Yield Crystallization Problem

Problems on `

ChE 411: Heat and Mass Transfer Transfer Department of Chemical Engineering University of Santo Tomas  – Faculty Faculty of Engineering Engineering

Crystallization Problem Set Problem #8

Statement of the Problem Modified Problem from Geankoplis 12.11-1

A batch of 1000 kg of KCl is dissolved in sufficient water to make a saturated solution at 370 K.The solution is cooled to 298 K. (Use the values in the Handbook for the Solubilities at given condition) (a) What is the weight of water required for solution and the weight of crystals of KCl obtained? (b) What is the weight of crystals obtained if 10% of the original water evaporates on cooling?

Given Data and Diagram H2O

1000 kg KCl Dilution

370 K

Cooling

298 K

Solution Solving for Solubilities at given conditions (using Perry’s Chemical Engineering Handbook)

Tdilution = 370 K T/K

370

T / °C

g KCL / 100 g H2O

100

56.7

97

?

90

54.0

Solving by interpolation gives the Solubility of

Solution Solving for Solubilities at given conditions (using Perry’s Chemical Engineering Handbook)

Tcooling = 298 K T/K

298

T / °C

g KCL / 100 g H2O

30

37.0

25

?

20

34.0

Solving by interpolation gives the Solubility of

Solution Solving for (a) Wt of Water required WtWater= 1000 kg KCl 55.89 g KCl 100 g H2O WtWater= 1789.23 kg H 2O of H2O is required to make a saturated KCl – H2O solution at 370 K

Solution Solving for (a) Wt of crystals formed When cooled at 298 K, WtCrystals Dissolved = 1789.23 kg x 35.5 g KCl 100 g H2O H2O WtCrystals Dissolved = 635.18 kg 1000 kg - 635.18 kg =

Solution Solving for (b) Wt of crystals formed if 10% of water evaporated while cooling WtWater = 1789.23 kg (0.9) = 1610.31 kg WtCrystals Dissolved = 1610.31 kg x 35.5 g KCl 100 g H2O H2O WtCrystals Dissolved = 571.66 kg 1000 kg – 571.66 kg =