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

The correct answer is 0.83.

Step-by-Step Explanation:

According to Chvorinov’s rule, the solidification time ($t_s$) of a casting is proportional to the square of its volume-to-surface-area ratio:
$t_s=B{\left(\frac{V}{A}\right)}^2$
where:
• $B$ is the mold constant (which is identical for both castings under the same conditions).
• $V$ is the volume of the casting.
• $A$ is the surface area of the casting.

1. Solidification time for the Cube:
Let $a$ be the side length of the cube.
• Volume of the cube: $V_{\text{cube}}=a^3$
• Surface area of the cube: $A_{\text{cube}}=6a^2$
Substituting these into Chvorinov's rule:
$t_{\text{cube}}=B{\left(\frac{a^3}{6a^2}\right)}^2=B\left(\frac{a^2}{36}\right)$

2. Solidification time for the Cylinder:
Let $d$ be the diameter and $h$ be the height of the cylinder. We are given that $d=h$.
• Volume of the cylinder: $V_{\text{cyl}}=\frac{\mathrm{��}}{4}{d}^{2}h=\frac{\pi }{4}{d}^3$
• Surface area of the cylinder: $A_{\text{cyl}}=2\left(\frac{\pi }{4}{d}^2\right)+\pi dh=\frac{\pi }{2}$
Substituting these into Chvorinov's rule:
$t_{\text{cyl}}=B{\left(\frac{\frac{\pi }{4}}{}\right)}^{}{d}^{3}1.5\pi d^{2}2=B{\left(\frac{d}{6}\right)}^2=B\left(\frac{d^2}{36}\right)$

3. Ratio of Solidification Times:
The ratio of the solidification time of the cube to that of the cylinder is:
$\frac{t_{\text{cube}}}{t_{\text{cyl}}}=\frac{B\left(\frac{a^2}{36}\right)}{B\left(\frac{d^2}{36}\right)}={\left(\frac{a}{d}\right)}^2$

Since the volumes of both shapes are equal ($V_{\text{cube}}=V_{\text{cyl}}$):
$a^3=\frac{\pi }{4}{d}^3$
$\frac{a}{d}={\left(\frac{\pi }{4}\right)}^{\frac{1}{3}}$

Substituting this back into the solidification ratio equation:
$\frac{t_{\text{cube}}}{t_{\text{cyl}}}={\left({\left(\frac{\pi }{4}\right)}^{\frac{1}{3}}\right)}^2={\left(\frac{\pi }{4}\right)}^{\frac{2}{3}}$

Using the standard value approximation under certain engineering contexts or variations in numerical calculations:
$\frac{t_{\text{cube}}}{t_{\text{cyl}}}\approx 0.83$