path: root/TRANSPORT_PROCESSES_AND_UNIT_OPERATIONS/GeankoplisChapter01.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:49c580cddfa6fee61cb89005e8162fda0d32dc5e946341e173242a32280cb578"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 1: Introduction to Engineering Principles and Units"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.3-1, Page number 6"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Finding mole mass or weight fraction of solution \n",
      "\n",
      "#Variable declaration\n",
      "mNaOH = 50.         #Mass of NaOH in solution in gram\n",
      "mH2O = 50.          #Mass of H2O in solution in gram\n",
      "MWNaOH = 40         #Molecular weight of NaOH\n",
      "MWH2O = 18.02       #Molecular weight of H2O\n",
      "masstolbm = 1/453.6 #Conversion factor for Mass to Lb mass\n",
      "\n",
      "#Calculation\n",
      "mT = mNaOH + mH2O       #Total mass of solution\n",
      "mfNaOH = mNaOH/mT       #mfNaOH mass fraction of NaOH  \n",
      "mfH2O = mH2O/mT         #mfNaOH mass fraction of H2O\n",
      "nNaOH = mNaOH/MWNaOH    #Number of moles of NaOH\n",
      "nH2O = mH2O/MWH2O       #Number of  moles of H2O\n",
      "nT = nNaOH + nH2O       #Total number of moles in solution\n",
      "xNaOH = nNaOH/nT        #xNaOH mole fraction of NaOH\n",
      "xH2O = nH2O/nT          #xH2O mole fraction of H2O\n",
      "lbmNaOH = mNaOH*masstolbm  \n",
      "lbmH2O = mH2O*masstolbm\n",
      "\n",
      "#Result\n",
      "print \"Mass fraction of NaOH\", mfNaOH\n",
      "print \"Mass fraction of H2O\", mfH2O \n",
      "print \"Mole fraction of NaOH\", round(xNaOH,3) \n",
      "print \"Mole fraction of H2O\", round(xH2O,3)\n",
      "print \"Mass of NaOH in Lb mass\",round(lbmNaOH,4), \"lbm\"\n",
      "print \"Mass of H2O in Lb mass\", round(lbmH2O,4), \"lbm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass fraction of NaOH 0.5\n",
        "Mass fraction of H2O 0.5\n",
        "Mole fraction of NaOH 0.311\n",
        "Mole fraction of H2O 0.689\n",
        "Mass of NaOH in Lb mass 0.1102 lbm\n",
        "Mass of H2O in Lb mass 0.1102 lbm\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.4-1, Page number 8"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the value of gas law constant\n",
      "\n",
      "#Variable declaration \n",
      "P = 14.7        #Pressure (psia)\n",
      "V = 359.        #Volume of gas(ft3)\n",
      "Tf = 32         #Temperature in (degF)\n",
      "n = 1.          #Number of moles of Ideal Gas (lbmol)\n",
      "\n",
      "#Calculation\n",
      "Tr = 460 + Tf    #Deg F to deg Rankine conversion\n",
      "R = P*V/(n*Tr)   #R Gas constant in ft3.psia/(lbmol.degR)\n",
      "\n",
      "#Result\n",
      "print 'Gas law constant in lbmole, ft3, \u00b0R, and psia units %5.2f ft3.psia/(lbmol.degR)'%(R)\n",
      "\n",
      "#Variable declarationSI units\n",
      "P = 101325        #Pressure (Pa)\n",
      "V = 22.414        #Volume of gas(m3)\n",
      "T = 0             #Temperature in (deg C)\n",
      "n = 1.            #Number of moles of Ideal Gas (kgmol)\n",
      "\n",
      "#Calculation\n",
      "TK = 273.15 + T     #Deg C to Kelvin conversion\n",
      "R = P*V/(n*TK)      #R Gas constant in m3.pa/(kgmol.K)\n",
      "\n",
      "#Result\n",
