path: root/Power_Plant_Engineering_by_P._K._Nag/Ch9.ipynb

{
 "cells": [
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 9 : Nuclear Power Plant"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.1 Pg: 648"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The mass defect = 0.13703 amu \n",
      " The binding energy per nucleon = 7.97 MeV \n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "mp=1.007277##Atomic Mass of proton in amu\n",
    "mn=1.008665##Atomic Mass of neutron in amu\n",
    "me=0.00055##Atomic Mass of electron in amu\n",
    "mo=15.99491##Atomic Mass of oxygen in amu\n",
    "np=8##Number of protons in oxygen\n",
    "ne=8##Number of electrons in oxygen\n",
    "nn=8##Number of neutrons in oxygen\n",
    "a=931##One amu in MeV\n",
    "No=16##Number of nucleons in oxygen\n",
    "\n",
    "#Calculations\n",
    "m=(np*mp)+(ne*me)+(nn*mn)-mo##The mass defect in amu\n",
    "B=m*a##Binding energy in MeV\n",
    "Bn=B/No##Binding energy per nucleon\n",
    "\n",
    "#Output\n",
    "print \" The mass defect = %3.5f amu \\n The binding energy per nucleon = %3.2f MeV \"%(m,Bn)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.2 Pg: 649"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The decay constant = 1.36e-11 s**-1\n",
      "The initial activity of 1 g of radium 226 = 3.61e+10 dis/s\n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "amr=226.095##Atomic mass of radium in amu\n",
    "AC=6.023*10**23##Avogadro constant in molecules/g.mol\n",
    "h=1620##Half life of radium in years\n",
    "\n",
    "#Calculations\n",
    "D=(0.6931/(h*365*24*3600))##The decay constant in 1/s\n",
    "Na=AC/amr##Number of atoms per gram of radium \n",
    "Ao=D*Na##Initial activity in dis/s\n",
    "\n",
    "#Output\n",
    "print \"The decay constant = %0.2e s**-1\"%D\n",
    "print \"The initial activity of 1 g of radium 226 = %0.2e dis/s\"%Ao,"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.3 Pg: 649"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The fuel consumed of U-235 per day = 3.9 g/day \n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "#Input data\n",
    "F=190##Each fission of U-235 yeilds in MeV\n",
    "a=85##Assuming the Neutrons absorbed by U-235 cause fission in percentage\n",
    "b=15##Non fission capture to produce an isotope U-236 in percentage\n",
    "Q=3000##The amount of thermal power produced in MW\n",
    "\n",
    "#Calculations\n",
    "E=F*1.60*10**-13##Each fission yields a useful energy in J\n",
    "N=1/E##Number of fissions required \n",
    "B=((10**6)*(N*86400))/(a/100)##One day operation of a reactor the number of U-235 nuclei burned is in absorptions per day\n",
    "M=(B*235)/(6.023*10**23)##Mass of U-235 consumed to produce one MW power in g/day\n",
    "M1=M*3##Mass of U-235 consumed to produce 3000 MW power in g/day\n",
    "\n",
    "#Output\n",
    "print \"The fuel consumed of U-235 per day = %3.1f g/day \"%(M1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.4 Pg: 650"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The cross section of neutrons = 5.02 barns \n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "sa1=10##Cross section of nucleus in barns\n",
    "N=2200##Neutrons in m/s\n",
    "En1=0.1##Kinetic energy of neutrons increases in eV\n",
    "En2=0.02525##Kinetic energy of neutron in eV\n",
    "\n",
    "#Calculations\n",
    "sa2=sa1/((En1/En2)**0.5)##The cross section of neutrons in barns\n",
    "\n",
    "#Output\n",
    "print \"The cross section of neutrons = %3.2f barns \"%(sa2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.5 Pg: 650"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The microscopic absorption cross section of natural uranium = 7.6 barns \n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "U1=99.285##Uranium consists of U-238 in percentage \n",
    "U2=0.715##Uranium consists of U-235 in Percentage\n",
    "E=0.025##The energy of neutrons in eV\n",
    "sc=2.72##Capture cross section for U-238 in barns\n",
    "sf=0##fission cross section for U-238 in barns\n",
    "sc1=101##Capture cross section for U-235 in barns\n",
    "sf1=579##fission cross section for U-235 in barns\n",
    "\n",
    "#Calculations\n",
    "sa=(U1/100)*(sc+sf)+(U2/100)*(sc1+sf1)##The microscopic absorption cross section of natural uranium in barns\n",
    "\n",
    "#Output\n",
    "print \"The microscopic absorption cross section of natural uranium = %3.1f barns \"%(sa)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.6 Pg: 650"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The microscopic capture cross section of water = 0.0222 cm**-1 \n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "p=1##The density of water in g/cm**3\n",
    "sch=0.332##The microscopic capture cross section of hydrogen in barn\n",
    "sco=0.0002##The microscopic capture cross section of oxygen in barn\n",
    "\n",
    "#Calculations\n",
    "N=(6.023*10**23)*p/18##Number of molecules of water per cm**3\n",
    "scw=(2*N*sch*10**-24)+(N*sco*10**-24)##The microscopic capture cross section of water in cm**-1\n",
    "\n",
    "#output\n",
    "print \"The microscopic capture cross section of water = %3.4f cm**-1 \"%(scw)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.7 Pg: 650"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The thermal neutron flux = 9.10e+11 cm**-2s**-1\n",
