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249518991 crystallization notes 1

CHEMICAL ENGINEERING SERIES

CRYSTALLIZATION
Compilation of Lectures and Solved Problems


CHEMICAL ENGINEERING SERIES 2
CRYSTALLIZATION

CRYSTALLIZATION
Refers to a solid-liquid separation process in which solid particles are formed within a homogenous phase.
It can occur as:
(1) formation of solid particles in a vapor
(2) formation of solid particles from a liquid melt
(3) formation of solid crystals from a solution
The process usually involves two steps:
(1) concentration of solution and cooling of solution until the solute concentration becomes greater than its
solubility at that temperature
(2) solute comes out of the solution in the form of pure crystals
Crystal Geometry
A crystal is highly organized type of matter, the constituent particles of which are arranged in an orderly and

repetitive manner; they are arranged in orderly three dimensional arrays called SPACE LATTICES
Supersaturation
Supersaturation is a measure of the quantity of solids actually present in solution as compared to the
quantity that is in equilibrium with the solution


Crystallization cannot occur without supersaturation.
supersaturation

There are 5 basic methods of generating

(1) EVAPORATION – by evaporating a portion of the solvent
(2) COOLING – by cooling a solution through indirect heat exchange
(3) VACUUM COOLING – by flashing of feed solution adiabatically to a lower temperature and inducing
crystallization by simultaneous cooling and evaporation of the solvent
(4) REACTION – by chemical reaction with a third substance
(5) SALTING – by the addition of a third component to change the solubility relationship


CHEMICAL ENGINEERING SERIES 3
CRYSTALLIZATION
Mechanism of Crystallization Process
There are two basic steps in the over-all process of crystallization from supersaturated solution:
(1) NUCLEATION’
a. Homogenous or Primary Nucleation – occurs due to rapid local fluctuations on a molecular scale in
a homogenous phase; it occurs in the bulk of a fluid phase without the involvement of a solid-fluid
interface
b. Heterogeneous Nucleation – occurs in the presence of surfaces other than those of the crystals
such as the surfaces of walls of the pipe or container, impellers in mixing or foreign particles; this is
dependent on the intensity of agitation
c. Secondary Nucleation – occurs due to the presence of crystals of the crystallizing species
(2) CRYSTAL GROWTH – a layer-by-layer process
a. Solute diffusion to the suspension-crystal interface
b. Surface reaction for absorbing solute into the crystal lattice

Crystallization Process
SOLUTION

WATER


CRYSTALS

Solution is concentrated
by evaporating water

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

Important Factors in a Crystallization Process
(1) Yield
(2) Purity of the Crystals
(3) Size of the Crystals – should be uniform to minimize caking in the package, for ease in pouring, ease
in washing and filtering and for uniform behaviour when used
(4) Shape of the Crystals
Magma
It is the two-phase mixture of mother liquor and crystals of all sizes, which occupies the crystallizer and is
withdrawn as product


CHEMICAL ENGINEERING SERIES 4
CRYSTALLIZATION

Types of Crystal Geometry
(1)
(2)
(3)
(4)

CUBIC SYSTEM – 3 equal axes at right angles to each other
TETRAGONAL – 3 axes at right angles to each other, one axis longer than the other 2
ORTHOROMBIC – 3 axes at right angles to each other, all of different lengths
HEXAGONAL – 3 equal axes in one plane at 60° to each other, and a fourth axis at a right angle to
this plane and not necessarily at the same length
(5) MONOCLINIC – 3 unequal axes, two at a right angles in a plane, and a third at some angle to this
plane
(6) TRICLINIC – 3 unequal axes at unequal angles to each other and not 30°, 60°, or 90°
(7) TRIGONAL – 3 unequal and equally inclined axes

Classification of Crystallizer
(1) May be classified according to whether they are batch or continuous in operation
(2) May be classified according on the methods used to bring about supersaturation
(3) Can also be classified according on the method of suspending the growing product crystals
Equilibrium Data (Solubilities)





Either tables or curves
Represent equilibrium conditions
Plotted data of solubilities versus temperatures
In general, solubility is dependent mainly on temperature although sometimes on size of materials and
pressure

