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+ "metadata": {
+  "name": "",
+  "signature": "sha256:cf9c0d91c794ab27af5e7da425a28b836daf2f988940f28c4a10f4b42bda95d5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Fundamentals of Material Balances"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2-1, page no. 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "input=50000 #ppl/yr\n",
+      "generation=22000 #ppl/yr\n",
+      "consumption=19000 #ppl/yr\n",
+      "output=75000 #ppl/yr\n",
+      "print (\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "#Calculations and printing:\n",
+      "accumulation = input+generation-output-consumption\n",
+      "print (\" We know that accumulation=input+generation-output-consumption\")\n",
+      "print '%s %d %s' %('Hence, Each year population decreases by', -accumulation, 'people')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        " We know that accumulation=input+generation-output-consumption\n",
+        "Hence, Each year population decreases by 22000 people\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2-2, page no. 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "inputBenzene=500 #kg/h\n",
+      "inputToluene=500 #kg/h\n",
+      "UpStreamBenzene=450 #kg/h\n",
+      "DownStreamToluene=475 #kg/h\n",
+      "#Calculations and printing:\n",
+      "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Using benzene balance,\")\n",
+      "m2dot=inputBenzene-UpStreamBenzene\n",
+      "print ' %s %d %s' %(' m2dot= ', m2dot, '  kg B/h')\n",
+      "print(\"Using Toluene balance\")\n",
+      "m1dot=inputToluene-DownStreamToluene\n",
+      "print ' %s %d %s' %(' m1dot= ', m1dot, '  kg T/h')\n",
+      "print(\"To check we can perform Overall mass balance\")\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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 benzene balance,\n",
+        "  m2dot=  50   kg B/h\n",
+        "Using Toluene balance\n",
+        "  m1dot=  25   kg T/h\n",
+        "To check we can perform Overall mass balance\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 2,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2-3, page no. 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "m1=200 #g\n",
+      "m2=150 #g\n",
+      "x1=0.4  #methanol/g\n",
+      "x2=0.7  #methanol/g\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "m=m1+m2\n",
+      "print '%s %d %s' %(' Total mass after mixing m=',m, 'g',)\n",
+      "x=(m1*x1 + m2*x2)/m\n",
+      "print'%s  %.2f %s  %.2f' %(' The composition of the methanol in the product is', x ,' and water is', 1-x)\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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 after mixing m= 350 g\n",
+        " The composition of the methanol in the product is  0.53  and water is  0.47\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 3,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2-4, page no. 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "rate=0.1 #kmol/min\n",
+      "x1=0.1 #mole fraction hexane vapour\n",
+      "vol=10 #m^3\n",
+      "d=0.659 #kg/L\n",
+      "M=86.2 #Kg/kmol\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "ndot=rate/(1-x1)\n",
+      "deltaN= -vol*d*1000 /M\n",
+      "tf=deltaN/(-0.1 * ndot)\n",
+      "print '%s  %.2f %s' %(' The time Required for the Total process= ',tf, 'min')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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 time Required for the Total process=   6880.51 min\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 5,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3-1, page no. 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "Vdot=20 #CC/min\n",
+      "x=0.015\n",
+      "MH2O=18.02 #g\n",
+      "DH2O=1 #g/CC\n",
+      "x1=0.2\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "n2=Vdot*DH2O/MH2O\n",
+      "print'%s  %.2f %s' %('n2=', n2, ' mol/min')\n",
+      "print(\"Using Water Balance,\")\n",
+      "n3=n2/x\n",
+      "print'%s  %.2f %s' %('n3=', n3, ' mol/min')\n",
+      "print(\"Using total mole balance,\")\n",
+      "n1=(n3-n2)/(1+x1)\n",
+      "print'%s  %.2f %s' %('n1=', n1, ' mol/min')\n",
+      "print(\"Using N2 balance,\")\n",
+      "y=1-x-0.79*n1/n3\n",
