# Tutorial 3 - Crystallization

August 7, 2017 | Author: Ahmad Muzammil | Category: Crystallization, Chemistry, Physical Chemistry, Physical Sciences, Science

#### Short Description

crystallization tutorials...

#### Description

Mass Transfer 2

CPB 20103

CPB 20103 MASS TRANSFER 2 TUTORIAL 3 CRYSTALLIZATION OPERATIONS 1. A hot solution of FeSO4 from an evaporator contains 30.6 kg FeSO4/100 kg H2O and goes to a crystallizer where the solution is cooled and FeSO4 crystallizes. On cooling, 10% of the original water present evaporates. For the feed solution of 100 kg total, calculate the following: a) The yield of FeSO4.7H2O crystals if the solution is cooled to 290 K where the solubility is 8.6 kg FeSO4/100 kg H2O. b) The yield of FeSO4.7H2O crystals if the solution is cooled instead to 283 K where the solubility is 7.0 kg FeSO4/100 kg H2O. 2. A hot solution containing 1000 kg of MgSO4 and water having a concentration of 30 wt% MgSO4 is cooled to 288.8 K where crystals of MgSO4.7H2O are precipitate. The solubility at 288.8 K is 24.5 wt% anhydrous MgSO4 in the solution. Calculate the yield of crystals obtained a) if 5% of the original water in the system is evaporates on cooling. b) if 5% of the total weight of solution is evaporates on cooling. 3. Feed to a cooling crystallizer is 1000 lb/hr of 32.5 wt% MgSO4 in water at 120oF. This solution is cooled to 70oF to form crystals of heptahydrate. Estimate the heat of removal rate in Btu/hr. Given the specific heat of solution is 0.72 Btu/lb.oF and the heat of solution is – 23.2 Btu/lb of heptahydrate. 4. A solution consisting of 30% MgSO4 and 70% H2O is cooled to 60oC. During cooling, 5% of the total water in the system evaporates. Calculate the amount of crystals obtained per 1000 kg of original mixture. 5. Copper sulfate is crystallized as CuSO4.5H2O by combined evaporative/cooling crystallization. 1000 kg/hr of water is mixed with 280 kg/hr of anhydrous copper sulfate at 40oC. The solution is cooled to 10oC and 38 kg/hr of water is evaporated in the process. Calculate the amount of crystals can be collected theoretically. 6. A batch of 1000 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) Calculate the weight of water required for solution b) Calculate the weight of crystals obtained if 3% of the original water evaporates on cooling. 7. Determine the effect of crystal diameter on the solubility of KCl in water at 25oC. Given: Molecular weight of KCl = 74.6 Density of KCl = 1980 kg/m3

FG/T3/July2011

Mass Transfer 2

CPB 20103

Average interfacial tension = 0.028 J/m2 8. Using the data in Question (4), estimate the effects of supersaturation ratio, α on the primary homogeneous nucleation of KCl from an aqueous solution at 25oC for values of α of 2.0, 1.5 and 1.1. 9. Determine the supersaturation ratio required to permit 0.5µm diameter crystals of sucrose to grow if average interfacial tension is 0.01 J/m2. Given: Molecular weight = 342 Density = 1590 kg/m3 10. If the desired production rate in the crystallizer is 4200 kg/hr of MgSO4.7H2O, the volume of liquid in the crystallizer is 7.7 m3 and the nucleation rate is 2.7 x 109 nuclei/m3.hr, estimate the predominant crystal size. What would it be if the nucleation rate were 2.0 x 108 nuclei/m3.hr? (assume a = 1)

FG/T3/July2011

Mass Transfer 2

CPB 20103

Molecular weights: MgSO4 MgSO4.7H2O

= 120.38 = 246.49

Molecular weights: FeSO4 FeSO4.7H2O

= 151.8 = 277.8

Molecular weight:

= 160 = 250

CuSO4 CuSO4.5 H2O

FG/T3/July2011