# Velocity of a liquid leaving a tank with an angle

1. Nov 8, 2009

### fluidistic

1. The problem statement, all variables and given/known data
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 $$\alpha$$?

2. Relevant equations
None given.

3. 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.

2. Nov 8, 2009

### 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.

3. Nov 8, 2009

### fluidistic

Ah thanks a lot!

4. Nov 8, 2009

### 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 $$x$$?

5. Nov 8, 2009

### 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.

6. Nov 8, 2009

### fluidistic

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