Question

A cast product of a particular material has dimensions 75 mm × 125 mm × 20 mm. The total solidification time for the cast product is found to be 2.0 minutes as calculated using Chvorinov's rule having the index, n = 2. If under the identical casting conditions, the cast product shape is changed to a cylinder having diameter = 50 mm and height = 50 mm, the total solidification time will be in minutes (round off to two decimal places).

• A. 2.83

Answer 1

The correct option is A 2.83
$\text{Casting dimensions = 75 mm × 125 mm × 20 mm}$

$\text{Total solidification time}{\left(t_s\right)}_{cu}=2.0min\text{.}$

$\text{Cylindrical casting = H = D = 50 min.}$

$\text{Total solidification Time}{\left(t_s\right)}_{cy}\text{= ?}$

$\text{Chvorinov's rule,}$

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

$\frac{{\left(t_s\right)}_{cu}}{{\left(t_s\right)}_{cy}}=\frac{{\left(\frac{V}{A}\right)}_{cu}^2}{{\left(\frac{V}{A}\right)}_{cy}^2}$

${\left(\frac{V}{A}\right)}_{cy}=\frac{\frac{\pi }{4}{d}^{2}h}{2\frac{\pi }{4}{d}^{2}h+\pi dh}$

$h=d$

${\left(\frac{V}{d}\right)}_{cy}\propto \frac{d}{6}$

$\frac{\left(t_s\right)cu}{{\left(t_s\right)}_{cy}}=\frac{{\left[\frac{75\times 125\times 20}{2(75\times 125+125\times 20+20\times 75)}\right]}^2}{{\left(\frac{50}{6}\right)}^2}$

$\frac{2}{{\left(t_s\right)}_{cy}}=\frac{49.131}{69.44}=0.7075$

${\left(t_s\right)}_{cy}=2.83min.$