{ "metadata": { "name": "", "signature": "sha256:98e5c2c383a1a1550ca177061033a45bcfa7bc464bb9def6e21c0e4e3bbd4374" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 13 : Thermodynamics in Phase Equilibria" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.1 Page No : 238" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "#N2 obeys the relation : Z = 1+(2.11*10**-4*P)\n", "Tc = 126;#Critical temperature in K\n", "Pc = 33.5;#Critical pressure in atm\n", "T = 373;#in K\n", "P = 100;#in atm\n", "\n", "#To Calculate the fugacity of N2 at 373K and 100 atm\n", "#(i)Umath.sing the Z relation given above\n", "#From equation 13.12 (page no 239)\n", "phi = math.e**(2.11*10**-4*(P-0));#fugacity coefficient\n", "f = phi*P;\n", "print \"i)The fugacity of N2 umath.sing the given Z relation is %f atm\"%(f);\n", "\n", "#(ii)Umath.sing the fugacity chart given in figure A.2.9\n", "Pr = P/Pc;#Reduced pressure in atm\n", "Tr = T/Tc;#Reduced temperature in K\n", "#From figure A.2.9,\n", "phi = 1.04\n", "f = phi*P;\n", "print \" ii)The fugacity of N2 umath.sing the fugacity chart is %f atm\"%(f);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)The fugacity of N2 umath.sing the given Z relation is 102.132418 atm\n", " ii)The fugacity of N2 umath.sing the fugacity chart is 104.000000 atm\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.2 Page No : 241" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "P1 = 50*1.03*10**4;#Initial pressure in Kgf/sq m\n", "T = 373.0;#Temperature in K\n", "P2 = 1.03*10**4;#Final pressure in Kgf/sq m\n", "V = 0.001*18;#Volume in cubic meter\n", "R = 848.0;#gas consmath.tant in m Kgf/Kgmole K\n", "\n", "#To Calculate the fugacity of liquid water\n", "#From equation 13.13(page no 240)\n", "del_u = (V/(R*T))*(P2-P1);#del_u = ln(f2/f1); Change in chemical potential\n", "f1 = P2;#in Kgf/sq m\n", "f2 = f1*(math.e**del_u);\n", "print \"The fugacity of the liquid water at 50 atm is %4.2e Kgf/sq m\"%(f2);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The fugacity of the liquid water at 50 atm is 1.00e+04 Kgf/sq m\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.3 Page No : 242" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "import math\n", "import matplotlib.pyplot as plt\n", "import numpy\n", "\n", "#Given\n", "x1 = 0.1;#mole fraction of methane\n", "x2 = 0.9;#mole fraction of propane\n", "P = [28.1,31.6,35.1];#Pressure in Kgf/sq cm are\n", "K1 = [5.8,5.10,4.36];#Vapourisation consmath.tants of methane at the corresponding presssures\n", "K2 = [0.61,0.58,0.56];#Vapourisation consmath.tants of propane at the correspondig pressures\n", "\n", "#To Calculate the bubble point pressure of the solution\n", "#From equation 13.27 (page no 245)\n", "y1 = []\n", "y2 = []\n", "y = []\n", "for i in range(0,3):\n", " y1.append(K1[i]*x1);#mole fraction of methane in the vapour phase\n", " y2.append(K2[i]*x2);#mole fraction of propane in the vapour phase\n", " y.append(y1[i]+y2[i]);#sum of the mole fraction in the vapour phase\n", " \n", "\n", "plt.plot(P,y)\n", "plt.title(\"y vs pressure\")\n", "plt.xlabel(\"P\")\n", "plt.ylabel(\"y\")\n", "plt.show()\n", "P1 = numpy.interp(1,y,P)\n", "print \"The bubble point pressure of the solution is %f Kgf/sq cm\"%(P1);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "The bubble point pressure of the solution is 35.100000 Kgf/sq cm\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.4 Page No : 244" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "T = [80.6,79.0,77.3,61.4];#Various temperature in deg cel\n", "x1 = [0.0,15.0,29.0,100.0];#mole fraction of CHCl3 in liquid phase\n", "y1 = [0.0,20.0,40.0,100.0];#mole fraction of CHCl3 in vapour phase\n", "P1 = [1370,1310,1230,700];#Vapour pressure of CHCl3 in mm Hg\n", "P = 760.0;#Total pressure in mm Hg\n", "\n", "#To Calculate the equilibrium data i.e y/x and compare with the experimental values\n", "#From equation 13.27 (page no 245);K = y1/x1 = Pi/P\n", "print \"Temperature Experimental Calculated\";\n", "Ke_x = []\n", "K_c = []\n", "for i in range(0,4):\n", " print \" %f\"%(T[i]),\n", " if x1[i] == 0 :\n", " print \" Not defined\",\n", " Ke_x.append(0)\n", " else:\n", " Ke_x.append(y1[i]/x1[i]);\n", " print \" %f\"%(Ke_x[i]),\n", " K_c.append(P1[i]/P);\n", " print \" %f\"%(K_c[i])\n", "\n", "if Ke_x[i] == K_c[i]:\n", " print ' The liquid solution is perfect';\n", "else:\n", " print \" The liquid solution is imperfect\";\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature Experimental Calculated\n", " 80.600000 Not defined 1.802632\n", " 79.000000 1.333333 1.723684\n", " 77.300000 1.379310 1.618421\n", " 61.400000 1.000000 0.921053\n", " The liquid solution is imperfect\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.6 Page No : 246" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "x1 = 0.1;#Mole fraction of dichloromethane (CCl2H2)\n", "x2 = 0.9;#Mole fraction of methyl acetate (C3H6O2)\n", "M1 = 85.0;#Molecular weight of CCl2H2\n", "M2 = 74.0;#Molecular weight of C3H602\n", "D1 = 1.3163;#Density of CCl2H2 in gm/cc\n", "D2 = 0.9279;#Density of C3H6O2 in gm/cc\n", "\n", "#To Calculate the volume of 10% dichloromethane solution\n", "V1 = M1/D1;#Specific volume of pure CCL2H2 in cc/gmole\n", "V2 = M2/D2;#Specific volume of C3H6O2 in cc/gmole\n", "#From equation 13.62(page no 256)& 13.78 (page no 257)\n", "V_e = x1*x2*(1.2672-0.771*x1);#excess volume in cc/gmole\n", "V = V1*x1+V2*x2+V_e;\n", "print \"The volume of 10 percent dichloromethane is %f cc/gmole\"%(V);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The volume of 10 percent dichloromethane is 78.339579 cc/gmole\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.7 Page No : 249" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "import math\n", "import matplotlib.pyplot as plt\n", "import numpy\n", "\n", "#Given\n", "x_T = 0.957;#mole fraction of Toluene\n", "x_D = 0.043;#mole fraction of 1,2-dichloroethane\n", "t = [90, 100, 110];#temperature in deg cel\n", "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", "\n", "#To Calculate the vapour pressure of the solution, bubble point at 686 mm Hg and the vapour composition at equilibrium,\n", "#compare the experimental value of 91.2% toluene in vapour with the calculated value & calculate the free energy of mixing\n", "#(1)Calculation of vapour pressure\n", "print \"1Tempdeg cel P_TmmHg P_DmmHg P_smmHg\";\n", "P_T = []\n", "P_D = []\n", "P_s = []\n", "for i in range(0,3):\n", " P_T.append(10**(6.95464-(1344.8/(219.482+t[i]))));#Given as equation(a)(page no 260)\n", " P_D.append(10**(7.03993-(1274.079/(223+t[i]))));#Given as equation(b)(page no 260)\n", " P_s.append(x_T*P_T[i]+x_D*P_D[i]);#pressure of the solution in mm Hg\n", " print ' %f'%(t[i]),\n", " print ' %f'%(P_T[i]),\n", " print ' %f'%(P_D[i]),\n", " print ' %f'%(P_s[i])\n", "\n", "#(2)Calculation of bubble point and comparison of values\n", "plt.plot(t,P_s)\n", "plt.title(\"t vs P_s\")\n", "plt.xlabel(\"t\")\n", "plt.ylabel(\"P_s\")\n", "plt.show()\n", "T = numpy.interp(686,P_s,t)\n", "P = 686.0;#pressure of solution in mm Hg\n", "y_T_e = 0.912;#experimental value of mole fraction of toluene\n", "#From the graph we found that the temperature at P = 686 mm Hg is\n", "#t = 105.3;#in deg cel\n", "print '2)The bubble point is %f deg cel'%(T);\n", "#From equation (a)(page no 260)\n", "P_T = 10**(6.95464-(1344.8/(219.482+T)));#vapour pressure of Toluene in mmHg\n", "#From equation 13.27 (page no 245)\n", "y_T_c = (x_T*P_T)/P;\n", "y_D_c = 1-y_T_c;\n", "print ' The vapour composition of toluene is %f'%(y_T_c);\n", "print ' The vapour composition of 1,(2-dichloroethane is %f'%(y_D_c);\n", "e = ((y_T_e-y_T_c)/y_T_e)*100;\n", "print ' The percentage error is %f percent'%(e);\n", "\n", "#(3)Calculation of free energy\n", "del_F = R*(T+273)*((x_T*math.log(x_T))+(x_D*math.log(x_D)));\n", "print '3)The free energy of mixing is %f Kcal/Kgmole'%(del_F);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1Tempdeg cel P_TmmHg P_DmmHg P_smmHg\n", " 90.000000 406.737847 931.944531 429.321734\n", " 100.000000 556.321921 1245.698588 585.965118\n", " 110.000000 746.589152 1636.315096 784.847367\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "2)The bubble point is 105.029855 deg cel\n", " The vapour composition of toluene is 0.901898\n", " The vapour composition of 1,(2-dichloroethane is 0.098102\n", " The percentage error is 1.107696 percent\n", "3)The free energy of mixing is -132.756668 Kcal/Kgmole\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.8 Page No : 252" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "#Consider the diagram shown in page no 263\n", "w1 = 100.0;#weight of LiBr entered as feed in the evaporator per hour in Kg\n", "x1 = 0.45;#weight fraction of LiBr entered as feed\n", "x2 = 0.0;#weight fraction of steam in the LiBr soln\n", "x3 = 0.65;#weight fraction of LiBr formed as product\n", "H1 = -39.0;#Enthalpy of 45% solution at 25 deg cel in Kcal/Kg\n", "H3 = -4.15;#Enthalpy of 65% solution at 114.4 deg cel in Kcal/Kg\n", "H2 = 649.0;#Enthalpy of superheated steam at 100 mmHg and 114.4 deg cel in Kcal/Kg\n", "\n", "#To Calculate the heating load required for the process\n", "#According to material balance\n", "w3 = (w1*x1)/x3;#weight of LiBr solution formed after evaporation per hour in Kg\n", "w2 = w1-w3;# weight of steam formed in Kg/hr\n", "#According to energy balance\n", "Q = (w2*H2)+(w3*H3)-(w1*H1);\n", "print 'The heat that has to be supplied for this concentration process is %f Kcal/hr'%(Q);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The heat that has to be supplied for this concentration process is 23581.923077 Kcal/hr\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.12 Page No : 253" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "import math\n", "import matplotlib.pyplot as plt\n", "import numpy\n", "\n", "#Given\n", "x_A = [0.0, 0.0435, 0.0942, 0.1711, 0.2403, 0.3380, 0.5981];#mole fraction of acetic acid\n", "p_A = [0.0, 17.2, 30.5, 46.5, 57.8, 69.3, 95.7];#partial pressure of acetic acid in mmHg\n", "P_T1 = 202.0;#vapour pressure of toulene in mmHg\n", "P_T2_ex = 167.3;#experimental partial pressure