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In order to minimize neutron leakage from a reactor, the ratio of the surface ar

ID: 1510199 • Letter: I

Question

In order to minimize neutron leakage from a reactor, the ratio of the surface area to the volume must be as small as possible. Assume that a sphere of radius a and a cube both have the same volume Find the surface-to-volume ratio for the sphere. (Use any variable or symbol stated above as necessary.) A_surface, sphere/V_sphere = Find the surface-to-volume ratio for the cube. (Use any variable or symbol stated above as necessary.) A_surface, cube/V_cube= Which of these reactor shapes would have the minimum leakage the sphere the cube

Explanation / Answer

volume of sphere = 4 * pi * a^3 / 3 = 4.2*a^3

volume of a cube = b^3 = 4 * pi * a^3 / 3 (given)

b =a* ( 4 * pi / 3 )^(1/3) = a * 1.6

so surface area of sphere = 4*pi*a^2 = 12.6 a^2

surface area of cube = b^2 = (1.6a)^2 = 2.56 a^2

a) for sphere

ratio (area to volume) = (4 * pi*a^2) / (4 * pi * a^3 / 3) = 3 / a

b) for cube

ratio (area to volume) = (a^2 * (4*pi/3)^(2/3)) / (4*pi*a^3/3) = 0.62 / a

c)

cube has lower surface to volume ratio

so reactor should be in cubic shape