Crystallization Tutorial

7/31/2019 Crystallization Tutorial

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CAPEC, Department of Chemical Engineering, DTU, Lyngby, Denmark

Generation of a batch recipe for acrystallization process and verification

through dynamic simulation

The following tutorial contains the steps throughICAS for obtaining the solid-liquid phase diagram,

generating the batch recipe based on it and verifyingthe recipe through dynamic simulation set up in ICAS

Irene Papaeconomou & Rafiqul Gani


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Application of the operational design algorithm for

the generation of a batch recipe to separate NaCland KCl from a mixture with water.

Problem Definition:

A mixture of water, NaCl and KCl with a feed composition

(on a 100 kg feed basis): water 80%, NaCl 15% and KCl5% is to be separated.

Objective:

To recover 95% of the dissolved sodium chloride (NaCl).

The operating range of temperature for the crystallizers is 0oC to 100 oC.


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Given:Identities of the solvent and the solids

Composition of the feedTemperature range

Draw the solubil ity curves for the range oftemperature on the ternary equilibrium diagram

According to feed location on the ternary diagram

and the specified product, check the feasibil ity ofthe precipitation of the solid

i=1, , N

Solid crystallized:Compute the product yieldDesired yield achieved?

Identify the solid to precipitate and select the operation task(choose temperature, evaporation/dilution ratio (or mixing ratio)

Solid/Liquidseparation

Treat MotherLiquor as newfeed

YES

Above operation path is converted tothe operational steps for a batch

crystallizer

NO

Solid/Liquidseparation

Algorithm for the operational modelling of batch crystallization


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How to produce a phase diagram with thehelp of ICAS

Step 1: Select thecompounds for the

specific problem.

Choose the compounds from thesalts database.

The first part of the algorithm is the generation of the phase diagram for therelevant system.


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General Note: Even though the main components are water, sodium-chlorideand potassium-chloride, some extra compounds should be included in thecomponent list, since they might precipitate at various temperatures.

These extra compounds are the double salts of the two single salts and theirhydrates (mono-, dihydrates).If these compounds are not included in the compound list, then even if they doprecipitate at various temperatures, it wont be possible to be identified.

Step 2: Select the thermodynamic model (use the default model for salts)Adjust the measure units to your problem.(kg for mass and oC for temperature)

Step 3: Draw the stream

Doubleclicking on a feed stream brings up the mixture specification dialog

box, where one defines the feed stream. The temperature and pressure are100 oC and 1 atm, respectively. The composition of stream 1 is water, 80kg/hr, NaCl 15 kg/hr and KCl 5 kg/hr.


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From the mixture specification dialog box, one can be directed to thethermodynamic property calculations window. There, a number of calculationsare available for the mixture and solubility is one of them. This is how the phasediagram is created for a range of temperatures

Selecting water, NaCland KCl and plottingsolubility in a tringulardiagram at a range of0 to 100 degrees(multiple curves), we

only need to run thesimulation

To visualize the resultswe just need to press

on the plotting button


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From the plotting button we getthe diagram on the right.

The actual point values for thesolubility curves are hidden onthe left of the diagram as itappears below.

These values can be

copied and pasted in anexcel file and further onmanipulated.The result could be a plot

as the one showed in thenext slide.


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The point values for the solubility curves in ICAS are given in rectangularcoordinates X and Y and not in triangular coordinates x, y and z.

In order to transform triangular coordinates to rectangular, one needs to do thefollowing:

If x is the composition of the compound on the top of the triangle, y thecomposition of the compound on the left of the triangle and z the composition ofthe compound on the right of the triangle, then X, Y and x, y, z are connectedlike that.

