Crystallization Notes (1)

CRYSTALLIZATION

CHEMICAL ENGINEERING SERIESCRYSTALLIZATION

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CHEMICAL ENGINEERING SERIESCRYSTALLIZATIONCompilation 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 GeometryA 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 LATTICESSupersaturationSupersaturation 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(5) SALTING by the addition of a third component to change the solubility relationship

Mechanism of Crystallization ProcessThere are two basic steps in the over-all process of crystallization from supersaturated solution:(1) NUCLEATIONa. 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 interfaceb. 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 agitationc. Secondary Nucleation occurs due to the presence of crystals of the crystallizing species

(2) CRYSTAL GROWTH a layer-by-layer processa. Solute diffusion to the suspension-crystal interfaceb. 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 CrystalsMagmaIt is the two-phase mixture of mother liquor and crystals of all sizes, which occupies the crystallizer and is withdrawn as product

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 pressureExpressions 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

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 COOLINGCrystallizers 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 substancesPan CrystallizersBatch 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 productAgitated batch CrystallizersConsist 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 flexibilitySwenson Walker CrystallizerA 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

SUPERSATURATION BY EVAPORATION OF SOLVENTCrystallizers 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 EvaporatorThe 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 CrystallizerModern 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 bodyCALCULATIONS: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

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 SEEDINGL 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

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 200F is pumped continuously to a vacuum crystallizer from which the magma is pumped at 60F. What is the composition of this magma, and what percentage of Na2SO4 in the feed is recovered as Na2SO410H2O crystals after this magma is centrifuged?

SOLUTION:Basis: 100 lb feedFrom table 2-122 (CHE HB), solubility of Na2SO410H2OT,C101520

g/100 g H2O9.019.440.8

Consider over-all material balance:

Consider solute balance:

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

Substitute in

Magma composition:

% Recovery:

PROBLEM # 02:

A solution of 32.5% MgSO4 originally at 150F is to be crystallized in a vacuum adiabatic crystallizer to give a product containing 4,000 lb/h of MgSO47H2O crystals from 10,000 lb/h of feed. The solution boiling point rise is estimated at 10F. 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 TEMPERATURE1. Assume temperature of the solution2. From figure 27-3 (Unit Operations by McCabe and Smoth 7th edition), obtain mass fraction of MgSO4 at the assumed temperature of the solution3. Solve for L using equation 4. Solve for V using equation 5. Check if assumed temperature is correct by conducting enthalpy balancea. Obtain values of hF, hC and hL from figure 27-4 (Unit Operations by McCabe and Smith 7th edition) at the designated temperatures and concentrationsb. Compute for hVc. Using the enthalpy balance equation, compute for V using the value of L from step 36. Compare values of V from step 4 with that from step 5-c7. If not the same (or approximately the same), conduct another trial and error calculations

TRIAL 1: Assume temperature of the solution at 60FFrom figure 27-3 (Unit Operations by McCabe and Smith 7th edition)

Substitute to equation


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

Temperature of vapor is 60 10 = 50F

From steam table at 50F,

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

Product temperature

Operating PressureFrom steam table for vapor temperature of 50F

Ratio of mother liquor to crystalline product

PROBLEM # 03 :

A plant produces 30,000 MT of anhydrous sulfate annually by crystallizing sulfate brine at 0C, 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 0C

From table 2-120 (CHE HB), solubility at 0C:

PROBLEM # 04 :

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

PROBLEM # 05:

A Swenson-Walker crystallizer is to be used to produce 1 ton/h of copperas (FeSO47H2O) crystals. The saturated solution enters the crystallizer at 120F. The slurry leaving the crystallizer will be at 80F. Cooling water enters the crystallizer jacket at 60F and leaves at 70F. It may be assumed that the U for the crystallizer is 35 BTU/hFft2. There are 3.5 ft2 of cooling surface per ft of crystallizer length.a) Estimate the cooling water requiredb) Determine the number of crystallizer section to be used.Data: specific heat of solution = 0.7 BTU/lbF; heat of solution= 4400 cal/gmol copperas; solubility at 120F = 140 parts copperas/100 parts excess water; solubility at 80F = 74 parts copperas/100 parts excess water