+      "print '%s  %.2f %s' %('y=',y,' mol O2/mol')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "n2=  1.11  mol/min\n",
+        "Using Water Balance,\n",
+        "n3=  73.99  mol/min\n",
+        "Using total mole balance,\n",
+        "n1=  60.73  mol/min\n",
+        "Using N2 balance,\n",
+        "y=  0.34  mol O2/mol\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 6,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3-2, page no. 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100 #mol\n",
+      "FinalBasis=1250.0 #lb-moles/h\n",
+      "#Calulations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "multiply=FinalBasis/basis\n",
+      "Feed=100*multiply\n",
+      "TopStream=50*multiply\n",
+      "BottomStream1=12.5*multiply\n",
+      "BottomStream2=37.5*multiply\n",
+      "print '%s  %.2f %s' %(' Final Basis=',Feed,' lb-moles /h')\n",
+      "print '%s  %.2f %s' %(' Final Top Stream Feed=',TopStream, 'lb-moles /h')\n",
+      "print '%s  %.2f %s' %(' Final Bottom Stream Feed 1 =', BottomStream1,'lb-moles A/h')\n",
+      "print '%s  %.2f %s' %(' Final Bottom Stream Feed 2 =' ,BottomStream2,' lb-moles B/h')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        " Final Basis=  1250.00  lb-moles /h\n",
+        " Final Top Stream Feed=  625.00 lb-moles /h\n",
+        " Final Bottom Stream Feed 1 =  156.25 lb-moles A/h\n",
+        " Final Bottom Stream Feed 2 =  468.75  lb-moles B/h\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 7,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3-3, page no. 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100 #kg\n",
+      "inputx=0.2\n",
+      "outputx=0.08\n",
+      "D=1 #kg/L\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Using NaOH balance\")\n",
+      "m2=inputx*basis/outputx\n",
+      "print '%s  %.2f %s' %('m2=',m2 ,'Kg NaOH')\n",
+      "print(\"Using Total mass balance\")\n",
+      "m1=m2-basis\n",
+      "print '%s  %.2f %s' %('m1=',m1, 'Kg Water')\n",
+      "V1=m1/D\n",
+      "print '%s  %.2f %s' %(' V1=',V1 ,'Litres')\n",
+      "Ratio1=V1/basis\n",
+      "Ratio2=m2/basis\n",
+      "print '%s  %.2f %s' %(' Ratio of lt water/Kg Feed =', Ratio1,' lt water/Kg Feed')\n",
+      "print '%s  %.2f %s' %(' Ratio of Kg product/Kg Feed =', Ratio2 ,'Kg product/Kg Feed')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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 NaOH balance\n",
+        "m2=  250.00 Kg NaOH\n",
+        "Using Total mass balance\n",
+        "m1=  150.00 Kg Water\n",
+        " V1=  150.00 Litres\n",
+        " Ratio of lt water/Kg Feed =  1.50  lt water/Kg Feed\n",
+        " Ratio of Kg product/Kg Feed =  2.50 Kg product/Kg Feed\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 8,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3-5, page no. 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "inputx=0.45\n",
+      "outputx=0.95\n",
+      "basis=2000 #L/h\n",
+      "outputBasis=100 #Kmol\n",
+      "M1=78.11\n",
+      "M2=92.13\n",
+      "D=0.872\n",
+      "z=0.08\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "mass1=outputBasis*M1*outputx\n",
+      "mass2=outputBasis*M2*(1-outputx)\n",
+      "mass=mass1+mass2\n",
+      "yB2=mass1/mass\n",
+      "m1=basis*D\n",
+      "print '%s  %.2f %s' %(' m1=',m1,' Kg/h')\n",
+      "mB3=z*inputx*m1\n",
+      "print '%s  %.2f %s' %(' mB3=',mB3,' Kg/h')\n",
+      "print(\"Using Benzene balance,\")\n",
+      "m2=(inputx*m1-mB3)/yB2\n",
+      "print '%s  %.2f %s' %(' m2=',m2, 'Kg/h')\n",
+      "print(\"Using Toluene balance,\")\n",
+      "mT3=(1-inputx)*m1-(1-yB2)*m2\n",
+      "print '%s  %.2f %s' %(' mT3=',mT3,' Kg/h')\n",
+      "m3=mB3+mT3\n",
+      "print '%s  %.2f %s' %(' m3=',m3,' Kg/h')\n",
+      "yB3=mB3/m3\n",
+      "print '%s  %.2f %s' %(' yB3=',yB3,' kg B/kg')\n",
+      "yT3=1-yB3\n",
+      "print '%s  %.2f %s' %(' yT3=',yT3, 'kg T/kg')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        " m1=  1744.00  Kg/h\n",
+        " mB3=  62.78  Kg/h\n",
+        "Using Benzene balance,\n",
+        " m2=  766.84 Kg/h\n",
+        "Using Toluene balance,\n",
+        " mT3=  914.38  Kg/h\n",
+        " m3=  977.16  Kg/h\n",
+        " yB3=  0.06  kg B/kg\n",
+        " yT3=  0.94 kg T/kg\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 10,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4-1, page no. 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "inputMass1=100 #Kg/h\n",
