{"id":15356,"date":"2017-09-01T15:22:58","date_gmt":"2017-09-01T15:22:58","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=15356"},"modified":"2022-10-28T08:07:15","modified_gmt":"2022-10-28T08:07:15","slug":"diffusion-equation","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-equation\/","title":{"rendered":"Diffusion Equation"},"content":{"rendered":"<div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-60 lgc-tablet-grid-60 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:10px\"><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-quote su-quote-style-default\"><div   id="8n9s8rpp6h"    class=\"su-quote-inner su-u-clearfix su-u-trim\">The diffusion equation, based on Fick\u2019s law, provides an analytical solution of spatial neutron flux distribution in the multiplying system.\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15204\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general.png\" alt=\"diffusion equation - general\" width=\"485\" height=\"206\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general.png 485w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general-300x127.png 300w\" sizes=\"(max-width: 485px) 100vw, 485px\" \/><\/a><\/p>\n<p>To solve the diffusion equation, which is a second-order partial differential equation throughout the reactor volume, it is necessary to specify certain <strong>boundary conditions<\/strong>.<\/p><\/div><\/div>\n<p>In previous chapters, we introduced <strong>two bases for the derivation<\/strong> of the diffusion equation:<\/p>\n<p><strong><a title=\"Fick\u2019s Law\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/ficks-law\/\">Fick\u2019s law<\/a>:<\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Ficks-Law-3D.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15179\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Ficks-Law-3D.png\" alt=\"Ficks Law - 3D\" width=\"139\" height=\"42\" \/><\/a><\/p>\n<p>which states that neutrons diffuse from high concentration (high flux) to low concentration.<\/p>\n<p><strong><a title=\"Neutron Balance \u2013 Continuity Equation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/continuity-equation-neutron-balance\/\">Continuity equation<\/a>:<\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/neutron-balance-continuity-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15202\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/neutron-balance-continuity-equation.png\" alt=\"neutron balance - continuity equation\" width=\"199\" height=\"65\" \/><\/a><\/p>\n<p>which states that rate of change of neutron density = production rate \u2013 absorption rate \u2013 leakage rate.<\/p>\n<p>We return now to the neutron balance equation and <strong>substitute<\/strong> the neutron current density vector by <strong>J = -D\u2207\u0424<\/strong>. Assuming that \u2207.\u2207 = \u2207<sup>2<\/sup> = \u0394 \u00a0(therefore <strong>div J = <\/strong>-D div (\u2207\u0424) = <strong>-D\u0394\u0424<\/strong>) we obtain the <strong>diffusion equation<\/strong>.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15204\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general.png\" alt=\"diffusion equation - general\" width=\"485\" height=\"206\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general.png 485w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-general-300x127.png 300w\" sizes=\"(max-width: 485px) 100vw, 485px\" \/><\/a><\/p>\n<p>See also: <a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\">Diffusion Coefficient<\/a><\/p>\n<p>See also: <a title=\"Neutron Cross-section\" href=\"http:\/\/sitepourvtc.com\/neutron-cross-section\/\">Neutron Cross-section<\/a><\/p>\n<p>See also: <a title=\"Neutron Flux Density \u2013 Neutron Intensity\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-flux-neutron-intensity\/\">Neutron Flux Density<\/a><\/p>\n<p>The derivation of the diffusion equation is based on Fick\u2019s law which is derived under <strong>many assumptions<\/strong>. Therefore, the diffusion equation cannot be exact or valid at places with strongly differing <a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\">diffusion coefficients<\/a> or in strongly absorbing media. This implies that the diffusion theory may show deviations from a more accurate solution of the transport equation in the proximity of external neutron sinks, sources, and media interfaces.<\/p>\n<figure id=\"attachment_15207\" aria-describedby=\"caption-attachment-15207\" style=\"width: 373px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/parameters-of-diffusion-table.png\"><img loading=\"lazy\" class=\"size-full wp-image-15207\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/parameters-of-diffusion-table.png\" alt=\"Table of diffusion parameters\" width=\"383\" height=\"138\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/parameters-of-diffusion-table.png 383w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/parameters-of-diffusion-table-300x108.png 300w\" sizes=\"(max-width: 383px) 100vw, 383px\" \/><\/a><figcaption id=\"caption-attachment-15207\" class=\"wp-caption-text\">Diffusion parameters for thermal neutrons of 0.025 eV in some materials<\/figcaption><\/figure>\n<\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-40 lgc-tablet-grid-40 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:10px\"><\/div>\n<h2>Physical Interpretation of Fick&#8217;s Law<\/h2>\n<canvas id=\"canvas\"><\/canvas>\r\n<script src=\"http:\/\/d3js.org\/d3.v3.min.js\"><\/script>\r\n<script>\r\nvar num = 10000;\r\n\r\nvar canvas = document.getElementById(\"canvas\");\r\nvar width = canvas.width = 300;\r\nvar height = canvas.height = 400;\r\nvar ctx = canvas.getContext(\"2d\");\r\nvar radius = 70;\r\n\r\nvar particles = d3.range(num).map(function(i) {\r\n  return [Math.random(), Math.round(height*Math.random())];\r\n}); \r\n\r\nd3.timer(step);\r\n\r\nfunction step() {\r\n  ctx.fillStyle = \"rgba(255,255,255,0.3)\";\r\n  ctx.fillRect(0,0,width,height);\r\n  ctx.fillStyle = \"rgba(0,0,0,0.5)\";\r\n  particles.forEach(function(p) {\r\n    p[0] += Math.round(2*Math.random()-1);\r\n    p[1] += Math.round(2*Math.random()-1);\r\n    if (p[0] < 0) p[0] = 0;\r\n    if (p[0] > width) p[0] = 0;\r\n    if (p[1] < 0) p[1] = 0;\r\n    if (p[1] > height) p[1] = height;\r\n    drawPoint(p);\r\n  });\r\n};\r\n\r\nfunction drawPoint(p) {\r\n  ctx.fillRect(p[0],p[1],1,1);\r\n};\r\n<\/script>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Ficks-Law-physical-interpretation.png\"><img loading=\"lazy\" class=\"alignleft wp-image-15180 size-medium\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Ficks-Law-physical-interpretation-300x166.png\" alt=\"Ficks Law - physical interpretation\" width=\"300\" height=\"166\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Ficks-Law-physical-interpretation-300x166.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Ficks-Law-physical-interpretation.png 389w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>The physical interpretation is similar to the fluxes of gases. The <a title=\"Neutron\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/\">neutrons <\/a>exhibit a net flow in the direction of least density. This is a natural consequence of <strong>greater collision densities<\/strong> at positions of <strong>greater neutron densities<\/strong>.<\/p>\n<p>Consider neutrons passing through the plane at x=0 from left to right due to collisions to the plane\u2019s left. Since the concentration of neutrons and the flux is larger for negative values of x, there are <strong>more collisions per cubic centimeter on the left<\/strong>. Therefore more neutrons are scattered from left to right, then the other way around. Thus the neutrons naturally diffuse toward the right.<\/p><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:10px\"><\/div>\n<h2>Boundary Conditions<\/h2>\n<p>To solve the diffusion equation, which is a second-order partial differential equation throughout the reactor volume, it is necessary to specify certain <strong>boundary conditions<\/strong>. It is very dependent on the complexity of certain problem. <strong>One-dimensional problems<\/strong> solutions of diffusion equation contain <strong>two arbitrary constants<\/strong>. Therefore, in order to solve one-dimensional one-group diffusion equation, we need two boundary conditions to determine these coefficients. The most convenient boundary conditions are summarized in following few points:<div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:15px\"><\/div><\/p><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\u00a0 <div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Vacuum Boundary Condition<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<figure id=\"attachment_15245\" aria-describedby=\"caption-attachment-15245\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition.png\"><img loading=\"lazy\" class=\"size-medium wp-image-15245\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition-300x224.png\" alt=\"extrapolated length - boundary condition\" width=\"300\" height=\"224\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition-300x224.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition.png 516w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-15245\" class=\"wp-caption-text\">Neutron flux as a function of position near a free surface according to diffusion theory and transport theory.<\/figcaption><\/figure>\n<p>The diffusion equation is mostly solved in media with high densities, such as <a title=\"Neutron Moderator\" href=\"http:\/\/sitepourvtc.com\/neutron-moderator\/\">neutron moderators<\/a> (H<sub>2<\/sub>O, D<sub>2<\/sub>O, or graphite). The problem is usually bounded by air. The <a title=\"Mean Free Path\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/mean-free-path\/\"><strong>mean free path<\/strong><\/a> of a <a title=\"Neutron\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/\">neutron<\/a> in the air is <strong>much larger<\/strong> than in the moderator so that it is possible to treat it<strong> as a vacuum<\/strong> in neutron flux distribution calculations. The vacuum boundary condition supposes that<strong> no neutrons are entering a surface<\/strong>.