{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 9 : Temperature and Pressure Effects" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.1 page no : 209" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "# Variables\n", "CpA = 35. # j/mol.K\n", "CpB = 45. # j/mol.K\n", "CpR = 70. # j/mol.K\n", "T1 = 25. # C\n", "T2 = 1025. # C\n", "Hr = -50000.\n", "\n", "# Calculations\n", "#Enthalpy balance for 1mol A,1 mol B,2 mol R\n", "nA = 1.\n", "nB = 1.\n", "nR = 2.\n", "dH = nA*CpA*(T1-T2)+nB*CpB*(T1-T2)+(Hr)+nR*CpR*(T2-T1);\n", "\n", "# Results\n", "print \" dHJ at temperature 1025C is %.f \"%(dH)\n", "if dH>0 :\n", " print \"Reaction is Exothermic\"\n", "else:\n", " print \"Reaction is endothermic at 1025OC\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " dHJ at temperature 1025C is 10000 \n", "Reaction is Exothermic\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.2 page no : 213" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "%matplotlib inline\n", "\n", "import math \n", "from matplotlib.pyplot import *\n", "from numpy import *\n", "\n", "Ho = -75300. # J/mol\n", "Go = -14130. # J/mol\n", "R = 8.3214\n", "T1 = 298.\n", "\n", "# Calculations\n", "#With all specific heais alike,dCp = 0\n", "Hr = -Ho;\n", "K298 = math.exp(-Go/(R*T1));\n", "\n", "\n", "#Taking different values of T\n", "T1 = array([2,15,25,35,45,55,65,75,85,95]) #degree celcius\n", "T = array([278,288,298,308,318,328,338,348,358,368]) #kelvin\n", "\n", "XAe = zeros(10)\n", "\n", "for i in range(10):\n", " K = K298*math.exp((Hr/R)*((1./T[i])-(1./298)));\n", " XAe[i] = K/(K+1);\n", "\n", "# Results\n", "plot(T1,XAe)\n", "xlabel(\"Temperature, C\")\n", "ylabel(\"XAe\")\n", "print (\" From the graph we see temp must stay below 78 C if conversion of 75% or above is expected\")\n", "show()\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n", " From the graph we see temp must stay below 78 C if conversion of 75% or above is expected" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "\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" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.3 page no : 217" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "# Variables \n", "\n", "# from example 9.2\n", "XA = 0.581; \n", "t = 1. #min\n", "XAe = 0.89;\n", "XAe1 = 0.993;\n", "T1 = 338.\n", "T2 = 298.\n", "R = 8.314\n", "\n", "\n", "# Calculations\n", "k1_338 = -(XAe/t)*math.log(1-(XA/XAe));\n", "XA1 = 0.6;\n", "t1 = 10. #min\n", "\n", "k1_298 = -(XAe1/t1)*math.log(1-(XA1/XAe1));\n", "E1 = (R*math.log(k1_338/k1_298))*(T1*T2)/(T1-T2)\n", "ko = k1_338/(math.exp(-E1/(R*T1)))\n", "\n", "# Results\n", "print \" The rate constants are k = exp[75300/RT-24.7] min-1\"\n", "print \" k1 = exp[17.2-48900/RT] min-1\"\n", "print \" k2 = exp[41.9-123800/RT] min-1 \"\n", "print \" E1 = %.2f\"%E1\n", "print \" K0 = %.2f\"%ko" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The rate constants are k = exp[75300/RT-24.7] min-1\n", " k1 = exp[17.2-48900/RT] min-1\n", " k2 = exp[41.9-123800/RT] min-1 \n", " E1 = 48679.81\n", " K0 = 31411847.30\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.4 page no : 229" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "CAo = 4. #mol/litre\n", "FAo = 1000. #mol/min\n", "A = 0.405 #litre/mol.min\n", "\n", "# Calculations\n", "t = CAo*A;\n", "V = FAo*A;\n", "\n", "# Results\n", "print \" Part a\"\n", "print \" The space time needed is %.2f min\"%(t)\n", "print \" The Volume needed is %.f litres\"%(V)\n", "\n", "\n", "# Note : We do not have value of 'rA'. so part B can not be calculated and plotted" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Part a\n", " The space time needed is 1.62 min\n", " The Volume needed is 405 litres\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.5 pageno : 231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "CAo = 4. # mol/liter\n", "FAo = 1000. # mol/min\n", "XA = 0.8; # %\n", "Cp = 250. #cal/molA.K\n", "Hr = 18000. #cal/molA\n", "rA = 0.4;\n", "\n", "# Calculations and Results\n", "V = FAo*XA/rA;\n", "\n", "print \" Part a\"\n", "print \" The size of reactorlitres) needed is %.f litres\"%(V)\n", "\n", "slope = Cp/Hr;\n", "#Using graph\n", "Qab1 = Cp*20; #cal/molA\n", "Qab = Qab1*1000; #cal/min\n", "Qab = Qab*0.000070; #KW\n", "\n", "print \" Part b\"\n", "print \" Heat Duty of precooler is %.2f kW\"%(Qab)\n", "\n", "Qce1 = Cp*37; #cal/molA fed\n", "Qce = Qce1*1000; #cal/min\n", "Qce = Qce*0.000070; #KW\n", "\n", "print \" Heat Duty of postcooler is %.2f kW\"%(Qce)\n", "\n", "# answers may vary because of rounding error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Part a\n", " The size of reactorlitres) needed is 2000 litres\n", " Part b\n", " Heat Duty of precooler is 350.00 kW\n", " Heat Duty of postcooler is 647.50 kW\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.6 pageno : 233" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "FAo = 1000. #mol/min\n", "Area = 1.72;\n", "\n", "# Calculations\n", "V = FAo*Area;\n", "\n", "# Results\n", "print \" The volume of adiabatic plug flow reactor is %.f litres\"%(V),\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The volume of adiabatic plug flow reactor is 1720 litres\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.7 page no : 234" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "FAo = 1000. #mol/min\n", "Area = (0.8-0)*1.5;\n", "\n", "# Calculations\n", "V = FAo*Area;\n", "\n", "# Results\n", "print \" The volume required is %.f litres\"%(V),\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The volume required is 1200 litres\n" ] } ], "prompt_number": 13 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }