{ "metadata": { "name": "", "signature": "sha256:11ca0d89ced68d0cbbe63c7f792aae8768450c19ca2bb47ac7ca1a7c91117b32" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 13 : Chemical Reaction Equilibria" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.1 page no : 216" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "#CH4 + H2O --> CO + 3H2\n", "n_CH4 = 2;\t\t\t#Moles of CH4\n", "n_H2O = 1;\t\t\t#Moles of H20\n", "n_CO = 1;\t\t\t#Moles of CO\n", "n_H2 = 4;\t\t\t#Moles of H2\n", "\n", "v_CH4 = -1;\n", "v_H2O = -1\n", "v_CO = 1;\n", "v_H2 = 3;\n", "\n", "# Calculations and Results\n", "v = v_CH4+v_H2O+v_CO+v_H2;\n", "n = n_CH4+n_H2O+n_CO+n_H2;\n", "\n", "#y_CH4 = (n_CH4+(v_CH4e)/n+(v*e))\n", "#y_H2O = (n_H2O+(v_H2Oe)/n+(v*e))\n", "#y_CO = (n_CO+(v_CO*e)/n+(v*e))\n", "#y_H2 = (n_H2+(v_H2*e)/n+(v*e))\n", "\n", "y_CH4 = '(n_CH4+(v_CH4e)/n+(v*e))';\n", "y_H2O = '(n_H2O+(v_H2Oe)/n+(v*e))';\n", "y_CO = '(n_CO+(v_CO*e)/n+(v*e))';\n", "y_H2 = '(n_H2+(v_H2*e)/n+(v*e))';\n", "\n", "#Hence\n", "\n", "y_CH4 = '(2-e/8+2e)';\n", "y_H2O = '(1-e/8+2e)';\n", "y_CO = '(1+e/8+2e)';\n", "y_H2 = '(4+3e/8+2e)';\n", "\n", "print 'y_CH4 = ',y_CH4\n", "print 'y_H2O = ',y_H2O\n", "print 'y_CO = ',y_CO\n", "print 'y_H2 = ',y_H2\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "y_CH4 = (2-e/8+2e)\n", "y_H2O = (1-e/8+2e)\n", "y_CO = (1+e/8+2e)\n", "y_H2 = (4+3e/8+2e)\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.3 page no : 218" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Program to Determine the Expression for yi for two reactions\n", "\n", "#CH4 + H2O --> CO + 3H2 (A)\n", "\n", "#CH4 + 2H2O --> CO2 + 4H2 (B)\n", "\n", "# Variables\n", "Species = ['CH4','H2O','CO','CO2','H2','sum'];\n", "v_A = ['-1','-1','1','0','3','2'];\n", "v_B = ['-1','-2','0','1','4','2'];\n", "m_CH4 = 2;\n", "m_H2O = 3;\n", "\n", "# Calculations\n", "mt = m_CH4+m_H2O;\n", "y = ['(m_CH4+vie1+vje2)/(mt+vie1+vje2)','(m_H2O+vie1+vje2)/(mt+vie1+vje2)','(vie1)/(mt+vie1+vje2)','(vje2)/(mt+vie1+vje2)','(vie1+vje2)/(mt+vie1+vje2)',' '];\n", "\n", "#Hence\n", "yf = ['(2-e1-e2)/(5+2e1+2e2)','(3-e1-2e2)/(5+2e1+3e2)','e1/(5+2e1+2e2)','e2/(5+2e1+2e2)','(3e1+4e2)/(5+2e1+2e2)',' '];\n", "Ans = [Species,v_A,v_B,y,yf];\n", "\n", "# Results\n", "print ' i v_A v_B y(Before substitution) y_species'\n", "print Ans\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " i v_A v_B y(Before substitution) y_species\n", "[['CH4', 'H2O', 'CO', 'CO2', 'H2', 'sum'], ['-1', '-1', '1', '0', '3', '2'], ['-1', '-2', '0', '1', '4', '2'], ['(m_CH4+vie1+vje2)/(mt+vie1+vje2)', '(m_H2O+vie1+vje2)/(mt+vie1+vje2)', '(vie1)/(mt+vie1+vje2)', '(vje2)/(mt+vie1+vje2)', '(vie1+vje2)/(mt+vie1+vje2)', ' '], ['(2-e1-e2)/(5+2e1+2e2)', '(3-e1-2e2)/(5+2e1+3e2)', 'e1/(5+2e1+2e2)', 'e2/(5+2e1+2e2)', '(3e1+4e2)/(5+2e1+2e2)', ' ']]\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.4 page no : 219" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "\n", "def IDCPH(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return (dA+((dB/2)*T0*(t+1))+((dC/3)*T0*T0*((t**2)+t+1))+(dD/(t*T0*T0)))*(T-T0)\n", "\n", "def IDCPS(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return ((dA)*math.log(t))+(((dB*T0)+(((dC*T0*T0)+(dD/(t*t*T0*T0)))*(t+1)/2))*(t-1))\n", "\n", "# Variables\n", "T0 = 298.16;\t\t\t#[K]\n", "T1 = 418.15;\t\t\t#[K]\n", "T2 = 593.15;\t\t\t#[K]\n", "R = 8.314;\n", "\n", "#C2H4(g) + H2O(g) --> C2H5OH(g)\n", "#Values From Table C.1 At T = 298.15K\n", "\n", "A_ethanol = 3.518;\n", "A_ethene = 1.424;\n", "A_water = 3.470;\n", "\n", "# Calculations\n", "B_ethanol = 20.001*10**-3;\n", "B_ethene = 14.394*10**-3;\n", "B_water = 1.450*10**-3;\n", "\n", "C_ethanol = -6.002*10**-6;\n", "C_ethene = -4.392*10**-6;\n", "C_water = 0;\n", "\n", "D_ethanol = 0;\n", "D_ethene = 0;\n", "D_water = 0.121*10**5;\n", "\n", "dA = A_ethanol-A_ethene-A_water\n", "dB = B_ethanol-B_ethene-B_water\n", "dC = C_ethanol-C_ethene-C_water\n", "dD = D_ethanol-D_ethene-D_water\n", "\n", "# Values from Table C.4 at T = 298.15K\n", "H_ethanol = -235100;\t\t\t#[J/mol]\n", "H_ethene = 52510;\t\t\t#[J/mol]\n", "H_water = -241572;\t\t\t#[J/mol]\n", "\n", "G_ethanol = -168490;\t\t\t#[J/mol]\n", "G_ethene = 68460;\t\t\t#[J/mol]\n", "G_water = -228572;\t\t\t#[J/mol]\n", "\n", "dHo = H_ethanol-H_ethene-H_water\n", "dGo = G_ethanol-G_ethene-G_water\n", "\n", "I1 = round(IDCPH(T0,T1,dA,dB,dC,dD),3)\n", "I2 = round(IDCPS(T0,T1,dA,dB,dC,dD),5)\n", "\n", "#Using Eqn 13.18\n", "#dG_418/RT = ((dGo - dHo)/RTo)+(dHo/RT)+((1/T)*I1)-I2 c1 = dG_418/RT\n", "\n", "c1 = round(((dGo-dHo)/(R*T0))+(dHo/(R*T1))+((1/T1)*I1)-I2,4)\n", "I3 = round(IDCPH(T0,T2,dA,dB,dC,dD),3)\n", "I4 = round(IDCPS(T0,T2,dA,dB,dC,dD),5)\n", "\n", "#Using Eqn 13.18\n", "#dG_593/RT = ((dGo - dHo)/RTo)+(dHo/RT)+((1/T)*I1)-I2 c2 = dG_593/RT\n", "\n", "c2 = round(((dGo-dHo)/(R*T0))+(dHo/(R*T2))+((1/T2)*I3)-I4,4)\n", "\n", "K_413 = round(math.exp(-c1),4);\n", "K_593 = math.exp(-c2);\n", "\n", "# Results\n", "print 'Equilibrium Consmath.tant at T = 413.15K is ',K_413*10,'X 10**-1'\n", "print 'Equilibrium Constant at T = 593.15K is ',round(K_593*1000,3),'X 10**-3'\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Equilibrium Consmath.tant at T = 413.15K is 1.404 X 10**-1\n", "Equilibrium Constant at T = 593.15K is 2.802 X 10**-3\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.4 page no : 220" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import array,round,ones,exp,log\n", "import math \n", "\n", "#Example 13.4 Alternate\n", "# Alternate Program to 13.4\n", "\n", "# Variables\n", "T = array([298.15,418.15,593.15]);\n", "t = round(T/T[0],4);\n", "R = 8.314;\n", "\n", "\n", "#C2H4(g) + H2O(g) --> C2H5OH(g)\n", "#Values From Table C.1 At T = 298.15K\n", "\n", "A_ethanol = 3.518;\n", "A_ethene = 1.424;\n", "A_water = 3.470;\n", "\n", "# Calculations\n", "B_ethanol = 20.001*10**-3;\n", "B_ethene = 14.394*10**-3;\n", "B_water = 1.450*10**-3;\n", "\n", "C_ethanol = -6.002*10**-6;\n", "C_ethene = -4.392*10**-6;\n", "C_water = 0;\n", "\n", "D_ethanol = 0;\n", "D_ethene = 0;\n", "D_water = 0.121*10**5;\n", "\n", "dA = A_ethanol-A_ethene-A_water\n", "dB = B_ethanol-B_ethene-B_water\n", "dC = C_ethanol-C_ethene-C_water\n", "dD = D_ethanol-D_ethene-D_water\n", "\n", "# Values from Table C.4 at T = 298.15K\n", "H_ethanol = -235100;\t\t\t#[J/mol]\n", "H_ethene = 52510;\t\t\t#[J/mol]\n", "H_water = -241572;\t\t\t#[J/mol]\n", "\n", "G_ethanol = -168490;\t\t\t#[J/mol]\n", "G_ethene = 68460;\t\t\t#[J/mol]\n", "G_water = -228572;\t\t\t#[J/mol]\n", "\n", "dHo = H_ethanol-H_ethene-H_water\n", "dGo = G_ethanol-G_ethene-G_water\n", "\n", "#Umath.sing Eqn(13.21)\n", "Ko = round(exp(-dGo/(R*T[0])),3)\n", "K0 = Ko*ones((1,3));\n", "#Umath.sing Eqn(13.22)\n", "K1 = exp((dHo/(R*T0))*(1-(T[0]/T)));\n", "#Umath.sing Eqn(13.24)\n", "K2 = round(exp((dA*(log(t)-((t-1)/t)))+(0.5*dB*T[0]*((t-1)**2)/t)+((1/6.)*dC*T[0]*T[0]*((t-1)**2)*(t+2)/t)+(0.5*dD*((t-1)**2)/((T[0]**2)*(t)**2))),4);\n", "\n", "K = K0*K1*K2;\n", "\n", "Ans = [T,t,K0,K1,K2,K];\n", "\n", "# Results\n", "print ' T/K t K0 K1 K2 K',Ans\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " T/K t K0 K1 K2 K [array([ 298.15, 418.15, 593.15]), array([ 1. , 1.4025, 1.9894]), array([[ 29.366, 29.366, 29.366]]), array([ 1.00000000e+00, 4.84531951e-03, 9.74032575e-05]), array([ 1. , 0.9862, 0.9794]), array([[ 2.93660000e+01, 1.40324083e-01, 2.80142097e-03]])]\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.5 page no : 222" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import array,round,ones,exp,log,poly1d,roots\n", "\n", "# Variables\n", "#CO(g) + H2O(g) --> CO2(g) + H2(g)\n", "v_CO = -1;\n", "v_H2O = -1;\n", "v_CO2 = 1;\n", "v_H2 = 1;\n", "v = v_CO+v_H2O+v_CO2+v_H2;\n", "\n", "#Calculate e(Fraction of Stream) in each case\n", "#(a) \n", "n_H2O_a = 1;\t\t\t#mol\n", "n_CO_a = 1;\t\t\t#mol\n", "T_a = 1100;\t\t\t#[K]\n", "P_a = 1;\t\t\t#[bar]\n", "\n", "x = 10**4/T_a;\n", "#at this x the value of ln K = 0 form Graph\n", "y = 0;\n", "nt = n_H2O_a+n_CO_a;\n", "K = exp(y);\n", "\n", "#y_H2O = (n_H2O+(v_H2O*e))/nt\n", "#y_CO = (n_CO+(v_CO*e))/nt\n", "#y_H2 = (v_H2*e)/nt\n", "#y_CO2 = (v_CO2*e)/nt\n", "\n", "\n", "y_H2O = '(1-e)/2'\n", "y_CO = '(1-e)/2'\n", "y_H2 = 'e/2'\n", "y_CO2 = 'e/2'\n", "\n", "\n", "#K = (y_H2*y_CO2)/(y_CO*y_H2O)\n", "\n", "K = 'e**2/((1-e)**2) = 1';\n", "\n", "\n", "#Solving \n", "e_a = 0.5;\n", "\n", "#(b) same as in (a) P_b = 10bar\n", "\n", "#Pressure has no effect on fraction of stream\n", "e_b = e_a;\n", "\n", "#(c) same as in (a) N2 = 2mols included\n", "\n", "#Since N2 just act as diluent\n", "#It only changes the total moles fraction remains the same\n", "e_c = e_a;\n", "\n", "#(d)\n", "n_H2O_d = 2;\t\t\t#mol\n", "n_CO_d = 1;\t\t\t#mol\n", "T_d = 1100;\t\t\t#[K]\n", "P_d = 1;\t\t\t#[bar]\n", "\n", "x = 10**4/T_d;\n", "#at this x the value of ln K = 0 form Graph\n", "y = 0;\n", "nt = n_H2O_d+n_CO_d;\n", "K = math.exp(y);\n", "\n", "#y_H2O = (n_H2O+(v_H2O*e))/nt\n", "#y_CO = (n_CO+(v_CO*e))/nt\n", "#y_H2 = (v_H2*e)/nt\n", "#y_CO2 = (v_CO2*e)/nt\n", "\n", "y_H2O = '(1-e)/3'\n", "y_CO = '(2-e)/3'\n", "y_H2 = 'e/3'\n", "y_CO2 = 'e/3'\n", "\n", "#K = (y_H2*y_CO2)/(y_CO*y_H2O)\n", "\n", "K = 'e**2/((1-e)(2-e)) = 1';\n", "\n", "#Solving \n", "e = round(2./3,3);\n", "e_d = round(e/2.,3);\n", "\n", "#(e) \n", "#Here the y_CO and y_H2O are interchanged\n", "\n", "#No change in Fraction of stream\n", "e_e = e_d;\n", "\n", "#(f) \n", "n_H2O_f = 1;\t\t\t#mol\n", "n_CO_f = 1;\t\t\t#mol\n", "n_CO2_f = 1;\t\t\t#[mol]\n", "T_f = 1100;\t\t\t#[K]\n", "P_f = 1;\t\t\t#[bar]\n", "\n", "x = 10**4/T_f;\n", "#at this x the value of ln K = 0 form Graph\n", "y = 0;\n", "nt = n_H2O_f+n_CO_f+n_CO2_f;\n", "K = math.exp(y);\n", "\n", "#y_H2O = (n_H2O+(v_H2O*e))/nt\n", "#y_CO = (n_CO+(v_CO*e))/nt\n", "#y_CO2 = (n_CO2+(v_CO2*e))/nt\n", "#y_H2 = (v_H2*e)/nt\n", "\n", "\n", "y_H2O = '(1-e)/3'\n", "y_CO = '(1-e)/3'\n", "y_H2 = '(1+e)/3'\n", "y_CO2 = 'e/2'\n", "\n", "\n", "#K = (y_H2*y_CO2)/(y_CO*y_H2O)\n", "\n", "K = '(e*(e+1))/((1-e)**2) = 1';\n", "\n", "#Solving \n", "e_f = round(1./3,3);\n", "\n", "#(g)\n", "n_H2O_g = 1;\t\t\t#mol\n", "n_CO_g = 1;\t\t\t#mol\n", "T_g = 1650;\t\t\t#[K]\n", "P_g = 1;\t\t\t#[bar]\n", "\n", "x = 10**4/T_g;\n", "#at this x the value of ln K = 0 form Graph\n", "y = -1.15;\n", "nt = n_H2O_g+n_CO_g;\n", "K = math.exp(y);\n", "\n", "#y_H2O = (n_H2O+(v_H2O*e))/nt\n", "#y_CO = (n_CO+(v_CO*e))/nt\n", "#y_H2 = (v_H2*e)/nt\n", "#y_CO2 = (v_CO2*e)/nt\n", "\n", "\n", "y_H2O = '(1-e)/2'\n", "y_CO = '(1-e)/2'\n", "y_H2 = 'e/2'\n", "y_CO2 = 'e/2'\n", "\n", "\n", "#K = (y_H2*y_CO2)/(y_CO*y_H2O)\n", "\n", "Exp = 'e**2/((1-e)**2) = 0.316';\n", "\n", "#Solving \n", "p = poly1d([K, -2*K, K-1],'e','c');\n", "\n", "root = roots(p);\n", "e_g = round(root[0],2);\n", "\n", "#Other Root is negative and the Fraction of steam cannot be negative\n", "\n", "\n", "# Results\n", "print ('(a)1mol H20 and 1 mol CO T = 1100K P = 1bar')\n", "print ('(b)1mol H20 and 1 mol CO T = 1100K P = 10bar')\n", "print ('(c)1mol H20 and 1 mol CO 2mol N2 T = 1100K P = 1bar')\n", "print ('(d)2mol H20 and 1 mol CO T = 1100K P = 1bar')\n", "print ('(e)1mol H20 and 2 mol CO T = 1100K P = 1bar')\n", "print ('(f)1mol H20 and 1 mol CO 1mol CO2 T = 1100K P = 1bar')\n", "print ('(g)1mol H20 and 1 mol CO T = 1650K P = 1bar')\n", "e = [e_a, e_b, e_c, e_d, e_e, e_f, e_g];\n", "print 'Fraction Of Steam Reacted in each case',e\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)1mol H20 and 1 mol CO T = 1100K P = 1bar\n", "(b)1mol H20 and 1 mol CO T = 1100K P = 10bar\n", "(c)1mol H20 and 1 mol CO 2mol N2 T = 1100K P = 1bar\n", "(d)2mol H20 and 1 mol CO T = 1100K P = 1bar\n", "(e)1mol H20 and 2 mol CO T = 1100K P = 1bar\n", "(f)1mol H20 and 1 mol CO 1mol CO2 T = 1100K P = 1bar\n", "(g)1mol H20 and 1 mol CO T = 1650K P = 1bar\n", "Fraction Of Steam Reacted in each case [0.5, 0.5, 0.5, 0.33400000000000002, 0.33400000000000002, 0.33300000000000002, -0.68000000000000005]\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.6 page no : 224" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "from numpy import array,round,ones,exp,log\n", "\n", "def IDCPH(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return (dA+((dB/2)*T0*(t+1))+((dC/3)*T0*T0*((t**2)+t+1))+(dD/(t*T0*T0)))*(T-T0)\n", "\n", "def IDCPS(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return ((dA)*math.log(t))+(((dB*T0)+(((dC*T0*T0)+(dD/(t*t*T0*T0)))*(t+1)/2))*(t-1))\n", "\n", "def PHIB(Tr,Pr,omega):\n", " B0 = 0.083-(0.422/(Tr**1.6));\n", " B1 = 0.139-(0.172/(Tr**4.2));\n", " return exp((Pr/Tr)*(B0+(omega*B1)));\n", "\n", "\n", "# Variables\n", "T0 = 298.16;\t\t\t#[K]\n", "T1 = 523.15;\t\t\t#[K]\n", "P = 35.; \t\t\t#[bar]\n", "R = 8.314;\n", "\n", "#C2H4(g) + H2O(g) --> C2H5OH(g)\n", "#Values From Table C.1 At T = 298.15K\n", "\n", "A_ethanol = 3.518;\n", "A_ethene = 1.424;\n", "A_water = 3.470;\n", "\n", "B_ethanol = 20.001*10**-3;\n", "B_ethene = 14.394*10**-3;\n", "B_water = 1.450*10**-3;\n", "\n", "C_ethanol = -6.002*10**-6;\n", "C_ethene = -4.392*10**-6;\n", "C_water = 0;\n", "\n", "D_ethanol = 0;\n", "D_ethene = 0;\n", "D_water = 0.121*10**5;\n", "\n", "# Calculations and Results\n", "dA = A_ethanol-A_ethene-A_water\n", "dB = B_ethanol-B_ethene-B_water\n", "dC = C_ethanol-C_ethene-C_water\n", "dD = D_ethanol-D_ethene-D_water\n", "\n", "# Values from Table C.4 at T = 298.15K\n", "H_ethanol = -235100;\t\t\t#[J/mol]\n", "H_ethene = 52510;\t\t\t#[J/mol]\n", "H_water = -241572;\t\t\t#[J/mol]\n", "\n", "G_ethanol = -168490;\t\t\t#[J/mol]\n", "G_ethene = 68460;\t\t\t#[J/mol]\n", "G_water = -228572;\t\t\t#[J/mol]\n", "\n", "dHo = H_ethanol-H_ethene-H_water\n", "dGo = G_ethanol-G_ethene-G_water\n", "\n", "I1 = round(IDCPH(T0,T1,dA,dB,dC,dD),3)\n", "I2 = round(IDCPS(T0,T1,dA,dB,dC,dD),5)\n", "\n", "#Umath.sing Eqn 13.18\n", "#dG_418/RT = ((dGo - dHo)/RTo)+(dHo/RT)+((1/T)*I1)-I2 c1 = dG_418/RT\n", "\n", "c1 = round(((dGo-dHo)/(R*T0))+(dHo/(R*T1))+((1/T1)*I1)-I2,4)\n", "K_523 = round(exp(-c1),4);\n", "print 'Equilibrium Consmath.tant at T = 523.15K is ',round(K_523*1000,3),'X 10**-3'\n", "\n", "#Values Frm App B\n", "Tc = array([282.3,647.1,513.9]);\n", "Pc = array([50.4,220.55,61.48]);\n", "omega = array([0.087,0.345,0.645]);\n", "\n", "Tr = round(T1/Tc,3);\n", "Pr = round(P/Pc,3);\n", "si = round(PHIB(Tr,Pr,omega),3);\n", "\n", "#Umath.sing eqn\n", "#(y_ETOH*si_ETOH)/(y_C2H4*si_C2H4*y_H20*si_H2O) = (P/Po)K\n", "#y_ETOH/(y_C2H4*y_H20) = c = ((si_C2H4*si_H2O)/si_ETOH)(P/Po)K\n", "c = round(((si[0]*si[1])/si[2])*(P*K_523),3)\n", "\n", "#y_C2H4 = (1-e)/(6-e) \n", "#y_ETOH = (5-e)/(6-e)\n", "#y_H2O = (e)/(6-e)\n", "\n", "#Solving we get a Eqn \n", "#poly([1.342 -6 1],'e','c')\n", "root = round(roots(poly1d([1.342, -6, 1],'e','c')),3)\n", "\n", "r = root[0]*100;\n", "#Since e > 1 not possible so e = 0.233\n", "\n", "print 'The Maximum Conversion of ethylene to ethanol by Vapor-Phase Hydration is ',r,'%'\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Equilibrium Consmath.tant at T = 523.15K is 9.8 X 10**-3\n", "The Maximum Conversion of ethylene to ethanol by Vapor-Phase Hydration is -600.0 %\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.7 page no : 226" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import roots, poly1d\n", "import math \n", "\n", "# Variables\n", "T = 1393.15;\t\t\t#K\n", "P = 1.;\t\t\t#[bar]\n", "x = 10.