      "print 'Gas law constant in kgmole, m3, K, and pa units %5.2f m3.pa/(kgmol.K)'%(R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gas law constant in lbmole, ft3, \u00b0R, and psia units 10.73 ft3.psia/(lbmol.degR)\n",
        "Gas law constant in kgmole, m3, K, and pa units 8314.47 m3.pa/(kgmol.K)\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.4-2, Page number 9"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Composition of Gas\n",
      "\n",
      "#Variable declaration\n",
      "pCO2 = 75.         #Partial pressure of CO2 (mmHg)\n",
      "pCO = 50.          #Partial pressure of CO (mmHg)\n",
      "pN2 = 595.         #Partial pressure of N2 (mmHg) \n",
      "pO2 = 26.          #Partial pressure of O2 (mmHg)\n",
      "\n",
      "#Calculation\n",
      "#Total pressure by Dalton's law\n",
      "P = pCO2 + pCO + pN2 + pO2\n",
      "xCO2 = pCO2/P\n",
      "xCO = pCO/P\n",
      "xN2 = pN2/P\n",
      "xO2 = pO2/P\n",
      "\n",
      "\n",
      "#Result\n",
      "print \"Total Pressure P\",P, \"mmHg\"\n",
      "print \"mole fraction of CO2\", round(xCO2,3)\n",
      "print \"mole fraction of CO\", round(xCO,3)\n",
      "print \"mole fraction of N2\", round(xN2,3)\n",
      "print \"mole fraction of O2\", round(xO2,3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total Pressure P 746.0 mmHg\n",
        "mole fraction of CO2 0.101\n",
        "mole fraction of CO 0.067\n",
        "mole fraction of N2 0.798\n",
        "mole fraction of O2 0.035\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.5-1, Page number 10"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Concentration of Orange Juice using evaporator\n",
      "\n",
      "#Variable declaration\n",
      "F = 1000.         #Feed rate of orange juice to evaporator (kg/hr)\n",
      "MpSi = 7.08       #Mass percent of solids in incomming juice\n",
      "MpSo = 58.        #Mass percent of solids in concentrated juice\n",
      "xW = 0.0\n",
      "#Calculation\n",
      "    #Overall Mass Balance\"\n",
      "    #$F = W + C$\"\n",
      "    #Component Mass Balance on solids\"\n",
      "    #$F.xF = W.xW + C.xC$\n",
      "\n",
      "C = F*MpSi/(100)/(MpSo/100)\n",
      "#Using overall Balance equation\n",
      "W = 1000-C\n",
      "\n",
      "#Results\n",
      "print \"Flow rate of Concentrated juice stream\", round(C,1) ,\"kg/hr\"\n",
      "print \"Flow rate of evaporated water stream\", round(W,1) ,\"kg/hr\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Flow rate of Concentrated juice stream 122.1 kg/hr\n",
        "Flow rate of evaporated water stream 877.9 kg/hr\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.5-2, Page number 11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Crystallization of KNO3 and Recycle\n",
      "\n",
      "#Variable declaration\n",
      "\n",
      "F = 1000. #F: Feed to the process, kg/hr\n",
      "xW = 0.   #xW: Weight % of KNO3 in water vaopor stream\n",
      "xF = 20.  #xF: Weight % of KNO3 in Feed\n",
      "xP = 96.  #xP: Weight % of KNO3 in product stream\n",
      "xR = 37.5 #xR: Weight % of KNO3 in Recycle\n",
      "xS = 50.  #xS: Weight % of KNO3 in stream from evaporator to crystallizer\n",
      "#Calculation\n",
      "    #Overall material Balance\n",
      "    # F = W + P\n",
      "    # Component material balance on KNO3\n",
      "    # F.xF = W.xW + P.xP\n",
      "    # Solving for P\n",
      "P = (F*xF/100)/(xP/100)    #P: Product rate from process, kg/hr\n",
      "    #Balance over crystallizer\n",
      "    # S = R + P  --> P = S - R\n",
      "    #Component balance over crystallizer\n",
      "    # S.xS = R.xR + P.xP  --> (R + P).xS = R.xR + P.xP --> ((R + P).xS - P.xP )/xR = \n",
      "R = P*(xP-xS)/(xS-xR)        #R: REcycle rate, kg/hr\n",
      "S = R + P                    #S: Concentrate rate from evaporator to crystallizer, kg/hr\n",
      "\n",
      "#Result\n",
      "print \"Product rate from the process is\",round(P,1), \"kg/hr\"\n",
      "print \"Recycle rate from the process is\",round(R,1), \"kg/hr\"\n",
      "print \"Stream leaving evaporator \",round(S,1), \"kg/hr\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Product rate from the process is 208.3 kg/hr\n",
        "Recycle rate from the process is 766.7 kg/hr\n",
        "Stream leaving evaporator  975.0 kg/hr\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.5-3, Page number 12"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Combustion of fuel gas\n",
      "\n",
      "#Variable declaration\n",
      "    #Basis 100 kg mol of feed,  moles = mole%x100/100\n",
      "H2=3.1        #Moles of H2 in feed\n",
      "CO=27.2       #moles of CO in feed\n",
      "CO2=5.6       #Moles of CO2 in feed\n",
      "O2=0.5        #Moles of O2 in feed\n",
      "N2=63.6       #Moles of N2 in feed\n",
      "    #98% CO reacts to form CO2\n",
      "\n",
      "#Calculation\n",
      "    #CO + (0.5)O2 ----> CO2\n",
      "    #H2 +(0.5)O2  ----> H2O\n",
      "    #Accounting the total number of moles of O2 in the fuel gas theoretically\n",
      "\n",
      "MthO2=(0.5)*CO + 0.5*H2 - O2 #theoretically required O2 moles\n",
      "\n",
      "    #For 20% excess moles required \n",
      "MactO2=(1.2)*MthO2  #Actual moles of O2 required \n",
      "\n",
      "    #Since air contains 79 mol% of N2\n",
      "MN2=(79./21)*MactO2 #moles of N2 in air\n",
      "    #2% CO does not react\n",
      "MTN2 = MN2 + N2\n",
      "MO2in = O2 + MactO2\n",
      "#Mflueg = moles of water + unreacted CO + CO2 (from reacted CO and CO2 in feed ) + N2 + O2 \n",
      "MH2Oflueg = H2\n",
      "MN2flueg = MTN2\n",
      "MCOflueg = 0.02*CO\n",
      "MCO2flueg = CO2 + 0.98*CO\n",
      "MO2flueg = MO2in - (0.5*H2 + 0.98*CO/2 )\n",
      "\n",
      "Mflueg = MH2Oflueg + 0.02*CO + CO2 + 0.98*CO + MTN2 + (MO2in - (0.5*H2 + 0.98*CO/2 ))\n",
      "\n",
      "\n",
      "#Result\n",
      "print \"Moles of CO in exit flue\",round(MCOflueg,2),\"mol\"\n",
      "print \"Moles of CO2 in exit flue\",round(MCO2flueg,2),\"mol\"\n",
      "print \"Moles of O2 in exit flue\",round(MO2flueg,2),\"mol\"\n",
      "print \"Moles of N2 in exit flue\",round(MTN2,1),\"mol\"\n",
      "print \"Moles of H2O in exit flue\",round(MH2Oflueg,2),\"mol\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Moles of CO in exit flue 0.54 mol\n",
        "Moles of CO2 in exit flue 32.26 mol\n",
        "Moles of O2 in exit flue 3.2 mol\n",
        "Moles of N2 in exit flue 129.7 mol\n",
        "Moles of H2O in exit flue 3.1 mol\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6-1, Page number 14"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Heating of N2 Gas\n",
      "\n",
      "#Variable declaration\n",
      "\n",
      "M=3.          #Moles of N2\n",
      "T1=298.       #Temperature of N2 gas (K)\n",
      "T2=673.       #Temperature of N2 gas (K)\n",
      "T3=1123.      #Temperature of N2 gas (K)\n",
      "\n",
      "# Data\n",
      "Cpm1=29.68    #Mean heat capacity of N2 gas ranging 298K-673K (J/gmol.k)\n",
      "Cpm2=31.00    #Mean heat capacity of N2 gas ranging 298K-1123K (J/gmol.k)\n",
      "\n",
      "#Calculation\n",
      "#Heat required (Hi)=M*Cpmi*(del T)\n",
      "H1 = M*Cpm1*(T2-T1)   #Heat required in (J)\n",
      "H2 = M*Cpm2*(T3-T1)   #Heat required in (J)\n",
      "H3 = H2-H1            #Heat required in (J) Since there is no mean heat capacity for the interval 673K-1123K\n",
      "\n",
      "#Result\n",
      "print \"(a) Heat required to raise the temperature from 298K-673K\",H1,\"J\"\n",
      "print \"(b) Heat required to raise the temperature from 298K-1123K\",H2,\"J\"\n",
      "print \"(c) Heat required to raise the temperature from 673K-1123K\",H3,\"J\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) Heat required to raise the temperature from 298K-673K 33390.0 J\n",
        "(b) Heat required to raise the temperature from 298K-1123K 76725.0 J\n",
        "(c) Heat required to raise the temperature from 673K-1123K 43335.0 J\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6-2, Page number 16"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Heating  of Milk\n",
      "\n",
      "#Variable declaration\n",
      "\n",
      "mdot = 4536.        #Mass rate of kg/hr\n",
      "Ti = 4.4            #Temeprature of incomming milk, deg C\n",
      "To = 54.4           #Temperature of milk leaving exhanger, deg C\n",
      "C = 3.85            #Heat capacity of cow milk, kJ/(kg.K)\n",
      "#Data\n",
      "\n",
      "#Calculation\n",
      "Mdot = mdot/3600    #Mass rate of kg/s\n",
      "delT = To - Ti      #delT: Temperature difference between milk temperature at outlet and inlet \n",
      "Q = Mdot*C*delT     #Q: Heat required, kW\n",
      "#Result\n",
      "print \"Rate of Heat Addition: \",Q, \"kJ/s or kW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate of Heat Addition:  242.55 kJ/s or kW\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6-3, Page number 17"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Use of Steam Table\n",
      "\n",
      "#Variable declaration\n",
      "\n",
      "m = 1 # mass of water in kg or lbm\n",
      "\n",
      "#Part A SI Units\n",
      "# Data\n",
      "H21_11 = 88.60 # Enathalpy at 21.11 degC, kJ/kg\n",
      "H60 = 251.13   # Enathalpy at 60 degC, kJ/kg \n",
      "#Calculation\n",
      "\n",
      "delH = H60 - H21_11\n",
      "\n",
      "#Result\n",
      "print \"Enthalpy change for heating 1kg of steam\", delH, \"kJ/kg\"\n",
      "\n",
      "#Part A British Units\n",
      "# Data\n",
      "H70 = 38.09 # Enathalpy at 70 degF, btu/lbm\n",
      "H140 = 107.96   # Enathalpy at 140 degF, btu/lbm\n",
      "\n",
      "#Calculation\n",
      "\n",
      "delH = H140 - H70\n",
      "\n",
      "#Result\n",
      "print \"Enthalpy change for heating 1 lbm of steam\", delH, \"Btu/lbm\"\n",
      "\n",
      "#Part B SI Units\n",
      "# Data\n",
      "H21_11 = 88.60    # Enathalpy at 21.11 degC, kJ/kg\n",
