path: root/Chemical_Engineering_Thermodynamics/ch16_1.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 16 : Other Phase Equilibria"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.1  Page Number : 564"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of solubility\n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\t\t\t\n",
      "T = 0 + 273.15;\t\t\t#[K] - Temperature\n",
      "P = 20*10**(5);\t\t\t#[Pa] - Pressure\n",
      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
      "\n",
      "\t\t\t#componenet 1 : methane (1)\n",
      "\t\t\t#componenet 2 : methanol (2)\n",
      "\n",
      "H_constant = 1022;\t\t\t#[bar] - Henry's law constant\n",
      "H_constant = H_constant*10**(5);\t\t\t#[Pa]\n",
      "\n",
      "\t\t\t# The second virial coefficients are\n",
      "B_11 = -53.9;\t\t\t#[cm**(3)/mol]\n",
      "B_11 = B_11*10**(-6);\t\t\t#[m**(3)/mol]\n",
      "B_12 = -166;\t\t\t#[cm**(3)/mol]\n",
      "B_12 = B_12*10**(-6);\t\t\t#[m**(3)/mol]\n",
      "B_22 = -4068;\t\t\t#[cm**(3)/mol]\n",
      "B_22 = B_22*10**(-6);\t\t\t#[m**(3)/mol]\n",
      "\n",
      "den_meth = 0.8102;\t\t\t#[g/cm**(3)] - Density of methanol at 0 C\n",
      "Mol_wt_meth = 32.04;\t\t\t# Molecular weight of methanol\n",
      "P_2_sat = 0.0401;\t\t\t#[bar] - Vapour pressure of methanol at 0 C\n",
      "\n",
      "# Calculations\n",
      "\t\t\t#The molar volume of methanol can be calculated as\n",
      "V_2_liq = (1/(den_meth/Mol_wt_meth))*10**(-6);\t\t\t#[m**(3)/mol]\n",
      "\n",
      "\t\t\t#The phase equilibrium equation of the components at high pressure\n",
      "\t\t\t#y1*phi_1*P = x_1*H_1\n",
      "\t\t\t#y2*phi_2*P = x_2*H_2\n",
      "\n",
      "\t\t\t#Since methane follows Henry's law therefore methanol follows the lewis-Rnadall rule\n",
      "\t\t\t#f_2 is the fugacity of the compressed liquid which is calculated using\n",
      "\t\t\t#f_2 = f_2_sat*math.exp[V_2_liq*(P - P_sat_2)/(R*T)]\n",
      "\t\t\t#where f_2_sat can be calculated using virial equation \n",
      "\t\t\t# math.log(phi_2_sat) = math.log(f_2_sat/P_2_sat) = (B_22*P_2_sat)/(R*T)\n",
      "\n",
      "f_2_sat = P_2_sat*math.exp((B_22*P_2_sat*10**(5))/(R*T));\t\t\t#[bar]\n",
      "\n",
      "\t\t\t#Putting the value of 'f_2_sat' in the math.expression of f_2 , we get\n",
      "f_2 = f_2_sat*math.exp(V_2_liq*(P - P_2_sat*10**(5))/(R*T));\t\t\t#[bar]\n",
      "\n",
      "\t\t\t#Now let us calculate the fugacity coefficients of the species in the vapour mixture\n",
      "del_12 = 2*B_12 - B_11 - B_22;\t\t\t#[m**(3)/mol]\n",
      "\n",
      "\t\t\t#math.log(phi_1) = (P/(R*T))*(B_11 + y2**(2)*del_12)\n",
      "\t\t\t#math.log(phi_2) = (P/(R*T))*(B_22 + y1**(2)*del_12)\n",
      "\n",
      "\n",
      "\t\t\t#The calculation procedure is to assume a value of y1, calculate 'phi_1' and 'phi_2' and calculate 'x_1' and 'x_2' from the phase equilibrium equations and see whether x_1 + x_2 = 1,if not then another value of y1 is assumed.\n",
      "\n",
      "y2 = 0.1;\n",
      "error=10;\n",
      "\n",
      "while(error>0.001):\n",
      "    y1=1-y2;\n",
      "    phi_1 = math.exp((P/(R*T))*((B_11 + y2**(2)*del_12)));\n",
      "    phi_2 = math.exp((P/(R*T))*((B_22 + y1**(2)*del_12)));\n",
      "    x_1 = (y1*phi_1*P)/H_constant;\n",
      "    x_2 = (y2*phi_2*P)/(f_2*10**(5));\n",
      "    x = x_1 + x_2;\n",
      "    error=abs(1-x);\n",
      "    y2=y2 - 0.000001;\n",
