summaryrefslogtreecommitdiff
path: root/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb
diff options
context:
space:
mode:
Diffstat (limited to 'Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb')
-rwxr-xr-xElementary_Principles_of_Chemical_Processes/Chapter6.ipynb1137
1 files changed, 1137 insertions, 0 deletions
diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb
new file mode 100755
index 00000000..e1ddd7be
--- /dev/null
+++ b/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb
@@ -0,0 +1,1137 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:27df12dd9928a7234ef95007770cbb19da31332c34bb007953024d3646517da2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6: Multiphase Systems"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1-1, page no. 244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "T2=15.4+273.2 #K\n",
+ "T1=7.6+273.2 #K\n",
+ "P1=40.0 #mm of Hg\n",
+ "P2=60.0 #mm of Hg\n",
+ "T=42.2+273.2 #K\n",
+ "R=8.314 #J/mol.k\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "print(\"let deltaHv/R = S\")\n",
+ "S= - (T1*T2* math.log(P2/P1))/(T1-T2)\n",
+ "deltahv=S*R\n",
+ "print '%s %d' %(\" \\n Latent Heat of Vaporization (J/mol) =\",deltahv)\n",
+ "B=math.log(P1) + S/T1\n",
+ "print '%s %.3f' %(\"\\n B=\",B)\n",
+ "P=math.exp(-S/T + B)\n",
+ "print '%s %.3f %s %.3f' %(\"\\n P* at\",T,\" K is (mm Hg)= \",P)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "let deltaHv/R = S\n",
+ " \n",
+ " Latent Heat of Vaporization (J/mol) = 35023\n",
+ "\n",
+ " B= 18.691\n",
+ "\n",
+ " P* at 315.400 K is (mm Hg)= 207.400\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 1,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3-1, page no. 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "P=760.0 #mm of Hg\n",
+ "Pstar=289.0 #mm of Hg\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "y=Pstar/P\n",
+ "print '%s %.3f %s %.3f' %(\" \\n Molar composition of Water is\",y,\" and Air is\",1-y)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ " \n",
+ " Molar composition of Water is 0.380 and Air is 0.620\n"
+ ]
+ }
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3-2, page no. 251"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "T=100.0+273.2 #K\n",
+ "PT=5260.0 #mm of Hg\n",
+ "y=0.1 #by volume\n",
+ "basis= 100 #mol of feed gas\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "P=y*PT\n",
+ "if(P<760):\n",
+ " print(\"The Vapour is Super heated\")\n",
+ "elif(P==760): \n",
+ " print(\"The vapour is At Dew point\")\n",
+ "else:\n",
+ " print(\"The vapour is not Super heated\")\n",
+ "\n",
+ "print(\"From tables Tdp=90 C\")\n",
+ "print(\"Superheat = 100-90=10 C \")\n",
+ "print(\"Using Raoult law at the outlet\")\n",
+ "y1=355/PT\n",
+ "print '%s %.3f' %(\"y1=\",y1)\n",
+ "print(\"Balance on Dry Air\")\n",
+ "n2=basis*(1-y)/(1-y1)\n",
+ "print '%s %.3f' %(\"n2 (mol) = \",n2)\n",
+ "print(\"Total mole balance\")\n",
+ "n1=basis-n2\n",
