Question

In a sand casting operation, the total liquid head is maintained constant such that it is equal to the mould height. The time taken to fill the mould with a top gate is $t_A.$ If the same mould is filled with a bottom gate, then the time taken is $t_B$. Ignore the time required to fill the runner and frictional effects. Assume atmospheric pressure at the top of molten metal surfaces. The relation between $t_A\text{and}{t}_B$ is

• A. $t_B=2t_A$
• B. $t_B=\sqrt{2}{t}_A$
• C. $t_B=\frac{t_A}{\sqrt{2}}$
• D. $t_B=2\sqrt{2}{t}_A$

Answer 1

The correct option is A

$t_B=2t_A$

$\text{Time taken to fill the mould with top gate}$

$t_A=\frac{AH}{A_g\sqrt{2g{h}_m}}\text{... (i)}$

$\text{Where,}$

$\text{A = Area of mould}$

$\text{H = Height of mould}$

$A_g\text{= Area of Gate}$

$h_m\text{= Gate height}$

$\text{Given data:}$

$h_m=H,\text{so Eq. (i) become:}$

$t_A=\frac{A\sqrt{h_m}}{A_g\sqrt{2g}}$ .... (ii)

$\text{Time taken to fill the mould with bottom gate,}$

$t_B=\frac{2A}{Ag\sqrt{2g}}(\sqrt{h_m}-\sqrt{h_m-H})$

$t_B=\frac{2A}{A_g\sqrt{2g}}\sqrt{h_m}\text{... (iii)}$

$\text{From Eqs. (ii) and (iii), we get}$

$\frac{t_B}{t_A}=2$

$t_B=2t_A$