Question

A company is producing a disc-shaped product of 50 mm thickness and 1.0 m diameter using sand casting process. The solidification time of the above casting process is estimated by Chvorinov’s equation \(t = B\left[ {\frac{V}{A}} \right]^2\) , where B is the mold constant, and V and A are the volume and surface area of the casting, respectively. It is decided to modify both the thickness and diameter of the disc to 25 mm and 0.5 m, respectively, maintaining the same casting condition. The percentage reduction in solidification time of the modified disc as compared to that of the bigger disc is _________.[round off to one decimal place]

Answer 1

Concept:

Solidification time: The time taken for the liquid molten metal into the solid is known as solidification time

Chvorinov's equation for solidification time is given by:

\({\bf{t}} = {\bf{B}} \times \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}\)

where, t = solidification time, B = mould constant, V = volume of the casting, A = surface area of the casting

Calculation:

Given:

\({\bf{t}} = {\bf{B}} \times \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}\) ⇒ \({\bf{t}}\; \propto \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}\)

\({\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2} = \;\frac{{\frac{{\bf{\pi }}}{4} \times {{\bf{D}}^2} \times {\bf{h}}}}{{\left[ {\left( {2 \times \frac{{\bf{\pi }}}{4} \times {{\bf{D}}^2}} \right) + \left( {{\bf{\pi }} \times {\bf{D}} \times {\bf{h}}} \right)} \right]}}\)

Initial casting:

Diameter (\(D_i\)) = 1 m, thickness (\(h_i\)) = 50 mm = 0.05 m

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{i}}}^2 = {\rm{\;}}\frac{{\frac{{\rm{\pi }}}{4} \times {1^2} \times 0.05}}{{\left[ {\left( {2 \times \frac{{\rm{\pi }}}{4} \times {1^2}} \right) + \left( {{\rm{\pi }} \times 1 \times 0.05} \right)} \right]}}\)

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{i}}}^2 = 0.0005165\)

Modified casting:

Diameter (\(D_m\)) = 0.5 m, thickness (\(h_m\)) = 25 mm = 0.0025 m

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{m}}}^2 = {\rm{\;}}\frac{{\frac{{\rm{\pi }}}{4} \times {{0.5}^2} \times 0.025}}{{\left[ {\left( {2 \times \frac{{\rm{\pi }}}{4} \times {{0.5}^2}} \right) + \left( {{\rm{\pi }} \times 0.5 \times 0.025} \right)} \right]}}\)

\({\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]_{\rm{m}}}^2 = 0.0001291\)

% reduction of solidification time is:

\(\frac{{{{\rm{t}}_{\rm{i}}} - {{\rm{t}}_{\rm{m}}}}}{{{{\rm{t}}_{\rm{i}}}}} \times 100 = {\rm{\;}}\frac{{{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}_{\rm{i}}}^2 - {\rm{\;}}{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}_{\rm{m}}}^2}}{{{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}_{\rm{i}}}^2}} = {\rm{\;}}\frac{{0.0005165{\rm{\;}} - {\rm{\;}}0.0001291}}{{0.0001565}} \times 100 = 75{\rm{\% }}\)

∴ the % reduction in solidification time is 75%