      "The average distance that a neutron travels before it is absorbed = 0.010 cm\n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "m=230##The amount of boron piece in g\n",
    "mw=10##The molecular weight of boron \n",
    "R=9.57*10**13##Reaction rate in cm**-3s**-1\n",
    "d=2.3##Density of boron in g/cm**3\n",
    "sa=755##Absorbption cross section in barns\n",
    "ss=4##Elastic scattering cross section in barns\n",
    "\n",
    "#Calculations\n",
    "st=sa+ss##The total cross section in barns\n",
    "N=(d*6.023*10**23)/mw##The number density of neutrons in cm**-3\n",
    "S=N*st*10**-24##Number density of neutrons for total in cm**-1\n",
    "F=R/S##Neutron flux in cm**-2s**-1\n",
    "L=1/S##Average distance a neutron travels before it is absorbed in cm\n",
    "\n",
    "#Output\n",
    "print \"The thermal neutron flux = %0.2e cm**-2s**-1\"%F\n",
    "print \"The average distance that a neutron travels before it is absorbed = %0.3f cm\"%L"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.8 Pg: 651"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The logarithmic energy decrement representing the neutron energy loss per elastic collision = 0.158 \n",
      " The number of collisions necessary = 121 \n"
     ]
    }
   ],
   "source": [
    "from math import log\n",
    "#Input data\n",
    "Eni=4.8##The energy of the newly born electron in MeV\n",
    "Enf=0.025##The energy of the electron after slow down in eV\n",
    "A=12##The mass number of the graphite (carbon)\n",
    "\n",
    "#Calculations\n",
    "L=1-(((A-1)**2/(2*A))*log((A+1)/(A-1)))##The logarithmic energy decrement\n",
    "n=(log(Eni*10**6/Enf))/L##The number of collisions required to slowdown the neutron \n",
    "\n",
    "#Output\n",
    "print \" The logarithmic energy decrement representing the neutron energy loss per elastic collision = %3.3f \\n The number of collisions necessary = %3.0f \"%(L,n)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.9 Pg: 651"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " (a) The rating of the reactor = 3.36 MW/tonne \n",
      " (b)The rate of consumption of U-235 per day = 0.353 kg/day (or) 353 g/day \n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "f=100##The reactor is fuelled of natural uranium in tonnes\n",
    "A=238.05##The atomic mass of natural uranium \n",
    "F=10**13##The average thermal neutron flux in neutrons/cm**2s\n",
    "A1=235.04##The atomic mass of U-235\n",
    "sf=579##The fission cross section of U-235 in barns\n",
    "sc=101##The capture cross section of U-235 in barns\n",
    "E=200##The energy released per fission in MeV\n",
    "P=0.715##U-235 in natural uranium in percentage\n",
    "N=2200##The average thermal neutron in m/s\n",
    "\n",
    "#Calculations\n",
    "n=((f*1000)*6.023*10**26*(P/100))/A##The number of U-235 atoms in the reactor in atoms\n",
    "R=(sf*10**-24)*F*n##The rate of fission in the reactor in fissions/s\n",
    "T=R*E*1.602*10**-19##Thermal power of the reactor in MW\n",
    "Rr=T/f##Rating the reactor MW/tonne\n",
    "Rc=(((R*A1*60*60*24))/(6.023*10**26))##The rate of consumption of U-235 by fission in kg/day\n",
    "Rcc=Rc*1000##The rate of consumption of U-235 by fission in g/day\n",
    "\n",
    "#Output\n",
    "print \" (a) The rating of the reactor = %3.2f MW/tonne \\n (b)The rate of consumption of U-235 per day = %3.3f kg/day (or) %3.0f g/day \"%(Rr,Rc,Rcc)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.10 Pg: 652"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The specific energy release rate for a light water uranium reactor = 138.13 W/cm**3\n"
     ]
    }
   ],
   "source": [
    "#Input data\n",
    "f=3.5##Mass fraction of U-235 in the fuel in percentage\n",
    "G=180##Energy per fission in Mev\n",
    "F=10**13##The neutron flux in neutrons/cm**2s\n",
    "sf=577##Fission cross section of U-235 in barns\n",
    "M=1.602*10**-13##One MeV in J\n",
    "\n",
    "#Calculations\n",
    "N=2.372*(f/100)*10**22##The fuel density for a uranium oxide fuel in nuclei/cm**3\n",
    "q=G*N*sf*10**-24*F##The rate of energy release in MeV/cm**3s\n",
    "qg=q*M##The rate of energy release in W/cm**3\n",
    "\n",
    "#Output\n",
    "print \"The specific energy release rate for a light water uranium reactor = %3.2f W/cm**3\"%(qg)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex: 9.11 Pg: 653"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The reactor power level at the end of 1s is 3.196 MW\n"
     ]
    }
   ],
   "source": [
    "from math import exp\n",
    "#Input data\n",
    "P=1##The operating power of a reactor in W\n",
    "K=1.0015##The effective multiplication factor of Reactor becomes suppercritical \n",
    "t=0.0001##The average neutron life in s\n",
    "t1=1.0001##Neutron life time in s\n",
    "\n",
    "#Calculations \n",
    "d=(K-1)/K##The reactivity \n",
    "Z=(d*P)/t##The number of neutrons\n",
    "n=exp(Z)/10**6##Neutron density * 10**6\n",
    "\n",
    "#Output\n",
    "print \"The reactor power level at the end of 1s is %3.3f MW\"%(n)"
   ]
  }
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