+      "inputMass2=30 #Kg/h\n",
+      "outputMass1=40 #Kg/h\n",
+      "outputMass2=30 #Kg/h\n",
+      "inputx1=0.5\n",
+      "inputx2=0.3\n",
+      "outputx1=0.9\n",
+      "outputx2=0.6\n",
+      "#Calculations and printing\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"using Overall Mass balance, \")\n",
+      "m3=inputMass1+inputMass2-outputMass1-outputMass2\n",
+      "print '%s %d %s' %('m3=',m3,' Kg/h')\n",
+      "print(\"using Overall balance on A, \")\n",
+      "x3=(inputMass1*inputx1+inputMass2*inputx2-outputMass1*outputx1-outputMass2*outputx2)/m3\n",
+      "print '%s  %.2f %s' %('x3=',x3,' Kg A/kg')\n",
+      "print(\"using Mass balance on Unit 1, \")\n",
+      "m1=inputMass1-outputMass1\n",
+      "print '%s %d %s' %('m1=',m1,' Kg/h')\n",
+      "print(\"using A balance on Unit 1, \")\n",
+      "x1=(inputMass1*inputx1-outputMass1*outputx1)/m1\n",
+      "print '%s  %.2f %s' %('x1=',x1,' Kg A/kg')\n",
+      "print(\"using Mass balance on mixing point, \")\n",
+      "m2=inputMass2+m1\n",
+      "print '%s %d %s' %('m2=',m2,'Kg/h')\n",
+      "print(\"using A balance on mixing point, \")\n",
+      "x2=(inputMass2*inputx2+m1*x1)/m2\n",
+      "print '%s  %.2f %s' %('x2=',x2 ,'Kg A/kg')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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 Overall Mass balance, \n",
+        "m3= 60  Kg/h\n",
+        "using Overall balance on A, \n",
+        "x3=  0.08  Kg A/kg\n",
+        "using Mass balance on Unit 1, \n",
+        "m1= 60  Kg/h\n",
+        "using A balance on Unit 1, \n",
+        "x1=  0.23  Kg A/kg\n",
+        "using Mass balance on mixing point, \n",
+        "m2= 90 Kg/h\n",
+        "using A balance on mixing point, \n",
+        "x2=  0.26 Kg A/kg\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 12,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4-2, page no. 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "from numpy import matrix, array, dot\n",
+      "from numpy.linalg import inv\n",
+      "import math\n",
+      "massin=100 #kg\n",
+      "M1=100 #kg\n",
+      "M2=75 #Kg\n",
+      "massout=43.1 #kg\n",
+      "inputx=0.5\n",
+      "outputxA=0.053\n",
+      "outputxM=0.016\n",
+      "m1xA=0.275\n",
+      "m1xM=0.725\n",
+      "m3xW=0.03\n",
+      "m3xA=0.09\n",
+      "m3xM=0.88\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Using Balances around two-extraxtor system,\")\n",
+      "print(\"Balance on total mass,\")\n",
+      "\n",
+      "A=array([[1,1],[m1xA,m3xA]])\n",
+      "b=array([[massin+M1+M2-massout],[massin*inputx - massout*outputxA]])\n",
+      "C=dot(inv(A),b)\n",
+      "m1=C[0,0]\n",
+      "m3=C[1,0]\n",
+      "print '%s  %.2f %s' %(' m1=',m1,' Kg')\n",
+      "print '%s  %.2f %s' %(' m3=',m3,' Kg')\n",
+      "print(\"Balance on M\")\n",
+      "xM1=(massin+M2-massout*outputxM-m3*m3xM)/m1\n",
+      "print '%s  %.2f %s' %(' xM1=',xM1,' kg MIBK/kg')\n",
+      "print(\"Balances around Extract mixing point, \")\n",
+      "print(\"Balance on A\")\n",
+      "mA4=m1*m1xA+m3*m3xA\n",
+      "print '%s  %.2f %s' %(' mA4=',mA4 ,'Kg Acetone')\n",
+      "print(\"Balance on M\")\n",
+      "mM4=m1*xM1+m3*m3xM\n",
+      "print '%s  %.2f %s' %(' mM4=',mM4,' Kg MIBK')\n",
+      "print(\"Balance on W\")\n",
+      "mW4=m1*(m1xM-xM1) + m3*m3xW\n",
+      "print '%s  %.2f %s' %(' mW4=',mW4 ,'Kg Water')\n",
+      "print(\"Balances around the First extractor\")\n",
+      "print(\"Balance on  A\")\n",
+      "mA2=massin*inputx-m1*m1xA\n",
+      "print '%s  %.2f %s' %(' mA2=',mA2,' Kg Acetone')\n",
+      "print(\"Balance on  M\")\n",
+      "mM2=massin-m1*xM1\n",
+      "print '%s  %.2f %s' %(' xM1=',xM1, 'Kg MIBK')\n",
+      "print(\"Balance on  W\")\n",
+      "mW2=massin*inputx- m1*(m1xM-xM1)\n",
+      "print '%s  %.2f %s' %(' mW2=',mW2,'Kg Water')\n",
+      "\n"
+     ],
+     "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 Balances around two-extraxtor system,\n",
+        "Balance on total mass,\n",
+        " m1=  145.11  Kg\n",
+        " m3=  86.79  Kg\n",
+        "Balance on M\n",
+        " xM1=  0.67  kg MIBK/kg\n",
+        "Balances around Extract mixing point, \n",
+        "Balance on A\n",