Expressions of Solubilities



Parts by mass of anhydrous materials per 100 parts by mass of total solvent
Mass percent of anhydrous materials or solute which ignores water of crystallization


CHEMICAL ENGINEERING SERIES 5
CRYSTALLIZATION

(1) TYPE I: Solubility increases with temperature
and there are no hydrates or water of
crystallization

Solubility, gram per 100 gram water

Types of Solubility Curve
300
250
200

150
100
50
0
0

20

40

60

80

100

80

100

(2) TYPE II: Solubility increases with temperature
but curve is marked with extreme flatness

Solubility, gram per 100 gram water

Temperature, °C
250
200

150
100
50
0
0

20

40

60

Temperature, °C

(3) TYPE III: Solubility increasing fairly rapid with
temperature but is characterized by “breaks”
and indicates different “hydrates” or water of
crystallization

Solubility, gram per 100 gram water

Solubility of NaCl (CHE HB 8th edition)

250
200
150
Na2HPO4·2H2O

Na2HPO4·7H2O

100

Na2HPO4

Na2HPO4·12H2O

50
0
0

20

40

60

80

100

(4) TYPE IV: Unusual Curve; Solubility increases
at a certain transition point while the solubility of
the hydrate decreases as temperature
increases

Solubility, gram per 100 gram water

Temperature, °C
Solubility of Na2HPO4 (CHE HB 8th edition)
60

50
40
Na2CO3·H2O

30
20

Na2CO3·10H2O

10
0
0

20

40

60

80

100

Temperature, °C
Solubility of Na2CO3 (CHE HB 8th edition)

SUPERSATURATION BY COOLING
Crystallizers that obtain precipitation by cooling a concentrated hot solution; applicable for substance that
have solubility curve that decreases with temperature; for normal solubility curve which are common for
most substances


CHEMICAL ENGINEERING SERIES 6
CRYSTALLIZATION
Pan Crystallizers
Batch operation; seldom used in modern practice, except in small scale operations, because they are
wasteful of floor space and of labor; usually give a low quality product
Agitated batch Crystallizers
Consist of an agitated tank; usually cone-bottomed, containing cooling coils. It is convenient in small scale
or batch operations because of their low capital costs, simplicity of operation and flexibility
Swenson Walker Crystallizer
A continuous crystallizer consist of an open round bottomed-trough, 24-in wide by 10 ft long, and containing
a long ribbon mixer that turns at about 7 rpm.
CALCULATIONS:
L
XL
hL
tL

F
XF
hf
tF

W
t1

W
t2

C
XC
hC
tC

Over-all material Balance:
Solute Balance:
Enthalpy Balance:
Heat Balance:
(

)
(

)

Heat Transfer Equation
[

(

)

(

)
]

where:
= mass of the feed solution
= mass of the mother liquor, usually saturated solution
= mass of the crystals
= mass of the cooling water
= mass solute (salt) in the feed solution per mass of feed solution
= mass of solute (salt) in the mother liquor per mass of mother liquor
= mass of solute (salt) in the srystals per mass of crystals
= enthalpy of the feed solution
= enthalpy of the mother liquor
= enthalpy of the crystals
= heat absorbed by the cooling water
= heat loss by the crystals
= specific heat of the feed solution
= specific heat of cooling water
= heat of crystallization
= over-all heat transfer coefficient
= heat transfer area
= temperature of the feed solution
= temperature of the mother liquor
= inlet temperature of cooling water
= outlet temperature of cooling water


CHEMICAL ENGINEERING SERIES 7
CRYSTALLIZATION

SUPERSATURATION BY EVAPORATION OF SOLVENT
Crystallizers that obtain precipitation by evaporating a solution; applicable for the substance whose solubility
curve is flat that yield of solids by cooling is negligible; acceptable to any substance whose solubility curve is
not to steep
Salting Evaporator
The most common of the evaporating crystallizers; in older form, the crystallizer consisted of an evaporator
below which were settling chambers into which the salt settled
Oslo Crystallizer
Modern form of evaporating crystallizer; this unit is particularly well adopted to the production of large-sized
uniform crystals that are usually rounded; it consists essentially of a forced circulation evaporator with an
external heater containing a combination of salt filter and particle size classifier on the bottom of the
evaporator body
CALCULATIONS:
V
hV
F
XF
hf
tF

