(ck = dtdN

rate Nucleation'*

n

Where c : 용액의농도c* : 포화농도kn and I : empirical parameter B : nucleation rate

4. Single crystal growth

5)-(1 )c-kA(c = dt

dM

as expressed is "controlleddifussion "*

Where, k : 물질전달계수 A : 결정면적c : 용액의농도 c* : 포화농도

교반에의해결정과주변용액간의상대속도가증가하면서성장속도는종종최대치에이른다. 이러한경향은작은상대속도로제어되는확산이나높은상대속도에서제어되는표면반응에서나타난다.

6)-(1 )c-(c1/ 1/k

A dt

dM

bygiven isrategrowth the

*

κ+=

반응속도가과포화에선형적으로의존한다고가정하면성장속도는

7)-(1 A

6M lρ

=

9)-(1 l6 A

8)-(1 l M2

A

3v

Φ=

Φρ=

)(

10)-(1G )c-(c k

)c-(c/1k/1

/2 dtdl

)c-(c/1k/1

l6 ldtd

*g

*vA

*2

A3

==

κ+ρΦΦ=

κ+

Φ=Φρ

For a cubic of side sM=ρs3, A=6s2, l=s

Example 1. Crystallization of Adipic Acid. 10kg of adipic acid is slurriedin 13.1 kg of water and heated to 90oC to solubilize the acid. The solution is thenfiltered to remove insoluble imputities. During the heating and filtration, 10% of the water is evaporated. The clarified solution is cooled to 35℃ and filtered. Thesolubility at 35℃ is 0.05 kg acid per kg water.

Determine the weight of crystals recovered in this operation.

Solution. We need two mass balances to solve this example. The first is a waterbalance:

water in = water in liquor ＋ water evaporated13.1 kg = water in liquor ＋ 0.1 (13.1 kg )

water in liquor = 11.79 kgThe second balance, for the adipic acid, isweight crystals in = weight of crystals ＋ weight remaining in mother liquid

10 kg = weight of crystals ＋ 0.05(11.79 kg)weight of crystals = 9.41 kg

The recovery can be increased by lowering the final temperature. Note that if thecrystals deposited were hydrated, their water of hydration must be included in themass balances.

Example 2. Separation of Soy Sterols. A mixture of stigmasterol and sitosterolweighing 2040 kg is divided into two fractions by crystallization. The original mixtu-re contains 86.5 % stigmasterol. The recovered crystals are 96.6% stigmasterol and weigh 1137 kg. The solids in the liquor contain 74.6% stigmasterol, found by evapo-ration to dryness.

Determine the β value for this separation.

Solution. We first calculate the E values for stigmasterol and sitosterol using Eqs. (10.1-1) and (10.1-2):

9.6 = 0.171.63

= E

E =β find weThus

168.0=229.438.66

=

/kgsitosterol 0.746)kg-000.1(×1137)kg-(2040/kgsitosterol 0.966)kg-(1.000 × 1137kg

=E

63.1=673.61098

=

ol/kgstigmaster 0.746kg×1137)kg-(2040ol/kgstigmaster 0.966kg × 1137kg

=E

sitosterol

olstigmaster

sitosterol

olstigmaster

Example 3. Streptomycin Growth.

As part of a fundamental study of crystallization, you have mounted a single crystal of this antibiotic in a clean, supersaturated solution which flows past the crystal at an adjustable rate. By examining it microscopically, you find that the crystal grows at a rate almost independent of crystal size.

At 10% supersaturation, this rate is 0.02 cm/hr at a flow of 5.0cm/sec and is 0.04cm/hr. When the flow is six times larger. From mass transfer correlations, you expect that the mass transfer coefficient varies with the square root of the fluid velocity.

Estimate k and κ from these observations.

Solution. We see that the growth rate is expected to be independent of crystal size l, so that our observation is consistent with the approximate analysis given in this section. We are also told that

=

ρΦΦκ

=

ρΦΦ

hr cm065.0

)c-c( and

hrsec cm0055.0

)c-c(a

find we,Solving

v

*A

2/1

v

*A

]κ/1+)30cm/sec[1/(a

)c-c)(ρΦ/Φ(2=

hrcm

0.04

]κ/1+)5.0cm/sec[1/(a)c-c)(ρΦ/Φ(2

=hrcm

0.02 find we

10)-Eq.(1 into valuesinsert the Thus constant. a is a whereav =k

*vA

*vA

1/2

2 CRYSTAL SIZE DISTRIBUTIONS

1. Population density

1)-(2 size in change

numberin change lim n

:n density population the

0 → lΔ=

4)-(2 dl )l(n l dl )l(n l 3,j If

3)-(2 dl )l(n dl )l(n 0,j if ,First

2)-(2 dl )l(n l dl )l(n l

∫∫∫∫

∫∫

∞0

3v

l0

3v

j

∞0

l0

j

∞0

j

l0

j

j

ΦρΦρ=μ=

=μ=

=μ

Fig. 3. The cumulative number of crystals versus size. The derivative of this curve, which has dimensions of (length)-1, is the population density of the crystals n (l,t) .

