Hello everyone, this is a cross-section diagram of a core tank. It is cylindrical in shape, with a cone-shaped bottom. We know the slope of the cone is 4°. I would like to ask for some advice on how to build this model. I have been stuck on this for some time, and I have noticed that when I define a cone surface, the surface’s region includes the top and bottom of the vertex, which feels different from a cylinder.
Thank you very much for your help.
You will want to create a ZCone
surface, which requires that you specify coefficients x0, y0, z0 (the apex) and r2, which is related to the aperature. Assuming the picture here is showing the z-axis in the vertical direction, if the center of this cylindrical tank has x=0 and y=0, that also means that the apex of the cone will be positioned at x0=0 and y0=0. The z0 coefficient will correspond to wherever the cone would intersect with the z axis. To determine r2, you can take a cut through the cone at y=0, which changes the full equation of the cone:
(x - x_0)^2 + (y - y_0)^2 = r^2 (z - z_0)^2
to the more simple:
(x - x_0)^2 = r^2 (z - z_0)^2
Take the square root of both sides and setting x0=0:
x = r (z - z_0)
Thus, r = x/(z - z_0) and since \tan(\theta) = (z - z_0)/x, this gives you:
r^2 = 1/\tan^2(\theta) where \theta is your 4 degree angle. In Python, you’d have
from math import tan, pi
r2 = 1 / tan(4 * pi/180)**2
The only thing this doesn’t give you is z0, which you’ll need to figure out based on where the apex of the cone should be in your geometry. One last note — you may want to use openmc.model.ZConeOneSided
to constrain the cone to a single side.
Thank you very much, @paulromano ,I comprehend