      "\n",
      "# Results\n",
      "print \" The solubility of methane in methanol is given by x1 = %f\"%(x_1);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The solubility of methane in methanol is given by x1 = 0.018614\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.2  Page Number : 566"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of solubility\n",
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "# Variables\n",
      "x_C2H6_1 = 0.33*10**(-4);\t\t\t# Solubility of ethane in water at 25 C and 1 bar\n",
      "\n",
      "\t\t\t#componenet 1 : ethane (1)\n",
      "\t\t\t#componenet 2 : water (2)\n",
      "\n",
      "\t\t\t# Z = 1 - 7.63*10**(3)*P - 7.22*10**(-5)*P**(2)\n",
      "\n",
      "\t\t\t#The phase equilibrium equation of ethane is\n",
      "\t\t\t#f_1_V = x_1*H_1\n",
      "\t\t\t#since vapour is pure gas,  f_1_V = x_1*H_1   or, phi_1*P = x_1*H_1,  where 'phi_1' is fugacity coefficient of pure ethane\n",
      "\t\t\t# math.log(phi) = integral('Z-1)/P) from limit '0' to 'P'\n",
      "\n",
      "P1 = 0;\n",
      "P2 = 1;\n",
      "P3 = 35;\n",
      "\n",
      "# Calculations\n",
      "def f51(P): \n",
      "\t return (1-7.63*10**(-3)*P-7.22*10**(-5)*P**(2)-1)/P\n",
      "\n",
      "intgral =  quad(f51,P1,P2)[0]\n",
      "\n",
      "phi_1_1 = math.exp(intgral);\t\t\t# - Fugacity coefficient of ethane at 1 bar\n",
      "f_1_1 = phi_1_1*P2;\t\t\t#[bar] - Fugacity of ethane at 1 bar\n",
      "\n",
      "\t\t\t#Similarly\n",
      "\n",
      "def f52(P): \n",
      "\t return (1-7.63*10**(-3)*P-7.22*10**(-5)*P**(2)-1)/P\n",
      "\n",
      "intgral_1 =  quad(f52,P1,P3)[0]\n",
      "\n",
      "phi_1_35 = math.exp(intgral_1);\t\t\t# Fugacity coefficient of ethane at 35 bar\n",
      "f_1_35 = phi_1_35*P3;\t\t\t#[bar] - Fugacity of ethane at 35 bar\n",
      "\n",
      "\t\t\t# At ethane pressure of 1 bar ,  x_C2H6_1*H_1 = phi_1_1\n",
      "H_1 = phi_1_1/x_C2H6_1;\t\t\t#[bar] - Henry's constant\n",
      "\n",
      "\t\t\t# At ethane pressure of 35 bar ,  x_C2H6_35*H_1 = phi_1_35\n",
      "x_C2H6_35 = f_1_35/H_1;\t\t\t# Solubility of ethane at 35 bar pressure\n",
      "\n",
      "# Results\n",
      "print \"The solubility of ethane at 35 bar is given by   x_C2H6 = %e \"%(x_C2H6_35);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The solubility of ethane at 35 bar is given by   x_C2H6 = 8.525648e-04 \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.4  Page Number : 571"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of composition\n",
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "T = 200;\t\t\t#[K]\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "\t\t\t# G_E = A*x_1*x_2\n",
      "A = 4000;\t\t\t#[J/mol]\n",
      "x_1 = 0.6;\t\t\t# Mle fraction of feed composition\n",
      "\n",
      "# Calculations and Results\n",
      "\t\t\t# Since A is given to be independent of temperature\n",
      "UCST = A/(2*R);\t\t\t#[K] - Upper critical solution temperature\n",
      "print \" The UCST of the system is %f K\"%(UCST);\n",
      "\n",
      "\t\t\t# Since the given temperature is less than UCST therefore two phase can get formed at the given temperature.\n",
      "\n",
      "\t\t\t# x1_alpha = 1 - x1_beta\n",
      "\t\t\t# We know that,  x1_alpha*Y_1_alpha = x2_alpha*Y_2_alpha\n",
      "\t\t\t# x1_alpha*math.exp[(A/(R*T))*(x2_alpha)**(2)] = (1 - x1_alpha)*math.exp[(A/(R*T))*(x1_alpha)**(2)]\n",
