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bleedpurple

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The engineering problem of calculating the time to drain a pipeline, tank, or vessel through an orifice is fairly straightforward using the orifice equation.\(Q=CA_{o}\sqrt{2gh}\)

With C being the coefficient of discharge for the orifice, Ao being the area of the orifice, g is the acceleration of gravity, and h the energy 'head' or elevation of the fluid above the orifice all in feet.

This equation can then be integrated by expressing Q as the change in volume over time.

\(Q=\frac{dv}{dt}\)

And expressing V as a function of head, h.

For a pipeline with constant gradient or slope the volume equals the cross sectional area of the pipe, Ap x length, and the length is the elevation change or head, h divided by the slope.

Therefore:

\(\frac{dv}{dt}=\frac{A_{p}}{slope} \frac{d_h}{dt} =C\,A_{o}\sqrt{2gh}\)and

\(\int{h^{-1/2}}dh=C\,A_{o}\sqrt{2gh} \frac{slope}{A_{p}} \int{dt}\)

or

\(\int{dt} = \frac{A_{p}}{slope} \frac{1}{CA_{o}\sqrt{2g}}\int{h^{-1/2}}dh\)

from this the result is

\(t=\frac{2 A_{p}}{slope} \frac{1}{CA_{o}\sqrt{2g}}(h_{1}^{1/2}-h_{0}^{1/2})\)

Now the case where there is also a constant flow in addition to the orifice flow.

\(Q=CA_{o}\sqrt{2gh}+K\)

resolves to the integral

\(\int{dt} = \frac{A_{p}}{slope CA_{o}\sqrt{2g}} \int\frac{1}{h^{1/2}+\frac{K}{CA_{o}\sqrt{2g}}}dh\)

from this the result is ?