L
XL
hL
tL

W
t1

W
t2

C
XC
hC
tC

Over-all material Balance:
Solute Balance:
Solvent Balance:
(
)

(

)

Enthalpy Balance:
Heat Balance:
(

)
(

)

(

)

where:
= mass of the feed solution
= mass of the mother liquor, usually saturated solution
= mass of the crystals
= mass of the cooling water
= mass of the evaporated solvent
= mass solute (salt) in the feed solution per mass of feed
solution
= mass of solute (salt) in the mother liquor per mass of
mother liquor
= mass of solute (salt) in the srystals per mass of crystals
= enthalpy of the feed solution
= enthalpy of the mother liquor
= enthalpy of the crystals
= enthalpy of the vapor
= heat absorbed by the cooling water
= heat loss by the crystals
= specific heat of the feed solution
= specific heat of cooling water
= heat of crystallization
= latent heat of vaporization
= over-all heat transfer coefficient
= heat transfer area
= temperature of the feed solution
= temperature of the mother liquor
= inlet temperature of cooling water
= outlet temperature of cooling water


CHEMICAL ENGINEERING SERIES 8
CRYSTALLIZATION

SUPERSATURATION BY ADIABATIC EVAPORATION OF SOLVENT
V
hV

F
XF
hf

L
XL
hL

M

C
XC
hC

Over-all material Balance:
Solute Balance:
Solvent Balance:
(
)

(

)

(

)

where:
= mass of the feed solution
= mass of the mother liquor, usually saturated solution
= mass of the crystals
= mass of the cooling water
= mass of the evaporated solvent
= mass solute (salt) in the feed solution per mass of feed
solution
= mass of solute (salt) in the mother liquor per mass of
mother liquor
= mass of solute (salt) in the srystals per mass of crystals
= enthalpy of the feed solution
= enthalpy of the mother liquor
= enthalpy of the crystals
= enthalpy of the vapor
= heat of crystallization
= temperature of the feed solution
= temperature of the mother liquor
= inlet temperature of cooling water
= outlet temperature of cooling water

Enthalpy Balance:

CRYSTALLIZATION BY SEEDING
ΔL Law of Crystals


States that if all crystals in magma grow in a supersaturation field and at the same temperature and if
all crystal grow from birth at a rate governed by the supersaturation, then all crystals are not only
invariant but also have the same growth rate that is independent of size



The relation between seed and product particle sizes may be written as

Where:
= characteristic particle dimension of the product
= characteristic particle dimension of the seed
= change in size of crystals and is constant throughout the range of size present


CHEMICAL ENGINEERING SERIES 9
CRYSTALLIZATION

Since the rate of linear crystal growth is independent of crystal size, the seed and product masses may
be related for
(
(

)
)

(

)
[

(
(

]
)

)

All the crystals in the seed have been assumed to be of the same shape, and the shape has been assumed
to be unchanged by the growth process. Through assumption is reasonably closed to the actual conditions
in most cases. For differential parts of the crystal masses, each consisting of crystals of identical
dimensions:






(

(

)

)


CHEMICAL ENGINEERING SERIES 10
CRYSTALLIZATION
PROBLEM # 01:
A 20 weight % solution of Na2SO4 at
200°F is pumped continuously to a
vacuum crystallizer from which the
magma is pumped at 60°F. What is
the composition of this magma, and
what percentage of Na2SO4 in the
feed is recovered as Na2SO4·10H2O
crystals after this magma is
centrifuged?

Na2SO4 solution
xF = 0.20
tF = 200°F

Na2SO4 ·10H2O
C

Magma, M
tM = 60°F

L

SOLUTION:
Basis: 100 lb feed
From table 2-122 (CHE HB), solubility of Na2SO4·10H2O
T,°C
10
15
20
g/100 g H2O
9.0
19.4
40.8
Consider over-all material balance:

Consider solute balance:

At 60°F, solubility is 21.7778 g per 100 g water

)(

(

Substitute

)

)( )

(

)( )

(

in
(

)

Magma composition:

% Recovery:
)(

(
(

)(

)
)


CHEMICAL ENGINEERING SERIES 11
CRYSTALLIZATION

PROBLEM # 02:
A solution of 32.5% MgSO4 originally
at 150°F is to be crystallized in a
vacuum adiabatic crystallizer to give
a product containing 4,000 lb/h of
MgSO4·7H2O crystals from 10,000
lb/h of feed. The solution boiling
point rise is estimated at 10°F.
Determine the product temperature,
pressure and weight ratio of mother
liquor to crystalline product.

