CHEMICAL ENGINEERING SERIES
CRYSTALLIZATION
Compilation of Lectures and Solved Problems
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. There are 5 basic methods of generating
supersaturation
(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
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
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
SOLUTION WATER
Solution is concentrated by evaporating water
The concentrated solution is cooled until
the concentration becomes greater than
its solubility at that temperature
Types of Crystal Geometry
(1) CUBIC SYSTEM – 3 equal axes at right angles to each other
(2) TETRAGONAL – 3 axes at right angles to each other, one axis longer than the other 2
(3) ORTHOROMBIC – 3 axes at right angles to each other, all of different lengths
(4) 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
Types of Solubility Curve
(1) TYPE I: Solubility increases with temperature
and there are no hydrates or water of
crystallization
(2) TYPE II: Solubility increases with temperature
but curve is marked with extreme flatness
(3) TYPE III: Solubility increasing fairly rapid with
temperature but is characterized by “breaks”
and indicates different “hydrates” or water of
crystallization
(4) TYPE IV: Unusual Curve; Solubility increases
at a certain transition point while the solubility of
the hydrate decreases as temperature
increases
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
0 50 100 150 200 250 300 0 20 40 60 80 100 Sol ubi lit y, gr am per 1 00 gr am w at er Temperature, °C 0 50 100 150 200 250 0 20 40 60 80 100 Sol ubi lit y, gr am per 1 00 gr am w at er Temperature, °CSolubility of NaCl (CHE HB 8thedition)
0 50 100 150 200 250 0 20 40 60 80 100 Sol ub ili ty , gr am per 10 0 gr am w at er Temperature, °C
Solubility of Na2HPO4 (CHE HB 8thedition)
Na2HPO4 Na2HPO4·12H2O Na2HPO4·7H2O Na2HPO4·2H2O 0 10 20 30 40 50 60 0 20 40 60 80 100 Sol ubi lit y, gr am per 1 00 gr am w at er Temperature, °C
Solubility of Na2CO3(CHE HB 8thedition) Na2CO3·10H2O
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:
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
F XF hf tF L XL hL tL C XC hC tC W t1 W t2
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:
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
F XF hf tF L XL hL tL C XC hC tC W t1 W t2 V hV
SUPERSATURATION BY ADIABATIC EVAPORATION OF SOLVENT
Over-all material Balance:
Solute Balance:
Solvent Balance:
(
) (
) (
)
Enthalpy 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
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
F XF hf V hV L XL hL C XC hC M
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:
∫
∫
(
)
∫
(
)
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? 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: ( ) ( ) ( ) ( ) Na2SO4solution xF= 0.20 tF= 200°F Na2SO4·10H2O C L Magma, M tM= 60°F
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.
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
2. From figure 27-3 (Unit Operations by McCabe and Smoth 7th 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
Smith 7th 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 MgSO4solution F = 10,000 lb/h xF= 0.325 tF= 150°F MgSO4·7H2O C = 4,000 lb/h L V
TRIAL 1: Assume temperature of the solution at 60°F
From figure 27-3 (Unit Operations by McCabe and Smith 7th edition) Substitute to equation Substitute to equation
From figure 27-4 (Unit Operations by McCabe and Smith, 7th 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
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
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:
CALCINATION CRYSTALLIZATION P Na2SO4 30,000 MT/yr T = 0 C YIELD = 95% YIELD = 90% F
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 CHE BP July 1968 SOLUTION: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Consider over-all material balance around the crystallizer
( )
Consider Ba(NO3)2 balance
( )( ) ( )( ) Substitute in ( ) [( )( )] CRYSTALLIZER C T = 20 C T = 90 C F 1,200 lb BaNO3 V L T = 20 C
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 for the crystallizer is 35 BTU/h·°F·ft2. There are 3.5 ft2 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: ( ) SWENSON-WALKER CRYSTALLIZER F tF= 120 F L tL= 80 F C, 1 ton/h Fe2SO4·7H2O tC= 80 F W t1= 60 F t2= 70 F
[( ) ( ) ( ) ] [( ) ( )] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )
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% Na2CO3;
molecular weight of Na2CO3 is 106
SOLUTION:
Assume 100 g of Na2CO3·10H2O added into the saturated solution
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?
