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Deriving Torricelli's law

  1. Apr 15, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must derive Torricelli's law.



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


    3. 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.
     
  2. jcsd
  3. Apr 17, 2009 #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: Apr 18, 2009
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