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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 7 : Fugacity Ideal Solutions Activity Activity Coefficient"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.1  Page: 134"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "\n",
      "T = 220+459.67          #[R] Temperature in Rankine\n",
      "P = 500.                #[psia] Pressure\n",
      "R = 10.73               #[(psi*ft**(3)/(lbmol*R))] Gas consmath.tant\n",
      "\n",
      "\n",
      "a = 4.256           #[ft**(3)/lbmol]\n",
      "\n",
      "def f6(p): \n",
      "\t return a*p**(0)\n",
      "\n",
      "I =  quad(f6,0,P)[0]\n",
      "\n",
      "\n",
      "f = P*math.exp((-1/(R*T))*I)            #[psia]\n",
      "\n",
      "print \"Fugacity of propane gas at the given condition is %.0f psia\"%(round(f,1))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Fugacity of propane gas at the given condition is 374 psia\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.2  Page: 138\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "T = 100. + 460          #[R] Temperature of the system in Rankine\n",
      "P = 1.                  # [psia]\n",
      "R = 10.73               #[(psi*ft**(3)/(lbmol*R))] Gas consmath.tant\n",
      "\n",
      "v = 0.016136*18         #[ft**(3)/lbmol]\n",
      "z = round((P*v)/(R*T),5)\n",
      "\n",
      "a = int(((R*T)/P))*(1-z)     #[ft**(3)/lbmol]\n",
      "\n",
      "print \" Compresssibility factor the liquid water at the given condition is %.5f \"%(z)\n",
      "print \"Volume residual for the liquid water at the given condition is     %0.1f cubic feet/lbmol\"%(a)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Compresssibility factor the liquid water at the given condition is 0.00005 \n",
        "Volume residual for the liquid water at the given condition is     6007.7 cubic feet/lbmol\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.3  Page: 138\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "T = 100+460.            #[R] Temperature\n",
      "P = 1000.               #[psia] Pressure\n",
      "R = 10.73               #[(psi*ft**(3)/(lbmol*R))] Gas consmath.tant\n",
      "\n",
      "f_b = 0.95              #[psia]\n",
      "f_c = f_b               #[psia]\n",
      "v = 0.016136*18         #[ft**(3)/lbmol]\n",
      "\n",
      "P_d = 1000.             #[psia]\n",
      "P_c = 1.                #[psia]\n",
      "\n",
      "def f4(p): \n",
      "\t return p**(0)\n",
      "\n",
      "f_d = f_c*math.exp((v/(R*T))* (quad(f4,P_c,P_d))[0])\n",
      "\n",
      "print \"Fugacity of the pure liquid water at the given condition is %0.1f psia\"%(f_d)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Fugacity of the pure liquid water at the given condition is 1.0 psia\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.4  Page: 145\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "T = 78.15           #[C]\n",
      "P = 1.0             #[atm]\n",
      "p_a_0 = 0.993       #[atm] Pure ethanol vapor pressure at 78.15C\n",
      "p_b_0 = 0.434       #[atm] Pure water vapor pressure at 78.15C\n",
      "\n",
      "x_a = 0.8943        # Amount of ethanol in the liquid phase \n",
      "x_b = 0.1057        # Amount of water in liquid phase \n",
      "\n",
      "y_a = x_a           # Amount of ethanol in vapor phase \n",
      "y_b = x_b           # Amount of water in the vapor phase \n",
      "\n",
      "\n",
      "\n",
      "Y_a_1 = 1.0\n",
      "Y_b_1 = 1.0\n",
      "\n",
      "Y_a_2 = ((y_a*P)/(x_a*p_a_0))\n",
      "Y_b_2 = ((y_b*P)/(x_b*p_b_0))\n",
      "\n",
      "f_a_1 = (y_a*Y_a_1*P)           #[atm]\n",
      "f_b_1 = (y_b*Y_b_1*P)           #[atm]\n",
      "f_a_2 = f_a_1                   #[atm]\n",
      "f_b_2 = f_b_1                   #[atm]\n",
      "\n",
      "f_a_1_0 = P                     #[atm]\n",
      "f_b_1_0 = P                     #[atm]\n",
      "\n",
      "f_a_2_0 = p_a_0                 #[atm]\n",
      "f_b_2_0 = p_b_0                 #[atm]\n",
      "\n",
