CHEMICAL ENGINEERING SERIES
CRYSTALLIZATION
Compilation of Lectures and Solved Problems
CHEMICAL ENGINEERING SERIES 2
CRYSTALLIZATION
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
𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
⁄100 𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑆=
𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
⁄100 𝑝𝑎𝑟𝑡𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Crystallization cannot occur without supersaturation.
supersaturation
There are 5 basic methods of generating
(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 3
CRYSTALLIZATION
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
SOLUTION
WATER
CRYSTALS
Solution is concentrated
by evaporating water
The concentrated
solution is cooled until
the concentration
becomes greater than
its solubility at that
temperature
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
CHEMICAL ENGINEERING SERIES 4
CRYSTALLIZATION
Types of Crystal Geometry
(1)
(2)
(3)
(4)
CUBIC SYSTEM – 3 equal axes at right angles to each other
TETRAGONAL – 3 axes at right angles to each other, one axis longer than the other 2
ORTHOROMBIC – 3 axes at right angles to each other, all of different lengths
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
(1) TYPE I: Solubility increases with temperature
and there are no hydrates or water of
crystallization
Solubility, gram per 100 gram water
CHEMICAL ENGINEERING SERIES 5
CRYSTALLIZATION
300
250
200
150
100
50
0
0
20
40
60
80
100
80
100
(2) TYPE II: Solubility increases with temperature
but curve is marked with extreme flatness
Solubility, gram per 100 gram water
Temperature, °C
250
200
150
100
50
0
0
20
40
60
Temperature, °C
(3) TYPE III: Solubility increasing fairly rapid with
temperature but is characterized by “breaks”
and indicates different “hydrates” or water of
crystallization
Solubility, gram per 100 gram water
Solubility of NaCl (CHE HB 8th edition)
250
200
150
Na2HPO4·2H2O
Na2HPO4·7H2O
100
Na2HPO4
Na2HPO4·12H2O
50
0
0
20
40
60
80
100
(4) TYPE IV: Unusual Curve; Solubility increases
at a certain transition point while the solubility
of the hydrate decreases as temperature
increases
Solubility, gram per 100 gram water
Temperature, °C
Solubility of Na2HPO4 (CHE HB 8th edition)
60
50
40
Na2CO3·H2O
30
20
Na2CO3·10H2O
10
0
0
20
40
60
80
100
Temperature, °C
Solubility of Na2CO3 (CHE HB 8th edition)
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
CHEMICAL ENGINEERING SERIES 6
CRYSTALLIZATION
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:
L
XL
hL
tL
F
XF
hf
tF
W
t2
W
t1
C
XC
hC
tC
Over-all material Balance:
𝐹 = 𝐿+𝐶
Solute Balance:
𝑋𝐹 𝐹 = 𝑋𝐿 𝐿 + 𝑋𝐶 𝐶
Enthalpy Balance:
ℎ𝑓 𝐹 = ℎ 𝐿 𝐿 + ℎ 𝑐𝐶 + 𝑞
Heat Balance:
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑝𝐹 (𝑡𝐹 − 𝑡𝐿 ) + 𝐶𝐻𝐶
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑊𝐶𝑝 𝐻2 𝑂 (𝑡2 − 𝑡1 )
Heat Transfer Equation
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚
( 𝑡𝐹 − 𝑡2 ) − (𝑡𝐿 − 𝑡1 )
𝑞 = 𝑈𝐴 [
]
𝑡 − 𝑡2
ln 𝐹
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
ℎ 𝐹 = enthalpy of the feed solution
ℎ 𝐿 = enthalpy of the mother liquor
ℎ 𝐶 = 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
CHEMICAL ENGINEERING SERIES 7
CRYSTALLIZATION
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:
V
hV
F
XF
hf
tF
W
t2
L
XL
hL
tL
W
t1
C
XC
hC
tC
Over-all material Balance:
𝐹 = 𝐿+𝐶 +𝑉
Solute Balance:
𝑋𝐹 𝐹 = 𝑋𝐿 𝐿 + 𝑋𝐶 𝐶
Solvent Balance:
(1 − 𝑋𝐹 )𝐹 = 𝑉 + (1 − 𝑋𝐿 )𝐿 + (1 − 𝑋𝐶 )𝐶
Enthalpy Balance:
ℎ𝑓 𝐹 = ℎ 𝑉 𝑉 + ℎ 𝐿 𝐿 + ℎ 𝑐𝐶 + 𝑞
Heat Balance:
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑝𝐹 (𝑡𝐹 − 𝑡𝐿 ) + 𝐶𝐻𝐶 − 𝑉𝜆𝑉
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑊𝐶𝑝 𝐻2 𝑂 (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
ℎ 𝐹 = enthalpy of the feed solution
ℎ 𝐿 = enthalpy of the mother liquor
ℎ 𝐶 = enthalpy of the crystals
ℎ 𝑉 = 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
CHEMICAL ENGINEERING SERIES 8
CRYSTALLIZATION
SUPERSATURATION BY ADIABATIC EVAPORATION OF SOLVENT
V
hV
F
XF
hf
M
L
XL
hL
C
XC
hC
Over-all material Balance:
𝐹 = 𝐿+𝐶 +𝑉
Solute Balance:
𝑋𝐹 𝐹 = 𝑋𝐿 𝐿 + 𝑋𝐶 𝐶
Solvent Balance:
(1 − 𝑋𝐹 )𝐹 = 𝑉 + (1 − 𝑋𝐿 )𝐿 + (1 − 𝑋𝐶 )𝐶
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
ℎ 𝐹 = enthalpy of the feed solution
ℎ 𝐿 = enthalpy of the mother liquor
ℎ 𝐶 = enthalpy of the crystals
ℎ 𝑉 = 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
Enthalpy Balance:
ℎ𝑓 𝐹 = ℎ 𝑉 𝑉 + ℎ 𝐿 𝐿 + ℎ 𝑐𝐶
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
CHEMICAL ENGINEERING SERIES 9
CRYSTALLIZATION
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
𝑊𝑆
(𝐷𝑆 + ∆𝐷) 3
𝑊𝑃 =
𝐷𝑆 3
𝐷𝑆 + ∆𝐷 3
)
𝑊𝑃 = 𝑊𝑆 (
𝐷𝑆
3
𝐷𝑆 + [𝐷𝑃 − 𝐷𝑆]
)
𝑊𝑃 = 𝑊𝑆 (
𝐷𝑆
𝐷𝑃 3
𝑊𝑃 = 𝑊𝑆 ( )
𝐷𝑆
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:
𝑊𝑃
∫
0
𝑊𝑆
𝑑𝑊𝑃 = ∫
0
𝑊𝑆
𝑊𝑃 = ∫
(1 +
0
𝐶 = 𝑊𝑃 − 𝑊𝑆
PROBLEM # 01:
(1 +
∆𝐷 3
) 𝑑𝑊𝑆
𝐷𝑆
∆𝐷 3
) 𝑑𝑊𝑆
𝐷𝑆
CHEMICAL ENGINEERING SERIES 10
CRYSTALLIZATION
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?
