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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

CHEMICAL ENGINEERING SERIES 26 CRYSTALLIZATION Consider heat balance: ( ) ( )

(

) th

From Table 2-194 (CHE HB 8 edition)

[(

)(

)

[(

(

)(

)(

)( [(

(

(

(

)

)(

) )(

)

(

)]

)

(

)

)

] )]

CHEMICAL ENGINEERING SERIES 27 CRYSTALLIZATION

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?

V

F, 20,000 lb/h xF = 0.3000 tF = 160 F

ADIABATIC VACUUM CRYSTALLIZER

C = 6,000 lb/h MgSO4·7H2O

L BPE = 10 F

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 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

CHEMICAL ENGINEERING SERIES 28 CRYSTALLIZATION 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 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 about 5%, assumed value can be considered correct.

CHEMICAL ENGINEERING SERIES 29 CRYSTALLIZATION

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 2 edition (Foust, et al)

V

F CaCl2 = 35% Inert = 10% H2O = 55% tF = 100 F

OSLO CRYSTALLIZER

M (magma) C Inert L tF = 40 F

L

CENTRIFUGE

C’’ CaCl2·6H2O

DRYER

nd

SOLUTION: Basis: 1 lb of inert soluble-free feed th from table 2-120 (CHE HB 8 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 H 2O 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)

̅

CHEMICAL ENGINEERING SERIES 30 CRYSTALLIZATION

For the feed (

)(

(

) (

)(

)

)

th

From table 2-224 (CHE HB 8 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 ( (

) )

)

)( )

CHEMICAL ENGINEERING SERIES 31 CRYSTALLIZATION

Composition of the liquor (including the inert soluble) (

CaCl2·6H2O H2O inerts

lb 0.0056 0.0090 0.1000 0.1146

% 4.89 7.85 87.26 100.00

For the crystals leaving the centrifuge:

Composition of crystals leaving the centrifuge lb CaCl2·6H2O crystallized from liquor

0.0762 x 0.0489

0.7620 0.0037

H2O inerts

0.0762 x 0.0785 0.0762 x 0.8726

0.0060 0.0665

In the dryer, assume all free water has been removed Composition of dried crystals lb CaCl2·6H2O 0.7657 inerts 0.0665 0.8322

% 92.01 7.99 100.00

0.7657 0.0060 0.0665 0.8382

)

CHEMICAL ENGINEERING SERIES 32 CRYSTALLIZATION 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 4 heat loss trough the vat walls is 4 x 10 kJ. Calculate the heat to be removed in the cooling process.

V

F 2,500 kg 8 g lactose per 10 g water tF = 57 C

L 1.5 g lactose per 10 g water

SOLUTION: Consider over-all material balance

( (

)

Consider lactose balance

(

Equate

)(

)

and

(

)( )

(

)( )

OSLO CRYSTALLIZER

)

C tC = 10 C

CHEMICAL ENGINEERING SERIES 33 CRYSTALLIZATION

Consider heat balance: ( ) At 10°C (50°F),

)(

[( [(

)( )(

) )]

]

[(

)(

)]

CHEMICAL ENGINEERING SERIES 34 CRYSTALLIZATION 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 Na 2CO3 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. 2 The Swenson-walker crystallizer is built in units 10 ft long, containing 3 ft of heating surface per 2 foot of length. An over-all heat transfer coefficient of 35 BTU/ft ·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. F (Soda Ash) W (Water) V

A DISSOLVER

B

CRYSTALLIZER 45C

D

FILTER

DRYER

15C

R (remainder mother liquor) C (Sal Soda)

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

(

)(

)

CHEMICAL ENGINEERING SERIES 35 CRYSTALLIZATION

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

(

)(

(

)

)

CHEMICAL ENGINEERING SERIES 36 CRYSTALLIZATION Equate

and

Consider heat balance around the crystallizer ( ) (

[(

)(

)(

(

)

[(

(

)

(

)

)(

Refrigeration capacity:

)

(

)

)]

)

]

[(

)(

)]

