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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter15-Nuclear fission and fusion"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg652"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.1 : : Page-652 (2011)\n",
      "#find Total leakage factor\n",
      "import math\n",
      "N_0_235 = 1.;        ## Number of uranium atom\n",
      "N_0_c = 10**5;        ## Number of graphite atoms per uranium atom\n",
      "sigma_a_235 = 698.;    ## Absorption cross section for uranium, barns\n",
      "sigma_a_c = 0.003;        ## Absorption cross section for graphite, barns\n",
      "f = N_0_235*sigma_a_235/(N_0_235*sigma_a_235+N_0_c*sigma_a_c );    ## Thermal utilization factor\n",
      "eta = 2.08;        ## Number of fast fission neutron produced\n",
      "k_inf = eta*f;        ## Multiplication factor\n",
      "L_m = 0.54;            ## Material length, metre\n",
      "L_sqr = ((L_m)**2.*(1.-f));   ## diffusion length, metre\n",
      "tau = 0.0364;                ## Age of the neutron\n",
      "B_sqr = 3.27;            ## Geometrical buckling\n",
      "k_eff = round (k_inf*math.exp(-tau*B_sqr)/(1+L_sqr*B_sqr));   ## Effective multiplication factor\n",
      "N_lf = k_eff/k_inf;    ## Non leakage factor\n",
      "lf = (1-N_lf)*100.;        ## Leakage factor, percent\n",
      "print'%s %.2f %s'%(\"\\n Total leakage factor = \",lf,\" percent\")\n",
      "\n",
      "## Result\n",
      "##  Total leakage factor = 31.3 percent  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " Total leakage factor =  31.26  percent\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2-pg652"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.2 : : Page-652 (2011)\n",
      "#find Neutron multiplication factor\n",
      "import math\n",
      "N_m = 50.;        ## Number of molecules of heavy water per uranium molecule\n",
      "N_u = 1.;        ## Number of uranium molecules \n",
      "sigma_a_u = 7.68;        ## Absorption cross section for uranium, barns\n",
      "sigma_s_u = 8.3;        ## Scattered cross section for uranium, barns\n",
      "sigma_a_D = 0.00092;    ## Absorption cross section for heavy water, barns\n",
      "sigma_s_D = 10.6;        ## Scattered cross section for uranium, barns \n",
      "f = N_u*sigma_a_u/(N_u*sigma_a_u+N_m*sigma_a_D );        ## Thermal utilization factor\n",
      "zeta = 0.570;        ## Average number of collisions\n",
      "N_0 = N_u*139./140.;        ## Number of U-238 atoms per unit volume \n",
      "sigma_s = N_m/N_0*sigma_s_D;    ## Scattered cross section, barns\n",
      "sigma_a_eff = 3.85*(sigma_s/N_0)**0.415;    ## Effective absorption cross section, barns\n",
      "p = math.exp(-sigma_a_eff/sigma_s);        ## Resonance escape probablity\n",
      "eps = 1.;                ## Fast fission factor\n",
      "eta = 1.34;            ## Number of fast fission neutron produced\n",
      "k_inf = eps*eta*p*f;        ## Effective multiplication factor\n",
      "print'%s %.2f %s'%(\"\\nNeutron multiplication factor =  \", k_inf,\"\");\n",
      "\n",
      "## Result\n",
      "## Neutron multiplication factor =  1.2  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Neutron multiplication factor =   1.21 \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg652"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.3 : : Page-652 (2011)\n",
      "#find The required multiplication factor\n",
      "import math\n",
      "## For graphite\n",
      "sigma_a_g = 0.0032;        ## Absorption cross section for graphite, barns\n",
      "sigma_s_g = 4.8;        ## Scattered cross section for graphite, barns\n",
      "zeta = 0.158;        ## Average number of collisions\n",
      "N_m = 50.;        ## Number of molecules of graphite per uranium molecule\n",
      "## For uranium\n",
      "sigma_f = 590.;     ## Fissioning cross section, barns\n",
      "sigma_a_u = 698.;        ## Absorption cross section for U-235, barns\n",
      "sigma_a_238 = 2.75;        ## Absorption cross section for U-238, barns\n",
      "v = 2.46;            ## Number of fast neutrons emitted\n",
      "N_u = 1            ## Number of uranium atoms \n",
      "f = N_u*sigma_a_u/(N_u*sigma_a_u+N_m*sigma_a_g );        ## Thermal utilization factor\n",
      "N_0 = N_u*(75./76.);        ## Number of U-238 atoms per unit volume\n",
      "sigma_s = N_m*76./75.*sigma_s_g/N_u;        ## Scattered cross section, barns\n",
      "sigma_eff = 3.85*(sigma_s/N_0)**0.415;        ## Effective cross section, barns\n",
      "p = math.exp(-sigma_eff/sigma_s);        ## Resonance escape probability, barns\n",
      "eps = 1.;        ## Fast fission factor\n",
      "eta = 1.34;        ## Number of fast fission neutron produced\n",
      "k_inf = eps*eta*p*f;        ## Multiplication factor\n",
      "print'%s %.2f %s'%(\"\\nThe required multiplication factor =  \", k_inf,\"\");\n",
      "\n",
      "## Result\n",
      "## The required multiplication factor = 1.1  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The required multiplication factor =   1.15 \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg653"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.4 : : Page-653 (2011)\n",
      "#find The ratio of number of uranium atoms to graphite atoms \n",
      "import math\n",
      "eta = 2.07;        ## Number of fast fission neutron produced\n",
      "x = 1./(eta-1.);     \n",
      "sigma_a_u = 687.;   ## Absorption cross section for uranium, barns\n",
