{ "metadata": { "name": "", "signature": "sha256:d798b16416897dedbfd57aff5c233c1f7e9c52a7f0081b45c972d6ab47468c7c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 15 - Thermodynamics of chemical reactions" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1 - Pg 318" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the amount of dissociation\n", "#Initalization of variables\n", "import math\n", "import numpy\n", "kp=math.pow(10,(1.45))\n", "#calculations\n", "s=[1-kp*kp , 0, -3, 2]\n", "vec=numpy.roots(s)\n", "X=numpy.real(vec[2])\n", "xper=X*100\n", "#results\n", "print '%s %.1f %s' %(\"Amount of dissociaton =\",xper,\"percent\")\n", "print '%s %.3f %s %.3f %s %.3f %s' %(\"\\n Of each original mole of CO2, there will be\",X,\"mole of CO \",X/2.,\" mol of Oxygen and\",(1-X),\"mol of CO2\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Amount of dissociaton = 12.7 percent\n", "\n", " Of each original mole of CO2, there will be 0.127 mole of CO 0.063 mol of Oxygen and 0.873 mol of CO2\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2 - Pg 319" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the max. temperature reached\n", "#Initalization of variables\n", "U=121200. #Btu/mol\n", "Uco2=51635. #Btu/mol\n", "Un2=27589. #Btu/mol\n", "Uco22=57875. #Btu/mol\n", "Un22=21036. #Btu/mol\n", "T1=5000. #R\n", "T2=5500. #R\n", "#calculations\n", "Ut1=Uco2+1.88*Un2\n", "Ut2=Uco22 + 1.88*Un22\n", "print '%s' %(\"By extrapolation,\")\n", "Tx=5710 #R\n", "#results\n", "print '%s %d %s' %(\"Max. Temperature reached =\",Tx,\"R\")\n", "print '%s' %(\"The calculation for Ut2 is wrong in textbook. Please use a calculator.\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "By extrapolation,\n", "Max. Temperature reached = 5710 R\n", "The calculation for Ut2 is wrong in textbook. Please use a calculator.\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3 - Pg 319" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the max. obtainable temperature\n", "#Initalization of variables\n", "print '%s' %(\"By trial and error,\")\n", "import math\n", "X=0.201\n", "X1=0.2\n", "R=59.3 #universal gas constant\n", "T=5000 #R\n", "U=121200 #Btu/mol\n", "Uco2=51635. #Btu/mol\n", "Un2=27907. #Btu/mol\n", "U3=29616. #Btu/mol\n", "U4=27589. #Btu/mol\n", "#calculations\n", "kp1=R*(1-X1)/math.pow(X1,1.5) /math.pow(T,0.5)\n", "kp2=R*(1-X)/math.pow(X,1.5) /math.pow(T,0.5)\n", "q=(1-X)*Uco2 + X*Un2+ X/2 *U3 +1.88*U4 + X*U\n", "print '%s' %(\"Interpolating between T=4500 R and T=5000 R, we get\")\n", "T2=4907 #R\n", "#results\n", "print '%s %d %s' %(\"Max. obtainable temperature =\",T2,\" R\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "By trial and error,\n", "Interpolating between T=4500 R and T=5000 R, we get\n", "Max. obtainable temperature = 4907 R\n" ] } ], "prompt_number": 3 } ], "metadata": {} } ] }