in mmmHg\n", "\n", "#To Calculate the partial pressure of toulene in the solution and check with the experimental value\n", "#From the equation 13.95,\n", "#ln(P_T2/P_T1) = -intg(x_A/((1-x_A)*p_A))\n", "x = []\n", "for i in range(0,7):\n", " if (p_A[i] != 0):\n", " x.append((x_A[i]/((1-x_A[i])*p_A[i]))*10**4)\n", " else:\n", " x.append(0)\n", "\n", "\n", "plt.plot(x,p_A)\n", "plt.title(\" \")\n", "plt.xlabel(\"(x_A/((1-x_A)*p_A))*10**4\")\n", "plt.ylabel(\"p_A\")\n", "plt.show()\n", " \n", "#Area of the graph drawn is\n", "A = -0.138;\n", "P_T2 = (math.e**A)*P_T1;\n", "e = ((P_T2-P_T2_ex)*100)/P_T2_ex;\n", "print 'The partial pressure of toulene is %f mmHg'%(P_T2);\n", "print ' This deviates %i percent from the reported value'%(e);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "The partial pressure of toulene is 175.961936 mmHg\n", " This deviates 5 percent from the reported value\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.13 Page No : 254" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "\n", "import math\n", "import matplotlib.pyplot as plt\n", "import numpy\n", "from numpy.linalg import solve\n", "\n", "#Given\n", "P = 760.0;#pressure at maximum boiling azeotrope of A and B in mmHg\n", "x_A = 0.6;#mole fraction of A in liquid phase\n", "x_B = 0.4;#mole fraction of B in liquid phase\n", "p_A = 600.0;#vapour pressure of A at 90 deg cel\n", "p_B = 300.0;#vapour pressure of B at 90 deg cel\n", "\n", "#To Check whether the activity coefficient of the solution can be represented by the Margules equation\n", "y_A = P/p_A;#Activity coefficient of A\n", "y_B = P/p_B;#Activity coefficient of B\n", "#From the Margules equation or equation (a) & (b)\n", "U = [[((x_B**2)-(2*(x_B**2)*x_A)), (2*(x_B**2)*x_A)], [(2*(x_A**2)*x_B), ((x_A**2)-(2*(x_A**2)*x_B))]];\n", "V = [math.log(y_A), math.log(y_B)];\n", "W = solve(U,V);\n", "#Now the value of consmath.tants A and B in equations(a)&(b) are given as\n", "\n", "A = W[0];\n", "B = W[1];\n", "#let us assume \n", "x_A = [0.0,0.2,0.4,0.6,0.8,1.0];\n", "x_B = [1.0,0.8,0.6,0.4,0.2,0.0];\n", "#C = lny_A; D = lny_B; E = ln(y_A/y_B)\n", "C = []\n", "D = []\n", "E = []\n", "for i in range(6):\n", " C.append((x_B[i]**2)*(2*(B-A)*x_A[i]+A));\n", " D.append((x_A[i]**2)*(2*(A-B)*x_B[i]+B));\n", " E.append(C[i]-D[i]);\n", " \n", "plt.plot(x_A,E)\n", "plt.title(\" \")\n", "plt.xlabel(\"x_A\")\n", "plt.ylabel(\"ln(y_A/y_B)\")\n", "plt.show()\n", "#Since the graph drawn is approximately symmetrical.Thus it satisfies the Redlich-Kister Test\n", "print 'The actvity coefficients of the system can be represented by Margules equation';\n", "#end\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "The actvity coefficients of the system can be represented by Margules equation\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.14 Page No : 259" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "import math\n", "import matplotlib.pyplot as plt\n", "import numpy\n", "\n", "#Given\n", "P = 760.0#Total pressure of the mixture in mmHg\n", "T = [80, 90, 95, 100];#Temperature in deg celsius\n", "P1 = [87.4, 129.0, 162.0, 187.0];#vapour pressure of 1,1,2,2-tetrachloroethane in mmHg\n", "P2 = [356, 526, 648, 760];#Vapour pressure of water in mmHg\n", "\n", "#To Calculate the composition of the vapour evolved\n", "plt.plot(T,P1,\"green\",T,P2,\"red\")\n", "plt.title(\" \")\n", "plt.xlabel(\"Temp