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"markdown": "# Report on Thermal Neutron Diffusion Length Measurement in Reactor Grade Graphite Using MCNP and COMSOL Multiphysics \n\nS. R. Mirfayzi
School of Physics and Astronomy University of Birmingham Birmingham B15 2TT United Kingdom
E-mail: srm105@bham.ac.uk\n\n\n#### Abstract\n\nNeutron diffusion length in reactor grade graphite is measured both experimentally and theoretically. The experimental work includes Monte Carlo (MC) coding using 'MCNP' and Finite Element Analysis (FEA) coding suing 'COMSOL Multiphysics' and Matlab. The MCNP code is adopted to simulate the thermal neutron diffusion length in a reactor moderator of $2 \\mathrm{~m} \\times 2 \\mathrm{~m}$ with slightly enriched uranium $\\left({ }^{235} U\\right)$, accompanied with a model designed for thermal hydraulic analysis using point kinetic equations, based on partial and ordinary differential equation. The theoretical work includes numerical approximation methods including transcendental technique to illustrate the iteration process with the FEA method. Finally collision density of thermal neutron in graphite is measured, also specific heat relation dependability of collision density is also calculated theoretically, the thermal neutron diffusion length in graphite is evaluated at $50.85 \\pm 0.3 \\mathrm{~cm}$ using COMSOL Multiphysics and $50.95 \\pm 0.5 \\mathrm{~cm}$ using MCNP. Finally the total neutron cross-section is derived using FEA in an inverse iteration form.\n\n\n## 1. Introduction\n\nThis work demonstrates an analytic approach accompanied with models of Finite Element Analysis (FEA) and Monte Carlo (MC) with an experimental measure on neutron cross-section and slowing down process. In MC approach Monte Carlo N-Particle Transport Code (MCNP) is used to simulate the simplified version of reactor moderation process. Similarly in FEA the moderator modelled (Assuming a symmetrical distribution) using point kinetic equations, based on partial and ordinary differential equation in software package.\n\n## 2. Theoretical Calculations\n\nHaving the number of particles found in a volume element dr where $d r=d x d y d z$ at $r$ with a vector with solid angle $d \\Omega$ at $\\Omega$ be donated by [1]:\n\n$$\nN(r, \\Omega, t) d r d \\Omega\n$$\n\nTherefore can have:",
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"markdown": "$$\n\\frac{d N}{d t}=-N V \\sigma+\\int N\\left(r, \\Omega^{\\prime}, t\\right) V \\sigma_{s} f\\left(\\Omega . \\Omega^{\\prime}\\right) d \\Omega^{\\prime}+S(r, \\Omega, t)\n$$\n\nWhere the first term $\\frac{d N}{d t}$ donates the number of particles present in given volume (particle density) and second term $-N V \\sigma$ represents the total number of particles removed from the given volume by scattering and capture. $\\sigma$ is representing the total cross-section. The third term represents the total number of particles scattered into the given volume, and $f\\left(\\Omega . \\Omega^{\\prime}\\right)$ represents the relative probability of scattering through an angle whose cosine is $\\Omega . \\Omega^{\\prime}$, where $\\Omega^{\\prime}$ is a unit vector in the direction of the initial velocity and $\\Omega$ is unit vector in the final direction. Finally $S(r, \\Omega, t)$ is the external source term available in the system and is given by:\n\n$$\nN(r, \\Omega, t) d r d \\Omega=N(Z, \\varphi) d Z d \\varphi d \\phi\n$$\n\nWhere $\\varphi$ is the cosine of the velocity vector in the Z direction and $\\phi$ is the longitude of velocity vector.