# 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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if you are not getting it then ask i am glad to help