TOLEDANO, Anthon Jay B.
4ChEB
1. Statement of the Problem
Epsom salt (MgSO4. 7H2O) is a common mineral medicinal preparation of magnesium. It
is a natural exfoliant remedy that can be used to treat dry skin, sore muscles, and small
wounds. It can also be added to any bath or foot soak to create a lavish home spa experience
[1]. In a certain plant, Epsom salt is produced by crystallization in continuous adiabatic vacuum
crystallizer with 30,000 lb per hour of aqueous MgSO 4 solution at 87.8°C. The crystallizer is to
be controlled so that the percent recovery of MgSO 4 in the feed as MgSO4.7H2O crystals is
47.2%. The boiling point elevation of the solution in the crystallizer is 8°F, based on the
average of previous productions. The weight fraction of the feed can be determined by the
relation xF = 2.5(xC)-3.821(xC)2. Determine the following:
a. the product temperature and pressure
b. the composition of the magma
c. Mother liquor produced (in kg) per one week (7 days) of production?
d. the weight ratio of mother liquor to crystalline product
2. Analysis
About continuous adiabatic vacuum crystallizer
In the problem, aqueous MgSO4 solution, saturated at 87.8oC, is fed in continuous
adiabatic vacuum crystallizer. Continuous adiabatic vacuum crystallizers function on the
principle of ‘adiabatic boiling’ wherein the feed liquor is exposed to high vacuum
adiabatically resulting in evaporation. At the same time, the temperature is lowered,
inducing super saturation by simultaneous cooling and evaporation of the solvent [2]. At the
point of super saturation, the Epsom salt present in the liquor starts crystallizing. The Epsom
salt formed was recovered and collected.
The Problem
In the problem, certain values essential for the solution were not simply given. It
would require a little “figuring-out” to obtain these values. The weight fraction of MgSO 4 in
the feed can be determined by the equation x F = 2.5(xC)-3.821(xC)2. However, the weight
fraction of MgSO4 in the crystals was not given; therefore it is yet to be calculated by getting
the molecular weight ratio of MgSO 4 to Epsom salt. The crystallization rate was not given as
well, but the given percent recovery can be used to solve for it with the formula
X CC
% recovery= x 100 . The temperature of the resulting magma was deduced
XFF
assumption and trial and error. More trials are conducted until the values obtained for
consecutive trials results to less than 5% of margin or error.
3. Diagram representation of the given data
VAPOR
CONTINUOUS ADIABATIC
FEED BPR = 8oC
30, 000 lb/hr VACUUM CRYSTALLIZER
aq. MgSO4 solution
tF = 87.8oC or 190oF
xF = 2.5(xC)-3.821(xC)2
MOTHER CRYSTALS
LIQUOR MgSO4 .7H2O crystals
tL = ?
47.2% recovery
xL = ?
tc= ?
xF = ?
4. Required
a. the product temperature and pressure (in kPa)
b. the composition of the magma
c. Mother liquor produced (in kg) per one week (5 working days) of production?
d. the weight ratio of mother liquor to crystalline product
5. Solution
lb
MgS O 4=120
mol
lb
MgS O 4 .7 H 2 O=246
mol
Trial 1:
Assume tL = 80oF ∴ xL = 0.275 (from H-x-T diagram for MgSO4)
Overall Mass Balance: F=V + L+C
30,000=V + L+ 9,000→ Equation ①
Solute Balance: F x F =L x L +C xC
120 lb MgSO 4
x C= =0.488 x F =2.5 ( x C )−3.821 ( x C )2=0.31
246 lb MgS O 4 .7 H 2 O
% recovery=0.472=
C ( 120
246 )
∴ C=8,998.68 lb ≈ 9,000 lb
0.31(30,000 lb)
(30,000) ( 0.31 )=L(0.275)+(9,000)(0.488) → Equation ②
lb lb
L=17,847.2727 V =3,152.7272
hr hr
Enthalpy Balance: F h F =V H V + L hL + C hC
Hv (steam) at 80oC =1096.11 Btu/lb H V =1096.11+0.45 ( 8 )=1099.71
T (oF) x (mass fraction) H (Btu/lb)
Feed 190 0.31 25
Liquor 80 0.275 -45
Crystals 80 0.488 -150
Vapor 80 - 1099.71
( 30,000 ) ( 25 )=V ( 1099.11 ) + L (−45 ) + ( 9,000 )(−150 )→ Equation ③
Solving equations 1 & 3lb lb
L=18,338.526 V =2,661. 4574
simultaneously, hr hr
%error =2.75 % %error=15.28 % (reject )
Substituting L to ( 30,000 ) ( 0.31 )=( 18,388.526 ) ( x L ) + ( 9,000 ) ( 0.488 )❑
