Question

The solidification time of a casting is proportional to \({\left( {\frac{V}{A}} \right)^2}\), where V is the volume of the casting and A is the total casting surface area losing heat. Two cubes of same material and size are cast using sand casting process. The top face of one of the cubes is completely insulated. The ratio of the solidification time for the cube with top face insulated to that of the other cube is
1. \(\frac{{25}}{{36}}\)
2. \(\frac{{36}}{{25}}\)
3. 1
4. \(\frac{6}{5}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{36}}{{25}}\)

Concept:

The total solidification time is the time required for the casting to solidify after pouring. This time is dependent on the size and shape of the casting by an empirical relationship known as Chvorinov’s rule.

\(T = {C_m}{\left( {\frac{V}{{{A_s}}}} \right)^2}\)

Calculation:

Solidification time \(t\propto {\left( {\frac{V}{A}} \right)^2}\)

For cube:

\(\frac{V}{A} = \frac{{{a^3}}}{{6{a^2}}} = \frac{a}{6}\)

Cube 2: Top face insulated

\(\begin{array}{l} \Rightarrow {t_1}\propto {\left( {\frac{a}{6}} \right)^2}\\ {t_2}\propto {\left( {\frac{{{V_2}}}{{{A_2}}}} \right)^2} \Rightarrow {t_2}\propto {\left( {\frac{{a^3}}{{5{a^2}}}} \right)^2} \Rightarrow {t_2}\propto {\left( {\frac{a}{5}} \right)^2} \end{array}\)

Then \(\frac{{{t_2}}}{{{t_1}}} = \frac{{{{\left( {\frac{a}{5}} \right)}^2}}}{{{{\left( {\frac{a}{6}} \right)}^2}}} = \frac{{36}}{{25}}\)