V

MgSO4 solution
F = 10,000 lb/h
xF = 0.325
tF = 150°F

MgSO4 ·7H2O
C = 4,000 lb/h

L

SOLUTION:
Consider over-all material balance:

Consider solute balance:

(

)(

)

( )

(

)(

)

Consider enthalpy balance:

THE PROBLEM CAN BE SOLVED BY TRIAL AND ERROR SINCE TEMPERATURE OF THE
SOLUTION AFTER CRYSTALLIZATION IS UNKNOWN AND ENTHALPIES ARE DEPENDENT
ON TEMPERATURE
1. Assume temperature of the solution
th
2. From figure 27-3 (Unit Operations by McCabe and Smoth 7 edition), obtain mass fraction of
MgSO4 at the assumed temperature of the solution
3. Solve for “L” using equation
4. Solve for “V” using equation
5. Check if assumed temperature is correct by conducting enthalpy balance
a. Obtain values of hF, hC and hL from figure 27-4 (Unit Operations by McCabe and
th
Smith 7 edition) at the designated temperatures and concentrations
b. Compute for hV
c. Using the enthalpy balance equation, compute for “V” using the value of “L” from step
3
6. Compare values of “V” from step 4 with that from step 5-c
7. If not the same (or approximately the same), conduct another trial and error calculations


CHEMICAL ENGINEERING SERIES 12
CRYSTALLIZATION
TRIAL 1: Assume temperature of the solution at 60°F
th
From figure 27-3 (Unit Operations by McCabe and Smith 7 edition)

Substitute to equation

Substitute to equation

th

From figure 27-4 (Unit Operations by McCabe and Smith, 7 edition)

Temperature of vapor is 60 – 10 = 50°F
From steam table at 50°F,
)(

[(

(

)(

)

(

)( )

(

)]

)(

)

(

)(

)

Since % error is less than 5%, assumed value can be considered correct.
Product temperature

Operating Pressure
From steam table for vapor temperature of 50°F

Ratio of mother liquor to crystalline product


CHEMICAL ENGINEERING SERIES 13
CRYSTALLIZATION
PROBLEM # 03 :
A plant produces 30,000 MT of anhydrous
sulfate annually by crystallizing sulfate brine at
0°C, yields of 95% and 90% in the
crystallization and calcinations operations are
obtained respectively. How many metric tons
of liquor are fed to the crystallizer daily? Note:
300 working days per year

F

CALCINATION

YIELD = 90%

CHE BP January 1970
SOLUTION:
Assume that the liquor entering the crystallizer is a saturated solution at 0°C
From table 2-120 (CHE HB), solubility at 0°C:

CRYSTALLIZATION
T=0C
YIELD = 95%
P
Na2SO4
30,000 MT/yr


CHEMICAL ENGINEERING SERIES 14
CRYSTALLIZATION
PROBLEM # 04 :
1,200 lb of barium nitrate are dissolved in
sufficient water to form a saturated solution at
90°C. Assuming that 5% of the weight of the
original solution is lost through evaporation,
calculate the crop of the crystals obtained
when cooled to 20°C. solubility data of barium
nitrate at 90°C = 30.6 lb/100 lb water; at 20°C =
9.2 lb/100 lb water

V

C
T = 20 C

F
1,200 lb BaNO3

CRYSTALLIZER
T = 90 C
L
T = 20 C

CHE BP July 1968
SOLUTION:
(

)
(
(

(

)

(

)

(

)

Consider Ba(NO3)2 balance

Substitute

)( )

(

)( )

in
(
[(

)
)(

(

)

)

)

)

Consider over-all material balance around the crystallizer

(

)

)

(

(

(

)]