X = x/2 + zY = x

So for the Feed (x=80, y=15, z=5), it isX = 80/2 + 5 = 45 and Y = 80


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0

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90

100

0 10 20 30 40 50 60 70 80 90 100NaClKC

H2O

0oC

100oC

Feed1

20oC

60 oC

NaCl-KCl-H2O phase diagram (various temperatures)


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0

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50

60

70

80

90

100

0 10 20 30 40 50 60 70 80 90 100NaClKCl

H2O

100oC

Feed1

Invariant point

The position of the feed is in theunsaturated region, which is theupper part of the triangle defined

by the solubility curves. Below thesolid-liquid equilibrium curves, oneor two salts precipitate dependingon the position in the triangulardiagram. In the area defined by thNaCl vertex of the triangle, the

invariant point, part of thesolubility curve and the NaCl-

H2O axis, the only salt

precipitating is NaCl. In a similaarea on the right side of the

triangle, the only saltprecipitating is KCl. In the

area defined by theNaCl, KCl vertices andthe invariant point,both salts precipitat

NaCl

precipitates

KClprecipitates

Both NaCl and KCl precipitate

Unsaturated region


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Check of the feasibility of the precipitation of a specified salt

Let (XF, YF) be the rectangular coordinates of the feed and(Xinv, Yinv) the coordinates of the invariant point at a specific temperature.

Then for the case whena) YF > Yinv

if < (1)

then the salt placed in the bottom left vertex of the triangle will precipitate.

If the X described above is greater than Xinv, then the salt in the bottom right vertex ofthe triangle will precipitate.If X equals Xinv, then both salts will precipitate.

and when

b) YF < Yinv

In addition to the constraint above, the following constraint has to be satisfied:

> (2)

for the salt in the bottom right vertex of the triangle to precipitate. Otherwise both saltswill precipitate.

F

invFFinv

Y

YXYYX

=

100

)100(*)(*50inv

X

FY

F

inv

invX

X

Y*


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For the specific problem, the precipitation of the desired salt (NaCl) is feasible for all thetemperatures in the given operating range (constraint (1) is satisfied).So crystallization at any temperature in the range will give NaCl.

Since the feed lies in the unsaturated region, evaporation will have to be applied. Aswater is removed, the composition of the feed moves away from the H2O vertex alongthe line connecting the water vertex and the feed.

0

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50

60

70

80

90

100

0 10 20 30 40 50 60 70 80 90 100NaClKCl

H2O

0oC

100oC

Feed1

SL1


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As it can be seen from the phase diagram, the area where NaCl solely precipitates islarger for the temperature of 100 oC. For that temperature, evaporation can be moreextensive and the resulting slurry has a larger density (slurry density = mass of salts /

mass of slurry = length Slurry_Mother Liquor / length NaCl_Mother Liquor).

So, the operating temperature for the crystallization is chosen to be a 100 oC, for theabove mentioned reasons.

The evaporation ratio is chosen as follows:As mentioned previously, as water is removed, the composition of the feed moves awayfrom the H2O vertex along the line connecting the water vertex and the feed. However,there is a maximum evaporation that can be achieved without crossing the boundary

into the area where both salts precipitate (unwanted).The composition of the slurry on that boundary is found as the intersection of theevaporation line and the line connecting the NaCl vertex and the invariant point at a 100oC.

The actual slurry composition is found for about 99 % of the maximum evaporation.

The amount of water evaporated is found as described next.


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0

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100

0 10 20 30 40 50 60 70 80 90 100NaCl

KCl

H2O

0 oC

100 oC

Feed1

1

2

3

NaCl-KCl-H2O in triangular coordinates: Evaporative crystallization

at 100 oCOn a 100 kg feed basis

Evaporate feed solution(Feed1). Saturationoccurs at pt 1, on curve.Continue evaporation

until slurry compositionat pt 2.Amount of solventevaporated:

Length Feed1_pt.2 /Length H20_pt.2 * 100

kg feed= 65.278 kg H20

Remaining at point 2:100-65.278=34.722 kgslurry.Reaching pt. 2 , we haveonly removed water.

