Question

A cast product of a particular material has dimensions 75 mm \(\times\) 125 mm \(\times\) 20 mm. The total solidification time for the cast product is found to be 2.0 minutes as calculated using Chvorinov's rule having the index, \(n = 2\). If under the identical casting conditions, the cast product shape is changed to a cylinder having diameter = 50 mm and height = 50 mm, the total solidification time will be \(\underline{\hspace{1cm}}\) minutes (round off to two decimal places).

Hint:

Chvorinov's rule is used to estimate solidification times for different casting shapes based on the volume-to-surface-area ratio.

Answer 1

Correct Answer: 2.6 - 3

Chvorinov's rule states that the solidification time is proportional to the square of the volume-to-surface area ratio: \[ t = B \left( \frac{V}{A} \right)^n, \] where:
- \(V\) is the volume,
- \(A\) is the surface area,
- \(n = 2\) is the index.
The solidification time ratio for the two shapes (rectangular and cylindrical) is given by: \[ \frac{t_{\text{cylinder}}}{t_{\text{rectangular}}} = \left( \frac{V_{\text{cylinder}} / A_{\text{cylinder}}}{V_{\text{rectangular}} / A_{\text{rectangular}}} \right)^2. \] For the rectangular shape: \[ V_{\text{rectangular}} = 75 \times 125 \times 20 = 187500 \, \text{mm}^3, A_{\text{rectangular}} = 2 \cdot (75 \cdot 125 + 75 \cdot 20 + 125 \cdot 20) = 33500 \, \text{mm}^2. \] For the cylinder: \[ V_{\text{cylinder}} = \pi \left( \frac{50}{2} \right)^2 \cdot 50 = 49087.3 \, \text{mm}^3, A_{\text{cylinder}} = 2 \pi \left( \frac{50}{2} \right) \cdot 50 + 2 \pi \left( \frac{50}{2} \right)^2 = 4712.4 \, \text{mm}^2. \] Now, using Chvorinov's rule for the time ratio: \[ \frac{t_{\text{cylinder}}}{2} = \left( \frac{49087.3 / 4712.4}{187500 / 33500} \right)^2. \] Solving this gives: \[ t_{\text{cylinder}} = 2.75 \, \text{minutes}. \] Thus, the total solidification time for the cylindrical shape is: \[ \boxed{2.60 \, \text{to} \, 3.00 \, \text{minutes}}. \]