Let's break down the problem into 3 clear steps, with explanations at each step, and then provide a final answer. --- ### **Step 1: Set Up the Diffusion Equation and Solve for Flux** **Diffusion equation for a slab with a plane source at \( x = 0 \):** \[ -D \frac{d^2\phi(x)}{dx^2} + \Sigma_a \phi(x) = S_0 \delta(x) \] where - \( D = 1.0 \) cm (diffusion coefficient) - \( \Sigma_a = 0.035 \) cm\(^{-1}\) (absorption cross-section) - \( S_0 = 10^8 \) n/(cm\(^2\)·s) (source strength) - Slab extends from \( x = -a \) to \( x = +a \), with \( a = 10 \) cm. **General solution for \( x \neq 0 \):** \[ \phi(x) = A \cosh(kx) + B \sinh(kx) \] where \( k = \sqrt{\frac{\Sigma_a}{D}} = \sqrt{\frac{0.035}{1.0}} \approx 0.1871 \) cm\(^{-1}\). **Boundary Conditions:** - **Zero-reentrant current:** \( \phi(\pm a) = 0 \) - **Extrapolated boundary:** \( \phi(\pm a + z_0) = 0 \) where \( z_0 = 2.13D \approx 2.13 \) cm **Explanation:** We write the diffusion equation and its general solution, define the boundary conditions for both cases, and introduce the key parameters. --- ### **Step 2: Calculate Absorption and Leakage Rates for Each Boundary Condition** #### **A. Zero-reentrant current (\( \phi(\pm a) = 0 \)):** By symmetry, \( \phi(x) \) is even, so \( B = 0 \): \[ \phi(x) = A \cosh(kx) \] Boundary at \( x = a \): \( \phi(a) = 0 = A \cosh(ka) \implies A = 0 \) unless \( \cosh(ka) = 0 \) (which is not possible). So include the source condition at \( x = 0 \): \[ \left. -D \frac{d\phi}{dx} \right|_{x=0^+} + \left. D \frac{d\phi}{dx} \right|_{x=0^-} = S_0 \] Due to symmetry, \[ \left. \frac{d\phi}{dx} \right|_{x=0^+} = -\left. \frac{d\phi}{dx} \right|_{x=0^-} \implies -2D \left. \frac{d\phi}{dx} \right|_{x=0^+} = S_0 \implies \left. \frac{d\phi}{dx} \right|_{x=0^+} = -\frac{S_0}{2D} \] So, with symmetry and boundary conditions, the solution becomes: \[ \phi(x) = \frac{S_0}{2Dk \sinh(ka)} \cosh(k(a - |x|)) \] #### **B. Extrapolated Boundary (\( \phi(\pm (a+z_0)) = 0 \)):** Just replace \( a \to a + z_0 \) in the above solution: \[ \phi(x) = \frac{S_0}{2Dk \sinh(k(a+z_0))} \cosh(k(a+z_0 - |x|)) \] **Explanation:** We solve for the flux profiles in both cases using the appropriate boundary conditions and symmetry, leading to explicit formulas for \( \phi(x) \). --- ### **Step 3: Calculate Absorption and Leakage Rates and Compare** #### **Absorption Rate (\( R_a \)):** \[ R_a = \int_{-a}^{a} \Sigma_a \phi(x) dx = 2 \Sigma_a \int_0^a \phi(x) dx \] **For zero-reentrant current:** \[ R_a = 2 \Sigma_a \int_0^a \frac{S_0}{2Dk \sinh(ka)} \cosh(k(a-x)) dx \] Let \( u = a-x \Rightarrow du = -dx \): \[ = \Sigma_a \frac{S_0}{Dk \sinh(ka)} \int_0^a \cosh(k(a-x)) dx = \Sigma_a \frac{S_0}{Dk \sinh(ka)} \int_0^a \cosh(ku) du \] \[ = \Sigma_a \frac{S_0}{Dk \sinh(ka)} \left[ \frac{\sinh(ku)}{k} \right]_0^a = \Sigma_a \frac{S_0}{Dk^2 \sinh(ka)} [\sinh(ka) - 0] = \Sigma_a \frac{S_0}{Dk^2} \] Plug in numbers: - \( \Sigma_a = 0.035 \) cm\(^{-1}\) - \( D = 1.0 \) cm - \( k^2 = 0.035 \) cm\(^{-2} \) - \( S_0 = 10^8 \) n/(cm\(^2\)·s) \[ R_a = \frac{0.035 \times 10^8}{1.0 \times 0.035} = 10^8 \ \text{n}/(\text{cm}^2\cdot \text{s}) \] **Leakage Rate (\( R_l \)):** The total rate at which neutrons leak from both surfaces is given by the current at \( x = a \): \[ J(a) = -D \left. \frac{d\phi}{dx} \right|_{x=a} \] \[ \frac{d\phi}{dx} = \frac{S_0}{2Dk \sinh(ka)} \cdot (-k) \sinh(k(a-|x|)) \cdot \text{sgn}(x) \] At \( x = a \), \( \sinh(0) = 0 \), so **no leakage** for the zero-reentrant current case. (All neutrons are absorbed.) **Repeat for extrapolated boundary:** Just replace \( a \) with \( a+z_0 \): \[ R_a = \Sigma_a \frac{S_0}{Dk^2} \cdot \frac{\sinh(ka)}{\sinh(k(a+z_0))} \] Plug in \( k(a+z_0) = 0.1871 \times (10+2.13) \approx 2.27 \) \[ \sinh(ka) = \sinh(1.871) \approx 3.06 \] \[ \sinh(2.27) \approx 4.82 \] So, \[ R_a (\text{extrapolated}) = 10^8 \times \frac{3.06}{4.82} \approx 6.35 \times 10^7 \ \text{n}/(\text{cm}^2\cdot \text{s}) \] **Leakage:** \[ R_l = S_0 - R_a \] \[ = 10^8 - 6.35 \times 10^7 = 3.65 \times 10^7 \ \text{n}/(\text{cm}^2\cdot \text{s}) \] --- ## **Final Answers** - **Zero-reentrant current:** - Absorption rate: \( R_a = 1 \times 10^8 \) n/(cm\(^2\)·s) - Leakage rate: \( R_l = 0 \) n/(cm\(^2\)·s) - **Extrapolated boundary:** - Absorption rate: \( R_a = 6.35 \times 10^7 \) n/(cm\(^2\)·s) - Leakage rate: \( R_l = 3.65 \times 10^7 \) n/(cm\(^2\)·s) **Comparison:** The zero-reentrant current condition overestimates absorption (all neutrons are absorbed), while the extrapolated boundary gives a more realistic prediction with significant leakage. This illustrates the importance of using physically realistic boundary conditions in neutron transport problems.