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{
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"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter9-Energy from the Oceans"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.3.5.1-pg 527"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex9.3.5.1.;Calculate Energy generated\n",
"import math\n",
"R=12.;##unit=m; R is the range\n",
"r=3.;##unit=m; the head below turbine stops operating\n",
"time=(44700./2.);\n",
"A=30*10**6;\n",
"g=9.80;\n",
"p=1025.;\n",
"##The total theoretical work W=integrate('1','w',R,r);\n",
"W=(g*p*A*((R**2)-(r**2)))/2.;\n",
"print'%s %.2f %s'%(\" W=\",W,\" \");\n",
"##The average power generated\n",
"Pav=W/time;##unit=watts\n",
"print'%s %.2f %s'%(\"\\n The average power generated=\",Pav,\" watts\");\n",
"pav=(Pav/1000.)*3600.;##unit=kWh\n",
"print'%s %.2f %s'%(\"\\n The average power generated=\",pav,\" kWh\")\n",
"##the energy generated\n",
"Energy_generated=pav*0.73\n",
"print'%s %.2f %s'%(\"\\n Energy generated=\",Energy_generated,\" kWh\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" W= 20341125000000.00 \n",
"\n",
" The average power generated= 910117449.66 watts\n",
"\n",
" The average power generated= 3276422818.79 kWh\n",
"\n",
" Energy generated= 2391788657.72 kWh\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.3.6.1-pg529"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex9.3.6.1;calculate power in h.p. at any instant and the yearly power output\n",
"import math\n",
"A=0.5*10**6;##unit=m\n",
"h0=8.5;##unit=m\n",
"t=3*3600.##unit=s; since t=3 hr\n",
"p=1025.;##unit=kg/m^3\n",
"h=8.;##unit=m\n",
"n0=0.70;##efficiency of the generator;70%\n",
"##volume of the basin=Ah0\n",
"volume_of_the_basin=A*h0;\n",
"##Average discharge Q=volume/time period\n",
"Q=(A*h0)/t;\n",
"print'%s %.2f %s %.2f %s '%(\" volume of the basin=\",volume_of_the_basin,\" m^3\"and \" \\n Average discharge Q=\",Q,\" m^3 /s\")\n",
"##power at any instant\n",
"P=((Q*p*h)/75)*n0;\n",
"print'%s %.2f %s'%(\"\\n power at any instant P=\",P,\" h.p.\");\n",
"##The total energy in kWh/tidal cycle\n",
"E=P*0.736*3;\n",
"print'%s %.2f %s'%(\"\\n The total energy in kWh/tidal cycle E=\",E,\"\");\n",
"##Total number of tidal cycle in a year=705\n",
"print(\"\\n Total number of tidal cycle in a year=705\");\n",
"##Therefore Total output per annum\n",
"Total_output_per_annum=E*705;\n",
"print'%s %.2f %s'%(\"\\n Total output per annum= \",Total_output_per_annum,\"kWh/year\");\n",
"\n",
"##The value of \"power of instant\" in a text book is misprinted.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" volume of the basin= 4250000.00 \n",
" Average discharge Q= 393.52 m^3 /s \n",
"\n",
" power at any instant P= 30117.28 h.p.\n",
"\n",
" The total energy in kWh/tidal cycle E= 66498.96 \n",
"\n",
" Total number of tidal cycle in a year=705\n",
"\n",
" Total output per annum= 46881768.89 kWh/year\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