      "sigma_a_g = 0.0045; ## Absorption cross section for graphite, barns\n",
      "N_ratio = x*sigma_a_g/sigma_a_u;    ## Ratio of number of uranium atoms to graphite atoms\n",
      "print'%s %.2e %s'%(\"\\nThe ratio of number of uranium atoms to graphite atoms = \", N_ratio,\"\");\n",
      "\n",
      "## Result\n",
      "## The ratio of number of uranium atoms to graphite atoms = 6.12e-006  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The ratio of number of uranium atoms to graphite atoms =  6.12e-06 \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg653"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.5 : : Page-653 (2011)\n",
      "import math \n",
      "#find The multiplication factor for LOPO reactor\n",
      "f = 0.754;        ## Thermal utilization factor\n",
      "sigma_s_o = 4.2;        ## Scattered cross section for oxygen, barns\n",
      "sigma_s_H = 20.;        ## Scattered cross section for hydrogen, barns\n",
      "N_O = 879.25;        ## Number of oxygen atoms\n",
      "N_238 = 14.19;        ## Number of uranium atoms\n",
      "N_H = 1573.;            ## Number of hydrogen atoms\n",
      "sigma_s = N_O/N_238*sigma_s_o+N_H/N_238*sigma_s_H;        ## Scattered cross section, barns\n",
      "N_0 = 14.19;        ## Number of U-238 per unit volume\n",
      "zeta_o = 0.120;    ## Number of collision for oxygen\n",
      "zeta_H = 1.;        ## Number of collision for hydrogen\n",
      "sigma_eff = (N_0/(zeta_o*sigma_s_o*N_O+zeta_H*sigma_s_H*N_H ));        ## Effective cross section, barns\n",
      "p = math.exp(-sigma_eff/sigma_s);        ## Resonance escape probablity\n",
      "eta = 2.08;        ## Number of fission neutron produced.\n",
      "eps = 1;        ## Fission factor\n",
      "K_inf = eps*eta*p*f;    ## Multiplication factor\n",
      "print'%s %.2f %s'%(\"\\nThe multiplication factor for LOPO reactor = \", K_inf,\"\");\n",
      "\n",
      "## Result\n",
      "## The multiplication factor for LOPO reactor = 1.6  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The multiplication factor for LOPO reactor =  1.57 \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg654"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.6 : : Page-654 (2011)\n",
      "#find The required controlled cross section\n",
      "import math\n",
      "r = 35;        ## Radius of the reactor, centi metre\n",
      "B_sqr = (math.pi/r)**2;    ## Geometrical buckling, per square centi metre\n",
      "D = 0.220;        ## Diffusion coefficient, centi metre\n",
      "sigma_a_f = 0.057;    ## Rate of absorption of thermal neutrons\n",
      "v = 2.5;        ## Number of fast neutrons emitted\n",
      "tau = 50.;        ## Age of the neutron\n",
      "sigma_f = 0.048;    ## Rate of fission\n",
      "sigma_a_c = -1/(1+tau*B_sqr)*(-v*sigma_f+sigma_a_f+B_sqr*D+tau*B_sqr*sigma_a_f);        ## Controlled cross section\n",
      "print'%s %.2f %s'%(\"\\nThe required controlled cross section =  \", sigma_a_c,\"\");\n",
      "\n",
      "## Result\n",
      "## The required controlled cross section = 0.0273  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The required controlled cross section =   0.03 \n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex7-pg655"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.7 : : Page-655 (2011)\n",
      "#find side of the cubical reactor nd critical radius of the reactor\n",
      "import math\n",
      "B_sqr = 65.;        ## Geometrical buckling\n",
      "a = math.sqrt(3*math.pi**2/B_sqr)*100.;    ## Side of the cubical reactor, centi metre\n",
      "R = round(math.pi/math.sqrt(B_sqr)*100.); ## Radius of the cubical reactor,centi metre\n",
      "print'%s %.2f %s %.2f %s '%(\"\\nThe side of the cubical reactor =\",a,\" cm\"and\" \\nThe critical radius of the reactor =\",R,\" cm\");\n",
      "\n",
      "## Result\n",
      "## The side of the cubical reactor = 67.5 cm\n",
      "## The critical radius of the reactor = 39 cm "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The side of the cubical reactor = 67.49  \n",
        "The critical radius of the reactor = 39.00  cm \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex8-pg655"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Exa15.8 : : Page-655 (2011)\n",
      "#find The critical volume of the reactor\n",
      "import math\n",
      "sigma_a_u = 698.;        ## Absorption cross section for uranium, barns\n",
      "sigma_a_M = 0.00092;        ## Absorption cross section for heavy water, barns\n",
      "N_m = 10**5;        ## Number of atoms of heavy water\n",
      "N_u = 1.;        ## Number of atoms of uranium\n",
      "f = sigma_a_u/(sigma_a_u+sigma_a_M*N_m/N_u);   ## Thermal utilization factor\n",
      "eta = 2.08;        ## Number of fast fission neutron produced\n",
      "k_inf = eta*f;        ## Multiplication factor\n",
      "L_m_sqr = 1.70;        ## Material length, metre\n",
      "L_sqr = L_m_sqr*(1-f);    ## Diffusion length, metre\n",
      "B_sqr = 1.819/0.30381*math.exp(-1/12.)-1./0.3038;    ## Geometrical buckling, per square metre\n",
      "V_c = 120./(B_sqr*math.sqrt(B_sqr));        ## Volume of the reactor, cubic metre\n",
      "print'%s %.2f %s'%(\"\\nThe critical volume of the reactor = \",V_c,\" cubic metre\");\n",
      "\n",
      "## Result\n",
      "## The critical volume of the reactor = 36.4 cubic metre "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The critical volume of the reactor =  36.35  cubic metre\n"
       ]
      }
     ],
     "prompt_number": 8
    }
   ],
   "metadata": {}
  }
 ]
}