<\/p>\n<p>If we consider that no neutrons are reflected from the vacuum back to the volume, the following condition can be derived from <a title=\"Fick\u2019s Law\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/ficks-law\/\"><strong>Fick\u2019s law<\/strong><\/a>:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15246\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-equation.png\" alt=\"extrapolated length - equation\" width=\"264\" height=\"85\" \/><\/a><\/p>\n<p>Where <strong>d \u2248 \u2154 \u03bb<sub>tr<\/sub><\/strong> is known as the <strong>extrapolated length<\/strong>, for\u00a0homogeneous, weakly absorbing media, an exact solution of the mono-energetic transport equation in this case yields <strong>d \u2248 0.7104 \u03bb<sub>tr<\/sub><\/strong>. The geometric interpretation of the previous equation is that the relative neutron flux near the boundary has a slope of <strong>-1\/d<\/strong>, i.e., the flux would extrapolate linearly to 0 at a distance d beyond the boundary. This <strong>zero flux boundary<\/strong> condition is more straightforward and can be written mathematically as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-equation2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15247\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-equation2.png\" alt=\"extrapolated length - equation2\" width=\"117\" height=\"79\" \/><\/a><\/p>\n<p>If d is not negligible, physical dimensions of the reactor are increased by d, and extrapolated boundary is formulated with dimension <strong>R<sub>e<\/sub> = R + d,<\/strong> and this condition can be written as <strong>\u03a6(R + d) = \u00a0\u03a6(R<sub>e<\/sub>) = 0.<\/strong><\/p>\n<p>It may seem the flux goes to 0 at an <strong>extrapolated length<\/strong> beyond the boundary. <strong>This interpretation is not correct.<\/strong> The flux <strong>cannot go to zero<\/strong> in a vacuum because there are no absorbers to absorb the neutrons. The flux only appears to be heading to the zero value at the extrapolation point.<\/p>\n<p>Note that the equation <strong>d \u2248 0.7104 \u03bb<sub>tr<\/sub><\/strong> is applicable to plane boundaries only. The formulas for curved boundaries can differ slightly. However, the difference is small unless the radius of curvature of the boundary is of the same order of magnitude as the extrapolated length.<\/p>\n<p><strong>Typical values of the extrapolated length:<\/strong><\/p>\n<p>The most common moderators have following <a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\">diffusion coefficients<\/a> (for thermal neutrons):<\/p>\n<p>D(H<sub>2<\/sub>O) = 0.142 cm<\/p>\n<p>D(D<sub>2<\/sub>O) = 0.84 cm<\/p>\n<p>D(Be) = 0.416 cm<\/p>\n<p>D(C) = 0.916 cm<\/p>\n<p>The thermal neutron <strong>extrapolated lengths<\/strong> are given by:<\/p>\n<p>d \u2248 0.7104 \u03bb<sub>tr<\/sub> = 0.7104 x 3 x D<\/p>\n<p><strong>therefore:<\/strong><\/p>\n<p><strong>H<sub>2<\/sub>O: d \u2248 0.30 cm<\/strong><\/p>\n<p><strong>D<sub>2<\/sub>O: d \u2248 1.79 cm<\/strong><\/p>\n<p><strong>Be: d \u2248 0.88 cm<\/strong><\/p>\n<p><strong>C: d \u2248 1.95 cm<\/strong><\/p>\n<p>As can be seen, this approximation is valid when the dimension L of the diffusing\u00a0medium is much larger than the extrapolated length, <strong>L &gt;&gt; d<\/strong>.<\/p><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Finite Flux Condition<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">This condition is determined from obvious requirement. The <a title=\"Neutron Flux Density \u2013 Neutron Intensity\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-flux-neutron-intensity\/\">flux <\/a>in the medium can reach only <strong>reasonable values,<\/strong> i.e., it must be real, non-negative, and single-valued. Also, the solution must be finite in those regions where the equation is valid, except perhaps at artificial singular points of a source distribution. This boundary condition can be written mathematically as:\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/finite-flux-condition-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15248\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/finite-flux-condition-equation.png\" alt=\"finite flux condition - equation\" width=\"192\" height=\"47\" \/><\/a><\/p>\n<p><strong><span style=\"font-weight: 400;\">These conditions are often used to eliminate <\/span><\/strong><span style=\"font-weight: 400;\">unnecessary functions<\/span> <strong><span style=\"font-weight: 400;\">from solutions.<\/span><\/strong><\/p><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Interface Conditions<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">It is also necessary to specify boundary conditions at an <strong>interface<\/strong> between<strong> two different media<\/strong>. At interfaces between two different diffusion media (such as between the <a title=\"Reactor Core\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor-core\/\">reactor core<\/a> and the <a title=\"Neutron Reflector\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor\/neutron-reflector\/\">neutron reflector<\/a>), on physical grounds, the <a title=\"Neutron Flux Density \u2013 Neutron Intensity\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-flux-neutron-intensity\/\">neutron flux<\/a> and the normal component of the <a title=\"Neutron Current Density\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/neutron-current-density\/\">neutron current <\/a>must be <strong>continuous<\/strong>. In other words, <strong>\u03c6 and J<\/strong> are not allowed to show a jump.\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/interface-condition-equations.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15249\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/interface-condition-equations.png\" alt=\"interface condition - equations\" width=\"268\" height=\"120\" \/><\/a><\/p>\n<p>It must be added, as <strong>J<\/strong> must be continuous, the flux gradient will show a jump if the <a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\">diffusion coefficients<\/a> in both media differ from each other.<\/p>\n<figure id=\"attachment_15221\" aria-describedby=\"caption-attachment-15221\" style=\"width: 509px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media.png\"><img loading=\"lazy\" class=\"size-full wp-image-15221\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media.png\" alt=\"diffusion equation - two media\" width=\"519\" height=\"374\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media.png 519w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media-300x216.png 300w\" sizes=\"(max-width: 519px) 100vw, 519px\" \/><\/a><figcaption id=\"caption-attachment-15221\" class=\"wp-caption-text\">Solution of the diffusion equation in two different non-multiplying diffusion media<\/figcaption><\/figure>\n<\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Source Condition<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">It was stated the diffusion equation is not valid near the <a title=\"Neutron Sources\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/neutron-sources\/\"><strong>neutron source<\/strong><\/a>. But the presence of the neutron source can be used as a <strong>boundary condition<\/strong> because all neutrons flowing through the <strong>bounding area<\/strong> of the source must come from the neutron source. This boundary condition depends on the source geometry and can be written mathematically as:\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-diffusion.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15250\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-diffusion.png\" alt=\"source condition - diffusion\" width=\"343\" height=\"206\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-diffusion.png 343w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-diffusion-300x180.png 300w\" sizes=\"(max-width: 343px) 100vw, 343px\" \/><\/a><\/p><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Albedo Boundary Condition<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">It is well known that each <a title=\"Reactor Core\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor-core\/\">reactor core<\/a> is surrounded by a <a title=\"Neutron Reflector\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor\/neutron-reflector\/\"><strong>neutron reflector<\/strong><\/a>. The reflector <strong>reduces the non-uniformity<\/strong> of the power distribution in the peripheral fuel assemblies, <strong>reduces neutron leakage,<\/strong> and reduces a coolant flow bypass of the core. The neutron reflector is a <strong>non-multiplying medium<\/strong>, whereas the reactor core is a multiplying medium.\n<p>On this special interface, we shall apply an <strong>albedo boundary condition<\/strong> to represent the neutron reflector. Albedo, the latin word for \u201cwhiteness\u201d, was defined by Lambert as the fraction of the incident light reflected diffusely by a surface.<\/p>\n<p>In reactor engineering, <strong>albedo<\/strong>, or the <strong>reflection coefficient<\/strong>, is defined as the <strong>ratio<\/strong> of <strong>exiting to entering neutrons,<\/strong> and we can express it in terms of <a title=\"Neutron Current Density\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/neutron-current-density\/\"><strong>neutron currents<\/strong><\/a> as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/albedo-reflection-coefficient-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15255\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/albedo-reflection-coefficient-equation.png\" alt=\"albedo - reflection coefficient - equation\" width=\"164\" height=\"66\" \/><\/a><\/p>\n<p>For sufficiently thick reflectors, it can be derived that albedo becomes<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/albedo-reflection-coefficient-equation2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15256\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/albedo-reflection-coefficient-equation2.png\" alt=\"albedo - reflection coefficient - equation2\" width=\"173\" height=\"56\" \/><\/a><\/p>\n<p>where <strong>D<sub>refl<\/sub><\/strong> is the <strong><a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\">diffusion coefficient <\/a><\/strong>in the reflector and the <strong>L<sub>refl<\/sub><\/strong> is the <strong>diffusion length<\/strong> in the reflector.<\/p>\n<p>If we are not interested in the neutron flux distribution in the reflector (let say in the slab B) but only in the effect of the reflector on the neutron flux distribution in the medium (let say in slab A), the albedo of the reflector can be used as a boundary condition for the diffusion equation solution. This boundary condition is similar to the <strong>vacuum boundary condition<\/strong>, i.e., <strong>\u03a6(R<sub>albedo<\/sub>) = 0<\/strong>, where <strong>R<sub>albedo<\/sub> = R + d<sub>e<\/sub><\/strong> and<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/albedo-boundary-condition-extrapolated.png\"><img loading=\"lazy\" class=\" size-full wp-image-15257 aligncenter\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/albedo-boundary-condition-extrapolated.png\" alt=\"albedo boundary condition - extrapolated\" width=\"186\" height=\"155\" \/><\/a><\/p><\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div><\/div><\/div>\n<div   id="8n9s8rpp6h"    class=\"collapsible-container\">\n<h2>Solutions of the Diffusion Equation &#8211; Non-multiplying Systems<\/h2>\n<p>As was previously discussed, the <strong>diffusion theory<\/strong> is widely used in the core design of the current Pressurized Water Reactors (PWRs) or Boiling Water Reactors (BWRs). This section is not about such calculations but provides illustrative insights, how can be the neutron flux distributed in any diffusion medium. In this section, we will solve diffusion equations:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-equations.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15270\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-equations.png\" alt=\"solution of diffusion equation - equations\" width=\"484\" height=\"204\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-equations.png 484w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-equations-300x126.png 300w\" sizes=\"(max-width: 484px) 100vw, 484px\" \/><\/a><\/p>\n<p>in various geometries that satisfy the <strong>boundary conditions<\/strong> discussed in the previous section.<\/p>\n<figure id=\"attachment_15285\" aria-describedby=\"caption-attachment-15285\" style=\"width: 198px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Solutions-of-the-Diffusion-Equation-Non-multiplying-Systems-min.png\"><img loading=\"lazy\" class=\"wp-image-15285 size-medium\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Solutions-of-the-Diffusion-Equation-Non-multiplying-Systems-min-208x300.png\" alt=\"Solution of diffusion equation\" width=\"208\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Solutions-of-the-Diffusion-Equation-Non-multiplying-Systems-min-208x300.png 208w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Solutions-of-the-Diffusion-Equation-Non-multiplying-Systems-min.png 586w\" sizes=\"(max-width: 208px) 100vw, 208px\" \/><\/a><figcaption id=\"caption-attachment-15285\" class=\"wp-caption-text\">Neutron flux distribution in various geometries of non-multiplying systems.<\/figcaption><\/figure>\n<p>We will start with simple systems and increase complexity gradually. The most important assumption is that <strong>all neutrons<\/strong> are lumped into a<strong> single energy group<\/strong>. They\u00a0are emitted and diffuse at <strong>thermal energy (0.025 eV)<\/strong>.<\/p>\n<p>In the first section, we will deal with neutron diffusion in a <strong>non-multiplying system<\/strong>, i.e., in the system where<a title=\"Fissile Material\" href=\"http:\/\/sitepourvtc.com\/glossary\/fissile-material\/\"> fissile isotopes<\/a> are missing, the <strong>fission cross-section is zero<\/strong>. An external neutron source emits the neutrons. We will assume that the system is uniform outside the source, i.e., <a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\"><strong>D<\/strong><\/a> and <a title=\"Neutron Absorption Cross-section\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-absorption\/neutron-absorption-cross-section\/\"><strong>\u03a3<sub>a <\/sub><\/strong><\/a>are constants.<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Solution for the Infinite Planar Source<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-planar-source-min.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15336\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-planar-source-min-300x157.png\" alt=\"Neutron Diffusion - planar source-min\" width=\"300\" height=\"157\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-planar-source-min-300x157.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-planar-source-min.png 527w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>Let assume the <a title=\"Neutron Sources\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/neutron-sources\/\"><strong>neutron source<\/strong><\/a> (with strength <strong>S<sub>0<\/sub><\/strong>) as an <strong>infinite plane source<\/strong> (in y-z plane geometry). In this geometry, the flux varies slowly in y and z allowing us to eliminate the y and z derivatives from \u2207<sup>2<\/sup>. The flux is then a <strong>function of x only<\/strong>, and therefore the <strong>Laplacian<\/strong> and <strong>diffusion equation<\/strong> (outside the source) can be written as (everywhere except x = 0):<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-equations2.png\"><img loading=\"lazy\" class=\" size-full wp-image-15271 aligncenter\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-equations2.png\" alt=\"solution of diffusion equation - equations2\" width=\"207\" height=\"219\" \/><\/a><\/p>\n<p>For x &gt; 0, this diffusion equation has <strong>two possible solutions<\/strong> <strong>exp(x\/L)<\/strong> and <strong>exp(-x\/L)<\/strong>, which give a general solution:<\/p>\n<p><strong>\u03a6(x) = Aexp(x\/L) + Cexp(-x\/L)<\/strong><\/p>\n<p>Note that B is not usually used as a constant because B is reserved for a <strong>buckling parameter<\/strong>. To determine the coefficients A and C, we must apply boundary conditions.<\/p>\n<p>From <strong>finite flux condition<\/strong> (0\u2264 \u03a6(x) &lt; \u221e), which required only reasonable values for the flux, it can be derived that <strong>A must be equal to zero<\/strong>. The term exp(x\/L) goes to \u221e as x \u279d\u221e \u00a0and therefore cannot be part of a physically acceptable solution for x &gt; 0. The physically acceptable solution for x &gt; 0 must then be:<\/p>\n<p><strong>\u03a6(x) = \u00a0Ce<sup>-x\/L<\/sup><\/strong><\/p>\n<p>where C is a constant that can be determined from the source condition at x \u279d0.<\/p>\n<p>If S<sub>0<\/sub> is the plane\u2019s source strength per unit area, then the number of neutrons crossing outwards per unit area in the positive x-direction <strong>must tend to S<sub>0<\/sub> \/2 as x \u279d0<\/strong>.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-planar-source.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15272\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-planar-source.png\" alt=\"source condition - planar source\" width=\"228\" height=\"286\" \/><\/a><\/p>\n<p>So that the solution may be written:<br \/>\n<a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-planar-source.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15273\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-planar-source.png\" alt=\"solution of diffusion equation - planar source\" width=\"199\" height=\"66\" \/><\/a><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span> Solution for the Point Source<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-point-source-min.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15335\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-point-source-min-300x160.png\" alt=\"Neutron Diffusion - point source-min\" width=\"300\" height=\"160\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-point-source-min-300x160.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-point-source-min.png 571w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>Let assume the <a title=\"Neutron Sources\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/neutron-sources\/\"><strong>neutron source<\/strong><\/a> (with strength S<sub>0<\/sub>) as an <strong>isotropic point source<\/strong> situated in spherical geometry. This point source is placed at the origin of coordinates. To solve the <strong>diffusion equation<\/strong>, we have to replace the <strong>Laplacian<\/strong> with its <strong>spherical form<\/strong>:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15274\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry.png\" alt=\"laplacian - spherical geometry\" width=\"493\" height=\"163\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry.png 493w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry-300x99.png 300w\" sizes=\"(max-width: 493px) 100vw, 493px\" \/><\/a><\/p>\n<p>We can replace the <strong>3D Laplacian<\/strong> with its <strong>one-dimensional spherical form<\/strong> because there is no dependence on the angle (whether polar or azimuthal). The source is assumed to be an <strong>isotropic source<\/strong> (there is the spherical symmetry). \u00a0The flux is then a function of <strong>radius &#8211; r<\/strong> only, and therefore the diffusion equation (outside the source) can be written as (everywhere except r = 0):<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-geometry.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15275\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-geometry.png\" alt=\"solution of diffusion equation - spherical geometry\" width=\"332\" height=\"223\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-geometry.png 332w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-geometry-300x202.png 300w\" sizes=\"(max-width: 332px) 100vw, 332px\" \/><\/a><\/p>\n<p>If we make the <strong>substitution<\/strong> <strong>\u03a6(r) = 1\/r \u03c8(r)<\/strong>, the equation simplifies to:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/substitution-diffusion.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15276\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/substitution-diffusion.png\" alt=\"substitution - diffusion\" width=\"227\" height=\"75\" \/><\/a><\/p>\n<p>For r &gt; 0, this differential equation has<strong> two possible solutions<\/strong> <strong>exp(r\/L)<\/strong> and <strong>exp(-r\/L)<\/strong>, which give a general solution:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-point-source.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15277\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-point-source.png\" alt=\"solution of diffusion equation - point source\" width=\"283\" height=\"147\" \/><\/a><\/p>\n<p>Note that B is not usually used as a constant because B is reserved for a <strong>buckling parameter<\/strong>. To determine the coefficients A and C, we must apply <strong>boundary conditions<\/strong>.<\/p>\n<p>To find constants A and C, we use the identical procedure as in the case of the infinite planar source. From<strong> finite flux condition<\/strong> (0\u2264 \u03a6(r) &lt; \u221e), which required only reasonable values for the flux, it can be derived that <strong>A must be equal to zero<\/strong>. The term exp(r\/L)\/r goes to \u221e as r \u279d\u221e \u00a0and therefore cannot be part of a physically acceptable solution for r &gt; 0. The physically acceptable solution for r &gt; 0 must then be:<\/p>\n<p>\u03a6(r) = \u00a0Ce<sup>-r\/L<\/sup>\/r<\/p>\n<p>where C is a constant that can be determined from <strong>source condition<\/strong> at x \u279d0.<\/p>\n<p>If S<sub>0<\/sub> is the source strength, then the number of neutrons crossing a sphere\u00a0outwards in the positive r-direction <strong>must tend to S<sub>0<\/sub> as r \u279d0<\/strong>.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-point-source.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15278\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-point-source.png\" alt=\"source condition - point source\" width=\"439\" height=\"368\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-point-source.png 439w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-point-source-300x251.png 300w\" sizes=\"(max-width: 439px) 100vw, 439px\" \/><\/a><\/p>\n<p>So that the solution may be written:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-point-source2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15279\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-point-source2.png\" alt=\"solution of diffusion equation - point source2\" width=\"190\" height=\"61\" \/><\/a><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span> Solution for the Line Source<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-line-source-min.