Consider copperas (FeSO47H2O) balance:

Equate and

Consider heat balance:

PROBLEM # 06:

Crystals of Na2CO310H2O are dropped into a saturated solution of Na2CO3 in water at 100C. What percent of the Na2CO3 in the Na2CO3H2O is recovered in the precipitated solid? The precipitated solid is Na2CO3H2O. Data at 100C: the saturated solution is 31.2% Na2CO3; molecular weight of Na2CO3 is 106

SOLUTION:Assume 100 g of Na2CO310H2O added into the saturated solution

PROBLEM # 07:

A solution of MgSO4 at 220F containing 43 g MgSO4 per 100 g H2O is fed into a cooling crystallizer operating at 50F. 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 MgSO47H2O per hour?


Consider MgSO4 balance

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

Equate and

PROBLEM # 08:

The solubility of sodium bicarbonate in water is 9.6 g per 100 g water at 20C and 16.4 g per 100 g water at 60C. If a saturated solution of NaHCO3 at 60C is cooled to 20C, what is the percentage of the dissolved salt that crystallizes out?

SOLUTION:Basis: 100 kg feed


Consider NaHCO3 balance

Equate and

PROBLEM # 09:Glaubers salt is made by crystallization from a water solution at 20C. The aqueous solution at 20C 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 20C, there will be crystallized out 80% of the original sodium sulfate as Glaubers 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 Na2SO4 balance


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 waterb) 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 290KConsider over-all material balance:

If water evaporated is 10% of the original water present


Equate and

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

If water evaporated is 10% of the original water present


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

d) Crystallization with evaporationConsider 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 30C, and 13.5 parts at 15C. The latent heat of crystallization (liberated when crystals form) is 18,000 g-cal per gmol Na2SO4. Glaubers salt (Na2SO410H2O) is to be made in a Swenson-Walker crystallizer by cooling a solution, saturated at 30C, to 15C. Cooling water enters at 10C and leaves at 20C. The over-all heat transfer coefficient in the crystallizer is 25 BTU/hft2F 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 Glaubers Salt

Source: Unit Operations (Brown)


Consider Na2SO4 balance

Equate and


From Table 2-194 (CHE HB 8th edition)

PROBLEM # 13:

A continuous adiabatic vacuum crystallizer is to be used for the production of MgSO47H2O crystals from 20,000 lb/h of solution containing 0.300 weight fraction MgSO4. The solution enters the crystallizer at 160F. The crystallizer is to be operated so that the mixture of mother liquor and crystals leaving the crystallizer contains 6,000 lb/h of MgSO47H2O crystals. The estimated boiling point elevation of the solution in the crystallizer is 10F. How many pounds of water are vaporized per hour?

Source: Unit Operations (Brown)



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 TEMPERATURE1. Assume temperature of the solution2. From figure 27-3 (Unit Operations by McCabe and Smoth 7th edition), obtain mass fraction of MgSO4 at the assumed temperature of the solution3. Solve for L using equation 4. Solve for V using equation 5. Check if assumed temperature is correct by conducting enthalpy balancea. Obtain values of hF, hC and hL from figure 27-4 (Unit Operations by McCabe and Smith 7th edition) at the designated temperatures and concentrationsb. Compute for hVc. Using the enthalpy balance equation, compute for V using the value of L from step 36. Compare values of V from step 4 with that from step 5-c7. If not the same (or approximately the same), conduct another trial and error calculations

TRIAL 1: Assume temperature of the solution at 60FFrom figure 27-3 (Unit Operations by McCabe and Smith 7th edition)



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

Temperature of vapor is 60 10 = 50F

From steam table at 50F,

Since % error is about 5%, assumed value can be considered correct.