CHEMICAL ENGINEERING SERIES 37 CRYSTALLIZATION 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 Na 2S2O3 is 83 parts per 100 parts water Source: Unit Operations (Brown, et al)

SOLUTION: ∫

∫

(

) th

From table 19-6 (CHE HB 8 edition) ( (

∫

∫

Equate

(

)

and

Consider Na2S2O3 balance:

(

)( )

(

)

(

)( )

(

Consider over-all material balance

Equate

and

)(

)

) )

CHEMICAL ENGINEERING SERIES 38 CRYSTALLIZATION

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 th

From figure 27-3 (Unit Operation 7 edition, McCabe and Smith) at 110°F

th

From figure 27-3 (Unit Operations 7 edition, McCabe and Smith) at 40°F

(

)(

)

Equate

(

(

)( )

and

∫

∫ [

)( )

(

)

] th

From table 19-6 (CHE HB 8 edition) ( (

)√

th

From table 19-6 (CHE HB 8 edition)

)

CHEMICAL ENGINEERING SERIES 39 CRYSTALLIZATION 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: Solve for required

)

) = fractional weight range :

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 5. If ∑

, then assumed

is correct; if not, redo another trial

CHEMICAL ENGINEERING SERIES 40 CRYSTALLIZATION TRIAL 1: Assume

̅

(

̅

)

(

̅

Since % error is less than 5%, assumed value can be considered For particle size distribution:

(

̅ )

Size Range, in

Size Range, in

Wt %

Consider over-all material balance:

Consider Na3PO4 balance:

(

)(

)

( )(

)

(

)(

)

Wt %

)

CHEMICAL ENGINEERING SERIES 41 CRYSTALLIZATION (

)

th

From table 2-120 (CHE HB 8 edition) 50°C 60°C

43 lb/100 lb H2O 55 lb/100 lb H2O

Cooling Duty: Consider heat balance: ( ) [(

)(

)(

)

]

[(

)(

)]

CHEMICAL ENGINEERING SERIES 42 CRYSTALLIZATION 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)

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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

CHEMICAL ENGINEERING SERIES 26 CRYSTALLIZATION Consider heat balance: ( ) ( )

(

) th

From Table 2-194 (CHE HB 8 edition)

[(

)(

)

[(

(

)(

)(

)( [(

(

(

(

)

)(

) )(

)

(

)]

)

(

)

)

] )]

CHEMICAL ENGINEERING SERIES 27 CRYSTALLIZATION

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?

V

F, 20,000 lb/h xF = 0.3000 tF = 160 F

ADIABATIC VACUUM CRYSTALLIZER

C = 6,000 lb/h MgSO4·7H2O

L BPE = 10 F

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 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

CHEMICAL ENGINEERING SERIES 28 CRYSTALLIZATION 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 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 about 5%, assumed value can be considered correct.

CHEMICAL ENGINEERING SERIES 29 CRYSTALLIZATION

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 2 edition (Foust, et al)

V

F CaCl2 = 35% Inert = 10% H2O = 55% tF = 100 F

OSLO CRYSTALLIZER

M (magma) C Inert L tF = 40 F

L

CENTRIFUGE

C’’ CaCl2·6H2O

DRYER

nd

SOLUTION: Basis: 1 lb of inert soluble-free feed th from table 2-120 (CHE HB 8 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 H 2O 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)

̅

CHEMICAL ENGINEERING SERIES 30 CRYSTALLIZATION

For the feed (

)(

(

) (

)(

)

)

th

From table 2-224 (CHE HB 8 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 ( (

) )

)

)( )

CHEMICAL ENGINEERING SERIES 31 CRYSTALLIZATION

Composition of the liquor (including the inert soluble) (

CaCl2·6H2O H2O inerts

lb 0.0056 0.0090 0.1000 0.1146

% 4.89 7.85 87.26 100.00

For the crystals leaving the centrifuge:

Composition of crystals leaving the centrifuge lb CaCl2·6H2O crystallized from liquor

0.0762 x 0.0489

0.7620 0.0037

H2O inerts

0.0762 x 0.0785 0.0762 x 0.8726

0.0060 0.0665

In the dryer, assume all free water has been removed Composition of dried crystals lb CaCl2·6H2O 0.7657 inerts 0.0665 0.8322

% 92.01 7.99 100.00

0.7657 0.0060 0.0665 0.8382

)

CHEMICAL ENGINEERING SERIES 32 CRYSTALLIZATION 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 4 heat loss trough the vat walls is 4 x 10 kJ. Calculate the heat to be removed in the cooling process.

