3 crystallization
DESCRIPTION
crystallizationTRANSCRIPT
OBJECTIVES
At the end of this lesson, students should be able to: Define crystallization process Explain the equilibrium solubility in crystallization Describe the supersaturation in crystallization Explain theory of crystallization Describe equipments in crystallization
12.11 INTRODUCTION (page 817)
Crystallization is a solid-liquid process, in which mass transfer of a solute from the liquid solution to a pure solid crystalline occurs.
The formation of solid particles within a homogeneous phase.
It can occurs in:
- the freezing of water to form ice
- the formation of snow particles from a vapor
- the formation of solid particles from a liquid melt
- the formation of solid crystals from a liquid solution
INTRODUCTION
In industrial crystallization from solution, the two phase mixture of mother liquor and crystals which occupies the crystallizer is withdrawn as product called magma.
In commercial crystallization, it is important to get good yield, high purity of crystals and uniform shape and size of the crystals.
Size uniformity is desirable:- to minimize caking in the package- for ease of pouring- for ease in washing and filtering- for uniform behavior when used
TYPES OF CRYSTAL GEOMETRY
A crystal is a solid composed of atoms, ions or molecules which are arranged in an orderly and repetitive manner.
It is highly organized type of matter. The atoms, ions or molecules are located in 3-dimentional
arrays (space lattices). Crystals appear as polyhedrons having flat faces and sharp
corners.
TYPES OF CRYSTAL GEOMETRY
The relative sizes of the faces and edges of different crystals of the same material may differ greatly.
The angles between the corresponding faces of all crystals of the same material are equal and are characteristic of that particular material.
Crystals are classified on the basis of these interfacial angles.
CRYSTAL GEOMETRY
7 classes of crystals depending on the arrangement ofthe axes to which the angles are referred:
1. Cubic system2. Tetragonal system3. Orthohombic system4. Hexagonal system5. Monoclinic system6. Triclinic system7. Trigonal system
Triclinic e.g. microcline
Orthorhombic e.g. aragonite
12.11B - EQUILIBRIUM SOLUBILITY IN CRYSTALLIZATION
Equilibrium is attained when the solution (mother liquor) is saturated and represented by a solubility curve.
Solubility is dependent mainly on temperature and pressure has a negligible effect on solubility.
Curve showing solubility as a function of temperature. Solubility of most salt increase slightly or markedly with
increasing of temperature.
EQUILIBRIUM SOLUBILITY IN CRYSTALLIZATION
Figure 3.1: Example of solubility curve
A general solubility curve for a solid that forms hydrate (a compound that has one or more water molecules attached) as it cools is shown in Figure 3.1.
EQUILIBRIUM SOLUBILITY IN CRYSTALLIZATION
Figure 12.11-1: Solubility of sodium thiosulfate in water
EQUILIBRIUM SOLUBILITY IN CRYSTALLIZATION
Figure 3.3: Solubility of several solids
12.11C - YIELDS IN CRYSTALLIZATION
In industrial crystallization, the solution (mother liquor) and the solid crystals are in contact for a long enough time to reach equilibrium.
The mother liquor is saturated at the final temperature of the process.
The final concentration of the solute in the solution can be obtained from the solubility curve.
YIELDS IN CRYSTALLIZATION
The yield of the process can be calculated from the initial concentration of solute and the solubility at the final temperature.
When the rate of crystal growth is slow (due to the very viscous solution or small surface of crystals exposed to the solution), considerable time is required to reach equilibrium
The final mother liquor may retain supersaturation and the actual yield will be less than calculated from the solubility curve.
YIELDS IN CRYSTALLIZATION
If the crystals are anhydrous, the calculation of the yield is simple.
When the crystals are hydrated, some of the water in the solution is removed with the crystals as a hydrate.
Solubility data are given either in
- parts by mass of anhydrous material per 100 parts by mass of total solvent
- mass fraction anhydrous solute These data ignore water of crystallization.