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15334\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-line-source-min-300x180.png\" alt=\"Neutron Diffusion - line source-min\" width=\"300\" height=\"180\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-line-source-min-300x180.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Neutron-Diffusion-line-source-min.png 583w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>Let assume the <a title=\"Neutron Sources\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/neutron-sources\/\"><strong>neutron source<\/strong><\/a> (with strength S<sub>0<\/sub>) as an isotropic line source situated in an infinite <strong>cylindrical geometry<\/strong>. This line source is placed at r = 0. To solve the <strong>diffusion equation<\/strong>, we have to replace the Laplacian by its <strong>cylindrical form<\/strong>:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-cylindrical-geometry.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15280\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-cylindrical-geometry.png\" alt=\"laplacian - cylindrical geometry\" width=\"284\" height=\"161\" \/><\/a><\/p>\n<p>Since there is no dependence on angle \u0398 and z-coordinate, we can replace the 3D Laplacian with its <strong>one-dimensional form<\/strong>, and we can solve the problem only in the <strong>radial direction<\/strong>. The source is assumed to be an isotropic source. Since the flux is a function of radius &#8211; r only, the diffusion equation (outside the source) can be written as (everywhere except r = 0):<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-geometry.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15281\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-geometry.png\" alt=\"solution of diffusion equation - cylindrical geometry\" width=\"335\" height=\"221\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-geometry.png 335w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-geometry-300x198.png 300w\" sizes=\"(max-width: 335px) 100vw, 335px\" \/><\/a><br \/>\n<a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/modified-bessel-functions-first-and-second-kind.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15308\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/modified-bessel-functions-first-and-second-kind-209x300.png\" alt=\"modified bessel functions - first and second kind\" width=\"209\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/modified-bessel-functions-first-and-second-kind-209x300.png 209w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/modified-bessel-functions-first-and-second-kind.png 453w\" sizes=\"(max-width: 209px) 100vw, 209px\" \/><\/a>This differential equation is called <strong>Bessel\u2019s equation,<\/strong> and it is well known to mathematicians. In this case, the solutions to the Bessel\u2019s equation are called the <strong>modified Bessel functions<\/strong> (or occasionally the <strong>hyperbolic Bessel functions<\/strong>) <strong>of the first and second kind, <\/strong>I<sub>\u03b1<\/sub>(x) and K<sub>\u03b1<\/sub>(x), respectively.<\/p>\n<p>For r &gt; 0, this differential equation has two possible solutions, <strong>I<sub>0<\/sub>(r\/L)<\/strong> and <strong>K<sub>0<\/sub>(r\/L)<\/strong>, the modified Bessel functions of order zero, which give a general solution:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-bessel-functions.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15282\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-bessel-functions.png\" alt=\"solution of diffusion equation - bessel functions\" width=\"300\" height=\"60\" \/><\/a><\/p>\n<p>To find constants A and C, we use the identical procedure as in the case of the infinite planar source. From<strong> finite flux condition<\/strong> (0\u2264 \u03a6(r) &lt; \u221e), which required only reasonable values for the flux, it can be derived that A must be equal to zero. The term I<sub>0<\/sub>(r\/L) goes to \u221e as r \u279d\u221e \u00a0and therefore cannot be part of a physically acceptable solution for r &gt; 0. The physically acceptable solution for r &gt; 0 must then be:<\/p>\n<p><strong>\u03a6(r) = \u00a0C.K<sub>0<\/sub>(r\/L)<\/strong><\/p>\n<p>where C is a constant that can be determined from<strong> the source condition<\/strong> at r\u00a0\u279d0.<\/p>\n<p>If S<sub>0<\/sub> is the source strength, then the number of neutrons crossing a cylinder\u00a0outwards in the positive r-direction must tend to S<sub>0<\/sub> as r \u279d0.<\/p>\n<p>So that the solution may be written:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-line-source.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15283\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-line-source.png\" alt=\"solution of diffusion equation - line source\" width=\"207\" height=\"65\" \/><\/a><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Solution in two different diffusion media<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">A diffusion environment may consist of <strong>various zones<\/strong> of different composition. The consequence is that the <a title=\"Diffusion Coefficient\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-coefficient\/\"><strong>diffusion coefficient<\/strong><\/a>,\u00a0\u00a0<strong><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-absorption\/neutron-absorption-cross-section\/\">absorption macroscopic<\/a><\/strong><a title=\"Neutron Absorption Cross-section\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-absorption\/neutron-absorption-cross-section\/\"><strong>\u00a0cross-section<\/strong><\/a>, and, therefore, the <a title=\"Neutron Flux Density \u2013 Neutron Intensity\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-flux-neutron-intensity\/\">neutron flux distribution<\/a>\u00a0will vary per zone. For the determination of the flux distribution in various zones, the <strong>diffusion equations<\/strong> in zone 1 and zone 2 need to be solved:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-different-media.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15309\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-different-media.png\" alt=\"diffusion equation - two different media\" width=\"360\" height=\"131\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-different-media.png 360w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-different-media-300x109.png 300w\" sizes=\"(max-width: 360px) 100vw, 360px\" \/><\/a><\/p>\n<p>where <strong>a<\/strong> is the real width of zone 1 and<strong> b<\/strong> is the outer dimension of the diffusion environment, including the <strong>extrapolated distance<\/strong>. With problems involving two different diffusion media, the interface boundary conditions play a crucial role and must be satisfied:<\/p>\n<p><em>At interfaces between two different diffusion media (such as between the reactor core and the neutron reflector), on physical grounds, the <strong>neutron flux<\/strong> and the normal component of the <strong>neutron current<\/strong> must be <strong>continuous<\/strong>. In other words, \u03c6 and J are not allowed to show a jump.<\/em><\/p>\n<p>1., 2. Interface Conditions<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/interface-condition-equations.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15249\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/interface-condition-equations.png\" alt=\"interface condition - equations\" width=\"268\" height=\"120\" \/><\/a><\/p>\n<p>It must be added, as <strong>J<\/strong> must be continuous, the flux gradient will show a jump if the <strong>diffusion coefficients<\/strong> in both media differ from each other. Since the solution of these two diffusion equations requires four boundary conditions, we have to use two boundary conditions more.<\/p>\n<p>3. Finite Flux Condition<\/p>\n<p>The solution must be finite in those regions where the equation is valid, except perhaps at artificial singular points of a source distribution. \u00a0This boundary condition can be written mathematically as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/finite-flux-condition-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15248\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/finite-flux-condition-equation.png\" alt=\"finite flux condition - equation\" width=\"192\" height=\"47\" \/><\/a>4. Source Condition<\/p>\n<p>The presence of the <strong>neutron source<\/strong> can be used as a boundary condition because it is necessary that all neutrons flowing through the bounding area of the source must come from the neutron source. This boundary condition depends on the source geometry, and for planar source can be written mathematically as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-planar-source2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15310\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-planar-source2.png\" alt=\"source condition - planar source2\" width=\"222\" height=\"46\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-planar-source2.png 222w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/source-condition-planar-source2-220x46.png 220w\" sizes=\"(max-width: 222px) 100vw, 222px\" \/><\/a><\/p>\n<p>For x &gt; 0, these diffusion equations have the following appropriate solutions:<\/p>\n<p><strong>\u03a6<sub>1<\/sub>(x) = A<sub>1<\/sub>exp(x\/L<sub>1<\/sub>) + C<sub>1<\/sub>exp(-x\/L<sub>1<\/sub>)<\/strong><\/p>\n<p>and<\/p>\n<p><strong>\u03a6<sub>2<\/sub>(x) = A<sub>2<\/sub>exp(x\/L<sub>2<\/sub>) + C<sub>2<\/sub>exp(-x\/L<sub>2<\/sub>)<\/strong><\/p>\n<p>where the four constants must be determined with the use of the four boundary conditions. The typical neutron flux distribution in a simple two-region diffusion problem is shown in the picture below.