Question

Two metal parts (a cylinder and a cube) of same volume are cast under identical conditions. The diameter of the cylinder is equal to its height. The ratio of the solidification time of the cube to that of the cylinder is _________ (rounded off to 2 decimal places).
Assume that solidification time follows Chvorinov’s rule with an exponent of 2.

Hint:

For solidification time calculations, use Chvorinov’s rule. The ratio of solidification times depends on the geometry of the part (surface area and volume).

Answer 1

The solidification time follows Chvorinov's rule, which is given by: \[ T = \frac{C \cdot V^2}{A} \] The ratio of the solidification times for the cube and cylinder is: \[ \frac{T_{{cube}}}{T_{{cylinder}}} = \frac{s^6 / 6s^2}{4\pi r^2 / r^6} \] Since the volumes of the cube and the cylinder are the same: \[ s^3 = \pi r^3 \] Thus, the relationship between \( s \) and \( r \) is: \[ s = \left( \pi r^3 \right)^{1/3} \] Substituting this into the ratio, we find: \[ \frac{T_{{cube}}}{T_{{cylinder}}} = 0.83 \]