PROBLEM # 14:

Crystals of CaCl26H2O 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 100F and receives 250 BTU/lb of feed from the external heater. Products are withdrawn from the crystallizer at 40F.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 CaCl26H2O 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 feedfrom table 2-120 (CHE HB 8th edition), solubilities of CaCl26H2O0C59.5 lb/100 lb H2O

10C65 lb/100 lb H2O

20C74.5 lb/100 lb H2O

30C102 lb/100 lb H2O

At 100F (37.8C), solubility is (by extrapolation), 123.45 lb/100 lb H2OAt 40F (4.4C), solubility is 61.92 lb/100 lb H2OSince 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/Kmol

where T is in KAt 100F (310.93 K)

At 40F (277.59 K)

For the feed

From table 2-224 (CHE HB 8th edition), heat of solution of CaCl26H2O = -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 40F,


Substitute in

Consider solute (CaCl26H2O) balance, inert soluble-free

Equate and

Composition of the liquor (including the inert soluble)

lb%

CaCl26H2O0.00564.89

H2O0.00907.85

inerts0.100087.26

0.1146100.00

For the crystals leaving the centrifuge:

Composition of crystals leaving the centrifugelb

CaCl26H2O

crystallized0.7620

from liquor0.0762 x 0.0489 0.00370.7657

H2O0.0762 x 0.07850.00600.0060

inerts0.0762 x 0.87260.06650.0665

0.8382

In the dryer, assume all free water has been removed

Composition of dried crystalslb%

CaCl26H2O0.765792.01

inerts0.06657.99

0.8322100.00

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 57C to 10C. Lactose crystallizes with one molecule of water of crystallization. The specific heat of the lactose solution is 3470 J/kgC. 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 10C 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


At 10C (50F),

PROBLEM # 16:

Sal soda (Na2CO310H2O) is to be made by dissolving soda ash in a mixture of mother liquor and water to form a 30% solution by weight at 45C and then cooling to 15C. 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 15C, 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 10C, and sufficient cooling water is to be used to ensure that the exit water will not be over 20C. 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/ft2hF is expected.

The latent heat of crystallization of sal soda at 15C is approximately 25,000 cal/mol. The specific heat of the solution is 0.85 BTU/lbF. 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 sodaConsider over-all material balance of the system

Consider Na2CO3 balance around the system


Consider solute (Na2CO3) balance around the dryer

Consider over-all material balance around the dryer


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 Na2S2O35H2O is to be crystallized per hour by cooling a solution containing 56.5% Na2S2O3 to 30C 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 30C, solubility of Na2S2O3 is 83 parts per 100 parts waterSource: 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 110F. The solution and its crystalline crop are cooled to 40F. 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 feedConsider over-all material balance


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

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

Equate and



PROBLEM # 19:

Trisodium phosphate is to be recovered as Na3PO412H2O from a 35 weight % solution originally at 190F 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 RangeSize 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/lbF.a) Estimate the product particle size distributionb) 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 error1. Assume value of 2. Solve for for each size range, use the mean for each size range3. Solve for 4. Get the total 5. If , then assumed is correct; if not, redo another trial

TRIAL 1: Assume

Since % error is less than 5%, assumed value can be considered

For particle size distribution:

Size Range, inWt %Size Range, inWt %


Consider Na3PO4 balance:

From table 2-120 (CHE HB 8th edition)50C43 lb/100 lb H2O

60C55 lb/100 lb H2O

Cooling Duty:


PROBLEM # 20:How much CaCl26H2O must be dissolved in 100 kg of water at 20C to form a saturated solution? The solubility of CaCl2 at 20C 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 CaCl26H2O required

4. Weight CaCl26H2O required

5. Composition of the solution in terms of CaCl26H2O

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 CaCl26H2O required for every 100 kg free water (net of water of hydration)

Crystallization Notes (1)

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