+        " mA4=  47.72 Kg Acetone\n",
+        "Balance on M\n",
+        " mM4=  174.31  Kg MIBK\n",
+        "Balance on W\n",
+        " mW4=  9.87 Kg Water\n",
+        "Balances around the First extractor\n",
+        "Balance on  A\n",
+        " mA2=  10.10  Kg Acetone\n",
+        "Balance on  M\n",
+        " xM1=  0.67 Kg MIBK\n",
+        "Balance on  W\n",
+        " mW2=  42.73 Kg Water\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5-1, page no. 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "x1=0.960\n",
+      "x2=0.977\n",
+      "x3=0.983\n",
+      "basis=100 #mol\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Overall dry Air balance,\")\n",
+      "n1=x3*basis/x1\n",
+      "print '%s  %.2f %s' %('n1=',n1,' mol Fresh feed')\n",
+      "print(\"Overall mole balance,\")\n",
+      "n3=n1-basis\n",
+      "print '%s  %.2f %s' %('n3=',n3,' mol Water condensed')\n",
+      "print(\"Mole balance on mixing point,\")\n",
+      "print(\"n1+n5=n2\")\n",
+      "print(\"Water balance on mixing point\")\n",
+      "print ' %.2f %s  %.2f %s  %.2f %s' %(1-x1,' n1+ ', 1-x3,' n5 =', 1-x2,'n2')\n",
+      "A=([[1 ,-1],[1-x2,-(1-x3)]])\n",
+      "b=([[n1],[(1-x1)*n1]])\n",
+      "C=numpy.dot(linalg.inv(A),b)\n",
+      "n2=C[0,0]\n",
+      "n5=C[1,0]\n",
+      "print '%s  %.2f %s' %(' \\n n2=',n2,' mol')\n",
+      "print '%s  %.2f %s' %(' \\n n5=',n5,' mol Recycled')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "Overall dry Air balance,\n",
+        "n1=  102.40  mol Fresh feed\n",
+        "Overall mole balance,\n",
+        "n3=  2.40  mol Water condensed\n",
+        "Mole balance on mixing point,\n",
+        "n1+n5=n2\n",
+        "Water balance on mixing point\n",
+        " 0.04  n1+   0.02  n5 =  0.02 n2\n",
+        " \n",
+        " n2=  392.52  mol\n",
+        " \n",
+        " n5=  290.12  mol Recycled\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 15,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5-2, page no. 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "feed=4500 #kg/h\n",
+      "feedx=0.333\n",
+      "m3x=0.494\n",
+      "m5x=0.364\n",
+      "x=0.95\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "m6x=m5x\n",
+      "print '%s  %.2f %s' %('Given, m4=',x,' (m4+m5)')\n",
+      "print(\"using Overall K2CrO4 balance, \")\n",
+      "print ' %.2f %s  %.2f %s  %.2f %s' %(feedx,' *', feed ,'= m4+', m6x,'m5')\n",
+      "A=([[1-x,-x],[1,m6x]])\n",
+      "b=([[0],[feedx*feed]])\n",
+      "C=numpy.dot(linalg.inv(A),b)\n",
+      "m4=C[0,0]\n",
+      "m5=C[1,0]\n",
+      "print '%s  %.2f %s' %(' \\n m4=',m4,' K2CrO4 crystals/h')\n",
+      "print '%s  %.2f %s' %(' \\n m5=',m5,' entrained solution/h')\n",
+      "print(\"Overall Total mass balance , \")\n",
+      "m2=feed-m4-m5\n",
+      "print '%s  %.2f %s' %('m2=',m2,' Kg H2O evaporated/h')\n",
+      "print(\"Mass balance around the crystallizer,\")\n",
+      "print(\"m3=m4+m5+m6\")\n",
+      "print(\"Water balance around the crystallizer, \")\n",
+      "print ' %.2f %s  %.2f %s  %.2f %s' %(1-m3x, 'm3=', 1-m5x,' m5 + ',1-m6x,' m6')\n",
+      "D=([[1,-1],[1,(-1+m6x)/(1-m3x)]])\n",
+      "e=([[m4+m5],[(1-m5x)*m5/(1-m3x)]])\n",
+      "F=numpy.dot(linalg.inv(D),e)\n",
+      "m3=F[0,0]\n",
+      "m6=F[1,0]\n",
+      "print '%s  %.2f %s' %(' \\n m3=',m3,' Kg/h fed to the crystallizer')\n",
+      "print '%s  %.2f %s' %(' \\n m6=',m6,' Kg/h')\n",
+      "ratio=m6/feed\n",
+      "print  '%s  %.2f %s' %(' \\n ratio=',ratio,' Kg recycle/ Kg fresh feed')\n",
+      "print(\"mass balance around Recycle-fresh feed mixing point, \")\n",
+      "m1=feed+m6\n",
+      "print '%s  %.2f %s' %('m1=',m1,' kg/h feed to the evaporator')\n",
+      "print(\"With out recycle, \")\n",
+      "print(\"m3=622 Kg/h\")\n",
+      "print(\"m5=2380 Kg/h\")\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "Given, m4=  0.95  (m4+m5)\n",
+        "using Overall K2CrO4 balance, \n",
+        " 0.33  *  4500.00 = m4+  0.36 m5\n",
+        " \n",
+        " m4=  1470.33  K2CrO4 crystals/h\n",
+        " \n",
+        " m5=  77.39  entrained solution/h\n",
+        "Overall Total mass balance , \n",
+        "m2=  2952.28  Kg H2O evaporated/h\n",
+        "Mass balance around the crystallizer,\n",
+        "m3=m4+m5+m6\n",
+        "Water balance around the crystallizer, \n",