Na2SO4 solution
xF = 0.20
tF = 200°F
Na2SO4 ·10H2O
C
Magma, M
tM = 60°F
L
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
𝑀𝑁𝑎 2𝑆𝑂4
142
𝑙𝑏 𝑁𝑎2 𝑆𝑂4
𝑋𝐶 =
=
= 0.4410
𝑀𝑁𝑎2𝑆𝑂4∙10𝐻2 𝑂 322
𝑙𝑏 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
At 60°F, solubility is 21.7778 g per 100 g water
21.7778
𝑙𝑏 𝑁𝑎2 𝑆𝑂4
𝑋𝐿 =
= 0.1788
100 + 21.7778
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑙𝑏 𝑁𝑎2 𝑆𝑂4
𝑙𝑏 𝑁𝑎2 𝑆𝑂4
𝑙𝑏 𝑁𝑎2 𝑆𝑂4
(0.20
) (100 𝑙𝑏 𝑓𝑒𝑒𝑑 ) = (0.1788
) (𝐿 ) + (0.4410
) (𝐶 )
𝑙𝑏 𝑓𝑒𝑒𝑑
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑙𝑏 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
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:
8.0854
%𝐶=
𝑥 100 = 𝟖. 𝟎𝟖𝟓𝟒 %
100
91.9146
%𝐿=
𝑥 100 = 𝟗𝟏. 𝟗𝟏𝟒𝟔 %
100
% Recovery:
% 𝑟𝑒𝑐𝑜𝑣𝑒𝑟𝑦 =
𝑋𝐶 𝐶
𝑋𝐹 𝐹
(0.4410
𝑥 100 =
% 𝒓𝒆𝒄𝒐𝒗𝒆𝒓𝒚 = 𝟏𝟕. 𝟖𝟑 % 𝐴𝑁𝑆𝑊𝐸𝑅
𝑙𝑏 𝑁𝑎2 𝑆𝑂4
) (8.0854𝑙𝑏 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂)
𝑙𝑏 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻 2 𝑂
𝑥100
𝑙𝑏 𝑁𝑎2 𝑆𝑂4 (
(0.20
) 100 𝑙𝑏 𝑓𝑒𝑒𝑑 )
𝑙𝑏 𝑓𝑒𝑒𝑑
CHEMICAL ENGINEERING SERIES 11
CRYSTALLIZATION
PROBLEM # 02:
A solution of 32.5% MgSO4
originally at 150°F is to be
crystallized in a vacuum adiabatic
crystallizer to give a produc t
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:
V
MgSO4 solution
F = 10,000 lb/h
xF = 0.325
tF = 150°F
MgSO4 ·7H2O
C = 4,000 lb/h
L
Consider over-all material balance:
𝐹 = 𝑉+𝐿+𝐶
𝑉 = 10,000 − 𝐿 − 4,000
𝑉 = 6,000 − 𝐿 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider solute balance:
𝑥 𝐹 𝐹 = 𝑥 𝐶𝐶 + 𝑥 𝐿 𝐿
𝑀𝑀𝑔 𝑆𝑂4
120 .38
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥𝐶 =
=
= 0.4884
𝑀𝑀𝑔 𝑆𝑂4∙7𝐻2 𝑂 246 .49
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑓𝑒𝑒𝑑
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏
(0.325
) (10,000
) = 𝑋𝐿 (𝐿 ) + (0.4884
) (4,000 )
𝑙𝑏 𝑓𝑒𝑒𝑑
ℎ
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
ℎ
𝑥 𝐿 𝐿 = 1,296.4 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
Consider enthalpy balance:
ℎ𝑓 𝐹 = ℎ𝑉 𝑉 + ℎ𝐿 𝐿 + ℎ𝑐 𝐶
THE PROBLEM CAN BE SOLVED BY TRIAL AND ERROR SINCE TEMPERATURE OF THE
SOLUTION AFTER CRYSTALLIZA TION IS UNKNOWN AND ENTHALPIES ARE DEPENDENT
ON TEMPERATURE
1.
2.
3.
4.
5.
6.
7.
Assume temperature of the solution
From figure 27-3 (Unit Operations by McCabe and Smoth 7 th edition), obtain mass fraction of
MgSO4 at the assumed temperature of the solution
Solve for “L” using equation 2
Solve for “V” using equation 1
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
Compare values of “V” from step 4 with that from step 5-c
If not the same (or approximately the same), conduct another trial and error calculations
CHEMICAL ENGINEERING SERIES 12
CRYSTALLIZATION
TRIAL 1: Assume temperature of the solution at 60°F
From figure 27-3 (Unit Operations by McCabe and Smith 7th edition)
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥 𝐿 = 0.245
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
Substitute to equation 2
1,296.4
𝐿=
= 5,291.43 𝑙𝑏
0.245
Substitute to equation 1
𝑉 = 6,000 − 5,291.43 = 708 .57 𝑙𝑏
From figure 27-4 (Unit Operations by McCabe and Smith, 7 th edition)
𝐵𝑇𝑈
ℎ𝐹 𝑎𝑡 150°𝐹 𝑎𝑛𝑑 32.5% 𝑀𝑔𝑆𝑂4 = −10
𝑙𝑏
𝐵𝑇𝑈
ℎ𝐶 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 48.84% 𝑀𝑔𝑆𝑂4 = −158
𝑙𝑏
𝐵𝑇𝑈
ℎ𝐿 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 24.5% 𝑀𝑔𝑆𝑂4 = −50
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
𝐵𝑇𝑈
) (10°𝐹 ) ]
ℎ𝑉 = 1,083.3
+ [(0.45
𝑙𝑏
𝑙𝑏 ∙ °𝐹
𝐵𝑇𝑈
ℎ𝑉 = 1,087.8
𝑙𝑏
𝐵𝑇𝑈
𝑙𝑏
ℎ𝑓 𝐹 = ℎ𝑉 𝑉 + ℎ𝐿 𝐿 + 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 5,291.43
=
𝐶
4,000
𝑳
𝑪
= 𝟏. 𝟑𝟐 𝐴𝑁𝑆𝑊𝐸𝑅
CHEMICAL ENGINEERING SERIES 13
CRYSTALLIZATION
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
F
CALCINATION
CRYSTALLIZATION
YIELD = 90%
T=0C
YIELD = 95%
P
Na2SO4
30,000 MT/yr
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 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
100 𝑔 𝐻2 𝑂
𝑚𝑎𝑠𝑠 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
𝑀𝑇 𝑁𝑎2 𝑆𝑂4
1
1 𝑀𝑇𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4 1 𝑀𝑇𝑚𝑜𝑙𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂 322 𝑀𝑇𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
= 30,000
𝑥
𝑥
𝑥
𝑥
𝑦𝑟
0.95
142 𝑀𝑇𝑁𝑎2 𝑆𝑂4
1 𝑀𝑇𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
𝑀𝑇𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
𝑚𝑎𝑠𝑠 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂 = 71,608.60 𝑀𝑇 𝑥
𝑚𝑎𝑠𝑠 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂 =
𝐹=
𝑑𝑎𝑦
𝑑𝑎𝑦
𝑴𝑻
𝒅𝒂𝒚
300 𝑑𝑎𝑦𝑠
238.6953𝑀𝑇
238.6953𝑀𝑇 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
𝑭 = 𝟓, 𝟎𝟏𝟐. 𝟔𝟎
1 𝑦𝑟
𝐴𝑁𝑆𝑊𝐸𝑅
𝑥
105 𝑀𝑇 𝑓𝑒𝑒𝑑
5 𝑀𝑇 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
CHEMICAL ENGINEERING SERIES 14
CRYSTALLIZATION
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
V
C
T = 20 C
F
1,200 lb BaNO3
CRYSTALLIZER
T = 90 C
CHE BP July 1968
SOLUTION:
𝑥 𝐹 = 0.306
𝑙𝑏 𝐵𝑎 (𝑁𝑂3 ) 2
𝑥
𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
𝑥 𝐹 𝐹 = 1,200 𝑙𝑏 𝐵𝑎( 𝑁𝑂3 )2
𝐹 = 1,200 𝑙𝑏 𝐵𝑎(𝑁 𝑂3 ) 2 𝑥
100 𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
𝑙𝑏 𝐵𝑎( 𝑁𝑂3 ) 2
= 0.2343
(100 + 30.6) 𝑙𝑏 𝑓𝑒𝑒𝑑
𝑙𝑏 𝑓𝑒𝑒𝑑
𝑙𝑏 𝑓𝑒𝑒𝑑
0.2343 𝑙𝑏 𝐵𝑎(𝑁 𝑂3 ) 2
𝐹 = 5,121.5686 𝑙𝑏
𝑙𝑏 𝐵𝑎 (𝑁𝑂3 )2
100 𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
𝑙𝑏 𝐵𝑎( 𝑁𝑂3 ) 2
𝑥 𝐿 = 0.092
𝑥
= 0.0842
(100 + 9.2) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑙𝑏 𝑤𝑎𝑡𝑒𝑟
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
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
L
T = 20 C
CHEMICAL ENGINEERING SERIES 15
CRYSTALLIZATION