equation 2 (SB) x L =0.2676 ≈ 0.27 ( Next assumption)
Trial 2:
Assume xL = 0.27 ∴ tL = 75oF
Enthalpy Balance: F h F =V H V + L hL + C hC
Hv (steam) at 75oC =1093.95 Btu/lb H V =1093.95+0.45 ( 8 )=1097.55
T (oF) x (mass fraction) H (Btu/lb)
Feed 190 0.31 25
Liquor 75 0.27 -48
Crystals 75 0.488 -150
Vapor 75 - 1097.55
( 30,000 ) ( 25 )=V ( 1097.55 ) + L (−48 ) + ( 9,000 )(−150 ) → Equation ③
Solving equations 1 & 3 lb lb
L=18,286.8928 V =2,713.1072
simultaneously, hr hr
%error=0.282 % %error=1.904 %(accept)
Substituting L to
( 30,000 ) ( 0.31 )=( 18,286.8928 ) ( x L )+ ( 9,000 ) ( 0.488 )❑
equation 2 (SB)
x L =0.268 ≈ 0.27
a. ∴ tL = 75 oC
Pressure (steam) at 75oC = 2.966 kPa Abs
b. Magma=Mother Liquor ( L )+Crystals (C )
lb lb lb
¿ 18,286.8928 +9,000 =27,286.8928
hr hr hr
18,286.8928 lb 9,000 lb
L= =67.02 % C= =32.98 %
27,286.8928 lb 27,286.8928 lb
lb 24 hr days lb
c. L=18,286.8928 x x5 =2,194,427.136
hr day week week
lb Mother Liquor 18,286.8928 lb Liquor
d. = =2.032lb
lb crystals 9,000 lb Crystals
6. Answers
a. 75 oC, 2.966 kPa Abs
b. L = 67.02 %, C = 32.98 %
c. 2,194,427.136 lb/week
d. 2.032 lb Liquor/Crystals
7. Reflection
At the end of this chapter, students were able to understand and study the process of
crystallization. This includes the mechanisms, apparatuses, applications in the industry, and theories
involving crystallization. Crystallization, in general, is the process of solid-liquid separation in which
solid particles are formed from a homogenous solution. The resulting product of this process is
called magma, which is a two-phase mixture of mother liquor and crystals. Crystals are the essential
products of this operation and can be varied into different types such as salts, sugar, and some of
the most precious gem stones. In order for crystals to be produced artificially, certain equipment
and mechanisms are to be considered. Some of the equipment discussed in this chapter are the
vacuum crystallizer and Swenson-Walker crystallizer, which are widely used in the industry. As for
mechanism of the crystallization itself, it is important to note that a solution must be supersaturated
first before undergoing crystallization. Hence, different methods such as cooling, evaporation, and
vacuum cooling must be done to supersaturate the solution. Once the solution is supersaturated, it
may now undergo the two-step mechanism of crystallization: nucleation, and crystal growth.
Solubility of a substance also plays a major role in crystallization as it stimulates crystallization of a
substance at a specific temperature. It also helps to choose a suitable crystallization process for a
solution.
Overall, it is no doubt that crystallization is an important separation technique. It is also
apparent in the world way back until present. In fact it is one of the oldest unit operations of
chemical engineering. For instance, sodium chloride has been mass-produced this way since the
dawn of civilization [3]. Today, crystallization is one of the most widely used technologies in
chemical industry. It may be used purification of substances, changing texture and size of
commercial products such as ice cream, chocolate, tablet medicines, and many more important
functions [4]. Without crystallization, some substances or objects may be found lacking of the
benefits crystallization grants.
References
[1] Amazing Uses for Epsom Salt. (n.d.). Retrieved November 8, 2018, from Doctor Oz:
https://www.doctoroz.com/slideshow/epsom-salt-uses
[2] Adiabatic Vacuum. (n.d.). Retrieved November 8, 2018, from Chem Process Systems:
http://www.chemprosys.com/products/crystallizer/adiabetic-vacuum/
[3] What is Crystallization? (n.d.). Retrieved November 8, 2018, from Syrris:
https://syrris.com/applications/what-is-crystallization-and-what-are-the-methods-of-
crystallization/
[4] Crystallization. (n.d.). Retrieved November 8, 2018, from Your Mom Was A Chemist:
http://kitchenscience.sci-toys.com/crystals
