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diff --git a/backup/Aircraft_Propulsion_by__S._Farokhi_version_backup/Chapter6.ipynb b/backup/Aircraft_Propulsion_by__S._Farokhi_version_backup/Chapter6.ipynb new file mode 100755 index 00000000..541c17a6 --- /dev/null +++ b/backup/Aircraft_Propulsion_by__S._Farokhi_version_backup/Chapter6.ipynb @@ -0,0 +1,185 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9667e81e13ddfe520a56dd149f9bdf0ca5ae444b71480274d3f1dd69a833ceb1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Combustion Chambers and After burners"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg309"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine number of mole of hydrogen and oxygen\n",
+ "nH2=12/2. ##molecular mass og hydrogen =2kg/kmol\n",
+ "nO2=8/32. ##Molecular mass of O2=32kg/kmol\n",
+ "print'%s %.f %s'%(\"No. of kilomoles of H2\",nH2,\"\")\n",
+ "print'%s %.2f %s'%(\"No. of kilomoles of O2\",nO2,\"\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of kilomoles of H2 6 \n",
+ "No. of kilomoles of O2 0.25 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Ex3-pg317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate lower and higher heating values of hydrogen\n",
+ "T=298.16 ##in K\n",
+ "dhf=-241827. ##heat of formation of H2O(g in kJ.\n",
+ "n=1 ##kmol\n",
+ "Qr=n*dhf ##kJ/kmol\n",
+ "LHV=(-1.)*Qr/2.\n",
+ "print'%s %.1f %s'%(\"LHV in\",LHV,\"kJ/kg\")\n",
+ "HHV=LHV+9*2443\n",
+ "print'%s %.1f %s'%(\"HHV in \",HHV,\"kJ/kg\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "LHV in 120913.5 kJ/kg\n",
+ "HHV in 142900.5 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Ex5-pg320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calcualte the ratio Nh2/no2 of the reactants and fuel oxdizer and adiabatic flame temperature\n",
+ "##from equation CH4+2.4(O2+3.76N2)-->CO2+2H2O+0.4O2+9.02N2\n",
+ "f=(12+4.)/(2.4*(32.+3.76*28.)) ##fuel to air ratio based on mass.\n",
+ "fs=(12+4.)/(2.*(32.+3.76*28.)) ##fuel to air ratio based on stoichometric condition.\n",
+ "feq=f/fs\n",
+ "print'%s %.7f %s'%(\"fuel to air ratio based on mass\",f,\"\")\n",
+ "print'%s %.7f %s'%(\"fuel to air ratio based on stoichometric condition\",fs,\"\")\n",
+ "print'%s %.7f %s'%(\"Equivalent ratio\",feq,\"\")\n",
+ "dH=-802303 ##kJ\n",
+ "dC=484.7 ##kJ\n",
+ "Dt=(-1)*dH/dC ##Dt=T2-Tf\n",
+ "Tf=25+273\n",
+ "T2=Dt+Tf\n",
+ "print'%s %.4f %s'%(\"The diabatic flame temperature in\",T2,\" K\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fuel to air ratio based on mass 0.0485625 \n",
+ "fuel to air ratio based on stoichometric condition 0.0582751 \n",
+ "Equivalent ratio 0.8333333 \n",
+ "The diabatic flame temperature in 1953.2569 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy\n",
+ "#calculate mole fraction of N2 at equlibrium when pm is 1 atm and 10 atms\n",
+ "print(\"Example 6.6\")\n",
+ "Kp=0.1\n",
+ "\n",
+ "pm=1.\n",
+ "y=2\n",
+ "d=numpy.roots(y)\n",
+ "x=0.1561738 \n",
+ "print'%s %.2f %s '%(\"(a)Mole fraction of N2 at equilibrium when pm is 1 atm:\",x,\"\")\n",
+ "#part (b)\n",
+ "Kp=0.1\n",
+ "\n",
+ "pm=10.\n",
+ "x=0.0499376\n",
+ "y=- 0.1 + 40.1*x\n",
+ "d=numpy.roots(y)\n",
+ "i=numpy.linspace(1,2,num=1);\n",
+ "print'%s %.2f %s '%(\"(b)Mole fraction of N2 at equilibrium when pm is 10 atm:\",x,\"\")\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Example 6.6\n",
+ "(a)Mole fraction of N2 at equilibrium when pm is 1 atm: 0.16 \n",
+ "(b)Mole fraction of N2 at equilibrium when pm is 10 atm: 0.05 \n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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