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: Understanding the Question:
We need to find the ratio of the solidification time of a cube to that of a cylinder, given they have the same volume and are cast under identical conditions. The cylinder's diameter is equal to its height. We must use Chvorinov's rule.
Step 2: Key Formula or Approach:
Chvorinov's rule states that the solidification time (\(t_s\)) is proportional to the square of the ratio of the volume (\(V\)) to the surface area (\(A\)) of the casting. \[ t_s = C \left( \frac{V}{A} \right)^n \] Given that the exponent \(n=2\) and the casting conditions are identical (so the mold constant \(C\) is the same for both), the formula is: \[ t_s = C \left( \frac{V}{A} \right)^2 \] The ratio of solidification times is: \[ \frac{t_{\text{cube}}}{t_{\text{cylinder}}} = \frac{C (V_{\text{cube}}/A_{\text{cube}})^2}{C (V_{\text{cylinder}}/A_{\text{cylinder}})^2} \] Since the volumes are the same (\(V_{\text{cube}} = V_{\text{cylinder}}\)), the equation simplifies to: \[ \frac{t_{\text{cube}}}{t_{\text{cylinder}}} = \left( \frac{A_{\text{cylinder}}}{A_{\text{cube}}} \right)^2 \] Step 3: Detailed Explanation:
Let the side of the cube be \(a\).

• Volume of cube: \( V_{\text{cube}} = a^3 \)
• Surface area of cube: \( A_{\text{cube}} = 6a^2 \)

Let the height of the cylinder be \(h\) and its diameter be \(d\). Given \(d=h\), so the radius \(r = d/2 = h/2\).

• Volume of cylinder: \( V_{\text{cylinder}} = \pi r^2 h = \pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{4} \)
• Surface area of cylinder: \( A_{\text{cylinder}} = 2\pi r^2 (\text{top/bottom}) + 2\pi r h (\text{side}) = 2\pi \left(\frac{h}{2}\right)^2 + 2\pi \left(\frac{h}{2}\right) h = \frac{\pi h^2}{2} + \pi h^2 = \frac{3\pi h^2}{2} \)

Now, we equate the volumes to find a relationship between \(a\) and \(h\): \[ a^3 = \frac{\pi h^3}{4} \implies a = h \left(\frac{\pi}{4}\right)^{1/3} \] Now we can express \(A_{\text{cube}}\) in terms of \(h\): \[ A_{\text{cube}} = 6a^2 = 6 \left( h \left(\frac{\pi}{4}\right)^{1/3} \right)^2 = 6h^2 \left(\frac{\pi}{4}\right)^{2/3} \] Now we find the ratio of the areas: \[ \frac{A_{\text{cylinder}}}{A_{\text{cube}}} = \frac{\frac{3\pi h^2}{2}}{6h^2 \left(\frac{\pi}{4}\right)^{2/3}} = \frac{3\pi}{12} \frac{1}{(\pi/4)^{2/3}} = \frac{\pi}{4} (\pi/4)^{-2/3} = \left(\frac{\pi}{4}\right)^{1 - 2/3} = \left(\frac{\pi}{4}\right)^{1/3} \] Finally, we calculate the ratio of solidification times: \[ \frac{t_{\text{cube}}}{t_{\text{cylinder}}} = \left( \frac{A_{\text{cylinder}}}{A_{\text{cube}}} \right)^2 = \left( \left(\frac{\pi}{4}\right)^{1/3} \right)^2 = \left(\frac{\pi}{4}\right)^{2/3} \] \[ \frac{t_{\text{cube}}}{t_{\text{cylinder}}} \approx (0.7854)^{2/3} \approx 0.8533 \] Rounding off to 2 decimal places, the ratio is 0.85.
Step 4: Final Answer:
The ratio of the solidification time of the cube to that of the cylinder is 0.85.