+        " 0.51 m3=  0.64  m5 +   0.64  m6\n",
+        " \n",
+        " m3=  7193.31  Kg/h fed to the crystallizer\n",
+        " \n",
+        " m6=  5645.60  Kg/h\n",
+        " \n",
+        " ratio=  1.25  Kg recycle/ Kg fresh feed\n",
+        "mass balance around Recycle-fresh feed mixing point, \n",
+        "m1=  10145.60  kg/h feed to the evaporator\n",
+        "With out recycle, \n",
+        "m3=622 Kg/h\n",
+        "m5=2380 Kg/h\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 16,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6-1, page no. 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100 #mol\n",
+      "xP=0.1\n",
+      "xN=0.12\n",
+      "xA=0.78\n",
+      "x=0.3\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "nP=basis*xP\n",
+      "nN=basis*xN\n",
+      "nO2=basis*xA*0.21\n",
+      "nN2=basis-nP-nN-nO2\n",
+      "if nN/nP>1:\n",
+      "    print(\"NH3 is in excess\")\n",
+      "else:\n",
+      "    print(\"Propene is in excess\")\n",
+      "\n",
+      "if nO2/nP>1:\n",
+      "    print(\"O2 is in excess\")\n",
+      "else:\n",
+      "    print(\"propene is in excess\")\n",
+      "\n",
+      "nO2reacted=nP*1.5\n",
+      "nNreacted=nP*1\n",
+      "ExcessAmmonia=(nN-nNreacted)*100/nNreacted\n",
+      "ExcessO2=(nO2-nO2reacted)*100/nO2reacted\n",
+      "print '%s  %.2f' %(' \\n percentage excess Ammonia=',ExcessAmmonia)\n",
+      "print '%s  %.2f' %(' \\n percentage excess Oxygen=',ExcessO2)\n",
+      "nPout=(1-x)*nP\n",
+      "print '%s  %.2f %s' %(' \\n no.of moles of Propylene left=', nPout,' mol')\n",
+      "E=nP-nPout\n",
+      "nNout=nN-E\n",
+      "nO2out=nO2-1.5*E\n",
+      "nAc=E\n",
+      "nW=3*E\n",
+      "print'%s  %.2f %s' %(' \\n no.of moles of Ammonia left=', nNout,' mol')\n",
+      "print'%s  %.2f %s' %(' \\n no.of moles of oxygen left=' ,nO2out,' mol')\n",
+      "print'%s  %.2f %s' %(' \\n no.of moles of ACN formed= ',nAc ,'mol')\n",
+      "print'%s  %.2f %s' %(' \\n no.of moles of water formed=', nW ,'mol')\n",
+      "print'%s  %.2f %s' %(' \\n no.of moles of Nirtogen present= ',nN2 ,'mol')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "NH3 is in excess\n",
+        "O2 is in excess\n",
+        " \n",
+        " percentage excess Ammonia=  20.00\n",
+        " \n",
+        " percentage excess Oxygen=  9.20\n",
+        " \n",
+        " no.of moles of Propylene left=  7.00  mol\n",
+        " \n",
+        " no.of moles of Ammonia left=  9.00  mol\n",
+        " \n",
+        " no.of moles of oxygen left=  11.88  mol\n",
+        " \n",
+        " no.of moles of ACN formed=   3.00 mol\n",
+        " \n",
+        " no.of moles of water formed=  9.00 mol\n",
+        " \n",
+        " no.of moles of Nirtogen present=   61.62 mol\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 17,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6-2, page no. 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "print(\"Given, Reaction is CO(g)+ H2O(g) --> CO2(g)+ H2(g)\")\n",
+      "nCO=1.\n",
+      "nH2O=2.\n",
+      "#Calculations and printing:\n",
+      "nt=nCO+nH2O\n",
+      "print(\"\\n It is found that Ee=0.667\")\n",
+      "Ee=0.667\n",
+      "yCO=(nCO-Ee)/nt\n",
+      "yH2O=(nH2O-Ee)/nt\n",
+      "yCO2=Ee/nt\n",
+      "yH2=Ee/nt\n",
+      "print(\"\\n Ee is obtained by solving the quadratic yCO2*yH2/(yCO*yH2O)=1\")\n",
+      "print(\"\\n Limiting reactant is CO\")\n",
+      "print '%s %.3f' %(\"\\n yCO = \",yCO)\n",
+      "print '%s %.3f' %(\"\\n yCO2 = \",yCO2)\n",
+      "print '%s %.3f' %(\"\\n yH2O = \",yH2O)\n",
+      "print '%s %.3f' %(\"\\n yH2 = \",yH2)\n",
+      "nCOF=yCO*nt\n",
+      "fCO=(nCO-nCOF)/nCO\n",
+      "print '%s %.3f' %(\"\\n Fractional conversion of CO at equilibrium is \", fCO)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Given, Reaction is CO(g)+ H2O(g) --> CO2(g)+ H2(g)\n",
+        "\n",
+        " It is found that Ee=0.667\n",
+        "\n",
+        " Ee is obtained by solving the quadratic yCO2*yH2/(yCO*yH2O)=1\n",
+        "\n",
+        " Limiting reactant is CO\n",
+        "\n",
+        " yCO =  0.111\n",
+        "\n",
+        " yCO2 =  0.222\n",
+        "\n",
+        " yH2O =  0.444\n",
+        "\n",
+        " yH2 =  0.222\n",
+        "\n",
