{
 "metadata": {
  "name": "",
  "signature": "sha256:154cdfdbffe4cd226ec00ca571625579dcd760fa5b25d6fc7f505d761a2beb76"
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 13 : Diffusion Mass Transfer"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.1  Page No : 544"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "ro2 = 0.21;\t\t\t#Ratio of O2 in the mixture  \n",
      "rn2 = 0.79;\t\t\t#Ratio of N2 in the mixture  \n",
      "T = (25+273);\t\t\t#Temperature of container in degree C\n",
      "p = 1;\t\t\t#Total pressure in atm\n",
      "\n",
      "# Calculations\n",
      "Co2 = (ro2*10**5)/(8314*T);\t\t\t#Molar concentration of O2 in K.mol/m**3\n",
      "Cn2 = (rn2*10**5)/(8314*T);\t\t\t#Molar concentration of N2 in K.mol/m**3\n",
      "po2 = (32*Co2);\t\t\t#Mass density in kg/m**3\n",
      "pn2 = (28*Cn2);\t\t\t#Mass density in kg/m**3\n",
      "p = (po2+pn2);\t\t\t#Overall mass density in kg/m**3\n",
      "mo2 = (po2/p);\t\t\t#Mass fraction of O2\n",
      "mn2 = (pn2/p);\t\t\t#Mass fraction of N2\n",
      "M = (ro2*32)+(rn2*28);\t\t\t#Average molecular weight \n",
      "\n",
      "# Results\n",
      "print 'Molar concentration of O2 is %3.4f K.mol/m**3  \\n \\\n",
      "Molar concentration of N2 is %3.3f K.mol/m**3  \\n \\\n",
      "Mass density of O2 is %3.3f kg/m**3  \\n \\\n",
      "Mass density of N2 is %3.3f kg/m**3  \\n \\\n",
      "Mole fraction of O2 is %3.2f  \\n \\\n",
      "Mole fraction of N2 is %3.2f  \\n \\\n",
      "Mass fraction of O2 is %3.3f  \\n \\\n",
      "Mass fraction of N2 is %3.3f  \\n \\\n",
      "Average molecular weight is %3.2f'%(Co2,Cn2,po2,pn2,ro2,rn2,mo2,mn2,M)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Molar concentration of O2 is 0.0085 K.mol/m**3  \n",
        " Molar concentration of N2 is 0.032 K.mol/m**3  \n",
        " Mass density of O2 is 0.271 kg/m**3  \n",
        " Mass density of N2 is 0.893 kg/m**3  \n",
        " Mole fraction of O2 is 0.21  \n",
        " Mole fraction of N2 is 0.79  \n",
        " Mass fraction of O2 is 0.233  \n",
        " Mass fraction of N2 is 0.767  \n",
        " Average molecular weight is 28.84\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.2  Page No : 545"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "yh2 = 0.4;\t\t\t#Mole fraction og H2\n",
      "yo2 = 0.6;\t\t\t#Mole fraction of O2\n",
      "vh2 = 1;\t\t\t#velocity of H2 in m/s\n",
      "vo2 = 0;\t\t\t#velocity of O2 in m/s\n",
      "\n",
      "# Calculations\n",
      "V = (yh2*vh2)+(yo2*vo2);\t\t#Molar average velocity in m/s\n",
      "M = (yh2*2)+(yo2*32);\t\t\t#Molecular weight of the mixture\n",
      "mh2 = (yh2*2)/M;\t    \t\t#Mass fraction of H2 \n",
      "mo2 = (yo2*32)/M;\t\t    \t#Mass fraction of O2\n",
      "v = (mh2*vh2)+(mo2*vo2);\t\t#Mass average velocity in m/s\n",
      "x1 = (mh2*vh2);\t\t\t        #Mass flux \n",
      "x2 = (mo2*vo2);\t\t\t        #Mass flux\n",
      "y1 = (v*vh2);\t\t    \t    #Molar flux\n",
      "y2 = (yo2*vo2);\t    \t\t    #Molar flux\n",
      "jh2 = (mh2*(vh2-v));\t\t\t#Mass diffusion flux\n",
      "jo2 = (mo2*(vo2-v));\t\t\t#Mass diffusion flux\n",
      "Jh2 = (yh2*(vh2-V));\t\t\t#Molar diffusion flux\n",
      "Jo2 = (yo2*(vo2-V));\t\t\t#Molar diffusion flux\n",
      "\n",
      "# Results\n",
      "print 'Molar average velocity is %3.1f m/s \\n \\\n",
      "Mass average velocity is %3.2f m/s  \\n \\\n",
      "Mass flux of H2 when it is stationary is %3.2fp kg/m2.s3 \\n \\\n",
      "Mass flux of O2 when it is stationary is %3.0f kg/m**2.s \\n \\\n",
      "Molar flux of H2 when it is stationary is %3.2fC k.mol/m**2.s \\n \\\n",
      "Molar flux of O2 when it is stationary is %3.0f k.mol/m**2.s \\n \\\n",
      "Mass diffusion flux of H2 across a surface moving with mass average velocity is %3.4fp kg/m**2.s \\n \\\n",
      "Mass diffusion flux of O2 across a surface moving with mass average velocity is %3.4fp kg/m**2.s  \\n \\\n",
      "Molar diffusion flux across a surface moving with molar average velociy for H2 is %3.2fC k.mol/m**2.s \\n \\\n",