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·ft 2. There are 3.5 ft 2 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; heat of 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
F
tF = 120 F
L
tL = 80 F
SWENSON-WALKER
CRYSTALLIZER
W
t1 = 60 F
t2 = 70 F
SOLUTION:
Consider over-all material balance:
𝐹 = 𝐿+𝐶
𝐿 = 𝐹 − 2,000 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider copperas (FeSO4·7H2O) balance:
𝑥 𝐹 𝐹 = 𝑥 𝐶𝐶 + 𝑥 𝐿 𝐿
𝑥 𝐶 = 1.0
74 𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2 𝑂
100 𝑙𝑏 𝐻2 𝑂
𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2 𝑂
𝑥𝐿 =
𝑥
= 0.4253
100 𝑙𝑏 𝐻2 𝑂
174 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
140 𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2 𝑂
100 𝑙𝑏 𝐻2 𝑂
𝑙𝑏 𝐹𝑒𝑆𝑂4 ∙ 7𝐻2 𝑂
𝑥𝐹 =
𝑥
= 0.5833
100 𝑙𝑏 𝐻2 𝑂
240 𝑙𝑏 𝑓𝑒𝑒𝑑
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
𝑙𝑏
𝐹 = 7,274.73
ℎ
𝑙𝑏
𝐿 = 5,274.73
ℎ
Consider heat balance:
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐹𝐶𝑝𝐹 (𝑡𝐹 − 𝑡𝐿 ) + 𝐶𝐻𝐶
C, 1 ton/h
Fe2SO4·7H2O
tC = 80 F
CHEMICAL ENGINEERING SERIES 16
CRYSTALLIZATION
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = [(7,274.73
𝑙𝑏
ℎ
) (0.70
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
) (120 − 80) °𝐹 ]
𝐵𝑇𝑈
𝑙𝑏
)]
+ [(2,000 ) ( 4,400
𝑥
𝑥
ℎ
𝑔𝑚𝑜𝑙 277 .85 𝑔 0.55556 𝑐𝑎𝑙
𝑔
𝐵𝑇𝑈
= 260,701.1615
ℎ
𝑙𝑏
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠
𝑐𝑎𝑙
𝑔𝑚𝑜𝑙
𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑊𝐶𝑝 𝐻2𝑂 ( 𝑡2 − 𝑡1 )
𝐵𝑇𝑈
260,701.1615
ℎ
𝑊=
𝐵𝑇𝑈 (
(1.0
) 70 − 60) °𝐹
𝑙𝑏 ∙ °𝐹
𝑙𝑏
1 𝑓𝑡 3
7.481 𝑔𝑎𝑙
1ℎ
𝑊 = 26,070.1162
𝑥
𝑥
𝑥
3
ℎ
62.335 𝑙𝑏
𝑓𝑡
60 𝑚𝑖𝑛
𝑾 = 𝟓𝟐. 𝟏𝟒
𝒈𝒂𝒍
𝒎𝒊𝒏
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚
𝐴𝑁𝑆𝑊𝐸𝑅
(𝑡𝐹 − 𝑡2 ) − ( 𝑡𝐿 − 𝑡1 )
𝑡 − 𝑡2
ln 𝐹
𝑡𝐿 − 𝑡1
(120 − 70) − (80 − 60)
=
120 − 70
ln
80 − 60
= 32.7407°𝐹
∆𝑇𝑙𝑚 =
∆𝑇𝑙𝑚
∆𝑇𝑙𝑚
260,701.1615
𝐴=
(35
𝐵𝑇𝑈
ℎ
𝐵𝑇𝑈
) (32.7407°𝐹 )
ℎ ∙ 𝑓𝑡 2 ∙ °𝐹
𝐴 = 227.5029 𝑓𝑡 2
# 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 = 227.5029 𝑓𝑡 2 𝑥
1 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
3.5 𝑓𝑡 2
# 𝒐𝒇 𝒖𝒏𝒊𝒕𝒔 = 𝟔. 𝟓 ≈ 𝟕 𝒖𝒏𝒊𝒕𝒔 𝐴𝑁𝑆𝑊𝐸𝑅
𝑥
1 𝑢𝑛𝑖𝑡
10 𝑓𝑡
1
CHEMICAL ENGINEERING SERIES 17
CRYSTALLIZATION
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% Na 2CO3 ;
molecular weight of Na2CO3 is 106
SOLUTION:
Assume 100 g of Na2CO3·10H2O added into the saturated solution
124 𝑔 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2 𝑂
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2 𝑂 = 100 𝑔𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂 𝑥
286 𝑔 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2 𝑂 = 43.3566 𝑔
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 = 100 𝑔𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂 𝑥
106 𝑔 𝑁𝑎2 𝐶𝑂3
286 𝑔 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 = 37.0629 𝑔
𝑤𝑡 𝐻2 𝑂 = 100 𝑔𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂 𝑥
180 𝑔 𝐻2 𝑂
286 𝑔 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
𝑤𝑡 𝐻2 𝑂 = 62.9371 𝑔
% 𝑁𝑎2 𝐶𝑂3 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 ∙ 𝐻2 𝑂 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 = 8.5217 𝑔𝑁𝑎2 𝐶𝑂3 𝑥
124 𝑔 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2 𝑂
𝑤𝑡 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2 𝑂 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 = 9.9688 𝑔
% 𝑁𝑎2 𝐶𝑂3 ∙ 𝐻2 𝑂 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒𝑑 =
9.9688
43.3566
𝑥 100
% 𝑵𝒂𝟐 𝑪𝑶 𝟑 ∙ 𝑯𝟐 𝑶 𝒑𝒓𝒆𝒄𝒊𝒑𝒊𝒕𝒂𝒕𝒆𝒅 = 𝟐𝟐. 𝟗𝟗 % 𝐴𝑁𝑆𝑊𝐸𝑅
106 𝑔 𝑁𝑎2 𝐶𝑂3
CHEMICAL ENGINEERING SERIES 18
CRYSTALLIZATION
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?
F
tF = 220 F
43 g MgSO4/100 g H2O
L
tL = 50 F
COOLING CRYSTALLIZER
SOLUTION:
Consider over-all material balance:
𝐹 = 𝐿+𝐶
𝐿 = 𝐹 − 1 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider MgSO4 balance
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
43 𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
100 𝑡𝑜𝑛 𝐻2 𝑂
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑥𝐹 =
𝑥
= 0.3007
(
)
100 𝑡𝑜𝑛 𝐻2 𝑂
100 + 43 𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
120 .38 𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑥𝐶 =
= 0.4884
246.49 𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
From table 27-3 (Unit Operations by McCabe and Smith, 7 th edition), at 50°F
𝑡𝑜𝑛 𝑀𝑔𝑆𝑂4
𝑥 𝐿 = 0.23
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
C, 1 ton/h
MgSO4·7H2O
tC = 50 F
CHEMICAL ENGINEERING SERIES 19
CRYSTALLIZATION
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?
F
tF = 60 F
16.4 g
NaHCO3 /100 g
H2O
L
tL = 20 F
COOLING CRYSTALLIZER
SOLUTION:
Basis: 100 kg feed
Consider over-all material balance:
𝐹 = 𝐿+𝐶
𝐿 = 100 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider NaHCO3 balance
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑥𝐹 =
16.4 𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
𝑥
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
= 0.1409
(100 + 16.4) 𝑘𝑔 𝑓𝑒𝑒𝑑
𝑘𝑔 𝑓𝑒𝑒𝑑
100 𝑘𝑔 𝐻2 𝑂
𝑥 𝐶 = 1.0
9.6 𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝑁𝑎𝐻𝐶𝑂3
𝑥𝐿 =
𝑥
= 0.0876
(100 + 9.6) 𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
100 𝑘𝑔 𝐻2 𝑂
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 =
% 𝑁𝑎𝐻𝐶𝑂3 𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑙𝑖𝑧𝑒𝑑 =
𝐶
𝑥𝐹 𝐹
𝑥 100
5.8417 𝑘𝑔
𝑥 100
(0.1409) (100 𝑘𝑔)
% 𝑵𝒂𝑯𝑪𝑶 𝟑 𝒄𝒓𝒚𝒔𝒕𝒂𝒍𝒍𝒊𝒛𝒆𝒅 = 𝟒𝟏. 𝟒𝟔 % 𝐴𝑁𝑆𝑊𝐸𝑅
C,
9.6 g NaHCO3
per 100 g H2O
tC = 20 F
CHEMICAL ENGINEERING SERIES 20
CRYSTALLIZATION
V
F
tF = 20 C
8.4% Na2SO4
L
tL = 20 C
CRYSTALLIZER
C,
tC = 20 C
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 Na 2SO4 per 100 g
H2O.
