Question

(5) Consider the figure below, showing the front of a brick-shaped object, with each mono-velocity beam completely covering the face of the object that it strikes. The object has depth Δ. Beam 1 travels in the -ex direction. Beam 2 travels in the +êy direction. You are given Σ,-Σ,-0.6 cm-1, 10 n/cm n2 105 n/em V1 = V2 = 1of cm/s (a) [10 points] "Calculate" the scalar flux at the center of the object. (b) [5 points] "Calculate" the net current density (usually denoted J) at the center of the object x = 1 cm, y = 1.5 cm inside the object (at y-2 cm) (c) [5 points] "Calculate" the capture rate density (neutrons captured per cm3) at the point (d) [10 points] "Calculate" the rate at which neutrons exiting the object through the top face x-6 cm y-2 cm ni, V n, V Extra Credit [5 points| Return to problem 5. "Calculate" the rate at which neutrons are being absorbed in the object.

Answer 1

Given that - Brick - shaped object having two beam. Beam 1 travel - e^x & Beam 2 in + e^y direction. Sigma_* = sigma_r = 0.6 cm^-1 delta z = 5 cm n_1 = 10^4 n/cm^3 n_2 = 10^5 n/cm^3 v_1 = v_2 = 10^6 cm/s Scalar flux at the center of the object - phi = phi_1 + phi_2 ....(i) phi_1 = n_1 v_1 = 10^4 times 10^6 = 10^10 n/cm^2 ....(ii) phi_2 = n_2 v_2 = 10^5 times 10^6 = 10^11 n/cm^2 ....(iii) phi = phi_1 + phi_2 = 10^10 + 10^11 = 11 times 10^10 n/cm^2 current density - J = phi_1 (-e^_x) + phi_2 e^_y J = -10^10 e^_x + 10^11 e^_y n/cm^2 The capture rate density - at the center of onject R = sigma_i phi_i = sigma_t phi_1 + sigma_r phi_2 R = 0.6 times 10^10 + 0.6 times 10^11 = 6.6 times 10^10 n/cm^3 \r\n Rate at