Bernoulli Equation and Velocities

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XanMan
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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]

20170125_183248_001.jpg


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!
 
on Phys.org
Looks good to me! But I assume they want the answers in terms of ##v_0##, not ##h##. (You can convert between the two.)

The tap being at an angle should not matter.
 
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Doc Al said:
Looks good to me! But I assume they want the answers in terms of ##v_0##, not ##h##. (You can convert between the two.)

The tap being at an angle should not matter.

Ah, how silly of me...I neglected to use ##v_0 = \sqrt{2gh}##! Thanks DocAl : - )
 
Doc Al said:
Looks good to me! But I assume they want the answers in terms of ##v_0##, not ##h##. (You can convert between the two.)

The tap being at an angle should not matter.
sir I have a doubt why component of gravitational force doesn't matter here when tap is at an angle
considering a fluid part it has 2 forces on it 1 by the pressure or weight created by fluid part above it and gravitational force on it.
please coeerct me if I am wrong
 
akshay86 said:
sir I have a doubt why component of gravitational force doesn't matter here when tap is at an angle
considering a fluid part it has 2 forces on it 1 by the pressure or weight created by fluid part above it and gravitational force on it.
please coeerct me if I am wrong

Where do forces come in however?
 
Also, why does the angle of the tap not matter?
 
akshay86 said:
sir I have a doubt why component of gravitational force doesn't matter here when tap is at an angle
considering a fluid part it has 2 forces on it 1 by the pressure or weight created by fluid part above it and gravitational force on it.
please coeerct me if I am wrong
The effect of gravity is included in the potential energy term.

XanMan said:
Also, why does the angle of the tap not matter?
All that matters is the height difference (review the derivation). Of course, the subsequent motion of the fluid does depend on whether it is sent out at an angle (like any other projectile).
 
Doc Al said:
The effect of gravity is included in the potential energy term.All that matters is the height difference (review the derivation). Of course, the subsequent motion of the fluid does depend on whether it is sent out at an angle (like any other projectile).

Yes, that's what I was thinking as well - it's the velocity *just* as it is leaving the pipe. Sorry for all the questions lately, but this was in my first university physics exam and I'm a bit scared about it to be honest! Never actually properly had physics before - quite a challenge but I'm enjoying it! :-)
 
XanMan said:
Yes, that's what I was thinking as well - it's the velocity *just* as it is leaving the pipe.
Exactly.

XanMan said:
Sorry for all the questions lately, but this was in my first university physics exam and I'm a bit scared about it to be honest! Never actually properly had physics before - quite a challenge but I'm enjoying it!
Never apologize for asking questions! I'm glad you're enjoying your physics journey and I'm sure you are up to the challenge. :smile:
 
Doc Al said:
Exactly.Never apologize for asking questions! I'm glad you're enjoying your physics journey and I'm sure you are up to the challenge. :smile:

Doc Al, did anyone tell you how awesome you are? Thanks for the support! :D