in deg cel\")\n", "plt.ylabel(\"Vapour pressure in mmHg\")\n", "plt.show()\n", "#From the graph we conclude that at 93.8 deg cel\n", "P1 = 155.0;#in mm Hg\n", "P2 = 605.0;#in mm Hg\n", "y_1 = P1/P;\n", "y_2 = P2/P;\n", "print 'Mole fraction of 1,(1,(2,(2-tetrachloroethane in vapour is %f'%(y_1);\n", "print ' Mole fraction of water in vapour is %f'%(y_2);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "Mole fraction of 1,(1,(2,(2-tetrachloroethane in vapour is 0.203947\n", " Mole fraction of water in vapour is 0.796053\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.15 Page No : 263" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "import math\n", "import matplotlib.pyplot as plt\n", "import numpy\n", "\n", "#Given\n", "T = [146.2, 142.3, 126.1, 115.9, 95.0, 98.0, 100.0];#Temperature in deg cel\n", "P1 = [760.0, 685.0, 450.3, 313.0];#Vapour pressure of 1,1,2,2-tetrachloroethane at the coressponding temperature in mm Hg\n", "P2_5 = 648.0;#Vapour pressure of water at 95 deg cel in mm Hg\n", "P2_6 = 711.0;#Vapour pressure of water at 98 deg cel in mm Hg\n", "P = 760.0;#Total pressure of mixture in mm Hg\n", "\n", "x1 = [0, 0, 0, 0, 0, 0, 0];\n", "#To plot a graph between temperature and vapour phase composition\n", "for i in range(0,4):\n", " x1[i] = P1[i]/P;#mole fraction of 1,1,2,2-tetrachloroethane\n", "x2_5 = P2_5/P;#mole fraction of water at 95 deg cel\n", "x2_6 = P2_6/P;#mole fraction of water at 98 deg cel\n", "x1[4] = 1-x2_5;\n", "x1[5] = 1-x2_6;\n", "\n", "plt.plot(x1,T)\n", "plt.title(\"\")\n", "plt.xlabel(\"mole fraction of 1,1,2,2-tetrachloroethane\")\n", "plt.ylabel(\"Temperature in deg cel\")\n", "plt.show()\n", "print 'The required graph has been ploted in the graphic window';\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "The required graph has been ploted in the graphic window\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.16 Page No : 266" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "#B = -(1.203*10**10)*(T**2.7); second virial coefficient, T is in K\n", "#math.log P = 6.95464-(1344.8/(219.482+t))...(a);Vapour pressure of toulene\n", "t = 107.2;#Temperature in deg cel\n", "T = t+273.16;#in K\n", "H_ex = 7964.0;#experimental value of heat of vapourisation in Kcal/Kgmole\n", "d = 800.0;#density of liquid toulene in Kg/cubic meter\n", "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", "M = 92.14;#molecular weight of toulene\n", "\n", "#To Calculate the heat of vapourization of toulene by umath.sing ideal gas law, second virial coefficient but neglecting vl and including vl\n", "#From equation (a), let K = dmath.logP/dT\n", "K = 1344.8/(219.482+t)**2;\n", "#(i)Umath.sing ideal gas behaviour\n", "#From equation 13.112(page no 286)\n", "H_c = (2.303*R*(T**2))*K;\n", "print 'i)The heat of vapourization umath.sing ideal gas behaviour is %f Kcal/Kgmole'%(H_c);\n", "D = ((H_c-H_ex)/H_c)*100;\n", "print ' The deviation is %f percent'%(D);\n", "\n", "#(ii)Umath.sing second virial coeff but neglecting vl\n", "#From equation(a)\n", "P = 10**(6.95464-1344.8/(219.482+t));#in mm Hg\n", "P1 = P*1.033*10**4/760;#in Kgf/sq m\n", "B = -((1.203*10**10)/(T**2.7))*10**-3;#in cubic meter/Kgmole\n", "#From equation 13.111 (page no 286) neglecting vl,\n", "l = (R*T)+((B*P1)/427);#in Kcal/Kgmole\n", "H_c = K*2.303*T*l;\n", "print 'ii)The heat of vapourisation umath.sing second virial