V

F 2,500 kg 8 g lactose per 10 g water tF = 57 C

L 1.5 g lactose per 10 g water

SOLUTION: Consider over-all material balance

( (

)

Consider lactose balance

(

Equate

)(

)

and

(

)( )

(

)( )

OSLO CRYSTALLIZER

)

C tC = 10 C

CHEMICAL ENGINEERING SERIES 33 CRYSTALLIZATION

Consider heat balance: ( ) At 10°C (50°F),

)(

[( [(

)( )(

) )]

]

[(

)(

)]

CHEMICAL ENGINEERING SERIES 34 CRYSTALLIZATION 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 Na 2CO3 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. 2 The Swenson-walker crystallizer is built in units 10 ft long, containing 3 ft of heating surface per 2 foot of length. An over-all heat transfer coefficient of 35 BTU/ft ·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. F (Soda Ash) W (Water) V

A DISSOLVER

B

CRYSTALLIZER 45C

D

FILTER

DRYER

15C

R (remainder mother liquor) C (Sal Soda)

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

(

)(

)

CHEMICAL ENGINEERING SERIES 35 CRYSTALLIZATION

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

(

)(

(

)

)

CHEMICAL ENGINEERING SERIES 36 CRYSTALLIZATION Equate

and

Consider heat balance around the crystallizer ( ) (

[(

)(

)(

(

)

[(

(

)

(

)

)(

Refrigeration capacity:

)

(

)

)]

)

]

[(

)(

)]

CHEMICAL ENGINEERING SERIES 37 CRYSTALLIZATION 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 Na 2S2O3 is 83 parts per 100 parts water Source: Unit Operations (Brown, et al)

SOLUTION: ∫

∫

(

) th

From table 19-6 (CHE HB 8 edition) ( (

∫

∫

Equate

(

)

and

Consider Na2S2O3 balance:

(

)( )

(

)

(

)( )

(

Consider over-all material balance

Equate

and

)(

)

) )

CHEMICAL ENGINEERING SERIES 38 CRYSTALLIZATION

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 th

From figure 27-3 (Unit Operation 7 edition, McCabe and Smith) at 110°F

th

From figure 27-3 (Unit Operations 7 edition, McCabe and Smith) at 40°F

(

)(

)

Equate

(

(

)( )

and

∫

∫ [

)( )

(

)

] th

From table 19-6 (CHE HB 8 edition) ( (

)√

th

From table 19-6 (CHE HB 8 edition)

)

CHEMICAL ENGINEERING SERIES 39 CRYSTALLIZATION 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: Solve for required

)

) = fractional weight range :

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 5. If ∑

, then assumed

is correct; if not, redo another trial

CHEMICAL ENGINEERING SERIES 40 CRYSTALLIZATION TRIAL 1: Assume

̅

(

̅

)

(

̅

Since % error is less than 5%, assumed value can be considered For particle size distribution:

(

̅ )

Size Range, in

Size Range, in

Wt %

Consider over-all material balance:

Consider Na3PO4 balance:

(

)(

)

( )(

)

(

)(

)

Wt %

)

CHEMICAL ENGINEERING SERIES 41 CRYSTALLIZATION (

)

th

From table 2-120 (CHE HB 8 edition) 50°C 60°C

43 lb/100 lb H2O 55 lb/100 lb H2O

Cooling Duty: Consider heat balance: ( ) [(

)(

)(

)

]

[(

)(

)]

CHEMICAL ENGINEERING SERIES 42 CRYSTALLIZATION 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)

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