Question

A cylindrical casting has 10 cm diameter and a mass of 12.56 kg. The material density is \( 7.85 \times 10^{-3} \, \text{kg/cm}^3 \). The value of exponent \(n\) is 2 and solidification time is 12 min. The Chvorinov's constant, in min/cm\(^2\), is ................. (round off to 2 decimal places).

Hint:

To calculate Chvorinov's constant, use the relationship between the volume and surface area of the casting, then apply the solidification time and exponent in the formula.

Answer 1

- The volume \(V\) of the cylindrical casting can be calculated using the formula for the volume of a cylinder: \[ V = \pi r^2 h \] Where the radius \(r = \frac{10}{2} = 5 \, \text{cm}\), and the height \(h\) is calculated using the mass and density: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} → \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{12.56}{7.85 \times 10^{-3}} = 1600 \, \text{cm}^3 \] Now, the volume is used to calculate the height of the cylinder: \[ V = \pi r^2 h → 1600 = \pi (5)^2 h → h = \frac{1600}{25\pi} \approx 2.04 \, \text{cm} \] - The Chvorinov's rule for solidification time \(t_s\) is given by: \[ t_s = C \cdot (V/A)^n \] Where \(C\) is Chvorinov’s constant, \(V\) is the volume of the casting, and \(A\) is the surface area. The surface area of a cylinder is: \[ A = 2 \pi r h + 2 \pi r^2 = 2 \pi (5) (2.04) + 2 \pi (5)^2 \approx 64.2 + 157.1 = 221.3 \, \text{cm}^2 \] We know the solidification time is 12 minutes, so: \[ 12 = C \cdot \left(\frac{1600}{221.3}\right)^2 \] Solving for \(C\): \[ C = \frac{12}{\left(\frac{1600}{221.3}\right)^2} \approx 2.94 \, \text{min/cm}^2 \] Thus, the Chvorinov’s constant is approximately \(2.94 \, \text{min/cm}^2\).