{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 3 : Interpretation of Batch Reactor Data" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1 page no : 60" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "%pylab inline\n", "\n", "import math \n", "from numpy import *\n", "from matplotlib.pyplot import *\n", "from scipy import stats\n", "\n", "# Variables\n", "#Given\n", "t = [0, 20, 40 ,60, 120 ,180, 300]; # time\n", "C_A = [10 ,8, 6, 5, 3, 2, 1]; # concentration\n", "CAo = 10.;\n", "k = zeros(7)\n", "CA_inv = zeros(7)\n", "\n", "# Calculations\n", "#Guesmath.sing 1st order kinetics\n", "for i in range(7):\n", " k[i] = math.log(CAo/C_A[i]);\n", " CA_inv[i] = 1/C_A[i];\n", "\n", "T = array([18.5,23,35]);\n", "CAo = array([10,5,2]);\n", "CA = zeros(3)\n", "log_Tf = zeros(3)\n", "log_CAo = zeros(3)\n", "\n", "for i in range(3):\n", " CA[i] = 0.8*CAo[i];\n", " log_Tf[i] = math.log10(T[i]);\n", " log_CAo[i] = math.log10(CAo[i]);\n", "\n", "# Results\n", "plot(log_CAo,log_Tf)\n", "plot(log_CAo,log_Tf,\"go\")\n", "xlabel(\"Ln CAO\")\n", "ylabel(\"log r\")\n", "#plot(log_Tf,log_CAo)\n", "#coeff1 = linalg.lstsq(log_CAo,log_Tf);\n", "slope, intercept, r_value, p_value, std_err = stats.linregress(log_CAo,log_Tf)\n", "coeff1 = stats.linregress(log_CAo,log_Tf)\n", "n = 1-coeff1[0];\n", "print \"From graph we get slope and intercept for calculating rate eqn\"\n", "k1 = ((0.8**(1-n))-1)*(10.**(1-n))/(18.5*(n-1));\n", "print \" The rate equation is given by %.3f\"%(k1),\n", "print \"CA**1.4 mol/litre.sec\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n", "From graph we get slope and intercept for calculating rate eqn" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "\n", " The rate equation is given by 0.005 CA**1.4 mol/litre.sec\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2 page no : 65" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%pylab inline\n", "\n", "import math\n", "# Variables\n", "CA = array([10,8,6,5,3,2,1]); # concentration\n", "T = array([0,20,40,60,120,180,300]); # time\n", "y = array([-0.1333,-0.1031,-0.0658,-0.0410,-0.0238,-0.0108,-0.0065]); # slope\n", "\n", "log_y = zeros(7)\n", "log_CA = zeros(7)\n", "\n", "# Calculations\n", "for i in range(7):\n", " log_y[i] = log10(complex(y[i]));\n", " log_CA[i] = log10(CA[i]);\n", "\n", "\n", "# Results\n", "plot(log_CA,log_y)\n", "plot(log_CA,log_y,\"go\")\n", "xlabel(\"log10 CA\")\n", "ylabel(\"log10 (-dCA/dt)\")\n", "show()\n", "coeff1 = stats.linregress(log_CA,log_y);\n", "n = coeff1[0];\n", "k = -10**(coeff1[1]);\n", "print \" After doing linear regression, the slope and intercept of the graph is %.0f, %.0f\"%(-coeff1[1],coeff1[0])\n", "print \" The rate equation is therefore given by %.3f\"%(-k),\n", "print \"CA**1.375 mol/litre.sec\"\n", "print ('The answer slightly differs from those given in book as regress fn is used for \\\n", " calculating slope and intercept')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n" ] }, { "output_type": "stream", "stream": "stderr", "text": [ "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n", "`%pylab --no-import-all` prevents importing * from pylab and numpy\n", "-c:14: ComplexWarning: Casting complex values to real discards the imaginary part\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] }, { "output_type": "stream", "stream": "stdout", "text": [ " After doing linear regression, the slope and intercept of the graph is 2, 1\n", " The rate equation is therefore given by 0.005 CA**1.375 mol/litre.sec\n", "The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.4 page no : 73" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "k1 = 2.3 # temperatures\n", "k2 = 2.3\n", "T1 = 400. # K\n", "T2 = 500. # K\n", "\n", "# Calculations\n", "R = 82.06*10**-6;\n", "R1 = 8.314\n", "E = (math.log(k2/k1)*R)/(1./T1-1./T2)\n", "\n", "# Results\n", "print \"using pressure units is %.0f EJ/mol\"%(E)\n", "\n", "#pA = CA*RT\n", "#-rA = 2.3(RT)**2*CA**2\n", "k1 = 2.3*(R*T1)**2\n", "k2 = 2.3*(R*T2)**2\n", "E = (math.log(k2/k1)*R1)/(1./T1-1./T2)\n", "print \"using concentration units is %.f J/mol\"%(E)\n", "\n", "# Answers might be different because of Rounding error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "using pressure units is 0 EJ/mol\n", "using concentration units is 7421 J/mol\n" ] } ], "prompt_number": 3 }, { "cell_type": "code", "collapsed": false, "input": [ "\n" ], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }