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diff --git a/Stoichiometry_And_Process_Calculations/ch5.ipynb b/Stoichiometry_And_Process_Calculations/ch5.ipynb new file mode 100755 index 00000000..8cc243fb --- /dev/null +++ b/Stoichiometry_And_Process_Calculations/ch5.ipynb @@ -0,0 +1,497 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Properties of Real Gases" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 page no: 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "V = 0.6 #m**3 vessel\n", + "T = 473. #K temperature\n", + "N = 1. * 10 ** 3 #mol\n", + "R = 8.314 #Pa * m**3/molK\n", + "\n", + "# Calculation \n", + "P = N * R * T / (V * 10**5) \n", + "\n", + "# Result\n", + "print \"(a)Pressure calculated using ideal gas equation = %.2f bar\"%P\n", + "a = 0.4233 #N * m**4 / mol**2\n", + "b = 3.73 * 10**-5 #m**3/mol\n", + "P1 = (R*T/(V/N - b)-a/(V/N)**2)/10**5 \n", + "print \"(a)Pressure calculated using van der waals equation = %.2f bar\"%P1\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Pressure calculated using ideal gas equation = 65.54 bar\n", + "(a)Pressure calculated using van der waals equation = 58.13 bar\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 pageno : 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P = 10.**7 #Pa \n", + "T = 500. #K molar volume\n", + "R = 8.314 #Pa * L / mol K\n", + "N = 1000.\n", + "\n", + "# Calculation \n", + "V = N * R * T / ( P * 1000) \n", + "\n", + "# Result\n", + "print \"(a)Volume of CO2 calculated using ideal gas equation = %.4e m**3\"%V\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Volume of CO2 calculated using ideal gas equation = 4.1570e-04 m**3\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 pageno : 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "V = 0.6 * 10**-3 #m**3\n", + "T = 473. #K vellel of volume\n", + "Tc = 405.5 #K temperature\n", + "Pc = 112.8 * 10 ** 5 #Pa pressure\n", + "R = 8.314 \n", + "\n", + "# Calculation \n", + "a = 0.4278 * (R**2) * (Tc ** 2.5)/Pc \n", + "b = 0.0867 * R * Tc / Pc \n", + "P1 = (R*T/(V - b) - a/((T**0.5)*V*(V + b)))/10**5 \n", + "\n", + "# Result\n", + "print \"Pressure developed by gas = %.2f bar\"%P1\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure developed by gas = 57.87 bar\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 pageno : 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P = 10.**6 #Pa\n", + "T = 373. #K molar volume\n", + "Tc = 405.5 #K temperature\n", + "Pc = 112.8 * 10 ** 5 #Pa pressure\n", + "R = 8.314 \n", + "\n", + "# Calculation \n", + "a = 0.4278 * (R**2) * (Tc ** 2.5)/Pc \n", + "b = 0.0867 * R * Tc / Pc \n", + "#P1 = (R*T/(V - b) - a/((T**0.5)*V*(V + b)))/10**5 \n", + "#10**6=((8.314*373)/(V-2.59*10**-5))-8.68/((373**0.5)*V*(V+2.59*10**-5) \n", + "#solving this we get,\n", + "V = 3.0 * 10**-3 #m**3/mol\n", + "\n", + "# Result\n", + "print \"molar volume of gas = %.e m**3/mol\"%V\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "molar volume of gas = 3e-03 m**3/mol\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5 pageno : 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "B = -2.19 * 10**-4 #m**3/mol virial coefficients\n", + "C = -1.73 * 10**-8 #m**6/mol**2\n", + "P = 10. #bar molar volume\n", + "T = 500. #K\n", + "\n", + "# Calculation \n", + "#virial equation is given as, Z = PV/RT = 1 + B/V + C/V**2\n", + "#V = (RT/P)*(1 + B/V + C/V**2)\n", + "# now by assuming different values for V on RHS and checking for corresponding V on LHS, we have to assume such value of V on RHS by which we get the same value for LHS V\n", + "#by trial and error we get,\n", + "V = 3.92 * 10**-3 #m**3\n", + "\n", + "# Result\n", + "print \"Molar volume of methanol = %.3e m**3\"%V\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molar volume of methanol = 3.920e-03 m**3\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 pageno : 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "T = 510. #K molar volume\n", + "P = 26.6 #bar \n", + "Tc = 425.2 #K temperature\n", + "Pc = 38. #bar pressure\n", + "Zc = 0.274 # compressibility factor\n", + "R = 8.314 \n", + "\n", + "# Calculation \n", + "Pr = P / Pc \n", + "Tr = T / Tc \n", + "\n", + "# Result\n", + "print \"Pr = \",Pr\n", + "print \"Tr = %.2f\"%Tr\n", + "\n", + "#From fig. 5.4 and 5.5 from the text book\n", + "Z = 0.865 \n", + "D = 0.15 \n", + "Z1 = Z + D * ( Zc - 0.27) \n", + "V = R * T * Z1 / (P * 10**5) \n", + "print \"Molar volume of n-butane = %.4e m**3/mol\"%V\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pr = 0.7\n", + "Tr = 1.20\n", + "Molar volume of n-butane = 1.3798e-03 m**3/mol\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 pageno : 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "T = 510. #K molar volumes\n", + "P = 26.6 #bar\n", + "Tc = 425.2 #K temperature\n", + "Pc = 38. #bar pressure\n", + "w = 0.193 # acentric factor\n", + "R = 8.314 \n", + "\n", + "# Calculation \n", + "Pr = P / Pc \n", + "Tr = T / Tc \n", + "\n", + "# Result\n", + "print \"Pr = \",Pr\n", + "print \"Tr = %.2f\"%Tr\n", + "\n", + "#From fig. 5.6 and 5.7 from the text book\n", + "Z0 = 0.855 \n", + "Z1 = 0.042 \n", + "Z = Z0 + w*Z1 \n", + "print \"Z = %.4f\"%Z\n", + "V = R * T * Z / (P * 10**5) \n", + "\n", + "print \"Molar volume of n-butane = %.4e m**3/mol\"%V\n", + "\n", + "# note : answer may vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pr = 0.7\n", + "Tr = 1.20\n", + "Z = 0.8631\n", + "Molar volume of n-butane = 1.3758e-03 m**3/mol\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8 pageno : 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import *\n", + "# variables \n", + "P = 6000. #kPa ethane\n", + "T = 325. #K ethane\n", + "xn2 = 0.4 # gas mixture\n", + "xethane = 0.6 \n", + "an2 = 0.1365 #N m**4 / mol**2 nitrogen\n", + "\n", + "# Calculation \n", + "bn2 = 3.86 * 10**-5 #m**3/mol\n", + "aethane = 0.557 #N m**4 / mol**2\n", + "bethane = 6.51 * 10**-5 #m**3/mol\n", + "Pcn2 = 3394. #kPa\n", + "Tcn2 = 126.2 #K\n", + "Pcethane = 4880. #kPa\n", + "Tcethane = 305.4 #K\n", + "R = 8.314 \n", + "\n", + "\n", + "# Result\n", + "V = R * T / (P*1000) \n", + "print \"(a)Molar volume by ideal gas equation = %.3e m**3/mol\"%V\n", + "\n", + "\n", + "a = (xn2 * (an2**0.5) + xethane * (aethane**0.5))**2 \n", + "b = (xn2*bn2 + xethane*bethane) \n", + "#substituting the above values in van der waals equation, and solving, we get\n", + "#V^3 - 5.0484*10^-4V^2+5.9117*10^-8V-3.2214*10^-12=0\n", + "#Using