Chvorinov's rule and riser design

Chvorinov's rule: the solidification time of a casting is proportional to the square of its volume-to-surface-area ratio (modulus):

$$t_s = C\left(\frac{V}{A}\right)^{2}$$

where C is the mould constant (min/cm$^2$), depending on the mould material, metal properties and pouring temperature.

(a) Mould constant

For the cube, side $a = 80$ mm $= 8$ cm:

$$\frac{V}{A} = \frac{a^3}{6a^2} = \frac{a}{6} = \frac{8}{6} = 1.333\ \text{cm}$$ $$C = \frac{t_s}{(V/A)^2} = \frac{3.2}{(1.333)^2} = 1.8\ \text{min/cm}^2$$

(b) Solidification time of the plate

$$V = 200\times100\times30 = 600\,000\ \text{mm}^3$$ $$A = 2(200\times100 + 200\times30 + 100\times30) = 2(29\,000) = 58\,000\ \text{mm}^2$$ $$\frac{V}{A} = \frac{600\,000}{58\,000} = 10.34\ \text{mm} = 1.034\ \text{cm}$$ $$t_{plate} = 1.8\,(1.034)^2 = 1.93\ \text{min}$$

(c) Riser design

Required riser time:

$$t_r = 1.25\times 1.926 = 2.41\ \text{min}$$

Required riser modulus:

$$\left(\frac{V}{A}\right)_r = \sqrt{\frac{2.41}{1.8}} = 1.157\ \text{cm}$$

For a cylinder with $H = 1.5D$:

$$V = \frac{\pi}{4}D^2(1.5D) = 0.375\pi D^3,\qquad A = \pi D(1.5D) + 2\cdot\frac{\pi}{4}D^2 = 2\pi D^2$$ $$\frac{V}{A} = \frac{0.375\pi D^3}{2\pi D^2} = 0.1875\,D$$ $$D = \frac{1.157}{0.1875} = 6.17\ \text{cm},\qquad H = 1.5\times 6.17 = 9.25\ \text{cm}$$

Check of feeding

Riser volume $= 0.375\pi(6.17)^3 = 276.5\ \text{cm}^3$. The casting volume is $600\ \text{cm}^3$. Even with 6 percent solidification shrinkage ($36\ \text{cm}^3$) the riser holds enough liquid, so the riser also satisfies the volume requirement.

Answer: $C = 1.8$ min/cm$^2$; plate solidification time $= 1.93$ min; riser diameter $= 61.7$ mm and height $= 92.5$ mm.