Table 1. Fractional moment of distribution. The quantity χ (= lQ / GV),a dimensionless length.

Moment Meaning BasicDefinition

Total forContinuousCrystallizer

Fraction forContinuousCrystallizer

0 Number ofCrystals ∫∞0

l0

n(l)dl n(l)dl ∫

QGVn0 1- e-χ

1 Size ofCrystals ∫

∫∞0

l0

n(l)dl l n(l)dl l 2

0 QGVn

1- (1+χ)e-χ

2 Area ofCrystals ∫

∫∞Φ

Φ

02

A

l0

2A

n(l)dll 6n(l)dll 6 3

0A QGVn12

Φ χ−χ+χ+− e )21(1 1 2

3 Mass ofCrystals ∫

∫∞ρΦ

ρΦ

03

v

l0

3v

n(l)dll n(l)dll 4

0V QGVn6

Φ χ−χ+χ+χ+− e )61

21(1 1 32

Crystal from continuous process

Fig. 4.Continuous crystallization. In this case, a supersaturated feedis always entering the well stirred crystallizer. A product stream containing thesame crystal concentration as in the bulk is withdrawn at the feed rate.

We can now make a balance on the number of crystals within a given size Δl[accumulation of crystals] = [crystals growing into range]

- [crystals growing out of range]- [crystals of given range flowing out] (2-5)

At steady state,0=[VnG]|l - [VnG]|l+Δl -QnΔl (2-6)

7)-(2 0Qndl

Vd(Gn)

becomes balance this0,l aslimit theIn

=+

→Δ

8)-(2 0GVnQ

dldn

find wel, size theoffunction anot isG rategrowth that assume wefI

=+

9)-(2 GB

dl/dtdN/dtn

0,l as s,other wordIn crystals. new of nucleation by the dominated be willndensity population thesmall, is l When

0

0

==

→

10)-(2 enn 8)-Eq.(10.2for condition boundary a as thisgsinU

lQ/GV-0=

Dominant crystal size

14)-(2 Q

3GVl

: l sizedominant or maximum obtain the wel,for solveand zero toequal l respect to with 13)-Eq.(10.2 of derivative set the weIf

13)-(2 e GVQ

6l

nn

GVQ

6l

dldw

bygiven is )M/M( wmass crystal offraction in the change theTherefore,

12)-(2 Q

GVn6M

is crystals of mass total the1,-10.2 tableFrom 11)-(2 ndlldM

is rangesizegiven afor mass the,definitionBy

D

D

lQ/GV-43

0

43

T

4

0vT

3v

=

=

=

=

Φρ=

Φρ=

Example 4. Characterizing an Ammonium sulfate crystallization. A continuous crystallization vessel containing 100liters of ammonium sulfate slurry isfed with 50 liters/hr of supersaturated solution. The withdrawal rate of product slurryis also 50 liters/hr. A nucleation rate B of 7.18* 107nuclei/liters hr, and growth rate G of 0.056 mm/hr are expected. Determine the following:

(a) the dominant crystal size,(b) the number of crystals equal to or smaller than this size,(c) the fraction of crystals in this range, and(d) the product slurry concentration.

In these calculations, assume cubic crystals with a density of 1.769 g/cm3.

Solution.(a) From Eq.(2-14), the dominant crystal size is

0.336mm

r50liters/h

100liters hr

mm0.0563 Q

3GVl D

=

==

(b) The number of crystals N equal to or smaller than this size I found fromTable 10.2-1 :

( )

( )

liter1 101.36

)e-(1 liters/hr 50

liters 100 hr liter

1 107.18N

isliter per crystals ofnumber theTherefore, (a).part in as 3GV/Ql But

e-1 Q

BV

e-1 Q

GVn dln N

8

3-7

D

GV/Ql-

GV/Ql-l

0 0

D

DD

×=

×=

=

=

==

(c) The fraction of the crystals in this size range is(1-e-3) = 0.95

g/liter 210

liter/hr 50

liter100hr

mm56.0 mm/hr 0.56

hr liter/107.18 )cm/(1000mm

1769g16

Q

GVn6 M

:1-10.2Tablefromalsofoundision concentratproduct The(d)

47

33

4

0vT

=

×

××=

ρΦ=

0.972 651.7

251.751.71 3 -1

62

1 e-1

: 1-10.2 Table from found is size this to0 from crystals offraction weight The

7.510.112mm0.841mm

so mm, 0.841 is sizemesh themesh, 20 ofscreen aFor

mm112.0

liters100

hr/liters50 0.056mm/hr

GV

Q

327.51-

32-

3

=

+++=

χ+χ+χ+=μ

==χ

=

==χ

χ

Example 5. Crystal size distribution for Ammonium sulfate. Using thegrowth rate and other conditions given in Example 4, calculate the anticipated crystal size distribution of the slurry. In particular, estimate the mass fraction ofcrystals retained on different standard screens.