      "Molar diffusion flux across a surface moving with molar average velociy for O2\\\n",
      " is %3.2fC k.mol/m**2.s'%(V,v,x1,x2,y1,y2,jh2,jo2,Jh2,Jo2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Molar average velocity is 0.4 m/s \n",
        " Mass average velocity is 0.04 m/s  \n",
        " Mass flux of H2 when it is stationary is 0.04p kg/m2.s3 \n",
        " Mass flux of O2 when it is stationary is   0 kg/m**2.s \n",
        " Molar flux of H2 when it is stationary is 0.04C k.mol/m**2.s \n",
        " Molar flux of O2 when it is stationary is   0 k.mol/m**2.s \n",
        " Mass diffusion flux of H2 across a surface moving with mass average velocity is 0.0384p kg/m**2.s \n",
        " Mass diffusion flux of O2 across a surface moving with mass average velocity is -0.0384p kg/m**2.s  \n",
        " Molar diffusion flux across a surface moving with molar average velociy for H2 is 0.24C k.mol/m**2.s \n",
        " Molar diffusion flux across a surface moving with molar average velociy for O2 is -0.24C k.mol/m**2.s\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.3  Page No : 557"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "t = 0.001;\t\t\t#Thickness of the membrane in m\n",
      "CA1 = 0.02;\t\t\t#Concentration of helium in the membrane at inner surface in k.mol/m**3\n",
      "CA2 = 0.005;\t\t\t#Concentration of helium in the membrane at outer surface in k.mol/m**3\n",
      "DAB = 10**-9;\t\t\t#Binary diffusion coefficient in m**2/s\n",
      "\n",
      "# Calculations\n",
      "NAx = ((DAB*(CA1-CA2))/t)/10**-9;\t\t\t#Diffusion flux of helium through the plastic in k.mol/sm**2 *10**-9\n",
      "\n",
      "# Results\n",
      "print 'Diffusion flux of helium through the plastic is %3.0f*10**-9 k.mol/sm**2'%(NAx)\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diffusion flux of helium through the plastic is  15*10**-9 k.mol/sm**2\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.4  Page No : 557"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "T = 273+25;\t\t\t#Temperature of Helium gas in K\n",
      "p = 4;\t\t\t#Pressure of helium gas in bar\n",
      "Di = 0.1;\t\t\t#Inner diamter of wall in m\n",
      "Do = 0.003;\t\t\t#Outer diamter of wall in m\n",
      "DAB = (0.4*10**-13);\t\t\t#Binary diffusion coefficient in m**2/s\n",
      "S = (0.45*10**-3);\t\t\t#S value for differentiation\n",
      "\n",
      "# Calculations\n",
      "A = (3.14*Di**2);\t\t\t#Area in m**2\n",
      "V = (3.14*Di**3)/6;\t\t\t#Volume in m**3\n",
      "R = 0.08316\t\t\t#Gas consmath.tant in m**3 bar/kmol.K\n",
      "d = ((-6*R*T*DAB*S*p)/(Do*Di))/10**-11;\t\t\t#Decrease of pressure with time in bar/s*10**-11\n",
      "\n",
      "# Results\n",
      "print 'Initial rate of leakage for the system is provided by the decrease of pressure \\\n",
      " with time which is %3.2f*10**-11 bar/s'%(d)\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Initial rate of leakage for the system is provided by the decrease of pressure  with time which is -3.57*10**-11 bar/s\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.5  Page No : 558"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "# Variables\n",
      "po2 = 2;\t\t\t#Pressure of O2 in bar\n",
      "Di = 0.025;\t\t\t#inside diamter of the pipe in m\n",
      "L = 0.0025;\t\t\t#Wall thickness in m\n",
      "a = (0.21*10**-2);\t\t\t#Diffusivity of O2 in m**2/s\n",
      "S = (3.12*10**-3);\t\t\t#Solubility of O2 in k.mol/m**3.bar\n",
      "DAB = (0.21*10**-9);\t\t\t#Binary diffusion coefficient in m**2/s\n",
      "\n",
      "# Calculations\n",
      "CAi = (S*po2);\t\t\t#Concentration of O2 on inside surface in kmol/m**3\n",
      "RmA = ((math.log((Di+(2*L))/Di))/(2*3.14*DAB));\t\t\t#Diffusion resismath.tance in sm**2\n",
      "Loss = (CAi/RmA)/10**-11;\t\t\t#Loss of O2 by diffusion per meter length of pipe *10**-11\n",
      "\n",
      "# Results\n",