SOLUTION:
Consider over-all material balance:
Consider MgSO4 balance
( )
From table 27-3 (Unit Operations by McCabe and Smith, 7th edition), at 50°F ( )( ) ( )( ) ( )( ) Equate and COOLING CRYSTALLIZER F tF= 220 F 43 g MgSO4/100 g H2O C, 1 ton/h MgSO4·7H2O tC= 50 F L tL= 50 F
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 ( )( ) COOLING CRYSTALLIZER F tF= 60 F 16.4 g NaHCO3 /100 g H2O C, 9.6 g NaHCO3 per 100 g H2O tC= 20 F L tL= 20 F CRYSTALLIZER F tF= 20 C 8.4% Na2SO4 C, tC= 20 C L tL= 20 C V
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
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(NO3)2/100 kg total
water
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 CRYSTALLIZER F 30.6 kg Ba(NO3)2/100 kg H2O C L V
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
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?
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 ( ) ( ) CRYSTALLIZER F 1,000 kg KCl 363K C 293K L 293K V
d) Crystallization with evaporation Consider over-all material balance: ( ) Consider KCl balance ( )( ) ( )( ) Equate and
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
Swenson-Walker crystallizer by cooling a solution, saturated at 30°C, to 15°C. Cooling water enters at 10°C and leaves at 20°C. The over-all heat transfer coefficient in the crystover-allizer is 25 BTU/h·ft2·°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 SWENSON-WALKER CRYSTALLIZER F tF= 30 C L tL= 15 C C, 1 ton/h Na2SO4·10H2O tC= 15 C W t1= 10 C t2= 20 C
Consider heat balance: ( )
( ) ( )
From Table 2-194 (CHE HB 8th edition) [( )( ) ( )( )] [( ) ( ) ( ) ] [( ) ( )] ( ) ( ) ( ) ( ) ( )( )
PROBLEM # 13:
A continuous adiabatic vacuum crystallizer is to be used for the production of MgSO4·7H2O
crystals from 20,000 lb/h of solution containing 0.300 weight fraction MgSO4. The solution
enters the crystallizer at 160°F. The crystallizer is to be operated so that the mixture of mother liquor and crystals leaving the crystallizer contains 6,000 lb/h of MgSO4·7H2O crystals. The estimated boiling
point elevation of the solution in the crystallizer is 10°F. How many pounds of water are vaporized per hour?
Source: Unit Operations (Brown)
SOLUTION:
Consider over-all material balance:
Consider MgSO4 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
2. From figure 27-3 (Unit Operations by McCabe and Smoth 7th 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
Smith 7th edition) at the designated temperatures and concentrations
ADIABATIC VACUUM CRYSTALLIZER F, 20,000 lb/h xF= 0.3000 tF= 160 F L BPE = 10 F C = 6,000 lb/h MgSO4·7H2O V
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
TRIAL 1: Assume temperature of the solution at 60°F
From figure 27-3 (Unit Operations by McCabe and Smith 7th edition) Substitute to equation Substitute to equation
From figure 27-4 (Unit Operations by McCabe and Smith, 7th edition)
Temperature of vapor is 60 – 10 = 50°F
From steam table at 50°F, [( ) ( )] ( )( ) ( )( ) ( )( ) ( )( )
Since % error is about 5%, assumed value can be considered correct.
PROBLEM # 14:
Crystals of CaCl2·6H2O are to be obtained
from a solution of 35 weight % CaCl2, 10
weight % inert soluble impurity, and 55 weight % water in an Oslo crystallizer. The solution is fed to the crystallizer at 100°F and receives 250 BTU/lb of feed from the external heater. Products are withdrawn from the crystallizer at 40°F.
a) What are the products from the crystallizer?
b) The magma is centrifuged to a moisture content of 0.1 lb of liquid per lb of CaCl2·6H2O crystals and then dried in a
conveyor drier. What is the purity of the final dried crystalline product?