Solution. For these conditions, the generalized dimensionless length χ is

Table 2. Crystal size distribution.

The fraction retained is1-μ3 = 0.054

The results for this and other screen sizes are given in Table 10.2-2

Ammonium Sulfate Crystallization Sucrose Crystallization

ScreenSize Mesh

ScreenSize

(mm)

DimensionlessSize

χ

FractionPassed

μ3

FractionRetained

1-μ3

MeshSize

AverageSize

l (mm)

PopulationDensity as

ln n

20 0.841 7.51 0.946 0.054 20b 1.02b 9.8

28 0.595 5.31 0.777 0.223 28 0.72 12.45

35 0.420 3.75 0.517 0.483 35 0.51 14.61

48 0.297 2.65 0.274 0.726 48 0.36 16.46

100 0.149 1.33 0.046 0.954 65 0.25 16.98

Example 6. The analysis of Sucrose crystallization. The following screen analysis is for a product prepared during a study of sucrose crystallization :

Product Size, mesh Cumulative Percent+ 20 3+ 28 14+ 35 38+ 48 76+ 65 92

Sucrose has density of 1.588 g/ cm3. The slurry density and retention time were given as 355 g/ liter and 2.5 hr, respectively. From those data, determine :

(a) the crystal growth rate(b) the nucleation rate(c) the dominant crystal size, and (d) the slurry concentration.

Solution.The sieve opening for a 20 mesh screen is 0.841 mm and for a 28 mesh screen is 0.595mm. Therefore, the size range Δl of crystals on 28 mesh is

Δl = 0.841-0.595 = 0.246mm

The average size l on the 28 mesh screen is

0.718mm 2

0.5950.841l =+=

The population density of the 28 mesh crystals can be estimated from Eq.(10.2-1)

[ ]

12.45 nln

liter mm

crystals 255,000

)mm246.0(rystals)(0.718mm/c )#(crystals/ mm1000/cm1 g/cm 1.588g/liter 335 0.03)-(0.14n

28

333328

=

=

=

hrliter 1101.3

hrmm 040.0

liter mm1 e GnB

bygiven is rate nucleation The )b(

hr/mm040.09.9(2.5hr)-

1 VQ

slope1 G

is rategrowth The (a)

719.60 ×=

==

=−=

−=

Fig. 5. The population density in sucrose crystallization. The logarithm ofthe population density is proportional to crystal size, as predicted by Eq. (2-10)

19.6

Slope =-9.9

g/liter 320

hr5.2hr

mm040.0 mmliter

crystalse 1000mm

1cm cm

g588.1 6

QGV n6M

bygiven ision concentratslurry The )d(

mm30.0hr]2.5hr3[0.040mm/ QV3G l

is size crystaldominant The (c)

419.6

3

3

3

4

0vT

D

=

=

ρΦ=

==

=

3. BATCH CRYSTALLIZATION

Fig. 6. Batch crystallizers. These are nothing more than stirred tanks which are slowly cooled. They are often baffled for better mixing.

2)-(3 )cc(k tlG

10)-Eq.(10.1by given is rategrowth heIn which t

1)-(3 (nG)l

V- tnV

systembatch ain operation unsteady For the

*g −=

∂∂=

∂∂=

∂∂

1. The cooling curve

4)-(3 nucleation

from change

growth crystalfrom change

re temperatualtered

from change

ationsupersaturin change

:unit batch in theation supersatur ofamount on the balance The

3)-(3 )cc(kGntN B

B, rate nucleation For the

i*n0

+

+

=

−==∂∂=

많은 batch crystallization은 metastable zone에서일어나므로, Eq.(.3-4)를Eq. (1-6)와결합시켜서

5)-(3 )c-(c )1/ (1/k

A dt

dcV 0 **

κ++=

Where A : the total crystal areac* : the saturation concentration (온도의함수)

6)-(3 dtdT

dTdc

dtdc **

=

{ } 7)-(3 )Gtl(6 l

M crystalarea crystals] of number[A

G rategrowth linear on the andM seeds of mass on the depends area crystal totalThe

