path: root/TRANSPORT_PROCESSES_AND_UNIT_OPERATIONS/GeankoplisChapter06.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 6: Principles of Mass Transfer"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.1-1, Page number 384"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Molecular Diffusion of Helium in Nitrogen\n",
      "\n",
      "#Variable declaration\n",
      "z1 = 0.0                             #Location of one end of pipe, m\n",
      "z2 = 0.2                             #Location of other end of pipe, m\n",
      "T = 298                              #Temeperature of gas, K\n",
      "pA1 = 0.6                            #Partial pressure of Helium at end 1, atm\n",
      "pA2 = 0.2                            #Partial pressure of Helium at end 2, atm\n",
      "DAB = 0.687e-4                       #Diffusivity of Helium in Nitrogen,  m2/s\n",
      "P = 1.                               #Total pressure, atm\n",
      "R = 82.057e-3                        #Gas Constant,m3.atm/(kmol.K)\n",
      "\n",
      "#Calculation SI Units\n",
      "JAz = DAB*(pA1-pA2)/(R*T*(z2-z1))\n",
      "\n",
      "#Result\n",
      "print 'Flux of Helium through the Nitrogen in SI units: %5.2e'%(JAz), \"kmol/(m2.s)\"\n",
      "\n",
      "#Variable declaration cgs Units\n",
      "z1 = 0.0                             #Location of one end of pipe, m\n",
      "z2 = 20                              #Location of other end of pipe, m\n",
      "DAB = 0.687                          #Diffusivity of Helium in Nitrogen,  cm2/s\n",
      "R = 82.057                           #Gas Constant,cm3.atm/(gmol.K)\n",
      "\n",
      "#Calculation cgs Units\n",
      "JAz = DAB*(pA1-pA2)/(R*T*(z2-z1))\n",
      "\n",
      "#Result\n",
      "print 'Flux of Helium through the Nitrogen in cgs units: %5.2e'%(JAz), \"gmol/(cm2.s)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Flux of Helium through the Nitrogen in SI units: 5.62e-06 kmol/(m2.s)\n",
        "Flux of Helium through the Nitrogen in cgs units: 5.62e-07 gmol/(cm2.s)\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.2-1, Page Number 386"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Equimolar Counterdiffusion\n",
      "\n",
      "#Variable declaration\n",
      "z1 = 0.0                             #Location of one end of pipe, m\n",
      "z2 = 0.1                             #Location of other end of pipe, m\n",
      "T = 298                              #Temeperature of gas, K\n",
      "pA1 = 1.013e4                        #Partial pressure of Ammonia at end 1, Pa\n",
      "pA2 = 0.507e4                        #Partial pressure of Ammonia at end 2, Pa\n",
      "DAB = 0.23e-4                        #Diffusivity of Ammonia in Nitrogen,  m2/s\n",
      "P = 1.0132e5                         #Total pressure, Pa\n",
      "R = 8314.3                           #Gas Constant,m3.Pa/(kmol.K)\n",
      "\n",
      "#Calculation\n",
      "\n",
      "JAz = DAB*(pA1-pA2)/(R*T*(z2-z1))\n",
      "pB1 = P - pA1\n",
      "pB2 = P - pA2\n",
      "JBz = DAB*(pB1-pB2)/(R*T*(z2-z1))\n",
      "#Result\n",
      "\n",
      "print 'Flux of Ammonia through the Nitrogen:%10.2e '%(JAz), \"kmolA/(m2.s)\"\n",
      "print 'Flux of Nitrogen through the Ammonia:%10.2e '%(JBz), \"kmolB/(m2.s)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Flux of Ammonia through the Nitrogen:  4.70e-07  kmolA/(m2.s)\n",
        "Flux of Nitrogen through the Ammonia: -4.70e-07  kmolB/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.2-2, Page number 389"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Diffusion of Water Through Stagnant, Nondiffusing Air\n",
      "from math import log\n",
      "\n",
      "#Variable declaration English Units\n",
      "z1 = 0.0                             #Location of one end of pipe, ft\n",
      "z2 = 0.5                             #Location of other end of pipe, ft\n",