Source: Principles of Unit Operations 2nd edition (Foust, et al)
SOLUTION:
Basis: 1 lb of inert soluble-free feed
from table 2-120 (CHE HB 8th edition), solubilities of CaCl2·6H2O
0°C 59.5 lb/100 lb H2O
10°C 65 lb/100 lb H2O
20°C 74.5 lb/100 lb H2O
30°C 102 lb/100 lb H2O
At 100°F (37.8°C), solubility is (by extrapolation), 123.45 lb/100 lb H2O
At 40°F (4.4°C), solubility is 61.92 lb/100 lb H2O
Since the equipment is Oslo crystallizer, there the process is supersaturation by evaporation By heat balance around the crystallizer
( )
From table 2-194, specific heat of CaCl2, cal/K·mol
where T is in K At 100°F (310.93 K) At 40°F (277.59 K) ̅ OSLO CRYSTALLIZER F CaCl2= 35% Inert = 10% H2O = 55% tF= 100 F C’’ CaCl2·6H2O V CENTRIFUGE DRYER L M (magma) C Inert L tF= 40 F
For the feed
( ) ( ) ( ) ( ) ( )
From table 2-224 (CHE HB 8th edition), heat of solution of CaCl2·6H2O = -4,100 cal/mol;
in the absence of data on heat of crystallization, heat of solution can be used instead but of opposite sign
From the steam table, at 40°F,
( )( ) ( )( )( ) ( )( ) ( )( )
Consider over-all material balance: Substitute in ( )
Consider solute (CaCl2·6H2O) balance, inert soluble-free
( ) ( ) ( )( ) ( )( ) ( )( ) Equate and ( ) ( )
Composition of the liquor (including the inert soluble) ( ) lb % CaCl2·6H2O 0.0056 4.89 H2O 0.0090 7.85 inerts 0.1000 87.26 0.1146 100.00
For the crystals leaving the centrifuge:
Composition of crystals leaving the centrifuge
lb CaCl2·6H2O crystallized 0.7620 from liquor 0.0762 x 0.0489 0.0037 0.7657 H2O 0.0762 x 0.0785 0.0060 0.0060 inerts 0.0762 x 0.8726 0.0665 0.0665 0.8382 In the dryer, assume all free water has been removed
Composition of dried crystals
lb %
CaCl2·6H2O 0.7657 92.01
inerts 0.0665 7.99
0.8322 100.00
PROBLEM # 15:
Lactose syrup is concentrated to 8 g lactose per 10 g of water and then run into a crystallizing vat which contains 2,500 kg of the syrup. In this vat, containing 2,500 kg of syrup, it is cooled from 57°C to 10°C. Lactose crystallizes with one molecule of water of crystallization. The specific heat of the lactose solution is 3470 J/kg·°C. The heat of solution for lactose monohydrate is -15,500 kJ/kmol. The molecular weight of lactose monohydrate is 360 and the solubility of lactose at 10°C is 1.5 g/10 g water. Assume that 1% of the water evaporates and that the heat loss trough the vat walls is 4 x 104 kJ. Calculate the heat to be removed in the cooling process.
SOLUTION:
Consider over-all material balance
( ) ( )
Consider lactose balance
( )( ) ( )( ) ( )( ) Equate and OSLO CRYSTALLIZER F 2,500 kg 8 g lactose per 10 g water tF= 57 C V L 1.5 g lactose per 10 g water C tC= 10 C
Consider heat balance: ( ) At 10°C (50°F), [( ) ( ) ( ) ] [( ) ( )] [( ) ( )]
PROBLEM # 16:
Sal soda (Na2CO3·10H2O) is to be made by dissolving soda ash in a mixture of mother liquor and
water to form a 30% solution by weight at 45°C and then cooling to 15°C. The wet crystals removed from the mother liquor consist of 90% sal soda and 10% mother liquor by weight. The mother liquor is to be dried on the crystals as additional sal soda. The remainder of the mother liquor is to be returned to the dissolving tanks. At 15°C, the solubility of Na2CO3 is 14.2 parts per
100 parts water.
Crystallization is to be done in a Swenson-Walker crystallizer. This is to be supplied with water at 10°C, and sufficient cooling water is to be used to ensure that the exit water will not be over 20°C. The Swenson-walker crystallizer is built in units 10 ft long, containing 3 ft2 of heating surface per foot of length. An over-all heat transfer coefficient of 35 BTU/ft2·h·°F is expected.
The latent heat of crystallization of sal soda at 15°C is approximately 25,000 cal/mol. The specific heat of the solution is 0.85 BTU/lb·°F. A production of 1 ton/h of dried crystals is desired. Radiation losses and evaporation from the crystallizer are negligible.
a) What amounts of water and sal soda are to be added to the dissolver per hour? b) How many units of crystallizer are needed?
c) What is to be the capacity of the refrigeration plant, in tons of refrigeration, if the cooling water is to be cooled and recycled? One ton of refrigeration is equivalent to 12,000 BTU/h.