      "\t\t\t# where use has beeen made of the fact that x1_alpha = 1 - x1_beta and x2_beta = 1 - x1_beta = x1_alpha .Taking math.logarithm of both side we get\n",
      "\t\t\t# math.log(x1_alpha) + (A/(R*T))*(1 - x1_alpha)**(2) = math.log(1 - x1_alpha) + (A/(R*T))*x1_alpha**(2)\n",
      "\t\t\t# math.log(x1_alpha/(1-x1_alpha)) = (A/(R*T))*(2*x1_alpha - 1)\n",
      "\n",
      "def f(x1_alpha): \n",
      "\t return  math.log(x1_alpha/(1-x1_alpha)) - (A/(R*T))*(2*x1_alpha - 1)\n",
      "x1_alpha = fsolve(f,0.1)\n",
      "x1_beta = fsolve(f,0.9)\n",
      "\t\t\t# Because of symmetry 1 - x1_beta = x1_alpha\n",
      "\n",
      "\t\t\t# It can be seen that the equation,  math.log(x1/(1-x1)) = (A/(R*T))*(2*x1 - 1) has two roots.\n",
      "\t\t\t# The two roots acn be determined by taking different values \n",
      "\t\t\t# Starting with x1 = 0.1, we get x1 = 0.169 as the solution and starting with x1 = 0.9,we get x1 = 0.831 as the solution.\n",
      "\t\t\t# Thus x1 = 0.169 is the composition of phase alpha and x1 = 0.831 is of phase beta\n",
      "print \" The composition of two liquid phases in equilibrium is given by x1_alpha = %f and x1_beta = %f\"%(x1_alpha,x1_beta);\n",
      "\n",
      "\t\t\t# From the equilibrium data it is seen that if the feed has composition x1 less than 0.169 or more than 0.831 the liquid mixture is of single phase\n",
      "\t\t\t# whereas if the overall (feed) composition is between 0.169 and 0.831 two phases shall be formed.\n",
      "\t\t\t# The amounts of phases can also be calculated. The feed composition is given to be z1 = 0.6\n",
      "z1 = 0.6;\n",
      "\t\t\t# z1 = x1_alpha*alpha + x1_beta*beta\n",
      "\t\t\t# beta = 1 - alpha\n",
      "alpha = (z1-x1_beta)/(x1_alpha-x1_beta);\t\t\t#[mol]\n",
      "Beta = 1 - alpha;\t\t\t#[mol]\n",
      "print \" The relative amount of phases is given by alpha = %f mol and beta = %f mol\"%(alpha,Beta);\n",
      "\n",
      "\t\t\t# the relative amounts of the phases changes with the  feed composition \n",
      "\n",
      "\t\t\t#math.log(x1/(1-x1)) = (A/(R*T))*(2*x1 - 1)\n",
      "\t\t\t# If the above equation has two real roots of x1 (one for phase alpha and the other for phase beta) then two liquid phases get formed\n",
      "\t\t\t# and if it has no  real roots then a homogeneous liquid mixtures is obtained.\n",
      "\n",
      "print \" math.logx1/1-x1 = A/R*T*2*x1 - 1\";\n",
      "print \" If the above equation has two real roots of x1 one for phase alpha and the other for phase beta then two liquid phases get formed\";\n",
      "print \" and if it has no  real roots then a homogeneous liquid mixture is obtained\";\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The UCST of the system is 240.558095 K\n",
        " The composition of two liquid phases in equilibrium is given by x1_alpha = 0.169102 and x1_beta = 0.830898\n",
        " The relative amount of phases is given by alpha = 0.348896 mol and beta = 0.651104 mol\n",
        " math.logx1/1-x1 = A/R*T*2*x1 - 1\n",
        " If the above equation has two real roots of x1 one for phase alpha and the other for phase beta then two liquid phases get formed\n",
        " and if it has no  real roots then a homogeneous liquid mixture is obtained\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.5  Page Number : 573"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of equilibrium composition\n",
      "\n",
