Question

A company is producing a disc-shaped product of 50 mm thickness and 1.0 m diameter using sand casting process. The solidification time of the above casting process is estimated by Chvorinov’s equation \[ t = B \left( \frac{V}{A} \right)^2, \] \text{where \( B \) is the mold constant, and \( V \) and \( A \) are the volume and surface area of the casting, respectively. It is decided to modify both the thickness and diameter of the disc to 25 mm and 0.5 m, respectively, maintaining the same casting condition. The percentage reduction in solidification time of the modified disc as compared to that of the bigger disc is _________.}
\text{[round off to one decimal place]}

Hint:

To calculate the reduction in solidification time, use Chvorinov’s equation and compare the volume-to-surface-area ratios of the modified and original discs.

Answer 1

Correct Answer: 74.5

The volume and surface area of a disc are: \[ V = \frac{\pi D^2 h}{4}, \quad A = \pi D h. \] For the bigger disc: \[ V_{\text{big}} = \frac{\pi (1)^2 (50)}{4} = 39.269 \, \text{m}^3, \quad A_{\text{big}} = \pi (1) (50) = 157.080 \, \text{m}^2. \] For the smaller disc: \[ V_{\text{small}} = \frac{\pi (0.5)^2 (25)}{4} = 4.909 \, \text{m}^3, \quad A_{\text{small}} = \pi (0.5) (25) = 39.269 \, \text{m}^2. \] The solidification time is proportional to \( \left( \frac{V}{A} \right)^2 \), so the percentage reduction is: \[ \text{Percentage reduction} = \left( \frac{V_{\text{small}}}{A_{\text{small}}} \right)^2 \times 100 - \left( \frac{V_{\text{big}}}{A_{\text{big}}} \right)^2 \times 100 = \left( \frac{4.909}{39.269} \right)^2 \times 100 \approx 74.5%. \] Thus, the percentage reduction in solidification time is approximately \( 74.5 \).