CHEMICAL_ENGINEERING_SERIES_CRYSTALLIZAT

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CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
Compilation of Lectures and Solved Problems 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
2 
 
 
CRYSTALLIZATION 
Refers to a solid-liquid separation process in which solid particles are formed within a homogenous phase. It 
can occur as: 
(1) formation of solid particles in a vapor 
(2) formation of solid particles from a liquid melt 
(3) formation of solid crystals from a solution 
 
The process usually involves two steps: 
 
(1) concentration of solution and cooling of solution until the solute concentration becomes greater than its 
solubility at that temperature 
(2) solute comes out of the solution in the form of pure crystals 
 
Crystal Geometry 
A crystal is highly organized type of matter, the constituent particles of which are arranged in an orderly and 
repetitive manner; they are arranged in orderly three dimensional arrays called SPACE LATTICES 
Supersaturation 
Supersaturation is a measure of the quantity of solids actually present in solution as compared to the quantity 
that is in equilibrium with the solution 
𝑆 = 
𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
100 𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡⁄
𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
100 𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡⁄
 
Crystallization cannot occur without supersaturation. There are 5 basic methods of generating 
supersaturation 
(1) EVAPORATION – by evaporating a portion of the solvent 
(2) COOLING – by cooling a solution through indirect heat exchange 
(3) VACUUM COOLING – by flashing of feed solution adiabatically to a lower temperature and inducing 
crystallization by simultaneous cooling and evaporation of the solvent 
(4) REACTION – by chemical reaction with a third substance 
(5) SALTING – by the addition of a third component to change the solubility relationship 
 
 
Mechanism of Crystallization Process 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
3 
 
There are two basic steps in the over-all process of crystallization from supersaturated solution: 
(1) NUCLEATION’ 
a. Homogenous or Primary Nucleation – occurs due to rapid local fluctuations on a molecular scale in 
a homogenous phase; it occurs in the bulk of a fluid phase without the involvement of a solid-fluid 
interface 
b. Heterogeneous Nucleation – occurs in the presence of surfaces other than those of the crystals such 
as the surfaces of walls of the pipe or container, impellers in mixing or foreign particles; this is 
dependent on the intensity of agitation 
c. Secondary Nucleation – occurs due to the presence of crystals of the crystallizing species 
 
(2) CRYSTAL GROWTH – a layer-by-layer process 
a. Solute diffusion to the suspension-crystal interface 
b. Surface reaction for absorbing solute into the crystal lattice 
 
Crystallization Process 
 
Important Factors in a Crystallization Process 
(1) Yield 
(2) Purity of the Crystals 
(3) Size of the Crystals – should be uniform to minimize caking in the package, for ease in pouring, ease in 
washing and filtering and for uniform behaviour when used 
(4) Shape of the Crystals 
Magma 
It is the two-phase mixture of mother liquor and crystals of all sizes, which occupies the crystallizer and is 
withdrawn as product 
 
SOLUTION
WATER
Solution is concentrated 
by evaporating water
The concentrated 
solution is cooled until 
the concentration 
becomes greater than 
its solubility at that 
temperature
CRYSTALS
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
4 
 
Types of Crystal Geometry 
(1) CUBIC SYSTEM – 3 equal axes at right angles to each other 
(2) TETRAGONAL – 3 axes at right angles to each other, one axis longer than the other 2 
(3) ORTHOROMBIC – 3 axes at right angles to each other, all of different lengths 
(4) HEXAGONAL – 3 equal axes in one plane at 60° to each other, and a fourth axis at a right angle to this 
plane and not necessarily at the same length 
(5) MONOCLINIC – 3 unequal axes, two at a right angles in a plane, and a third at some angle to this plane 
(6) TRICLINIC – 3 unequal axes at unequal angles to each other and not 30°, 60°, or 90° 
(7) TRIGONAL – 3 unequal and equally inclined axes 
 
Classification of Crystallizer 
(1) May be classified according to whether they are batch or continuous in operation 
(2) May be classified according on the methods used to bring about supersaturation 
(3) Can also be classified according on the method of suspending the growing product crystals 
 
Equilibrium Data (Solubilities) 
 Either tables or curves 
 Represent equilibrium conditions 
 Plotted data of solubilities versus temperatures 
 In general, solubility is dependent mainly on temperature although sometimes on size of materials and 
pressure 
Expressions of Solubilities 
 Parts by mass of anhydrous materials per 100 parts by mass of total solvent 
 Mass percent of anhydrous materials or solute which ignores water of crystallization 
 
 
 
 
 
 
 
Types of Solubility Curve 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
5 
 
(1) TYPE I: Solubility increases with temperature 
and there are no hydrates or water of 
crystallization 
 
 
(2) TYPE II: Solubility increases with temperature 
but curve is marked with extreme flatness 
 
 
(3) TYPE III: Solubility increasing fairly rapid with 
temperature but is characterized by “breaks” 
and indicates different “hydrates” or water of 
crystallization 
 
 
(4) TYPE IV: Unusual Curve; Solubility increases 
at a certain transition point while the solubility 
of the hydrate decreases as temperature 
increases 
 
 
 
SUPERSATURATION BY COOLING 
Crystallizers that obtain precipitation by cooling a concentrated hot solution; applicable for substance that 
have solubility curve that decreases with temperature; for normal solubility curve which are common for most 
substances 
Pan Crystallizers 
0
50
100
150
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250
300
0 20 40 60 80 100
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 1
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 g
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Temperature, °C0
50
100
150
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0 20 40 60 80 100
So
lu
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lit
y,
 g
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 1
00
 g
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Temperature, °C
Solubility of NaCl (CHE HB 8th edition)0
50
100
150
200
250
0 20 40 60 80 100
So
lu
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lit
y,
 g
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 1
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 g
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er
 
Temperature, °C
Solubility of Na2HPO4 (CHE HB 8
th edition)
Na2HPO4
Na2HPO4·12H2O
Na2HPO4·7H2O
Na2HPO4·2H2O
0
10
20
30
40
50
60
0 20 40 60 80 100
So
lu
bi
lit
y,
 g
ra
m
 p
er
 1
00
 g
ra
m
 w
at
er
 
Temperature, °C
Solubility of Na2CO3 (CHE HB 8
th edition)
Na2CO3·10H2O
Na2CO3·H2O
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
6 
 
Batch operation; seldom used in modern practice, except in small scale operations, because they are wasteful 
of floor space and of labor; usually give a low quality product 
Agitated batch Crystallizers 
Consist of an agitated tank; usually cone-bottomed, containing cooling coils. It is convenient in small scale or 
batch operations because of their low capital costs, simplicity of operation and flexibility 
Swenson Walker Crystallizer 
A continuous crystallizer consist of an open round bottomed-trough, 24-in wide by 10 ft long, and containing 
a long ribbon mixer that turns at about 7 rpm. 
CALCULATIONS: 
 
 
Over-all material Balance: 
𝐹 = 𝐿 + 𝐶 
 
Solute Balance: 
𝑋𝐹𝐹 = 𝑋𝐿𝐿 + 𝑋𝐶𝐶 
 
Enthalpy Balance: 
ℎ𝑓𝐹 = ℎ𝐿𝐿 + ℎ𝑐𝐶 + 𝑞 
 
Heat Balance: 
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑝𝐹(𝑡𝐹 − 𝑡𝐿) + 𝐶𝐻𝐶 
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑊𝐶𝑝 𝐻2𝑂(𝑡2 − 𝑡1) 
 
Heat T ransfer Equation 
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚 
𝑞 = 𝑈𝐴 [
(𝑡𝐹 − 𝑡2) − (𝑡𝐿 − 𝑡1)
ln
𝑡𝐹 − 𝑡2
𝑡𝐿 − 𝑡1
] 
 
 
 
where: 
𝐹 = mass of the feed solution 
𝐿 = mass of the mother liquor, usually saturated solution 
𝐶 = mass of the crystals 
𝑊 = mass of the cooling water 
𝑋𝐹 = mass solute (salt) in the feed solution per mass of feed solution 
𝑋𝐿 = mass of solute (salt) in the mother liquor per mass of mother liquor 
𝑋𝐶 = mass of solute (salt) in the srystals per mass of crystals 
ℎ𝐹 = enthalpy of the feed solution 
ℎ𝐿 = enthalpy of the mother liquor 
ℎ𝐶 = enthalpy of the crystals 
𝑞𝑤𝑎𝑡𝑒𝑟 = heat absorbed by the cooling water 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = heat loss by the crystals 
𝐶𝑝𝐹 = specific heat of the feed solution 
𝐶𝑝𝐻2𝑂 = specific heat of cooling water 
𝐻𝐶 = heat of crystallization 
𝑈 = over-all heat transfer coefficient 
𝐴 = heat transfer area 
𝑡𝐹 = temperature of the feed solution 
𝑡𝐿 = temperature of the mother liquor 
𝑡1 = inlet temperature of cooling water 
𝑡2 = outlet temperature of cooling water 
 
 
 
F
XF
hf
tF
L
XL
hL
tL
C
XC
hC
tC
W
t1
W
t2
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
7 
 
SUPERSATURATION BY EVAPORATION OF SOLVENT 
Crystallizers that obtain precipitation by evaporating a solution; applicable for the substance whose solubility 
curve is flat that yield of solids by cooling is negligible; acceptable to any substance whose solubility curve is 
not to steep 
Salting Evaporator 
The most common of the evaporating crystallizers; in older form, the crystallizer consisted of an evaporator 
below which were settling chambers into which the salt settled 
Oslo Crystallizer 
Modern form of evaporating crystallizer; this unit is particularly well adopted to the production of large -sized 
uniform crystals that are usually rounded; it consists essentially of a forced circulation evaporator with an 
external heater containing a combination of salt filter and particle size classifier on the bottom of the evaporator 
body 
CALCULATIONS: 
 
Over-all material Balance: 
𝐹 = 𝐿 + 𝐶 + 𝑉 
 
Solute Balance: 
𝑋𝐹𝐹 = 𝑋𝐿𝐿 + 𝑋𝐶𝐶 
 
Solvent Balance: 
(1 − 𝑋𝐹)𝐹 = 𝑉 + (1 − 𝑋𝐿)𝐿 + (1 − 𝑋𝐶)𝐶 
 
Enthalpy Balance: 
ℎ𝑓𝐹 = ℎ𝑉𝑉 + ℎ𝐿𝐿 + ℎ𝑐𝐶 + 𝑞 
 
Heat Balance: 
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑝𝐹(𝑡𝐹 − 𝑡𝐿) + 𝐶𝐻𝐶 − 𝑉𝜆𝑉 
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑊𝐶𝑝 𝐻2𝑂(𝑡2 − 𝑡1) 
 
where: 
𝐹 = mass of the feed solution 
𝐿 = mass of the mother liquor, usually saturated solution 
𝐶 = mass of the crystals 
𝑊 = mass of the cooling water 
𝑉 = mass of the evaporated solvent 
𝑋𝐹 = mass solute (salt) in the feed solution per mass of feed 
solution 
𝑋𝐿 = mass of solute (salt) in the mother liquor per mass of 
mother liquor 
𝑋𝐶 = mass of solute (salt) in the srystals per mass of crystals 
ℎ𝐹 = enthalpy of the feed solution 
ℎ𝐿 = enthalpy of the mother liquor 
ℎ𝐶 = enthalpy of the crystals 
ℎ𝑉 = enthalpy of the vapor 
𝑞𝑤𝑎𝑡𝑒𝑟 = heat absorbed by the cooling water 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = heat loss by the crystals 
𝐶𝑝𝐹 = specific heat of the feed solution 
𝐶𝑝𝐻2𝑂 = specific heat of cooling water 
𝐻𝐶 = heat of crystallization 
𝜆𝑉 = latent heat of vaporization 
𝑈 = over-all heat transfer coefficient 
𝐴 = heat transfer area 
𝑡𝐹 = temperature of the feed solution 
𝑡𝐿 = temperature of the mother liquor 
𝑡1 = inlet temperature of cooling water 
𝑡2 = outlet temperature of cooling water 
 
