{"id":13493,"date":"2017-01-02T10:14:14","date_gmt":"2017-01-02T10:14:14","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=13493"},"modified":"2022-10-19T16:16:36","modified_gmt":"2022-10-19T16:16:36","slug":"macroscopic-cross-section","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/macroscopic-cross-section\/","title":{"rendered":"Macroscopic Cross-section"},"content":{"rendered":"
The macroscopic cross-section<\/strong> represents the effective target area of all of the nuclei<\/strong> contained in the volume of the material (such as fuel pellet). The units are given in cm-1<\/sup><\/strong>. It is the probability of neutron-nucleus interaction per centimeter of neutron travel. Codes commonly use these data for reactor core analyses and design<\/a>. These codes are based on pre-computed assembly homogenized macroscopic cross-sections.<\/strong><\/p>\n<\/div><\/div>\n

Macroscopic Cross-section<\/h2>\n

The difference between the microscopic cross-section<\/strong> and macroscopic cross-section<\/strong>\u00a0is very important and is restated for clarity. The microscopic cross-section<\/strong> represents the effective target area of a single target nucleus<\/strong> for an incident particle. The units are given in barns or cm2<\/sup><\/strong>.<\/p>\n

While the macroscopic cross-section<\/strong> represents the effective target area of all of the nuclei<\/strong> contained in the volume of the material, the units are given in cm-1<\/sup><\/strong>.<\/p>\n

A macroscopic cross-section is derived from microscopic cross-section<\/strong> and the atomic number density<\/strong><\/a>:<\/p>\n

\u03a3=\u03c3.N<\/strong><\/p>\n

Here \u03c3<\/strong>, which has units of m2<\/sup>, is the microscopic cross-section. Since the units of N (nuclei density) are nuclei\/m3<\/sup>, the macroscopic cross-section \u03a3 has units of m-1<\/sup>. Thus, it is an incorrect name because it is not a correct unit of cross-sections. In terms of \u03a3t<\/sub> (the total cross-section), the equation for the intensity of a neutron beam can be written as<\/p>\n

-dI = N.\u03c3.\u03a3t<\/sub>.dx<\/strong><\/p>\n

Dividing this expression by I(x) gives<\/p>\n

-d\u0399(x)\/I(x) = \u03a3t<\/sub>.dx<\/strong><\/p>\n

Since dI(x) is the number of neutrons that collide in dx, the quantity –d\u0399(x)\/I(x)<\/strong> represents the probability that a neutron that has survived without colliding until x will collide in the next layer dx. It follows that the probability P(x) that a neutron will travel a distance x without any interaction in the material, which \u03a3t characterizes, is:<\/p>\n

P(x) = e-\u03a3t<\/sub>.x<\/sup><\/strong><\/p>\n

We can derive the probability that a neutron will make its first collision in dx from this equation<\/strong>. It will be the quantity P(x)dx<\/strong>. Suppose the probability of the first collision in dx is independent of its history. In that case, the required result will be equal to the probability that a neutron survives up to layer x without any interaction (~\u03a3t<\/sub>dx) times the probability that the neutron will interact in the additional layer dx (i.e., ~e-\u03a3t<\/sub>.x<\/sup>).<\/p>\n

P(x)dx = \u03a3t<\/sub>dx . e-\u03a3t<\/sub>.x<\/sup> = \u03a3t<\/sub> e-\u03a3t<\/sub>.x<\/sup> dx<\/strong><\/p>\n

\n

Mean Free Path<\/h2>\n

From the equation for the probability of the first collision in dx,<\/strong> we can calculate the mean free path<\/strong> traveled by a neutron between two collisions. The symbol \u03bb usually designates this quantity. It<\/b>\u00a0is equal to the average value of x, the distance traveled by a neutron without any interaction, over the interaction probability distribution.<\/p>\n

\"mean<\/a><\/p>\n

whereby one can distinguish \u03bbs<\/sub>, \u03bba<\/sub>, \u03bbf<\/sub><\/strong>, etc. This quantity is also known as the relaxation length<\/strong>\u00a0because it is the distance in which the intensity of the neutrons that have not caused a reaction has decreased with a factor e.<\/p>\n

For materials with a high absorption cross-section<\/a>, the mean free path is very short,<\/strong> and neutron absorption occurs mostly on the material\u2019s surface<\/strong>. This surface absorption is called self-shielding<\/strong> because the outer layers of atoms shield the inner layers.<\/p>\n

Macroscopic Cross-section of Mixtures and Molecules<\/h2>\n

Most materials are composed of several chemical elements and compounds. Most chemical elements contain several isotopes<\/strong> of these elements (e.g.,, gadolinium<\/a> with its six stable isotopes). For this reason, most materials involve many cross-sections. Therefore, to include all the isotopes within a given material, it is necessary to determine the macroscopic cross-section for each isotope and then sum all the individual macroscopic cross-sections.<\/p>\n

In this section, both factors (different atomic densities<\/strong> and cross-sections<\/strong>) will be considered in calculating the macroscopic cross-section of mixtures<\/strong>.<\/p>\n

First, consider Avogadro\u2019s number N0<\/sub> = 6.022 x 1023<\/sup><\/strong>, which is the number of particles (molecules, atoms) contained in the amount of substance given by one mole. Thus if M is the molecular weight<\/strong>, the ratio N0<\/sub>\/M<\/strong> equals the number of molecules in 1g of the\u00a0mixture. The number of molecules per cm3<\/sup> in the material of density \u03c1 and the macroscopic cross-section for mixtures is given by the following equations:<\/p>\n