coefficient but neglecting vl is %f Kcal/Kgmole'%(H_c);\n", "D = ((H_c-H_ex)/H_c)*100;\n", "print ' The deviation in this case is %f percent'%(D);\n", "\n", "#(iii)Umath.sing second virial coeff including vl\n", "vl = M/d;#Liquid specific volume in cubic meter/Kgmole\n", "n = P1*vl/427;#in Kcal/Kgmole\n", "H_c = K*2.303*T*(l-n);\n", "print 'iii)The heat of vapourisation umath.sing second virial coefficient including vl is %f Kcal/Kgmole'%(H_c);\n", "D = ((H_c-H_ex)/H_c)*100;\n", "print ' The deviation in this case is %f'%(D);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)The heat of vapourization umath.sing ideal gas behaviour is 8312.967730 Kcal/Kgmole\n", " The deviation is 4.197872 percent\n", "ii)The heat of vapourisation umath.sing second virial coefficient but neglecting vl is 7998.487742 Kcal/Kgmole\n", " The deviation in this case is 0.431178 percent\n", "iii)The heat of vapourisation umath.sing second virial coefficient including vl is 7970.612948 Kcal/Kgmole\n", " The deviation in this case is 0.082967\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.17 Page No : 269" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "H_ex = 539.0;#Heat of vapoization of water in Kcal/Kg\n", "Tc = 647.0;#Critical temperature in K\n", "Pc = 218.0;#Critical pressure in atm\n", "Tb = 373.0;#Boiling point of water in K\n", "t = 100.0;#temperature in deg cel\n", "M = 18.0;#Molecular weight of water\n", "P = 1.0;#pressure at boiling point in atm\n", "P1 = 1.033*10**4;#pressure in Kgf/sq m\n", "\n", "#To Calculate the heat of vapourisation of water by Vishwanath and Kuloor method and by Riedel's method and compare with the experimental value\n", "#(i) Umath.sing Vishwanath and Kuloor method\n", "H_c = (4.7*Tc*((1-(P/Pc))**0.69)*math.log(P/Pc))/((1-(Tc/Tb))*18);\n", "print 'i)The heat of vapourisation of water umath.sing Vishwanath and Kuloor method is %f Kcal/Kg'%(H_c);\n", "D = (H_c-H_ex)*100/H_c;\n", "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", "\n", "#(ii)Umath.sing Riedel's method\n", "H_c = (Tb*2.17*(math.log(218)-1))/((0.93-(Tb/Tc))*18);\n", "print 'ii)The heat of vapourisation of water umath.sing Riedel method is %f Kcal/Kg'%(H_c);\n", "D = (H_c-H_ex)*100/H_c;\n", "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", "\n", "#(iii)By umath.sing given vapour equation; math.logP = 8.2157-(2218.8537/(273.16+t)), t is in deg cel\n", "#From steam table,\n", "Vv = 1.673;#in cubic meter/Kg\n", "Vl = 0.001;#in cubic meter/Kg\n", "H_c = (2218.8/(273.16+t)**2)*(2.3*Tb*P1*(Vv-Vl)/427);\n", "print 'iii)The heat of vapourisation umath.sing the given vapour equation is %f Kcal/Kg'%(H_c);\n", "D = (H_c-H_ex)*100/H_c;\n", "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)The heat of vapourisation of water umath.sing Vishwanath and Kuloor method is 1234.397750 Kcal/Kg\n", " The deviation occurs umath.sing this method is 56.334982 percent\n", "ii)The heat of vapourisation of water umath.sing Riedel method is 557.743828 Kcal/Kg\n", " The deviation occurs umath.sing this method is 3.360652 percent\n", "iii)The heat of vapourisation umath.sing the given vapour equation is 552.934102 Kcal/Kg\n", " The deviation occurs umath.sing this method is 2.520029 percent\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.18 Page No : 270" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "T1 = 273-87.0;#temp in K\n", "T2 = 