What Is the Correct Derivation of Torricelli's Law?

[fluidistic]

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.

 

[fluidistic]

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.

 

[nlightner]

Dear student,

Thank you for your question. To derive Torricelli's law, we must first understand the physical principles behind it. Torricelli's law states that the velocity of a liquid flowing out of a small opening at the bottom of a container is equal to the velocity that a freely falling object would have at the same height. This is known as the Torricelli's theorem.

To derive this law, we can use the Bernoulli's equation, which states that the total energy of a fluid remains constant along a streamline. This equation can be written as:

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

Where P is the pressure, \rho is the density, g is the acceleration due to gravity, h is the height and v is the velocity of the fluid.

Now, let us consider a small opening at the bottom of a container filled with liquid. We can assume that the pressure at the surface of the liquid (P_0) is equal to the pressure at the opening (P_1). This is because the liquid is at rest and the pressure is the same at all points at the same height.

Using this information, we can write the Bernoulli's equation for the surface of the liquid (P_0) and the opening (P_1) as:

P_0 + \rho gh_0 + \frac{\rho v_0^2}{2} = P_1 + \rho gh_1 + \frac{\rho v_1^2}{2}

Where h_0 is the height of the surface of the liquid and h_1 is the height of the opening.

Since P_0 = P_1, we can simplify the equation to:

\rho gh_0 + \frac{\rho v_0^2}{2} = \rho gh_1 + \frac{\rho v_1^2}{2}

Now, we can rearrange this equation to solve for the velocity at the opening (v_1):

v_1^2 = v_0^2 + 2g(h_0 - h_1)

Since the liquid is at rest at the surface (v_0 = 0) and the opening is at the same height as the surface (h_0 = h_1), the equation simplifies to:

v_1^2 = 2gh_0

This is the same result that