CHEMICAL ENGINEERING SERIES 15
CRYSTALLIZATION
PROBLEM # 05:
A Swenson-Walker crystallizer is to be used
to produce 1 ton/h of copperas (FeSO4·7H2O)
crystals. The saturated solution enters the
crystallizer at 120°F. The slurry leaving the
crystallizer will be at 80°F. Cooling water
enters the crystallizer jacket at 60°F and
leaves at 70°F. It may be assumed that the U
2
for the crystallizer is 35 BTU/h·°F·ft . There
2
are 3.5 ft of cooling surface per ft of
crystallizer length.
a) Estimate the cooling water required
b) Determine the number of crystallizer
section to be used.
Data:
specific heat of solution = 0.7
BTU/lb·°F; heat of solution= 4400 cal/gmol
copperas; solubility at 120°F = 140 parts
copperas/100 parts excess water; solubility at
80°F = 74 parts copperas/100 parts excess
water

SOLUTION:
Consider over-all material balance:

Consider copperas (FeSO4·7H2O) balance:

(

Equate

)( )

(

)(

)

and

Consider heat balance:
(

)

(

)( )

F
tF = 120 F

L
tL = 80 F
SWENSON-WALKER
CRYSTALLIZER

W
t1 = 60 F

t2 = 70 F

C, 1 ton/h
Fe2SO4·7H2O
tC = 80 F


CHEMICAL ENGINEERING SERIES 16
CRYSTALLIZATION
[(

)(

)(
[(

)(

(

(

)

(

)

(

(

(
)

)(

)
(

)

)

]
)]

)

)(

)


CHEMICAL ENGINEERING SERIES 17
CRYSTALLIZATION
PROBLEM # 06:
Crystals of Na2CO3·10H2O are dropped into a saturated solution of Na2CO3 in water at 100°C.
What percent of the Na2CO3 in the Na2CO3·H2O is recovered in the precipitated solid? The
precipitated solid is Na2CO3·H2O. Data at 100°C: the saturated solution is 31.2% Na 2CO3;
molecular weight of Na2CO3 is 106

SOLUTION:
Assume 100 g of Na2CO3·10H2O added into the saturated solution


CHEMICAL ENGINEERING SERIES 18
CRYSTALLIZATION
PROBLEM # 07:
A solution of MgSO4 at 220°F containing 43 g
MgSO4 per 100 g H2O is fed into a cooling
crystallizer operating at 50°F. If the solution
leaving the crystallizer is saturated, what is the
rate at which the solution must be fed to the
crystallizer to produce one ton of MgSO4·7H2O
per hour?

F
tF = 220 F
43 g MgSO4/100 g H2O

L
tL = 50 F

COOLING CRYSTALLIZER

C, 1 ton/h
MgSO4·7H2O
tC = 50 F

SOLUTION:
Consider over-all material balance:

Consider MgSO4 balance

(

)

th

From table 27-3 (Unit Operations by McCabe and Smith, 7 edition), at 50°F

(

Equate

)( )

and

(

)( )

(

)( )


CHEMICAL ENGINEERING SERIES 19
CRYSTALLIZATION
PROBLEM # 08:
The solubility of sodium bicarbonate in water
is 9.6 g per 100 g water at 20°C and 16.4 g
per 100 g water at 60°C. If a saturated
solution of NaHCO3 at 60°C is cooled to 20°C,
what is the percentage of the dissolved salt
that crystallizes out?

SOLUTION:
Basis: 100 kg feed
Consider over-all material balance:

Consider NaHCO3 balance

(

)

(
(

)(

Equate

)

(

)

)( )

( )( )

(

)(

and

)

V
F
tF = 20 C
8.4% Na2SO4

L
tL = 20 C

CRYSTALLIZER

C,
tC = 20 C

F
tF = 60 F
16.4 g
NaHCO3 /100 g
H2O

L
tL = 20 F
COOLING CRYSTALLIZER

C,
9.6 g NaHCO3
per 100 g H2O
tC = 20 F


CHEMICAL ENGINEERING SERIES 20
CRYSTALLIZATION

PROBLEM # 09:
Glauber’s salt is made by crystallization from a water solution at 20°C. The aqueous solution at
20°C contains 8.4% sodium sulfate. How many grams of water must be evaporated from a liter of
such solution whose specific gravity is 1.077 so that when the residue solution after evaporation
is cooled to 20°C, there will be crystallized out 80% of the original sodium sulfate as Glauber’s
salt. The solubility of sodium sulfate in equilibrium with the decahydrate is 19.4 g Na2SO4 per 100
g H2O.
SOLUTION:
Basis: 1 L feed