F
XF
hf
tF
L
XL
hL
tL
C
XC
hC
tC
W
t1
W
t2
V
hV
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
8 
 
SUPERSATURATION BY ADIABATIC EVAPORATION OF SOLVENT 
 
 
Over-all material Balance: 
𝐹 = 𝐿 + 𝐶 + 𝑉 
 
Solute Balance: 
𝑋𝐹𝐹 = 𝑋𝐿𝐿 + 𝑋𝐶𝐶 
 
Solvent Balance: 
(1 − 𝑋𝐹)𝐹 = 𝑉 + (1 − 𝑋𝐿)𝐿 + (1 − 𝑋𝐶)𝐶 
 
Enthalpy Balance: 
ℎ𝑓𝐹 = ℎ𝑉𝑉 + ℎ𝐿𝐿 + ℎ𝑐𝐶 
 
 
where: 
𝐹 = mass of the feed solution 
𝐿 = mass of the mother liquor, usually saturated solution 
𝐶 = mass of the crystals 
𝑊 = mass of the cooling water 
𝑉 = mass of the evaporated solvent 
𝑋𝐹 = mass solute (salt) in the feed solution per mass of feed 
solution 
𝑋𝐿 = mass of solute (salt) in the mother liquor per mass of 
mother liquor 
𝑋𝐶 = mass of solute (salt) in the srystals per mass of crystals 
ℎ𝐹 = enthalpy of the feed solution 
ℎ𝐿 = enthalpy of the mother liquor 
ℎ𝐶 = enthalpy of the crystals 
ℎ𝑉 = enthalpy of the vapor 
𝐻𝐶 = heat of crystallization 
𝑡𝐹 = temperature of the feed solution 
𝑡𝐿 = temperature of the mother liquor 
𝑡1 = inlet temperature of cooling water 
𝑡2 = outlet temperature of cooling water 
 
 
CRYSTALLIZATION BY SEEDING 
ΔL Law of Crystals 
 States that if all crystals in magma grow in a supersaturation field and at the same temperature and if all 
crystal grow from birth at a rate governed by the supersaturation, then all crystals are not only invariant 
but also have the same growth rate that is independent of size 
 
 The relation between seed and product particle sizes may be written as 
 
𝐿𝑃 = 𝐿𝑆 + ∆𝐿 
𝐷𝑃 = 𝐷𝑆 + ∆𝐷 
Where: 
𝐿𝑃 𝑜𝑟 𝐷𝑃 = characteristic particle dimension of the product 
𝐿𝑆 𝑜𝑟 𝐷𝑆 = characteristic particle dimension of the seed 
∆𝐿 𝑜𝑟 ∆𝐷 = change in size of crystals and is constant throughout the range of size present 
 
 
 
F
XF
hf
V
hV
L
XL
hL
C
XC
hC
M
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
9 
 
Since the rate of linear crystal growth is independent of crystal size, the seed and product masses may 
be related for 
 
𝑊𝑃 = 𝑎𝜌𝐷𝑃
3 = 𝑎𝜌(𝐷𝑆 + ∆𝐷)
3 
𝑊𝑆 = 𝑎𝜌𝐷𝑆
3
 
𝑊𝑃 = 
𝑊𝑆
𝐷𝑆
3 (𝐷𝑆 + ∆𝐷)
3 
𝑊𝑃 = 𝑊𝑆 (
𝐷𝑆 + ∆𝐷
𝐷𝑆
)
3
 
𝑊𝑃 = 𝑊𝑆 (
𝐷𝑆 + [𝐷𝑃 − 𝐷𝑆]
𝐷𝑆
)
3
 
𝑊𝑃 = 𝑊𝑆 (
𝐷𝑃
𝐷𝑆
)
3
 
All the crystals in the seed have been assumed to be of the same shape, and the shape has been assumed 
to be unchanged by the growth process. Through assumption is reasonably closed to the actual conditions 
in most cases. For differential parts of the crystal masses, each consisting of crystals of identical dimensions: 
∫ 𝑑𝑊𝑃 = ∫ (1 +
∆𝐷
𝐷𝑆
)
3
𝑑𝑊𝑆
𝑊𝑆
0
𝑊𝑃
0
 
𝑊𝑃 = ∫ (1 +
∆𝐷
𝐷𝑆
)
3
𝑑𝑊𝑆
𝑊𝑆
0
 
𝐶 = 𝑊𝑃 − 𝑊𝑆 
 
 
 
 
 
 
 
 
 
PROBLEM # 01: 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
10 
 
A 20 weight % solution of Na2SO4 
at 200°F is pumped continuously 
to a vacuum crystallizer from which 
the magma is pumped at 60°F. 
What is the composition of this 
magma, and what percentage of 
Na2SO4 in the feed is recovered as 
Na2SO4·10H2O crystals after this 
magma is centrifuged? 
 
SOLUTION: 
Basis: 100 lb feed 
From table 2-122 (CHE HB), solubility of Na2SO4·10H2O 
T,°C 10 15 20 
g/100 g H2O 9.0 19.4 40.8 
 
Consider over-all material balance: 
𝐹 = 𝐶 + 𝐿 
𝐿 = 100 − 𝐶 𝑒𝑞𝑛 1 
Consider solute balance: 
𝑋𝐹 𝐹 = 𝑋𝐶 𝐶 + 𝑋𝐿 𝐿 
𝑋𝐶 =
𝑀𝑁𝑎2𝑆𝑂4
𝑀𝑁𝑎2𝑆𝑂4∙10𝐻2 𝑂
=
142
322
= 0.4410
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2𝑂
 
At 60°F, solubility is 21.7778 g per 100 g water 
𝑋𝐿 =
21.7778
100 + 21.7778
= 0.1788
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
(0.20
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑓𝑒𝑒𝑑
) (100 𝑙𝑏 𝑓𝑒𝑒𝑑) = (0.1788
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
) (𝐿) + (0.4410
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2𝑂
) (𝐶) 
20 = 0.1788 𝐿 + 0.4410 𝐶 𝑒𝑞𝑛 2 
Substitute 1 in 2 
20 = 0.1788(100 −𝐶) + 0.4410 𝐶 
𝐶 = 8.0854 𝑙𝑏 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 
𝐿 = 100 − 8.0854 
𝐿 = 91.9146 𝑙𝑏 
 
Magma composition: 
% 𝐶 = 
8.0854
100
 𝑥 100 = 𝟖. 𝟎𝟖𝟓𝟒 % 
% 𝐿 = 
91.9146
100
 𝑥 100 = 𝟗𝟏. 𝟗𝟏𝟒𝟔 % 
% Recovery: 
% 𝑟𝑒𝑐𝑜𝑣𝑒𝑟𝑦 =
𝑋𝐶 𝐶
𝑋𝐹 𝐹
 𝑥 100 =
(0.4410
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
) (8.0854𝑙𝑏 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2 𝑂)
(0.20
𝑙𝑏 𝑁𝑎2𝑆𝑂4
𝑙𝑏 𝑓𝑒𝑒𝑑
) (100 𝑙𝑏 𝑓𝑒𝑒𝑑)
𝑥100 
% 𝒓𝒆𝒄𝒐𝒗𝒆𝒓𝒚 = 𝟏𝟕.𝟖𝟑 % 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Na2SO4 solution
xF = 0.20
tF = 200°F
Na2SO4 ·10H2O
C
L
Magma, M
tM = 60°F
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
11 
 
 
PROBLEM # 02: 
 
A solution of 32.5% MgSO4 
originally at 150°F is to be 
crystallized in a vacuum adiabatic 
crystallizer to give a product 
containing 4,000 lb/h of 
MgSO4·7H2O crystals from 10,000 
lb/h of feed. The solution boiling 
point rise is estimated at 10°F. 
Determine the product temperature, 
pressure and weight ratio of mother 
liquor to crystalline product. 
SOLUTION: 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝑉 = 10,000 − 𝐿 − 4,000 
𝑉 = 6,000 − 𝐿 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
Consider solute balance: 
𝑥𝐹 𝐹 = 𝑥𝐶𝐶 + 𝑥𝐿𝐿 
𝑥𝐶 =
𝑀𝑀𝑔 𝑆𝑂4
𝑀𝑀𝑔 𝑆𝑂4∙7𝐻2 𝑂
=
120 .38
246 .49
= 0.4884
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
 
(0.325
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑓𝑒𝑒𝑑
) (10,000 
𝑙𝑏 𝑓𝑒𝑒𝑑
ℎ
) = 𝑋𝐿 (𝐿) + (0.4884
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
) (4,000
𝑙𝑏
ℎ
) 
𝑥𝐿𝐿 = 1,296.4 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Consider enthalpy balance: 
ℎ𝑓 𝐹 = ℎ𝑉𝑉 + ℎ𝐿𝐿 + ℎ𝑐 𝐶 
THE PROBLEM CAN BE SOLVED BY TRIAL AND ERROR SINCE TEMPERATURE OF THE 
SOLUTION AFTER CRYSTALLIZATION IS UNKNOWN AND ENTHALPIES ARE DEPENDENT 
ON TEMPERATURE 
1. Assume temperature of the solution 
2. From figure 27-3 (Unit Operations by McCabe and Smoth 7th edition), obtain mass fraction of 
MgSO4 at the assumed temperature of the solution 
3. Solve for “L” using equation 2 
4. Solve for “V” using equation 1 
5. Check if assumed temperature is correct by conducting enthalpy balance 
a. Obtain values of hF, hC and hL from figure 27-4 (Unit Operations by McCabe and 
Smith 7th edition) at the designated temperatures and concentrations 
b. Compute for hV 
c. Using the enthalpy balance equation, compute for “V” using the value of “L” from step 
3 
6. Compare values of “V” from step 4 with that from step 5-c 
7. If not the same (or approximately the same), conduct another trial and error calculations 
 
MgSO4 solution
F = 10,000 lb/h
xF = 0.325
tF = 150°F
MgSO4 ·7H2O
C = 4,000 lb/h
L
V
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
12 
 
TRIAL 1: Assume temperature of the solution at 60°F 
From figure 27-3 (Unit Operations by McCabe and Smith 7th edition) 
𝑥𝐿 = 0.245
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
Substitute to equation 2 
𝐿 =
1,296.4
0.245
= 5,291.43 𝑙𝑏 
Substitute to equation 1 
𝑉 = 6,000 − 5,291.43 = 708 .57 𝑙𝑏 
 
From figure 27-4 (Unit Operations by McCabe and Smith, 7th edition) 
ℎ𝐹 𝑎𝑡 150°𝐹 𝑎𝑛𝑑 32.5% 𝑀𝑔𝑆𝑂4 = −10
𝐵𝑇𝑈
𝑙𝑏
 
ℎ𝐶 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 48.84% 𝑀𝑔𝑆𝑂4 = −158
𝐵𝑇𝑈
𝑙𝑏
 
ℎ𝐿 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 24.5% 𝑀𝑔𝑆𝑂4 = −50
𝐵𝑇𝑈
𝑙𝑏
 
Temperature of vapor is 60 – 10 = 50°F 
ℎ𝑉 = 𝐻𝑉 + 𝐶𝑃 𝑥 𝐵𝑃𝐸 
From steam table at 50°F, 𝐻𝑉 = 1,083.3
𝐵𝑇𝑈
𝑙𝑏
 
ℎ𝑉 = 1,083.3
𝐵𝑇𝑈
𝑙𝑏
+ [(0.45
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (10°𝐹 )] 
ℎ𝑉 = 1,087.8
𝐵𝑇𝑈
𝑙𝑏
 
 
ℎ𝑓 𝐹 = ℎ𝑉𝑉 + ℎ𝐿𝐿 + ℎ𝑐 𝐶 
(−10)(10,000) = (1087.8)(𝑉) + (−50)(5,291.43) + (−158) (4,000) 
𝑉 = 732.28 𝑙𝑏 
 
Since % error is less than 5%, assumed value can be considered correct. 
 