\n\n\nFigure 1. The Velocity Vector: Where $\\varphi$ is the cosine of the velocity vector in the Z direction and $\\phi$ is the longitude of velocity vector and $\\Omega$ is unit vector in the final direction.\n\nNow an assumption can be made such:\n\n$$\nN_{0}(Z)=2 \\pi \\int_{-1}^{+1} N(Z, \\varphi) d \\varphi\n$$\n\nRewrite the equation 1 as:\n\n$$\nV_{\\varphi} \\frac{\\delta N(Z, \\varphi)}{d Z}=-N(Z, \\varphi) V \\sigma+\\int N\\left(Z, \\varphi^{\\prime}\\right) V \\sigma_{s} f\\left(\\varphi_{0}\\right) d \\Omega+S(z)\n$$\n\nWhere $\\varphi_{0}$ is the cosine of the angle between initial and final velocities and it can be found by:\n\n$$\n\\cos \\theta \\cos \\theta^{\\prime}+\\sin \\theta \\sin \\theta^{\\prime} \\cos \\left(\\phi-\\phi^{\\prime}\\right)\n$$",
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"markdown": "Or using:\n\n$$\n\\varphi_{0}=\\varphi \\varphi^{\\prime}+\\sqrt{1-\\varphi^{2}} \\sqrt{1-\\varphi^{\\prime 2}} \\cos \\left(\\phi-\\phi^{\\prime}\\right)\n$$\n\nNow if the collision function of $F\\left(\\varphi_{0}\\right)$ expanded in spherical harmonics:\n\n$$\nF\\left(\\varphi_{0}\\right)=\\sum_{0}^{\\infty} \\frac{2 l+1}{4 \\pi} F_{1} P_{1}\\left(\\varphi_{0}\\right)\n$$\n\nWith $F_{1}=\\int f\\left(\\varphi_{0}\\right) P_{1}\\left(\\varphi_{0}\\right) d \\Omega$. Using similar expansion the phase density function will be:\n\n$$\nN(Z, \\varphi)=\\sum_{0}^{\\infty} \\frac{2 l+1}{4 \\pi} N_{1}(Z) P_{1}(\\varphi)\n$$\n\nWhere $N(Z, \\varphi)=\\int N(Z, \\varphi) P_{1}(\\varphi) d \\Omega$, with assumption that $N(Z, \\varphi)$ is isotropic, three conditions must be satisfied all the times: first, it is far from the source (equal to Mean Free Path (MFP)), second, it is far from the boundaries; third, the probability of capture is small compared to probability of scattering. Having all the conditions satisfied, the following can be assumed:\n\n$$\nN(Z, \\varphi) \\cong \\frac{1}{4 \\pi}\\left(N_{0}(Z)+3 \\varphi N_{1}(Z)\\right)\n$$\n\nWhere the second term in the bracket donates the particle flux (J). Here $N_{1}=\\int \\varphi N(Z, \\varphi) d \\Omega=$ $J / V$. For simplicity we choose our unit such that $V=1$ and $\\sigma=1$, hence:\n\n$$\n1-f=\\frac{\\sigma_{s}}{\\sigma}\n$$\n\nNow the Boltzmann equation takes the form of:\n\n$$\nN \\frac{d N}{d Z}=-N+(1-f) \\int N\\left(Z, \\varphi^{\\prime}\\right) f\\left(N_{0}\\right) d \\Omega^{\\prime}+S(Z)\n$$\n\nBy integrating the equation over all possible angles $(d \\Omega)$ we have:\n\n$$\n\\frac{d N_{1}}{d Z}=-N_{0}+(1-f) N_{0}+4 \\pi S(Z)\n$$\n\n$f$ is normalized in such a way that $\\int f\\left(\\Omega . \\Omega^{\\prime}\\right) d \\Omega^{\\prime}=\\int f(\\Omega . \\Omega) d \\Omega=1$, hence going back to Eq. 9 for the case $l=0$ we have:\n\n$$\nF_{0}=\\int f\\left(\\varphi_{0}\\right) d \\Omega=1\n$$\n\nHence by integrating over all angles and Multiplying by $\\varphi$ we have:\n\n$$\n\\frac{1}{3} \\frac{d N_{0}}{d Z}=-N_{1}+(1-f) F_{1} N_{1}\n$$",
