Chapter 7 : Crystallisation

Example 7.1

In [16]:
 
c=.498;                #solute content afetr crystallisation
W1=111.;                #molecular weight of CaCl2
W2=219.;                #molecular weight of CaCl2.6H2O

# Calculation  
M1=(108./W2)*100;       #water present in 100kg of CaCl2.6H2O
M2=(W1/W2)*100;       #CaCl2 present in 100kg of CaCl2.6H20
#t=M2+c*x;           #total weight entering the solubility
#x+49.3;             total water solubility used
#s*(x+49.3)/100      #total Cacl2 after solubility
#x=poly([0],'x');        #calc. x the weight of crystal
t= 32.112871 #roots((M2+c*x)-(x+49.3)*.819);             

# Result
print "\nthe weight of water in the quantity of solution needed :%f kg"%t

h=(c)*t;           #weight of CaCl2 corresponding to weight water
tw=t+h;              # total weight of the solution
print "\nthe  total weight of the solution is :%f kg"%tw
#end

the weight of water in the quantity of solution needed :32.112871 kg

the  total weight of the solution is :48.105081 kg

Example 7.2

In [17]:
 

#part(i)
w1=1000.;               #weight of solution to be cooled
s1=104.1;              #solubility at 50 degree per 100 kg of water
s2=78.0;              #solubility at 10 degree per 100 kg of water
a2=45.;               #percentage of sodium nitrate in the solution per 100kg of solution 

# Calculation  
x1=s1/(100+s1)*100;                #percentage of saturated solution at 50 degree
tw=(a2/(100-a2))/(x1/(100-x1));     #the  percentage saturation
print "\nthe  percentage saturation is :%f percent"%(tw*100)

#part(ii)
#let x be the weight of NaNO3 crystal formed after crystallisation
#x=poly([0],'x');        #calc. x the weight of crystal
t=21.000000  #roots((w1*a2/100)-(x+(w1-x)*s2/(100+s2)));

# Result
print "\n the weight of NaNO3 crystal formed after crystallisation :%f kg"%t

#part(iii)
y=t/(a2*w1/100);          #yield = weight of NaNO3 crystal formed/weight of NaNO3
print "\n the percentage yield is:%f percent"%(y*100)
#end

the  percentage saturation is :78.595756 percent

 the weight of NaNO3 crystal formed after crystallisation :21.000000 kg

 the percentage yield is:4.666667 percent

Example 7.3

In [18]:
   

s1=19.75;              #solubility at 70 degree per 100 gm of water
s2=16.5;              #solubility at 50 degree per 100 gm of water
s3=12.97;              #solubility at 30 degree per 100 gm of water
s4=9.22;              #solubility at 10 degree per 100 gm of water
s5=7.34;              #solubility at 0 degree per 100 gm of water

# Calculation  
        #basis is 1000kg of saturated solution
w1=1000.*(s1/(s1+100));            #weight of K2SO4 in the original solution
w2=1000.-w1;                      #weight of water in kg
w3=w1*.5;                       #weight of K2SO4 in the solution
wp=w3/(w3+w2);              #weight percent of K2SO4 in the solution after crystallistion

# Result
print "\n for the corresponding temperature to :%f percent of K2SO4  is 15 degree (by linear interpolation between 10 to 30 degree) "%(wp*100)

#end

 for the corresponding temperature to :8.987486 percent of K2SO4  is 15 degree (by linear interpolation between 10 to 30 degree)

Example 7.4

In [19]:
   

#part(i)
a1=146.;                #solubility at 70 degree
a2=121.;                #solubility at 10 degree
t1=58.;                 # percentage of solute content
t2=40.66;

# Calculation  and  Result
x1=a1/(100+a1) *100;        #percentage of saturated solution at 50 degree
tw=(t1/42.)/(x1/t2);    #the  percentage saturation
print "\nthe  percentage saturation is :%f percent"%(tw*100)

#part(ii)
p1=2000*.58;              #weight of solute in 200kg of solution 2000*.58
#let x be the weight of  crystal formed after crystallisation
#x=poly([0],'x');         #calc. x the weight of crystal
t=231.743929  #roots((1160)-(x+(1055.02-.547*x)));             
print "\n the weight of NaNO3 crystal formed after crystallisation :%f kg"%t

#part(iii)
y=t/p1;            #yield = weight of NaNO3 crystal formed/weight of NaNO3
print "\n the percentage yield is:%f percent"%(y*100)
#end

the  percentage saturation is :94.608102 percent

 the weight of NaNO3 crystal formed after crystallisation :231.743929 kg

 the percentage yield is:19.977925 percent

Example 7.5

In [20]:
  

p1=.3;                   #percentage of the solute in the solution
w1=1000.;                 #weight of the solution taken
w2=142.;                 #molecular weight of Na2SO4.