      "print 'Loss of O2 by diffusion per meter length of pipe is %3.2f*10**-11 kmol/s'%(Loss)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Loss of O2 by diffusion per meter length of pipe is 4.51*10**-11 kmol/s\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.6  Page No : 560"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "p = 1;\t\t\t#Pressure of system in atm\n",
      "T = 25+273;\t\t\t#Temperature of system in K\n",
      "pco2 = (190./760);\t\t\t#Partial pressure of CO2 at one end in atm\n",
      "pco2o = (95./760);\t\t\t#Partial pressure of CO2 at other end in atm\n",
      "DAB = (0.16*10**-4);\t\t\t#Binary diffusion coefficient in m**2/s from Table 13.3\n",
      "R = 0.08205\t\t\t#Gas constant in m**3 atm/kmol.K\n",
      "\n",
      "# Calculations\n",
      "NAx = (DAB*(pco2-pco2o))/(R*T*p);\t\t\t#Equimolar counter diffusion in kmol/m**2s\n",
      "M = (NAx*3.14*(0.05**2/4)*3600);\t\t\t#Mass transfer rate in kmol/h\n",
      "MCO2 = (M*44)/10**-5;\t\t\t#Mass flow rate of CO2 in kg/h *10**-5\n",
      "Mair = (29*-M)/10**-5;\t\t\t#Mass flow rate of air in kg/h *10**-5\n",
      "\n",
      "# Results\n",
      "print 'Mass transfer rate of CO2 is %3.2f*10**-5 kg/h Mass transfer rate of air is %3.2f*10**-5 kg/h'%(MCO2,Mair)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass transfer rate of CO2 is 2.54*10**-5 kg/h Mass transfer rate of air is -1.68*10**-5 kg/h\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.7  Page No : 563"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "T = 27+273;\t\t\t#Temperature of water in K\n",
      "D = 0.02;\t\t\t#Diameter of the tube in m\n",
      "L = 0.4;\t\t\t#Length of the tube in m\n",
      "DAB = (0.26*10**-4);\t\t\t#Diffusion coefficient in m**2/s\n",
      "\n",
      "# Calculations\n",
      "p = 1.0132;\t\t\t#Atmospheric pressure in bar\n",
      "pA1 = 0.03531;\t\t\t#Vapour pressure in bar\n",
      "m = ((p*10**5*3.14*(D/2)**2*18*DAB)/(8316*T*L))*(1000*3600)*math.log(p/(p-pA1));\t\t\t#Diffusion rate of water in gram per hour\n",
      "\n",
      "# Results\n",
      "print 'Diffusion rate of water is %3.4f gram per hour'%(m)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diffusion rate of water is 0.0019 gram per hour\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.8  Page No : 564"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "T = 25+273;\t\t\t#Temperature of water in K\n",
      "D = 0.02;\t\t\t#Diameter of the tube in m\n",
      "L = 0.08;\t\t\t#Length of the tube in m\n",
      "m = (8.54*10**-4);\t\t\t#Diffusion coefficient in kg/h\n",
      "\n",
      "# Calculations\n",
      "p = 1.0132;\t\t\t#Atmospheric pressure in bar\n",
      "pA1 = 0.03165;\t\t\t#Vapour pressure in bar\n",
      "DAB = (((m/3600)*8316*T*L)/(p*10**5*3.14*(D/2)**2*18*math.log(p/(p-pA1))*10**2))/10**-4;\t\t\t#Diffusion coefficient of water in m**2/s *10**-4\n",
      "\n",
      "# Calculations\n",
      "print 'Diffusion coefficient of water is %3.3f*10**-4 m**2/s'%(DAB)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diffusion coefficient of water is 0.259*10**-4 m**2/s\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.9  Page No : 569"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Variables\n",
      "CAs = 0.02;\t\t\t#Carbon mole fraction\n",
      "CAo = 0.004;\t\t#Content of steel\n",
      "CA = 0.012;\t\t\t#Percet of depth\n",
      "d = 0.001;\t\t\t#Depth in m\n",
      "H = (6*10**-10);\t#Diffusivity of carbon in m**2/s\n",
      "\n",
      "# Calculations\n",
      "X = (CA-CAs)/(CAo-CAs);\t\t\t#Calculation for erf function\n",
      "n = 0.48;           \t\t\t#erf(n) = 0.5; n = 0.48\n",
      "t = ((d/(n*2.))**2/(3600.*H))*3600;\t\t\t#Time required to elevate the carbon content of steel in s\n",
      "\n",
      "\n",
      "# Results\n",
      "print 'Time required to elevate the carbon content of steel is %3.2f s'%(t)\n",
      "\n",
      "\n",
      "# note : rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Time required to elevate the carbon content of steel is 1808.45 s\n"
       ]
      }
     ],
     "prompt_number": 1
    }
   ],
   "metadata": {}
  }
 ]
}