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"markdown": "Now the second order differential equation gives:\n\n$$\n-\\frac{1}{3\\left(1-f_{1}\\right.} \\frac{d^{2} \\varphi_{0}}{d Z^{2}}=-f N_{0}+S_{0}(Z)\n$$\n\nThis also can be written as:\n\n$$\n\\nabla^{2} N_{0}-\\frac{1}{L^{2}} N_{0}+\\frac{1}{D} S_{0}=0\n$$\n\nEqs. 16 and 17 are known as diffusion equation. Here $L$ is diffusion Length abd D is diffusion coefficient and it is equal to $\\frac{1}{3} \\frac{\\lambda_{s}}{1-\\bar{L}_{t}}=\\frac{\\lambda_{t r}}{3} V$, where $\\lambda_{s}$ and $\\lambda_{t r}$ are the scattering and transport mean free path. $\\lambda_{t r}$ can be calculated from:\n\n$$\n\\lambda_{t r}=\\frac{\\lambda_{s}}{1-(\\cos \\theta)_{a v}}\n$$\n\nand $(\\cos \\theta)_{a v}$ is equal to $\\int f\\left(\\varphi_{0}\\right) \\varphi_{0} d \\Omega=f_{1}$. Also $L^{2}$ can be measured using following relation:\n\n$$\nL^{2}=\\frac{\\lambda_{c} \\lambda_{t r}}{3}\n$$\n\nHere $\\lambda_{c}$ is the capture mean free path.\n\n# MAXIMUM ENERGY LOSS \n\nIf a neutron with initial velocity $V_{0}$ collides with a nucleus of mass M (at rest), then in the Centre of Mass (CoM) system, the initial velocity is $M V_{0} / M+1$ after collision. The momentum of of neutron and the nucleus will be equal to magnitude oppositely directed vector. Figure 2 demonstrates the collision in CoM system.\nAs demonstrated in Fig. 2 the $\\theta$ is the deflection angle and $\\Theta$ is angle on the final velocity $v$. The $v^{2}$ in this case is given by:\n\n$$\n\\begin{gathered}\n\\frac{M v_{0}}{M+1} \\cos \\theta+\\frac{v_{0}}{M+1}=v \\cos \\Theta \\\\\n\\left(\\frac{M v_{0}}{M+1}\\right)^{2}+\\left(\\frac{v_{0}}{M+1}\\right)^{2}-\\frac{2 M v_{0}^{2}}{M+1} \\cos \\theta=v^{2}\n\\end{gathered}\n$$\n\nso\n\n$$\n\\cos \\theta=1-\\frac{(M+1)^{2}}{2 M} 1-\\frac{v}{v_{0}})^{2}\n$$\n\nsince $u=\\log \\frac{F_{0}}{E}$ then:\n\n$$\n\\cos \\theta=1-\\frac{(M+1)^{2}}{2 M} 1-e^{-u}\n$$",
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"markdown": "\n\nFigure 2. The Collision in Centre-of-Mass System:If a neutron with initial velocity $V_{0}$ collides with a nucleus of mass M at rest, then in the Centre of Mass (CoM)system the initial velocity is $M V_{0} / M+1$ after collision. The momentum of of neutron and the nucleus will be equal to magnitude oppositely directed vector. Here $\\theta$ is the deflection angle and $\\Theta$ is angle on the final velocity $v$.\nnow differential cross-section gives:\n\n$$\n\\frac{d \\cos \\theta}{d u}=-\\frac{(M+1)^{2}}{2 M} e^{-u}\n$$\n\nHence:\n\n$$\n\\cos \\Theta=-\\frac{(M+1)^{2}}{2} e^{\\frac{-u}{2}}-\\frac{M-1}{2} e^{\\frac{u}{2}}\n$$\n\nTherefore the maximum logarithmic energy loss can be calculated from:\n\n$$\nq_{M}=\\log \\left(\\frac{M+1}{M-1}\\right)^{2}\n$$\n\nThe $q_{M}$ is at most when $\\Theta=\\pi$. Now going back to the problem we can redefine the collision density function as:\n\n$$\nF\\left(\\varphi_{0}, u\\right)=\\frac{(M+1)^{2}}{8 \\pi M} e^{-u} \\times \\delta\\left(\\varphi_{0}-\\left(\\frac{(M+1)}{2} e^{\\frac{-u}{2}}-\\frac{M-1}{2} e^{\\frac{u}{2}}\\right)\\right.\n$$\n\nThe term $\\frac{(M+1)^{2}}{8 \\pi M} e^{-u}$ is the normalization constant chosen to satisfy $\\int d \\Omega \\int d u f\\left(\\varphi_{0}, u\\right)=1$ and",
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"markdown": "$\\delta$ is the Dirac $\\delta$ function. So that $\\delta(x-a)=0$ when $x \\neq 0$ and $\\int d(x-a) F(x) d x=F(a)$. Now the average logarithmic loss $(\\xi)$ can be calculated from:\n\n$$\n\\xi=1-\\frac{(M+1)^{2}}{4 M} q_{m} e^{-q_{m}}\n$$\n\nand $(\\cos )_{a} v$ is $2 / 3 M$.