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 molten metal surfaces. The relation between t_A and t_B is:

1. \({t_B} = \sqrt 2 {t_A}\)
2. \({t_B} = 2{t_A}\)
3. \({t_B} = \frac{{{t_A}}}{{\sqrt 2 }}\)
4. \({t_B} = 2\sqrt 2 {t_A}\)

Answer 1

Correct Answer - Option 2 : \({t_B} = 2{t_A}\)

Explanation:

Time taken to fill the mould with top gate:

\({t_A} = \frac{{A.H}}{{{A_g}\sqrt {2g{h_m}} }}\ \ \ \ \ \ldots \ldots \ldots \ldots .\left( 1 \right)\)

Where, A = Area of mould, H = Height of mould, A_g = Area of gate, h_m = Gate height

Given that, h_m = H, So eq. (1) becomes

\({t_A} = \frac{{A\sqrt {{h_m}} }}{{{A_g}\sqrt {2g} }}\ \ \ \ \ \ldots \ldots \ldots \ldots \ldots \left( 2 \right)\)

Time taken to fill the mould with bottom gate:

\(\begin{array}{l} {t_B} = \frac{{2A}}{{{A_g}\sqrt {2g} }}\left( {\sqrt {{h_m}} - \sqrt {H- {h_m\;} } } \right)\\ {t_B} = \frac{{2A}}{{{A_g}\sqrt {2g} }}\sqrt {{h_m}}\ \ \ \ \ \ \ \ldots \ldots \ldots \ldots \ldots \left( 3 \right) \end{array}\)

From eq. (2) & (3)

\(\frac{{{t_B}}}{{{t_A}}} = 2 \Rightarrow {t_B} = 2{t_A}\)