# Time take to fill Bottom Gating Casting formula derivation

Apply Bernoullis equation between point 1 and 2 , we get

\frac{p_{1}}{\rho g} + \frac{v_{1}^{2}}{2g} + h_{t} = \frac{p_{2}}{\rho g} + \frac{v_{2}^{2}}{2g} + 0              .......(1)

v_{1} can be neglected because pouring basin is very large and p_{1} is atmospheric pressure . Putting these value in above equation we get

h_{t} = \frac{p_{2}}{\rho g} + \frac{v_{2}^{2}}{2g}  ......(2)

From the diagram we can say pressure at point 2 and 2' will remain same. So if we apply bernoulli equation at point 2' and 3 we get

\frac{p_{2}}{\rho g} = \frac{p_{3}}{\rho g} + h    ............(3)

But p_{3} is atmospheric pressure and cross sectional area of casting is same v_{2} = v_{3}

\frac{p_{2}}{\rho g} =  + h ............(4)

put this value in equation 2 we get

\frac{v_{2}^{2}}{2 g} =  h_{t}-h

v_{2} =\sqrt{2g(h_{t}-h)} = v_{g}

Now consider that in time dt_{f} ,the liquid metal comes out of gate increasing level of metal in the cavity by dh.

Equating the volume , we get

A_{g} v_{g} dt_{f} = A dh

dt_{f} = \frac{A}{A_{g}} = \frac{dh}{\sqrt{2g(h_{t}-h)}}

Total time intepate this exposing from 0 to H

t_{f_{2}} = \int_{0}^{H} \frac{A}{A_{g}} \frac{dh}{\sqrt{2g(h_{t}-h)}} = \frac{A}{A_{g} \sqrt{2g}} \int_{0}^{H}\frac{dh}{\sqrt{h_{t}-h}}

= \frac{A}{A_{g} \sqrt{2g}} *(-2[\sqrt{h_{t}-H} -\sqrt{h_{t}}])

t_{f_{2}} = \frac{2A}{A_{g} \sqrt{2g}} *([ \sqrt{h_{t}}-\sqrt{h_{t}-H}])

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