SOLUTION:
Basis: 1 L feed
1.077 𝑘𝑔
𝐹 = 1𝐿 𝑥
= 1.077 𝑘𝑔
𝐿
Consider over-all material balance:
𝐹 = 𝑉+𝐿+𝐶
𝐿 = 1.077 − 𝑉 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
𝑥 𝐶 𝐶 = 0.80𝑥 𝐹 𝐹
8.4 𝑘𝑔 𝑁𝑎2 𝑆𝑂4
) = 0.0905 𝑘𝑔 𝑁𝑎2 𝑆𝑂4
𝑥 𝐹 𝐹 = (1.077 𝑘𝑔 𝑓𝑒𝑒𝑑 ) (
100 𝑘𝑔 𝑓𝑒𝑒𝑑
𝑥 𝐶 𝐶 = (0.80)(0.0905 𝑘𝑔 𝑁𝑎2 𝑆𝑂4 ) = 0.0724 𝑘𝑔 𝑁𝑎2 𝑆𝑂4
𝑀𝑁𝑎 2𝑆𝑂4
142
𝑘𝑔 𝑁𝑎2 𝑆𝑂4
𝑥𝐶 =
=
= 0.4410
𝑀𝑁𝑎2𝑆𝑂4∙10𝐻2 𝑂 322
𝑘𝑔 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
0.0724 𝑘𝑔 𝑁𝑎2 𝑆𝑂4
𝐶=
= 0.1642 𝑘𝑔
𝑘𝑔 𝑁𝑎2 𝑆𝑂4
0.4410
𝑘𝑔 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
Substitute to equation 1
𝐿 = 1.077 − 𝑉 − 0.1642
𝐿 = 0.9128 − 𝑉 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
Consider Na2SO4 balance
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
19.4 𝑘𝑔 𝑁𝑎2 𝑆𝑂4
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝑁𝑎2 𝑆𝑂4
𝑥𝐿 =
𝑥
= 0.1625
(
)
100 𝑘𝑔 𝐻2 𝑂
100 + 19.4 𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
0.0905 = (0.1625) (𝐿 ) + 0.0724
𝐿 = 0.1114 𝑘𝑔
CHEMICAL ENGINEERING SERIES 21
CRYSTALLIZATION
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(NO 3)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(NO 3)2/100
kg total water
V
F
30.6 kg Ba(NO3)2/100 kg H2O
L
CRYSTALLIZER
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 − 𝑥 𝐹 )𝐹
30.6 𝑘𝑔 𝐵𝑎( 𝑁𝑂3 )2
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝐵𝑎 (𝑁𝑂3 )2
𝑥𝐹 =
𝑥
= 0.2343
(100 + 30.6) 𝑘𝑔 𝑓𝑒𝑒𝑑
100 𝑘𝑔 𝐻2 𝑂
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
8.6 𝑘𝑔 𝐵𝑎 (𝑁𝑂3 )2
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝐵𝑎 (𝑁𝑂3 ) 2
𝑥𝐿 =
𝑥
= 0.0792
(100 + 8.6)𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
𝑥 𝐶 = 1.0
(0.2343) (100) = (0.0792) (𝐿 ) + (1.0)(𝐶 )
C
CHEMICAL ENGINEERING SERIES 22
CRYSTALLIZATION
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 − 𝑥 𝐹 )𝐹
30.6 𝑘𝑔 𝐵𝑎( 𝑁𝑂3 )2
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝐵𝑎 (𝑁𝑂3 )2
𝑥𝐹 =
𝑥
= 0.2343
(100 + 30.6) 𝑘𝑔 𝑓𝑒𝑒𝑑
100 𝑘𝑔 𝐻2 𝑂
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
7.0 𝑘𝑔 𝐵𝑎 (𝑁𝑂3 )2
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝐵𝑎( 𝑁𝑂3 ) 2
𝑥𝐿 =
𝑥
= 0.0654
(100 + 7) 𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
100 𝑘𝑔 𝐻2 𝑂
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
𝑥 𝐶 = 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 23
CRYSTALLIZATION
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?
V
F
1,000 kg KCl
363K
Source:
Transport Processes and Unit
Operations (Geankoplis)
SOLUTION:
c) Assume crystallization by cooling (without evaporation)
Consider over-all material balance:
𝐹 = 𝐿+𝐶
100 𝑘𝑔 𝑠𝑜𝑙𝑛
𝐹 = 1,000 𝑘𝑔 𝐾𝐶𝑙 𝑥
= 2,857.14 𝑘𝑔 𝑓𝑒𝑒𝑑
35 𝑘𝑔 𝐾𝐶𝑙
𝐿 = 2,857.14 − 𝐶 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider KCl balance
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑘𝑔 𝐾𝐶𝑙
𝑥 𝐿 = 0.254
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
𝑥 𝐶 = 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
% 𝐻2 𝑂 𝑖𝑛 𝑓𝑒𝑒𝑑 = 100 − %𝐾𝐶𝑙 = 100 − 35 = 65%
L
293K
CRYSTALLIZER
C
293K
CHEMICAL ENGINEERING SERIES 24
CRYSTALLIZATION
% 𝐻2 𝑂 𝑖𝑛 𝑓𝑒𝑒𝑑 =
𝑤𝑡 𝐻2 𝑂
𝑤𝑡 𝑓𝑒𝑒𝑑
𝑥 100
𝑤𝑡 𝐻2 𝑂 = (2,857.14 𝑘𝑔 𝑓𝑒𝑒𝑑 ) (
65 𝑘𝑔 𝐻2 𝑂
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑥 𝐿 = 0.254
𝑥 𝐶 = 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 25
CRYSTALLIZATION
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·ft 2·°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
F
tF = 30 C
L
tL = 15 C
SWENSON-WALKER
CRYSTALLIZER
W
t1 = 10 C
t2 = 20 C
Source: Unit Operations (Brown)
SOLUTION:
Consider over-all material balance:
𝐹 = 𝐿+𝐶
𝐿 = 𝐹 − 1 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider Na2SO4 balance
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
40 𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4
100 𝑡𝑜𝑛 𝐻2 𝑂
𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4
𝑥𝐹 =
𝑥
= 0.2857
(
)
100 𝑡𝑜𝑛 𝐻2 𝑂
100 + 40 𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
𝑡𝑜𝑛 𝑓𝑒𝑒𝑑