# Calculation  
M1=(w2/(180+w2));        #solute (Na2SO4) present in the Na2CO3.10H2O solution
s1=40.8;              #solubility of Na2SO4 at 30 degree per 100 gm of water
s2=9.0;              #solubility of Na2SO4 at 10 degree per 100 gm of water
#percent weight of solute in Na2SO4.10H2O= 144/322
#let x be the weight of crystal formed 
#x=poly([0],'x');         #calc. x the weight of crystal
t= 576.477290 #roots((w1*40.8/140.8)-(.442*x+(w1-x)*(s2/(100+s2))));             

# Result
print "\n the weight of  crystal formed after crystallisation :%f kg"%t

#end

 the weight of  crystal formed after crystallisation :576.477290 kg

Example 7.6

In [21]:
   

s1=12.5;                  #solubility of Na2CO3 at 10 degree per 100 gm of water
p1=.3;                   #percentage of the solute in the solution
w1=2000.;                 #weight of the solution taken
w2=106.;                 #molecular weight of Na2CO3.

# Calculation  
M1=(w2/(180+w2));        #solute (Na2CO3) present in the Na2CO3.10H2O solution
#let x be the quantity of Na2CO3.10H2O crystal formed
#x=poly([0],'x');         #calc. x the weight of crystal
t= 1455.688623 #roots(w1*p1-M1*x-(w1-x)*(s1/(100+s1)));    

# Result
print "\n the weight of quantity of Na2CO3.10H2O  :%f kg"%t
#in the book the ans is wrong, they have calculated 2000*0.3-2000*12.5/112.5 as =x(miscalculation)

p=(286./106)*w1*p1;           #weight of Na2CO3.10H2O crystal present in the original solution
y=t/p;                 #percentage yield 
print "\n percentage yield :%f percent"%(y*100)
#end

 the weight of quantity of Na2CO3.10H2O  :1455.688623 kg

 percentage yield :89.920160 percent

Example 7.7

In [14]:
 
s1=139.8;              #solubility at 80 degree per 100 gm of water
s2=110.5;              #solubility at 20 degree per 100 gm of water
w2=174.2;                 #molecular weight of K2CO3.10H2O

# Calculation  
M1=(138/w2)*100;        #water present in 100kg of K2CO3.10H2O
#let x be the quantity of Na2CO3.10H2O
#x=poly([0],'x');         #calc. x the weight of crystal
t=108.634036 #roots(500*(139.8/239.8)-.7921*x-(500-x)*110.5/210.5);             

# Result
print "\n the weight of quantity of K2CO3.10H2O formed  :%f kg"%t

p=(174./138)*500*(139.8/239.8);       #weight of crystal present in the original solution
y=t/p;                 #percentage yield 
print "\n percentage yield :%f percent"%(y*100)
#end

 the weight of quantity of K2CO3.10H2O formed  :108.634036 kg

 percentage yield :29.557504 percent

Example 7.8

In [22]:
  

s1=20.51;                  #solubility at 10 degree per 100 gm of water
w2=277.85;                 #molecular weight of FeSO4.7H2O

# Calculation  and  Result
#let x be the quantity of Na2CO3.10H2O
#x=poly([0],'x');         #calc. x the weight of crystal
t= 549.684973 #roots(900*.4-.5465*x-(900-x)*20.5/120.5);             
print "\n the weight of quantity of FeSO4.7H2O formed  :%f kg"%t

p=(277.85/151.85)*900*(0.4);       #weight of crystal present in the original solution
y=t/p;                             #percentage yield 
print "\n percentage yield :%f percent"%(y*100)
#end

 the weight of quantity of FeSO4.7H2O formed  :549.684973 kg

 percentage yield :83.447967 percent

Example 7.9

In [23]:
 
#part(i)
a1=229.7;              #solubility at 60 degree
a2=174.7;              #solubility at 60 degree
t1=68;                # percentage of sodium nitrate
t2=30.34;

# Calculation and  Result
x1=a1/329.7 *100;        #percentage of saturated solution at 50 degree
tw=(t1/32.)/(x1/t2);    #the  percentage saturation
print "\nthe  percentage saturation is :%f percent"%(tw*100)

#part(ii)
#let x be the weight of Cesium chloride crystal formed after crystallisation
#x=poly([0],'x');        #calc. x the weight of crystal
t=120.960000 #roots(1000*.68-(x+(1000-x)*174.7/274.7));             
print "\n the weight of CaCl2 crystal formed after crystallisation :%f kg"%t

#part(iii)
y=t/680.;          #yield = weight of CaCl2 crystal formed/weight of CaCl2
print "\n the percentage yield of Cesium chloride  is:%f percent"%(y*100)
#end

the  percentage saturation is :92.540632 percent

 the weight of CaCl2 crystal formed after crystallisation :120.960000 kg

 the percentage yield of Cesium chloride  is:17.788235 percent

Example 7.10

In [24]:
  

s1=38.8;              #solubility at 30 degree per 100 gm of water
s2=12.5;              #solubility at 10 degree per 100 gm of water
w2=296;                 #molecular weight of Na2CO3.10H2O
per=116./w2 *100;        #percentage solute in Na2CO3.10H2O

#let x be the quantity of Na2CO3.10H2O
w=200.;                    #original solotion weight