+ "print '%s %.3f' %(\"n1 (mol) = \",n1)\n",
+ "print '%s %.3f' %(\" \\n Percentage condesation=\",n1*100/(y*basis))\n",
+ "Psaturation=760/y\n",
+ "print '%s %.3f %s' %(\"\\n Any increase in pressure above\", Psaturation,\" mm of Hg must cause condensation \")\n",
+ "print(\" \\n For the next part of the problem use the same code by modifying PT to be 8500 mm of Hg\")\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "The Vapour is Super heated\n",
+ "From tables Tdp=90 C\n",
+ "Superheat = 100-90=10 C \n",
+ "Using Raoult law at the outlet\n",
+ "y1= 0.067\n",
+ "Balance on Dry Air\n",
+ "n2 (mol) = 96.514\n",
+ "Total mole balance\n",
+ "n1 (mol) = 3.486\n",
+ " \n",
+ " Percentage condesation= 34.862\n",
+ "\n",
+ " Any increase in pressure above 7600.000 mm of Hg must cause condensation \n",
+ " \n",
+ " For the next part of the problem use the same code by modifying PT to be 8500 mm of Hg\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 1,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3-3, page no. 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "T=75.0 +273 #K\n",
+ "P75=289.0 #mm of Hg\n",
+ "hr=0.3\n",
+ "Porig=825.0 #mm of Hg\n",
+ "PorigBar=1.1 #bar\n",
+ "Vdot=1000.0 #M^3/h\n",
+ "R=0.0831\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print (\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "P=hr*P75\n",
+ "y=P/Porig\n",
+ "ndot=PorigBar*Vdot/(R*T)\n",
+ "ndotWater=ndot*y\n",
+ "print '%s %.3f' %(\" \\n Molar flowrate of Water (Kmol/h) = \",ndotWater)\n",
+ "ndotBDA=ndot*(1-y)\n",
+ "print '%s %.3f' %(\" \\n Molar flowrate of Dry Air (Kmol/h) = \",ndotBDA)\n",
+ "ndotO2=ndotBDA*0.21\n",
+ "print '%s %.3f' %(\" \\n Molar flowrate of Oxygen (Kmol/h) = \",ndotO2)\n",
+ "hm=P/(Porig-P)\n",
+ "ha=hm*18/29\n",
+ "hmdot=P75/(Porig-P75)\n",
+ "hp=100*hm/hmdot\n",
+ "print '%s %.3f' %(\" \\n Molal Humidity (mol water/mol BDA) = \",hm)\n",
+ "print '%s %.3f' %(\" \\n Absolute Humidity (kg water/kg BDA) = \",ha)\n",
+ "print '%s %.3f' %(\" \\n Percentage Humidity=\",hp)\n",
+ "print(\"\\n From table B.3, Tdp=48.7 C\")\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ " \n",
+ " Molar flowrate of Water (Kmol/h) = 3.997\n",
+ " \n",
+ " Molar flowrate of Dry Air (Kmol/h) = 34.040\n",
+ " \n",
+ " Molar flowrate of Oxygen (Kmol/h) = 7.148\n",
+ " \n",
+ " Molal Humidity (mol water/mol BDA) = 0.117\n",
+ " \n",
+ " Absolute Humidity (kg water/kg BDA) = 0.073\n",
+ " \n",
+ " Percentage Humidity= 21.780\n",
+ "\n",
+ " From table B.3, Tdp=48.7 C\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 2,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4-1, page no. 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "basis=100.0 #lb-mole/h\n",
+ "x=0.45\n",
+ "PH2O=31.6 #mm of Hg\n",
+ "PSO2=176.0 #mm of Hg\n",
+ "P=760.0 #mm of Hg\n",
+ "y=2.0\n",
+ "M1=64.0 \n",
+ "M2=18.0\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "yH2O=PH2O/P\n",
+ "ySO2=PSO2/P\n",
+ "yAir=1-yH2O-ySO2\n",