13.5 𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4
100 𝑡𝑜𝑛 𝐻2 𝑂
𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4
𝑥𝐿 =
𝑥
= 0.1189
(100 + 13.5) 𝑡𝑜𝑛 𝑙𝑖𝑞𝑢𝑜𝑟
100 𝑡𝑜𝑛 𝐻2 𝑂
𝑡𝑜𝑛 𝑙𝑖𝑞𝑢𝑜𝑟
142 𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4
𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4
𝑥𝐶 =
= 0.4410
322 𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
𝑡𝑜𝑛 𝑁𝑎2 𝑆𝑂4 ∙ 10𝐻2 𝑂
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
𝑡𝑜𝑛
𝐹 = 1.931
ℎ
C, 1 ton/h
Na2SO4·10H2O
tC = 15 C
CHEMICAL ENGINEERING SERIES 26
CRYSTALLIZATION
Consider heat balance:
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙 = 𝐹𝐶𝑃 (𝑡𝐹 − 𝑡𝐶 ) + 𝐶𝐻𝐶
(𝑚𝐶𝑃 )𝑁𝑎 𝑆𝑂 + (𝑚𝐶𝑃 )𝐻 𝑂
2 4
2
𝐶𝑃 =
𝐹
From Table 2-194 (CHE HB 8th edition)
𝑐𝑎𝑙
𝐵𝑇𝑈
𝐶𝑃 𝑁𝑎 𝑆𝑂 = 32.8
= 0.231
2 4
°𝐶 ∙ 𝑚𝑜𝑙
𝑙𝑏 ∙ °𝐹
[(0.2857) (0.231) + (0.7143) (1.000)]
𝐵𝑇𝑈
𝐶𝑃 =
= 0.7803
1
𝑙𝑏 ∙ °𝐹
𝐵𝑇𝑈
1.8°𝐹
) (0.7803
) (30 − 15) °𝐶 𝑥
]
𝑡𝑜𝑛
𝑙𝑏 ∙ °𝐹
°𝐶
𝑡𝑜𝑛 2,000 𝑙𝑏
𝑙𝑏𝑚𝑜𝑙
454 𝑔𝑚𝑜𝑙 18000 𝑐𝑎𝑙
𝐵𝑇𝑈
)(
)]
+ [(1
𝑥
𝑥
𝑥
𝑥
ℎ
𝑡𝑜𝑛
322 𝑙𝑏
𝑙𝑏𝑚𝑜𝑙
𝑔𝑚𝑜𝑙
252.16 𝑐𝑎𝑙
𝐵𝑇𝑈
= 282 ,656.8961
ℎ
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙 = [(1.931
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚
𝑡𝑜𝑛
ℎ
𝑥
2,000 𝑙𝑏
(𝑡𝐹 − 𝑡2 ) − ( 𝑡𝐿 − 𝑡1 )
𝑡 − 𝑡2
ln 𝐹
𝑡𝐿 − 𝑡1
𝑡𝐹 = 30°𝐶 = 86°𝐹
𝑡𝐿 = 15°𝐶 = 59°𝐹
𝑡1 = 10°𝐶 = 50°𝐹
𝑡2 = 20°𝐶 = 68°𝐹
(86 − 68) − (59 − 50)
=
= 12.9842°𝐹
86 − 68
ln
59 − 50
∆𝑇𝑙𝑚 =
∆𝑇𝑙𝑚
262 ,656.8961
𝐴=
(25
𝐵𝑇𝑈
ℎ
𝐵𝑇𝑈
) (12.9842°𝐹 )
ℎ ∙ 𝑓𝑡 2 ∙ °𝐹
𝐴 = 870.7718𝑓𝑡 2
# 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 = 880.7718 𝑓𝑡 2 𝑥
1 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
3 𝑓𝑡 2
𝑥
1 𝑢𝑛𝑖𝑡
10 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
# 𝒐𝒇 𝒖𝒏𝒊𝒕𝒔 = 𝟐𝟗. 𝟎𝟑 ≈ 𝟑𝟎 𝒖𝒏𝒊𝒕𝒔 𝐴𝑁𝑆𝑊𝐸𝑅
CHEMICAL ENGINEERING SERIES 27
CRYSTALLIZATION
PROBLEM # 13:
A continuous adiabatic vacuum crystallizer is
to be used for the production of MgSO 4·7H2 O
crystals from 20,000 lb/h of solution
containing 0.300 weight fraction MgSO 4. 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?
V
F, 20,000 lb/h
xF = 0.3000
tF = 160 F
ADIABATIC VACUUM
CRYSTALLIZER
C = 6,000 lb/h
MgSO4·7H2O
L
BPE = 10 F
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥 𝐹 = 0.3000
𝑙𝑏 𝑓𝑒𝑒𝑑
120 𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥𝐶 =
= 0.4878
246 𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
(0.30)(20,000) = ( 𝑥 𝐿)(𝐿 ) + (0.4878) (6,000)
3,073.2
𝐿=
⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
𝑥𝐿
Consider enthalpy balance:
ℎ𝑓 𝐹 = ℎ𝑉 𝑉 + ℎ𝐿 𝐿 + ℎ𝑐 𝐶
THE PROBLEM CAN BE SOLVED BY TRIAL AND ERROR SINCE TEMPERATURE OF THE
SOLUTION AFTER CRYSTALLIZA TION IS UNKNOWN AND ENTHALPIES ARE DEPENDENT
ON TEMPERATURE
CHEMICAL ENGINEERING SERIES 28
CRYSTALLIZATION
1.
2.
3.
4.
5.
6.
7.
Assume temperature of the solution
From figure 27-3 (Unit Operations by McCabe and Smoth 7 th edition), obtain mass fraction of
MgSO4 at the assumed temperature of the solution
Solve for “L” using equation 2
Solve for “V” using equation 1
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
Compare values of “V” from step 4 with that from step 5-c
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)
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥 𝐿 = 0.245
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
Substitute to equation 2
3,073.2
𝐿=
= 12,543.67 𝑙𝑏
0.245
Substitute to equation 1
𝑉 = 14,000 − 12,543.67 = 1,456.33 𝑙𝑏
From figure 27-4 (Unit Operations by McCabe and Smith, 7 th edition)
𝐵𝑇𝑈
ℎ𝐹 𝑎𝑡 160°𝐹 𝑎𝑛𝑑 30% 𝑀𝑔𝑆𝑂4 = 5
𝑙𝑏
𝐵𝑇𝑈
ℎ𝐶 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 48.78% 𝑀𝑔𝑆𝑂4 = −158
𝑙𝑏
𝐵𝑇𝑈
ℎ𝐿 𝑎𝑡 60°𝐹 𝑎𝑛𝑑 24.5% 𝑀𝑔𝑆𝑂4 = −50
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
𝐵𝑇𝑈
) (10°𝐹 ) ]
ℎ𝑉 = 1,083.3
+ [(0.45
𝑙𝑏
𝑙𝑏 ∙ °𝐹
𝐵𝑇𝑈
ℎ𝑉 = 1,087.8
𝑙𝑏
𝐵𝑇𝑈
𝑙𝑏
ℎ𝑓 𝐹 = ℎ𝑉 𝑉 + ℎ𝐿 𝐿 + 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 29
CRYSTALLIZATION
PROBLEM # 14:
Crystals of CaCl2·6H2O are to be obtained
from a solution of 35 weight % CaCl 2, 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?