Pt. 2: 80-65.278=14.722kg H2O, 15 kg NaCl and5 kg KCl


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Slurry at 2 is NaCl +liquor at pt 3.Pt. 3 is found as theintersection of the lineconnecting NaCl and the

slurry (pt.2) and thesolubility curve at a 100oCSlurry density:Length 2_3 / Length

NaCl_3 = 0.3176Solids precipitated:0.3176 * 34.722 =11.028kg NaClYield =11.028/15=73.52%< desired yield = 95%

Pt 3: 34.722 - 11.028 =23.694 kg Mother Liquor.

Moving from pt.2 to pt. 3we have only removedNaCl.0

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100

0 10 20 30 40 50 60 70 80 90 100NaCl

KCl

H2O

0 oC

100 oC

Feed1

1

2

3

NaCl-KCl-H2O in triangular coordinates: Evaporative crystallization

at 100 oC


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0

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100

0 10 20 30 40 50 60 70 80 90 100NaCl

KCl

H2O

0 oC

100 oC

Feed1

1

2

3

4

NaCl-KCl-H2O in triangular coordinates: Cooling crystallization at 0 o

Second crystallizer at 0oC.

Mother Liquor (pt 3) fromfirst crystallizer is in thesaturated region for KCl at0 oC. (no evaporation

needed)

Slurry at 3 is KCl + liquorat pt 4.

Pt 4 is found similarly to pt3.Slurry density:Length 3_4 / Length KCl_4= 0.1322

Solids precipitated:0.1322 * 23.694 =3.132 kg

KClPoint 4: 23.694 - 3.132 =

20.562 kg mother liquor

Th th li i ilib i ith th i it t d lt i th d t lli ( t 4)


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The mother liquor in equilibrium with the precipitated salt in the second crystallizer (pt. 4)is treated as a new feed.The precipitation of NaCl is feasible for a temperature range of 20 to 100 oC. For thesame reasons as in the first sub-task, the operating temperature is chosen to be T=100o

The new feed (pt 4) lies in the unsaturated region and evaporation needs to be applied.The evaporation ratio is calculated such, so that the resulting slurry is in equilibrium with

mother liquor of a composition ofpt 3 (ML1).

This means that the resultingslurry has a composition of theintersection of evaporation lineconnecting points 4 and the

H2O vertex and of the line

connecting the NaCl vertexand point 3 (ML1).


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Amount of solvent evaporated: Length pt.4_pt.5 / Length H20_pt.5 * 20.562 kg feed (pt4)= 9.209 kg H20(From 4 --> 5): Remaining at point 5: 20.562 - 9.209 = 11.353 kg slurry.

Slurry density: Length 5_3 / Length NaCl_3 = 0.2179Solids precipitated: 0.2179 * 11.353 =2.474 kg NaClYield =(11.028+2.474)/15=90.01% < desired yield = 95%


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A repeated procedure, moving on the phase diagram from point 3 to 4 to 5 and again to3, resulting in an alternate precipitation of NaCl and KCl at a 100 oC and 0 oCrespectively, is applied.

In summary, the operational path generated to recover 95 % of the dissolved NaCl is thefollowing:

Crystallizer TemperatureoC Operational task Predicted Prod. yield

1 100 Evaporation

NaCl 65.278 kg 73.60%

2 0 Cooling

KCl crystallization 62.55%

1 100 Evaporation

NaCl 9.209 kg 90.09%

2 0 Cooling

KCl crystallization 85.95%1 100 Evaporation

NaCl 3.145 kg 96.28%>95%

V ifi ti f th t d i th h d i i l ti


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Verification of the generated recipe through dynamic simulationset up in ICAS

A crystallizer needs to be added to the icas file that was created in order toobtain the solid-liquid phase diagram for the ternary system.The unit CRYST is addedand 3 streams are taken from the top, right and bottom of the unit.

The top stream is the vapour leaving the crystallizer (solvent removed). Thestream leaving from the right side of the unit is in the liquid form, which is themother liquor in equilibrium with the precipitated solid of the bottom streamleaving the crystallizer.

