Question

How to find time in case of torricelli's theorem (fluids)?

Answer 1

I'm guessing you want to start with Toricelli Theorem/Law

• `v = sqrt(2 gh)`

and end up with something that says

• `t = ...`

Looking at continuity, at any point in time:

• The volume flow rate out of the small aperture of area, a, is: `dot V_(out) = v * a`

• The flow into the aperture, assuming a container of uniform cross section area , `A`, is: `dot V_(i n) = dot h * A`

`dot V_(out)= dot V_(i n)`

`v * a = dot h * A implies v = dot h A/a`

However, setting the co-ordinate system so the aperture is to be at `h = 0`, then `dot h le 0` and so:

`v = color(red)(-) dot h A/a`

Putting that back into the Law:

• `dot h A/a = - sqrt(2 gh)`

That separates for integration:

• `( dh)/sqrth = -a/Asqrt(2 g) * dt`

Integrating:

• `2 [ sqrth]_(H_o)^(h) = -a/Asqrt(2 g) \ [t]_0^t`

`:. \ t = (sqrt2 A)/( a sqrt( g)) ( sqrtH_o - sqrth )`

Because `H_o ge h`, you can see what that `color(red)(-)` adjustment was necessary.

That is the time that will be required for the container to drain from `h = H_o` to `h(t)`