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| author | hardythe1 | 2015-04-07 15:58:05 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-04-07 15:58:05 +0530 |
| commit | c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131 (patch) | |
| tree | 725a7d43dc1687edf95bc36d39bebc3000f1de8f /Thermodynamics_for_Engineers/Chapter_15_3.ipynb | |
| parent | 62aa228e2519ac7b7f1aef53001f2f2e988a6eb1 (diff) | |
| download | Python-Textbook-Companions-c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131.tar.gz Python-Textbook-Companions-c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131.tar.bz2 Python-Textbook-Companions-c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131.zip | |
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diff --git a/Thermodynamics_for_Engineers/Chapter_15_3.ipynb b/Thermodynamics_for_Engineers/Chapter_15_3.ipynb new file mode 100755 index 00000000..25cf3bae --- /dev/null +++ b/Thermodynamics_for_Engineers/Chapter_15_3.ipynb @@ -0,0 +1,158 @@ +{
+ "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": {}
+ }
+ ]
+}
\ No newline at end of file |