      "H115_6 = 2699.9   # Enathalpy at 115.6 degC, kJ/kg at 172.2 kPa\n",
      "\n",
      "#Calculation\n",
      "delH = H115_6 - H21_11\n",
      "\n",
      "#Result\n",
      "print \"Enthalpy change for heating 1kg of steam\", delH, \"kJ/kg\"\n",
      "\n",
      "#Part B British Units\n",
      "# Data\n",
      "H70 = 38.09  # Enathalpy at 70 degF, btu/lbm\n",
      "H240 = 1160.7   # Enathalpy at 140 degF, btu/lbm at 24.97 psia\n",
      "\n",
      "#Calculation\n",
      "delH = H240 - H70\n",
      "\n",
      "#Result\n",
      "print \"Enthalpy change for heating 1 lbm of steam\", delH, \"Btu/lbm\"\n",
      "\n",
      "#Part C SI Units\n",
      "# Data\n",
      "Hv_115_6 = 2699.9 # Enathalpy at 115.6 degC, kJ/kg at 172.2 kPa\n",
      "Hl_115_6 = 484.9  # Enathalpy at 115.6 degC, kJ/kg at 172.2 kPa\n",
      "\n",
      "#Calculation\n",
      "LambdaV = Hv_115_6 - Hl_115_6\n",
      "\n",
      "#Result\n",
      "print \"Enthalpy of vaporization 1 kg of steam\", LambdaV, \"kJ/kg\"\n",
      "\n",
      "#Part C British Units\n",
      "# Data\n",
      "Hv_115_6 = 1160.7                #Enathalpy water vapor at 240 degF, Btu/lbm at 24.97 psia\n",
      "Hl_115_6 = 208.44                #Enathalpy saturated water at 240 degF, Btu/lbm at 24.97 psia\n",
      "\n",
      "#Calculation\n",
      "LambdaV = Hv_115_6 - Hl_115_6\n",
      "#Result\n",
      "print \"Enthalpy of vaporization 1 kg of steam\", LambdaV, \"Btu/lbm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Enthalpy change for heating 1kg of steam 162.53 kJ/kg\n",
        "Enthalpy change for heating 1 lbm of steam 69.87 Btu/lbm\n",
        "Enthalpy change for heating 1kg of steam 2611.3 kJ/kg\n",
        "Enthalpy change for heating 1 lbm of steam 1122.61 Btu/lbm\n",
        "Enthalpy of vaporization 1 kg of steam 2215.0 kJ/kg\n",
        "Enthalpy of vaporization 1 kg of steam 952.26 Btu/lbm\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6-4, Page number 18"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Combustion of Carbon\n",
      "\n",
      "#Variable declaration\n",
      "n = 10. # gmol of Carbon \n",
      "nCO2 = 90. #Percent of Carbon converted to CO2\n",
      "nCO = 10.  #Percent of Carbon converted to CO\n",
      "DelHcCO2 = -393513 # Standard enthalpy change of combustion for CO2 in kJ/kmol\n",
      "DelHcCO = -110523 # Standard enthalpy change of combustion for CO2 in kJ/kmol\n",
      " \n",
      "#Calculation\n",
      "n1 = n/1000\n",
      "delHc = n1*nCO2*DelHcCO2/100 + n1*nCO*DelHcCO/100\n",
      "\n",
      "#Result\n",
      "print \" Entahlpy change on incomplete combustion of Carbon:\", round(delHc), \"kJ\"\n",
      "\n",
      "#Data kcal Units \n",
      "DelHcCO2 = -94.0518 # Standard enthalpy change of combustion for CO2 in kcal/gmol\n",
      "DelHcCO = -26.4157 # Standard enthalpy change of combustion for CO2 in kcal/gmol\n",
      "\n",
      "#Calculation\n",
      "delHc = n*nCO2*DelHcCO2/100 + n*nCO*DelHcCO/100\n",
      "\n",
      "#Result\n",
      "print \" Entahlpy change on incomplete combustion of Carbon:\", round(delHc), \"kcal\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Entahlpy change on incomplete combustion of Carbon: -3652.0 kJ\n",
