# Deriving Torricelli's law

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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?

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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.

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