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diff --git a/Elements_Of_Mass_Transfer_Part_1/ch7.ipynb b/Elements_Of_Mass_Transfer_Part_1/ch7.ipynb new file mode 100644 index 00000000..0c7c641e --- /dev/null +++ b/Elements_Of_Mass_Transfer_Part_1/ch7.ipynb @@ -0,0 +1,738 @@ +{ + "metadata": { + "name": "" + }, + "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": [ + "# total weight of the solution\n", + "\n", + "#let x be the weight of water in the quantity of solution needed\n", + "# Variable Declaration \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": [ + "# percentage saturation and yield\n", + "\n", + "# Variable Declaration \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": [ + "# temperature to which solution should be cooled\n", + "\n", + "# Variable Declaration \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": [ + "# percentage saturation and yield\n", + "\n", + "# Variable Declaration \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": [ + "# the weight of Na2SO4 hydrate crystal\n", + "\n", + "# Variable Declaration \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": [ + "# percentage yield of Na2CO3 hydrated crysta\n", + "\n", + "# Variable Declaration \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": [ + "# percentage yield of K2CO3 hydrated crystal\n", + "\n", + "# Variable Declaration \n", + "\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": [ + "# percentage yield FeSO4 hydrate crystal\n", + "\n", + "# Variable Declaration \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": [ + "# percentage saturation and weight of Cesium chloride crystal\n", + "\n", + "# Variable Declaration \n", + "\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": [ + "# weight of Na2CO3hydrate needed to dissolve Na2CO3\n", + "\n", + "# Variable Declaration \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": [ + "# weight of Na2CO3 hydrated formed\n", + "\n", + "# Variable Declaration \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": [ + "# feed rate of FeSO4 hydrated crystal produced per hour\n", + "\n", + "# Variable Declaration \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", + "# F*xf = M*xm + C*xc ------eqn 1st\n", + "# F = M + 10000 + V ------eqn 2nd\n", + "# F*Hf = V*Hv + M*Hm +C*Hc-----eqn 3rd\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": [ + "# cooling water requirement\n", + "\n", + "# Variable Declaration \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", + " #mass balance F = M+C ----eqn 1st\n", + " #sloute balance F*xf = M*xm + C*xc ----eqn 2nd\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 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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