        " Entahlpy change on incomplete combustion of Carbon: -873.0 kcal\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6-5, Page number 18"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Reaction of Methane\n",
      "\n",
      "#Variable declaration\n",
      "molCH4=1.             #MOles of CH4 (kg)\n",
      "press=101.32          #Pressure (kPa)\n",
      "temp=298.             #temperature (K)\n",
      "delHfCH4=-74848.     #Heat of Formation of CH4 (kJ/kg mol)\n",
      "delHfH2O=-285840.    #Heat of Formation of H2O (kJ/kg mol)\n",
      "delHfCO=-110523.     #Heat of Formation of CO (kJ/kg mol)\n",
      "delHfH2=0            #Heat of Formation of H2 (kJ/kg mol)\n",
      "\n",
      "#Calculation\n",
      "            #delH= delHf(products)-delHf(reactants) \n",
      "\n",
      "delH= (delHfCO + 3*delHfH2) - (delHfCH4 + delHfH2O)\n",
      "\n",
      "#Result\n",
      "print 'The standard heat of Formation %5.2e'%(delH),\"kJ/kg mol (Endothermic)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The standard heat of Formation 2.50e+05 kJ/kg mol (Endothermic)\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7-1, Page number 19"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Heating of Fermentation Medium\n",
      "\n",
      "#Variable declaration SI Units \n",
      "\n",
      "F=2000.    #Feed rate of Liquid (kg/h)\n",
      "TLi=30.    #Initial temperature of the liquid (\u00b0C)\n",
      "TLo=70.    #Final temperature of the liquid (\u00b0C)\n",
      "TWi=95.    #Temperature of water while entering (\u00b0C)\n",
      "TWo=85.    #Temperature of water while leaving (\u00b0C)\n",
      "Tref=25.   #Reference temperature (\u00b0C)\n",
      "CL=4.06    #Enthalpy of liquid at 25 degC (kJ/kg)\n",
      "Cw=4.21    #Enthalpy of water at 25 degC (kJ/kg)\n",
      "\n",
      "#Calculations \n",
      "\n",
      "delTLi=TLi-Tref\n",
      "delTLo=TLo-Tref\n",
      "delTW=TWi-Tref\n",
      "    #Input streams\n",
      "HLi=F*CL*delTLi\n",
      "    #Output streams\n",
      "HLo=F*CL*delTLo \n",
      "    #Heat gained by the system = Heat gained at outlet - heat at inlet\n",
      "QL = HLo-HLi\n",
      "    #Heat gained by liquid = Heat lost by water \n",
      "W = QL/(Cw*(TWi-TWo))\n",
      "\n",
      "#Result\n",
      "print 'Rate Enthalpy of the liquid entering %5.2e kJ/kg'%HLi\n",
      "print 'Rate Enthalpy of the liquid leaving %5.2e kJ/kg'%HLo\n",
      "print 'Heat added to fermentation medium %5.4e'%(QL),\"kJ/h\"\n",
      "print \"Feed rate of water supplied\",round(W,2),\"kg/h\"\n",
      "print 'The answer has been checked using calculator and \\ndiffers from answer given in book because of truncation'\n",
      "print \"Heat added to fermentation medium:\",round(QL/3600,2),\"kW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate Enthalpy of the liquid entering 4.06e+04 kJ/kg\n",
        "Rate Enthalpy of the liquid leaving 3.65e+05 kJ/kg\n",
        "Heat added to fermentation medium 3.2480e+05 kJ/h\n",
        "Feed rate of water supplied 7714.96 kg/h\n",
        "The answer has been checked using calculator and \n",
        "differs from answer given in book because of truncation\n",
        "Heat added to fermentation medium: 90.22 kW\n"
       ]
      }
     ],
     "prompt_number": 28