273.0;#temp in K\n", "H1 = 115.0;#Latent heat of saturated ethane at 1 atm and -87 deg cel in Kcal/Kg\n", "H2_ex = 72.44;#Experimental value of latent heat at 0 deg cel in Kcal/Kg\n", "Tc = 306.0;#Critical temperature in K\n", "M = 30.0;#Molecular weight of ethane\n", "\n", "#To Calculate the latent heat of saturated ethane at 0 deg cel\n", "Tr1 = T1/Tc;#reduced temp in K\n", "Tr2 = T2/Tc;#reduced temp in K\n", "#(i)Umath.sing Waton's method:\n", "H2_c = H1*((1-Tr2)/(1-Tr1))**0.38;\n", "print 'i)The latent heat of saturated ethane at 0 deg cel umath.sing Waton method is %f Kcal/Kg'%(H2_c);\n", "D = (H2_ex-H2_c)*100/H2_ex;\n", "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", "\n", "#(ii)Umath.sing Vishwanath and Kuloor method\n", "#From equation 13.117 (page no 289)\n", "n = (0.00133*(H1*M/T1)+0.8794)**(1/0.1);\n", "H2_c = H1*((1-Tr2)/(1-Tr1))**n;\n", "print 'ii)The latent heat of saturated ethane at 0 deg cel umath.sing Vishwanath and Kuloor method is %f Kcal/Kg'%(H2_c);\n", "D = (H2_ex-H2_c)*100/H2_ex;\n", "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)The latent heat of saturated ethane at 0 deg cel umath.sing Waton method is 70.411608 Kcal/Kg\n", " The deviation occurs umath.sing this method is 2.800099 percent\n", "ii)The latent heat of saturated ethane at 0 deg cel umath.sing Vishwanath and Kuloor method is 71.809846 Kcal/Kg\n", " The deviation occurs umath.sing this method is 0.869898 percent\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.19 Page No : 272" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "H_s_ex = 32.7;#experimental value of latent heat of the solution in KJ/mole\n", "x1 = 0.536;#mole percent of toulene in the solution\n", "x2 = 1-0.536;#mole percent of 1,1,1-trichloroethane in the solution\n", "H1 = 33.34;#Latent heat of toulene in KJ/gmole\n", "H2 = 29.72;#Latent heat of 1,1,1-trichloroethane in KJ/gmole\n", "He = 0.0;#excess enthalpy is neglected\n", "Cp1 = 39.55;#Specific heat of toulene in cal/gmole deg cel\n", "Cp2 = 24.62;#Specific heat of 1,1,1-trichloroethane in cal/gmole deg cel\n", "T_D = 100.0;#dew point temperature in deg cel\n", "T_B = 92.6;#bubble point temperature in deg cel\n", "\n", "#To calculate the latent heat of the solution and compare it with the one which calculated from the given vapour pressure equation\n", "#(i)Calculation of latent heat of the solution\n", "#From equation 13.118 (page no 291)\n", "H_s = H1*x1+H2*x2+He+(Cp1*x1+Cp2*x2)*10**-3*4.17*(T_D-T_B);\n", "print 'i)The latent heat of the solution is %f KJ/gmole'%(H_s);\n", "D = ((H_s_ex-H_s)*100)/H_s_ex;\n", "print ' The deviation occurs using this method is %f percent'%(D);\n", "\n", "#(ii)Calculation of latent heat from the vapour pressure equation\n", "#From equation (a) (page no 291)\n", "K = 1657.599/((273.16+5)**2);\n", "H_s = (K*2.303*8.314*(273.16+5)**2)*10**-3;\n", "print 'ii)The latent heat of the solution is %f KJ/gmole'%(H_s);\n", "D = ((H_s_ex-H_s)*100)/H_s_ex;\n", "print ' The deviation occurs using this method is %f percent'%(D);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)The latent heat of the solution is 32.666984 KJ/gmole\n", " The deviation occurs using this method is 0.100965 percent\n", "ii)The latent heat of the solution is 31.738283 KJ/gmole\n", " The deviation occurs using this method is 2.941029 percent\n" ] } ], "prompt_number": 23 } ], "metadata": {} } ] }