- #1
XanMan
- 14
- 1
Homework Statement
A cubic wine box of dimensional length ##h## has a small tap at an angle at the bottom. When the box is full and is lying on a horizontal plane with the tap open, the wine comes out with a speed ##v_0##.
i) What is the speed of the wine if the box is half empty? (Neglect the speed of the liquid at the top of the box.)
ii) What is the speed of the wine if the box is tilted by 45 degrees? (See attached figure)
[Assume the pressure at the top and bottom is equal]
Homework Equations
Bernoulli Equation: ##\cfrac{p}{\rho} + \cfrac{v^2}{2} + \phi = ## const.
where ##\phi## is the potential energy for a unit mass as a function of the height ##z##.
It follows from the Bernoulli Equation that is ##p_0## is constant at the top and bottom, and the initial height is ##h##, we get: $$v_0 = \sqrt{2gh}$$
The Attempt at a Solution
[/B]
i) By Bernoulli Equation, we get:
$$\cfrac{p_0}{\rho} + \cfrac{0}{2} + \cfrac{gh}{2} = \cfrac{p_0}{\rho} + \cfrac{v_1^2}{2} + 0$$
(since at the bottom ##z = 0##, and if I understand right, the velocity of the liquid at ##h/2## is 0 (?)).
Simplifying, we get:
$$\cfrac{gh}{2} = \cfrac{v_1^2}{2}$$
Answer: $$v_1 = \sqrt{gh}$$
ii) I am not quite sure about this part of the question. I tried using Pythagoras' Theorem to find the height of the liquid in terms of ##h##, and got ##\cfrac{\sqrt{2}h}{2}##.
Following a similar procedure as in i), I got the following result, which I think is incorrect, and would like your help on it (thanks!):
Answer: $$v_2 = \sqrt{\sqrt{2}gh}$$
Note: I am unsure where to use ##v_0## in the problem, or the fact that the tap is "at an angle"!
Cheers in advance!