<\/p>\n<figure id=\"attachment_15221\" aria-describedby=\"caption-attachment-15221\" style=\"width: 509px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media.png\"><img loading=\"lazy\" class=\"size-full wp-image-15221\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media.png\" alt=\"diffusion equation - two media\" width=\"519\" height=\"374\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media.png 519w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-two-media-300x216.png 300w\" sizes=\"(max-width: 519px) 100vw, 519px\" \/><\/a><figcaption id=\"caption-attachment-15221\" class=\"wp-caption-text\">Solution of the diffusion equation in two different non-multiplying diffusion media<\/figcaption><\/figure>\n<\/div><\/div><\/div>\n<h2>Solutions of the Diffusion Equation &#8211; Multiplying Systems<\/h2>\n<p>In the previous section, it has been considered that the environment is <strong>non-multiplying<\/strong>. In a non-multiplying environment, <a title=\"Neutron\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/\">neutrons<\/a> are emitted by a neutron source situated in the center of the coordinate system and then freely diffuse through media. We are now prepared to consider <strong>neutron diffusion<\/strong> in the <strong>multiplying system <\/strong>containing\u00a0<a title=\"Fissionable Material\" href=\"http:\/\/sitepourvtc.com\/glossary\/fissionable-material\/\">fissionable nuclei<\/a> (i.e., <strong>\u03a3<\/strong><strong><sub>f <\/sub><\/strong><strong>\u2260 0<\/strong>). In this multiplying system, we will also study the spatial distribution of neutrons, but in contrast to the non-multiplying environment, these neutrons can trigger <a title=\"Nuclear Fission\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/fission\/\"><strong>nuclear fission reactions<\/strong><\/a>.<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Multiplying systems from criticality point of view<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<p>The required condition for a <strong>stable, <a title=\"Nuclear Fission Chain Reaction\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/\">self-sustained fission chain reaction<\/a><\/strong> in a multiplying system (in a <a title=\"Nuclear Reactor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor\/\">nuclear reactor<\/a>) is that <strong>exactly every\u00a0<a title=\"Nuclear Fission\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/fission\/\">fission <\/a>initiates another fission<\/strong>. The minimum condition is for each nucleus undergoing fission to produce, on average, at least one neutron that causes fission of another nucleus. Also, the number of fissions occurring per unit time (<a title=\"Reaction Rate\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/reaction-rate\/\">the reaction rate<\/a>) within the system must be constant.<\/p>\n<p>This condition can be expressed conveniently in terms of <strong>the multiplication factor<\/strong>. The infinite multiplication factor is the ratio of the <strong>neutrons produced by fission<\/strong> in one <a title=\"Neutron Generation \u2013 Neutron Population\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/neutron-generation-neutron-population\/\">neutron generation<\/a> to the number of <strong>neutrons lost through absorption<\/strong> in the preceding neutron generation. This can be mathematically expressed as shown below.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/02\/Multiplication-Factor.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-13628\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/02\/Multiplication-Factor.png\" alt=\"Multiplication Factor\" width=\"485\" height=\"71\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/02\/Multiplication-Factor.png 485w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/02\/Multiplication-Factor-300x44.png 300w\" sizes=\"(max-width: 485px) 100vw, 485px\" \/><\/a><\/p>\n<p>The<a title=\"Four Factor Formula \u2013 Infinite Multiplication Factor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/four-factor-formula-infinite-multiplication-factor\/\"><strong>\u00a0infinite multiplication factor<\/strong><\/a> in a multiplying system measures the change in the fission <a title=\"Neutron Generation \u2013 Neutron Population\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/neutron-generation-neutron-population\/\">neutron population<\/a> from one neutron generation to the subsequent\u00a0generation.<\/p>\n<ul>\n<li><strong>k<sub>\u221e<\/sub>\u00a0&lt; 1<\/strong>. Suppose the multiplication factor for a multiplying system is <strong>less than 1.0. In that case,<\/strong>\u00a0the <strong>number of neutrons decreases<\/strong> in time (with the <a title=\"Mean Generation Time with Delayed Neutrons\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/fission\/delayed-neutrons\/mean-generation-time-with-delayed-neutrons\/\">mean generation time<\/a>), and the chain reaction will never be self-sustaining. This condition is known as <strong>the subcritical state<\/strong>.<\/li>\n<\/ul>\n<ul>\n<li><strong>k<sub>\u221e<\/sub> = 1<\/strong>. If the multiplication factor for a multiplying system is <strong>equal to 1.0<\/strong>, then there is <strong>no change in neutron population<\/strong> in time, and the chain reaction will be<strong> self-sustaining<\/strong>. This condition is known as <strong>the critical state<\/strong>.<\/li>\n<\/ul>\n<ul>\n<li><strong>k<sub>\u221e<\/sub>\u00a0&gt; 1<\/strong>. If the multiplication factor for a multiplying system is <strong>greater than 1.0<\/strong>, then the multiplying system produces <strong>more neutrons<\/strong>\u00a0than are needed to be self-sustaining. The number of neutrons is exponentially increasing in time (with the mean generation time). This condition is known as <strong>the supercritical state<\/strong>.<\/li>\n<\/ul>\n<figure id=\"attachment_228\" aria-describedby=\"caption-attachment-228\" style=\"width: 377px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2014\/11\/reactor_criticality.png\"><img loading=\"lazy\" class=\"size-full wp-image-228\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2014\/11\/reactor_criticality.png\" alt=\"Reactor criticality\" width=\"387\" height=\"333\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2014\/11\/reactor_criticality.png 387w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2014\/11\/reactor_criticality-300x258.png 300w\" sizes=\"(max-width: 387px) 100vw, 387px\" \/><\/a><figcaption id=\"caption-attachment-228\" class=\"wp-caption-text\">Reactor criticality. A &#8211; a supercritical state; B &#8211; a critical state; C &#8211; a subcritical state<\/figcaption><\/figure>\n<\/div><\/div>\n<p>In this section, we will solve the following <strong>diffusion equation<\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-multiplying-system.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15314\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-multiplying-system.png\" alt=\"diffusion equation - multiplying system\" width=\"314\" height=\"52\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-multiplying-system.png 314w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-multiplying-system-300x50.png 300w\" sizes=\"(max-width: 314px) 100vw, 314px\" \/><\/a><\/p>\n<p>in various geometries that satisfy the <strong>boundary conditions<\/strong>. In this equation,<strong> \u03bd<\/strong> is the number of neutrons emitted in fission, and \u00a0<strong>\u03a3<sub>f<\/sub><\/strong> is the\u00a0<a title=\"Macroscopic Cross-section\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/macroscopic-cross-section\/\">macroscopic cross-section<\/a> of the fission reaction.<strong>\u00a0\u0424 <\/strong> denotes a\u00a0<strong><a title=\"Reaction Rate\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/reaction-rate\/\">reaction rate<\/a><\/strong>. For example, the fission of <a href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-fuel\/uranium\/uranium-235\/\"><sup>235<\/sup>U<\/a> by thermal neutron yields <strong>2.43 neutrons. <\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Multiplying-Systems-min.png\"><img loading=\"lazy\" class=\"size-medium wp-image-15340 alignright\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Multiplying-Systems-min-193x300.png\" alt=\"Diffusion Theory - Multiplying Systems\" width=\"193\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Multiplying-Systems-min-193x300.png 193w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Multiplying-Systems-min.png 609w\" sizes=\"(max-width: 193px) 100vw, 193px\" \/><\/a>It must be noted that we will solve the diffusion equation without any external source. This is very important because such an equation is a <strong>linear homogeneous equation<\/strong> in the flux. Therefore if we find one solution of the equation, then any multiple is also a solution. This means that the<strong> absolute value<\/strong> of the <a title=\"Neutron Flux Density \u2013 Neutron Intensity\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-flux-neutron-intensity\/\">neutron flux<\/a> <strong>cannot possibly be deduced<\/strong> from the diffusion equation. This is different from problems with external sources, which determine the absolute value of the neutron flux.<\/p>\n<p>We will start with simple systems (planar) and increase complexity gradually. The most important assumption is that all neutrons are lumped into a <strong>single energy group<\/strong>. They\u00a0are emitted and diffuse at<strong> thermal energy<\/strong> (<strong>0.025 eV<\/strong>). Solutions of diffusion equations, in this case, provide illustrative insights, how can be the neutron flux distributed in a <a title=\"Reactor Core\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor-core\/\">reactor core<\/a>.<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Solution for the Infinite Reactor<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">Let assume a <strong>uniform infinite reactor<\/strong>, i.e., uniform infinite multiplying system without an external neutron source. This system is in a Cartesian coordinate system, and under these assumptions (no neutron leakage, no changes in diffusion parameters), the <strong>neutron flux<\/strong> must be <strong>inherently constant<\/strong> throughout space.<\/p>\n<p>Since the <a title=\"Neutron Current Density\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/neutron-current-density\/\">neutron current<\/a> is equal to zero (<strong>J<\/strong> = -D\u2207\u0424, where \u0424 is constant), the diffusion equation in the infinite uniform multiplying system must be:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-infinite-system.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15315\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-infinite-system.png\" alt=\"diffusion equation - infinite system\" width=\"236\" height=\"47\" \/><\/a><\/p>\n<p>The only solution of this equation is a <strong>trivial solution<\/strong>, i.e., \u0424 = 0, <strong>unless<\/strong> <strong>\u03a3<\/strong><strong><sub>a<\/sub><\/strong><strong> = <\/strong>\u03bd<strong>\u03a3<\/strong><strong><sub>f<\/sub><\/strong><strong>. <\/strong>This equation (<strong>\u03a3<\/strong><strong><sub>a<\/sub><\/strong><strong> = <\/strong>\u03bd<strong>\u03a3<\/strong><strong><sub>f<\/sub><\/strong>) is the <strong>critical condition<\/strong> for an infinite reactor and expresses the perfect balance (critical state) between <a title=\"Neutron Absorption\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-absorption\/\">neutron absorption <\/a>and neutron production. This balance must be continuously maintained to have<strong> a steady-state neutron flux<\/strong>.