In addition t o tanks are added in the flo sheet to collect the liq id and the


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Doubleclicking on the crystallizer, brings upthe unit specification dialog box for this unit.From the Property values button: thetemperature of the crystallization is set (sub-task 1: precipitation of NaCl T=100 oC).

In addition, two tanks are added in the flowsheet to collect the liquid and thesolid, after their separation from the desired slurry.


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Before the dynamic simulation engine can be started the initial holdup file has


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Before the dynamic simulation engine can be started, the initial holdup file hasto be created. Invoking the dynamic simulator from the main frame of ICAS, theoption to generate holdup information is chosen first.

The generated dynsimb.dat file should look like that.

0H- H+

equivelent in kmoles

of 80 kg of H2O

Na+ K+Cl-


of 15 kg of NaCl


of 5 kg of KCl

initial tanks holdup

Cl- = Na+ + K+


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Before the simulation is run through DYNSIM, the flowrate of the feed stream inthe crystallizer is set to zero. This is because the unit is already filled with thefeed mixture (initial holdup file: dynsimb.dat).

The simulation time depends on the evaporation rate specified in the crystallizerSince the evaporation rate was specified to 3.6235 kmoles/hr = 65.278 kg/hr,we only need to run the simulation for one hour. (160 time steps, pfw=0.01)

After the end of this run, the composition in the crystallizer will be the desiredslurrys composition.

The precipitation of the salt (solid/liquid separation) is simulated by changing


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The precipitation of the salt (solid/liquid separation) is simulated by changingonly the following specifications in the crystallizer:Vaporization is set to 0 kmoles/hr

and the valve is opened (valve constant = 1500).The resulting holdup file from the previous vaporization is used as an initialholdup file for the precipitation.The simulation time is such that the content of the crystallizer (mother liquor +

salt) is transferred to the two tanks. (t=0.67 hr, No of steps 127, pfw=0.01)

For the precipitation of KCl (Cooling crystallization at 0 oC), one uses as initialholdup the resulting holdup file from the previous precipitation of NaCl, where

the previous mother liquor holdup is now the content of the crystallizer and thetanks are empty and T=0 oC.Vaporization is still set to 0 kmoles/hrand the valve is opened (valve constant = 1500). (Because the content of the

crystallizer (previous ML) is in the saturation region of KCl for 0oC)

The difference in the input is that the temperature of the crystallizer is 0 oC.The simulation time is such that the content of the crystallizer (mother liquor +salt) is transferred to the two tanks. (t=0.53 hr, No of steps 113, pfw=0.01)

The next step is crystallization at 100 oC which will result to the precipitation


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The next step is crystallization at 100 oC, which will result to the precipitationof NaCl.

The resulting holdup file from the previous precipitation of KCl (where the

previous mother liquor holdup is now the content of the crystallizer, thetanks are empty and T=100oC) is the initial holdup.Only the following specifications in the crystallizer are changed:The evaporation rate is specified to 0.511 kmoles/hr = 9.209 kg/hr and we only

need to run the simulation for one hour. (160 time steps, pfw=0.01)After the end of this run, the composition in the crystallizer will be the desiredslurrys composition.

The second time precipitation of NaCl is simulated similarly as described

before. The same applies for the following precipitation of KCl and the lastevaporation and precipitation of NaCl.

The icas files for this tutorial are gathered in 8 steps in the zip file that is

attached.1) Evaporation at 100 oC (Evap1.ics) 5) Precipitation of NaCl at 100 oC (NaCl2.ics)2) Precipitation of NaCl at 100 oC (NaCl1.ics) 6) Precipitation of KCl at 0 oC (KCl2.ics)3) Precipitation of KCl at 0 oC (KCl1.ics) 7) Evaporation at 100 oC (Evap3.ics)

4) Evaporation at 100 oC (Evap2.ics) 8) Precipitation of NaCl at 100 oC (NaCl3.ics)

To run the simulationyourself , youll have to replace the dynsimb.dat filepresent in the directory with the one from the corresponding directory Initialh ld

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Crystallization Tutorial