      "T = 68                               #Temeperature of gas, \u00b0F\n",
      "DAB = 0.25e-4                        #Diffusivity of Water in Air,  m2/s\n",
      "p0w = 17.54                          #Vapor pressure of Water at 68 \u00b0F, mmHg\n",
      "R = 0.73                             #Gas Constant, ft3.atm/(lbmol.\u00b0R)\n",
      "P = 1.\n",
      "\n",
      "#Calculation English units\n",
      "DAB = DAB*3.875e4\n",
      "p0w = p0w/760\n",
      "pA1 = p0w\n",
      "pA2 = 0.\n",
      "T = T + 460.\n",
      "\n",
      "pB1 = P - pA1\n",
      "pB2 = P - pA2\n",
      "pBM = (pB2-pB1)/log(pB2/pB1)\n",
      "NA = DAB*P*(pA1-pA2)/(R*T*(z2-z1)*pBM)\n",
      "#Result\n",
      "print 'Rate of evaporation of water at steady state:English Units %10.3e'%(NA), \"lbmol/(ft2.hr)\"\n",
      "\n",
      "#Variable declaration SI Units\n",
      "z1 = 0.0                             #Location of one end of pipe, m\n",
      "z2 = 0.1524                          #Location of other end of pipe, m\n",
      "T = 293                              #Temeperature of gas, K\n",
      "p0w = 17.54                          #Vapor pressure of Water at 293 K, mmHg\n",
      "DAB = 0.25e-4                        #Diffusivity of Water in Air,  m2/s\n",
      "P = 1.01325e5                        #Total pressure, Pa\n",
      "R = 8314.3                           #Gas Constant,m3.Pa/(kmol.K)\n",
      "\n",
      "#Calculation SI units\n",
      "\n",
      "p0wSI = p0w*P/760\n",
      "pA1 = p0wSI\n",
      "pA2 = 0.\n",
      "JAz = DAB*(pA1-pA2)/(R*T*(z2-z1))\n",
      "\n",
      "pB1 = P - pA1\n",
      "pB2 = P - pA2\n",
      "pBM = (pB2-pB1)/log(pB2/pB1)\n",
      "NA = DAB*P*(pA1-pA2)/(R*T*(z2-z1)*pBM)\n",
      "\n",
      "#Result\n",
      "print 'Rate of evaporation of water at steady state SI Units:%10.3e'%(NA), \"kmol/(m2.s)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate of evaporation of water at steady state:English Units  1.174e-04 lbmol/(ft2.hr)\n",
        "Rate of evaporation of water at steady state SI Units: 1.593e-07 kmol/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.2-4, Page number 392"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Evaporation of Napthalene Sphere\n",
      "\n",
      "# Variable declaration\n",
      "r = 2.            #Radius of Napthalene ball, mm\n",
      "Tair = 318.       #Ambient temperature of air, K\n",
      "Tball = 318.      #Temperature of the Napthalene ball, K\n",
      "p0N = 0.555       #Vapor pressure of Napthalene at 318K, mmHg\n",
      "DAB = 6.92e-6     #Diffusion Coefficient, m2/s\n",
      "P = 101325.       #Atmospheric pressure, Pa \n",
      "R = 8314.         #Gas constant, m3.Pa/(kmol.K)\n",
      "\n",
      "# Calculation\n",
      "pA1 = p0N*P/760\n",
      "r = r/1000\n",
      "pA2 = 0.\n",
      "pB1 = P - pA1\n",
      "pB2 = P- pA2\n",
      "pBM = (pB1+pB2)/2.\n",
      "NA = DAB*P*(pA1-pA2)/(R*Tair*r*pBM)\n",
      "   \n",
      "#Result\n",
      "print 'Rate of evaporation of Napthalene from surface is:%10.3e'%(NA),\"kmol A/(m2.s)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate of evaporation of Napthalene from surface is: 9.687e-08 kmol A/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.2-5, Page number 397"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Estimation of diffusivity of a Gas Mixture\n",
      "from math import sqrt\n",
      "\n",
      "# Variable declaration\n",
      "P = 1.                    #Pressure in atmosphere\n",
      "MA = 74.1                 #Molecular weight of Butanol\n",
      "MB = 29.                  #Molecular weight of Air\n",
      "T1 = 0.                   #Temperature, deg C\n",
      "T2 = 25.9                 #Temperature, deg C\n",
      "T3 = 0.                   #Temperature, deg C\n",
      "P3 = 2.\n",
      "# Calculation\n",
