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diff --git a/backup/Modern_Physics_version_backup/chapter15.ipynb b/backup/Modern_Physics_version_backup/chapter15.ipynb deleted file mode 100755 index 4227a2a8..00000000 --- a/backup/Modern_Physics_version_backup/chapter15.ipynb +++ /dev/null @@ -1,397 +0,0 @@ -{
- "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": {}
- }
- ]
-}
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