Radiation Shielding and Multilayered Build-up Factors

Uncollated Flux and Exponential Integral Functions

• Uncollated flux ($\phi_u$) for infinite planes is calculated as:     $\phi_u = \frac{S}{2\mu r}$     where $S$ represents the source and $r$ can also be interpreted as $r^2$ depending on geometry.

• The parameter $B = \mu r$ involves $\mu$, the linear recognition coefficient, and $r$, the distance from the source.

• For a planar source where distance is represented by $z$, the integration required to determine flux must be handled specifically. This necessitates the introduction of the exponential integral function ($E_n$).

• The exponential integral function $E_1(x)$ is often used to perform this integration. It can be written as a series or expansion:     $E_1(x) = e^{-x} \times (1 + x + ...)$

• The specific relation for the nth-order exponential integral can be defined where $n = 1$ if only distance $t$ is considered. It is also possible to read these values from plots of the exponential function versus the value of $\mu r$.

• Uncollated flux can be expressed in terms of the exponential integral function, sometimes referred to as $EI$. This is distinct from the logarithmic integral function ($LI$), which is not utilized in this shielding context.

Build-up Flux and the Build-up Factor

• Build-up flux ($\phi_b$) accounts for scattered radiation and is expressed as:     $\phi_b = B(\mu r) \times \phi_u$     where $B(\mu r)$ is the build-up factor.

• The build-up factor can be represented using the Taylor expansion (Taylor function) form:     B = A_1 e^{-\alpha_1 \mu r} + A_2 e^{-\alpha_2 ̇\mu r}

• The short-form summation for the build-up factor is:     $\sum A_n e^{-\alpha_n \mu r}$

• Key constraints and values for the build-up calculation:

  • $A_1 + A_2 = 1 \implies A_2 = 1 - A_1$.

  • The values for $A_1$, $\alpha_1$, and $\alpha_2$ are obtained from specific tables such as Table 10.3.

• Calculation of the build-up flux involves defining energy levels and related values for $\mu r$, where $t = \mu r$ and the final expression relates back to the exponential integral function as a function of $(1 + \alpha_n) \mu a$.

Design and Calculation of Concrete Shielding Thickness

• Numerical example for an isotropic planar source:

  • Source strength ($S_p$): $10^9 \, \text{gamma photons/cm}^2\text{/s}$.

  • Energy ($E$): $2 \, \text{MeV}$.

  • Intended exposure rate ($\dot{X}$): $2.5 \, \text{mR/h}$.

