Question

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.

Hint:

For problems involving Chvorinov's rule, remember that for a given volume, a sphere has the minimum surface area and will therefore take the longest to solidify. A shape with a higher surface area-to-volume ratio will cool and solidify faster.

Answer 1

Step 1: State Chvorinov's Rule. \(t_s = K(V/A)^2\). Given same volume and conditions, \(t_{cube}/t_{cyl} = (A_{cyl}/A_{cube})^2\).
Step 2: Relate dimensions for Cylinder (\(d=h\)). Volume \(V = \frac{\pi}{4}d^2 h = \frac{\pi d^3}{4}\). Surface Area \(A_{cyl} = \pi dh + 2(\frac{\pi d^2}{4}) = \pi d^2 + \frac{\pi d^2}{2} = 1.5\pi d^2\).
Step 3: Relate dimensions for Cube. Volume \(V = a^3\). Since Volumes are equal: \(a^3 = \frac{\pi d^3}{4} \implies a = d(\frac{\pi}{4})^{1/3}\). Surface Area \(A_{cube} = 6a^2 = 6d^2(\frac{\pi}{4})^{2/3}\).
Step 4: Calculate Area Ratio and Time Ratio. \[ \frac{A_{cyl}}{A_{cube}} = \frac{1.5\pi d^2}{6d^2(\pi/4)^{2/3}} = \frac{\pi}{4(\pi/4)^{2/3}} = (\frac{\pi}{4})^{1/3} \] Time ratio = \( [(\frac{\pi}{4})^{1/3}]^2 = (\frac{\pi}{4})^{2/3} \approx (0.785)^{0.666} \approx 0.85 \).