Product temperature 
𝑻 = 𝟔𝟎°𝑭 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Operating Pressure 
From steam table for vapor temperature of 50°F 
𝑷 = 𝟎.𝟏𝟕𝟖𝟎𝟑 𝒑𝒔𝒊 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Ratio of mother liquor to crystalline product 
𝐿
𝐶
=
5,291.43
4,000
 
 
𝑳
𝑪
= 𝟏. 𝟑𝟐 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
13 
 
PROBLEM # 03 : 
A plant produces 30,000 MT of anhydrous 
sulfate annually by crystallizing sulfate brine 
at 0°C, yields of 95% and 90% in the 
crystallization and calcinations operations 
are obtained respectively. How many metric 
tons of liquor are fed to the crystallizer daily? 
Note: 300 working days per year 
 
CHE BP January 1970 
SOLUTION: 
Assume that the liquor entering the crystallizer is a saturated solution at 0°C 
 
From table 2-120 (CHE HB), solubility at 0°C: 
5 𝑔 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
100 𝑔 𝐻2𝑂
 
 
𝑚𝑎𝑠𝑠 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
= 30,000
𝑀𝑇 𝑁𝑎2𝑆𝑂4
𝑦𝑟
 𝑥 
1
0.95
 𝑥 
1 𝑀𝑇𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂4 
142 𝑀𝑇𝑁𝑎2𝑆𝑂4
 𝑥
1 𝑀𝑇𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2𝑂 
1 𝑀𝑇𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
𝑥
322 𝑀𝑇𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2𝑂
𝑀𝑇𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
 
 
𝑚𝑎𝑠𝑠 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂 = 71,608.60 𝑀𝑇 𝑥 
1 𝑦𝑟
300 𝑑𝑎𝑦𝑠
 
 
𝑚𝑎𝑠𝑠 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂 =
238.6953𝑀𝑇
𝑑𝑎𝑦
 
 
 
𝐹 =
238.6953𝑀𝑇𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
𝑑𝑎𝑦
 𝑥 
105 𝑀𝑇 𝑓𝑒𝑒𝑑 
5 𝑀𝑇𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂 
 
 
𝑭 = 𝟓, 𝟎𝟏𝟐. 𝟔𝟎
𝑴𝑻
𝒅𝒂𝒚
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
 
 
 
CALCINATION CRYSTALLIZATION
P
Na2SO4
30,000 MT/yr
T = 0 C
YIELD = 95%
YIELD = 90%
F
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
14 
 
PROBLEM # 04 : 
1,200 lb of barium nitrate are dissolved in 
sufficient water to form a saturated solution at 
90°C. Assuming that 5% of the weight of the 
original solution is lost through evaporation, 
calculate the crop of the crystals obtained 
when cooled to 20°C. solubility data of 
barium nitrate at 90°C = 30.6 lb/100 lb water; 
at 20°C = 9.2 lb/100 lb water 
 
CHE BP July 1968 
SOLUTION: 
𝑥𝐹 = 0.306
𝑙𝑏 𝐵𝑎(𝑁𝑂3)2
𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
 𝑥 
100 𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
(100 + 30.6) 𝑙𝑏 𝑓𝑒𝑒𝑑
= 0.2343
𝑙𝑏 𝐵𝑎(𝑁𝑂3)2
𝑙𝑏 𝑓𝑒𝑒𝑑
 
𝑥𝐹 𝐹 = 1,200 𝑙𝑏 𝐵𝑎(𝑁𝑂3)2 
𝐹 = 1,200 𝑙𝑏 𝐵𝑎(𝑁 𝑂3)2 𝑥 
𝑙𝑏 𝑓𝑒𝑒𝑑
0.2343 𝑙𝑏 𝐵𝑎(𝑁 𝑂3)2
 
𝐹 = 5,121.5686 𝑙𝑏 
𝑥𝐿 = 0.092
𝑙𝑏 𝐵𝑎(𝑁𝑂3 )2
𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
 𝑥 
100 𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
(100 + 9.2) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.0842
𝑙𝑏 𝐵𝑎(𝑁𝑂3)2
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
 
Consider over-all material balance around the crystallizer 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝑉 = 0.05𝐹 
𝐿 = 0.95(5,121.5686) − 𝐶 
𝐿 = 4,865.4902 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider Ba(NO3)2 balance 
 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
1,200 = (0.0842)(𝐿) + (1.0)(𝐶) 
1,200 = 0.0842𝐿 + 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Substitute 1 in 2 
 
1,200 = 0.0842(4,865.4902 − 𝐶) + 𝐶 
 
𝐶 =
1,200 − [(0.0842)(4,865.4902)]
0.9158
 
 
𝑪 = 𝟖𝟔𝟐. 𝟗𝟖𝟗𝟒 𝒍𝒃 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
CRYSTALLIZER
C
T = 20 C
T = 90 C
F
1,200 lb BaNO3
V
L
T = 20 C
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
15 
 
PROBLEM # 05: 
 
A Swenson-Walker crystallizer is to be used 
to produce 1 ton/h of copperas 
(FeSO4·7H2O) crystals. The saturated 
solution enters the crystallizer at 120°F. The 
slurry leaving the crystallizer will be at 80°F. 
Cooling water enters the crystallizer jacket at 
60°F and leaves at 70°F. It may be assumed 
that the U for the crystallizer is 35 
BTU/h·°F·ft2. There are 3.5 ft2 of cooling 
surface per ft of crystallizer length. 
a) Estimate the cooling water required 
b) Determine the number of crystallizer 
section to be used. 
Data: specific heat of solution = 0.7 
BTU/lb·°F; heatof solution= 4400 cal/gmol 
copperas; solubility at 120°F = 140 parts 
copperas/100 parts excess water; solubility 
at 80°F = 74 parts copperas/100 parts 
excess water 
 
SOLUTION: 
Consider over-all material balance: 
𝐹 = 𝐿 + 𝐶 
𝐿 = 𝐹 − 2,000 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider copperas (FeSO4·7H2O) balance: 
𝑥𝐹 𝐹 = 𝑥𝐶𝐶 + 𝑥𝐿𝐿 
𝑥𝐶 = 1.0 
𝑥𝐿 =
74 𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2𝑂
100 𝑙𝑏 𝐻2𝑂
 𝑥 
100 𝑙𝑏 𝐻2𝑂
174 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.4253
𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2𝑂
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐹 =
140 𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2𝑂
100 𝑙𝑏 𝐻2𝑂
 𝑥 
100 𝑙𝑏 𝐻2𝑂
240 𝑙𝑏 𝑓𝑒𝑒𝑑
= 0.5833
𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2𝑂
𝑙𝑏 𝑓𝑒𝑒𝑑
 
(0.5833) (𝐹) = (1.0)(2,000) + (0.4253)(𝐿) 
𝐿 = 1.3715𝐹 − 4,702.5629 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
𝐹 − 2,000 = 1.3715 𝐹 − 4,702.5629 
𝐹 = 7,274.73 
𝑙𝑏
ℎ
 
𝐿 = 5,274.73 
𝑙𝑏
ℎ
 
 
Consider heat balance: 
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑝𝐹(𝑡𝐹 − 𝑡𝐿) + 𝐶𝐻𝐶 
SWENSON-WALKER 
CRYSTALLIZER
F
tF = 120 F
L
tL = 80 F
C, 1 ton/h
Fe2SO4·7H2O
tC = 80 F
W
t1 = 60 F
t2 = 70 F
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
16 
 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = [(7,274.73
𝑙𝑏
ℎ
) (0.70
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (120 − 80)°𝐹]
+ [(2,000
𝑙𝑏
ℎ
) (4,400
𝑐𝑎𝑙
𝑔𝑚𝑜𝑙
𝑥
𝑔𝑚𝑜𝑙
277 .85 𝑔
𝑥
1
𝐵𝑇𝑈
𝑙𝑏
0.55556
𝑐𝑎𝑙
𝑔
)] 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 260,701.1615
𝐵𝑇𝑈
ℎ
 
 
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑊𝐶𝑝 𝐻2𝑂
(𝑡2 − 𝑡1) 
𝑊 =
260,701.1615
𝐵𝑇𝑈
ℎ
(1.0
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (70 − 60)°𝐹
 
𝑊 = 26,070.1162
𝑙𝑏
ℎ
 𝑥 
1 𝑓𝑡3
62.335 𝑙𝑏
 𝑥 
7.481 𝑔𝑎𝑙
𝑓𝑡3
 𝑥 
1 ℎ
60 𝑚𝑖𝑛
 
 
𝑾 = 𝟓𝟐. 𝟏𝟒
𝒈𝒂𝒍
𝒎𝒊𝒏
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚 
∆𝑇𝑙𝑚 =
(𝑡𝐹 − 𝑡2) − (𝑡𝐿 − 𝑡1)
ln
𝑡𝐹 − 𝑡2
𝑡𝐿 − 𝑡1
 
∆𝑇𝑙𝑚 =
(120 − 70) − (80 − 60)
ln
120 − 70
80 − 60
 
∆𝑇𝑙𝑚 = 32.7407°𝐹 
 
𝐴 =
260,701.1615
𝐵𝑇𝑈
ℎ
(35
𝐵𝑇𝑈
ℎ ∙ 𝑓𝑡2 ∙ °𝐹
) (32.7407°𝐹)
 
 
𝐴 = 227.5029 𝑓𝑡2 
 
# 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 = 227.5029 𝑓𝑡2 𝑥 
1 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
3.5 𝑓𝑡2
 𝑥
1 𝑢𝑛𝑖𝑡
10 𝑓𝑡
 
 
# 𝒐𝒇 𝒖𝒏𝒊𝒕𝒔 = 𝟔. 𝟓 ≈ 𝟕 𝒖𝒏𝒊𝒕𝒔 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
17 
 
PROBLEM # 06: 
Crystals of Na2CO3·10H2O are dropped into a saturated solution of Na2CO3 in water at 100°C. 
What percent of the Na2CO3 in the Na2CO3·H2O is recovered in the precipitated solid? The 
precipitated solid is Na2CO3·H2O. Data at 100°C: the saturated solution is 31.2% Na2CO3; 
molecular weight of Na2CO3 is 106 
 
SOLUTION: 
Assume 100 g of Na2CO3·10H2O added into the saturated solution 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2𝑂 = 100 𝑔𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂 𝑥 
124 𝑔 𝑁𝑎2𝐶𝑂3 ∙ 𝐻2𝑂
286 𝑔 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2𝑂 = 43.3566 𝑔 
 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 = 100 𝑔𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂 𝑥 
106 𝑔 𝑁𝑎2𝐶𝑂3
286 𝑔 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2𝑂
 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 = 37.0629 𝑔 
 
𝑤𝑡 𝐻2𝑂 = 100 𝑔𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2𝑂 𝑥 
180 𝑔 𝐻2𝑂
286 𝑔 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
 
𝑤𝑡 𝐻2𝑂 = 62.9371 𝑔 
 
% 𝑁𝑎2𝐶𝑂3 𝑖𝑛 𝑠𝑎𝑡𝑑 𝑠𝑜𝑙𝑛 𝑎𝑡 100°𝐶 =
𝑋 
𝑋 + 62.9371
 𝑥 100 = 31.2 
𝑋 = 28.5412 𝑔 
 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 = 37.0629 − 28.5412 = 8.5217 𝑔 
 
 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2𝑂 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 = 8.5217 𝑔𝑁𝑎2𝐶𝑂3 𝑥 
124 𝑔 𝑁𝑎2𝐶𝑂3 ∙ 𝐻2𝑂
106 𝑔 𝑁𝑎2𝐶𝑂3
 
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2𝑂 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 = 9.9688 𝑔 
 
% 𝑁𝑎2𝐶𝑂3 ∙ 𝐻2𝑂 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 =
9.9688
43.3566
 𝑥 100 
 
% 𝑵𝒂𝟐𝑪𝑶𝟑 ∙ 𝑯𝟐𝑶 𝒑𝒓𝒆𝒄𝒊𝒑𝒊𝒕𝒂𝒕𝒆𝒅 = 𝟐𝟐.𝟗𝟗 % 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
18 
 
 
PROBLEM # 07: 
 
 
A solution of MgSO4 at 220°F containing 43 g 
MgSO4 per 100 g H2O is fed into a cooling 
crystallizer operating at 50°F. If the solution 
leaving the crystallizer is saturated, what is 
the rate at which the solution must be fed to 
the crystallizer to produce one ton of 
MgSO4·7H2O per hour? 
 