Consider over-all material balance:

(
(

)(

)(

)

)

Substitute to equation

Consider Na2SO4 balance

(
(

)( )

Substitute to equation

)


CHEMICAL ENGINEERING SERIES 21
CRYSTALLIZATION
PROBLEM # 10:
A hot solution of Ba(NO3)2 from an evaporator
contains 30.6 kg Ba(NO3)2/100 kg H2O and
goes to a crystallizer where the solution is
cooled and Ba(NO3)2 crystallizes. On cooling,
10% of the original water present evaporates.
For a feed solution of 100 kg total, calculate
the following:
a) The yield of crystals if the solution is
cooled to 290K, where the solubility is 8.6
kg Ba(NO3)2/100 kg total water
b) The yield if cooled instead to 283K, where
the solubility is 7 kg Ba(NO 3)2/100 kg total
water

V
L

F
30.6 kg Ba(NO3)2/100 kg H2O

CRYSTALLIZER

C

Source:
Transport Processes and Unit
Operations (Geankoplis)

SOLUTION:
a) If solution is cooled to 290K
Consider over-all material balance:

If water evaporated is 10% of the original water present
(
)
(
)
(
)
(
)(
)

(

)

Consider Ba(NO3)2 balance
(

)

(
(

(

Equate

)(

)

and

(

)( )

(

)

)( )

)


CHEMICAL ENGINEERING SERIES 22
CRYSTALLIZATION

b) If solution is cooled to 283 K
Consider over-all material balance:

If water evaporated is 10% of the original water present
(
)
(
)
(
)
(
)(
)

(

)

Consider Ba(NO3)2 balance
(

)

(
(

(

Equate

)(

)

and

(

)( )

(

)

)( )

)


CHEMICAL ENGINEERING SERIES 23
CRYSTALLIZATION

PROBLEM # 11:
A batch of 1,000 kg of KCl is dissolved in
sufficient water to make a saturated solution at
363 K, where the solubility is 35 wt % KCl in
water. The solution is cooled to 293 K, at
which temperature its solubility is 25.4 wt %.
a) What are the weight of water required for
the solution and the weight of KCl crystals
obtained?
b) What is the weight of crystals obtained if
5% of the original water evaporates on
cooling?

V
F
1,000 kg KCl
363K

Source:
Transport Processes and Unit
Operations (Geankoplis)

SOLUTION:
c) Assume crystallization by cooling (without evaporation)
Consider over-all material balance:

Consider KCl balance

(

)( )

Equate

(

)( )

and

(

)(

)

L
293K
CRYSTALLIZER

C
293K


CHEMICAL ENGINEERING SERIES 24
CRYSTALLIZATION

d) Crystallization with evaporation
Consider over-all material balance:
(

)

Consider KCl balance

(

Equate

)( )

and

(

)( )


CHEMICAL ENGINEERING SERIES 25
CRYSTALLIZATION

PROBLEM # 12:
The solubility of sodium sulfate is 40 parts
Na2SO4 per 100 parts of water at 30°C, and
13.5 parts at 15°C.
The latent heat of
crystallization (liberated when crystals form) is
18,000 g-cal per gmol Na2SO4. Glauber’s salt
(Na2SO4·10H2O) is to be made in a SwensonWalker crystallizer by cooling a solution,
saturated at 30°C, to 15°C. Cooling water
enters at 10°C and leaves at 20°C. The overall heat transfer coefficient in the crystallizer is
2
25 BTU/h·ft ·°F and each foot of crystallizer
has 3 sq ft of cooling surface. How many 10-ft
units of crystallizer will be required to produce
1 ton/h of Glauber’s Salt
Source: Unit Operations (Brown)

SOLUTION:
Consider over-all material balance:

Consider Na2SO4 balance

(

)
(

(

Equate

and

)

)

F
tF = 30 C

L
tL = 15 C
SWENSON-WALKER
CRYSTALLIZER

W
t1 = 10 C

t2 = 20 C

C, 1 ton/h
Na2SO4·10H2O
tC = 15 C


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