+        " Fractional conversion of CO at equilibrium is  0.667\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 18,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6-3, page no. 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100 #mol\n",
+      "x=0.850\n",
+      "conv1=0.501\n",
+      "conv2=0.471\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "n1=(1-conv1)*basis*x\n",
+      "n2=conv2*basis*x\n",
+      "E1=n2\n",
+      "E2=basis*x - E1-n1\n",
+      "n3=E1-E2\n",
+      "n4=2*E2\n",
+      "n5=basis*(1-x)\n",
+      "nt=n1+n2+n3+n4+n5\n",
+      "selectivity=n2/n4\n",
+      "print '%s  %.2f %s ' %('selectivity=',selectivity,' mol Ethene/mol methane')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "selectivity=  7.85  mol Ethene/mol methane \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 19,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7-2, page no. 135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100 #mol\n",
+      "x=0.850\n",
+      "conv1=0.501\n",
+      "conv2=0.471\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "n1=(1-conv1)*basis*x\n",
+      "n2=conv2*basis*x\n",
+      "E1=n2\n",
+      "E2=basis*x - E1-n1\n",
+      "n3=E1-E2\n",
+      "n4=2*E2\n",
+      "n5=basis*(1-x)\n",
+      "nt=n1+n2+n3+n4+n5\n",
+      "selectivity=n2/n4\n",
+      "print '%s  %.2f %s ' %('selectivity=',selectivity,' mol Ethene/mol methane')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "selectivity=  7.85  mol Ethene/mol methane \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 20,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7-3, page no. 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "x0=0.996\n",
+      "basis=100 #mol combined feed to the reactor\n",
+      "inputxH2=0.7\n",
+      "single_pass=0.6\n",
+      "inputxCO2=0.28\n",
+      "molI=2\n",
+      "Ix=0.004\n",
+      "final=155 #kmol/h\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Reactor analysis, \")\n",
+      "n2=(1-single_pass)*basis*inputxH2\n",
+      "print '%s  %.2f %s' %('n2=',n2,' mol H2')\n",
+      "print(\"H2 balance\")\n",
+      "consH2=basis*inputxH2-n2\n",
+      "print '%s  %.2f %s' %('H2 moles consumed=',consH2,' mol H2')\n",
+      "print(\"CO2 balance\")\n",
+      "n1=basis*inputxCO2-consH2/3\n",
+      "print '%s  %.2f %s' %('n1=',n1,'mol CO2')\n",
+      "print(\"Methanol balance\")\n",
+      "n3=consH2/3\n",
+      "print '%s  %.2f %s' %('n3=',n3,'mol Methanol')\n",
+      "print(\"H2O balance\")\n",
+      "n4=consH2/3\n",
+      "print '%s  %.2f %s' %('n4=',n4,' mol H2O')\n",
+      "print(\"condenser analysis\")\n",
+      "print(\"Total mole balance\")\n",
+      "n5=n1+n2+molI\n",
+      "print '%s  %.2f %s' %('n5=',n5 ,'mol')\n",
+      "print(\"CO2 balance\")\n",
+      "x5C=n1/n5\n",
+      "print '%s  %.2f %s' %('x5C=',x5C,' mol CO2/mol')\n",
+      "print(\"H2 balance\")\n",
+      "x5H=n2/n5\n",
+      "print '%s  %.2f %s' %('x5H=', x5H,'mol CO2/mol')\n",
+      "x1=1-x5C-x5H\n",
+      "print '%s  %.2f %s' %('x1=',x1,' mol I/mol')\n",
+      "print(\"Fresh Feed-Recycle mixing point analysis\")\n",
+      "print(\"Total mole balance\")\n",
+      "print '%s  %.2f' %('n0+nr=',basis)\n",
+      "print(\"I balance\")\n",
+      "print '%s  %.2f %s  %.2f %s  %.2f' %('n0', Ix,' + nr ',x1,' = ',molI)\n",
+      "A=([[1, 1],[Ix, x1]])\n",
+      "b=([[basis],[molI]])\n",
+      "C=numpy.dot(linalg.inv(A),b) #Here We solve two linear equations simultaneously\n",
+      "n0=C[0,0]\n",
+      "nr=C[1,0]\n",
+      "print '%s  %.2f %s' %(' \\n n0=',n0,' mol fresh feed')\n",
+      "print '%s  %.2f %s' %(' \\n nr=', nr ,'mol recycle')\n",
+      "x0C=(basis*inputxCO2-nr*x5C)/n0\n",
+      "print '%s  %.2f %s' %(' \\n x0C=',x0C,' mol CO2/mol')\n",
+      "x0H=1-x0C-Ix\n",
+      "print '%s  %.2f %s' %(' \\n x0H=',x0H ,'mol H2/mol')\n",
+      "print(\"Recycle-Purge splitting Analysis\")\n",
+      "print(\"Total mole balance\")\n",
+      "np=n5-nr\n",
+      "print '%s  %.2f %s' %('np=',np,' mol purge')\n",
+      "print(\"Flow chart scaling\")\n",
+      "Factor=final/n3\n",
+      "print '%s  %.2f %s' %('Factor for scaling=',Factor,' Kmol/h/mol')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "Reactor analysis, \n",
+        "n2=  28.00  mol H2\n",
+        "H2 balance\n",
+        "H2 moles consumed=  42.00  mol H2\n",
+        "CO2 balance\n",