SOLUTION: 
Consider over-all material balance: 
 
𝐹 = 𝐿 + 𝐶 
𝐿 = 𝐹 − 1 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider MgSO4 balance 
 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐹 =
43 𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
100 𝑡𝑜𝑛 𝐻2𝑂
 𝑥 
100 𝑡𝑜𝑛 𝐻2𝑂
(100 + 43)𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
= 0.3007
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
 
𝑥𝐶 =
120 .38 𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
246.49 𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
= 0.4884 
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
 
 
From table 27-3 (Unit Operations by McCabe and Smith, 7th edition), at 50°F 
𝑥𝐿 = 0.23
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑡𝑜𝑛 𝑙𝑖𝑞𝑢𝑜𝑟
 
(0.3007) (𝐹) = (0.23)(𝐿) + (0.4884) (1) 
𝐿 = 1.3074𝐹 − 2.1235 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
 
𝐹 − 1 = 1.3074 𝐹 − 2.1235 
 
𝑭 = 𝟑. 𝟔𝟓
𝒕𝒐𝒏
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
COOLING CRYSTALLIZER
F
tF = 220 F
43 g MgSO4/100 g H2O
C, 1 ton/h
MgSO4·7H2O
tC = 50 F
L
tL = 50 F
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
19 
 
PROBLEM # 08: 
The solubility of sodium bicarbonate in water 
is 9.6 g per 100 g water at 20°C and 16.4 g 
per 100 g water at 60°C. If a saturated 
solution of NaHCO3 at 60°C is cooled to 
20°C, what is the percentage of the 
dissolved salt that crystallizes out? 
 
SOLUTION: 
Basis: 100 kg feed 
 
Consider over-all material balance: 
𝐹 = 𝐿 + 𝐶 
𝐿 = 100 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider NaHCO3 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
 
𝑥𝐹 =
16.4 𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 16.4)𝑘𝑔 𝑓𝑒𝑒𝑑
= 0.1409
𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
𝑘𝑔 𝑓𝑒𝑒𝑑
 
𝑥𝐶 = 1.0 
𝑥𝐿 =
9.6 𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 9.6)𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.0876
𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
 
 
(0.1409) (100) = (0.0876) (𝐿) + (𝐶)(1) 
𝐿 = 160.8447 − 11.4155𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
 
100 − 𝐶 = 160.8447 − 11.4155𝐶 
𝐶 = 5.8417 𝑘𝑔 
 
% 𝑁𝑎𝐻𝐶𝑂3 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑙𝑖𝑧𝑒𝑑 =
𝐶
𝑥𝐹 𝐹
 𝑥 100 
 
% 𝑁𝑎𝐻𝐶𝑂3 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑙𝑖𝑧𝑒𝑑 =
5.8417 𝑘𝑔
(0.1409) (100 𝑘𝑔)
 𝑥 100 
 
% 𝑵𝒂𝑯𝑪𝑶𝟑 𝒄𝒓𝒚𝒔𝒕𝒂𝒍𝒍𝒊𝒛𝒆𝒅 = 𝟒𝟏. 𝟒𝟔 % 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
COOLING CRYSTALLIZER
F
tF = 60 F
16.4 g 
NaHCO3 /100 g 
H2O
C, 
9.6 g NaHCO3
per 100 g H2O 
tC = 20 F
L
tL = 20 F
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
20 
 
 
PROBLEM # 09: 
Glauber’s salt is made by crystallization from a water solution at 20°C. The aqueous solution at 
20°C contains 8.4% sodium sulfate. How many grams of water must be evaporated from a liter of 
such solution whose specific gravity is 1.077 so that when the residue solution after evaporation is 
cooled to 20°C, there will be crystallized out 80% of the original sodium sulfate as Glauber’s salt. 
The solubility of sodium sulfate in equilibrium with the decahydrate is 19.4 g Na2SO4 per 100 g 
H2O. 
SOLUTION: 
Basis: 1 L feed 
𝐹 = 1 𝐿 𝑥 
1.077 𝑘𝑔
𝐿
= 1.077 𝑘𝑔 
 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝐿 = 1.077 − 𝑉 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
𝑥𝐶𝐶 = 0.80𝑥𝐹 𝐹 
𝑥𝐹 𝐹 = (1.077 𝑘𝑔 𝑓𝑒𝑒𝑑) (
8.4 𝑘𝑔 𝑁𝑎2𝑆𝑂4
100 𝑘𝑔 𝑓𝑒𝑒𝑑
) = 0.0905 𝑘𝑔 𝑁𝑎2𝑆𝑂4 
𝑥𝐶𝐶 = (0.80)(0.0905 𝑘𝑔 𝑁𝑎2𝑆𝑂4) = 0.0724 𝑘𝑔 𝑁𝑎2 𝑆𝑂4 
𝑥𝐶 =
𝑀𝑁𝑎2𝑆𝑂4
𝑀𝑁𝑎2𝑆𝑂4∙10𝐻2 𝑂
=
142
322
= 0.4410
𝑘𝑔 𝑁𝑎2𝑆𝑂4
𝑘𝑔 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2𝑂
 
𝐶 =
0.0724𝑘𝑔 𝑁𝑎2𝑆𝑂4
0.4410
𝑘𝑔 𝑁𝑎2𝑆𝑂4
𝑘𝑔 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
= 0.1642 𝑘𝑔 
 
Substitute to equation 1 
𝐿 = 1.077 − 𝑉 − 0.1642 
𝐿 = 0.9128 − 𝑉 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Consider Na2SO4 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐿 =
19.4 𝑘𝑔 𝑁𝑎2𝑆𝑂4
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 19.4)𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.1625
𝑘𝑔 𝑁𝑎2𝑆𝑂4
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
 
 
0.0905 = (0.1625) (𝐿) + 0.0724 
𝐿 = 0.1114 𝑘𝑔 
CRYSTALLIZER
F
tF = 20 C
8.4% Na2SO4
C, 
tC = 20 C
L
tL = 20 C
V
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
21 
 
 
Substitute to equation 2 
0.1114 = 0.9128 − 𝑉 
𝑉 = 0.8014 𝑘𝑔 
 
𝑽 = 𝟖𝟎𝟏. 𝟒 𝒈 𝐴𝑁𝑆𝑊𝐸𝑅 
 
PROBLEM # 10: 
A hot solution of Ba(NO3)2 from an 
evaporator contains 30.6 kg Ba(NO3)2/100 
kg H2O and goes to a crystallizer where the 
solution is cooled and Ba(NO3)2 crystallizes. 
On cooling, 10% of the original water present 
evaporates. For a feed solution of 100 kg 
total, calculate the following: 
a) The yield of crystals if the solution is 
cooled to 290K, where the solubility is 
8.6 kg Ba(NO3)2/100 kg total water 
b) The yield if cooled instead to 283K, 
where the solubility is 7 kg Ba(NO3)2/100 
kg total water 
 
Source: Transport Processes and Unit 
Operations (Geankoplis) 
 
SOLUTION: 
a) If solution is cooled to 290K 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝐿 = 100 − 𝑉 − 𝐶 
𝐿 = 100 − 𝑉 − 𝐶 
If water evaporated is 10% of the original water present 
𝑉 = 0.10(1 − 𝑥𝐹)𝐹 
𝑥𝐹 =
30.6 𝑘𝑔 𝐵𝑎(𝑁𝑂3)2
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 30.6)𝑘𝑔 𝑓𝑒𝑒𝑑
= 0.2343
𝑘𝑔 𝐵𝑎(𝑁𝑂3 )2
𝑘𝑔 𝑓𝑒𝑒𝑑
 
𝑉 = 0.10(1 − 0.2343) (100 𝑘𝑔) 
𝑉 = 7.657 𝑘𝑔 
𝐿 = 100 − 7.657 − 𝐶 
𝐿 = 92.343 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider Ba(NO3)2 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐿 =
8.6 𝑘𝑔 𝐵𝑎(𝑁𝑂3 )2
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 8.6)𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.0792
𝑘𝑔 𝐵𝑎(𝑁𝑂3)2
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 = 1.0 
 
(0.2343) (100) = (0.0792) (𝐿) + (1.0)(𝐶) 
CRYSTALLIZER
F
30.6 kg Ba(NO3)2/100 kg H2O
C
L
V
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
22 
 
𝐿 = 295.8333 − 12.6263 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
92.343 − 𝐶 = 295 .8333 − 12.6263 𝐶 
𝑪 = 𝟏𝟕. 𝟓𝟎𝟐𝟔 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
b) If solution is cooled to 283 K 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝐿 = 100 − 𝑉 − 𝐶 
𝐿 = 100 − 𝑉 − 𝐶 
If water evaporated is 10% of the original water present 
𝑉 = 0.10(1 − 𝑥𝐹)𝐹 
𝑥𝐹 =
30.6 𝑘𝑔 𝐵𝑎(𝑁𝑂3)2
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 30.6)𝑘𝑔 𝑓𝑒𝑒𝑑
= 0.2343
𝑘𝑔 𝐵𝑎(𝑁𝑂3 )2
𝑘𝑔 𝑓𝑒𝑒𝑑
 
𝑉 = 0.10(1 − 0.2343) (100 𝑘𝑔) 
𝑉 = 7.657 𝑘𝑔 
𝐿 = 100 − 7.657 − 𝐶 
𝐿 = 92.343 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider Ba(NO3)2 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐿 =
7.0 𝑘𝑔 𝐵𝑎(𝑁𝑂3 )2
100 𝑘𝑔 𝐻2𝑂
 𝑥 
100 𝑘𝑔 𝐻2𝑂
(100 + 7)𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.0654
𝑘𝑔 𝐵𝑎(𝑁𝑂3)2
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 = 1.0 
 
(0.2343) (100) = (0.0654) (𝐿) + (1.0)(𝐶) 
𝐿 = 358.2569 − 15.2905 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
92.343 − 𝐶 = 358 .2569 − 15.2905 𝐶 
 
𝑪 = 𝟏𝟖. 𝟔𝟎𝟕𝟕 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
23 
 
 
 
 
 
 
PROBLEM # 11: 
A batch of 1,000 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) What are the weight of water required for 
the solution and the weight of KCl 
crystals obtained? 
b) What is the weight of crystals obtained if 
5% of the original water evaporates on 
cooling? 
 
