{
"metadata": {
"name": "ch7",
"signature": "sha256:9c280dff7f4e1d5ae99e0fbadb3fa83466c67b9a415e20bbef76992fab6df294"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 : Crystallisation"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.1 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"c=.498; #solute content afetr crystallisation\n",
"W1=111.; #molecular weight of CaCl2\n",
"W2=219.; #molecular weight of CaCl2.6H2O\n",
"\n",
"# Calculation \n",
"M1=(108./W2)*100; #water present in 100kg of CaCl2.6H2O\n",
"M2=(W1/W2)*100; #CaCl2 present in 100kg of CaCl2.6H20\n",
"#t=M2+c*x; #total weight entering the solubility\n",
"#x+49.3; total water solubility used\n",
"#s*(x+49.3)/100 #total Cacl2 after solubility\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t= 32.112871 #roots((M2+c*x)-(x+49.3)*.819); \n",
"\n",
"# Result\n",
"print \"\\nthe weight of water in the quantity of solution needed :%f kg\"%t\n",
"\n",
"h=(c)*t; #weight of CaCl2 corresponding to weight water\n",
"tw=t+h; # total weight of the solution\n",
"print \"\\nthe total weight of the solution is :%f kg\"%tw\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"the weight of water in the quantity of solution needed :32.112871 kg\n",
"\n",
"the total weight of the solution is :48.105081 kg\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.2 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"#part(i)\n",
"w1=1000.; #weight of solution to be cooled\n",
"s1=104.1; #solubility at 50 degree per 100 kg of water\n",
"s2=78.0; #solubility at 10 degree per 100 kg of water\n",
"a2=45.; #percentage of sodium nitrate in the solution per 100kg of solution \n",
"\n",
"# Calculation \n",
"x1=s1/(100+s1)*100; #percentage of saturated solution at 50 degree\n",
"tw=(a2/(100-a2))/(x1/(100-x1)); #the percentage saturation\n",
"print \"\\nthe percentage saturation is :%f percent\"%(tw*100)\n",
"\n",
"#part(ii)\n",
"#let x be the weight of NaNO3 crystal formed after crystallisation\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t=21.000000 #roots((w1*a2/100)-(x+(w1-x)*s2/(100+s2)));\n",
"\n",
"# Result\n",
"print \"\\n the weight of NaNO3 crystal formed after crystallisation :%f kg\"%t\n",
"\n",
"#part(iii)\n",
"y=t/(a2*w1/100); #yield = weight of NaNO3 crystal formed/weight of NaNO3\n",
"print \"\\n the percentage yield is:%f percent\"%(y*100)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"the percentage saturation is :78.595756 percent\n",
"\n",
" the weight of NaNO3 crystal formed after crystallisation :21.000000 kg\n",
"\n",
" the percentage yield is:4.666667 percent\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.3"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"s1=19.75; #solubility at 70 degree per 100 gm of water\n",
"s2=16.5; #solubility at 50 degree per 100 gm of water\n",
"s3=12.97; #solubility at 30 degree per 100 gm of water\n",
"s4=9.22; #solubility at 10 degree per 100 gm of water\n",
"s5=7.34; #solubility at 0 degree per 100 gm of water\n",
"\n",
"# Calculation \n",
" #basis is 1000kg of saturated solution\n",
"w1=1000.*(s1/(s1+100)); #weight of K2SO4 in the original solution\n",
"w2=1000.-w1; #weight of water in kg\n",
"w3=w1*.5; #weight of K2SO4 in the solution\n",
"wp=w3/(w3+w2); #weight percent of K2SO4 in the solution after crystallistion\n",
"\n",
"# Result\n",
"print \"\\n for the corresponding temperature to :%f percent of K2SO4 is 15 degree (by linear interpolation between 10 to 30 degree) \"%(wp*100)\n",
"\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" for the corresponding temperature to :8.987486 percent of K2SO4 is 15 degree (by linear interpolation between 10 to 30 degree) \n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.4 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"#part(i)\n",
