What Is the Correct Derivation of Torricelli's Law?

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SUMMARY

The discussion focuses on deriving Torricelli's Law using Bernoulli's equation, specifically the equation P + ρgh + (ρv²)/2 = constant. The user initially misapplies the pressure terms, leading to incorrect results for the velocity of fluid flow. The correct derivation requires setting P₀ equal to P₁ and properly defining the reference point for pressure. Ultimately, the correct expression for velocity is v₁ = √(2gh), which the user realizes after identifying their errors in sign and reference point selection.

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Homework Statement


I must derive Torricelli's law.



Homework Equations


P+\rho gh +\frac{\rho v^2}{2} = \text{constant}.


The Attempt at a Solution


I chose the origin of the system as being on the surface of the liquid.
I have that P_0 = P_1+ \frac{\rho v_1^2}{2}.
But P_1=P_0+\rho gh, so the equation is equivalent to 0 = \rho gh +\frac{ \rho v_1^2}{2}.
Hence v_1^2=- 2gh.
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.
 
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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 P_0 be the pressure on water' surface and P_1 be the pressure of the point underwater where the liquid flows.
Using Bernoulli's equation P+\rho gh +\frac{\rho v^2}{2} = \text{constant}, at water' surface we have that P_0+\rho g \times 0 + 0 = P_0.
At the point where the liquid flows : P_1+ \rho gh + \frac{\rho v_1^2}{2}.
But P_1=P_0+\rho gh.
Thus we have that 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.
I should reach v_1=\sqrt {2gh} 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 :) .
P_0=P_1. Also, I cannot chose the origin as being on water' surface so my expression get different.
 
Last edited:

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