Question 5 of 7: Chvorinov's Rule — Solidification Time of a Sphere, Cube, and Cylinder of Equal Volume

Question 5: Chvorinov's Rule — Solidification Time of a Sphere, Cube, and Cylinder of Equal Volume (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three castings of the same metal, poured in the same mold material, with equal volume $V$: a sphere, a cube, and a cylinder whose height equals its diameter ($h=d$).

Find. Which shape solidifies fastest, which solidifies slowest, using Chvorinov's Rule to verify.

Approach. Chvorinov's Rule states the total solidification time $TST = C_m\,(V/A)^n$, where $C_m$ is the mold constant (same metal poured into the same mold material for all three pieces, so $C_m$ is identical) and $n$ is an empirical exponent, commonly taken as $n=2$ for a first approximation (the exam gives neither $C_m$ nor $n$, so this standard value is used and flagged below). Because $C_m$ and $n$ are common to all three shapes, only the relative $V/A$ ratio — and hence the relative $(V/A)^2$ — needs to be compared: the shape with the SMALLEST $V/A$ solidifies fastest, and the shape with the LARGEST $V/A$ solidifies slowest.

1. Express $V/A$ for each shape in terms of its own size parameter, using a common volume $V$. Sphere (radius $r$): $V=\tfrac{4}{3}\pi r^3$, $A=4\pi r^2$, so $\dfrac{V}{A}=\dfrac{r}{3}$. Cube (side $s$): $V=s^3$, $A=6s^2$, so $\dfrac{V}{A}=\dfrac{s}{6}$. Cylinder ($h=d=2r_c$): $V=\pi r_c^2 h = 2\pi r_c^3$; $A=2\pi r_c^2\;(\text{ends}) + 2\pi r_c h\;(\text{side}) = 6\pi r_c^2$, so $\dfrac{V}{A}=\dfrac{r_c}{3}$.
2. Solve each shape's size parameter for the same volume $V=1$ (unit volume) and evaluate $V/A$. $r=\left(\dfrac{3V}{4\pi}\right)^{1/3}=0.6204 \Rightarrow V/A_{\text{sphere}}=0.2068$. $s=V^{1/3}=1.0000 \Rightarrow V/A_{\text{cube}}=0.1667$. $r_c=\left(\dfrac{V}{2\pi}\right)^{1/3}=0.5419 \Rightarrow V/A_{\text{cylinder}}=0.1806$. $$\boxed{V/A_{\text{cube}}=0.1667 \;<\; V/A_{\text{cylinder}}=0.1806 \;<\; V/A_{\text{sphere}}=0.2068}$$
3. Convert the $V/A$ ordering into a solidification-time ordering via $TST\propto(V/A)^2$. Taking the cube's $TST$ as the reference (ratio $=1$): $\dfrac{TST_{\text{cylinder}}}{TST_{\text{cube}}}=\left(\dfrac{0.1806}{0.1667}\right)^{2}=1.173$, and $\dfrac{TST_{\text{sphere}}}{TST_{\text{cube}}}=\left(\dfrac{0.2068}{0.1667}\right)^{2}=1.541$. $$\boxed{TST_{\text{cube}} \;<\; TST_{\text{cylinder}}\;(1.17\times) \;<\; TST_{\text{sphere}}\;(1.54\times)}$$

Physically, the sphere is the shape that minimizes surface area for a given volume, so it has the least surface through which to lose heat relative to how much metal it must freeze — it solidifies slowest. The cube, of the three shapes considered, has proportionally the MOST surface area exposed per unit volume (its flat faces and sharp corners give it more surface than a rounded shape of the same volume), so it loses heat fastest relative to its volume and solidifies fastest. The cylinder with $h=d$ sits between the two, closer to the cube's ratio than the sphere's.