"a1=146.; #solubility at 70 degree\n",
"a2=121.; #solubility at 10 degree\n",
"t1=58.; # percentage of solute content\n",
"t2=40.66;\n",
"\n",
"# Calculation and Result\n",
"x1=a1/(100+a1) *100; #percentage of saturated solution at 50 degree\n",
"tw=(t1/42.)/(x1/t2); #the percentage saturation\n",
"print \"\\nthe percentage saturation is :%f percent\"%(tw*100)\n",
"\n",
"#part(ii)\n",
"p1=2000*.58; #weight of solute in 200kg of solution 2000*.58\n",
"#let x be the weight of crystal formed after crystallisation\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t=231.743929 #roots((1160)-(x+(1055.02-.547*x))); \n",
"print \"\\n the weight of NaNO3 crystal formed after crystallisation :%f kg\"%t\n",
"\n",
"#part(iii)\n",
"y=t/p1; #yield = weight of NaNO3 crystal formed/weight of NaNO3\n",
"print \"\\n the percentage yield is:%f percent\"%(y*100)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"the percentage saturation is :94.608102 percent\n",
"\n",
" the weight of NaNO3 crystal formed after crystallisation :231.743929 kg\n",
"\n",
" the percentage yield is:19.977925 percent\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.5"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"p1=.3; #percentage of the solute in the solution\n",
"w1=1000.; #weight of the solution taken\n",
"w2=142.; #molecular weight of Na2SO4.\n",
"\n",
"# Calculation \n",
"M1=(w2/(180+w2)); #solute (Na2SO4) present in the Na2CO3.10H2O solution\n",
"s1=40.8; #solubility of Na2SO4 at 30 degree per 100 gm of water\n",
"s2=9.0; #solubility of Na2SO4 at 10 degree per 100 gm of water\n",
"#percent weight of solute in Na2SO4.10H2O= 144/322\n",
"#let x be the weight of crystal formed \n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t= 576.477290 #roots((w1*40.8/140.8)-(.442*x+(w1-x)*(s2/(100+s2)))); \n",
"\n",
"# Result\n",
"print \"\\n the weight of crystal formed after crystallisation :%f kg\"%t\n",
"\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the weight of crystal formed after crystallisation :576.477290 kg\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.6 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"s1=12.5; #solubility of Na2CO3 at 10 degree per 100 gm of water\n",
"p1=.3; #percentage of the solute in the solution\n",
"w1=2000.; #weight of the solution taken\n",
"w2=106.; #molecular weight of Na2CO3.\n",
"\n",
"# Calculation \n",
"M1=(w2/(180+w2)); #solute (Na2CO3) present in the Na2CO3.10H2O solution\n",
"#let x be the quantity of Na2CO3.10H2O crystal formed\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t= 1455.688623 #roots(w1*p1-M1*x-(w1-x)*(s1/(100+s1))); \n",
"\n",
"# Result\n",
"print \"\\n the weight of quantity of Na2CO3.10H2O :%f kg\"%t\n",
"#in the book the ans is wrong, they have calculated 2000*0.3-2000*12.5/112.5 as =x(miscalculation)\n",
"\n",
"p=(286./106)*w1*p1; #weight of Na2CO3.10H2O crystal present in the original solution\n",
"y=t/p; #percentage yield \n",
"print \"\\n percentage yield :%f percent\"%(y*100)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the weight of quantity of Na2CO3.10H2O :1455.688623 kg\n",
"\n",
" percentage yield :89.920160 percent\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.7 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"s1=139.8; #solubility at 80 degree per 100 gm of water\n",
"s2=110.5; #solubility at 20 degree per 100 gm of water\n",
"w2=174.2; #molecular weight of K2CO3.10H2O\n",
"\n",
"# Calculation \n",
"M1=(138/w2)*100; #water present in 100kg of K2CO3.10H2O\n",
"#let x be the quantity of Na2CO3.10H2O\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t=108.634036 #roots(500*(139.8/239.8)-.7921*x-(500-x)*110.5/210.5); \n",
"\n",
"# Result\n",