+ "print(\"Using Air balance, \")\n",
+ "nG2=(1-x)*basis/yAir\n",
+ "print '%s %.3f' %(\"nG2 (lbm/h) = \",nG2)\n",
+ "xSO2=y/102\n",
+ "xH2O=1-xSO2\n",
+ "print(\"Using SO2 balance, \")\n",
+ "nL2=(basis*x-nG2*ySO2)*M1/(xSO2)\n",
+ "print '%s %.3f' %(\"nL2 (lbm/h) = \",nL2)\n",
+ "print(\"Using H2O balance, \")\n",
+ "nL1=nG2*yH2O*M2 + nL2*xH2O\n",
+ "print '%s %.3f' %(\"nL1 (lbm H2O/h) = \",nL1)\n",
+ "SO2Absorbed=nL2*xSO2\n",
+ "SO2Fed=basis*x*M1\n",
+ "Fraction=SO2Absorbed/SO2Fed\n",
+ "print '%s %.3f' %(\" \\n Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = \",Fraction)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "Using Air balance, \n",
+ "nG2 (lbm/h) = 75.670\n",
+ "Using SO2 balance, \n",
+ "nL2 (lbm/h) = 89683.186\n",
+ "Using H2O balance, \n",
+ "nL1 (lbm H2O/h) = 87981.325\n",
+ " \n",
+ " Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = 0.611\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 3,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4-2, page no. 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "basis=100.0 #lb-mole/h\n",
+ "x=0.45\n",
+ "PH2O=31.6 #mm of Hg\n",
+ "PSO2=176.0 #mm of Hg\n",
+ "P=760.0 #mm of Hg\n",
+ "y=2.0\n",
+ "M1=64.0 \n",
+ "M2=18.0\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "yH2O=PH2O/P\n",
+ "ySO2=PSO2/P\n",
+ "yAir=1-yH2O-ySO2\n",
+ "print(\"Using Air balance, \")\n",
+ "nG2=(1-x)*basis/yAir\n",
+ "print '%s %.3f' %(\"nG2 (lbm/h) = \",nG2)\n",
+ "xSO2=y/102\n",
+ "xH2O=1-xSO2\n",
+ "print(\"Using SO2 balance, \")\n",
+ "nL2=(basis*x-nG2*ySO2)*M1/(xSO2)\n",
+ "print '%s %.3f' %(\"nL2 (lbm/h) = \",nL2)\n",
+ "print(\"Using H2O balance, \")\n",
+ "nL1=nG2*yH2O*M2 + nL2*xH2O\n",
+ "print '%s %.3f' %(\"nL1 (lbm H2O/h) = \",nL1)\n",
+ "SO2Absorbed=nL2*xSO2\n",
+ "SO2Fed=basis*x*M1\n",
+ "Fraction=SO2Absorbed/SO2Fed\n",
+ "print '%s %.3f' %(\" \\n Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = \",Fraction)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "Using Air balance, \n",
+ "nG2 (lbm/h) = 75.670\n",
+ "Using SO2 balance, \n",
+ "nL2 (lbm/h) = 89683.186\n",
+ "Using H2O balance, \n",
+ "nL1 (lbm H2O/h) = 87981.325\n",
+ " \n",
+ " Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = 0.611\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 4,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5-1, page no. 264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "basis=150.0 #kg feed\n",
+ "S100=0.905 #g AgNO3/g\n",
+ "S20=0.689 #g AgNO3/g\n",
+ "inputx=0.095 #kg water/kg\n",
+ "outputx=0.311 #kg water/kg\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "print(\"\\n Composition of Filter cake,\")\n",
+ "print(\"\\n m2=4m3\")\n",
+ "print(\"\\n Water balance around the crystallizer,\")\n",
+ "print '%s' %(\"\\n basis*inputx* kg H2O = outputx m1 + outputx m1 \\n\")\n",
+ "print(\"Mass balance around crystallizer, \\n\")\n",
+ "print '%d %s' %(basis,\"=m1+m2+m3\")\n",
+ "A=([[0, 1, -4],[outputx, 0, outputx],[1, 1, 1]])\n",