Source: Transport Processes and Unit 
Operations (Geankoplis) 
 
SOLUTION: 
c) Assume crystallization by cooling (without evaporation) 
Consider over-all material balance: 
𝐹 = 𝐿 + 𝐶 
𝐹 = 1,000 𝑘𝑔 𝐾𝐶𝑙 𝑥 
100 𝑘𝑔 𝑠𝑜𝑙𝑛
35 𝑘𝑔 𝐾𝐶𝑙
= 2,857.14 𝑘𝑔 𝑓𝑒𝑒𝑑 
𝐿 = 2,857.14 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider KCl balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐿 = 0.254
𝑘𝑔 𝐾𝐶𝑙
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 = 1.0 
1,000 = (0.254)(𝐿) + (1.0)(𝐶) 
𝐿 = 3,937 − 3.937 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
2,857.14 − 𝐶 = 3,937 − 3.937 𝐶 
 
𝑪 = 𝟑𝟔𝟕. 𝟔𝟕 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅 
 
% 𝐻2𝑂 𝑖𝑛 𝑓𝑒𝑒𝑑 = 100 − %𝐾𝐶𝑙 = 100 − 35 = 65% 
CRYSTALLIZER
F
1,000 kg KCl
363K
C
293K
L
293K
V
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
24 
 
% 𝐻2𝑂 𝑖𝑛 𝑓𝑒𝑒𝑑 =
𝑤𝑡 𝐻2𝑂
𝑤𝑡 𝑓𝑒𝑒𝑑
 𝑥 100 
𝑤𝑡 𝐻2𝑂 = (2,857.14 𝑘𝑔 𝑓𝑒𝑒𝑑 ) (
65 𝑘𝑔 𝐻2𝑂
100 𝑘𝑔 𝑓𝑒𝑒𝑑
) 
𝒘𝒕 𝑯𝟐 𝑶 = 𝟏, 𝟖𝟓𝟕. 𝟏𝟒 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
d) Crystallization with evaporation 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝑉 = 0.05(1,857.14 𝑘𝑔) 
𝑉 = 92.8571 𝑘𝑔 
𝐿 = 2,857.14 − 92.8571 − 𝐶 
𝐿 = 2,764.2829 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3 
 
Consider KCl balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐿 = 0.254 
𝑥𝐶 = 1.0 
 
1,000 = (0.254)(𝐿) + (1.0)(𝐶) 
𝐿 = 3,937 − 3.937 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4 
 
Equate 3 and 4 
2,764.2829 − 𝐶 = 3,937 − 3.937 𝐶 
𝑪 = 𝟑𝟗𝟗. 𝟐𝟗 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
25 
 
 
 
 
 
 
 
PROBLEM # 12: 
The solubility of sodium sulfate is 40 parts 
Na2SO4 per 100 parts of water at 30°C, and 
13.5 parts at 15°C. The latent heat of 
crystallization (liberated when crystals form) 
is 18,000 g-cal per gmol Na2SO4. Glauber’s 
salt (Na2SO4·10H2O) is to be made in a 
Swenson-Walker crystallizer by cooling a 
solution, saturated at 30°C, to 15°C. Cooling 
water enters at 10°C and leaves at 20°C. 
The over-all heat transfer coefficient in the 
crystallizer is 25 BTU/h·ft2·°F and each foot 
of crystallizer has 3 sq ft of cooling surface. 
How many 10-ft units of crystallizer will be 
required to produce 1 ton/h of Glauber’s Salt 
 
Source: Unit Operations (Brown) 
 
SOLUTION: 
Consider over-all material balance: 
𝐹 = 𝐿 + 𝐶 
𝐿 = 𝐹 − 1 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider Na2SO4 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐹 =
40 𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4
100 𝑡𝑜𝑛 𝐻2𝑂
 𝑥 
100 𝑡𝑜𝑛 𝐻2𝑂
(100 + 40) 𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
= 0.2857
𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4
𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
 
𝑥𝐿 =
13.5 𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4
100 𝑡𝑜𝑛 𝐻2𝑂
 𝑥 
100 𝑡𝑜𝑛 𝐻2𝑂
(100 + 13.5) 𝑡𝑜𝑛 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.1189
𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4
𝑡𝑜𝑛 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 =
142 𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4
322 𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
= 0.4410
𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4
𝑡𝑜𝑛 𝑁𝑎2𝑆𝑂4 ∙ 10𝐻2𝑂
 
0.2857 𝐹 = 0.1189 𝐿 + 0.4410(1.0) 
𝐿 = 2.4029 𝐹 − 3.709 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Equate 1 and 2 
𝐹 − 1 = 2.4029 𝐹 − 3.709 
𝐹 = 1.931 
𝑡𝑜𝑛
ℎ
 
SWENSON-WALKER 
CRYSTALLIZER
F
tF = 30 C
L
tL = 15 C
C, 1 ton/h
Na2SO4·10H2O
tC = 15 C
W
t1 = 10 C
t2 = 20 C
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
26 
 
 
 
 
 
 
 
 
 
 
Consider heat balance: 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙 = 𝐹𝐶𝑃(𝑡𝐹 − 𝑡𝐶 ) + 𝐶𝐻𝐶 
𝐶𝑃 =
(𝑚𝐶𝑃)𝑁𝑎2𝑆𝑂4 +
(𝑚𝐶𝑃 )𝐻2𝑂
𝐹
 
From Table 2-194 (CHE HB 8th edition) 
𝐶𝑃 𝑁𝑎2𝑆𝑂4
= 32.8
𝑐𝑎𝑙
°𝐶 ∙ 𝑚𝑜𝑙
= 0.231
𝐵𝑇𝑈𝑙𝑏 ∙ °𝐹
 
𝐶𝑃 =
[(0.2857) (0.231) + (0.7143) (1.000)]
1
= 0.7803
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
 
 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙 = [(1.931
𝑡𝑜𝑛
ℎ
 𝑥 
2,000 𝑙𝑏
𝑡𝑜𝑛
) (0.7803
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (30 − 15)°𝐶 𝑥 
1.8°𝐹
°𝐶
]
+ [(1
𝑡𝑜𝑛
ℎ
 𝑥 
2,000 𝑙𝑏
𝑡𝑜𝑛
 𝑥 
𝑙𝑏𝑚𝑜𝑙
322 𝑙𝑏
 𝑥 
454 𝑔𝑚𝑜𝑙
𝑙𝑏𝑚𝑜𝑙
) (
18000 𝑐𝑎𝑙
𝑔𝑚𝑜𝑙
 𝑥 
𝐵𝑇𝑈
252.16 𝑐𝑎𝑙
)] 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙 = 282 ,656.8961
𝐵𝑇𝑈
ℎ
 
 
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚 
∆𝑇𝑙𝑚 =
(𝑡𝐹 − 𝑡2) − (𝑡𝐿 − 𝑡1)
ln
𝑡𝐹 − 𝑡2
𝑡𝐿 − 𝑡1
 
𝑡𝐹 = 30°𝐶 = 86°𝐹 
𝑡𝐿 = 15°𝐶 = 59°𝐹 
𝑡1 = 10°𝐶 = 50°𝐹 
𝑡2 = 20°𝐶 = 68°𝐹 
∆𝑇𝑙𝑚 =
(86 − 68) − (59 − 50)
ln
86 − 68
59 − 50
= 12.9842°𝐹 
 
𝐴 =
262 ,656.8961 
𝐵𝑇𝑈
ℎ
(25
𝐵𝑇𝑈
ℎ ∙ 𝑓𝑡2 ∙ °𝐹
) (12.9842°𝐹)
 
 
𝐴 = 870.7718𝑓𝑡2 
# 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 = 880.7718 𝑓𝑡2 𝑥 
1 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
3 𝑓𝑡2
 𝑥 
1 𝑢𝑛𝑖𝑡
10 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
 
 
# 𝒐𝒇 𝒖𝒏𝒊𝒕𝒔 = 𝟐𝟗. 𝟎𝟑 ≈ 𝟑𝟎 𝒖𝒏𝒊𝒕𝒔 𝐴𝑁𝑆𝑊𝐸𝑅 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
27 
 
 
 
 
 
 
 
PROBLEM # 13: 
A continuous adiabatic vacuum crystallizer is 
to be used for the production of MgSO4·7H2O 
crystals from 20,000 lb/h of solution 
containing 0.300 weight fraction MgSO4. The 
solution enters the crystallizer at 160°F. The 
crystallizer is to be operated so that the 
mixture of mother liquor and crystals leaving 
the crystallizer contains 6,000 lb/h of 
MgSO4·7H2O crystals. The estimated boiling 
point elevation of the solution in the 
crystallizer is 10°F. How many pounds of 
water are vaporized per hour? 
 
Source: Unit Operations (Brown) 
 
SOLUTION: 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝐿 = 20,000 − 6,000 − 𝑉 
𝐿 = 14,000 − 𝑉 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider MgSO4 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐹 = 0.3000
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑓𝑒𝑒𝑑
 
𝑥𝐶 =
120 𝑙𝑏 𝑀𝑔𝑆𝑂4
246 𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
= 0.4878
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
 
(0.30)(20,000) = (𝑥𝐿)(𝐿) + (0.4878) (6,000) 
𝐿 =
3,073.2
𝑥𝐿
 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Consider enthalpy balance: 
ℎ𝑓 𝐹 = ℎ𝑉𝑉 + ℎ𝐿𝐿 + ℎ𝑐 𝐶 
THE PROBLEM CAN BE SOLVED BY TRIAL AND ERROR SINCE TEMPERATURE OF THE 
SOLUTION AFTER CRYSTALLIZATION IS UNKNOWN AND ENTHALPIES ARE DEPENDENT 
ON TEMPERATURE 
ADIABATIC VACUUM 
CRYSTALLIZER
F, 20,000 lb/h
xF = 0.3000
tF = 160 F
L
BPE = 10 F
C = 6,000 lb/h
MgSO4·7H2O
V
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
28 
 
1. Assume temperature of the solution 
2. From figure 27-3 (Unit Operations by McCabe and Smoth 7th edition), obtain mass fraction of 
MgSO4 at the assumed temperature of the solution 
3. Solve for “L” using equation 2 
4. Solve for “V” using equation 1 
5. Check if assumed temperature is correct by conducting enthalpy balance 
a. Obtain values of hF, hC and hL from figure 27-4 (Unit Operations by McCabe and 
Smith 7th edition) at the designated temperatures and concentrations 
b. Compute for hV 
c. Using the enthalpy balance equation, compute for “V” using the value of “L” from step 
3 
6. Compare values of “V” from step 4 with that from step 5-c 
7. If not the same (or approximately the same), conduct another trial and error calculations 
 
TRIAL 1: Assume temperature of the solution at 60°F 
From figure 27-3 (Unit Operations by McCabe and Smith 7th edition) 
𝑥𝐿 = 0.245
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
Substitute to equation 2 
𝐿 =
3,073.2
0.245
= 12,543.67 𝑙𝑏 
Substitute to equation 1 
𝑉 = 14,000 − 12,543.67 = 1,456.33 𝑙𝑏 
 
From figure 27-4 (Unit Operations by McCabe and Smith, 7th edition) 
ℎ𝐹 𝑎𝑡 160°𝐹 𝑎𝑛𝑑 30% 𝑀𝑔𝑆𝑂4 = 5
𝐵𝑇𝑈
𝑙𝑏
 
ℎ𝐶 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 48.78% 𝑀𝑔𝑆𝑂4 = −158
𝐵𝑇𝑈
𝑙𝑏
 
ℎ𝐿 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 24.5% 𝑀𝑔𝑆𝑂4 = −50
𝐵𝑇𝑈
𝑙𝑏
 
 
Temperature of vapor is 60 – 10 = 50°F 
ℎ𝑉 = 𝐻𝑉 + 𝐶𝑃 𝑥 𝐵𝑃𝐸 
From steam table at 50°F, 𝐻𝑉 = 1,083.3
𝐵𝑇𝑈
𝑙𝑏
 
ℎ𝑉 = 1,083.3
𝐵𝑇𝑈
𝑙𝑏
+ [(0.45
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (10°𝐹 )] 
ℎ𝑉 = 1,087.8
𝐵𝑇𝑈
𝑙𝑏
 
 
ℎ𝑓 𝐹 = ℎ𝑉𝑉 + ℎ𝐿𝐿 + ℎ𝑐 𝐶 
(5)(20 ,000) = (1087 .8)(𝑉) + (−50)(12,543.67) + (−158)(6,000) 
𝑉 = 1,539.97 𝑙𝑏 
 
Since % error is about 5%, assumed value can be considered correct. 
 
𝑽 = 𝟏, 𝟓𝟑𝟗. 𝟗𝟕 
𝒍𝒃
𝒉
 𝒐𝒓 𝟏, 𝟒𝟓𝟔. 𝟑𝟑 
𝒍𝒃
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
29 
 
 
 
 
 
 
PROBLEM # 14: 
Crystals of CaCl2·6H2O are to be obtained 
from a solution of 35 weight % CaCl2, 10 
weight % inert soluble impurity, and 55 
weight % water in an Oslo crystallizer. 
The solution is fed to the crystallizer at 
100°F and receives 250 BTU/lb of feed 
from the external heater. Products are 
withdrawn from the crystallizer at 40°F. 
a) What are the products from the 
crystallizer? 
b) The magma is centrifuged to a 
moisture content of 0.1 lb of liquid per 
lb of CaCl2·6H2O crystals and then 
dried in a conveyor drier. What is the 
purity of the final dried crystalline 
product? 
 
