Question 4.1: A reactor is to be built with fuel rods of 1.2 cm in diamete......
A reactor is to be built with fuel rods of 1.2 cm in diameter, and a liquid moderator with a 2:1 volume ratio of moderator to fuel. What will the distance between nearest fuel centerlines be
a. for a square lattice?
b. for a hexagonal lattice?
Note: solution is presently for
1:1 volume ratio
Let r be the radius of the fuel rod or r = 0.6 cm and V_f / V_m = 1.0 The fuel to moderator volume ratio.
a. Let l be the lattice pitch, which is also the distance between nearest centerline:
V_{f}\,/\,V_{m}={\frac{\pi r^{2}}{l^{2}-\pi r^{2}}} or l=\sqrt{\pi(V_{m},/V_{f}+1)}\;r=\sqrt{3.141(1+1)}\;0.6=1.504~c m
b. Let l be the base of an equilateral triangle. Then the height will be h={\frac{\sqrt{3}}{2}}\,l\,. The area of the hexagon will be that of six equilateral triangles or
A_{h e x}=6\cdot{\frac{1}{2}}l\cdot{\frac{\sqrt{3}}{2}}l={\frac{3{\sqrt{3}}}{2}}\,l^{2} while V_{f}{} / V_{m}={\frac{\pi r^{2}}{A_{h\mathrm{{ex}}}-\pi r^{2}}}={\frac{\pi r^{2}}{{\frac{3\sqrt{3}}{2}}l^{2}-\pi r^{2}}}
Hence l=3^{-3/4}\sqrt{2\pi(V_{m},/V_{f}+1)}\ r However the distance between centerlines is
\sqrt{3}l=3^{-1/4}\sqrt{2\pi(V_{m}/V_{f}+1)}\stackrel{}{r}=3^{-1/4}\sqrt{2\cdot3.141(1+1)}\ 0.6=1.616