+ "b=([[0],[basis*inputx],[basis]])\n",
+ "C=numpy.dot(linalg.inv(A),b)\n",
+ "#Here we solved two linear equations simultaneously\n",
+ "m1=C[0,0]\n",
+ "print '%s %.2f' %(\" \\n m1 (Kg) = \",m1)\n",
+ "m2=C[1,0]\n",
+ "print '%s %d' %(\" \\n m2 (Kg) = \",m2)\n",
+ "m3=C[2,0]\n",
+ "print '%s %d' %(\" \\n m3 (Kg) = \",m3)\n",
+ "print(\"\\n Overall AgNO3 balance,\")\n",
+ "m5=(1-inputx)*basis - (1-outputx)*m1\n",
+ "print '%s %d' %(\"\\n m5 (kg AgNO3 crystals recovered) = \",m5)\n",
+ "percentage=m5*100/(basis*(1-inputx))\n",
+ "print '%s %.3f' %(\" \\n Percentage recovery=\",percentage)\n",
+ "print(\"\\n Overall mass balance\")\n",
+ "m4=basis-m1-m5\n",
+ "print '%s %d' %(\"\\n m4 (Kg water removed in the Dryer) = \",m4)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "\n",
+ " Composition of Filter cake,\n",
+ "\n",
+ " m2=4m3\n",
+ "\n",
+ " Water balance around the crystallizer,\n",
+ "\n",
+ " basis*inputx* kg H2O = outputx m1 + outputx m1 \n",
+ "\n",
+ "Mass balance around crystallizer, \n",
+ "\n",
+ "150 =m1+m2+m3\n",
+ " \n",
+ " m1 (Kg) = 19.77\n",
+ " \n",
+ " m2 (Kg) = 104\n",
+ " \n",
+ " m3 (Kg) = 26\n",
+ "\n",
+ " Overall AgNO3 balance,\n",
+ "\n",
+ " m5 (kg AgNO3 crystals recovered) = 122\n",
+ " \n",
+ " Percentage recovery= 89.963\n",
+ "\n",
+ " Overall mass balance\n",
+ "\n",
+ " m4 (Kg water removed in the Dryer) = 8\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 5,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5-2, page no. 266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "inputx=0.6\n",
+ "basis=100.0 #kg Feed\n",
+ "S=63.0 #Kg KNO3/100 Kg H2O\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "outputx=S/(S+100)\n",
+ "print '%s %.3f' %(\"x (Kg KNO3/Kg) = \",outputx)\n",
+ "print(\"Water balance\")\n",
+ "m1=basis*(1-inputx)/(1-outputx)\n",
+ "print '%s %.3f' %(\" \\n m1 (Kg) = \",m1)\n",
+ "print(\"Mass balance\")\n",
+ "m2=basis-m1\n",
+ "print '%s %.3f' %(\" \\n m2 (kg) = \",m2)\n",
+ "percentage=m2*100/(basis*inputx)\n",
+ "print '%s %.3f' %(\" \\n Percentage of KNO3 in the feed that crystallizes is \",percentage)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "x (Kg KNO3/Kg) = 0.387\n",
+ "Water balance\n",
+ " \n",
+ " m1 (Kg) = 65.200\n",
+ "Mass balance\n",
+ " \n",
+ " m2 (kg) = 34.800\n",
+ " \n",
+ " Percentage of KNO3 in the feed that crystallizes is 58.000\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 6,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5-3, page no. 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "basis=1.0 #Tonne Epsom salt produced/h\n",
+ "inputx=0.301 #Tonne MgSO4/tonne\n",
+ "outputx=0.232 #Tonne MgSO4/tonne\n",
+ "M=120.4\n",
+ "M1=246.4\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "print(\"Total mass balance\")\n",
+ "print(\"m1=1+m2\")\n",
+ "print(\"MgSO4 balance\")\n",
+ "print(\" \\n inputx m1 = bassis* M/ M1 + m2 outputx\")\n",
+ "A=([[1, -1],[inputx, -outputx]])\n",
