Variation of Torriceli's theorem?

[JorgeM]

Hi there.
Everyone knows about Torricelli's theorem that says about , in a too big container (opened) the speed of the liquid is given by:

v=√(2gh)
This result is just for containers that have a hole in the side and the fluid goes out perpendicular to the gravity. And also this result is just for an specific period of time when the h is almost constant.

But, when I think about a little container (Where the variations of parameters can not be taken as zero) that has its hole in the bottom (Like a bottle of water almost cylindrical) parallel to the gravity, I would like to know how this is not constant and also how this goes out of the bottle. I get so confused with the thing I am supossed to solve (I imagine it is going to be a differential ecuation or something like that).

Do you know if someone has already solved it in this forum?

How is the flow in function of h,because if I Integer the flow as function of h for a period of time(Big enough not to consider h as constant),
I will know how much fluid has gone out in that period.
Anyways I can not figure out the solution and don't even know where to read for this topic.

If your could help me or say some books that could help to solve my problem.
Thanks.
JorgeM.

 

[dRic2]

If the height of water in the container can be considered constant you can still use torricelli's formula.

 

[JorgeM]

If the height of water in the container can be considered constant you can still use torricelli's formula.

I was thinking about an small container where the variations are not minimal and makes impossible to take Torricelli's theorem as true.

 

[dRic2]

I would refer everything to the volumetric flow rate instead of the velocity. Assuming the fluid is incompressible then volumetric flow rate has to be constant because of conservation of mass.

Using Bernoulli:

##\frac {v_0^2} 2 + gh_0 = \frac {v_1^2} 2 + gh_1##

##\frac {v_1^2} 2 - \frac {v_0^2} 2 = g(h_0 - h_1)##

Then, remember the formula for the volumetric flow rate ##Q= v A## so ##v = Q/A##:

##\frac {Q^2} {2 A_1^2} - \frac {Q^2} {2A_0^2} = g(h_0 - h_1)##

##Q^2 \left( \frac 1 {2 A_1^2} - \frac 1 {2 A_0^2} \right ) = g(h_0 - h_1) ##

If you want to know the time needed to empty the bottle, I would suggest a mass balance:

##\frac {dm} {dt} = -\dot m_{out}##

it should give a simple differential equation.

That's my suggestion, hope it may help you.