Source: Principles of Unit Operations 2nd 
edition (Foust, et al) 
 
SOLUTION: 
Basis: 1 lb of inert soluble-free feed 
from table 2-120 (CHE HB 8th edition), solubilities of CaCl2·6H2O 
0°C 59.5 lb/100 lb H2O 
10°C 65 lb/100 lb H2O 
20°C 74.5 lb/100 lb H2O 
30°C 102 lb/100 lb H2O 
At 100°F (37.8°C), solubility is (by extrapolation), 123.45 lb/100 lb H2O 
At 40°F (4.4°C), solubility is 61.92 lb/100 lb H2O 
Since the equipment is Oslo crystallizer, there the process is supersaturation by evaporation 
 
By heat balance around the crystallizer 
𝑞 = 𝐹𝐶𝑃(𝑡𝐹 − 𝑡𝐿) + 𝐶𝐻𝐶 − 𝑉𝜆 𝑉 
From table 2-194, specific heat of CaCl2, cal/K·mol 
𝐶𝑃 = 16.9 + 0.00386𝑇 
where T is in K 
At 100°F (310.93 K) 
OSLO CRYSTALLIZER
F
CaCl2 = 35%
Inert = 10%
H2O = 55%
tF = 100 F
C’’
CaCl2·6H2O
V
CENTRIFUGE
DRYER
L
M (magma)
C
Inert 
L
tF = 40 F
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
30 
 
𝐶𝑃 = 18.1
𝑐𝑎𝑙
𝑚𝑜𝑙 · 𝐾
𝑥 
1 𝑚𝑜𝑙
110.9 𝑔
 𝑥 
1
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
1
𝑐𝑎𝑙
𝑔 ∙ °𝐹
= 0.1632
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
 
At 40°F (277.59 K) 
𝐶𝑃 = 17.97
𝑐𝑎𝑙
𝑚𝑜𝑙 · 𝐾
𝑥 
1 𝑚𝑜𝑙
110.9 𝑔
 𝑥 
1
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
1
𝑐𝑎𝑙
𝑔 ∙ °𝐹
= 0.1620
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
 
𝐶̅𝑃 = 
0.1632 + 0.1620
2
= 0.1626
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
 
 
 
 
 
For the feed 
𝐶𝑃 =
(0.35 𝑙𝑏 𝐶𝑎𝐶𝑙2) (0.1626
𝐵𝑇𝑈
𝑙𝑏 𝐶𝑎𝐶𝑙2 ∙ °𝐹
) + (0.55 𝑙𝑏 𝐻2𝑂) (1
𝐵𝑇𝑈
𝑙𝑏 𝐻2𝑂 ∙ °𝐹
)
(0.35 + 0.55) 𝑙𝑏 𝑓𝑒𝑒𝑑
 
𝐶𝑃 = 0.6743
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
 
 
 
From table 2-224 (CHE HB 8th edition), heat of solution of CaCl2·6H2O = -4,100 cal/mol; in 
the absence of data on heat of crystallization, heat of solution can be used instead but of 
opposite sign 
𝐻𝐶 = 4,100
𝑐𝑎𝑙
𝑚𝑜𝑙
= 18.73
𝑐𝑎𝑙
𝑔
= 33.71
𝐵𝑇𝑈
𝑙𝑏
 
 
From the steam table, at 40°F, 𝜆 = 1,070.9 𝐵𝑇𝑈/𝑙𝑏 
 
 
(250)(1) = (1)(0.6743)(100 − 40) + (33.71)(𝐶) − (1,070.9)(𝑉) 
𝑉 = 0.0315𝐶 − 0.1957 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider over-all material balance: 
𝐹 = 𝑉 + 𝐿 + 𝐶 
𝐿 = 1 − 𝑉 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Substitute 1 in 2 
𝐿 = 1 − (0.0315𝐶 − 0.1994) − 𝐶 
𝐿 = 0.8006 − 1.0315𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3 
 
Consider solute (CaCl2·6H2O) balance, inert soluble-free 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐹 =
35 𝑙𝑏 𝐶𝑎𝐶𝑙2
(35 + 55) 𝑙𝑏 𝑓𝑒𝑒𝑑
 𝑥 
1 𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2∙ 6𝐻2𝑂
𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
 𝑥 
218 .9
𝑙𝑏 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 
𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂
110.9
𝑙𝑏 𝐶𝑎𝐶𝑙2
𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
= 0.7676 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
31 
 
𝑥𝐿 =
61.92 𝑙𝑏 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 
100 𝑙𝑏 𝐻2𝑂
 𝑥 
100 𝑙𝑏 𝐻2𝑂
(100 + 61.92) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.3824 
𝑥𝐶 = 1 
(0.7676) (1) = (0.3824)(𝐿) + (1)(𝐶) 
𝐿 = 2.0073 − 2.6151𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4 
 
Equate 3 and 4 
0.8006 − 1.0315 𝐶 = 2.0073 − 2.6151 𝐶 
𝐶 = 0.7620 𝑙𝑏 (𝑖𝑛𝑒𝑟𝑡 𝑠𝑜𝑙𝑢𝑏𝑙𝑒 𝑓𝑟𝑒𝑒 ) 
𝐿 = 0.0146 𝑙𝑏 (𝑖𝑛𝑒𝑟𝑡 𝑠𝑜𝑙𝑢𝑏𝑙𝑒 𝑓𝑟𝑒𝑒 ) 
𝑉 = 0.2234 𝑙𝑏 
 
Composition of the liquor (including the inert soluble) 
𝑤𝑡 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 = 0.0146 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟 𝑥 
61.92 𝑙𝑏 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 
100 𝑙𝑏 𝐻2𝑂
 𝑥 
100 𝑙𝑏 𝐻2𝑂
(100 + 61.92)𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑤𝑡 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 = 0.0056 𝑙𝑏 
𝑤𝑡 𝐻2𝑂 𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 = 0.0146 − 0.0056 = 0.0090 𝑙𝑏 
 
 lb % 
CaCl2·6H2O 0.0056 4.89 
H2O 0.0090 7.85 
inerts 0.1000 87.26 
 0.1146 100.00 
 
For the crystals leaving the centrifuge: 
𝑤𝑡 𝑙𝑖𝑞𝑢𝑜𝑟 𝑎𝑑ℎ𝑒𝑟𝑒𝑑 𝑖𝑛 𝑐𝑟𝑠𝑦𝑡𝑎𝑙𝑠 = 0.7620 𝑙𝑏 𝑐𝑟𝑠𝑦𝑡𝑎𝑙𝑠 𝑥 
0.1 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑙𝑏 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
= 0.0762 𝑙𝑏 
Composition of crystals leaving the centrifuge 
 lb 
CaCl2·6H2O 
 crystallized 0.7620 
 from liquor 0.0762 x 0.0489 0.0037 0.7657 
 
H2O 0.0762 x 0.0785 0.0060 0.0060 
inerts 0.0762 x 0.8726 0.0665 0.0665 
 0.8382 
 
In the dryer, assume all free water has been removed 
 
Composition of dried crystals 
 lb % 
CaCl2·6H2O 0.7657 92.01 
inerts 0.0665 7.99 
 0.8322 100.00 
 
 
𝑷𝒖𝒓𝒊𝒕𝒚 = 𝟗𝟐. 𝟎𝟏% 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
32 
 
 
 
 
 
 
 
PROBLEM # 15: 
 
Lactose syrup is concentrated to 8 g lactose 
per 10 g of water and then run into a 
crystallizing vat which contains 2,500 kg of the 
syrup. In this vat, containing 2,500 kg of syrup, 
it is cooled from 57°C to 10°C. Lactose 
crystallizes with one molecule of water of 
crystallization. The specific heat of the lactose 
solution is 3470 J/kg·°C. The heat of solution 
for lactose monohydrate is -15,500 kJ/kmol. 
The molecular weight of lactose monohydrate 
is 360 and the solubility of lactose at 10°C is 
1.5 g/10 g water. Assume that 1% of the water 
evaporates and that the heat loss trough the 
vat walls is 4 x 104 kJ. Calculate the heat to be 
removed in the cooling process. 
 
 
SOLUTION: 
Consider over-all material balance 
𝐹 = 𝐿 + 𝑉 + 𝐶 
𝑤𝑡 𝐻2𝑂 𝑖𝑛 𝑓𝑒𝑒𝑑 = 2,500 𝑘𝑔 𝑓𝑒𝑒𝑑 𝑥 
10 𝑘𝑔 𝐻2𝑂
(10 + 8) 𝑘𝑔 𝑓𝑒𝑒𝑑
= 1,388.89 𝑘𝑔 
𝑉 = 0.01(1,388.89 𝑘𝑔) = 13.89 𝑘𝑔 
𝐿 = 2,500 − 13.89 − 𝐶 
𝐿 = 2,486.11 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
Consider lactose balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐹 =
8 𝑘𝑔
10 + 8
= 0.4444
𝑘𝑔 𝐶12𝐻22𝑂11
𝑘𝑔 𝑓𝑒𝑒𝑑
 
𝑥𝐿 =
1.5
10 + 1.5
= 0.1304
𝑘𝑔 𝐶12𝐻22𝑂11
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 =
𝑀𝐶12𝐻22𝑂11
𝑀𝐶12𝐻22𝑂11∙𝐻2 𝑂
=
342
360
= 0.95
𝑘𝑔 𝐶12𝐻22𝑂11
𝑘𝑔 𝑐𝑟𝑦𝑠𝑡𝑎𝑙
 
(0.4444) (2,500) = (0.1304)(𝐿) + (0.95)(𝐶) 
𝐿 = 8,519.9386 − 7.2853 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
OSLO CRYSTALLIZER
F
2,500 kg
8 g lactose per 10 g 
water
tF = 57 C
V
L
1.5 g lactose 
per 10 g water
C
tC = 10 C
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
33 
 
 
Equate 1 and 2 
2,486.11 − 𝐶 = 8,519.9386 − 7.2853𝐶 
𝐶 = 959.99 𝑘𝑔 
𝐿 = 1,526.12 𝑘𝑔 
 
 
 
 
 
 
 
 
Consider heat balance: 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑃(𝑡𝐹 − 𝑡𝐿) + 𝐶𝐻𝐶 − 𝑉𝜆 𝑉 
At 10°C (50°F), 
𝜆 = 1,065.2
𝐵𝑇𝑈
𝑙𝑏
= 2,472.47
𝑘𝐽
𝑘𝑔
 
𝐻𝐶 = 15,500
𝑘𝐽
𝑘𝑚𝑜𝑙
 𝑥 
𝑘𝑚𝑜𝑙
360 𝑘𝑔
= 43.06
𝑘𝐽
𝑘𝑔
 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = [(2,500 𝑘𝑔) (3.47
𝑘𝐽
𝑘𝑔 ∙ °𝐶
) (57 − 10)°𝐶] + [(959 .99 𝑘𝑔) (43 .06
𝑘𝐽
𝑘𝑔
)]
− [(13.89 𝑘𝑔) (2,472.47 
𝑘𝐽
𝑘𝑔
)] 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 414.7196 𝑥 10
3 𝑘𝐽 
 
𝑞𝑇 = 414.7196 𝑥 10
3 𝑘𝐽 + 4 𝑥 104 𝑘𝐽 
 
𝒒𝑻 = 𝟒𝟓𝟒. 𝟕𝟐 𝒙 𝟏𝟎
𝟑 𝒌𝑱 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
 
 
 
 
 
 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
34 
 
 
 
 
 
 
 
PROBLEM # 16: 
Sal soda (Na2CO3·10H2O) is to be made by dissolving soda ash in a mixture of mother liquor and 
water to form a 30% solution by weight at 45°C and then cooling to 15°C. The wet crystals removed 
from the mother liquor consist of 90% sal soda and 10% mother liquor by weight. The mother liquor 
is to be dried on the crystals as additional sal soda. The remainder of the mother liquor is to be 
returned to the dissolving tanks. At 15°C, the solubility of Na2CO3 is 14.2 parts per 100 parts water. 
 
Crystallization is to be done in a Swenson-Walker crystallizer. This is to be supplied with water at 
10°C, and sufficient cooling water is to be used to ensure that the exit water will not be over 20°C. 
The Swenson-walker crystallizer is built in units 10 ft long, containing 3 ft2 of heating surface per 
foot of length. An over-all heat transfer coefficient of 35 BTU/ft2·h·°F is expected. 
 