      "# Variables\n",
      "T = 300;\t\t\t#[K]\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "A = 7000;\t\t\t#[J/mol]\n",
      "\n",
      "\t\t\t# math.log(x_1/(1-x_1)) = (A/(R*T))*(2*x_1-1)\n",
      "\n",
      "# Calculations\n",
      "def f(x_1): \n",
      "\t return math.log(x_1/(1-x_1))-((A/(R*T))*(2*x_1-1))\n",
      "\n",
      "x1_alpha=fsolve(f,0.1)\n",
      "\n",
      "x1_beta=1-x1_alpha;\n",
      "\n",
      "# Results\n",
      "print \"The equilibrium compositin of the two liquid phase system is given by x1_alpha \\t = %f  x1_beta \\t = %f\"%(x1_alpha,x1_beta);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The equilibrium compositin of the two liquid phase system is given by x1_alpha \t = 0.091897  x1_beta \t = 0.908103\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.7  Page Number : 579"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of freezing point depression\n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
      "M_wt_meth = 32;\t\t\t# Molecular weight of methanol \n",
      "M_wt_water = 18;\t\t\t# Molecular weight of water \n",
      "m_meth = 0.01;\t\t\t#[g] - Mass of methanol added per cm**(3) of solution\n",
      "\n",
      "\t\t\t#Since the concentration of methanol is very small therefore we can assume that the density of solution = pure water\n",
      "den_sol = 1;\t\t\t#[g/cm**(3)]\n",
      "\n",
      "# Calculations\n",
      "\t\t\t#The mole fraction of solute is given by\n",
      "\t\t\t#x_2 = (moles of solute in cm**(3) of solution)/(moles of solute + moles of water) in 1 cm**(3) of solution\n",
      "x_2 = (m_meth/M_wt_meth)/((m_meth/M_wt_meth)+((1-m_meth)/M_wt_water));\n",
      "\n",
      "\t\t\t#We know that heat of fusion of water is\n",
      "H_fus = -80;\t\t\t#[cal/g] - Enthalpy change of fusion at 0 C\n",
      "H_fus = H_fus*4.186*M_wt_water;\t\t\t#[J/mol]\n",
      "\n",
      "\t\t\t#Therefore freezing point depression is given by\n",
      "\t\t\t# T - T_m = (R*(T**(2))*x_2)/H_fus\n",
      "T_f = 273.15;\t\t\t#[K] - Freezing point of water\n",
      "delta_T_f = (R*(T_f**(2))*x_2)/H_fus;\t\t\t#[K]\n",
      "\n",
      "# Results\n",
      "print \"The depression in freezing point is given by  delta_T = %f K\"%(delta_T_f);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The depression in freezing point is given by  delta_T = -0.581403 K\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.8  Page Number : 580"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of freezing point\n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "T_f = 273.15;\t\t\t#[K] - Freezing point of water\n",
      "m_water = 100;\t\t\t#[g] - Mass of water\n",
      "m_NaCl = 3.5;\t\t\t#[g] - Mass of NaCl\n",
      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
      "M_wt_NaCl = 58.5;\t\t\t# Molecular weight of NaCl\n",
      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
      "mol_NaCl = m_NaCl/M_wt_NaCl;\t\t\t#[mol] - Moles of NaCl\n",
      "\n",
      "H_fus = -80;\t\t\t#[cal/g] - Enthalpy change of fusion at 0 C\n",
      "H_fus = H_fus*4.186*M_wt_water;\t\t\t#[J/mol]\n",
      "\n",
      "# Calculations\n",
      "\t\t\t#Mole fraction of the solute (NaCl) is given by\n",
      "x_2 = mol_NaCl/(mol_NaCl+mol_water);\n",
      "\n",
      "\t\t\t#But NaCl is compietely ionized and thus each ion acts independently to lower the water mole fraction.\n",
      "x_2_act = 2*x_2;\t\t\t# Actual mole fraction\n",