# Calculation  
m1=w*(s2/(s2+100));       #weight of Na2CO3.10H2O needed to dissolve Na2CO3 present in the original solotion 
w3=w-m1;                 #weight of water 
#w4=m1+per/100;            weight of Na2CO3 after dissolution
x1=s1/(s1+100);           #weight fraction of solute after dissolution 

# Result
print "\n the weight of quantity of Na2CO3.10H2O  formed  :%f kg"%w3

#for the total solution after dissolution
#x=poly([0],'x');        #calc. x the weight of crystal
t= 300.485784 #roots((m1+per*x/100)-((m1+per*x/100)+(w3+.609*x))*x1);             
print "\nweight of Na2CO3.10H2O needed to dissolve Na2CO3 present in the original solution  %f kg"%t

#end

 the weight of quantity of Na2CO3.10H2O  formed  :177.777778 kg

weight of Na2CO3.10H2O needed to dissolve Na2CO3 present in the original solution  300.485784 kg

Example 7.11

In [25]:
   

s1=35.;                #percentage of solution
x1=6000.;              #weight of Na2CO3 solution
s2=21.5;              #solubility at 20 degree per 100 gm of water
w2=296.;                 #molecular weight of Na2CO3.10H2O

# Calculation  
per=116./w2 *100;        #percentage solute in Na2CO3.10H2O
w1=s1*x1;            #weight of solute
w3=x1*0.04;          #weight of solution lost by vaporisation
#let x be the quantity of Na2CO3.10H2O formed 
#making material balance 
#x=poly([0],'x');        #calc. x the weight of crystal
t= 5049.122335 #roots(2100-(.391*x)-(6000-240-x)*(21.5/121.5));             

# Result
print "\n the weight of Na2CO3.10H2O crystal formed after crystallisation :%f kg"%t


#end

 the weight of Na2CO3.10H2O crystal formed after crystallisation :5049.122335 kg

Example 7.12

In [26]:
 

C=1000.;                      #crystal formed in kg
hf=26.002;                   #enthalpy of the feed at 80 degree in cal/g
hl=-1.33;                   #enthalpy of the saturated sol at 30 degree in cal/g
hc=-50.56;                 #enthalpy of crystal
xf=40./(100+40);
xm=30./(100+30);
xc=151.84/277.85;        #151.84 is the weight of FeSO4
        #component balance

Hf=26.002;                #enthalpy of the feed at 80 degree in cal/g
Hv=612.;                  #
Hm=-1.33;               #enthalpy of the saturated sol at 30 degree in cal/g
Hc=-50.56;             #enthalpy of crystal leaving the crystalliser
from numpy import matrix
from numpy import linalg
#solving these we gt 

# Calculation  
a=matrix([[1,-1,-1],[.286,-.231,0],[26.002,1.33,-612]])
b=matrix([[10000],[5470],[-505600]])
x=linalg.inv(a)*b;                    #solving out the values using matrices 
t1=x[0];                        #3 solution of the eqn
t2=x[1];
t3=x[2];

# Result
print "\n the feed rate F= : %f kg/hr \n value of M= : %f kg/hr\n value of V=: %f kg/hr"%(t1,t2,t3)
#end

 the feed rate F= : 45556.688340 kg/hr 
 value of M= : 32723.865218 kg/hr
 value of V=: 2832.823122 kg/hr

Example 7.13

In [27]:
 

import math
C=800.;                       #crystal formed in kg/hr
t2=49.;                        #temp. of the entering fed
t1=27.;                       #temp. of the product
t3=21.;                       #temp. of the leaving cooling water
t4=15.;                      #temp. of the enetring  cooling water
U=175.;                      #overall heat transfer coefficient
F=140*151.85/277.85;       #feed concentration 
xf=F/240;                  #concentration in feed solution
P=74*151.85/277.85;        #product concentration 
xm=P/174;                  #concentration of FeSO4 in product solution
xc=151.85/277.85;         #
       
# Calculation  
#solving these we get
F=800*.3141/0.0866;        #feed conc.
M=F-C;                     #product concentration 
        #making energy balance
        #heat to be removed by cooling water =heat to be removed from solution + heat of crystallization
cp=.7;                    #specific heat capacity
dt=(t2-t1);              #change in temp.
dh=15.8;                 #heat of crystallization
Q=F*cp*dt+dh*C;          #heat to be removed by cooling water
cp=1;                    #specific heat capacity of water
dt=(t3-t4);                   #change in temp.
mw=Q/(cp*dt);            #cooling water needed

# Result        
print "\n cooling water requiement is :%f kg/hr"%mw
    #Q=U*A*(dtlm)
dtlm=((t2-t3)-(t1-t4))/(math.log((t2-t3)/(t1-t4)));#log mean temp. difference
A=Q/(U*dtlm);                    #area of the crystallizer section
l=A/1.3;
print "\n length of crystallliser sections needed is :%f m"%l

#end

 cooling water requiement is :9554.149346 kg/hr

 length of crystallliser sections needed is :13.343753 m