+ "b=([[1],[basis*M/M1]])\n",
+ "C=numpy.dot(linalg.inv(A),b)\n",
+ "#Here we solved two linear equations simultaneously\n",
+ "m1=C[0,0]\n",
+ "m2=C[1,0]\n",
+ "print '%s %.3f' %(\" \\n m1 (Tonne/h) = \",m1)\n",
+ "print '%s %.3f' %(\" \\n m2 (Tonne/h) = \",m2)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "Total mass balance\n",
+ "m1=1+m2\n",
+ "MgSO4 balance\n",
+ " \n",
+ " inputx m1 = bassis* M/ M1 + m2 outputx\n",
+ " \n",
+ " m1 (Tonne/h) = 3.719\n",
+ " \n",
+ " m2 (Tonne/h) = 2.719\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 7,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5-4, page no. 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "m1=5.0 #g of solute\n",
+ "m2=100.0 #g of Water\n",
+ "P=1.0 #atm\n",
+ "Tf=100.421 #C\n",
+ "Ti=25.0 #C\n",
+ "R=8.314 #J/mol.K\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "x=(Tf-100.)*40656./(R*373.16*373.16)\n",
+ "y=m2/18.016\n",
+ "Ms=m1*(1-x)/(y*x)\n",
+ "print '%s %.3f' %(\" \\n Ms (g/mol) =\",Ms)\n",
+ "deltaTm=R*273.16*273.16*x/6009.5\n",
+ "Tms=0-deltaTm\n",
+ "print '%s %.3f' %(\" \\n Tms (C)=\",Tms)\n",
+ "Pstar=(1-x)*23.756\n",
+ "print '%s %.3f' %(\" \\n Solvent Vapour pressure (mm Hg) = \",Pstar)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ " \n",
+ " Ms (g/mol) = 60.028\n",
+ " \n",
+ " Tms (C)= -1.526\n",
+ " \n",
+ " Solvent Vapour pressure (mm Hg) = 23.405\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 8,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6-1, page no. 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "V1=200.0 #CC Acetone\n",
+ "x=0.1 #Wt acetone \n",
+ "V2=400.0 #CC chloroform\n",
+ "DA=0.792 #g/cc\n",
+ "DC=1.489 #g/cc\n",
+ "DW=1.0 #g/cc\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "Dbar=DA*DW/(x*DW + (1-x)*DA)\n",
+ "mass1=V1*Dbar\n",
+ "mass2=V2*DC\n",
+ "print(\"C balance\")\n",
+ "m4=mass2\n",
+ "print '%s %.3f' %(\" \\n m4=\",m4)\n",
+ "print(\"W balance\")\n",
+ "m2=(1-x)*mass1\n",
+ "print '%s %.3f' %(\" \\n m2=\",m2)\n",
+ "print(\"A balance\")\n",
+ "print(\"m1+m3=x * mass1\")\n",
+ "print(\"Distribution Cooefficient ,K=m3*(m1+m2)/m1*(m3+m4)\")\n",
+ "print(\"On solving, \")\n",
+ "m1=2.7\n",
+ "m3=16.8\n",
+ "percentage=m3*100/(x*mass1)\n",
+ "print '%s %.3f' %(\" \\n m1=\",m1)\n",
+ "print '%s %.3f' %(\" \\n m3=\",m3)\n",
+ "print '%s %.3f' %(\" \\n percentage of acetone transferred to chloroform=\",percentage)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "C balance\n",
+ " \n",
+ " m4= 595.600\n",
+ "W balance\n",
+ " \n",
+ " m2= 175.394\n",
+ "A balance\n",
+ "m1+m3=x * mass1\n",
+ "Distribution Cooefficient ,K=m3*(m1+m2)/m1*(m3+m4)\n",
+ "On solving, \n",
+ " \n",
+ " m1= 2.700\n",
+ " \n",
+ " m3= 16.800\n",
+ " \n",
+ " percentage of acetone transferred to chloroform= 86.206\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 9,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6-2, page no. 274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "basis=1000.0 #Kg of solution\n",
