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{
"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": {}
}
]
}
|