      "\n",
      "\t\t\t#Now depression in freezing point is given by\n",
      "\t\t\t# T - T_m = (R*(T**(2))*x_2_act)/H_fus\n",
      "delta_T_f = (R*(T_f**(2))*x_2_act)/H_fus;\t\t\t#[C]\n",
      "\n",
      "\t\t\t#Thus freezing point of seawater = depression in freezing point\n",
      "\n",
      "# Results\n",
      "print \"The freezing point of seawater is  %f C\"%(delta_T_f);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The freezing point of seawater is  -2.192853 C\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.10  Page Number : 583"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determination of boiling point elevation \n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "T_b = 373.15;\t\t\t#[K] - Boiling point of water\n",
      "m_water = 100.;\t\t\t#[g] - Mass of water\n",
      "m_C12H22 = 5.;\t\t\t#[g] - Mass of glucise (C12H22)\n",
      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
      "M_wt_C12H22 = 342.30;\t\t\t# Molecular weight of C12H22\n",
      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
      "mol_C12H22 = m_C12H22/M_wt_C12H22;\t\t\t#[mol] - Moles of C12H22\n",
      "\n",
      "# Calculations\n",
      "H_vap = 540.;\t\t\t#[cal/g] - Enthalpy change of vaporisation\n",
      "H_vap = H_vap*4.186*M_wt_water;\t\t\t#[J/mol]\n",
      "\n",
      "\t\t\t#Mole fraction of the solute (C12H22) is given by\n",
      "x_2 = mol_C12H22/(mol_C12H22+mol_water);\n",
      "\n",
      "\t\t\t#The boiling point elevation is given by\n",
      "\t\t\t# T - T_b = (R*T_b**(2)*x_2**(2))/H_vap**(2)\n",
      "\n",
      "delta_T_b = (R*T_b**(2)*x_2)/(H_vap);\n",
      "\n",
      "# Results\n",
      "print \"The elevation in boiling point is given by  delta_T = %f C\"%(delta_T_b)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The elevation in boiling point is given by  delta_T = 0.074611 C\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.11  Page Number : 584"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of osmotic pressure\n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
      "T = 25 + 273.15;\t\t\t#[K] - Surrounding temperature \n",
      "den_water = 1000.;\t\t\t#[kg/m**(3)] - Density of water\n",
      "m_water = 100.;\t\t\t#[g] - Mass of water\n",
      "m_C12H22 = 5.;\t\t\t#[g] - Mass of glucise (C12H22)\n",
      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
      "M_wt_C12H22 = 342.30;\t\t\t# Molecular weight of C12H22\n",
      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
      "mol_C12H22 = m_C12H22/M_wt_C12H22;\t\t\t#[mol] - Moles of C12H22\n",
      "\n",
      "# Calculations\n",
      "\t\t\t#Mole fraction of the water is given by\n",
      "x_1 = mol_water/(mol_C12H22+mol_water);\n",
      "\n",
      "\t\t\t#Molar volume of water can be calculated as\n",
      "V_l_water = (1./den_water)*M_wt_water*10**(-3);\t\t\t#[m**(3)/mol]\n",
      "\n",
      "\t\t\t#The osmotic pressure is given by\n",
      "pi = -(R*T*math.log(x_1))/V_l_water;\t\t\t#[N/m**(2)]\n",
      "pi = pi*10**(-5);\t\t\t#[bar]\n",
      "\n",
      "# Results\n",
      "print \"The osmotic pressure of the mixture is %f bar\"%(pi);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The osmotic pressure of the mixture is 3.616073 bar\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.12  Page Number : 585"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Determination of pressure\n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "T = 25 + 273.15;\t\t\t#[K] - Surrounding temperature\n",
      "den_water = 1000.;\t\t\t#[kg/m**(3)] - Density of water\n",
      "m_water = 100.;\t\t\t#[g] - Mass of water\n",
      "m_NaCl = 3.5;\t\t\t#[g] - Mass of NaCl\n",
      "M_wt_water = 18.015;\t\t\t# Molecular weight of water \n",
      "M_wt_NaCl = 58.5;\t\t\t# Molecular weight of NaCl\n",
      "mol_water = m_water/M_wt_water;\t\t\t#[mol] - Moles of water\n",
      "mol_NaCl = m_NaCl/M_wt_NaCl;\t\t\t#[mol] - Moles of NaCl\n",
      "\n",
      "# Calculations\n",
      "H_fus = -80.;\t\t\t#[cal/g] - Enthalpy change of fusion at 0 C\n",
      "H_fus = H_fus*4.186*M_wt_water;\t\t\t#[J/mol]\n",
      "\n",
      "\t\t\t#Mole fraction of the solute (NaCl) is given by\n",
      "x_2 = mol_NaCl/(mol_NaCl+mol_water);\n",
      "\n",
      "\t\t\t#But NaCl is compietely ionized and thus each ion acts independently to lower the water mole fraction.\n",
      "x_2_act = 2*x_2;\t\t\t# Actual mole fraction\n",
      "\n",
      "x_1 = 1 - x_2_act;\n",
      "\n",
      "\t\t\t#Molar volume of water can be calculated as\n",
      "V_l_water = (1/den_water)*M_wt_water*10**(-3);\t\t\t#[m**(3)/mol]\n",
      "\n",
      "\t\t\t#The osmotic pressure is given by\n",
      "pi = -(R*T*math.log(x_1))/V_l_water;\t\t\t#[N/m**(2)]\n",
      "pi = pi*10**(-5);\t\t\t#[bar]\n",
      "\t\t\t#The minimum pressure to desalinate sea water is nothing but the osmotic pressure\n",
      "\n",
      "# Results\n",
      "print \"The minimum pressure to desalinate sea water is %f bar\"%(pi);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum pressure to desalinate sea water is 29.662232 bar\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.13  Page Number : 586"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determination of amount of precipitate\n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "T = 173.15;\t\t\t#[K] - Surrounding temperature\n",
      "P = 60;\t\t\t#[bar]\n",
      "P = P*10**(5);\t\t\t#[Pa] \n",
      "\n",
      "\t\t\t#componenet 1 : CO2 (1)\n",
      "\t\t\t#componenet 2 : H2 (2)\n",
      "P_1_sat = 0.1392;\t\t\t#[bar] - Vapour pressre of pure solid CO2\n",
      "P_1_sat = P_1_sat*10**(5);\t\t\t#[bar]\n",
      "V_s_1 = 27.6;\t\t\t#[cm**(3)/mol] - Molar volume of solid CO2\n",
      "V_s_1 = V_s_1*10**(-6);\t\t\t#[m**(3)/mol]\n",
      "n_1 = 0.01;\t\t\t#[mol] - Initial number of moles of CO2\n",
      "n_2 = 0.99;\t\t\t#[mol] - Initial number of moles of H2\n",
      "\n",
      "\t\t\t#Let us determine the fugacity of solid CO2 (1) at 60 C and 173.15 K\n",
      "\t\t\t# f_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T))\n",
      "\n",
      "# Calculations\n",
      "\t\t\t#Since vapour pressure of pure solid CO2 is very small, therefore\n",
      "f_1_sat = P_1_sat;\n",
      "f_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T));\n",
      "\n",
      "\t\t\t#Since gas phase is ideal therefore\n",
      "\t\t\t# y1*P = f_1\n",
      "y1 = f_1/P;\n",
      "\n",
      "\t\t\t#Number of moles of H2 in vapour phase at equilibrium remains the same as initial number of moles.\n",
      "\t\t\t#Number of moles of CO2 in vapour phase at equilibrium can be calculated as\n",
      "\t\t\t#y1 = (n_1_eq/(n_1_eq + n_2)). Therefore\n",
      "n_1_eq = n_2*y1/(1-y1);\n",
      "\n",
      "\t\t\t#Therefore moles of CO2 precipitated is\n",
