path: root/Diffusion:_Mass_Transfer_In_Fluid_Systems_by_E._L._Cussler/ch18.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 18 : Membranes"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.1.1 pg : 519"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\n",
      "#initialization of variables\n",
      "d = 240.*10**-4 \t# cm\n",
      "D = 2.1*10**-5 \t# cm**2/sec\n",
      "v = 10.      \t# cm/sec\n",
      "Nu = 0.01 \t# cm**2/sec\n",
      "E = 0.5\n",
      "ka1 = 0.09 \t# sec**-1\n",
      "\t\n",
      "#Calculations\n",
      "k = 0.8*(D/d)*((d*v/Nu)**0.47)*((Nu/D)**0.33)\n",
      "a = 4*(1-E)/d \t# cm**2/cm**3\n",
      "ka2 = k*a\n",
      "ratio = ka2/ka1\n",
      "\t\n",
      "#Results\n",
      "print \"The rapidness is roughly %d times that of sparger\"%(ratio)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The rapidness is roughly 22 times that of sparger\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.2.1 pg : 524"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialization of variables\n",
      "p1 = 10.**-10 \t# cm**3(stp)cm/cm**2-sec-cm-Hg\n",
      "c = 1./(22.4*10**3) \t# mol at stp /cc\n",
      "P = p1*c \t# for proper units\n",
      "R = 6240. \t# cmHg cm**3 \t#mol-K (gas consmath.tant)\n",
      "T = 298. \t# Kelvin\n",
      "\t\n",
      "#Calculations\n",
      "DH = P*R*T \t# Permeability in x*10**-9 cm**2/sec\n",
      "\t\n",
      "#Results\n",
      "print \"The permeability is %.1e cm**2/sec\"%(DH)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The permeability is 8.3e-09 cm**2/sec\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.2.2 pg : 525"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\n",
      "#initialization of variables\n",
      "P = 1.*10**-4 \t# membrane permeability in cm**2/sec\n",
      "l = 2.3*10**-4 \t# membrane thickness in cm\n",
      "d = 320.*10**-4 \t# fiber dia in cm\n",
      "E = 0.5 \t# void fraction\n",
      "c0 = 1. \t# initial concentration\n",
      "c = 0.1\t    # final concentration\n",
      "\t\n",
      "#Calculations\n",
      "a = 4*(1-E)/d \t# surface area per module volume in cm**2/cm**3\n",
      "t = (math.log(c0/c))*(l/P)/a \t# t = z/v in seconds , time gas spends in the module in sec\n",
      "\t\n",
      "#Results\n",
      "print \"The gas spends %.2f sec in the module\"%(t)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The gas spends 0.08 sec in the module\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.3.1 pg : 532"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\n",
      "#initialization of variables\n",
      "R = 0.082 \t# litre-atm/mol-K\n",
      "T = 283. \t# Kelvin\n",
      "V2 = 0.018 \t# litre/mol\n",
      "\n",
      "#For first solution contents are sucrose and water\n",
      "w1 = 0.01 \t# gm of sucrose\n",
      "MW1 = 342. \t# MW of sucrose\n",
      "w2 = 0.09 \t# gm of water\n",
      "MW2 = 18. \t# MW of water\n",
      "n1 = 1.  \t# no of particles sucrose divides into in water\n",
      "\t\n",
      "#Calculations\n",
      "x1juice = (n1*w1/MW1)/((n1*w1/MW1)+(w2/MW2))\t# Mole fracion of sucrose\n",
      "#For second solution , contents are NaCl and water\n",
      "w1 = 35. \t# gm of NaCl\n",
      "MW1 = 58.5 \t# MW of Nacl\n",
      "w2 = 100. \t# gm of water\n",
      "MW2 = 18. \t# MW of water\n",
      "n1 = 2. \t# no of particles sucrose divides into in water\n",
      "\t\n",
      "#Calculations\n",
      "x1brine = (n1*w1/MW1)/((n1*w1/MW1)+(w2/MW2))\t# Mole fracion of sucrose\n",
      "#Calculation of difference in Osmotic pressure\n",
      "DeltaPi = (R*T/V2)*math.log((1-x1juice)/(1-x1brine))\t# atm\n",
