# Question Regarding torricelli's theorem/Law

I recently studied about bernoulli's equation/principle.

After the derivation of the said equation , my book gave some applications of the principle, which include torricelli's theorem/law.

In deriving torricelli's law from bernoulli's principle, the pressure at the opening of the tank in which the fluid is contained , is said to be equal to the same pressure which is applied at the the top surface of the applied fluid , namely the pressure of the atmosphere.

But my book also states that the pressure drops (according to bernoullis principle) when the fluid passes through a narrow pipe or opening and its velocity increases.

So why does the pressure remain the same in this situation ? Why doesn't it change? Any help would be much appreciated , THANKS.

Could you please answer in simple and easy to learn terms , Thanks AGAIN.

Here is a proof on wikipedia if any one else wants to follow along. The proof states that the pressure in the water is zero (I will take atmospheric pressure to be zero) after it has exited the hole. This is because there is no longer fluid on top of it after it goes out the hole. However, the proof does not say that the pressure is zero immediately inside the hole. In fact, the pressure there will be given by $\rho g h$, (where $\rho$ is the fluid mass density, $g$ is the acceleration of gravity, and $h$ is the depth in the water) as you would have expected.

So what happens is as the water moves through the hole it moves to a region of high pressure ($\rho g h$) to a region of zero pressure. This pressure gradient causes the water to accelerate to a speed given by $\dfrac{1}{2} v^2 = gh$.