+        "n1=  14.00 mol CO2\n",
+        "Methanol balance\n",
+        "n3=  14.00 mol Methanol\n",
+        "H2O balance\n",
+        "n4=  14.00  mol H2O\n",
+        "condenser analysis\n",
+        "Total mole balance\n",
+        "n5=  44.00 mol\n",
+        "CO2 balance\n",
+        "x5C=  0.32  mol CO2/mol\n",
+        "H2 balance\n",
+        "x5H=  0.64 mol CO2/mol\n",
+        "x1=  0.05  mol I/mol\n",
+        "Fresh Feed-Recycle mixing point analysis\n",
+        "Total mole balance\n",
+        "n0+nr=  100.00\n",
+        "I balance\n",
+        "n0  0.00  + nr   0.05  =   2.00\n",
+        " \n",
+        " n0=  61.40  mol fresh feed\n",
+        " \n",
+        " nr=  38.60 mol recycle\n",
+        " \n",
+        " x0C=  0.26  mol CO2/mol\n",
+        " \n",
+        " x0H=  0.74 mol H2/mol\n",
+        "Recycle-Purge splitting Analysis\n",
+        "Total mole balance\n",
+        "np=  5.40  mol purge\n",
+        "Flow chart scaling\n",
+        "Factor for scaling=  11.07  Kmol/h/mol\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 21,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8-1, page no. 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "xN2wet=0.6\n",
+      "xCO2wet=0.15\n",
+      "xO2wet=0.1\n",
+      "xH2O=0.15\n",
+      "basis=100 #mol Wet gas \n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"part 1\")\n",
+      "wet=xN2wet+xCO2wet+xO2wet\n",
+      "xN2dry=xN2wet/wet\n",
+      "xCO2dry=xCO2wet/wet\n",
+      "xO2dry=xO2wet/wet\n",
+      "print '%s %f %s' %(' \\n xN2 dry =', xN2dry,' mol N2/mol dry gas')\n",
+      "print '%s %f %s' %(' \\n xO2 dry =',xO2dry,' mol O2/mol dry gas')\n",
+      "print '%s %f %s' %(' \\n xCO2 dry =', xCO2dry ,'mol CO2/mol dry gas')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "part 1\n",
+        " \n",
+        " xN2 dry = 0.705882  mol N2/mol dry gas\n",
+        " \n",
+        " xO2 dry = 0.117647  mol O2/mol dry gas\n",
+        " \n",
+        " xCO2 dry = 0.176471 mol CO2/mol dry gas\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 22,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8-2, page no. 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basisButane=100. #mol/h butane\n",
+      "basisAir=5000. #mol/h\n",
+      "#calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "nO2Theoretical=basisButane*6.5\n",
+      "nAirTheoretical=nO2Theoretical*4.76\n",
+      "percent=(basisAir-nAirTheoretical)*100/nAirTheoretical\n",
+      "print '%s  %.2f' %(' \\n percent excess air=',percent)\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        " percent excess air=  61.60\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 23,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8-3, page no. 147"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100. #mol ethane feed\n",
+      "E1=0.9\n",
+      "E2=0.25\n",
+      "excess=0.5\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\" 50percent excess air\")\n",
+      "nO2theoretical=basis*3.5\n",
+      "n0=nO2theoretical*(1+excess)/0.21\n",
+      "print '%s  %.4f %s' %(' n0=',n0,' mol air fed')\n",
+      "print(\"90% ethane conversion\")\n",
+      "n1=(1-E1)*basis\n",
+      "print '%s  %.4f' %('No. of moles of ethane unreacted=',n1)\n",
+      "print(\"25% conversion to CO\")\n",
+      "n4=E2*(basis-n1)*2\n",
+      "print '%s  %.4f %s' %('n4=', n4 ,'mol CO')\n",
+      "print(\"nitrogen balance\")\n",
+      "n3=0.79*n0\n",
+      "print '%s  %.4f %s' %('n3=', n3,' mol N2')\n",
+      "print(\"Atomic carbon balance\")\n",
+      "n5=2*basis-2*n1-n4\n",
+      "print '%s  %.4f %s' %('n5=', n5 ,'mol CO2')\n",
+      "print(\"Atomic hydrogen balance\")\n",
+      "n6=(basis*6-n1*6)/2\n",
+      "print '%s  %.4f %s' %('n6=', n6,' mol H2O')\n",
+      "print(\"Atomic oxygen balance\")\n",
+      "n2=(nO2theoretical*1.5*2-n4-n5*2-n6)/2\n",
+      "print '%s  %.4f %s' %('n2=', n2 ,'mol O2')\n",
+      "dry=n1+n2+n3+n4+n5\n",
+      "wet=dry+n6\n",
+      "y1=n1/dry\n",
+      "print '%s  %.4f %s' %('\\n y1=', y1,' mol C2H6/mol')\n",
+      "y2=n2/dry\n",
+      "print '%s  %.4f %s' %('\\n y2=', y2 ,'mol O2/mol')\n",
+      "y3=n3/dry\n",
+      "print '%s  %.4f %s' %('\\n y3=', y3,' mol N2/mol')\n",
+      "y4=n4/dry\n",
+      "print '%s  %.4f %s' %('\\n y4=', y4 ,'mol CO/mol')\n",
