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{
"metadata": {
"name": "",
"signature": "sha256:d798b16416897dedbfd57aff5c233c1f7e9c52a7f0081b45c972d6ab47468c7c"
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"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": {}
}
]
}
|
