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{
"metadata": {
"name": "",
"signature": "sha256:a8391f5a05f23d60fe8f4f8931e579dcdcd76160c3d9f4755fb607b65fe83b63"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"15: Nuclear Energy Sources"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.1, Page number 290"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"#m is mass of neutron and M is mass of other neucleus\n",
"ma=1;\n",
"Ma=2;\n",
"mb=1;\n",
"Mb=12;\n",
"mc=1;\n",
"Mc=238; \n",
"\n",
"#Calculation\n",
"eeta1=(4*ma*Ma/((ma+Ma)**2))*100; #Maximum fraction of KE lost by a neutron for H2(%)\n",
"eeta2=(4*mb*Mb/((mb+Mb)**2))*100; #Maximum fraction of KE lost by a neutron for C12(%)\n",
"eeta3=(4*mc*Mc/((mc+Mc)**2))*100; #Maximum fraction of KE lost by a neutron for U238(%)\n",
"\n",
"#Result\n",
"print \"Maximum fraction of KE lost by a neutron for H2 is\",round(eeta1,1),\"%\"\n",
"print \"Maximum fraction of KE lost by a neutron for C12 is\",round(eeta2,1),\"%\"\n",
"print \"Maximum fraction of KE lost by a neutron for U238 is\",round(eeta3,2),\"%\"\n",
"print \"answer for eeta2 given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum fraction of KE lost by a neutron for H2 is 88.9 %\n",
"Maximum fraction of KE lost by a neutron for C12 is 28.4 %\n",
"Maximum fraction of KE lost by a neutron for U238 is 1.67 %\n",
"answer for eeta2 given in the book is wrong\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.2, Page number 291"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"E=200; #energy released per fission(MeV)\n",
"e=1.6*10**-19; #the charge on electron(C)\n",
"Na=6.02*10**26; #Avgraodo no.(per kg mole)\n",
"\n",
"#Calculation\n",
"CE=E*e*10**6; #conversion in J\n",
"RF=1/CE; #fission rate(fissions/second)\n",
"Ekg=CE*Na/235; #Energy realeased in complete fission of 1 kg(J)\n",
"\n",
"#Result\n",
"print \"fission rate is\",round(RF/10**10,1),\"*10**10 fissions/second\"\n",
"print \"Energy realeased in complete fission of 1 kg is\",round(Ekg/1e+13,1),\"*10**13 J\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"fission rate is 3.1 *10**10 fissions/second\n",
"Energy realeased in complete fission of 1 kg is 8.2 *10**13 J\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.3, Page number 291"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"R=3*10**7; #rate of energy development(Js)\n",
"E=200; #energy released per fission(MeV)\n",
"e=1.6*10**-19; #the charge on electron(C)\n",
"t=1000; #time(hours)\n",
"Ekg=8.2*10**13; #energy released per kg of U-235\n",
"\n",
"#Calculation\n",
"CE=E*e*10**6; #conversion in J\n",
"n=R/CE; #no of atoms undergoing fission/second\n",
"TE=R*t*3600; #energy produced in 1000 hours(J)\n",
"MC=TE/Ekg; #mass consumed(kg) \n",
"\n",
"#Result\n",
"print \"number of atoms undergoing fissions per second is\",round(n/1e+17,1),\"*10**17\"\n",
"print \"mass consumed is\",round(MC,2),\"kg\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of atoms undergoing fissions per second is 9.4 *10**17\n",
"mass consumed is 1.32 kg\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.4, Page number 292"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"EPF=180; #Energy consumed per disintegration(MeV)\n",
"E=1200; #average power(kW)\n",
"t=10; #time(hours)\n",
"Na=6.02*10**26; #Avgraodo no.(per kg mole)\n",
"e=1.6*10**-19; #the charge on electron(C)\n",
"\n",
"#Calculation\n",
"TE=E*t; #energy consumed(kWh)\n",
"TE=TE*36*10**5; #conversion(J)\n",
"EE=TE/0.2; #efficient energy\n",
"CE=EPF*e*10**6; #conversion in J\n",