<\/p>\n<h2><strong><span style=\"font-weight: 400;\">Infinite Multiplication Factor<\/span><\/strong><\/h2>\n<p>In this section, the <a title=\"Four Factor Formula \u2013 Infinite Multiplication Factor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/four-factor-formula-infinite-multiplication-factor\/\"><strong>infinite multiplication factor, k<sub>\u221e<\/sub><\/strong><\/a>, will be defined from another point of view than in section &#8211; <a title=\"Nuclear Fission Chain Reaction\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/\">Nuclear Chain Reaction<\/a>.<\/p>\n<p>As can be seen, we can rewrite the diffusion equation in the following way, and we can define a new factor &#8211; <strong>k<sub>\u221e<\/sub> = \u03bd\u03a3<sub>f <\/sub>\/ \u03a3<sub>a<\/sub><\/strong>:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-infinite-system2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15316\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/diffusion-equation-infinite-system2.png\" alt=\"diffusion equation - infinite system2\" width=\"219\" height=\"116\" \/><\/a><\/p>\n<p>A non-trivial solution of this equation is guaranteed when <strong>k<sub>\u221e<\/sub> = \u03bd\u03a3<sub>f <\/sub>\/ \u03a3<sub>a <\/sub>= 1<\/strong>. On the other hand, we have no information about the <strong>neutron flux<\/strong> in such a <a title=\"Critical Reactor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactor-criticality\/critical-reactor\/\">critical reactor<\/a>. The neutron flux can have <strong>any value,<\/strong> and the critical uniform infinite reactor can operate at any power level. It should be noted this theory can be used for a reactor <strong>at low power levels<\/strong>, hence \u201c<strong>zero power criticality<\/strong>\u201d.<\/p>\n<p>In a power reactor core, the power level does not influence the criticality of a reactor unless<a title=\"Reactivity Coefficients \u2013 Reactivity Feedbacks\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactivity-coefficients-reactivity-feedbacks\/\"><strong> thermal reactivity feedbacks<\/strong> <\/a>act (operation of a power reactor without reactivity feedbacks is between 10E-8% \u2013 1% of rated power).<\/p>\n<p>See also: <a title=\"Reactor Criticality\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactor-criticality\/\">Reactor Criticality\u00a0<\/a><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span> Solution for the Infinite Slab Reactor<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Slab-Reactor-min.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15339\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Slab-Reactor-min-300x194.png\" alt=\"Diffusion Theory - Slab Reactor-min\" width=\"300\" height=\"194\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Slab-Reactor-min-300x194.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Slab-Reactor-min.png 561w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>Let assume a <strong>uniform reactor<\/strong> (multiplying system) in the shape of a slab of physical <strong>width<\/strong> <strong>a<\/strong> in the x-direction and infinite in the y- and z-directions. This reactor is situated in the center at x=0. In this geometry, the flux does not vary in y and z allowing us to eliminate the y and z derivatives from \u2207<sup>2<\/sup>. The flux is then a <strong>function of x only<\/strong>, and therefore the Laplacian and diffusion equation can be written as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/solution-of-diffusion-equation-infinite-slab.png\"><img loading=\"lazy\" class=\"aligncenter size-medium wp-image-21140\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/solution-of-diffusion-equation-infinite-slab-300x295.png\" alt=\"solution of diffusion equation - infinite slab\" width=\"300\" height=\"295\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/solution-of-diffusion-equation-infinite-slab-300x295.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/solution-of-diffusion-equation-infinite-slab-52x50.png 52w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/solution-of-diffusion-equation-infinite-slab-66x66.png 66w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/solution-of-diffusion-equation-infinite-slab.png 391w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><\/p>\n<p>The quantity <strong>B<sub>g<\/sub><sup>2<\/sup><\/strong> is called the <strong>geometrical buckling<\/strong> of the reactor and depends only on the geometry. This term is derived from the notion that the <strong>neutron flux distribution<\/strong> is somehow <strong>\u201cbuckled\u201d<\/strong> in a homogeneous finite reactor. It can be derived the geometrical buckling is the negative relative curvature of the neutron flux (<strong>B<sub>g<\/sub><sup>2<\/sup> = \u2207<sup>2<\/sup>\u0424(x) \/ \u0424(x)<\/strong>). The <strong>neutron flux<\/strong> has more concave downward or \u201cbuckled\u201d curvature (<strong>higher B<sub>g<\/sub><sup>2<\/sup><\/strong>) in a small reactor than in a large one. This is a very important parameter, and it will be discussed in the following sections.<\/p>\n<p>For x &gt; 0, this diffusion equation has two possible solutions <strong>sin(B<sub>g<\/sub>x)<\/strong> and <strong>cos(B<sub>g<\/sub>x)<\/strong>, which give a general solution:<\/p>\n<p><strong>\u03a6(x) = A.sin(B<sub>g<\/sub>x) + C.cos(B<sub>g <\/sub>x)<\/strong><\/p>\n<p>From finite flux condition (<strong>0\u2264 \u03a6(x) &lt; \u221e<\/strong>), which required only reasonable values for the flux, it can be derived that A must be equal to zero. The term <strong>sin(B<sub>g<\/sub>x)<\/strong> goes to negative values as x goes to negative values, and therefore it cannot be part of a physically acceptable solution. The physically acceptable solution must then be:<\/p>\n<p><strong>\u03a6(x) = \u00a0C.cos(B<sub>g <\/sub>x)<\/strong><\/p>\n<p>where <strong>B<sub>g<\/sub><\/strong> can be determined from the <strong>vacuum boundary condition<\/strong>.<\/p>\n<figure id=\"attachment_15245\" aria-describedby=\"caption-attachment-15245\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition.png\"><img loading=\"lazy\" class=\"size-medium wp-image-15245\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition-300x224.png\" alt=\"extrapolated length - boundary condition\" width=\"300\" height=\"224\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition-300x224.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition.png 516w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-15245\" class=\"wp-caption-text\">Neutron flux as a function of position near a free surface according to diffusion theory and transport theory.<\/figcaption><\/figure>\n<p>The vacuum boundary condition requires the relative neutron flux near the boundary to have a <strong>slope<\/strong> of <strong>-1\/d<\/strong>, i.e., the flux would extrapolate linearly to<strong> 0 at a distance d<\/strong> beyond the boundary. This <strong>zero flux boundary condition<\/strong> is more straightforward and can be written mathematically as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-boundary-condition.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15322\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-boundary-condition.png\" alt=\"infinite slab reactor - boundary condition\" width=\"114\" height=\"70\" \/><\/a><\/p>\n<p>If d is not negligible, physical dimensions of the reactor are increased by d, and an extrapolated boundary is formulated with dimension <strong>a<sub>e<\/sub>\/2 = a\/2 + d<\/strong>. This\u00a0condition can be written as <strong>\u03a6(a\/2 + d) = \u00a0\u03a6(a<sub>e<\/sub>\/2) = 0<\/strong>.<\/p>\n<p>Therefore, the solution must be \u00a0<strong>\u03a6(a<sub>e<\/sub>\/2) = \u00a0C.cos(B<sub>g <\/sub>.a<sub>e<\/sub>\/2) = 0<\/strong> and the values of geometrical buckling, B<sub>g<\/sub>, are limited to <strong>B<sub>g<\/sub> = <sup>n\u03c0<\/sup>\/<sub>a_e<\/sub><\/strong>, where n is any <strong>odd integer<\/strong>. The only one physically acceptable odd integer is <strong>n=1<\/strong>\u00a0because higher values of n would give cosine functions which would become negative for some values of x. The solution of the diffusion equation is:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15323\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-solution.png\" alt=\"infinite slab reactor - solution\" width=\"182\" height=\"52\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-solution.png 182w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-solution-180x52.png 180w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/infinite-slab-reactor-solution-177x52.png 177w\" sizes=\"(max-width: 182px) 100vw, 182px\" \/><\/a><\/p>\n<p>It must be added the <strong>constant C cannot be obtained<\/strong> from this diffusion equation because this constant gives the absolute value of neutron flux. The<strong> neutron flux<\/strong> can have<strong> any value,<\/strong> and the <a title=\"Critical Reactor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactor-criticality\/critical-reactor\/\"><strong>critical reactor<\/strong><\/a> can operate at any power level. It should be noted the <strong>cosine flux shape<\/strong> is only in a hypothetical case in a uniform homogeneous reactor at low power levels (at \u201c<strong>zero power criticality<\/strong>\u201d).<\/p>\n<p>In the power reactor core (at full power operation), the neutron flux can reach, for example, about <strong>3.11 x 10<\/strong><strong><sup>13 <\/sup><\/strong><strong>neutrons.cm<\/strong><strong><sup>-2<\/sup><\/strong><strong>.s<\/strong><strong><sup>-1<\/sup><\/strong><strong>, <\/strong>but this value depends significantly on many parameters (type of fuel, fuel burnup, fuel enrichment, position in fuel pattern, etc.).<\/p>\n<p>The power level does not influence the <a title=\"Reactor Criticality\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactor-criticality\/\">criticality (k<sub>eff<\/sub>)<\/a> of a power reactor unless <a title=\"Reactivity Coefficients \u2013 Reactivity Feedbacks\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactivity-coefficients-reactivity-feedbacks\/\">thermal reactivity feedbacks<\/a> act (operation of a power reactor without reactivity feedbacks is between 10E-8% \u2013 1% of rated power).<\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Solution for the Finite Spherical Reactor<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<p style=\"text-align: left;\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Spherical-Reactor-min.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15337\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Spherical-Reactor-min-300x146.png\" alt=\"Diffusion Theory - Spherical Reactor-min\" width=\"300\" height=\"146\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Spherical-Reactor-min-300x146.