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Deriving Torricelli's law

  • Thread starter fluidistic
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fluidistic
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Homework Statement


I must derive Torricelli's law.



Homework Equations


[tex]P+\rho gh +\frac{\rho v^2}{2} = \text{constant}[/tex].


The Attempt at a Solution


I chose the origin of the system as being on the surface of the liquid.
I have that [tex]P_0 = P_1+ \frac{\rho v_1^2}{2}[/tex].
But [tex]P_1=P_0+\rho gh[/tex], so the equation is equivalent to [tex]0 = \rho gh +\frac{ \rho v_1^2}{2}[/tex].
Hence [tex]v_1^2=- 2gh[/tex].
I see that I made an error of sign, but I don't know where. The "x-axis"'s positive sense I considered was the one pointing to the ground.
What did I do wrong?
Thanks in advance.
 

Answers and Replies

  • #2
fluidistic
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I don't know what I was thinking about, but replanting and redoing the problem I now get an even worse answer.
Let [tex]P_0[/tex] be the pressure on water' surface and [tex]P_1[/tex] be the pressure of the point underwater where the liquid flows.
Using Bernoulli's equation [tex]P+\rho gh +\frac{\rho v^2}{2} = \text{constant}[/tex], at water' surface we have that [tex]P_0+\rho g \times 0 + 0 = P_0[/tex].
At the point where the liquid flows : [tex]P_1+ \rho gh + \frac{\rho v_1^2}{2}[/tex].
But [tex]P_1=P_0+\rho gh[/tex].
Thus we have that [tex]P_0=P_0+\rho gh + \rho gh + \frac{\rho v_1^2}{2} \Leftrightarrow 0=2 \rho gh + \frac{\rho v_1^2}{2} \Leftrightarrow 0=2gh+\frac{v_1^2}{2} \Leftrightarrow v_1^2=-4gh[/tex].
I should reach [tex]v_1=\sqrt {2gh}[/tex] but I'm not close to it.
I'm wondering if I'm using the right equations. I don't see any error but there is at least one.
Edit: I just found my error so don't lose your time helping me :) .
[tex]P_0=P_1[/tex]. Also, I cannot chose the origin as being on water' surface so my expression get different.
 
Last edited:

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