+ "inputxA=0.3 #Wt. fraction of acetone\n",
+ "outputxA1=0.05\n",
+ "outputxM1=0.02\n",
+ "outputxA2=0.1\n",
+ "outputxM2=0.87\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "print(\"Mass balance\")\n",
+ "print(\"ms + basis = mE + mR\")\n",
+ "print(\"acetone balance\")\n",
+ "print(\"inputxA *basis = outputxA2 *mE + outputxA1* mR\")\n",
+ "print(\"Water balance\")\n",
+ "print(\" \\n 1-inputxA *basis = 1-outputxA2-outputxM2 *mE + 1-outputxA1-outputxM1* mR\")\n",
+ "A=([[1, 1, -1],[outputxA2, outputxA1, 0],[1-outputxA2-outputxM2, 1-outputxA1-outputxM1, 0]])\n",
+ "b=([[basis],[inputxA*basis],[(1-inputxA)*basis]])\n",
+ "C=numpy.dot(linalg.inv(A),b)\n",
+ "#Here We solved three linear equations simultaneously\n",
+ "mE=C[0,0]\n",
+ "mR=C[1,0]\n",
+ "mS=C[2,0]\n",
+ "print '%s %.3f' %(\" \\n mE (Kg) = \",mE)\n",
+ "print '%s %.3f' %(\" \\n mR (Kg) = \",mR)\n",
+ "print '%s %.3f' %(\" \\n mS (Kg MIBK) = \",mS)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ "Mass balance\n",
+ "ms + basis = mE + mR\n",
+ "acetone balance\n",
+ "inputxA *basis = outputxA2 *mE + outputxA1* mR\n",
+ "Water balance\n",
+ " \n",
+ " 1-inputxA *basis = 1-outputxA2-outputxM2 *mE + 1-outputxA1-outputxM1* mR\n",
+ " \n",
+ " mE (Kg) = 2666.667\n",
+ " \n",
+ " mR (Kg) = 666.667\n",
+ " \n",
+ " mS (Kg MIBK) = 2333.333\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 10,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7-1, page no. 276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "V=50.0 #L\n",
+ "P=1.0 #atm\n",
+ "T=34.0+273.2 #K\n",
+ "y=0.3\n",
+ "xF=0.001\n",
+ "R=0.08206\n",
+ "Pstar=169.0 #mm of Hg\n",
+ "Pmm=760.0 #mm of Hg\n",
+ "\n",
+ "#Calculations and printing :\n",
+ "\n",
+ "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+ "n=P*V/(R*T)\n",
+ "print '%s %.3f' %(\" \\n No.of moles (mol) = \",n)\n",
+ "y0=y*Pstar/Pmm\n",
+ "print '%s %.3f' %(\" \\n Y0= (mol CCl4/mol) = \",y0)\n",
+ "Pfinal=xF*Pmm\n",
+ "b=0.096*Pfinal\n",
+ "Xstar=0.794*b/(1+b)\n",
+ "print '%s %.3f' %(\" \\n Mass of CCl4 adsorbed to Carbon at equilibrium (g CCl4 ads/g C) = \",Xstar)\n",
+ "Mads=(y0*n- xF*n)*154\n",
+ "print '%s %.3f' %(\" \\n Mass of CCl4 adsorbed (g) = \",Mads)\n",
+ "Mc=Mads/Xstar\n",
+ "print '%s %.3f' %(\" \\n Mass of carbon Required (g) = \",Mc)\n",
+ "raw_input('press enter key to exit')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+ " \n",
+ " No.of moles (mol) = 1.983\n",
+ " \n",
+ " Y0= (mol CCl4/mol) = 0.067\n",
+ " \n",
+ " Mass of CCl4 adsorbed to Carbon at equilibrium (g CCl4 ads/g C) = 0.054\n",
+ " \n",
+ " Mass of CCl4 adsorbed (g) = 20.071\n",
+ " \n",
+ " Mass of carbon Required (g) = 371.750\n"
+ ]
+ },
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "press enter key to exit\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "pyout",
+ "prompt_number": 11,
+ "text": [
+ "''"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-04-05 17:39:36 +0000