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Spherical-Reactor-min.png 604w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>Let us assume a <strong>uniform reactor<\/strong> (multiplying system) in the shape of a <strong>sphere<\/strong> of <strong>physical radius R. <\/strong>The spherical reactor is situated in <strong>spherical geometry<\/strong> at the origin of coordinates. To solve the diffusion equation, we have to replace the Laplacian with its spherical form:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15274\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry.png\" alt=\"laplacian - spherical geometry\" width=\"493\" height=\"163\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry.png 493w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/laplacian-spherical-geometry-300x99.png 300w\" sizes=\"(max-width: 493px) 100vw, 493px\" \/><\/a><\/p>\n<p>We can replace the 3D Laplacian with its one-dimensional spherical form because there is <strong>no dependence on an angle<\/strong> (whether polar or azimuthal). The source term is assumed to be isotropic (there is the spherical symmetry). The flux is then a <strong>function of radius &#8211; r only<\/strong>, and therefore the diffusion equation can be written as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15324\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor.png\" alt=\"solution of diffusion equation - spherical reactor\" width=\"436\" height=\"154\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor.png 436w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor-300x106.png 300w\" sizes=\"(max-width: 436px) 100vw, 436px\" \/><\/a><\/p>\n<p>The solution of the diffusion equation is based on a <strong>substitution \u03a6(r) = 1\/r \u03c8(r)<\/strong>, that leads to an equation for \u03c8(r):<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15325\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor2.png\" alt=\"solution of diffusion equation - spherical reactor2\" width=\"212\" height=\"165\" \/><\/a><\/p>\n<p>For r &gt; 0, this differential equation has two possible solutions, <strong>sin(B<sub>g<\/sub>r)<\/strong> and<strong> cos(B<sub>g<\/sub>r)<\/strong>, which give a general solution:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor3.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15326\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor3.png\" alt=\"solution of diffusion equation - spherical reactor3\" width=\"301\" height=\"118\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor3.png 301w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor3-300x118.png 300w\" sizes=\"(max-width: 301px) 100vw, 301px\" \/><\/a><\/p>\n<p>From finite flux condition (<strong>0\u2264 \u03a6(r) &lt; \u221e<\/strong>), which required only reasonable values for the flux, it can be derived that <strong>C must be equal to zero<\/strong>. The term <strong>cos(B<sub>g<\/sub>r)\/r<\/strong> goes to \u221e as r \u279d0 \u00a0and therefore cannot be part of a physically acceptable solution. The physically acceptable solution must then be:<\/p>\n<p><strong>\u03a6(r) = \u00a0A sin(B<sub>g<\/sub>r)\/r<\/strong><\/p>\n<figure id=\"attachment_15245\" aria-describedby=\"caption-attachment-15245\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition.png\"><img loading=\"lazy\" class=\"size-medium wp-image-15245\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition-300x224.png\" alt=\"extrapolated length - boundary condition\" width=\"300\" height=\"224\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition-300x224.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/extrapolated-length-boundary-condition.png 516w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-15245\" class=\"wp-caption-text\">Neutron flux as a function of position near a free surface according to diffusion theory and transport theory.<\/figcaption><\/figure>\n<p>The vacuum boundary condition requires the relative <strong>neutron flux<\/strong> near the boundary to have a <strong>slope<\/strong> of <strong>-1\/d<\/strong>, i.e., the flux would extrapolate linearly to <strong>0 at a distance d<\/strong> beyond the boundary. This <strong>zero flux boundary condition<\/strong> is more straightforward and can be written mathematically as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/spherical-reactor-boundarycondition.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15327\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/spherical-reactor-boundarycondition.png\" alt=\"spherical reactor - boundarycondition\" width=\"102\" height=\"67\" \/><\/a><\/p>\n<p>If d is not negligible, physical dimensions of the reactor are increased by d, and extrapolated boundary is formulated with dimension <strong>R<sub>e<\/sub> = R + d,<\/strong> and this condition can be written as <strong>\u03a6(R + d) = \u00a0\u03a6(R<sub>e<\/sub>) = 0<\/strong>.<\/p>\n<p>Therefore, the solution must be \u00a0<strong>\u03a6(R<sub>e<\/sub>) = \u00a0A sin(B<sub>g<\/sub>R<sub>e<\/sub>)\/R<sub>e<\/sub> = 0,<\/strong> and the values of geometrical buckling, B<sub>g<\/sub>, are limited to <strong>B<sub>g<\/sub> = <sup>n\u03c0<\/sup>\/R<sub>e<\/sub><\/strong>, where n is any <strong>odd integer<\/strong>. The only one physically acceptable odd integer is <strong>n=1<\/strong>\u00a0because higher values of n would give sine functions which would become negative for some values of x before returning to 0 at R<sub>e<\/sub>. The solution of the diffusion equation is:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor4.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15328\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-spherical-reactor4.png\" alt=\"solution of diffusion equation - spherical reactor4\" width=\"198\" height=\"82\" \/><\/a><\/p>\n<p>It must be added the constant <strong>A cannot be obtained<\/strong> from this diffusion equation because this constant gives the absolute value of neutron flux. The <strong>neutron flux can have any value,<\/strong> and the<a title=\"Critical Reactor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactor-criticality\/critical-reactor\/\"> critical reactor<\/a> can operate at any power level. It should be noted this flux shape is only in a hypothetical case in a uniform homogeneous spherical reactor at low power levels (at \u201c<strong>zero power criticality<\/strong>\u201d).<\/p>\n<p>In a power reactor core, the neutron flux can reach, for example, about <strong>3.11 x 10<\/strong><strong><sup>13 <\/sup><\/strong><strong>neutrons.cm<\/strong><strong><sup>-2<\/sup><\/strong><strong>.s<\/strong><strong><sup>-1<\/sup><\/strong><strong>, <\/strong>but this value depends significantly on many parameters (type of fuel, fuel burnup, fuel enrichment, position in fuel pattern, etc.).<\/p>\n<p>The power level does not influence the <a title=\"Reactor Criticality\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactor-criticality\/\">criticality (k<sub>eff<\/sub>)<\/a> of a power reactor unless <a title=\"Reactivity Coefficients \u2013 Reactivity Feedbacks\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactivity-coefficients-reactivity-feedbacks\/\">thermal reactivity feedbacks<\/a> act (operation of a power reactor without reactivity feedbacks is between 10E-8% \u2013 1% of rated power).<\/div><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span> Solution for the Finite Cylindrical Reactor<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Cylindrical-Reactor-min.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15338\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Cylindrical-Reactor-min-300x168.png\" alt=\"Diffusion Theory - Cylindrical Reactor-min\" width=\"300\" height=\"168\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Cylindrical-Reactor-min-300x168.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/Diffusion-Theory-Cylindrical-Reactor-min.png 600w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>Let assume a <strong>uniform reactor<\/strong> (multiplying system) in the shape of a cylinder of physical radius <strong>R and height H. <\/strong>This finite cylindrical reactor is situated in cylindrical geometry at the origin of coordinates. To solve the <strong>diffusion equation<\/strong>, we have to replace the Laplacian by its cylindrical form:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/cylindrical-coordinates-3D-2D.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15342\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/cylindrical-coordinates-3D-2D.png\" alt=\"cylindrical coordinates - 3D, 2D\" width=\"276\" height=\"165\" \/><\/a><\/p>\n<p><strong><span style=\"font-weight: 400;\">Since there is <strong>no dependence on angle \u0398<\/strong>, we can replace the 3D Laplacian with its <strong>two-dimensional form<\/strong>, and we can solve the problem in radial and axial directions. Since the flux is a function of radius &#8211; r and height &#8211; z only (<strong>\u03a6(r,z)<\/strong>), the diffusion equation can be written as:<\/span><\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-reactor.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15343\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-reactor.png\" alt=\"solution of diffusion equation - cylindrical reactor\" width=\"614\" height=\"244\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-reactor.png 614w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-of-diffusion-equation-cylindrical-reactor-300x119.png 300w\" sizes=\"(max-width: 614px) 100vw, 614px\" \/><\/a><\/p>\n<p>The solution of this diffusion equation is based on the use of the <strong>separation-of-variables technique<\/strong>, therefore:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separation-of-variables-diffusion.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15344\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separation-of-variables-diffusion.png\" alt=\"separation-of-variables - diffusion\" width=\"184\" height=\"47\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separation-of-variables-diffusion.png 184w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separation-of-variables-diffusion-180x47.png 180w\" sizes=\"(max-width: 184px) 100vw, 184px\" \/><\/a><\/p>\n<p>where R(r) and Z(z) are functions to be determined. Substituting this into the diffusion equation and dividing by <strong>R(r)Z(z)<\/strong>, we obtain:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separated-diffusion-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15345\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separated-diffusion-equation.png\" alt=\"separated diffusion equation\" width=\"506\" height=\"58\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separated-diffusion-equation.png 506w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separated-diffusion-equation-300x34.png 300w\" sizes=\"(max-width: 506px) 100vw, 506px\" \/><\/a><\/p>\n<p>Because the first term depends only on r and the second only on z, both terms must be <strong>constants<\/strong> for the equation to have a solution. Suppose we take the constants to be <strong>v<sup>2<\/sup><\/strong> and <strong>\u043a<sup>2<\/sup><\/strong>. The sum of these constants must be equal to <strong>B<sub>g<\/sub><sup>2<\/sup> = v<sup>2<\/sup> + \u043a<sup>2<\/sup><\/strong>. Now we can <strong>separate variables<\/strong>, and the <strong><a title=\"Neutron Flux Density \u2013 Neutron Intensity\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-flux-neutron-intensity\/\">neutron flux<\/a><\/strong> must satisfy the following differential equations:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separated-diffusion-equation2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15346\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/separated-diffusion-equation2.png\" alt=\"separated diffusion equation2\" width=\"298\" height=\"212\" \/><\/a><\/p>\n<p><strong>Solution for the radial direction<\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/bessel-functions-J-Y.