The latent heat of crystallization of sal soda at 15°C is approximately 25,000 cal/mol. The specific 
heat of the solution is 0.85 BTU/lb·°F. A production of 1 ton/h of dried crystals is desired. Radiation 
losses and evaporation from the crystallizer are negligible. 
a) What amounts of water and sal soda are to be added to the dissolver per hour? 
b) How many units of crystallizer are needed? 
c) What is to be the capacity of the refrigeration plant, in tons of refrigeration, if the cooling water 
is to be cooled and recycled? One ton of refrigeration is equivalent to 12,000 BTU/h. 
 
 
SOLUTION: 
Basis: 2,000 lb/h (1 ton/h) of sal soda 
Consider over-all material balance of the system 
𝑊 + 𝐹 = 𝑉 + 𝐶 
DISSOLVER CRYSTALLIZER DRYERFILTER
C (Sal Soda)
45C 15C
F (Soda Ash)
W (Water)
R (remainder 
mother liquor)
V
A B D
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
35 
 
𝑉 = 𝑊 + 𝐹 − 2,000 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
Consider Na2CO3 balance around the system 
𝑥𝐹 𝐹 = 𝑥𝐶𝐶 
𝑥𝐹 = 1.0 
𝑥𝐶 =
𝑀𝑁𝑎2𝐶𝑂3
𝑀𝑁𝑎2𝐶𝑂3∙10𝐻2𝑂
=
106
286
= 0.3706
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
 
𝐹 =
(0.3706
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
) (2,000 
𝑙𝑏 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂 
ℎ
)
1.0 
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝑠𝑜𝑑𝑎 𝑎𝑠ℎ
 
𝑭 = 𝟕𝟒𝟏. 𝟐 
𝒍𝒃
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Substitute to equation 1 
𝑉 = 𝑊 + 741.2 − 2,000 
𝑉 = 𝑊 − 1,258.8 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
 
Consider solute (Na2CO3) balance around the dryer 
𝑥𝐷𝐷 = 𝑥𝐶𝐶 
𝑥𝐷 =
(0.90 𝑙𝑏𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂) (
106 𝑙𝑏 𝑁𝑎2𝐶𝑂3
286 𝑙𝑏 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
) + (0.10 𝑙𝑏 𝐿) (
14.2 𝑙𝑏 𝑁𝑎2𝐶𝑂3
(100 + 14.2)𝑙𝑏 𝐿
)
1 𝑙𝑏 𝐷
 
𝑥𝐷 = 0.3460
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝐷
 
 
𝐷 =
(0.3706
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
) (2,000 
𝑙𝑏 𝑁𝑎2𝐶𝑂3 ∙ 10𝐻2𝑂
ℎ
)
0.3460
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝐷
 
𝐷 = 2,142.20 
𝑙𝑏
ℎ
 
 
Consider over-all material balance around the dryer 
𝐷 = 𝑉 + 𝐶 
𝑉 = 2,142.20 − 2,000 
𝑉 = 142.20 
𝑙𝑏
ℎ
 
Substitute to equation 2 
142.20 = 𝑊 − 1,258.8 
𝑾 = 𝟏, 𝟒𝟎𝟏 
𝒍𝒃
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Consider solute (Na2CO3) balance around the dissolver 
𝑥𝐹 𝐹 + 𝑥𝑅𝑅 = 𝑥𝐴𝐴 
𝑥𝐴 = 0.30
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝐴
 
𝑥𝑅 =
14.2 𝑙𝑏 𝑁𝑎2𝐶𝑂3
(100 + 14.2)𝑙𝑏𝑅
= 0.1243
𝑙𝑏 𝑁𝑎2𝐶𝑂3
𝑙𝑏 𝑅
 
(1.0)(741.2) + (0.1243) (𝑅) = (0.30)(𝐴) 
𝐴 = 2,470.67 + 0.4143𝑅 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
36 
 
 
Consider over-all material balance around the dissolver 
𝐹 + 𝑊 + 𝑅 = 𝐴 
𝐴 = 741.2 + 1,401 + 𝑅 
𝐴 = 2,142.2 + 𝑅 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4 
 
 
 
 
Equate 3 and 4 
2,470.67 + 0.4143𝑅 = 2,142.2 + 𝑅 
𝑅 = 560.8 
𝑙𝑏
ℎ
 
𝐴 = 2,973.0 
𝑙𝑏
ℎ
 
 
Consider heat balance around the crystallizer 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐴𝐶𝑃(𝑡𝐴 − 𝑡𝐵) + 𝐶′𝐻𝐶 
𝐶 ′ = 0.90𝐷 = 0.90 (2,142.20 
𝑙𝑏
ℎ
) = 1,928.0 
𝑙𝑏
ℎ
 
𝐻𝐶 = 25,000
𝑐𝑎𝑙
𝑚𝑜𝑙
 𝑥 
𝑚𝑜𝑙
286 𝑔
 𝑥 
1 
𝐵𝑇𝑈
𝑙𝑏
0.55556
𝑐𝑎𝑙
𝑔
= 157.34
𝐵𝑇𝑈
𝑙𝑏
 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = [(2,973.0
𝑙𝑏
ℎ
) (0.85
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (45 − 15) °𝐶 𝑥 
1.8°𝐹
°𝐶
] + [(1,928.0
𝑙𝑏
ℎ
) (157.34
𝐵𝑇𝑈
𝑙𝑏
 )] 
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 439,812.22
𝐵𝑇𝑈
ℎ
 
 
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚 
∆𝑇𝑙𝑚 =
(𝑡𝐴 − 𝑡2) − (𝑡𝐵 − 𝑡1)
ln
𝑡𝐴 − 𝑡2
𝑡𝐵 − 𝑡1
 
∆𝑇𝑙𝑚 =
[(45 − 20) − (15 − 10)]°𝐶 𝑥 
1.8°𝐹
°𝐶
ln
45 − 20
15 − 10
 
∆𝑇𝑙𝑚 = 22.37°𝐹 
𝐴 =
439,812.22
𝐵𝑇𝑈
ℎ
(35
𝐵𝑇𝑈
ℎ ∙ 𝑓𝑡2 ∙ °𝐹
) (22.37°𝐹)
 
𝐴 = 561.74 𝑓𝑡2 
# 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 = 561.74 𝑓𝑡2 𝑥 
1 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
3 𝑓𝑡2
 𝑥 
1 𝑢𝑛𝑖𝑡
10 𝑓𝑡
 
 
# 𝒐𝒇 𝒖𝒏𝒊𝒕𝒔 = 𝟏𝟖. 𝟕 ≈ 𝟏𝟗 𝒖𝒏𝒊𝒕𝒔 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
Refrigeration capacity: 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
37 
 
𝑅𝐶 = 439 ,812.22
𝐵𝑇𝑈
ℎ
 𝑥 
𝑡𝑜𝑛 𝑟𝑒𝑓𝑟𝑖𝑔𝑒𝑟𝑎𝑡𝑖𝑜𝑛
12,000
𝐵𝑇𝑈
ℎ
 
 
𝑹𝑪 = 𝟑𝟔. 𝟔𝟓 𝒕𝒐𝒏𝒔 𝐴𝑁𝑆𝑊𝐸𝑅 
 
 
 
PROBLEM # 17: 
 
One ton of Na2S2O3·5H2O is to be crystallized per hour by cooling a solution containing 56.5% 
Na2S2O3 to 30°C in a Swenson-Walker crystallizer. Evaporation is negligible. The product is to 
be sized closely to approximately 14 mesh. Seed crystals closely sized to 20 mesh are 
introduced with the solution as it enters the crystallizer. How many tons of seed crystals and 
how many tons of solutions are required per hour? At 30°C, solubility of Na2S2O3 is 83 parts per 
100 parts water 
Source: Unit Operations (Brown, et al) 
 
SOLUTION: 
∫ 𝑑𝑊𝑃 = ∫ (1 +
∆𝐷
𝐷𝑆
)
3
𝑑𝑊𝑆
𝑊𝑆
0
𝑊𝑃
0
 
From table 19-6 (CHE HB 8th edition) 
𝐷𝑃 = 𝑚𝑒𝑠ℎ 14 = 1.19 𝑚𝑚 (𝑠𝑖𝑒𝑣𝑒 𝑜𝑝𝑒𝑛𝑖𝑛𝑔) 
𝐷𝑆 = 𝑚𝑒𝑠ℎ 20 = 0.841 𝑚𝑚 (𝑠𝑖𝑒𝑣𝑒 𝑜𝑝𝑒𝑛𝑖𝑛𝑔) 
∆𝐷 = 𝐷𝑃 − 𝐷𝑆 
∆𝐷 = 1.19 − 0.841 = 0.349 𝑚𝑚 
∫ 𝑑𝑊𝑃 = ∫ (1 +
0.349
0.841
)
3
𝑑𝑊𝑆
𝑊𝑆
0
𝑊𝑃
0
 
𝑊𝑃 = 2.833 𝑊𝑆 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
 
𝑊𝑃 = 𝐶 + 𝑊𝑆 
𝑊𝑃 = 2,000 + 𝑊𝑆 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
Equate 1 and 2 
2.833𝑊𝑆 = 2,000 + 𝑊𝑆 
𝑾𝑺 = 𝟏, 𝟎𝟗𝟏. 𝟏𝟏
𝒍𝒃
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Consider Na2S2O3 balance: 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐹 = 0.565
𝑙𝑏 𝑁𝑎2𝑆2𝑂3
𝑙𝑏 𝑓𝑒𝑒𝑑
 
𝑥𝐿 =
83 𝑙𝑏 𝑁𝑎2𝑆2𝑂3
(100 + 83) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
= 0.4536
𝑙𝑏 𝑁𝑎2𝑆2𝑂3
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 =
𝑀𝑁𝑎2𝑆2𝑂3
𝑀𝑁𝑎2𝑆2𝑂3∙5𝐻2 𝑂
=
158
248
= 0.6371
𝑙𝑏 𝑁𝑎2𝑆2𝑂3 
𝑙𝑏 𝑁𝑎2𝑆2𝑂3 ∙ 5𝐻2𝑂
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
38 
 
(0.565)(𝐹) = (0.4536)(𝐿) + (0.6371)(2,000) 
𝐿 = 1.2456𝐹 − 2,809.08 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3 
 
Consider over-all material balance 
𝐹 = 𝐿 + 𝐶 
𝐿 = 𝐹 − 2,000 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4 
Equate 3 and 4 
1.2456𝐹 − 2,809.08 = 𝐹 − 2000 
𝑭 = 𝟑, 𝟐𝟗𝟒. 𝟑𝟏 
𝒍𝒃
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
PROBLEM # 18: 
 
A Swenson-Walker crystallizer is fed with a saturated solution of magnesium sulfate at 110°F. 
The solution and its crystalline crop are cooled to 40°F. The inlet solution contains 1 g of seed 
crystals per 100 g of solution. The seeds are 80 mesh. Assuming ideal growth, what is the mesh 
size of the crystals leaving with the cooled product? Evaporation may be neglected. 
 
