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 ∝ 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?

The correct answer is a/2.

For a riser to be design-wise acceptable, its solidification time must be greater than or equal to the solidification time of the casting to ensure it feeds molten metal to the casting as it shrinks.
According to the given solidification time relation:

$$t_s\propto \frac{V}{A}$$

The ratio $M=\frac{V}{A}$ is defined as the modulus. Therefore, for an acceptable riser design:

$$M_{\text{riser}}\ge M_{\text{casting}}$$

1. Modulus of the Riser (M_riser):
Let the cylindrical riser have a radius r and height h.
The volume of the riser is:

$$V_r=\pi r^{2}h$$

Since the top of the riser is insulated and the bottom is in contact with the casting, heat is dissipated only through the curved lateral surface area:

$$A_r=2\pi rh$$

Thus, the modulus of the riser is:

$$M_r=\frac{V_r}{}=\frac{\pi r^{2}h}{2\pi rh}=\frac{r}{2}$$

2. Modulus of the Casting (M_casting):
The casting is a cube of edge length a.
The volume of the casting is:

$$V_c=a^3$$

The total surface area of a standalone cube is 6a². However, the contact area between the riser and the top surface of the casting (which is a circle of area πr²) does not dissipate heat to the mold. Therefore, the heat-dissipating surface area of the casting is:

$$A_c=6a^2-\pi r^2$$

Thus, the modulus of the casting is:

$$M_c=\frac{a^3}{6a^2-\pi r^2}$$

3. Evaluation of Design Condition:
Setting up the inequality:

$$\frac{r}{2}\ge \frac{a^3}{6a^2-\pi r^2}$$

Multiplying both sides by the denominator:

$$6a^{2}r-\pi r^3\ge 2a^3$$

Let us evaluate this condition for the given options:

• For r = a/3:
LHS = $6a^2\left(\frac{a}{3}\right)-\pi {\left(\frac{a}{3}\right)}^3=2a^3-\frac{\pi }{27}$, which is less than 2a³. Thus, a/3 is not acceptable.

• For r = a/2:
LHS = $6a^2\left(\frac{a}{2}\right)-\pi {\left(\frac{a}{2}\right)}^3=3a^3-\frac{\pi }{8}$, which is greater than 2a³. Thus, a/2 is acceptable.