Ni<\/sub> = \u03c1i<\/sub>.N0<\/sub> \/ Mi<\/sub><\/strong><\/p>\n

\"Macroscopic<\/a><\/p>\n

\"Scattering<\/a>
The scattering of slow neutrons by molecules is greater than by free nuclei.<\/figcaption><\/figure>\n

Note that, in some cases, the cross-section of the molecule is not equal<\/strong> to the sum of cross-sections of its individual nuclei<\/strong>. For example, the cross-section of elastic neutron scattering<\/a> of water exhibits anomalies for thermal neutrons<\/a>. It occurs because the kinetic energy of an incident neutron is of the order or less than the chemical binding energy<\/strong>. Therefore,\u00a0the scattering of slow neutrons by water (H2<\/sub>O) is greater than by free nuclei (2H + O).<\/p>\n

<\/span>Example - Macroscopic cross-section for boron carbide in control rods<\/div>
A control rod<\/strong><\/a> usually contains solid boron carbide<\/strong> with natural boron. Natural boron<\/a> consists primarily of two stable isotopes,11<\/sup>B<\/strong> (80.1%) and 10<\/sup>B<\/strong> <\/a>(19.9%). Boron carbide has a density of 2.52 g\/cm3<\/sup><\/strong>.<\/p>\n

Determine the total macroscopic cross-section<\/strong> and the mean free path<\/strong>.<\/p>\n

Density:
\nMB<\/sub> = 10.8
\nMC<\/sub> = 12
\nMMixture<\/sub> = 4 x 10.8 + 1×12 g\/mol
\nNB4C<\/sub> = \u03c1 . Na<\/sub> \/ MMixture<\/sub>
\n= (2.52 g\/cm3<\/sup>)x(6.02×1023<\/sup> nuclei\/mol)\/ (4 x 10.8 + 1×12 g\/mol)
\n= 2.75×1022<\/sup> molecules of B4C\/cm3<\/sup><\/strong><\/p>\n

NB<\/sub> = 4 x 2.75×1022<\/sup> atoms of boron\/cm3<\/sup>
\nNC<\/sub> = 1 x 2.75×1022<\/sup> atoms of carbon\/cm3<\/sup><\/p>\n

NB10<\/sub> = 0.199 x 4 x 2.75×1022<\/sup> = 2.18×1022<\/sup> atoms of 10B\/cm3<\/sup>
\nNB11<\/sub> = 0.801 x 4 x 2.75×1022<\/sup> = 8.80×1022<\/sup> atoms of 11B\/cm3<\/sup>
\nNC<\/sub> = 2.75×1022<\/sup> atoms of 12C\/cm3<\/sup><\/p>\n

the microscopic cross-sections<\/strong><\/p>\n

\u03c3t<\/sub>10B<\/sup> = 3843 b of which \u03c3(n,alpha)<\/sub>10B<\/sup> = 3840 b
\n\u03c3t<\/sub>11B<\/sup> = 5.07 b
\n\u03c3t<\/sub>12C<\/sup> = 5.01 b<\/p>\n

the macroscopic cross-section<\/strong><\/p>\n

\u03a3t<\/sub>B4C<\/sup> <\/strong>= 3843×10-24<\/sup> x 2.18×1022<\/sup> + 5.07×10-24<\/sup> x 8.80×1022<\/sup> + 5.01×10-24<\/sup> x 2.75×1022<\/sup>
\n= 83.7 + 0.45 + 0.14 = 84.3 cm-1<\/sup><\/strong><\/p>\n

the mean free path<\/strong><\/p>\n

\u03bbt<\/sub> <\/strong>= 1\/\u03a3t<\/sub>B4C<\/sup> = 0.012 cm = 0.12 mm<\/strong> (compare with B4C pellets diameter in control rods which may be around 7mm)
\n\u03bba<\/sub> \u2248 0.12 mm<\/strong><\/p>\n<\/div><\/div>\n

<\/span>Example - Atomic number density of 235U in uranium fuel<\/div>
It was written the macroscopic cross-section<\/strong> is derived from microscopic cross-section<\/strong> and the atomic number density (N)<\/strong>:<\/p>\n

\u03a3=\u03c3.N<\/strong><\/p>\n

In this equation, the atomic number density<\/strong> plays a crucial role as the microscopic cross-section. In the reactor core<\/a>, the atomic number density of certain materials (e.g.,, water as the moderator<\/a>) can be changed, leading to certain reactivity changes<\/strong>. To understand the nature of these reactivity changes<\/strong>, we must understand the term the atomic number density.<\/p>\n

See theory: Atomic Number Density<\/a>.<\/p>\n

Most PWRs<\/a> use uranium fuel<\/b>, which is in the form of uranium dioxide <\/b>(UO2<\/sub>). Typically, the fuel has enrichment of \u03c9235<\/sub> = 4% [grams of 235<\/sup>U<\/a>\u00a0per gram of uranium<\/a>] of isotope 235<\/sup>U<\/a>.<\/p>\n

Calculate the atomic number density of 235<\/sup>U<\/a><\/strong>\u00a0(N235U), when:<\/strong><\/p>\n