+      "y5=n5/dry\n",
+      "print '%s  %.4f %s' %('\\n y5=',y5,' mol CO2/mol')\n",
+      "ratio=n6/dry\n",
+      "print '%s  %.4f %s' %(' \\n ratio=',ratio,' mol H2O/mol dry stack gas')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        " 50percent excess air\n",
+        " n0=  2500.0000  mol air fed\n",
+        "90% ethane conversion\n",
+        "No. of moles of ethane unreacted=  10.0000\n",
+        "25% conversion to CO\n",
+        "n4=  45.0000 mol CO\n",
+        "nitrogen balance\n",
+        "n3=  1975.0000  mol N2\n",
+        "Atomic carbon balance\n",
+        "n5=  135.0000 mol CO2\n",
+        "Atomic hydrogen balance\n",
+        "n6=  270.0000  mol H2O\n",
+        "Atomic oxygen balance\n",
+        "n2=  232.5000 mol O2\n",
+        "\n",
+        " y1=  0.0042  mol C2H6/mol\n",
+        "\n",
+        " y2=  0.0970 mol O2/mol\n",
+        "\n",
+        " y3=  0.8238  mol N2/mol\n",
+        "\n",
+        " y4=  0.0188 mol CO/mol\n",
+        "\n",
+        " y5=  0.0563  mol CO2/mol\n",
+        " \n",
+        " ratio=  0.1126  mol H2O/mol dry stack gas\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 24,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8-4, page no. 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "basis=100 #mol pf product gas\n",
+      "xCO=0.015\n",
+      "xCO2=0.060\n",
+      "xO2=0.082\n",
+      "xN2=0.843\n",
+      "#Calculations and printing:\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"N2 balance\")\n",
+      "na=basis*xN2/0.79\n",
+      "print '%s  %.2f %s' %('na=',na,' mol air')\n",
+      "print(\"Atomic C balance\")\n",
+      "nc=basis*xCO + basis*xCO2\n",
+      "print '%s  %.2f %s' %('nc=',nc ,'mol C')\n",
+      "print(\"Atomic O balance\")\n",
+      "nw=0.21*na*2-basis*(xCO + xCO2*2 + xO2*2)\n",
+      "print '%s  %.2f %s' %('nw=',nw,' mol oxygen')\n",
+      "print(\"Atomic H2 balance\")\n",
+      "nh=nw*2\n",
+      "print '%s  %.2f %s' %('nh=',nh ,'mol H2')\n",
+      "ratio=nh/nc\n",
+      "print '%s  %.2f %s' %('\\n C/H ratio in fuel=',ratio,' mol H/mol C')\n",
+      "print(\"percent excess air\")\n",
+      "nO2theoretical=nc + nh/4\n",
+      "print '%s  %.2f %s' %('nO2 theoretical=',nO2theoretical ,'mol O2')\n",
+      "nO2fed=0.21*na\n",
+      "print '%s  %.2f %s' %(' \\n nO2fed=',nO2fed,' mol O2')\n",
+      "percent=(nO2fed-nO2theoretical)*100/nO2theoretical\n",
+      "print '%s  %.2f %s' %('\\n percentage excess air=',percent ,'excess air')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "N2 balance\n",
+        "na=  106.71  mol air\n",
+        "Atomic C balance\n",
+        "nc=  7.50 mol C\n",
+        "Atomic O balance\n",
+        "nw=  14.92  mol oxygen\n",
+        "Atomic H2 balance\n",
+        "nh=  29.84 mol H2\n",
+        "\n",
+        " C/H ratio in fuel=  3.98  mol H/mol C\n",
+        "percent excess air\n",
+        "nO2 theoretical=  14.96 mol O2\n",
+        " \n",
+        " nO2fed=  22.41  mol O2\n",
+        "\n",
+        " percentage excess air=  49.80 excess air\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Press the Enter key to quit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 25,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9-1, page no. 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initilization of variables\n",
+      "feed=59.6 #mol/s\n",
+      "x=0.2\n",
+      "TopFlow1=48.7 #mol/s\n",
+      "outputx1=0.021\n",
+      "BottomFlow1=10.9\n",
+      "TopFlow2=48.3\n",
+      "outputx2=0.063\n",
+      "BottomFlow2=6.4\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Design\")\n",
+      "MEKin1=feed*x\n",
+      "MEKout1=TopFlow1*outputx1 + BottomFlow1\n",
+      "closure1=MEKout1*100/MEKin1\n",
+      "print '%s %d %s' %('closure1=',closure1,' percent')\n",
+      "print(\"Experiment\")\n",
+      "MEKin2=feed*x\n",
+      "MEKout2=TopFlow2*outputx2 + BottomFlow2\n",
+      "closure2=MEKout2*100/MEKin2\n",
+      "print '%s %d %s' %('closure2=',closure2, 'percent')\n",
+      "raw_input(\"Press the Enter key to quit\")"
+     ],
+     "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",
+        "Design\n",
+        "closure1= 100  percent\n",
+        "Experiment\n",
+        "closure2= 79 percent\n"
+       ]
+      }
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file