      "def BinaryDiffusivity(P,T):\n",
      "    dab =  1.0e-7*T**1.75*sqrt(1./MA+1./MB)/(P*(SvA**(1./3)+ SvB**(1./3))**2)\n",
      "    print \"The binary diffusivity Butanol in Air at\",round(P,2),\"&\",T,'K is %5.3e'%(dab),\"m2/s\"  \n",
      "    return dab\n",
      "\n",
      "#Atomic Diffusion volumes using table 6.2-2 pp-396\n",
      "SvA = 4*16.5+10*1.98+1*5.48\n",
      "SvB = 20.1\n",
      "T1 = 273 + T1\n",
      "DAB1 = BinaryDiffusivity(P,T1)\n",
      "T2 = 273 + T2\n",
      "DAB2 = BinaryDiffusivity(P,T2)\n",
      "DAB3 = DAB1*(1./2.)\n",
      "print 'The binary diffusivity Butanol in Air at %3.1f & %4.1f K is %5.3e m2/s'%(P3,T1,DAB3)\n",
      "#OR\n",
      "#DAB3 = BinaryDiffusivity(P3,T1)\n",
      "#Result\n",
      "print 'The answers different than book because of book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The binary diffusivity Butanol in Air at 1.0 & 273.0 K is 7.701e-06 m2/s\n",
        "The binary diffusivity Butanol in Air at 1.0 & 298.9 K is 9.025e-06 m2/s\n",
        "The binary diffusivity Butanol in Air at 2.0 & 273.0 K is 3.851e-06 m2/s\n",
        "The answers different than book because of book uses rounded numbers\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.3-1, Page number 399"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Diffusion of Ethanol(A) through Water(B)\n",
      "\n",
      "#Variable declaration\n",
      "\n",
      "T = 298                 #Temperature of solution, K\n",
      "rho1 = 972.8            #Density of solution at 16.8% wt, kg/m3\n",
      "rho2 = 998.1            #Density of solution at 6.8% wt, kg/m3\n",
      "DAB = 0.740e-9          #Diffusivity of Ethanol, m2/s\n",
      "MA = 46.05              #Molecular wt of ethanol\n",
      "MB = 18.02              #Molecular wt of Water\n",
      "xw1 = 16.8              #weight % of Ethanol at 1\n",
      "xw2 = 6.8               #weight % of Ethanol at 2\n",
      "z1 = 0.                 #Location 1, m\n",
      "z2 = 2.e-3              #Location 2, m\n",
      "#Calculation\n",
      "\n",
      "xmA1 = xw1/MA/(xw1/MA+(100-xw1)/MB)\n",
      "xmA2 = xw2/MA/(xw2/MA+(100-xw2)/MB)\n",
      "xmB1 = 1. - xmA1\n",
      "xmB2 = 1. - xmA2\n",
      "MW1 = MA*xmA1 + MB*xmB1\n",
      "MW2 = MA*xmA2 + MB*xmB2\n",
      "Cav = (rho1/MW1+rho2/MW2)/2.\n",
      "xBM = (xmB1+xmB2)/2.\n",
      "NA = DAB*Cav*(xmA1-xmA2)/(xBM*(z2-z1))\n",
      "\n",
      "#Result\n",
      "print 'Steady State flux of Ethanol: %4.3e'%(NA), \"kgmol/(m2.s)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Steady State flux of Ethanol: 8.998e-07 kmol/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.3-2, Page number 402"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Prediction of Liquid Diffusivity \n",
      "from math import sqrt\n",
      "#Variable declaration\n",
      "T1 = 25                #Temperature of solution, degC\n",
      "T2 = 50                #Temperature of solution, degC\n",
      "mu25B = 0.8937e-3      #Viscosity of Water at 25 degC, Pa.s\n",
      "mu50B = 0.5494e-3      #Viscosity of Water at 50 degC, Pa.s\n",
      "MB = 18.02\n",
      "\n",
      "\n",
      "#Calculation\n",
      "mvA = 3*0.0148 + 6*0.0037 + 1*0.0074\n",
      "si =  2.6\n",
      "\n",
      "def LiquidDiffusivity(muB,T):\n",
      "    dab = 1.173e-16*sqrt(si*MB)*T/(muB*mvA**0.6)\n",
      "    print 'Liquid diffusivity of Acetone in Water at %5.3f \u00b0C is %5.3e m2/s' %(T,dab)\n",
      "    return \n",
      "\n",
      "LiquidDiffusivity(mu25B,T1+273)\n",
      "LiquidDiffusivity(mu50B,T2+273)\n",
      "#Result\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Liquid diffusivity of Acetone in Water at 298.000 \u00b0C is 1.277e-09 m2/s\n",
        "Liquid diffusivity of Acetone in Water at 323.000 \u00b0C is 2.251e-09 m2/s\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.4-1, Page number 405"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Prediction of Diffusivity of Albumin\n",
