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).
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).
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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.
Updated On: Sep 4, 2025
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Solution and Explanation
- 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 \(2.94 \, \text{min/cm}^2\).
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