      "print \" The results are summarized in the following table: \\n\\tPhase\\t\\t\\t\\t Etahnol(i=a)\\t\\t\\t\\t Water,i=b\"\n",
      "print \" \\tVAPOR PHASE 1\"\n",
      "print \"  \\t  f_i_1 atm   \\t\\t\\t %.4f \\t\\t\\t\\t %.4f\"%(f_a_1,f_b_1)\n",
      "print \"  \\t  f_i_1_0 atm \\t\\t\\t %.4f \\t\\t\\t\\t %.4f\"%(f_a_1,f_b_1)\n",
      "print \"  \\t  Y_i_1 assumed  \\t\\t %f \\t\\t\\t\\t %f\"%(Y_a_1,Y_b_1)\n",
      "print \" \\tLIQUID PHASE 2\"\n",
      "print \"  \\t  f_i_2 atm   \\t\\t\\t %.4f \\t\\t\\t\\t %.4f\"%(f_a_2,f_b_2)\n",
      "print \"  \\t  f_i_2_0 atm \\t\\t\\t %.4f \\t\\t\\t\\t %.4f\"%(f_a_2,f_b_2)\n",
      "print \"  \\t  Y_i_2assumed  \\t\\t %.4f \\t\\t\\t\\t %.4f\"%(Y_a_2,Y_b_2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The results are summarized in the following table: \n",
        "\tPhase\t\t\t\t Etahnol(i=a)\t\t\t\t Water,i=b\n",
        " \tVAPOR PHASE 1\n",
        "  \t  f_i_1 atm   \t\t\t 0.8943 \t\t\t\t 0.1057\n",
        "  \t  f_i_1_0 atm \t\t\t 0.8943 \t\t\t\t 0.1057\n",
        "  \t  Y_i_1 assumed  \t\t 1.000000 \t\t\t\t 1.000000\n",
        " \tLIQUID PHASE 2\n",
        "  \t  f_i_2 atm   \t\t\t 0.8943 \t\t\t\t 0.1057\n",
        "  \t  f_i_2_0 atm \t\t\t 0.8943 \t\t\t\t 0.1057\n",
        "  \t  Y_i_2assumed  \t\t 1.0070 \t\t\t\t 2.3041\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.5  Page: 149\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "T = 220+460.            #[R] Temperature in rankine\n",
      "P = 1000.               #[psia] Pressure\n",
      "y_methane = 0.784       # Mol fraction of methane in the given mixture\n",
      "y_butane = (1-y_methane)       # Mol fraction of n-bumath.tane in the given mixture\n",
      "R = 10.73               #[(psia*ft**(3)/(lbmol*R))] gas consmath.tant\n",
      "\n",
      "\n",
      "Im = 290.               #[ft**(3)/lbmol]\n",
      "\n",
      "Jm = math.exp((-1/(R*T))*Im)\n",
      "\n",
      "f_methane = Jm*P*y_methane          #[psia] fugacity of methane\n",
      "\n",
      "Ib = 5859.                          #[ft**(3)/lbmol]\n",
      "Jb = math.exp((-1/(R*T))*Ib)\n",
      "f_butane = Jb*P*y_butane          #[psia] fugacity of bumath.tane\n",
      "\n",
      "print \" Fugacity of the methane in the gaseous mixture is %0.0f psia\"%(f_methane)\n",
      "print \" Fugacity of the butane in the gaseous mixture is  %0.1f psia\"%(f_butane)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Fugacity of the methane in the gaseous mixture is 753 psia\n",
        " Fugacity of the butane in the gaseous mixture is  96.8 psia\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.6  Page: 153\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math \n",
      "\n",
      "T = 220+460.            #[R] Temperature in rankine\n",
      "P = 1000.               #[psia] Pressure\n",
      "x_methane = 0.784       # Mol fraction of methane in the given mixture\n",
      "x_bumath_tane = (1-x_methane)       # Mol fraction of n-bumath_tane in the given mixture\n",
      "\n",
      "v_i_into_Y_i = 0.961\n",
      "phi_cap_i = 0.961\n",
      "\n",
      "v_i = 0.954\n",
      "phi_i = v_i\n",
      "Y_i = phi_cap_i/v_i\n",
      "\n",
      "print \" The value of v_i is %f\"%(v_i)\n",
      "print \" The value of Y_i is %f\"%(Y_i)\n",
      "print \" The value of phi_cap_i is %f\"%(phi_cap_i)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The value of v_i is 0.954000\n",
        " The value of Y_i is 1.007338\n",
        " The value of phi_cap_i is 0.961000\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 7.7  Page: 154\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "T_r = 0.889\n",
      "P_r = 1.815\n",
      "\n",
      "f_f = -0.48553\n",
      "\n",
      "v = math.exp((P_r/T_r)*f_f)\n",
      "phi = v\n",
      "\n",
      "print \" The value of v=phi for n-bumath.tane at given condition is %f\"%(v)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The value of v=phi for n-bumath.tane at given condition is 0.371106\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}