      "\n",
      "#Variable declaration\n",
      "T = 298            #Temperature, K\n",
      "MA = 67500         #Molecular Weight of Albumin\n",
      "muw298 = 0.8937e-3 #Viscosity of water at 298 K, Pa.s\n",
      "\n",
      "#Calculations\n",
      "DAB = 9.4e-15*T/(muw298*MA**(1./3))\n",
      "\n",
      "#Result\n",
      "print 'Diffusivity of Albumin in Water at 298 K %3.2e m2/s'%(DAB)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diffusivity of Albumin in Water at 298 K 7.70e-11 m2/s\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.4-2, Page number 407"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Diffusion of urea in Agar\n",
      "\n",
      "#Variable Declaration\n",
      "CA1 = 0.2               #Concentration at one end of tube\n",
      "CA2 = 0.0               #Concentration at other end of tube\n",
      "z1 = 0.0\n",
      "z2 = 0.04               #Location of other end of tube from end 1\n",
      "DAB = 0.727e-9          #Diffusivity of urea \n",
      "\n",
      "#Calculation\n",
      "NA = DAB*(CA1-CA2)/(z2-z1)\n",
      "\n",
      "#Result\n",
      "print 'Steady State Flux of urea through agar solution:%4.3e kgmol/(m2.s)'%(NA)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Steady State Flux of urea through agar solution:3.635e-09 kmol/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.5-1, Page number 409"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Diffusion of H2 through Neoprene Membrance\n",
      "\n",
      "#Variable Declaration\n",
      "\n",
      "T = 17                      #Temnperature of hydrogen, deg C\n",
      "pH21 = 0.01                 #Partial pressure of H2 on one end of membrane, atm\n",
      "z = 0.5                     #Membrane thickness, mm\n",
      "pH22 = 0.0                  #Partial pressure of H2 on other end of membrane, atm\n",
      "S = 0.051                   #Solubility of H2 in Neoprene, [m3 at STP/(m3solid.atm)]\n",
      "DAB = 1.03e-10              #Diffusivity at 17 deg C, m2/s\n",
      "\n",
      "#Calculations\n",
      "cA1 = S*pH21/22.414\n",
      "cA2 = S*pH22/22.414\n",
      "NA = DAB*(cA1-cA2)/(z/1000.)\n",
      "\n",
      "#Results\n",
      "print 'concentrations at face 1 and face 2 are %3.2e and %3.2e kgmol H2/m3 solid respectively'%(cA1,cA2)\n",
      "print 'Steady State Flux of Hydrogen through Neoprene membrane:%5.2e'%(NA),\"kgmol H2/(m2.s)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "concentrations at face 1 and face 2 are 2.28e-05 and 0.00e+00 kgmol H2/m3 solid respectively\n",
        "Steady State Flux of Hydrogen through Neoprene membrane:4.69e-12 kgmol H2/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.5-2, Page number 411"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Diffusion through a Packaging Film using Permeability\n",
      "\n",
      "#Variable Declaration\n",
      "z = 0.00015            #Thickness of the pkg film, m\n",
      "T = 30.                #Temperature of the film, deg C\n",
      "pO21 = 0.21            #Partial pressure of O2 outside the film, atm\n",
      "pO22 = 0.01            #Partial pressure of O2 inside the film, atm\n",
      "PM = 4.17e-12          #Permeability of the film m3 solute STP/(s.m2.atm/m)\n",
      "\n",
      "#Calcualtions\n",
      "NA = PM*(pO21-pO22)/(22.414*z)\n",
      "\n",
      "#Result\n",
      "print 'Diffusional flux of the Oxygen through the polyethylene film:%10.3e kgmol/(m2.s)'%(NA)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diffusional flux of the Oxygen through the polyethylene film: 2.481e-10 kgmol/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.5-3, Page number 412"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Diffusion of KCl in porous Silica\n",
      "\n",
      "#Variable Declaration\n",
