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{
"metadata": {
"name": "",
"signature": "sha256:c0cb9f8947e3855e41c033660942a2e70a38f8ec8f1601ec17db209e7514083a"
},
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"nbformat_minor": 0,
"worksheets": [
{
"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": {}
}
]
}
|