  • Density of concrete (\rho$):$2.35 \, \text{g/cm}^3.\n * Mass energy-absorption coefficient (\mu_{en}/\rho$):$0.0238 \, \text{cm}^2\text{/g}.\n * Mass attenuation coefficient (\mu/\rho$):$0.0445 \, \text{cm}^2\text{/g}.\n\n* Step 1: Calculate the required build-up flux (\phi_b):\n    \phi_b = \frac{\dot{X}}{0.0659 \times E \times (\mu_{en}/\rho)}\n    \phi_b = \frac{2.5}{0.0659 \times 2 \times 0.0238} = 797 \, \text{gamma photons/cm}^2\text{/s}\n    This indicates that concrete must block almost all photons, reducing the count from 1 billion to approximately 800.\n\n* Step 2: Determine Table 10.3 values for 2 \, \text{MeV}:\n * A_1 = 18.089\n * A_2 = 1 - 18.089 = -17.089\n * \alpha_1 = -0.04250 \implies 1 + \alpha_1 = 0.9575\n * \alpha_2 = 0.00849 \implies 1 + \alpha_2 = 1.00849\n\n* Step 3: Solve for shielding thickness (a) graphically or numerically:\n * Using the relation 1 = 1.13 \times 10^7 \times \text{RHS}.\n * Testing values for \mu a$: at 10 ($34.6$), at 12 ($4.79$), at 14 ($0.67).\n * The intersection where the ratio equals 1 occurs at \mu a = 13.6.\n\n* Step 4: Final thickness calculation:\n * \mu = (\mu/\rho) \times \rho = 0.0445 \times 2.35 = 0.1046 \, \text{cm}^{-1}.\n * a = \frac{13.6}{0.1046} \approx 130 \, \text{cm}.\n * The required thickness of concrete is 130 \, \text{cm}.\n\n# Line Source in Cylindrical Shielding\n\n* For a line source (such as a fuel rod) within a cylindrical shield, let L_1$and$L_2$be lengths at radial distance$R$. A small element$dz$is analyzed at distance$r$from point$P.\n\n* Uncollated flux (\phi_u) for a line source involves integrating over the length:\n    \phi_u = \int_{L_1}^{L_2} \frac{S_L}{4\pi r^2} dz\n\n* Transformation to angular coordinates:\n * r = R \sec(\theta)\n * z = R \tan(\theta)\n * dz = R \sec^2(\theta) d\theta\n\n* The uncollated flux is rewritten using the "c words integral function" (F(\theta, x), also referred to as the Sievert integral):\n    \phi_u = \frac{S_L}{4\pi R} [F(\theta_1, \mu R) + F(\theta_2, \mu R)]\n    F(\theta, x) = \int_0^{\theta} e^{-x \sec(\theta')} d\theta'\n\n* The c words integral is an even function where f(-\theta, x) = -f(\theta, x)$in some cases, though for standard integration from$-\theta_1$to$\theta_2, the sum is used.\n\n# Multilayered Shielding Principles\n\n* In multilayered shielding, the sequence of materials matters because build-up factors vary based on energy and the atomic number (Z) of the material.\n\n* Interchanging positions of material layers changes the build-up flux. For example, at 0.5 \, \text{MeV}$and$\mu x = 10$, the build-up factor for water is approximately 36, while for lead ($Pb) it is only 2.\n\n* **Case 1: Similar Atomic Numbers**\n * If \Delta Z < 10$, the overall build-up factor is determined by the total$\mu r$($\mu_1 a_1 + \mu_2 a_2) as if it were a single material.\n\n* **Case 2: Low-Z followed by High-Z**\n * If \Delta Z > 10 and the low-Z material is the first barrier, the build-up factor is determined primarily by the high-Z material. The high-Z material effectively shields the build-up produced in the low-Z material.\n\n* **Case 3: High-Z followed by Low-Z**\n * Sub-case 3.1 (E < 3 \, \text{MeV}$): The total build-up factor is the product of the individual factors:$B = B_1(\mu_1 a_1) \times B_2(\mu_2 a_2).\n * Sub-case 3.2 (E > 3 \, \text{MeV}$): The build-up factor for the first layer is taken at the original energy ($E_0$). The build-up factor for the second layer (low-Z) is taken at the minimum energy level, typically$3 \, \text{MeV}, regardless of the source energy. This is because the high-Z material shifts the radiation spectrum toward the minimum transmission energy.\n\n# Comparative Example: Water and Lead Shielding\n\n* Initial conditions: Mono-directional beam, 6 \, \text{MeV}$, intensity$10^6 \, \text{gamma/cm}^2\text{/s}$. Shielding consists of$100 \, \text{cm}$water and$8 \, \text{cm} lead.\n * (\mu a){\text{water}} = 0.0275 \times 100 = 2.75\n * (\mu a){\text{lead}} = 0.4944 \times 8 = 3.96\n\n* **Scenario A: Water placed before Lead**\n * Build-up factor (B_{net}$) is dominated by Lead at$3.96$, yielding$B \approx 1.87.\n * Exposure rate: 15.1 \, \text{mR/h}.\n\n* **Scenario B: Lead placed before Water**\n * B_{net} = B_{\text{lead}}(6 \, \text{MeV}) \times B_{\text{water}}(3 \, \text{MeV}) = 1.86 \times 2.72 = 5.06\n * Exposure rate: 41.1 \, \text{mR/h}.\n\n* Conclusion: Placing low-Z material first and high-Z material as the outer shield results in a significantly lower (3x) exposure rate.\n\n# Neutron Shielding Concepts and Materials\n\n* **Neutron Types:** Prompt neutrons involve the biggest shielding challenges. Delayed neutrons are less of a concern in light water reactors (LWRs) but may be an issue in fusion reactors where there is high circulation.\n\n* **Fast Neutron Shielding:** The best approach is to slow neutrons down to thermal energies (moderation) where absorption cross-sections are higher. \n * Low-Z materials like hydrogen (in water, concrete, or polymers) are superior moderators. \n * Hydrogen can slow a neutron in approximately 18 scattering events.\n\n* **Specific Materials:**\n * **Concrete:** Contains about 10% water by weight, which helps moderate fast neutrons.\n * **Steel:** Reduces neutron energy through inelastic scattering. This is most effective for very high energy neutrons (>10 \, \text{MeV}$, such as$14 \, \text{MeV} in fusion).\n * **Lead:** Good for gammas, but can be activated at energies above 1.2 \, \text{MeV}$, becoming Lead-210 ($210), which emits its own gammas.\n * **Tungsten:** High atomic number (Z=74$) and high density ($19 \, \text{g/cm}^3$). However, it is brittle and has a very high melting point ($>3000 \, \text{K}), requiring manufacturing via powder metallurgy.\n * **Paraffin Wax:** Often used as a neutron scatterer, sometimes in combination with lithium or boron.\n\n* **Inelastic Scattering Energy Reduction:** The average energy after scattering (E') is given by:\n    E' = 6.4 \times \frac{E}{A}\n    where E$is initial energy and$A$is atomic mass number. One scattering event in steel can reduce energy from$14 \, \text{MeV}$to$3 \, \text{MeV}.\n\n* **Secondary Radiation (Secondary Gammas):**\n * Hydrogen capture: releases 2.2 \, \text{MeV} gammas.\n * Boron-10: releases alpha particles and 0.5 \, \text{MeV} gammas.\n * Iron inelastic scattering: releases gammas at 7.6 \, \text{MeV}$and$9.3 \, \text{MeV}$$.

• Composite Structures: Utilizing layers (e.g., steel, graphite, concrete) allows the total thickness to be much smaller than a single material (like pure concrete) for the same exposure rate.

Questions & Discussion

• Participant: Can you slide that up a little bit, please?

• Response: (The visual material was adjusted accordingly to show the bottom of the calculation for the multilayered shield).