      "\t\n",
      "#Results\n",
      "print \"The osmotic pressure difference is %.f atm\"%(DeltaPi)\n",
      "\t#answer wrong in textbook\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The osmotic pressure difference is 244 atm\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.3.2 pg : 533"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\t\n",
      "#initialization of variables\n",
      "D1=0.0035\n",
      "l = 2.59 \t#cm\n",
      "t = 1.62  \t#hr\n",
      "C1 = 0.03 \t#mol/l\n",
      "T1 = 298. \t#K\n",
      "R = 0.0821 \t#arm/mol K\n",
      "D2 = 0.005\n",
      "t2 = 0.49 \t#hr\n",
      "Ps = 733.   \t#mm of Hg\n",
      "P = 760. \t#mm of Hg\n",
      "\t\n",
      "#Calculations\n",
      "Lps=D1*l/(t*3600) /(C1*R*T1)\n",
      "Lp=(D2*l/(t2*3600) + Lps*(C1*R*T1))/(Ps/P)\n",
      "Lp=2.4*10**-6\n",
      "sig=Lps/Lp\n",
      "sig2=0.95\n",
      "\t\n",
      "#Results\n",
      "print \"Transport coefficient for phase 1 = %.2f\"%(sig)\n",
      "print \" Transport coefficient for phase 2 = %.2f\"%(sig2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Transport coefficient for phase 1 = 0.88\n",
        " Transport coefficient for phase 2 = 0.95\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.4.1 pg : 538"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\t\n",
      "#initialization of variables\n",
      "D1 = 3.0*10**-7 \t# cm**2/sec\n",
      "H1 = 0.18 \t# mol/cc-atm\n",
      "D2 = 1.4*10**-6 \t# cm**2/sec\n",
      "H2 = 2.2*10**-3 \t# mol/cc-atm\n",
      "H11 = 13. \t# atm-cc/mol\n",
      "H21 = 0.6 \t# atm-cc/mol\n",
      "\t\n",
      "#Calculations\n",
      "Beta = (D1*H1/(D2*H2))*(H11/H21)\t# Membrane selectivity\n",
      "\t\n",
      "#Results\n",
      "print \"The membrane selectivity is %.f\"%(Beta)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The membrane selectivity is 380\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18.5.2 pg : 544"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Initialization of variables\n",
      "D = 2.*10**-5 \t# cm**2/sec\n",
      "l = 32.*10**-4 \t# cm\n",
      "c = 6.8*10**-6 \t# mol/cc\n",
      "C10 = 10.**-4 \t# mol/cc\n",
      "\n",
      "def Totalflux(H,K):\n",
      "    return (D*H*C10/l)+((D*H*K*c*C10)/(l*(1+(H*K*C10))))\n",
      "\n",
      "\n",
      "#For Lithium Chloride\n",
      "H1 = 4.5*10**-4 \t#Partition coefficient \n",
      "K1 = 2.6*10**5 \t# cc/mol association consmath.tant\n",
      "j1 = (Totalflux(H1,K1))        \t# TOtal flux in x*10**-10 mol/cm**2-sec\n",
      "print \"The total flux for Lithium Chloride is %.1e mol/cm**2-sec\"%(j1)\n",
      "\n",
      "#For Sodium Chloride\n",
      "H2 = 3.4*10**-4 \t#Partition coefficient \n",
      "K2 = 1.3*10**7 \t# cc/mol association consmath.tant\n",
      "j2 = (Totalflux(H2,K2)) \t# TOtal flux in x*10**-10 mol/cm**2-sec\n",
      "print \"The total flux for Sodium Chloride is %.2e  mol/cm**2-sec\"%(j2)\n",
      "\n",
      "#For potassium Chloride\n",
      "H3 = 3.8*10**-4 \t#Partition coefficient \n",
      "K3 = 4.7*10**9 \t# cc/mol association consmath.tant\n",
      "j3 = (Totalflux(H3,K3)) \t# TOtal flux in x*10**-10 mol/cm**2-sec\n",
      "print \"The total flux for Potassium Chloride is %.2e  mol/cm**2-sec\"%(j3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total flux for Lithium Chloride is 7.7e-10 mol/cm**2-sec\n",
        "The total flux for Sodium Chloride is 1.32e-08  mol/cm**2-sec\n",
        "The total flux for Potassium Chloride is 4.25e-08  mol/cm**2-sec\n"
       ]
      }
     ],
     "prompt_number": 12
    }
   ],
   "metadata": {}
  }
 ]
}