\n\n# Energy Distribution of Slowed Down Neutrons \n\n## I. Stationary Case\n\nThe average collision density per unit time, with logarithmic energy intervals is given by:\n\n$$\n\\psi_{0}(u)=\\int_{0}^{u} d u^{\\prime} \\psi_{0}\\left(u^{\\prime}\\right) h\\left(u^{\\prime}\\right) f_{0}\\left(u-u^{\\prime}\\right)+\\delta(u)\n$$\n\nwhere $f_{0}(u)$ is $(M+1)^{2} e^{-u} / 4 m$ for $u \\leq q_{m}$ and it is zero otherwise. In stationary case the total number of neutron produced is unity per unit in this case, i.e. for $M=12, f+0(u)$ becomes $3.5 e^{-u}$. Hence the equation 29 becomes:\n\n$$\n\\psi_{0}(u)=\\int_{u-q_{m}}^{u} d u^{\\prime} \\psi_{0}\\left(u^{\\prime}\\right) h\\left(u^{\\prime}\\right) 3.5 e^{-\\left(u-u^{\\prime}\\right)}+\\delta(u)\n$$\n\nwhere $q_{m}=0.72$\n\n## II. Time-dependent Case\n\nThe time dependent when there is no absorption in the system and source strength is unity and is given by:\n\n$$\n\\frac{l(u)}{v} \\frac{d \\psi_{0}}{\\delta t}+\\psi_{0}(u, t)=\\int_{0}^{u} d u^{\\prime} \\psi_{0}\\left(u^{\\prime}, t\\right) e^{-\\left(u-u^{\\prime}\\right)}+\\delta(u) \\delta(t)\n$$\n\nwhere $l(u)$ is the mean free path and if the mean free path is constant, the Laplacian form of the equation for $M \\neq 1$ becomes:\n\n$$\n1+\\frac{s l_{0}}{v} \\phi_{0}(u, s)=\\int_{0}^{u} d u^{\\prime} \\phi_{0}\\left(u^{\\prime}, s\\right) f_{0}\\left(u-u^{\\prime}\\right)+\\delta(u)\n$$\n\nnow:\n\n$$\n\\phi(w, s)=\\frac{2}{\\left(1-r^{2}\\right) w^{2}} \\int_{r w}^{w} d w^{\\prime} \\frac{w^{\\prime} \\phi\\left(w^{\\prime}, s\\right)}{\\left(1+w^{\\prime}\\right)}\n$$\n\nEq. 33 applies for $u>q_{m}$ where $w=l_{0} s / v, r=M-1 / M+1$ and $\\phi(w, s)=(1+w) \\phi_{0}(u, s)$, so that the mean free path is proportional to velocity.\n\n## III. Rigorous Numerical Solution\n\nThe slowing down process is not an easy approach, therefore a more discrete form of solution also could be defined using:\n\n$$\nF(E)=\\int_{E}^{\\infty} \\sum_{S}\\left(E^{\\prime} \\rightarrow E\\right) \\phi\\left(E^{\\prime}\\right) d E^{\\prime}+\\delta(u)\n$$",
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"markdown": "Where $\\sum_{S}\\left(E^{\\prime} \\rightarrow E\\right)$ is the scattering term between energies $E^{\\prime}$ and $E$. Recalling $\\sum_{S}$ :\n\n$$\n\\sum_{S}\\left(E^{\\prime} \\rightarrow E\\right)=\\sum_{S}\\left(E^{\\prime}\\right) P\\left(E^{\\prime} \\rightarrow E\\right)\n$$\n\nWhere $P\\left(E^{\\prime} \\rightarrow E\\right)$ is the probability of collision happens between $E^{\\prime}$ and $E$. It can be defined by:\n\n$$\nP\\left(E^{\\prime} \\rightarrow E\\right) \\cdot E^{\\prime}(1-\\alpha)=1\n$$\n\nHence:\n\n$$\nP\\left(E^{\\prime} \\rightarrow E\\right)=\\frac{1}{(1-\\alpha) E^{\\prime}}\n$$\n\nNow:\n\n$$\n\\sum_{S}\\left(E^{\\prime} \\rightarrow E\\right)=\\frac{\\sum_{S}\\left(E^{\\prime}\\right)}{(1-\\alpha) E^{\\prime}}\n$$\n\nFor $M=1$ the collision is defined as:\n\n$$\nF(E)=\\int_{E}^{\\infty} \\frac{\\sum_{S}\\left(E^{\\prime}\\right)}{E^{\\prime}} \\phi(E) d E^{\\prime}+\\delta(E)\n$$\n\nAlso the solution with capture process:\n\n$$\nF_{c}(E)=\\frac{\\sum_{S}\\left(E_{0}\\right)}{\\sum_{t}(E)} \\frac{S_{0}}{E} \\exp \\left(-\\int_{E}^{E_{0}} \\frac{\\sum_{S}\\left(E^{\\prime}\\right)}{\\sum_{t}\\left(E^{\\prime}\\right)} \\frac{d E^{\\prime}}{E^{\\prime}}\\right)\n$$\n\nAlso For $M \\neq 1$ the collision density can be found:\n\n$$\nF(E)=\\int_{E / \\alpha}^{E} \\frac{\\sum_{S}\\left(E^{\\prime}\\right)}{(1-\\alpha) E^{\\prime}} \\phi\\left(E^{\\prime}\\right) d E^{\\prime}+\\frac{\\delta(E)}{(1-\\alpha) E_{0}}\n$$\n\nThis only applicable if $\\alpha E_{0}