SOLUTION: 
Basis: 100 lb feed 
Consider over-all material balance 
𝐹 = 𝐿 + 𝐶 
𝐿 = 100 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 
Consider MgSO4 balance 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
From figure 27-3 (Unit Operation 7th edition, McCabe and Smith) at 110°F 
𝑥𝐹 = 0.32
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑓𝑒𝑒𝑑
 
From figure 27-3 (Unit Operations 7th edition, McCabe and Smith) at 40°F 
𝑥𝐿 = 0.21
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
 
𝑥𝐶 =
𝑀𝑀𝑔𝑆 𝑂4
𝑀𝑀𝑔𝑆 𝑂4∙7𝐻2 𝑂
=
120 .38
246 .49
= 0.4884
𝑙𝑏𝑀𝑔𝑆 𝑂4 
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2𝑂
 
(0.32)(100) = (0.21)(𝐿) + (0.4884) (𝐶) 
𝐿 = 152.38 − 2.3257𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 
Equate 1 and 2 
100 − 𝐶 = 152.38 − 2.3257 𝐶 
𝐶 = 39.51 𝑙𝑏 
𝑊𝑆 = 100 𝑙𝑏 𝑓𝑒𝑒𝑑 𝑥 
1 𝑙𝑏 𝑠𝑒𝑒𝑑𝑠
100 𝑙𝑏 𝑓𝑒𝑒𝑑
= 1 𝑙𝑏 
𝑊𝑃 = 𝐶 + 𝑊𝑆 = 39.51 + 1 = 40.51 𝑙𝑏 
∫ 𝑑𝑊𝑃 = ∫ (1 +
∆𝐷
𝐷𝑆
)
3
𝑑𝑊𝑆
𝑊𝑆
0
𝑊𝑃
0
 
𝑊𝑃 = [
𝐷𝑃
𝐷𝑆
]
3
𝑊𝑆 
From table 19-6 (CHE HB 8th edition) 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
39 
 
𝐷𝑆 = 𝑚𝑒𝑠ℎ 80 = 0.177 𝑚𝑚 (𝑠𝑖𝑒𝑣𝑒 𝑜𝑝𝑒𝑛𝑖𝑛𝑔) 
𝐷𝑃 = (0.177 𝑚𝑚) √
40.51 𝑙𝑏
1 𝑙𝑏
3
 
𝐷𝑃 = 0.6079 𝑚𝑚 
 
From table 19-6 (CHE HB 8th edition) 
 
𝑴𝑬𝑺𝑯 𝑺𝑰𝒁𝑬 = 𝟐𝟒 𝑴𝑬𝑺𝑯 𝐴𝑁𝑆𝑊𝐸𝑅 
PROBLEM # 19: 
 
Trisodium phosphate is to be recovered as Na3PO4·12H2O from a 35 weight % solution originally 
at 190°F by cooling and seeding in a Swenson-Walker crystallizer. From 20,000 lb/h feed, 7,000 
lb/h of product crystals in addition to the seed crystals are to be obtained. Seed crystals fed at 
a rate of 500 lb/h have the following size range: 
Weight Range Size Range, in 
10 % - 0.0200 + 0.0100 
20 % - 0.0100 + 0.0050 
40 % - 0.0050 + 0.0025 
30 % - 0.0025 + 0.0010 
Latent heat of crystallization of trisodium phosphate is 27,500 BTU/lbmol. Specific heat for the 
trisodium phosphate solution may be taken as 0.8 BTU/lb·°F. 
a) Estimate the product particle size distribution 
b) To what temperature must the solution be cooled, and what will be the cooling duty in BTU/h 
 
 
SOLUTION: 
∫ 𝑑𝑊𝑃
𝑊𝑃
0
= ∫ (1 +
∆𝐷
𝐷𝑆
)
3
𝑑𝑊𝑆
𝑊𝑆
0
 
𝑑𝑊𝑆 = 𝑊𝑆𝑑𝜙𝑆 
𝑊𝑃 = 𝑊𝑆 ∫ (1 +
∆𝐷
𝐷𝑆
)
31
0
𝑑𝜙𝑆 
𝑊𝑃
𝑊𝑆
= ∫ (1 +
∆𝐷
𝐷𝑆
)
31
0
𝑑𝜙𝑆 
𝑙𝑒𝑡, 𝑚 =
𝑊𝑃
𝑊𝑆
= ∫ (1 +
∆𝐷
𝐷𝑆
)
31
0
𝑑𝜙𝑆 
Δ𝑚 = (1 +
∆𝐷
𝐷𝑆
)
3
Δ𝜙𝑆 
Where: Δ𝜙𝑆 = fractional weight range 
 
Solve for required 𝑚: 
𝑚𝑟𝑒𝑞𝑑 =
𝑊𝑃
𝑊𝑆
=
7,000 𝑙𝑏
500 𝑙𝑏
= 14 
 
This problem can be solved by trial and error 
1. Assume value of Δ𝐷 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
40 
 
2. Solve for (1 +
∆𝐷
𝐷𝑆
)
3
for each size range, use the mean 𝐷𝑆 for each size range 
3. Solve for Δ𝑚 
4. Get the total Δ𝑚 
5. If ∑ Δ𝑚 = 𝑚𝑟𝑒𝑞𝑑 , then assumed Δ𝐷 is correct; if not, redo another trial 
 
 
 
TRIAL 1: Assume Δ𝐷 = 0.004 𝑖𝑛 
𝐷𝑆 Δ𝜙𝑆 (1 +
Δ𝐷
𝐷𝑆
)
3
 Δ𝑚 = Δ𝜙𝑆 (1 +
Δ𝐷
𝐷𝑆
)
3
 
0.0150 𝑖𝑛 0.10 2.0322 0.2032 
0.0075𝑖𝑛 0.20 3.6050 0.7210 
0.0038 𝑖𝑛 0.40 8.6483 3.4593 
0.0018 𝑖𝑛 0.30 33.4554 10.0366 
 1.00 14.4201 
Since % error is less than 5%, assumed value can be considered 
 
For particle size distribution: 
𝐷𝑃 = Δ𝐷 + 𝐷𝑆 
% 𝑤𝑡 = 100Δ𝜙𝑆 =
Δ𝑚
(1 +
Δ𝐷
𝐷𝑆
)
3 𝑥 100 
 
𝑆𝐸𝐸𝐷 𝐶𝑅𝑌𝑆𝑇𝐴𝐿𝑆 𝑷𝑹𝑶𝑫𝑼𝑪𝑻 𝑪𝑹𝒀𝑺𝑻𝑨𝑳𝑺 
Size Range, in Wt % Size Range, in Wt % 
− 0.0200 + 0.0100 10.00 − 𝟎. 𝟎𝟐𝟒𝟎 + 𝟎. 𝟎𝟏𝟒𝟎𝟏. 𝟒𝟏 
− 0.0100 + 0.0050 20.00 − 𝟎. 𝟎𝟏𝟒𝟎 + 𝟎. 𝟎𝟎𝟗𝟎 𝟓. 𝟎𝟎 
− 0.0050 + 0.0025 40.00 − 𝟎. 𝟎𝟎𝟗𝟎 + 𝟎. 𝟎𝟎𝟔𝟓 𝟐𝟑. 𝟗𝟗 
− 0.0025 + 0.0010 30.00 − 𝟎. 𝟎𝟎𝟔𝟓 + 𝟎. 𝟎𝟎𝟓𝟎 𝟔𝟗. 𝟔𝟎 
 100.00 𝟏𝟎𝟎. 𝟎𝟎 
 
 
Consider over-all material balance: 
𝐹 = 𝐿 + 𝐶 
𝐶 = 𝑊𝑃 − 𝑊𝑆 = 7,000 − 500 = 6,500 
𝑙𝑏
ℎ
 
𝐿 = 20,000 − 6,500 = 13,500 
𝑙𝑏
ℎ
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
41 
 
 
Consider Na3PO4 balance: 
𝑥𝐹 𝐹 = 𝑥𝐿𝐿 + 𝑥𝐶𝐶 
𝑥𝐶 =
𝑀𝑁𝑎3𝑃𝑂4
𝑀𝑁𝑎3𝑃𝑂4∙12𝐻2 𝑂
=
164
380
= 0.4316
𝑙𝑏 𝑁𝑎3𝑃𝑂4
𝑙𝑏 𝑁𝑎3𝑃𝑂4 ∙ 12𝐻2𝑂
 
(0.35)(20,000) = (𝑥𝐿)(13,500) + (0.4316)(6,500) 
 
𝑥𝐿 = 0.3107
𝑙𝑏 𝑁𝑎3𝑃𝑂4
𝑙𝑏 𝑠𝑜𝑙𝑛
 
𝑥𝐿 = 0.3107
𝑙𝑏 𝑁𝑎3𝑃𝑂4
𝑙𝑏 𝑠𝑜𝑙𝑛
 𝑥 
𝑙𝑏 𝑠𝑜𝑙𝑛
(1 − 0.3107) 𝑙𝑏 𝐻2𝑂
 
𝑥𝐿 = 0.4507
𝑙𝑏 𝑁𝑎3𝑃𝑂4
𝑙𝑏 𝐻2𝑂
 
 
From table 2-120 (CHE HB 8th edition) 
50°C 43 lb/100 lb H2O 
60°C 55 lb/100 lb H2O 
 
𝑻 = 𝟓𝟏. 𝟕𝟐𝟓°𝑪 ≈ 𝟏𝟐𝟓. 𝟏𝟏°𝑭 𝐴𝑁𝑆𝑊𝐸𝑅 
 
Cooling Duty: 
 
Consider heat balance: 
𝑞 = 𝐹𝐶𝑃(𝑡𝐹 − 𝑡𝑃 ) + 𝐶𝐻𝐶 
𝑞 = [(20,000
𝑙𝑏
ℎ
) (0.8
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (190 − 125.11)°𝐹] + [(6,500
𝑙𝑏
ℎ
) (27,500
𝐵𝑇𝑈
𝑙𝑏𝑚𝑜𝑙
 𝑥 
𝑙𝑏𝑚𝑜𝑙
380 𝑙𝑏
)] 
 
𝒒 = 𝟏, 𝟓𝟎𝟖, 𝟔𝟑𝟒. 𝟕𝟒
𝑩𝑻𝑼
𝒉
 𝐴𝑁𝑆𝑊𝐸𝑅 
 
CHEMICAL ENGINEERING SERIES 
CRYSTALLIZATION 
42 
 
PROBLEM # 20: 
How much CaCl2·6H2O must be dissolved in 100 kg of water at 20°C to form a saturated solution? 
The solubility of CaCl2 at 20°C is 6.7 gmol anhydrous salt (CaCl2) per kg of water. 
SOLUTION: 
For a saturated solution utilizing 100 kg water as solvent: 
1. Mole of CaCl2 required 
𝑛𝐶𝑎 𝐶𝑙2 = 100 𝑘𝑔 𝐻2𝑂 𝑥 
6.7 𝑔𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
𝑘𝑔 𝐻2𝑂
 𝑥 
1 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
1,000 𝑔𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
 
𝑛𝐶𝑎 𝐶𝑙2 = 0.67 𝑘𝑚𝑜𝑙 
 
2. Weight of CaCl2 required 
𝑊𝐶𝑎 𝐶𝑙2 = 0.67 𝑘𝑔 𝐶𝑎𝐶𝑙2 𝑥 
110.994 𝑘𝑔 𝐶𝑎𝐶𝑙2
𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
 
𝑊𝐶𝑎 𝐶𝑙2 = 74.36 𝑘𝑔 
 
3. Mole of CaCl2·6H2O required 
𝑛𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 0.67 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 𝑥 
1 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂
𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
 
𝑛𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 0.67 𝑘𝑚𝑜𝑙 
 
4. Weight CaCl2·6H2O required 
𝑊𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 0.67 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 𝑥 
218 .994 𝑘𝑔 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂
𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂
 
𝑊𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 146.72 𝑘𝑔 
 
5. Composition of the solution in terms of CaCl2·6H2O 
𝑊𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 146.72 𝑘𝑔 
 
Since there should only be total of 100 kg water in the solution, the amount of free water (net 
of water of hydration) 
 
𝑊𝑓𝑟𝑒𝑒 𝐻2 𝑂 = 100 𝑘𝑔 − 
(146 .72 − 74.36)𝑘𝑔 = 27.64 𝑘𝑔 
 
6. Amount of CaCl2·6H2O required for every 100 kg free water (net of water of hydration) 
 
𝑊𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 100 𝑘𝑔 𝑓𝑟𝑒𝑒 𝐻2𝑂 𝑥 
146.72 𝑘𝑔 𝐶𝑎𝐶𝑙2 ∙ 6𝐻2𝑂 
27.64 𝑘𝑔 𝑓𝑟𝑒𝑒 𝐻2𝑂
 
𝑾𝑪𝒂𝑪𝒍𝟐∙𝟔𝑯𝟐 𝑶 = 𝟓𝟑𝟎. 𝟖𝟐 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