co-efficients from the above equation,\n", + "coeff = [1, -5.0484e-4, 5.9117e-8, -3.2217e-12]\n", + "x = roots(coeff)\n", + "y = x[0]\n", + "print \"(b)Molar volume by van der waals equation =\",round((y.real)/1e-4,3),\"*10^-4 m**3/mol\"\n", + "\n", + "\n", + "Prin2 = P/Pcn2 \n", + "Trin2 = T/Tcn2 \n", + "Priethane = P/Pcethane \n", + "Triethane = T/Tcethane \n", + "# using compressibilty chart, \n", + "Zn2 = 1 \n", + "Zethane = 0.42 \n", + "Z = xn2 * Zn2 + xethane * Zethane \n", + "V2 = Z * R * T / P * (10**-3)\n", + "print \"(c)Molar volume based on compressibilty factor = %.2e m**3/mol\"%V2\n", + "\n", + "\n", + "Pri1n2 = xn2*P/Pcn2 \n", + "Tri1n2 = T/Tcn2 \n", + "Pri1ethane = xethane*P/Pcethane \n", + "Tri1ethane = T/Tcethane \n", + "# using compressibilty chart, \n", + "Zn21 = 1. \n", + "Zethane1 = 0.76 \n", + "Z1 = xn2 * Zn21 + xethane * Zethane1 \n", + "V3 = (Z1 * R * T / P ) * (10**-3)\n", + "print \"(c)Molar volume based on daltons law = %.3e m**3/mol\"%V3\n", + "\n", + "\n", + "Tc = xn2 * Tcn2 + xethane * Tcethane \n", + "Pc = xn2 * Pcn2 + xethane * Pcethane \n", + "Zc = 0.83 \n", + "V4 = Zc * R *T / P * (10**-3)\n", + "print \"(d)Molar volume by kays method = %.3e m**3/mol\"%V4\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Molar volume by ideal gas equation = 4.503e-04 m**3/mol\n", + "(b)Molar volume by van der waals equation = 3.68 *10^-4 m**3/mol\n", + "(c)Molar volume based on compressibilty factor = 2.94e-04 m**3/mol\n", + "(c)Molar volume based on daltons law = 3.855e-04 m**3/mol\n", + "(d)Molar volume by kays method = 3.738e-04 m**3/mol\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9 page no : 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P1 = 40. #% ( nitrogen )\n", + "P2 = 60. #% ( ethane )\n", + "T = 325. #K ethane\n", + "\n", + "# Calculation \n", + "V = 4.5 * 10**-4 #m**3/mol\n", + "a1 = 0.1365 #N*m**4/mol**2\n", + "b1 = 3.86 * 10 ** -5 #m**3/mol\n", + "a2 = 0.557 #N*m**4/mol**2\n", + "b2 = 6.51 * 10 ** -5 #m**3/mol\n", + "Pc1 = 3394. #kPa\n", + "Tc1 = 126.1 #K\n", + "Pc2 = 4880. #kPa\n", + "Tc2 = 305.4 #K\n", + "R = 8.314 \n", + "Pideal = R * T / (V * 1000) #kPa\n", + "\n", + "# Result\n", + "print \"(a)Pressure of Gas by the ideal gas equation = %.f kPa\"%Pideal\n", + "y1 = P1/100. \n", + "y2 = P2/100. \n", + "a = (y1 * (a1**(1./2)) + y2 * (a2**(1./2)))**2 \n", + "b = y1 * b1 + y2 * b2 \n", + "Pv = ((R * T / (V - b)) - a / (V**2))/1000 \n", + "print \"(b)Pressure of Gas by Van der waals equation = %.f kPa\"%Pv\n", + "Tc = y1*Tc1 + y2*Tc2 \n", + "Pc = y1*Pc1 + y2*Pc2 \n", + "Vc = R * Tc / Pc #Pseudo critical ideal volume\n", + "Vr = V / Vc #Pseudo reduced ideal volume\n", + "Tr = T / Tc #Pseudo reduced temperature\n", + "\n", + "#From fig 5.3, we get Pr = 1.2\n", + "Pr = 1.2 \n", + "Pk = Pr * Pc \n", + "print \"(c)Pressure of Gas by the Kays method = %.f kPa\"%Pk\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Pressure of Gas by the ideal gas equation = 6005 kPa\n", + "(b)Pressure of Gas by Van der waals equation = 5080 kPa\n", + "(c)Pressure of Gas by the Kays method = 5143 kPa\n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
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