V
F
CaCl2 = 35%
Inert = 10%
H2O = 55%
tF = 100 F
OSLO CRYSTALLIZER
M (magma)
C
Inert
L
tF = 40 F
L
CENTRIFUGE
C’’
CaCl2·6H2O
DRYER
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 CaCl 2·6H2O
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)
CHEMICAL ENGINEERING SERIES 30
CRYSTALLIZATION
𝐵𝑇𝑈
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
𝐶𝑃 = 18.1
𝑥
𝑥
= 0.1632
𝑐𝑎𝑙
𝑚𝑜𝑙 · 𝐾 110.9 𝑔
𝑙𝑏 ∙ °𝐹
1
𝑔 ∙ °𝐹
At 40°F (277.59 K)
𝐵𝑇𝑈
1
𝑐𝑎𝑙
1 𝑚𝑜𝑙
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
𝐶𝑃 = 17.97
𝑥
𝑥
= 0.1620
𝑐𝑎𝑙
𝑚𝑜𝑙 · 𝐾 110.9 𝑔
𝑙𝑏 ∙ °𝐹
1
𝑔 ∙ °𝐹
0.1632 + 0.1620
𝐵𝑇𝑈
𝐶̅𝑃 =
= 0.1626
2
𝑙𝑏 ∙ °𝐹
𝑐𝑎𝑙
1 𝑚𝑜𝑙
1
For the feed
(0.35 𝑙𝑏 𝐶𝑎𝐶𝑙 2 ) (0.1626
𝐶𝑃 =
𝐶𝑃 = 0.6743
𝐵𝑇𝑈
𝐵𝑇𝑈
) + (0.55 𝑙𝑏 𝐻2 𝑂) (1
)
𝑙𝑏 𝐶𝑎𝐶𝑙 2 ∙ °𝐹
𝑙𝑏 𝐻2 𝑂 ∙ °𝐹
(0.35 + 0.55) 𝑙𝑏 𝑓𝑒𝑒𝑑
𝐵𝑇𝑈
𝑙𝑏 ∙ °𝐹
From table 2-224 (CHE HB 8th edition), heat of solution of CaCl 2·6H2O = -4,100 cal/mol; in
the absence of data on heat of crystallization, heat of solution can be used instead but of
opposite sign
𝑐𝑎𝑙
𝑐𝑎𝑙
𝐵𝑇𝑈
𝐻𝐶 = 4,100
= 18.73
= 33.71
𝑚𝑜𝑙
𝑔
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 (CaCl 2·6H2O) balance, inert soluble-free
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑙𝑏 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
35 𝑙𝑏 𝐶𝑎𝐶𝑙 2
1 𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂 218 .9 𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
𝑥𝐹 =
𝑥
𝑥
= 0.7676
(35 + 55) 𝑙𝑏 𝑓𝑒𝑒𝑑
𝑙𝑏 𝐶𝑎𝐶𝑙 2
𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
110.9
𝑙𝑏𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
CHEMICAL ENGINEERING SERIES 31
CRYSTALLIZATION
𝑥𝐿 =
61.92 𝑙𝑏 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
100 𝑙𝑏 𝐻2 𝑂
100 𝑙𝑏 𝐻2 𝑂
= 0.3824
(100 + 61.92) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑥
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)
61.92 𝑙𝑏 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
100 𝑙𝑏 𝐻2 𝑂
𝑤𝑡 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂 𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 = 0.0146 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟 𝑥
𝑥
(100 + 61.92) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
100 𝑙𝑏 𝐻2 𝑂
𝑤𝑡 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂 𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 = 0.0056 𝑙𝑏
𝑤𝑡 𝐻2 𝑂 𝑖𝑛 𝑙𝑖𝑞𝑢𝑜𝑟 = 0.0146 − 0.0056 = 0.0090 𝑙𝑏
CaCl2·6H2O
H2O
inerts
lb
0.0056
0.0090
0.1000
0.1146
%
4.89
7.85
87.26
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
from liquor
0.0762 x 0.0489
0.7620
0.0037
H2O
inerts
0.0762 x 0.0785
0.0762 x 0.8726
0.0060
0.0665
In the dryer, assume all free water has been removed
Composition of dried crystals
lb
CaCl2·6H2O
0.7657
inerts
0.0665
0.8322
𝑷𝒖𝒓𝒊𝒕𝒚 = 𝟗𝟐. 𝟎𝟏% 𝐴𝑁𝑆𝑊𝐸𝑅
%
92.01
7.99
100.00
0.7657
0.0060
0.0665
0.8382
CHEMICAL ENGINEERING SERIES 32
CRYSTALLIZATION
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 monohydrat e
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.
V
F
2,500 kg
8 g lactose per 10 g
water
tF = 57 C
OSLO CRYSTALLIZER
L
1.5 g lactose
per 10 g water
SOLUTION:
Consider over-all material balance
𝐹 = 𝐿+𝑉+𝐶
𝑤𝑡 𝐻2 𝑂 𝑖𝑛 𝑓𝑒𝑒𝑑 = 2,500 𝑘𝑔 𝑓𝑒𝑒𝑑 𝑥
10 𝑘𝑔 𝐻2 𝑂
= 1,388.89 𝑘𝑔
(10 + 8) 𝑘𝑔 𝑓𝑒𝑒𝑑
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
8 𝑘𝑔
𝑘𝑔 𝐶12 𝐻22 𝑂11
𝑥𝐹 =
= 0.4444
10 + 8
𝑘𝑔 𝑓𝑒𝑒𝑑
1.5
𝑘𝑔 𝐶12 𝐻22 𝑂11
𝑥𝐿 =
= 0.1304
10 + 1.5
𝑘𝑔 𝑙𝑖𝑞𝑢𝑜𝑟
𝑀𝐶12𝐻22𝑂11
342
𝑘𝑔 𝐶12 𝐻22 𝑂11
𝑥𝐶 =
=
= 0.95
𝑀𝐶12𝐻22𝑂11∙𝐻2 𝑂 360
𝑘𝑔 𝑐𝑟𝑦𝑠𝑡𝑎𝑙
(0.4444) (2,500) = (0.1304 )(𝐿 ) + (0.95)(𝐶 )
𝐿 = 8,519.9386 − 7.2853 𝐶
⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
C
tC = 10 C
CHEMICAL ENGINEERING SERIES 33
CRYSTALLIZATION
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),
𝐵𝑇𝑈
𝑘𝐽
𝜆 = 1,065.2
= 2,472.47
𝑙𝑏
𝑘𝑔
𝑘𝐽
𝑘𝑚𝑜𝑙
𝑘𝐽
𝐻𝐶 = 15,500
𝑥
= 43.06
𝑘𝑚𝑜𝑙 360 𝑘𝑔
𝑘𝑔
𝑘𝐽
𝑘𝐽
) (57 − 10) °𝐶 ] + [(959 .99 𝑘𝑔) (43 .06 )]
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = [(2,500 𝑘𝑔) (3.47
𝑘𝑔 ∙ °𝐶
𝑘𝑔
𝑘𝐽
)]
− [(13.89 𝑘𝑔) (2,472.47
𝑘𝑔
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 414.7196 𝑥 103 𝑘𝐽
𝑞𝑇 = 414.7196 𝑥 103 𝑘𝐽 + 4 𝑥 104 𝑘𝐽
𝒒𝑻 = 𝟒𝟓𝟒. 𝟕𝟐 𝒙 𝟏𝟎𝟑 𝒌𝑱
𝐴𝑁𝑆𝑊𝐸𝑅
CHEMICAL ENGINEERING SERIES 34
CRYSTALLIZATION
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 Na 2CO3 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 ft 2 of heating surface per
foot of length. An over-all heat transfer coefficient of 35 BTU/ft 2·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.