      "n_ppt = n_1 - n_1_eq;\t\t\t#[mol]\n",
      "\n",
      "# Results\n",
      "print \"The moles of CO2 precipitated is %f mol\"%(n_ppt);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The moles of CO2 precipitated is 0.007417 mol\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 16.14  Page Number : 586"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Calculation of pressure\n",
      "\n",
      "# Variables\n",
      "R = 8.314;\t\t\t#[J/mol*K] - universal gas constant\n",
      "T = 350.;\t\t\t#[K] - Surrounding temperature\n",
      "\n",
      "\t\t\t#componenet 1 : organic solid (1)\n",
      "\t\t\t#componenet 2 : CO2 (2)\n",
      "\n",
      "P_1_sat = 133.3;\t\t\t#[N/m**(2)] - Vapour pressre of organic solid\n",
      "V_s_1 = 200.;\t\t\t#[cm**(3)/mol] - Molar volume of organic solid\n",
      "V_s_1 = V_s_1*10.**(-6);\t\t\t#[m**(3)/mol]\n",
      "\n",
      "\t\t\t#/At 350 K, the values of the coefficients \n",
      "B_11 = -500.;\t\t\t#[cm**(3)/mol]\n",
      "B_22 = -85.;\t\t\t#[cm****(3)/mol]\n",
      "B_12 = -430.;\t\t\t#[cm**(3)/mol]\n",
      "\n",
      "# Calculations\n",
      "\t\t\t#From phase equilibrium equation of component 1, we get\n",
      "\t\t\t# y1*P*phi_1 = f_1\n",
      "\t\t\t# f_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T))\n",
      "\n",
      "\t\t\t#Since vapour pressure of organic solid is very small, therefore\n",
      "f_1_sat = P_1_sat;\n",
      "\n",
      "\t\t\t# Now let us determine the fugacity coefficient of organic solid in the vapour mixture.\n",
      "\t\t\t# math.log(phi_1) = (P/(R*T))*(B_11 + y2**(2)*del_12) \n",
      "del_12 = (2*B_12 - B_11 - B_22)*10**(-6);\t\t\t#[m**(3)/mol]\n",
      "\n",
      "\t\t\t#It is given that the partial pressure of component 1 in the vapour mixture is 1333 N?m**(2)\n",
      "\t\t\t# y1*P = 1333 N/m**(2) or, y1 = 1333/P\n",
      "\t\t\t# y2 = 1- 1333/P\n",
      "\t\t\t# math.log(phi_1) = (P/(R*T))*(B_11 + (1- 1333/P)**(2)*del_12)\n",
      "\n",
      "\t\t\t#The phase equilibrium equation becomes\n",
      "\t\t\t# y1*P*phi_1 = f_1_sat*math.exp(V_s_1*(P-P_1_sat)/(R*T))\n",
      "\t\t\t#Taking math.log on both side we have\n",
      "\t\t\t# math.log(y1*P) + math.log(phi_1) = math.log(f_1_sat) + (V_s_1*(P-P_1_sat)/(R*T))\n",
      "\t\t\t# (V_s_1*(P-P_1_sat)/(R*T)) - math.log(phi_1) = math.log(1333/133.3) = math.log(10)\n",
      "\n",
      "\t\t\t#substituting for math.log(phi_1) from previous into the above equation we get\n",
      "\t\t\t# (V_s_1*(P-P_1_sat)/(R*T)) - (P/(R*T))*(B_11 + (1- 1333/P)**(2)*del_12) - math.log(10) = 0\n",
      "\t\t\t# On simplification we get,\n",
      "\t\t\t# 975*P**(2) - 6.7*10**(9)*P + 4.89*10**(8) = 0\n",
      "\t\t\t# Solving the above qyadratic equation using shreedharcharya rule\n",
      "\n",
      "P3 = (6.7*10**(9) + ((-6.7*10**(9))**(2)-4*975*4.98*10**(8))**(1./2))/(2*975);\t\t\t#[Pa]\n",
      "P4 = (6.7*10**(9) - ((-6.7*10**(9))**(2)-4*975*4.98*10**(8))**(1./2))/(2*975);\t\t\t#[Pa]\n",
      "\t\t\t# The second value is not possible, therefore pressure of the system is P3\n",
      "P3 = P3*10**(-5);\t\t\t#[bar]\n",
      "\n",
      "# Results\n",
      "print \" The total pressure of the system is %f bar\"%(P3);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The total pressure of the system is 68.717948 bar\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 11
    }
   ],
   "metadata": {}
  }
 ]
}