      "z = 0.002                  #Thickness of Silica, m\n",
      "DAB = 1.87e-9              #Diffusivity of KCl in water, m2/s\n",
      "cA1 = 0.1                  #Concentration of KCl, kmol/m3\n",
      "cA2 = 0.0                  #Concentration of KCl on other side of Silica\n",
      "epps = 0.3                 #Porosity of  Silica\n",
      "tau = 4.0                  #Tortuosity\n",
      "\n",
      "#Calculation\n",
      "NA = epps*DAB*(cA1-cA2)/(tau*z)\n",
      "\n",
      "#Results\n",
      "print 'Diffusional flux of the KCl through the Porous Silica: %5.2e kmol/(m2.s)'%(NA)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diffusional flux of the KCl through the Porous Silica: 7.01e-09 kmol/(m2.s)\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6.6-1, Page number 416"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Numerical Method for Convection and Steady State Diffusion\n",
      "import numpy as np\n",
      "\n",
      "#Variable Declaration\n",
      "ci = 6.00e-3           #Inside concentration, kmol/m3\n",
      "kc = 2.0e-7            #Outside convective coefficient, m/s\n",
      "cinf = 2.00e-3         #Outside concentration, kmol/m3\n",
      "DAB = 1.0e-9           #Diffusivity in solid, m2/s\n",
      "dx = dy = 0.005        #Grid size in x and y directions, m\n",
      "K = 1.0                #Distribution coefficient\n",
      "\n",
      "#Calculations\n",
      "kdxbyD = kc*dx/DAB\n",
      "\n",
      "#Index used are one less as in book\n",
      "c1 = np.zeros(5)\n",
      "c2 = np.zeros(5)\n",
      "c3 = np.zeros(5)\n",
      "\n",
      "\n",
      "cinf = cinf*1e3\n",
      "ci = ci*1e3\n",
      "np.set_printoptions(precision=2)\n",
      "\n",
      "#Initializations\n",
      "c1[2] = c1[3] = ci\n",
      "c2[1] = 3.8\n",
      "c1[1] = c2[2] = c2[4] = 4.2\n",
      "c2[3] = 4.4\n",
      "c3[0] = 2.5\n",
      "c3[1] = c2[0] = 2.7\n",
      "c3[2] = c3[4] = 3.0\n",
      "c3[3] = 3.2\n",
      "\n",
      "for j in range(3):\n",
      "    N22 = c1[1]+c3[1]+c2[0]+c2[2]-4*c2[1]\n",
      "    c2[1] = (c1[1]+ c3[1]+c2[0]+c2[2])/4\n",
      "    N23 = c2[1]+c2[3]+c1[2]+c3[2]-4*c2[2]\n",
      "    c2[2] = c2[4] = c1[1] = (c2[1]+c2[3]+c1[2]+c3[2])/4\n",
      "    N24 = c2[2]+c2[4]+c1[3]+c3[3]-4*c2[3]\n",
      "    c2[3] = (c2[2]+c2[4]+c1[3]+c3[3])/4\n",
      "    N31 = kdxbyD*cinf + (c2[0]+c3[1])/2 - (kdxbyD+1)*c3[0]  \n",
      "    c3[0] = (kdxbyD*cinf + (c2[0]+c3[1])/2)/(kdxbyD+1) \n",
      "    N32 = kdxbyD*cinf + (2*c2[1]+c3[0]+c3[2])/2 - (kdxbyD+2)*c3[1]\n",
      "    c3[1] = c2[0] =(kdxbyD*cinf + (2*c2[1]+c3[0]+c3[2])/2)/(kdxbyD+2)\n",
      "    N33 = kdxbyD*cinf + (2*c2[2]+c3[1]+c3[3])/2-(kdxbyD+2)*c3[2]\n",
      "    c3[2] = c3[4] = (kdxbyD*cinf + (2*c2[2]+c3[1]+c3[3])/2)/(kdxbyD+2)\n",
      "    N34 = kdxbyD*cinf + (2*c2[3]+c3[2]+c3[4])/2 -(kdxbyD+2)*c3[3]\n",
      "    c3[3] = (kdxbyD*cinf + (2*c2[3]+c3[2]+c3[4])/2)/(kdxbyD+2)\n",
      "    \n",
      "    \n",
      "c1[0] = c3[2]\n",
      "c1[4] = c1[2]\n",
      "\n",
      "No = kc*(dx*1.)*((c3[0]-cinf)/2+(c3[1]-cinf)+(c3[2]-cinf)+(c3[3]-cinf)/2)*1e-3\n",
      "Ni = DAB*(dx*1.)*((c1[2]-c2[2])+(c1[3]-c2[3])/2)*1e-3/dy\n",
      "#Results\n",
      "print \"Concentration Values at nodes are as follows\"\n",
      "print \"C    1     2     3     4 \"\n",
      "print 1,c1[:4]\n",
      "print 2,c2[:4]\n",
      "print 3,c3[:4]\n",
      "\n",
      "print '\\nThe average flux is %6.3e kmol/s'%((Ni+No)*.5)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Concentration Values at nodes are as follows\n",
        "C    1     2     3     4 \n",
        "1 [ 3.06  4.23  6.    6.  ]\n",
        "2 [ 2.73  3.48  4.23  4.41]\n",
        "3 [ 2.36  2.73  3.06  3.15]\n",
        "\n",
        "The average flux is 2.554e-12 kmol/s\n"
       ]
      }
     ],
     "prompt_number": 47
    }
   ],
   "metadata": {}
  }
 ]
}