F (Soda Ash)
W (Water)
V
A
DISSOLVER
B
CRYSTALLIZER
45C
D
FILTER
DRYER
15C
R (remainder
mother liquor)
C (Sal Soda)
SOLUTION:
Basis: 2,000 lb/h (1 ton/h) of sal soda
Consider over-all material balance of the system
𝑊+𝐹= 𝑉+𝐶
CHEMICAL ENGINEERING SERIES 35
CRYSTALLIZATION
𝑉 = 𝑊 + 𝐹 − 2,000 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Consider Na2CO3 balance around the system
𝑥 𝐹 𝐹 = 𝑥 𝐶𝐶
𝑥 𝐹 = 1.0
𝑀𝑁𝑎 2𝐶𝑂3
106
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
𝑥𝐶 =
=
= 0.3706
𝑀𝑁𝑎2𝐶𝑂3∙10𝐻2𝑂 286
𝑙𝑏 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
𝑙𝑏
𝑁𝑎
𝐶𝑂
𝑙𝑏
𝑁𝑎
2
3
2 𝐶𝑂3 ∙ 10𝐻2 𝑂 )
(0.3706
) (2,000
𝑙𝑏 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
ℎ
𝐹=
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
1.0
𝑙𝑏 𝑠𝑜𝑑𝑎 𝑎𝑠ℎ
𝒍𝒃
𝑭 = 𝟕𝟒𝟏. 𝟐
𝐴𝑁𝑆𝑊𝐸𝑅
𝒉
Substitute to equation 1
𝑉 = 𝑊 + 741.2 − 2,000
𝑉 = 𝑊 − 1,258.8 ⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
Consider solute (Na2CO3) balance around the dryer
𝑥 𝐷𝐷 = 𝑥 𝐶𝐶
106 𝑙𝑏 𝑁𝑎2 𝐶𝑂3
14.2 𝑙𝑏 𝑁𝑎2 𝐶𝑂3
(0.90 𝑙𝑏𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂) (
) + (0.10 𝑙𝑏 𝐿 ) (
)
(100 + 14.2)𝑙𝑏 𝐿
286 𝑙𝑏 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
𝑥𝐷 =
1 𝑙𝑏 𝐷
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
𝑥 𝐷 = 0.3460
𝑙𝑏 𝐷
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
) (2,000 𝑙𝑏 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂)
𝑙𝑏 𝑁𝑎2 𝐶𝑂3 ∙ 10𝐻2 𝑂
ℎ
𝐷=
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
0.3460
𝑙𝑏 𝐷
𝑙𝑏
𝐷 = 2,142.20
ℎ
(0.3706
Consider over-all material balance around the dryer
𝐷= 𝑉+𝐶
𝑉 = 2,142.20 − 2,000
𝑙𝑏
𝑉 = 142.20
ℎ
Substitute to equation 2
142.20 = 𝑊 − 1,258.8
𝒍𝒃
𝑾 = 𝟏, 𝟒𝟎𝟏
𝐴𝑁𝑆𝑊𝐸𝑅
𝒉
Consider solute (Na2CO3) balance around the dissolver
𝑥 𝐹 𝐹 + 𝑥 𝑅 𝑅 = 𝑥𝐴 𝐴
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
𝑥𝐴 = 0.30
𝑙𝑏 𝐴
14.2 𝑙𝑏 𝑁𝑎2 𝐶𝑂3
𝑙𝑏 𝑁𝑎2 𝐶𝑂3
𝑥𝑅 =
= 0.1243
(100 + 14.2)𝑙𝑏 𝑅
𝑙𝑏 𝑅
(1.0)(741.2) + (0.1243) (𝑅) = (0.30)(𝐴)
𝐴 = 2,470.67 + 0.4143𝑅
⟶ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3
CHEMICAL ENGINEERING SERIES 36
CRYSTALLIZATION
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
𝑙𝑏
𝑅 = 560.8
ℎ
𝑙𝑏
𝐴 = 2,973.0
ℎ
Consider heat balance around the crystallizer
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 𝐴𝐶𝑃 (𝑡𝐴 − 𝑡𝐵 ) + 𝐶′𝐻𝐶
𝑙𝑏
𝑙𝑏
𝐶 ′ = 0.90𝐷 = 0.90 (2,142.20 ) = 1,928.0
ℎ
ℎ
𝐵𝑇𝑈
1
𝑐𝑎𝑙
𝑚𝑜𝑙
𝐵𝑇𝑈
𝑙𝑏
𝐻𝐶 = 25,000
𝑥
𝑥
= 157.34
𝑚𝑜𝑙 286 𝑔 0.55556 𝑐𝑎𝑙
𝑙𝑏
𝑔
𝑙𝑏
𝐵𝑇𝑈
1.8°𝐹
𝑙𝑏
𝐵𝑇𝑈
) (45 − 15) °𝐶 𝑥
] + [(1,928.0 ) (157.34
)]
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = [(2,973.0 ) (0.85
ℎ
𝑙𝑏 ∙ °𝐹
°𝐶
ℎ
𝑙𝑏
𝐵𝑇𝑈
𝑞𝑐𝑟𝑦𝑠𝑡𝑎𝑙𝑠 = 439,812.22
ℎ
𝑞 = 𝑈𝐴∆𝑇𝑙𝑚
∆𝑇𝑙𝑚 =
(𝑡𝐴 − 𝑡2 ) − (𝑡𝐵 − 𝑡1 )
𝑡 − 𝑡2
ln 𝐴
𝑡𝐵 − 𝑡1
[(45 − 20) − (15 − 10)] °𝐶 𝑥 1.8°𝐹
°𝐶
∆𝑇𝑙𝑚 =
45 − 20
ln
15 − 10
∆𝑇𝑙𝑚 = 22.37°𝐹
𝐵𝑇𝑈
439,812.22
ℎ
𝐴=
𝐵𝑇𝑈
(35
) (22.37°𝐹 )
ℎ ∙ 𝑓𝑡 2 ∙ °𝐹
𝐴 = 561.74 𝑓𝑡 2
# 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 = 561.74 𝑓𝑡 2 𝑥
1 𝑓𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
3
𝑓𝑡 2
𝑥
1 𝑢𝑛𝑖𝑡
10 𝑓𝑡
# 𝒐𝒇 𝒖𝒏𝒊𝒕𝒔 = 𝟏𝟖. 𝟕 ≈ 𝟏𝟗 𝒖𝒏𝒊𝒕𝒔 𝐴𝑁𝑆𝑊𝐸𝑅
Refrigeration capacity:
CHEMICAL ENGINEERING SERIES 37
CRYSTALLIZATION
𝑅𝐶 = 439 ,812.22
𝐵𝑇𝑈
ℎ
𝑥
𝑡𝑜𝑛 𝑟𝑒𝑓𝑟𝑖𝑔𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝐵𝑇𝑈
12,000
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 Na 2S2O3 is 83 parts per
100 parts water
Source: Unit Operations (Brown, et al)
SOLUTION:
𝑊𝑃
∫
𝑊𝑆
𝑑𝑊𝑃 = ∫
(1 +
∆𝐷
3
) 𝑑𝑊𝑆
𝐷𝑆
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
𝑊𝑃
𝑊𝑆
0.349 3
∫ 𝑑𝑊𝑃 = ∫ (1 +
) 𝑑𝑊𝑆
0.841
0
0
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
𝒍𝒃
𝑾𝑺 = 𝟏, 𝟎𝟗𝟏. 𝟏𝟏
𝐴𝑁𝑆𝑊𝐸𝑅
𝒉
Consider Na2S2O3 balance:
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑙𝑏 𝑁𝑎2 𝑆2 𝑂3
𝑥 𝐹 = 0.565
𝑙𝑏 𝑓𝑒𝑒𝑑
83 𝑙𝑏 𝑁𝑎2 𝑆2 𝑂3
𝑙𝑏 𝑁𝑎2 𝑆2 𝑂3
𝑥𝐿 =
= 0.4536
(100 + 83) 𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑀𝑁𝑎 2𝑆2𝑂3
158
𝑙𝑏 𝑁𝑎2 𝑆2 𝑂3
𝑥𝐶 =
=
= 0.6371
𝑀𝑁𝑎2𝑆2𝑂3∙5𝐻2 𝑂 248
𝑙𝑏 𝑁𝑎2 𝑆2 𝑂3 ∙ 5𝐻2 𝑂
CHEMICAL ENGINEERING SERIES 38
CRYSTALLIZATION
(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
𝒍𝒃
𝑭 = 𝟑, 𝟐𝟗𝟒. 𝟑𝟏
𝐴𝑁𝑆𝑊𝐸𝑅
𝒉
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
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
From figure 27-3 (Unit Operation 7th edition, McCabe and Smith) at 110°F
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥 𝐹 = 0.32
𝑙𝑏 𝑓𝑒𝑒𝑑
From figure 27-3 (Unit Operations 7th edition, McCabe and Smith) at 40°F
𝑙𝑏 𝑀𝑔𝑆𝑂4
𝑥 𝐿 = 0.21
𝑙𝑏 𝑙𝑖𝑞𝑢𝑜𝑟
𝑀𝑀𝑔𝑆 𝑂4
120 .38
𝑙𝑏𝑀𝑔𝑆 𝑂4
𝑥𝐶 =
=
= 0.4884
𝑀𝑀𝑔𝑆 𝑂4∙7𝐻2 𝑂 246 .49
𝑙𝑏 𝑀𝑔𝑆𝑂4 ∙ 7𝐻2 𝑂
(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
1 𝑙𝑏 𝑠𝑒𝑒𝑑𝑠
𝑊𝑆 = 100 𝑙𝑏 𝑓𝑒𝑒𝑑 𝑥
= 1 𝑙𝑏
100 𝑙𝑏 𝑓𝑒𝑒𝑑
𝑊𝑃 = 𝐶 + 𝑊𝑆 = 39.51 + 1 = 40.51 𝑙𝑏
𝑊𝑃
𝑊𝑆
∆𝐷 3
∫ 𝑑𝑊𝑃 = ∫ (1 +
) 𝑑𝑊𝑆
𝐷𝑆
0
0
𝐷𝑃 3
𝑊𝑃 = [ ] 𝑊𝑆
𝐷𝑆
From table 19-6 (CHE HB 8th edition)
CHEMICAL ENGINEERING SERIES 39
CRYSTALLIZATION
𝐷𝑆 = 𝑚𝑒𝑠ℎ 80 = 0.177 𝑚𝑚 (𝑠𝑖𝑒𝑣𝑒 𝑜𝑝𝑒𝑛𝑖𝑛𝑔)
3
𝐷𝑃 = (0.177 𝑚𝑚) √
40.51 𝑙𝑏
1 𝑙𝑏
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:
𝑊𝑃
∫
𝑊𝑆
𝑑𝑊𝑃 = ∫
(1 +
∆𝐷
3
) 𝑑𝑊𝑆
𝐷𝑆
0
𝑑𝑊𝑆 = 𝑊𝑆 𝑑𝜙𝑆
1
∆𝐷 3
∫
(
) 𝑑𝜙𝑆
𝑊𝑃 = 𝑊𝑆
1+
𝐷𝑆
0
1
𝑊𝑃
∆𝐷 3
) 𝑑𝜙𝑆
= ∫ (1 +
𝑊𝑆
𝐷𝑆
0
1
𝑊𝑃
∆𝐷 3
) 𝑑𝜙𝑆
𝑙𝑒𝑡, 𝑚 =
= ∫ (1 +
𝑊𝑆
𝐷𝑆
0
0
Δ𝑚 = (1 +
∆𝐷
3
) Δ𝜙𝑆
𝐷𝑆
Where: Δ𝜙𝑆 = fractional weight range
Solve for required 𝑚:
𝑊𝑃 7,000 𝑙𝑏
𝑚𝑟𝑒𝑞𝑑 =
=
= 14
𝑊𝑆
500 𝑙𝑏
This problem can be solved by trial and error
1.