Example 12.11-1 (Geankoplis)Yields and mass balances in crystallization
A salt solution weighing 10 000kg with 30wt. % Na2CO3 is cooled to 293 K (200C). The salt crystallizes as the decahydrate. What will be the yield of Na2CO3.10H2O crystals if the solubility is 21.5 kg anhydrous Na2CO3/100 kg of total water? Do this for the following cases.
a) Assume that no water is evaporated.b) Assume that 3% of the total weight of the solution is lost by evaporation of water in cooling.
MW Na2CO3 =106.0 kg/kg mol
MW 10H2O = 180.2
Mass balance :Water balance : Input = OutputSolute balance : Input = Output
10 000 kg Hot solution
30% Na2CO3
C kg Crystals, Na2C03.10H2O
21.5 kg Na2C03/100 kg H2O
S kg solution
W kg H2O evaporate
SolutionSolution
12.11C HEAT BALANCE IN CRYSTALLIZATION
When a solubility of compound increases as temperature increases during dissolving, an absorption of heat is called heat of solution.
When a solubility of compound decrease as temperature increases during dissolving, an evolution of heat occurs.
When a solubility of compound does not changes as temperature increases during dissolving, there is no neat of evolution on dissolution.
In crystallization the opposite of dissolution occurs. At equilibrium, the heat of crystallization is equal to the negative of the heat of solution at the same concentration in solution.
The most satisfactory method of calculating heat effects during a crystallization process is to use the enthalpy-concentration chat. However only a few charts are available.
ENTHALPY BALANCES IN CRYSTALLIZATION
In heat balances calculation, the heat of crystallization is important.
This is the latent heat evolved when solid forms a solution.
The heat of crystallization is varies with temperature and concentration.
It is equal to the heat absorbed by crystals dissolving in a saturated solution.
ENTHALPY BALANCES IN CRYSTALLIZATION
The total heat absorbed is
Where q = total heat absorbed (kJ)
H1 = enthalpy of entering solution at the initial temperature (kJ/Btu)
H2 = enthalpy of the final mixture of crystals and mother liquor at the final temperature (kJ/Btu)
Hv =enthalpy of vaporizationHs = heat of solution (kJ/kg mol)
If q (+ ve) = heat must added to the systemq (-ve) = heat is evolved or given off.
12 )( HHHq v (12.11-4)
Tc
H2OTc,Hv (enthalpy of evaporation) kJ
Hot solutionTf (Initial temperature)
H1 (enthalpy of entering solution ) kJ
Solute/solvent Tc, H3
Total Heat Absorbed
q = (H2 + H3 + Hv) – H1
Product
Tc (final temperature)
H2 (enthalphy of the final mixture of crystals and mother liquor) kJ
Example 12.11-2
A feed solution of 2268 kg at 327.6 K (54.40C) containing 48.2 kg MgSO4/100 kg total water is cooled to 293.2 K (200C), where MgSO4.7H2O crystals are removed. The solubility of the salt is 35.5 kg MgSO4/100 kg total water (P1). The average heat capacity of the feed solution can be assumed as 2.93 kJ/kg.K (H1). The heat of solution at 291.2 K (180C) is -13.31 x 103 kJ/kg mol MgSO4.7H2O (P1). Calculate the yield of crystals and make a heat balance to determine the total heat absorbed, q, assuming that no water is vaporized.
MW MgSO4 = 120.35 kg/kgmolMW 7H2O = 126.14 kg/kgmol
Solution Make a water and a MgSO4 balances.
solutions kg 1664S
crystals kg 617C
usly,simultaneo equations theSolve
0.262
0.488C737.1S 2, From
2 0.488C0.262S)2268(0.325
C246.49
120.35S
10035.5
35.5)2268(0.325
balance MgSO
1 0.5117C0.7351530.4
C246.49
126.14S
10035.5
100)2268(0.674
balance OH
0.674fractionWater
0.32548.2100
48.2fraction Crystal
4
2
MW MgSO4 = 120.35 kg/kgmolMW 7H2O = 126.14 kg/kgmol
