Vdocuments.site_freetee.pptx

*

Crystallization: the formation of solid particles within a homogeneous phase. Example: 1.The freezing of water to form ice 2.The formation of snow particles from a vapor 3.The formation of solid particles from a liquid melt 4.The formation of solid crystals from a liquid solution Types of crystal geometry Crystal  a solid composed of atoms, ion, or molecules, which are arranged in an orderly and repetitive manner. Crystals are divided into seven classes based on the arrangement of the axes to which the angles are referred 2

3

4

Equilibrium Solubility in Crystallization

5

Yields and Heat and Material Balances in Crystallization In crystallization process solution (mother liquor) and the solid crystals are in contact for a long enough time to reach equilibrium

mother liquor is saturated at the final temperature of the process, and the final concentration of the solute in the solution can be obtained from the solubility curve.

By knowing the initial concentration of solute, the final temperature, and the solubility at this temperature, the yield of crystals from crystallization process6 can be calculated.

Example 12.11-1 A salt solution weighing 10000 kg with 30 wt% Na2CO3 is cooled to 293K (20 oC) 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.

Solution: W kg H2O

10,000 kg solution 30% Na2CO3

Cooler and crystallizer

S kg solution 21.5 kg Na2CO3/100 kg H2O

C kg crystals Na2CO3.10H2O 7

Solution example 12.11-1: (a) Assume that no water is evaporated  W=0 kg Material balance for water: 0.7(10,000) 

100 180.2 S C  xH 2OW 100  21.5 286.2

Material balance for Na2CO3:

0

0

21.5 106 0.3(10,000)  S C  xH 2OW 100  21.5 286.2

Solving Eq. 1 and 2 simultaneously, resulting: C = 6370 kg of Na2CO3.10H2O crystals S = 3630 kg solution

8

(1)

(2)

Solution example 12.11-1: (b) Assume that 3% of the total weight of the solution is lost by evaporation of water in cooling  W=0.03(10,000) = 300 kg H2O Material balance for water: 0.7(10,000) 

100 180.2 S C  300 100  21.5 286.2

Material balance for Na2CO3: 0.3(10,000) 

(1)

0 (no salt in stream W) (2) W

21.5 106 S C  xH 2O 100  21.5 286.2

Solving Eq. 1 and 2 simultaneously, resulting: C = 6630 kg of Na2CO3.10H2O crystals S = 3070 kg solution 9

Heat Effect and Heat Balances in Crystallization Heat of solution: absorbed heat during the dissolution of compound due to the increasing of solubility as temperature increases In crystallization  Heat of crystallization = - Heat of solution Total heat absorbed (q) in kJ is:

q  H 2  HV   H1 Enthalpy of water vapor Enthalpy of the final mixture of crystals and mother liquor at final temp

10

Enthalpy of the entering solution at the initial temp

System

q (+)

System

q (-)

Example 12.11-2 A feed solution of 2268 kg at 327.6 K (54.4 oC) containing 48.2 kg MgSO4/100 kg total water is cooled to 293.2 K (20 oC), where MgSO4.7H2O crystals are removed. The solubility of the salt is 35.5 kg MgSO4/100 kg total water. The average heat capacity of the feed solution can be assumed as 2.93 kJ/kg.K. The heat of solution at 291.2 K (18 oC) is -13.31×103 kJ/kg mol MgSO4.7H2O.  Calculate the yield of crystals and make a heat balance to determine the total heat absorbed, assuming that no water is 11 evaporaized!

Solution example 12.11-2: W kg H2O

2268 kg solution 48.2 kg MgSO4/100 kg total water T=327.6 K

Cooler and crystallizer

S kg solution 35.5 kg MgSO4/100 kg total water

C kg crystals MgSO4.7H2O T=293.2 K

Assume that no water is evaporated  W=0 kg Material balance for water: 100 2268  100 S  126 C  xH 2OW 100  48.2 100  35.5 246

0 (1)

Material balance for Na2CO3:

0 48.2 35.5 120 2268  S C  xH 2OW 100  48.2 100  35.5 246 12

(2)

Solution example 12.11-2: Solving Eq. 1 and 2 simultaneously, resulting: C = 616.9 kg of MgSO4.7H2O crystals S = 1651.1 kg solution H1  mc p T

Tref

H1  2268(2.93)(327.6  293.2)  228,600kJ

Heat of solution = -13.31×103 kJ/kg mol MgSO4.7H2O = -13.31×103 /246 = -54 kJ/kg crystals Heat of crystallization = - Heat of solution = + 54 kJ/kg crystals = 54 (616.9) = 33,312 kJ

Assume: heat of crystallization at 291.2 K = heat of crystalization at 293.2 K 13

Solution example 12.11-2: 0 (No water is vaporized)

q  H 2  HV   H1 0 (Tfinal=Tref)

q = -H1 - heat of crystallization q = -228,600 – 33,312 = -261,912 kJ

14