Assume value of Δ𝐷
CHEMICAL ENGINEERING SERIES 40
CRYSTALLIZATION
∆𝐷 3
̅𝑆 for each size range
) for each size range, use the mean 𝐷
2.
Solve for (1 +
3.
4.
5.
Solve for Δ𝑚
Get the total Δ𝑚
If ∑ Δ𝑚 = 𝑚𝑟𝑒𝑞𝑑 , then assumed Δ𝐷 is correct; if not, redo another trial
𝐷𝑆
TRIAL 1: Assume Δ𝐷 = 0.004 𝑖𝑛
3
Δ𝐷
)
̅𝑆
𝐷
3
̅𝑆
𝐷
Δ𝜙𝑆
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 +
Δ𝑚 = Δ𝜙𝑆 (1 +
1.00
Δ𝐷
)
̅𝑆
𝐷
14.4201
Since % error is less than 5%, assumed value can be considered
For particle size distribution:
𝐷𝑃 = Δ𝐷 + 𝐷𝑆
Δ𝑚
% 𝑤𝑡 = 100Δ𝜙𝑆 =
3 𝑥 100
Δ𝐷
(1 +
)
̅𝑆
𝐷
𝑆𝐸𝐸𝐷 𝐶𝑅𝑌𝑆𝑇𝐴𝐿𝑆
𝑷𝑹𝑶𝑫𝑼𝑪𝑻 𝑪𝑹𝒀𝑺𝑻𝑨𝑳𝑺
Size Range, in
Wt %
Size Range, in
Wt %
− 0.0200 + 0.0100
10.00
− 𝟎. 𝟎𝟐𝟒𝟎 + 𝟎. 𝟎𝟏𝟒𝟎
𝟏. 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
𝐿 = 20,000 − 6,500 = 13,500
𝑙𝑏
ℎ
𝑙𝑏
ℎ
CHEMICAL ENGINEERING SERIES 41
CRYSTALLIZATION
Consider Na3PO4 balance:
𝑥 𝐹 𝐹 = 𝑥 𝐿 𝐿 + 𝑥 𝐶𝐶
𝑀𝑁𝑎 3𝑃𝑂4
164
𝑙𝑏 𝑁𝑎3 𝑃𝑂4
𝑥𝐶 =
=
= 0.4316
𝑀𝑁𝑎3𝑃𝑂4∙12𝐻2 𝑂 380
𝑙𝑏 𝑁𝑎3 𝑃𝑂4 ∙ 12𝐻2 𝑂
(0.35)(20,000) = ( 𝑥 𝐿)(13,500) + (0.4316 )(6,500)
𝑥 𝐿 = 0.3107
𝑥 𝐿 = 0.3107
𝑥 𝐿 = 0.4507
𝑙𝑏 𝑁𝑎3 𝑃𝑂4
𝑙𝑏 𝑠𝑜𝑙𝑛
𝑙𝑏 𝑁𝑎3 𝑃𝑂4
𝑙𝑏 𝑠𝑜𝑙𝑛
𝑙𝑏 𝑁𝑎3 𝑃𝑂4
𝑥
𝑙𝑏 𝑠𝑜𝑙𝑛
(1 − 0.3107) 𝑙𝑏 𝐻2 𝑂
𝑙𝑏 𝐻2 𝑂
From table 2-120 (CHE HB 8th edition)
50°C
60°C
43 lb/100 lb H2O
55 lb/100 lb H2O
𝑻 = 𝟓𝟏. 𝟕𝟐𝟓°𝑪 ≈ 𝟏𝟐𝟓. 𝟏𝟏°𝑭 𝐴𝑁𝑆𝑊𝐸𝑅
Cooling Duty:
Consider heat balance:
𝑞 = 𝐹𝐶𝑃 (𝑡𝐹 − 𝑡𝑃 ) + 𝐶𝐻𝐶
𝑙𝑏
𝐵𝑇𝑈
𝑙𝑏
𝐵𝑇𝑈
𝑙𝑏𝑚𝑜𝑙
) (190 − 125.11) °𝐹 ] + [(6,500 ) (27,500
)]
𝑞 = [(20,000 ) (0.8
𝑥
ℎ
𝑙𝑏 ∙ °𝐹
ℎ
𝑙𝑏𝑚𝑜𝑙 380 𝑙𝑏
𝒒 = 𝟏, 𝟓𝟎𝟖, 𝟔𝟑𝟒. 𝟕𝟒
𝑩𝑻𝑼
𝒉
𝐴𝑁𝑆𝑊𝐸𝑅
CHEMICAL ENGINEERING SERIES 42
CRYSTALLIZATION
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 CaCl 2 at 20°C is 6.7 gmol anhydrous salt (CaCl 2) per kg of water.
SOLUTION:
For a saturated solution utilizing 100 kg water as solvent:
1.
Mole of CaCl2 required
6.7 𝑔𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
1 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
𝑛𝐶𝑎 𝐶𝑙2 = 100 𝑘𝑔 𝐻2 𝑂 𝑥
𝑥
𝑘𝑔 𝐻2 𝑂
1,000 𝑔𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
𝑛𝐶𝑎 𝐶𝑙2 = 0.67 𝑘𝑚𝑜𝑙
2.
Weight of CaCl2 required
𝑊𝐶𝑎 𝐶𝑙2 = 0.67 𝑘𝑔 𝐶𝑎𝐶𝑙 2 𝑥
110.994 𝑘𝑔 𝐶𝑎𝐶𝑙 2
𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
𝑊𝐶𝑎 𝐶𝑙2 = 74.36 𝑘𝑔
3.
Mole of CaCl2·6H2O required
𝑛 𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 0.67 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2 𝑥
1 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2
𝑛 𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 0.67 𝑘𝑚𝑜𝑙
4.
Weight CaCl2·6H2O required
𝑊𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 0.67 𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂 𝑥
218 .994 𝑘𝑔 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
𝑘𝑚𝑜𝑙 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
𝑊𝐶𝑎𝐶𝑙2 ∙6𝐻2 𝑂 = 146.72 𝑘𝑔
5.
Composition of the solution in terms of CaCl 2·6H2O
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 𝐻2 𝑂 𝑥
𝑾𝑪𝒂𝑪𝒍𝟐∙𝟔𝑯𝟐 𝑶
146.72 𝑘𝑔 𝐶𝑎𝐶𝑙 2 ∙ 6𝐻2 𝑂
27.64 𝑘𝑔 𝑓𝑟𝑒𝑒 𝐻2 𝑂
= 𝟓𝟑𝟎. 𝟖𝟐 𝒌𝒈 𝐴𝑁𝑆𝑊𝐸𝑅