"print \"\\n the weight of quantity of K2CO3.10H2O formed :%f kg\"%t\n",
"\n",
"p=(174./138)*500*(139.8/239.8); #weight of crystal present in the original solution\n",
"y=t/p; #percentage yield \n",
"print \"\\n percentage yield :%f percent\"%(y*100)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the weight of quantity of K2CO3.10H2O formed :108.634036 kg\n",
"\n",
" percentage yield :29.557504 percent\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.8 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"s1=20.51; #solubility at 10 degree per 100 gm of water\n",
"w2=277.85; #molecular weight of FeSO4.7H2O\n",
"\n",
"# Calculation and Result\n",
"#let x be the quantity of Na2CO3.10H2O\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t= 549.684973 #roots(900*.4-.5465*x-(900-x)*20.5/120.5); \n",
"print \"\\n the weight of quantity of FeSO4.7H2O formed :%f kg\"%t\n",
"\n",
"p=(277.85/151.85)*900*(0.4); #weight of crystal present in the original solution\n",
"y=t/p; #percentage yield \n",
"print \"\\n percentage yield :%f percent\"%(y*100)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the weight of quantity of FeSO4.7H2O formed :549.684973 kg\n",
"\n",
" percentage yield :83.447967 percent\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"#part(i)\n",
"a1=229.7; #solubility at 60 degree\n",
"a2=174.7; #solubility at 60 degree\n",
"t1=68; # percentage of sodium nitrate\n",
"t2=30.34;\n",
"\n",
"# Calculation and Result\n",
"x1=a1/329.7 *100; #percentage of saturated solution at 50 degree\n",
"tw=(t1/32.)/(x1/t2); #the percentage saturation\n",
"print \"\\nthe percentage saturation is :%f percent\"%(tw*100)\n",
"\n",
"#part(ii)\n",
"#let x be the weight of Cesium chloride crystal formed after crystallisation\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t=120.960000 #roots(1000*.68-(x+(1000-x)*174.7/274.7)); \n",
"print \"\\n the weight of CaCl2 crystal formed after crystallisation :%f kg\"%t\n",
"\n",
"#part(iii)\n",
"y=t/680.; #yield = weight of CaCl2 crystal formed/weight of CaCl2\n",
"print \"\\n the percentage yield of Cesium chloride is:%f percent\"%(y*100)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"the percentage saturation is :92.540632 percent\n",
"\n",
" the weight of CaCl2 crystal formed after crystallisation :120.960000 kg\n",
"\n",
" the percentage yield of Cesium chloride is:17.788235 percent\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.10"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"s1=38.8; #solubility at 30 degree per 100 gm of water\n",
"s2=12.5; #solubility at 10 degree per 100 gm of water\n",
"w2=296; #molecular weight of Na2CO3.10H2O\n",
"per=116./w2 *100; #percentage solute in Na2CO3.10H2O\n",
"\n",
"#let x be the quantity of Na2CO3.10H2O\n",
"w=200.; #original solotion weight\n",
"\n",
"# Calculation \n",
"m1=w*(s2/(s2+100)); #weight of Na2CO3.10H2O needed to dissolve Na2CO3 present in the original solotion \n",
"w3=w-m1; #weight of water \n",
"#w4=m1+per/100; weight of Na2CO3 after dissolution\n",
"x1=s1/(s1+100); #weight fraction of solute after dissolution \n",
"\n",
"# Result\n",
"print \"\\n the weight of quantity of Na2CO3.10H2O formed :%f kg\"%w3\n",
"\n",
"#for the total solution after dissolution\n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t= 300.485784 #roots((m1+per*x/100)-((m1+per*x/100)+(w3+.609*x))*x1); \n",
"print \"\\nweight of Na2CO3.10H2O needed to dissolve Na2CO3 present in the original solution %f kg\"%t\n",
"\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the weight of quantity of Na2CO3.10H2O formed :177.777778 kg\n",
"\n",
"weight of Na2CO3.10H2O needed to dissolve Na2CO3 present in the original solution 300.485784 kg\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.11"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"s1=35.; #percentage of solution\n",