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-15347\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/bessel-functions-J-Y-189x300.png\" alt=\"bessel functions - J - Y\" width=\"189\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/bessel-functions-J-Y-189x300.png 189w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/bessel-functions-J-Y.png 446w\" sizes=\"(max-width: 189px) 100vw, 189px\" \/><\/a>The differential equation for radial direction is called <strong>Bessel\u2019s equation,<\/strong> and it is well known to mathematicians. In this case, the Bessel\u2019s equation\u2019s solutions are called the <strong>Bessel functions<\/strong> <strong>of the first and second kind, <\/strong>J<sub>\u03b1<\/sub>(x) and Y<sub>\u03b1<\/sub>(x), respectively.<\/p>\n<p>For r &gt; 0, this differential equation has two possible solutions, <strong>J<sub>0<\/sub>(vr)<\/strong> and <strong>Y<sub>0<\/sub>(vr)<\/strong>, the Bessel functions of order zero, which give a general solution:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-for-radial-direction.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15348\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-for-radial-direction.png\" alt=\"solution for radial direction\" width=\"240\" height=\"35\" \/><\/a><\/p>\n<p>From finite flux condition (<strong>0\u2264 \u03a6(r) &lt; \u221e<\/strong>), which required only reasonable values for the flux, it can be derived that C must be equal to zero. The term <strong>Y<sub>0<\/sub>(vr)<\/strong> goes to -\u221e as r \u279d0 \u00a0and therefore cannot be part of a physically acceptable solution. The physically acceptable solution must then be:<\/p>\n<p><strong>R(r) = \u00a0AJ<sub>0<\/sub>(vr)<\/strong><\/p>\n<p>The <strong>vacuum boundary condition<\/strong> requires the relative neutron flux near the boundary to have a <strong>slope of -1\/d<\/strong>, i.e., the flux would extrapolate linearly to <strong>0 at a distance d<\/strong> beyond the boundary. This <strong>zero flux boundary condition<\/strong> is more straightforward and can be written mathematically as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/radial-direction-boundary-condition.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15349\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/radial-direction-boundary-condition.png\" alt=\"radial direction - boundary condition\" width=\"103\" height=\"61\" \/><\/a><\/p>\n<p>If d is not negligible, physical dimensions of the reactor are increased by d, and extrapolated boundary is formulated with dimension <strong>R<sub>e<\/sub> = R + d,<\/strong> and this condition can be written as <strong>\u03a6(R + d) = \u00a0\u03a6(R<sub>e<\/sub>) = 0<\/strong>.<\/p>\n<p>Therefore, the solution must be \u00a0<strong>R(R<sub>e<\/sub>) = \u00a0A J<sub>0<\/sub>(vR<sub>e<\/sub>) = 0<\/strong>. The function of <strong>J<sub>0<\/sub>(r)<\/strong> has several zeroes. The first is at<strong> r<sub>1<\/sub> = 2.405<\/strong>, and the second at r<sub>2<\/sub> = 5.6. However, because the neutron flux cannot have negative values, the only physically acceptable value for <strong>v<\/strong> is <strong>2.405\/R<sub>e<\/sub><\/strong>.\u00a0The solution of the diffusion equation is:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-for-radial-direction-diffusion.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15350\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-for-radial-direction-diffusion.png\" alt=\"solution for radial direction - diffusion\" width=\"196\" height=\"60\" \/><\/a><\/p>\n<p><strong>Solution for axial direction<\/strong><\/p>\n<p>The solution for axial direction has been solved in previous sections (<strong>Infinite Slab Reactor<\/strong>), and therefore it has the same solution. The solution in an axial direction is:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-for-axial-direction-diffusion.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15351\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/solution-for-axial-direction-diffusion.png\" alt=\"solution for axial direction - diffusion\" width=\"187\" height=\"59\" \/><\/a><\/p>\n<p><strong>Solution for radial and axial\u00a0directions<\/strong><\/p>\n<p>The <strong>full solution<\/strong> for the <strong>neutron flux distribution<\/strong> in the finite cylindrical reactor is, therefore:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/full-solution-of-diffusion-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15352\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/full-solution-of-diffusion-equation.png\" alt=\"full solution of diffusion equation\" width=\"333\" height=\"135\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/full-solution-of-diffusion-equation.png 333w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/08\/full-solution-of-diffusion-equation-300x122.png 300w\" sizes=\"(max-width: 333px) 100vw, 333px\" \/><\/a><\/p>\n<p>where <strong>B<sub>g<\/sub><sup>2<\/sup> <\/strong>is the total geometrical buckling.<\/p>\n<p>The constants A and C must be added that they cannot be obtained from the diffusion equation because they give the <strong>absolute value of neutron flux<\/strong>. The neutron flux can have <strong>any value,<\/strong> and the critical reactor can operate at any power level. It should be noted this flux shape is only in a hypothetical case in a uniform homogeneous cylindrical reactor at low power levels (at \u201c<strong>zero power criticality<\/strong>\u201d).<\/p>\n<p>In a power reactor core, the neutron flux can reach, for example, about <strong>3.11 x 10<\/strong><strong><sup>13 <\/sup><\/strong><strong>neutrons.cm<\/strong><strong><sup>-2<\/sup><\/strong><strong>.s<\/strong><strong><sup>-1<\/sup><\/strong><strong>, <\/strong>but this value depends significantly on many parameters (type of fuel, fuel burnup, fuel enrichment, position in fuel pattern, etc.).<\/p>\n<p>The power level does not influence the<a title=\"Six Factor Formula \u2013 Effective Multiplication Factor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/six-factor-formula-effective-multiplication-factor\/\"> criticality (k<sub>eff<\/sub>)<\/a> of a power reactor unless <a title=\"Reactivity Coefficients \u2013 Reactivity Feedbacks\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/reactivity-coefficients-reactivity-feedbacks\/\">thermal reactivity feedbacks<\/a> act (operation of a power reactor without reactivity feedbacks is between 10E-8% \u2013 1% of rated power).<\/div><\/div><\/div>\n<\/div>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:20px 0;border-width:2px;border-color:#999999\"><\/div>\n<div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\u00a0 <div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>References:<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><strong>Nuclear and Reactor Physics:<\/strong>\n<ol>\n<li>J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading,\u00a0MA (1983).<\/li>\n<li>J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.<\/li>\n<li>W. M. Stacey, Nuclear Reactor Physics, John Wiley &amp; Sons, 2001, ISBN: 0- 471-39127-1.<\/li>\n<li>Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering,\u00a0Springer; 4th edition, 1994, ISBN:\u00a0978-0412985317<\/li>\n<li>W.S.C. Williams. Nuclear and Particle Physics.\u00a0Clarendon Press; 1 edition, 1991, ISBN:\u00a0978-0198520467<\/li>\n<li>G.R.Keepin. Physics of Nuclear Kinetics.\u00a0Addison-Wesley Pub. Co; 1st edition, 1965<\/li>\n<li>Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.<\/li>\n<li>U.S. Department of Energy, Nuclear Physics and Reactor Theory.\u00a0DOE Fundamentals Handbook,\u00a0Volume 1 and 2.\u00a0January\u00a01993.<\/li>\n<\/ol>\n<p><strong><\/strong><strong>Advanced Reactor Physics:<\/strong><\/p>\n<ol>\n<li>K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.<\/li>\n<li>K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.<\/li>\n<li><span style=\"font-weight: 400;\">D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.\u00a0<\/span><\/li>\n<li>E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.<\/li>\n<\/ol>\n<\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-default\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\n<h2>See above:<\/h2>\n<p>Diffusion Theory<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/\" class=\"su-button su-button-style-flat\" style=\"color:#606060;background-color:#ffffff;border-color:#cccccc;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px\" target=\"_self\"><span style=\"color:#606060;padding:7px 20px;font-size:16px;line-height:24px;border-color:#ffffff;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px;text-shadow:0px 0px 0px #000000;-moz-text-shadow:0px 0px 0px #000000;-webkit-text-shadow:0px 0px 0px #000000\"><i class=\"sui sui-link\" style=\"font-size:16px;color:#5d5d5d\"><\/i> <\/span><\/a><\/p><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Solutions of the Diffusion Equation &#8211; Non-multiplying Systems As was previously discussed, the diffusion theory is widely used in the core design of the current Pressurized Water Reactors (PWRs) or Boiling Water Reactors (BWRs). This section is not about such calculations but provides illustrative insights, how can be the neutron flux distributed in any diffusion &#8230; <a title=\"Diffusion Equation\" class=\"read-more\" href=\"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/neutron-diffusion-theory\/diffusion-equation\/\" aria-label=\"More on Diffusion Equation\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":15150,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"generate_page_header":""},"_links":{"self":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15356"}],"collection":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/comments?post=15356"}],"version-history":[{"count":8,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15356\/revisions"}],"predecessor-version":[{"id":33559,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15356\/revisions\/33559"}],"up":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15150"}],"wp:attachment":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/media?parent=15356"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}