Velocity of a liquid leaving a tank with an angle

[fluidistic]

Homework Statement

A closed tank contains a liquid of density \rho and a gas (over the liquid) at a pressure P. Suppose there's a little hole (much smaller than the cross section of the tank so that when the liquid flows by the hole, the height of the liquid doesn't change with time) in the tank at a distance H under the liquid' surface. Now suppose I plug a small tube (at the orifice) forming an angle \alpha over the horizontal. How high will the liquid goes over the orifice, in terms of H and \alpha?

Homework Equations

None given.

The Attempt at a Solution

I've calculated the velocity of which the liquid leaves the tank : v=\sqrt{2 \left [ \frac{(P-P_{\text{atm}})}{\rho}+gH \right ] }.
Now how do I continue? I've tried something with conservation of energy but I'm sure I made an error since I get x=\frac{P-P_{\text{atm}}}{\rho g}+H which doesn't depend on \alpha as requested. I also notice that the height cannot be larger than H! So my result is wrong.
I'd like to have some guidance, like "Consider a small element dr and check out its velocity" or so.
Thanks.

 

[Delphi51]

That formula seems to give the height for the case when it is shot straight up.
To get the height in the angular case, won't you have to do a horizontal and vertical trajectory solution? As if a particle of the liquid is launched at the initial speed you found.

 

[fluidistic]

That formula seems to give the height for the case when it is shot straight up.
To get the height in the angular case, won't you have to do a horizontal and vertical trajectory solution? As if a particle of the liquid is launched at the initial speed you found.

Ah thanks a lot!

 

[fluidistic]

So the answer would be v'=\sin (\alpha ) \cdot v=\sqrt{2 \left [ \frac{(P-P_{\text{atm}})}{\rho}+gH \right ] } for the speed in the vertical axis and it remains to apply the conservation of energy, namely that \frac{mv'^2}{2}=mgx and get x?

 

[Delphi51]

I'm wary of using energy - not sure how energy splits between horizontal and vertical! Say you have a whole velocity of v, vertical component u = v*sinα. Then gravity will reduce the vertical speed to zero according to 0 = u-gt so t = u/g.
The distance traveled vertically in this time is
h = ut - .5*g*t² = u²/g -.5*u²/g = .5u²/g.
Ah, that is the same answer you got with energy.

 

[fluidistic]

I'm wary of using energy - not sure how energy splits between horizontal and vertical! Say you have a whole velocity of v, vertical component u = v*sinα. Then gravity will reduce the vertical speed to zero according to 0 = u-gt so t = u/g.
The distance traveled vertically in this time is
h = ut - .5*g*t² = u²/g -.5*u²/g = .5u²/g.
Ah, that is the same answer you got with energy.

I appreciate very much your time and effort.
Problem solved then. Thank you.