"n=EE/CE;\n",
"m=235*n/Na*1000; #mass consumed(gram)\n",
"\n",
"#Result\n",
"print \"mass consumed is\",round(m,2),\"gram\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mass consumed is 2.93 gram\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.5, Page number 292"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"OE=200; #output power(MW)\n",
"E=200; #energy released per fission(MeV)\n",
"WF=3.1*10**10; #fission rate(fissions/second)\n",
"Na=6.02*10**26; #Avagadro no.(per kg mole)\n",
"\n",
"#Calculation\n",
"IE=OE/0.3*10**6; #reactor input(W)\n",
"TFR=WF*IE;\n",
"n=TFR*24*3600; #no. of U-235 for one day\n",
"m=235*n/Na; #mass required(kg)\n",
" \n",
"#Result\n",
"print \"amount of natural uranium consumed per day is\",round(m*100/0.7,3),\"kg\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"amount of natural uranium consumed per day is 99.577 kg\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.6, Page number 292"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"AE=100; #electrical power(MW)\n",
"E=200; #energy released per fission(MeV)\n",
"e=1.6*10**-19; #the charge on electron(C)\n",
"Na=6.02*10**26; #Avagadro no.(per kg mole)\n",
"\n",
"#Calculation\n",
"TE=AE*10**6*24*3600; #energy consumed in city in one day(J)\n",
"EE=TE/0.2;\n",
"CE=E*e*10**6; #conversion in J\n",
"n=EE/CE; #no. of atoms to be fissioned \n",
"m=235*n/Na; #amount of fuel required(kg)\n",
"\n",
"#Result\n",
"print \"amount of fuel required is\",round(m,2),\"kg\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"amount of fuel required is 0.53 kg\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.7, Page number 293"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"OE=3000; #output power(MWh)\n",
"E=200; #energy released per fission(MeV)\n",
"e=1.6*10**-19; #the charge on electron(C)\n",
"Na=6.02*10**26; #Avagadro no.(per kg mole)\n",
"\n",
"#Calculation\n",
"IE=OE/0.2; #nuclear energy input(MWh)\n",
"TE=IE*36*10**8; #conversion in J\n",
"CE=E*e*10**6; #conversion in J\n",
"n=TE/CE; #number of nuclides required per day\n",
"m=235*n/Na; #daily fuel requirement(kg)\n",
"\n",
"#Result\n",
"print \"daily fuel requirement is\",round(m,3),\"kg or\",round(m,3)*1000,\"gram\"\n",
"print \"answer given in the book varies due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"daily fuel requirement is 0.659 kg or 659.0 gram\n",
"answer given in the book varies due to rounding off errors\n"
]
}
],
"prompt_number": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.8, Page number 293"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#import modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"OP=32824; #output power(kW)\n",
"E=200; #energy released per fission(MeV)\n",
"Ekg=8.2*10**13; #energy released per kg of U-235(J)\n",
"\n",
"#Calculation\n",
"DOP=OP*1000*24*3600; #daily output power(J)\n",
"IP=DOP/0.2; #nuclear energy input(J)\n",
"DFC=IP/Ekg; #daily fuel consumption(kg)\n",
"DI=DOP/(0.8*4186); #daily input at 80% efficiency(kcal)\n",
"Crqd=DI/(7*10**3); #Coal required per day(tonnes)\n",
"\n",
"#Result\n",
"print \"Daily fuel consumption is\",round(DFC,3)*1000,\"gram\"\n",
"print \"Coal required per day is\",int(Crqd),\"tonnes\"\n",
"print \"answer for coal required per day Crqd given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Daily fuel consumption is 173.0 gram\n",
"Coal required per day is 120981 tonnes\n",
"answer for coal required per day Crqd given in the book is wrong\n"
]
}
],
"prompt_number": 42
}
],
"metadata": {}
}
]
}
|