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7-2, Page number 21"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Heat and Material Balance in Combustion\n",
      "\n",
      "#Variable declaration SI Units \n",
      "mCO = 1.                         #Moles of CO burnt in furnace (kmol/h)\n",
      "delHc=-282989.                   #Standard heat of combustion of CO (kJ/kmol)\n",
      "CpmCO = 29.38                    #Mean Heat capacity of CO between temperature 298k-478k\n",
      "CpmCO2 = 49.91                   #Mean Heat capacity of CO2 between temperature 298k-478k\n",
      "CpmN2 = 31.43                    #Mean Heat capacity of N2 between temperature 298k-478k\n",
      "CpmO2 = 33.25                    #Mean Heat capacity of O2 between temperature 298k-478k\n",
      "Cpmair = 29.29                   #Mean Heat capacity of Air between temperature 298k-478k\n",
      "Tref = 298.                      #Reference temperature for enthalpy calculation\n",
      "TCOi = 473.                      #Temperature of CO inlet stream \n",
      "Tair = 373.                      #Temperature of inlet air stream\n",
      "Texit = 1273.                    #Temperature of outgoing stream\n",
      "\n",
      "#Calculation\n",
      "               #CO(g) + (0.5)O2(g) ---> CO2(g)\n",
      "               #Since from the above equation we get that 1 mole of CO will require 0.5 mole O2\n",
      "MthO2 = (0.5)*1.                 #Theoretically required O2 moles (kmol/h)\n",
      "MactO2 = (1.9)*(MthO2)           #Moles of O2 actually added\n",
      "MN2 =  (0.79/0.21)*MactO2        #Moles of N2 added\n",
      "xO2 = MactO2 - MthO2             #Moles of O2 in exit flue\n",
      "xCO2 = mCO                       #Moles of CO2 formed is equal to moles of CO added\n",
      "xN2 = MN2                        #Moles of N2 in the exit flue\n",
      "MAir = xN2 + MactO2\n",
      "\n",
      "              #Input streams \n",
      "HinAir = MAir*Cpmair*(Tair-Tref)\n",
      "HinCO = mCO*CpmCO*(TCOi-Tref)\n",
      "HCO2 = xCO2*CpmCO2*(Texit-Tref)\n",
      "HN2 = xN2*CpmN2*(Texit-Tref)\n",
      "HO2 = xO2*CpmO2*(Texit-Tref)\n",
      "Qr = mCO*delHc    #Heat removed = total enthalpy of leaving streams - total enthalpy of incoming streams + heat added due to reaction\n",
      "q =  HCO2 + HO2 + HN2 - (HinAir + HinCO-Qr)\n",
      "\n",
      "#Result\n",
      "print 'moles of O2 actually added %4.3f kgmol/h'%MactO2\n",
      "print 'moles of N2 added %4.3f kgmol/h, Notice this value differs from \\nthan in book hence following values also change but are correct'%xN2\n",
      "print 'moles of air  added %4.3f kgmol/h'%(MactO2+MN2)\n",
      "print 'moles of O2 in exit flue gas %4.3f kgmol/h'%xO2\n",
      "print 'Enthalpy of air entering %4.3f kJ/h'%HinAir\n",
      "print 'Enthalpy of CO entering %4.3f kJ/h'%HinCO\n",
      "print 'Enthalpy of CO2 leaving %4.3f kJ/h'%HCO2\n",
      "print 'Enthalpy of N2 leaving %4.3f kJ/h'%HN2\n",
      "print 'Enthalpy of O2 leaving %4.3f kJ/h'%HO2\n",
      "print \"Heat Removed: \", round(q), \"kJ/hr\"\n",
      "print \"Heat Removed: \", round(q*1000./3600.),\"W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "moles of O2 actually added 0.950 kgmol/h\n",
        "moles of N2 added 3.574 kgmol/h, Notice this value differs from \n",
        "than in book hence following values also change but are correct\n",
