Question

Consider sand casting of a cube of edge length $a$. A cylindrical riser is placed at the top of the casting. Assume solidification time, $t_s \propto V/A$, where $V$ is the volume and $A$ is the total surface area dissipating heat. If the top of the riser is insulated, which of the following radius/radii of riser is/are acceptable?

• A. $\frac{a}{3}$
• B. $\frac{a}{2}$
• C. $\frac{a}{4}$
• D. $\frac{a}{6}$

Hint:

A riser works only if it solidifies slower than the casting. To ensure this, choose a riser with a higher $V/A$ ratio than the casting.

Answer 1

The Correct Option is A, B

For directional solidification, the riser must freeze after the casting. According to Chvorinov’s rule, solidification time is proportional to the ratio $V/A$, where $V$ is volume and $A$ is surface area dissipating heat. A larger $V/A$ means slower cooling.
Step 1: $V/A$ for the cube.
The cube of edge $a$ has: \[ V_c = a^3,\quad A_c = 6a^2 \Rightarrow \frac{V_c}{A_c} = \frac{a}{6}. \]
Step 2: Compare with cylindrical riser.
For an insulated top, only side and bottom areas radiate heat. A riser with larger radius increases volume faster than area, giving higher $V/A$. Options $\frac{a}{3}$ and $\frac{a}{2}$ provide a sufficiently larger $V/A$ than the cube, ensuring they solidify later.
Thus acceptable riser radii are $(A)$ and $(B)$.