SOLUTION:
Basis: 2,000 lb/h (1 ton/h) of sal soda
Consider over-all material balance of the system
Consider Na2CO3 balance around the system
( ) ( )
DISSOLVER CRYSTALLIZER FILTER DRYER
C (Sal Soda) 45C 15C F (Soda Ash) W (Water) R (remainder mother liquor) V A B D
Substitute to equation
Consider solute (Na2CO3) balance around the dryer
( ) ( ) ( ) (( ) ) ( ) ( )
Consider over-all material balance around the dryer Substitute to equation
Consider solute (Na2CO3) balance around the dissolver
( ) ( )( ) ( )( ) ( )( )
Consider over-all material balance around the dissolver
Equate and
Consider heat balance around the crystallizer ( ) ( ) [( ) ( ) ( ) ] [( ) ( )] ( ) ( ) [( ) ( )] ( )( ) Refrigeration capacity:
PROBLEM # 17:
One ton of Na2S2O3·5H2O is to be crystallized per hour by cooling a solution containing 56.5%
Na2S2O3 to 30°C in a Swenson-Walker crystallizer. Evaporation is negligible. The product is to
be sized closely to approximately 14 mesh. Seed crystals closely sized to 20 mesh are introduced with the solution as it enters the crystallizer. How many tons of seed crystals and how many tons of solutions are required per hour? At 30°C, solubility of Na2S2O3 is 83 parts per 100
parts water
Source: Unit Operations (Brown, et al)
SOLUTION:
∫ ∫ ( )
From table 19-6 (CHE HB 8th edition) ( ) ( ) ∫ ∫ ( ) Equate and Consider Na2S2O3 balance: ( ) ( )( ) ( )( ) ( )( ) Consider over-all material balance
Equate and
PROBLEM # 18:
A Swenson-Walker crystallizer is fed with a saturated solution of magnesium sulfate at 110°F. The solution and its crystalline crop are cooled to 40°F. The inlet solution contains 1 g of seed crystals per 100 g of solution. The seeds are 80 mesh. Assuming ideal growth, what is the mesh size of the crystals leaving with the cooled product? Evaporation may be neglected.
SOLUTION:
Basis: 100 lb feed
Consider over-all material balance
Consider MgSO4 balance
From figure 27-3 (Unit Operation 7th edition, McCabe and Smith) at 110°F
From figure 27-3 (Unit Operations 7th edition, McCabe and Smith) at 40°F ( )( ) ( )( ) ( )( ) Equate and ∫ ∫ ( ) [ ]
From table 19-6 (CHE HB 8th edition) ( ) ( )√
From table 19-6 (CHE HB 8th edition)
PROBLEM # 19:
Trisodium phosphate is to be recovered as Na3PO4·12H2O from a 35 weight % solution originally
at 190°F by cooling and seeding in a Swenson-Walker crystallizer. From 20,000 lb/h feed, 7,000 lb/h of product crystals in addition to the seed crystals are to be obtained. Seed crystals fed at a rate of 500 lb/h have the following size range:
Weight Range Size Range, in
10 % - 0.0200 + 0.0100
20 % - 0.0100 + 0.0050
40 % - 0.0050 + 0.0025
30 % - 0.0025 + 0.0010
Latent heat of crystallization of trisodium phosphate is 27,500 BTU/lbmol. Specific heat for the trisodium phosphate solution may be taken as 0.8 BTU/lb·°F.
a) Estimate the product particle size distribution
b) To what temperature must the solution be cooled, and what will be the cooling duty in BTU/h
SOLUTION: ∫ ∫ ( ) ∫ ( ) ∫ ( ) ∫ ( ) ( )
Where: = fractional weight range Solve for required :
This problem can be solved by trial and error 1. Assume value of
2. Solve for ( ) for each size range, use the mean ̅ for each size range 3. Solve for
4. Get the total
TRIAL 1: Assume ̅ ( ̅ ) ( ̅ )
Since % error is less than 5%, assumed value can be considered For particle size distribution:
( ̅ )
Size Range, in Wt % Size Range, in Wt %
Consider over-all material balance: Consider Na3PO4 balance: ( )( ) ( )( ) ( )( )
( )
From table 2-120 (CHE HB 8th edition)
50°C 43 lb/100 lb H2O
60°C 55 lb/100 lb H2O
Cooling Duty:
Consider heat balance:
( ) [( ) ( ) ( ) ] [( ) ( )]
PROBLEM # 20:
How much CaCl2·6H2O must be dissolved in 100 kg of water at 20°C to form a saturated
solution? The solubility of CaCl2 at 20°C is 6.7 gmol anhydrous salt (CaCl2) per kg of water.
SOLUTION:
For a saturated solution utilizing 100 kg water as solvent: 1. Mole of CaCl2 required
2. Weight of CaCl2 required
3. Mole of CaCl2·6H2O required
4. Weight CaCl2·6H2O required
5. Composition of the solution in terms of CaCl2·6H2O
Since there should only be total of 100 kg water in the solution, the amount of free water (net of water of hydration)
( )
6. Amount of CaCl2·6H2O required for every 100 kg free water (net of water of hydration)