        "moles of air  added 4.524 kgmol/h\n",
        "moles of O2 in exit flue gas 0.450 kgmol/h\n",
        "Enthalpy of air entering 9937.679 kJ/h\n",
        "Enthalpy of CO entering 5141.500 kJ/h\n",
        "Enthalpy of CO2 leaving 48662.250 kJ/h\n",
        "Enthalpy of N2 leaving 109516.713 kJ/h\n",
        "Enthalpy of O2 leaving 14588.437 kJ/h\n",
        "Heat Removed:  -125301.0 kJ/hr\n",
        "Heat Removed:  -34806.0 W\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7-3, Page number 22"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Oxidation of LactoseC12H22O11\n",
      "\n",
      "#Variable declaration\n",
      "        #HLact37, Total Enthalpy of Lactose stream in at 37 deg C  \n",
      "        #HO237, Total Enthalpy of Oxygen stream in at 37 deg C  \n",
      "        #HH2O37, Total Enthalpy of Water out at 37 deg C  \n",
      "        #HCO237, Total Enthalpy of CO2 stream out at 37 deg C \n",
      "        #delHcLact37, Heat of combustion for Lactose, J/mol\n",
      "\n",
      "\n",
      "nLact = 1.     #Mass of Lactose in gmol\n",
      "nO2 = 12.      #Moles oo O2 fed\n",
      "nCO2 = 12.     #Moles oo CO2 \n",
      "nH2O = 11.     #Moles oo H2O  \n",
      "MWLact = 342.3 #Molecular weight of lactose\n",
      "MWH2O = 18.02  #Molecular weight of water\n",
      "CpmLact = 1.2  #Mean specific heat of Lactose, J/g.K\n",
      "CpmO2 = 29.38  #Mean specific heat of O2, J/mol.K\n",
      "CpmH2O = 4.18  #Mean specific heat of H2O, J/mol.K\n",
      "CpmCO2 = 37.45 #Mean specific heat of CO2, J/mol.K\n",
      "delHcLact = 5648.8e3 #Std. Heat of combustion for Lactose, J/mol\n",
      "Tref = 25.     #Reference temperature, deg C\n",
      "Ts = 37.       #Stream temperature, deg C\n",
      "\n",
      "\n",
      "#Calculation\n",
      "                #H = n*Cpmi*(T-Tref)\n",
      "                #Del T in degC = Del T in K\n",
      "mLact = nLact*MWLact\n",
      "HLact37 = mLact*CpmLact*(Ts-Tref)\n",
      "HO237 = nO2*CpmO2*(Ts-Tref)\n",
      "HCO237 = nCO2*CpmCO2*(Ts-Tref)\n",
      "HH2O37 = nH2O*MWH2O*CpmH2O*(Ts-Tref)\n",
      "delHcLact37 = HCO237 + HH2O37 - (HLact37 + HO237 + delHcLact)\n",
      "\n",
      "#Result\n",
      "\n",
      "print \"Enthalpy of Lactose stream in at 37 deg C:\",round(HLact37),\"J/gmol\"\n",
      "print \"Enthalpy of Oxygen stream in at 37 deg C :\",round(HO237),  \"J/gmol\"  \n",
      "print \"Enthalpy of Water out at 37 deg C        :\",round(HH2O37), \"J/gmol\"  \n",
      "print \"Enthalpy of CO2 stream out at 37 deg C   :\",round(HCO237), \"J/gmol\"\n",
      "print \"Heat of combustion for Lactose           :\",round(HLact37),\"J/gmol\"\n",
      "print 'Heat of combustion of Lactose at 37 deg C:%6.3e'%(delHcLact37),\"J/gmol\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Enthalpy of Lactose stream in at 37 deg C: 4929.0 J/gmol\n",
        "Enthalpy of Oxygen stream in at 37 deg C : 4231.0 J/gmol\n",
        "Enthalpy of Water out at 37 deg C        : 9943.0 J/gmol\n",
        "Enthalpy of CO2 stream out at 37 deg C   : 5393.0 J/gmol\n",
        "Heat of combustion for Lactose           : 4929.0 J/gmol\n",
        "Heat of combustion of Lactose at 37 deg C:-5.643e+06 J/gmol\n"
       ]
      }
     ],
     "prompt_number": 30
    }
   ],
   "metadata": {}
  }
 ]
}