"x1=6000.; #weight of Na2CO3 solution\n",
"s2=21.5; #solubility at 20 degree per 100 gm of water\n",
"w2=296.; #molecular weight of Na2CO3.10H2O\n",
"\n",
"# Calculation \n",
"per=116./w2 *100; #percentage solute in Na2CO3.10H2O\n",
"w1=s1*x1; #weight of solute\n",
"w3=x1*0.04; #weight of solution lost by vaporisation\n",
"#let x be the quantity of Na2CO3.10H2O formed \n",
"#making material balance \n",
"#x=poly([0],'x'); #calc. x the weight of crystal\n",
"t= 5049.122335 #roots(2100-(.391*x)-(6000-240-x)*(21.5/121.5)); \n",
"\n",
"# Result\n",
"print \"\\n the weight of Na2CO3.10H2O crystal formed after crystallisation :%f kg\"%t\n",
"\n",
"\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the weight of Na2CO3.10H2O crystal formed after crystallisation :5049.122335 kg\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"C=1000.; #crystal formed in kg\n",
"hf=26.002; #enthalpy of the feed at 80 degree in cal/g\n",
"hl=-1.33; #enthalpy of the saturated sol at 30 degree in cal/g\n",
"hc=-50.56; #enthalpy of crystal\n",
"xf=40./(100+40);\n",
"xm=30./(100+30);\n",
"xc=151.84/277.85; #151.84 is the weight of FeSO4\n",
" #component balance\n",
"\n",
"Hf=26.002; #enthalpy of the feed at 80 degree in cal/g\n",
"Hv=612.; #\n",
"Hm=-1.33; #enthalpy of the saturated sol at 30 degree in cal/g\n",
"Hc=-50.56; #enthalpy of crystal leaving the crystalliser\n",
"from numpy import matrix\n",
"from numpy import linalg\n",
"#solving these we gt \n",
"\n",
"# Calculation \n",
"a=matrix([[1,-1,-1],[.286,-.231,0],[26.002,1.33,-612]])\n",
"b=matrix([[10000],[5470],[-505600]])\n",
"x=linalg.inv(a)*b; #solving out the values using matrices \n",
"t1=x[0]; #3 solution of the eqn\n",
"t2=x[1];\n",
"t3=x[2];\n",
"\n",
"# Result\n",
"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)\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" the feed rate F= : 45556.688340 kg/hr \n",
" value of M= : 32723.865218 kg/hr\n",
" value of V=: 2832.823122 kg/hr\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.13 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"\n",
"import math\n",
"C=800.; #crystal formed in kg/hr\n",
"t2=49.; #temp. of the entering fed\n",
"t1=27.; #temp. of the product\n",
"t3=21.; #temp. of the leaving cooling water\n",
"t4=15.; #temp. of the enetring cooling water\n",
"U=175.; #overall heat transfer coefficient\n",
"F=140*151.85/277.85; #feed concentration \n",
"xf=F/240; #concentration in feed solution\n",
"P=74*151.85/277.85; #product concentration \n",
"xm=P/174; #concentration of FeSO4 in product solution\n",
"xc=151.85/277.85; #\n",
" \n",
"# Calculation \n",
"#solving these we get\n",
"F=800*.3141/0.0866; #feed conc.\n",
"M=F-C; #product concentration \n",
" #making energy balance\n",
" #heat to be removed by cooling water =heat to be removed from solution + heat of crystallization\n",
"cp=.7; #specific heat capacity\n",
"dt=(t2-t1); #change in temp.\n",
"dh=15.8; #heat of crystallization\n",
"Q=F*cp*dt+dh*C; #heat to be removed by cooling water\n",
"cp=1; #specific heat capacity of water\n",
"dt=(t3-t4); #change in temp.\n",
"mw=Q/(cp*dt); #cooling water needed\n",
"\n",
"# Result \n",
"print \"\\n cooling water requiement is :%f kg/hr\"%mw\n",
" #Q=U*A*(dtlm)\n",
"dtlm=((t2-t3)-(t1-t4))/(math.log((t2-t3)/(t1-t4)));#log mean temp. difference\n",
"A=Q/(U*dtlm); #area of the crystallizer section\n",
"l=A/1.3;\n",
"print \"\\n length of crystallliser sections needed is :%f m\"%l\n",
"\n",
"#end"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" cooling water requiement is :9554.149346 kg/hr\n",
"\n",
" length of crystallliser sections needed is :13.343753 m\n"
]
}
],
"prompt_number": 27
}
],
"metadata": {}
}
]
}