Unformatted text preview: H 2 O crystals formed
Analysis: At 100oC, have a solution of 3,750 lb/h of FeCl3 and 3,750 lb/h of water. Note that
this amount of FeCl3 is far below its solubility at that temperature.
The molecular weights are: 162.22 for FeCl3 and 270.32 for FeCl3 6 H 2 O
The final solution will contain less water than the initial solution because of the water of
hydration in the crystals.
Let x = lb/h of water in the final aqueous solution. Then, 3,750 – x = water in crystals.
162.22
The crystals are
(100%) = 60.0% FeCl3 and 40.0% water
270.32
A mass balance for FeCl3 at 20oC gives,
3, 750 = 91.8
162.22
x + (3, 750 − x)
100
270.32 − 162.22 (1) Solving (1), gives x = 3,222 lb/h of water in solution and, therefore, 3,750 – 3,222 = 528
lb/h of water in crystals.
Therefore, the FeCl3 still in the solution = 0.918(3,222) = 2,958 lb/h
Crystals of FeCl3 6 H 2 O = 3,750 – 2,958 + 528 = 1,320 lb/h Exercise 17.11
Subject: Continuous vacuum crystallization of MgSO4.
Given: Feed to the crystallizer consisting of a mixture of 5,870 lb/h of aqueous 35 wt% MgSO4
at 180oF and 25 psia as a concentrate from an evaporator, and 10,500 lb/h of saturated aqueous
recycle filtrate of MgSO4 at 80oF and 25 psia. The vacuum crystallizer operates at 85oF and 0.58
psia to produce steam and a magma of 25 wt% crystals and 75 wt% saturated solution.
Assumptions: Equilibrium in the crystallizer.
Find: Evaporation rate of water in the crystallizer in lb/h and the production rate of crystals in
tons/day (dry basis for 2,000 lb/ton)
Analysis: First compute the total feed to the crystallizer.
Concentrate: 0.35(5,870) = 2,055 lb/h MgSO4
5,870 – 2,055 = 3,815 lb/h water
At 189oF, from Fig. 17.2, solubility of MgSO4 = 39 wt%.
So, all of the sulfate is in solution.
Filtrate:
From Fig. 17.2, the solubility of MgSO4 in water at 80oF = 27.5 wt%
0.275(10,500) = 2,888 lb/h MgSO4
10,500 – 2,888 = 7,612 lb/h water
Total feed:
2,055 + 2,888 = 4,943 lb/h MgSO4
3,815 + 7,612 = 11,427 lb/h water
Compute the mass balance around the crystallizer.
From Fig. 17.2, the solubility of MgSO4 in water = 28 wt% and MgSO4 is crystallized as
the heptahydrate.
Let: L = lb/h of solution in the magma
V = lb/h of water vaporized in the crystallizer
S = lb/h of crystals in the magma
MW of MgSO4 = 120.4.
MW of MgSO 4 7H 2 O = 246.4
Crystals are 120.4/246.4 = 0.4886 mass fraction MgSO4 and 0.5114 mass fraction water.
Magma is 25 wt% crystals and 75 wt% of a solution of 28 wt% MgSO4.
MgSO4 mass balance around the crystallizer:
4,943 = 0.28 L + 0.4886 S
But, S = 0.25(S + L)
Solving these two equations, L = 11,160 lb/h
and
S = 3,720 lb/h
Therefore, the tons/day of crystals = 3,720(24)/2,000 = 44.6 tons/day
Total mass balance around the crystallizer:
4,943 + 11,427 = V + L + S = V + 11,160 + 3,720
Solving, V = 1,490 lb/h of water vapor produced in the crystallizer. Exercise 17.12
Subject: Crystallization of urea from an aqueous solution by cooling and evaporation.
Given: Feed is a 90% saturated solution of urea in water at 100...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details