2sA3

sv

s

s

+Φ

Φρ

=

=

Where Ms : the total mass of seedρ : density , ls : initial size of seed crystal

10)-(3 lGt

31

lGt1

l3Gt

dt/dcV/MTT

9)-(3 )Gt(l l

3G dt/dcV/M-

dtdT

8)-(10.3 5)-(10.3 Combine

8)-(3 )c-(c /1k/1

/2 dtdl G

10)-(10.1 Eq.by given rate,growth linear theisG

2

sss*s

0

2s3

s*s

*vA

++

−=

+

=

−

κ+ρΦΦ==

하면상수라용해도를대한온도에대해변화에미소온도의

Batch scale-up

11)-(3 )(311 )(3

MM

TTT-T

: variablesdifferent of in terms written sometimes isresult This

2

P

s

0P

0

ητ+ητ+ητ=

−

Example 7. Modified Tetracycline crystallization. You have been successfully running crystallizations of a substituted form of this antibiotic inmixed chlorinated solvents. Because some of these may be carcinogenic, you have been asked to explore the use of mixed alcohols.You plan to do this by cooling saturated solutions of this material which initiallyare at 20oC. Near this temperature the solubility changes 1.14 * 10-3g tetracyline/cm3

solvent/oK. The seed crystals are 0.01cm, roughly cubic, with a density of 1.06g/cm3. They are added at a concentration of only 35 ppm. The crystal size in this crop is about 0.088 cm. The supersaturaion is expected to be around 0.077 g/cm3 ; the growth is diffusion controlled, with a mass transfer coefficient of 6.5×10-5 cm/sec.Estimate how the crystallizer temperature should vary with time to achieve these goals.

Solution. To solve this problem, we need only calculate the various quantities in Eq.(10.3-11). First, we note that the increase in crystal size η is given by

7.8 = 0.010

0.010-0.088 =

ll-l

= s

sPη

Second, we find the time for crystal growth by calculating the growth rate G(=dl/dT) from Eq.(1-10) or (3-8), which for diffusion control is

hr3.2 t cm010.0 0.088cm tcm/hr) (0.034

ll G t from found is t timeprocess The

crystals. cubicfor case theis which one, are Φ and Φ that assumes This

0.034cm/hr hr

sec3600 cm

0.077g )cm/g06.1(1

1seccm 10 (2)6.5

)cc(Φ

Φ2kG

P

P

sPP

P

Av

335-

*

v

A

=−=

−=

=

×=

−

ρ

=

)t83.3t39.31(t312.020

hr3.2t8.7

31

hr3.2t8.71

hr3.27.8t(3)

C g/cm 1014.1g/cm 1035 20

)(311

dT/dcV/MTT

as written bemay which 11),-(10.3or 10)-(10.3 Eq.fromfoundbenowcan curve coolingThe

2o

2

o33

36o

2*S

0

++−=

++×

××−=

ητ+ητ−ητ

−=

−

−

Fig. 7. The cooling curve for a modifiedtetracycline. The temperature is initially nearly constant, but later drops abruptly. This general shape is characteristic of the cooling curves for batch crystallization.

4 RECRYSTALLIZATION

Fig. 8. Simple recrystallization. This scheme, which uses fresh solventfor each recrystallization, gives a high purity but a low yield.

S

S

S

L1

L2

L3

X1

X2

X3

ABS

S

S

L1

L2X3

L4X5

X1

X2

X4

AB

L3

L6 L2X3

Increasing Purity of A

Fig. 9. Fractional recrystallization. This scheme makes more efficient use ofthe various liquors, and hence gives a higher yield than that in the previous figure.

Fig. 10. Fractional crystallization as a stage operation. This scheme produces very pure crystals by a process equivalent to that of Craig extraction detailed inSection 5.5.

YA=(EA/(1+EA))N (4-1)

(4.2) p)-(1p r)!-(nr!

n! n)f(r, r-nr=

Example 8. Multiple simple crystallization of Soybean Sterols. A mixtureof soy sterols contains 20% stigmasterol and 80% sitosterol. This mixture and subsequent products are recrystallized six times following the scheme in Fig. 10.4-1. The crystallizations are conducted under conditions where the E values are 2.0 and 0.5 for stigmasterol and sitosterol, respectively.

(a) Calculate the yields of stigmasterol and sitosterol in the final product.(b) Determine the purity of the final product.

Solution. (a) The recovered yields of stigmasterol and sitosterol are calculatedusing Eq.(10.4-1) :

0.94 sitosterol 80)g(0.0014 olstigmaster g)20088.0(

olstigmaster g)20(0.088purity

bygiven is crystals final theofpurity The )b(

0014.015.0

5.0)sitosterol(Y

88.032

1EE)olstigmaster(Y

6

66

=×+×

×=

=

+=

=

=

+=

5 CONCLUSIONS

Crystallization is a powerful tool for the recovery of the final highly purified

product.

Crystallization enhances the performance of subsequent operations.

Crystallization improves the appearance of the final product.

There is a dearth of useful information for organic materials and for batch

operations.

Analyzing crystallization is complex and difficult.

Scale-up is risky and often compromise both product purity and crystal

characteristics.