**4/T;\n", "#C2H2 --> 2C + H2 (I)\n", "#2C + 2H2 --> C2H4 (II)\n", "\n", "# Calculations\n", "#Values Of ln K (at 1000/T )for Reactions I and II from Graph\n", "K_I = math.exp(12.9);\n", "K_II = math.exp(-12.9);\n", "K = K_I*K_II;\n", "\n", "#Application in Eqn (13.5)\n", "\n", "#y_C2H4/(y_C2H2*y_H2) = c = (P/Po)K\n", "c = P*K;\n", "#y_H2 = y_C2H2 = (1-e)/(2-e)\n", "#y_C2H4 = e/(2-e)\n", "\n", "\n", "#The Eqn comes out to be\n", "#poly([1 -4 2],'e','c')\n", "\n", "root = round(roots(poly1d([1, -4, 2],'e','c')),3)\n", "e = root[0];\n", "#Since e > 1 not possible so e = 0.293\n", "y_C2H2 = round((1-e)/(2-e),3);\n", "y_H2 = y_C2H2;\n", "y_C2H4 = round(e/(2-e),3);\n", "\n", "print 'Equilibrium Composition of H2 C2H2 and C2H4 Respectively ',y_C2H4,y_C2H2,y_H2\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Equilibrium Composition of H2 C2H2 and C2H4 Respectively -0.667 0.833 0.833\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.8 page no : 227" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import roots,poly1d\n", "import math \n", "\n", "# Variables\n", "T0 = 298.15;\n", "T = 373.15;\t\t\t#[K]\n", "R = 8.314;\n", "#CH3COOH(l)+C2H5OH(l) --> CH3COOC2H5(l) + H2O(l)\n", "\n", "#From Table C.4\n", "dHo_EtAc = -480000.;\t\t\t#[J]\n", "dHo_H2O = -285830.;\t\t\t#[J]\n", "dHo_EtOH = -277690.;\t\t\t#[J]\n", "dHo_AcH = -484500.;\t\t\t#[J]\n", "\n", "dGo_EtAc = -332200.;\t\t\t#[J]\n", "dGo_H2O = -237130.;\t\t\t#[J]\n", "dGo_EtOH = -174780.;\t\t\t#[J]\n", "dGo_AcH = -389900.;\t\t\t#[J]\n", "\n", "# Calculations\n", "dHo_298 = dHo_EtAc+dHo_H2O-dHo_EtOH-dHo_AcH;\n", "dGo_298 = dGo_EtAc+dGo_H2O-dGo_EtOH-dGo_AcH;\n", "\n", "K_298 = round(math.exp(-dGo_298/(R*T0)),4);\n", "\n", "#Umath.sing Eqn(13.15)\n", "#ln(K_373/K_298) = c = -(dHo_298/R)*((1/373.15)-(1/298.15))\n", "c = round(-(dHo_298/R)*((1/373.15)-(1/298.15)),4);\n", "K_373 = round(K_298*math.exp(c),4);\n", "\n", "#x_AcH = x_EtOH = (1-e)/2 and x_EtAc = x_H2O = e/2\n", "#K = (x_EtAc*x_H2O)/(x_AcH*x_EtOH)\n", "\n", "#Hence The Eqn is\n", "q = poly1d([K_373, -2*K_373, K_373-1],'e','c')\n", "root = round(roots(q),4)\n", "e = root[0];\n", "#Since Other Root is > 1 hence e = 0.6879\n", "x_EtAc = round(e/2,3);\n", "\n", "# Results\n", "print 'Composition of Ethyl Acetate in the Reacting Mixture',x_EtAc\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Composition of Ethyl Acetate in the Reacting Mixture -4.859\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.9 page no : 228" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import isreal\n", "\n", "def MCPH(T0,T,A,B,C,D):\n", " t = T/T0;\n", " return (A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t**2)+t+1))+(D/(t*T0*T0)))\n", "\n", "def IDCPH(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return (dA+((dB/2)*T0*(t+1))+((dC/3)*T0*T0*((t**2)+t+1))+(dD/(t*T0*T0)))*(T-T0)\n", "\n", "def IDCPS(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return ((dA)*math.log(t))+(((dB*T0)+(((dC*T0*T0)+(dD/(t*t*T0*T0)))*(t+1)/2))*(t-1))\n", "\n", "# Variables\n", "P = 1.;\t\t\t#[bar]\n", "T0 = 298.15;\t\t\t#[K]\n", "R = 8.314;\n", "\n", "#SO2 + 0.5O2 --> SO3\n", "dHo_298 = -98890.;\t\t\t#[J/mol]\n", "dGo_298 = -70866.;\t\t\t#[J/mol]\n", "\n", "# Calculations\n", "n_O2_i = 0.5*1.2;\t\t\t# Moles O2 Entering\n", "n_N2_i = n_O2_i*(79./21);\t\t\t#Moles N2 Entering\n", "\n", "#n_SO2 = 1-e\n", "#n_O2 = 0.6-(0.5*e)\n", "#n_SO3 = e\n", "#n_N2 = 2.257\n", "\n", "#By Energy Balance\n", "#(dHo_298*e)+dHo_P = dH = 0 (A)\n", "\n", "#dHo_P = Cp*(T-298.15) Cp = E nCp (B)\n", "\n", "#Cp_SO2 = R*MCPH(T0,T,5.699,0.801E-3,0,-1.015E+5) \n", "#Cp_O2 = R*MCPH(T0,T,3.639,0.506E-3,0,-0.227E+5)\n", "#Cp_SO3 = R*MCPH(T0,T,8.06,1.056E-3,0,-2.028E+5)\n", "#Cp_N2 = R*MCPH(T0,T,3.28,0.593E-3,0,0.04E+5)\n", "\n", "#T = (-(dHo_298*e)/Cp)+T0 (C)\n", "\n", "#K = (e/(1-e))*((3.857-(0.5*e))/(0.6-(0.5*e)))**0.5 (D)\n", "\n", "#ln K = ((dHo_298-dGo_298)/(R*T0))-(dHo_298/(R*T))+I1-(I2/T) (E)\n", "\n", "#I1 = IDCPS(T0,T,0.5415,0.002E-3,0,-0.8995E+5)\n", "#I2 = IDCPH(T0,T,0.5415,0.002E-3,0,-0.8995E+5)\n", "\n", "#Iteration\n", "A1 = 300.;\t\t\t#Initial\n", "i = -1.;\n", "\n", "while(i == -1):\n", " I1 = IDCPS(T0,A1,0.5415,0.002E-3,0,-0.8995E+5);\n", " I2 = IDCPH(T0,A1,0.5415,0.002E-3,0,-0.8995E+5);\n", " #Applying in Eqn (E)\n", " K = math.exp(((dHo_298-dGo_298)/(R*T0))-(dHo_298/(R*A1))+I1-(I2/A1));\n", " #Applying in Eqn (D)\n", " if(isreal(K)):\n", " x = 0;\n", " # p = poly([-0.6*(K**2) 1.7*(K**2) 3.857-(1.6*(K**2)) 0.5*((K**2)-1)],'e','c')\n", " # (0.5*((K**2)-1)*(x**3))+((3.857-(1.6*(K**2)))*(x**2))+(1.7*(K**2)*x)+(-0.6*(K**2))\n", " F_x = (0.5*((K**2)-1)*(x**3))+((3.857-(1.6*(K**2)))*(x**2))+(1.7*(K**2)*x)+(-0.6*(K**2));\n", " F_a = F_x;\n", "\n", " x = 1;\n", " F_x = (0.5*((K**2)-1)*(x**3))+((3.857-(1.6*(K**2)))*(x**2))+(1.7*(K**2)*x)+(-0.6*(K**2));\n", " #F_x = (x**3)-(4*x)+1;\n", " F_b = F_x;\n", " root = -100;\n", " A = 0;\n", " B = 1;\n", " i = 1;\n", " while(i == 1):\n", " a = A;\n", " F_a = (0.5*((K**2)-1)*(a**3))+((3.857-(1.6*(K**2)))*(a**2))+(1.7*(K**2)*a)+(-0.6*(K**2));\n", " #F_a = (a**3)-(4*a)+1;\n", " b = B;\n", " F_b = (0.5*((K**2)-1)*(b**3))+((3.857-(1.6*(K**2)))*(b**2))+(1.7*(K**2)*b)+(-0.6*(K**2));\n", " #F_b = (b**3)-(4*b)+1;\n", " x1 = ((a*F_b)-(b*F_a))/(F_b-F_a);\n", " F_x1 = (0.5*((K**2)-1)*(x1**3))+((3.857-(1.6*(K**2)))*(x1**2))+(1.7*(K**2)*x1)+(-0.6*(K**2));\n", " #F_x1 = (x1**3)-(4*x1)+1;\n", "\n", " if((F_a*F_x1)<0):\n", " flag = 1;\n", " A = a;\n", " B = x1;\n", " elif ((F_x1*F_b)<0):\n", " flag = 2;\n", " A = x1;\n", " B = b; \n", " x1_a = round(x1,4);\n", " b_a = round(b,4);\n", " a_a = round(a,4);\n", " if(x1_a == b_a):\n", " root = round(x1,5);\n", " i = 0;\n", " break;\n", " elif(x1_a == a_a):\n", " root = round(x1,5);\n", " i = 0;\n", " break; \n", " e = root;\n", " Cp_SO2 = R*MCPH(T0,A1,5.699,0.801E-3,0,-1.015E+5); \n", " Cp_O2 = R*MCPH(T0,A1,3.639,0.506E-3,0,-0.227E+5);\n", " Cp_SO3 = R*MCPH(T0,A1,8.06,1.056E-3,0,-2.028E+5);\n", " Cp_N2 = R*MCPH(T0,A1,3.28,0.593E-3,0,0.04E+5);\n", "\n", " n_SO2 = 1-e;\n", " n_O2 = 0.6-(0.5*e);\n", " n_SO3 = e;\n", " n_N2 = 2.257;\n", " if(n_SO2<0 or n_O2<0 or n_SO3<0):\n", " e = 0;\n", "\n", " Cp = (n_SO2*Cp_SO2)+(n_O2*Cp_O2)+(n_SO3*Cp_SO3)+(n_N2*Cp_N2);\n", " #Applying in Eqn (C)\n", " B = (-(dHo_298*e)/Cp)+T0;\n", " m = (A1+B)/2;\n", " dT = round(abs(m-A1),2);\n", " if(dT<0.1):\n", " i = 0;\n", " T = round(A1,1);\n", " e = round(e,2);\n", " break;\n", " A1 = m;\n", " i = -1;\n", " else:\n", " i = -1;\n", " A1 = A1+1; \n", "\n", "print 'Fraction',e\n", "\n", "n_SO2 = 1-e\n", "n_O2 = 0.6-(0.5*e)\n", "n_SO3 = e\n", "n_N2 = 2.257\n", "\n", "nt = n_SO2+n_O2+n_SO3+n_N2;\n", "\n", "y_SO2 = round(n_SO2/nt,4);\n", "y_O2 = round(n_O2/nt,4);\n", "y_SO3 = round(n_SO3/nt,4);\n", "y_N2 = round(n_N2/nt,4);\n", "\n", "# Results\n", "print 'Final Temperature',T\n", "print 'Composition of SO2',y_SO2\n", "print 'Composition of O2',y_O2\n", "print 'Composition of SO3',y_SO3\n", "print 'Composition of N2',y_N2\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Fraction 0.77\n", "Final Temperature 855.8\n", "Composition of SO2 0.0662\n", "Composition of O2 0.0619\n", "Composition of SO3 0.2218\n", "Composition of N2 0.6501\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.12 page no : 229" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "def IDCPH(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return (dA+((dB/2)*T0*(t+1))+((dC/3)*T0*T0*((t**2)+t+1))+(dD/(t*T0*T0)))*(T-T0)\n", "\n", "def IDCPS(T0,T,dA,dB,dC,dD):\n", " t = T/T0;\n", " return ((dA)*math.log(t))+(((dB*T0)+(((dC*T0*T0)+(dD/(t*t*T0*T0)))*(t+1)/2))*(t-1))\n", "\n", "# Variables\n", "T0 = 298.16;\t\t\t#[K]\n", "T1 = 750;\t\t\t#[K]\n", "R = 8.314;\n", "P = 1.2;\t\t\t#[bar]\n", "\n", "#C4H10 --> C2H4 + C2H6 (I)\n", "#C4H10 --> C3H6 + CH4 (II)\n", "\n", "#Values From Table C.1 At T = 298.15K\n", "\n", "A_butane = 1.935;\n", "A_ethene = 1.424;\n", "A_ethane = 1.131;\n", "A_propene = 1.637;\n", "A_methane = 1.702;\n", "\n", "B_butane = 36.915*10**-3;\n", "B_ethene = 14.394*10**-3;\n", "B_ethane = 19.225*10**-3;\n", "B_propene = 22.706*10**-3;\n", "B_methane = 9.081*10**-3;\n", "\n", "C_butane = -11.402*10**-6;\n", "C_ethene = -4.392*10**-6;\n", "C_ethane = -5.561*10**-6;\n", "C_propene = -6.915*10**-6;\n", "C_methane = -2.164*10**-6;\n", "\n", "D_butane = 0;\n", "D_ethene = 0;\n", "D_ethane = 0;\n", "D_propene = 0;\n", "D_methane = 0;\n", "\n", "dA_I = A_ethene+A_ethane-A_butane;\n", "dA_II = A_methane+A_propene-A_butane;\n", "\n", "dB_I = B_ethene+B_ethane-B_butane;\n", "dB_II = B_methane+B_propene-B_butane;\n", "\n", "dC_I = C_ethene+C_ethane-C_butane;\n", "dC_II = C_methane+C_propene-C_butane;\n", "\n", "dD_I = D_ethene+D_ethane-D_butane;\n", "dD_II = D_methane+D_propene-D_butane;\n", "\n", "# Values from Table C.4 at T = 298.15K\n", "H_butane = -125790.;\t\t\t#[J/mol]\n", "H_ethene = 52510.;\t\t\t#[J/mol]\n", "H_ethane = -83820.;\t\t\t#[J/mol]\n", "H_propene = 19710.;\t\t\t#[J/mol]\n", "H_methane = -74520.;\t\t\t#[J/mol]\n", "\n", "G_butane = -16570;\t\t\t#[J/mol]\n", "G_ethene = 68460.;\t\t\t#[J/mol]\n", "G_ethane = -31855.;\t\t\t#[J/mol]\n", "G_propene = 62205.;\t\t\t#[J/mol]\n", "G_methane = -50460.;\t\t\t#[J/mol]\n", "\n", "# Calculations\n", "dHo_I = H_ethene+H_ethane-H_butane\n", "dHo_II = H_methane+H_propene-H_butane\n", "\n", "dGo_I = G_ethene+G_ethane-G_butane\n", "dGo_II = G_methane+G_propene-G_butane\n", "\n", "I1_I = round(IDCPH(T0,T1,dA_I,dB_I,dC_I,dD_I),3)\n", "I1_II = round(IDCPH(T0,T1,dA_II,dB_II,dC_II,dD_II),3)\n", "I2_I = round(IDCPS(T0,T1,dA_I,dB_I,dC_I,dD_I),5)\n", "I2_II = round(IDCPS(T0,T1,dA_II,dB_II,dC_II,dD_II),5)\n", "\n", "#Using Eqn 13.18\n", "#dG_418/RT = ((dGo - dHo)/RTo)+(dHo/RT)+((1/T)*I1)-I2 c1 = dG_418/RT\n", "\n", "c1_I = round(((dGo_I-dHo_I)/(R*T0))+(dHo_I/(R*T1))+((1/T1)*I1_I)-I2_I,4)\n", "c1_II = round(((dGo_II-dHo_II)/(R*T0))+(dHo_II/(R*T1))+((1/T1)*I1_II)-I2_II,4)\n", "\n", "K_I = round(math.exp(-c1_I),4)\n", "K_II = round(math.exp(-c1_II),4)\n", "\n", "k = (K_II/K_I)**0.5;\n", "e_I = round(((K_I/P)/(1+(K_I*(1/P)*(1+k)*(1+k))))**0.5,4);\n", "\n", "e_II = round(k*e_I,4);\n", "\n", "n_C4H10 = 1-e_I-e_II;\n", "n_C2H4 = e_I;\n", "n_C2H6 = e_I;\n", "n_C3H6 = e_II;\n", "n_CH4 = e_II;\n", "nt = n_C4H10+n_C2H4+n_C2H6+n_C3H6+n_CH4;\n", "\n", "y_C4H10 = round(n_C4H10/nt,4);\n", "y_C2H4 = round(n_C2H4/nt,4);\n", "y_C2H6 = round(n_C2H6/nt,4);\n", "y_C3H6 = round(n_C3H6/nt,4);\n", "y_CH4 = round(n_CH4/nt,4);\n", "\n", "y = [y_C4H10, y_C2H4, y_C2H6, y_C3H6, y_CH4];\n", "\n", "# Results\n", "print ' Y_C4H10 Y_C2H4 Y_C2H6 Y_C3H6 Y_CH4'\n", "print y\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Y_C4H10 Y_C2H4 Y_C2H6 Y_C3H6 Y_CH4\n", "[0.0014, 0.049399999999999999, 0.049399999999999999, 0.45000000000000001, 0.45000000000000001]\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.13 page no : 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline\n", "import math \n", "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel,legend\n", "from numpy import array,linspace,exp,linalg,zeros\n", "\n", "\n", "# Variables\n", "n_air = 2.381\t\t\t#[mol]\n", "n_O2 = 0.21*n_air;\n", "n_N2 = 0.79*n_air;\n", "R = 8.314;\n", "\n", "P = 20.; \t\t\t#[bar]\n", "T = array([1000, 1100, 1200, 1300, 1400, 1500]);\n", "dG_H2O = array([-192420, -187000, -181380, -175720, -170020, -164310]);\n", "dG_CO = array([-200240, -209110, -217830, -226530, -235130, -243740]);\n", "dG_CO2 = array([-395790, -395960, -396020, -396080, -396130, -396160]);\n", "\n", "KI = 'y_H2O/((y_O2)**0.5*y_H2)(P/Po)**-0.5'\n", "KII = 'y_CO/((y_O2)**0.5)(P/Po)**0.5'\n", "KIII = 'y_CO2/y_O2'\n", "\n", "n = '3.38+((e2-e1)/2)'\n", "y_H2 = '-e1/n'\n", "y_CO = 'e2/n'\n", "y_O2 = '((0.5(1-e1-e2))-e3)/n'\n", "y_H2O = '(1+e1)/n'\n", "y_CO2 = 'e3/n'\n", "y_N2 = '1.88/n'\n", "\n", "KI = '(1+e1)(2n)**0.5*(P/Po)**-0.5'\n", "KII = '(e3*(P/Po)**0.5)/(1-e1-e2-2e3)**0.5*(n/2)**0.5'\n", "KIII = '2e3/(1-e1-e2-2e3)'\n", "\n", "# Calculations\n", "K_I = round(exp(-dG_H2O/(R*T)),1)\n", "K_II = round(exp(-dG_CO/(R*T)),1)\n", "K_III = round(exp(-dG_CO2/(R*T)),1)\n", "\n", "#Now math.since the values of KI KII KIII valyes are so High the mole fraction of O2 must be very small\n", "#Hence We eleminate O2,Hence 2 Eqns are,\n", "\n", "#C + CO2 --> 2CO (a)\n", "#H2O + C --> H2 + CO (b)\n", "\n", "Ka = '(y_CO**2/y_CO2)*(P/Po)'\n", "Kb = '((y_H2*y_CO)/y_H2O)*(P/Po)'\n", "\n", "n = '3.38+(e_a+e_b)'\n", "y_H2 = 'e_b/n'\n", "y_CO = '(2e_a+e_b)/n'\n", "y_H2O = '(1-e_b)/n'\n", "y_CO2 = '(0.5-e_a)/n'\n", "y_N2 = '1.88/n'\n", "\n", "Ka = '(2e_a+e_b)**2/((0.5-e_a)*n)*(P/Po)'\n", "Kb = 'e_b(2e_a+e_b)/((1-e_b)*n)*(P/Po)'\n", "\n", "dG_new_a = (2*dG_CO)-dG_CO2;\n", "dG_new_b = dG_CO-dG_H2O;\n", "\n", "Ka = round(exp((-dG_new_a/(R*T))),3);\n", "Kb = round(exp((-dG_new_b/(R*T))),3);\n", "\n", "#Calculation of e_a and e_b\n", "\n", "a = 0.1;\t\t\t#Initial Value\n", "\n", "b = 0.7;\t\t\t#Initial Value\n", "\n", "C1 = Ka/20;\n", "C2 = Kb/20;\n", "e_a = zeros(6)\n", "e_b = zeros(6)\n", "for i in range(6):\n", " c = -1;\n", " while(c == -1):\n", " fa = round((((a**2)*(4+C1[i]))+(b**2)+((4+C1[i])*(a*b))+(2.88*C1[i]*a)-(0.5*C1[i]*b)-(1.69*C1[i])),4);\n", " dfax = round(((2*a*(4+C1[i]))+((4+C1[i])*b)+(2.88*C1[i])),4);\n", " dfay = round((2*b)+((4+C1[i])*a)-(0.5*C1[i]),4);\n", "\n", " fb = round(((b**2*(1+C2[i]))+((2+C2[i])*a*b)-(C2[i]*a)+(2.38*C2[i]*b)-(3.38*C2[i])),4);\n", " dfbx = round((((2+C2[i])*b)-C2[i]),4);\n", " dfby = round(((2*b*(1+C2[i]))+((2+C2[i])*a)+(2.38*C2[i])),4);\n", "\n", " A = [[dfax ,dfay],[dfbx ,dfby]];\n", " B = [-fa,-fb];\n", " Ans = round(linalg.solve(A,B),4); \n", " da = Ans[0];\n", " db = Ans[1];\n", "\n", " if(da == 0 and db == 0):\n", " c = 0;\n", " e_a[i] = a;\n", " e_b[i] = b;\n", " break;\n", "\n", " a = a+da;\n", " b = b+db;\n", "\n", "\n", "n = 3.38+(e_a+e_b);\n", "y_H2 = round(e_b/n,3);\n", "y_CO = round(((2*e_a)+e_b)/n,3);\n", "y_H2O = round((1-e_b)/n,3);\n", "y_CO2 = round((0.5-e_a)/n,3);\n", "y_N2 = round(1.88/n,3);\n", "\n", "Ans = [T,Ka,Kb,e_a,e_b];\n", "Ans1 = [T,y_H2,y_CO,y_H2O,y_CO2,y_N2];\n", "\n", "# Results\n", "plot(T,y_H2,'r-')\n", "plot(T,y_CO,'b-')\n", "plot(T,y_H2O,'g-')\n", "plot(T,y_CO2,'m-')\n", "plot(T,y_N2,'y-')\n", "\n", "suptitle('Equllibrium Compositions')\n", "xlabel('T/K')\n", "ylabel('yi')\n", "\n", "\n", "print ' T/K Ka Kb e_a e_b'\n", "print Ans\n", "print ' T/K y_H2 y_CO y_H2O y_CO2 y_N2'\n", "print Ans1\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " T/K Ka Kb e_a e_b\n", "[array([1000, 1100, 1200, 1300, 1400, 1500]), array([ 1.758, 11.405, 53.155, 194.79 , 583.343, 1514.118]), array([ 2.561, 11.219, 38.609, 110.064, 268.764, 583.577]), array([-0.0506, 0.121 , 0.3168, 0.4302, 0.4738, 0.4895]), array([ 0.5336, 0.7124, 0.8551, 0.9357, 0.9713, 0.9863])]\n", " T/K y_H2 y_CO y_H2O y_CO2 y_N2\n", "[array([1000, 1100, 1200, 1300, 1400, 1500]), array([ 0.138, 0.169, 0.188, 0.197, 0.201, 0.203]), array([ 0.112, 0.227, 0.327, 0.378, 0.398, 0.405]), array([ 0.121, 0.068, 0.032, 0.014, 0.006, 0.003]), array([ 0.143, 0.09 , 0.04 , 0.015, 0.005, 0.002]), array([ 0.487, 0.446, 0.413, 0.396, 0.39 , 0.387])]\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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