Water in the bunker problem

Question

We have a bunker(Like a Cylinder), full of water with height $H$ initially, there is a whole at the bottom of the bunker. The pressure of the air outside is $P_0$. The surface of the bunker is $A$ and the surface of the whole is $a$. I want to calculate the decrease rate of the height of water in the bunker, with respect to time.

I tried using Bernoulli's theorem, but my answer is wrong. This is my attempt:

$P+\dfrac{1}{2}\rho v^2+\rho gh = const$

For the top section of the bunker I wrote:

$\dfrac{1}{2}\rho v^2+\rho gh+P_0$

And for the hole in the bottom:($n$ is the speed of the water going out of the hole)

$\dfrac{1}{2}\rho n^2+\rho g\times(0)+P_0=\dfrac{1}{2}\rho n^2+P_0$

So: $n^2=v^2+2gh$

Since: $n \times a = v \times A$, (I'm not sure about this part) so:

$gh=n^2-v^2=v^2 \times \dfrac{A^2-a^2}{a^2}$

I take the derivatives with respect to time:($w$ is the acceleration)

$gv = 2v \times w \times \dfrac{A^2-a^2}{a^2} => w = g \times \dfrac{a^2}{A^2-a^2}$

$\therefore v(t) = wt = gt \times \dfrac{a^2}{A^2-a^2}$

Answer 1

If you consider the hole to be very small ,i.e. $A>>a$

Then you can simply conclude:

$$-\frac{dy}{dt}=\sqrt{2gy}$$ where $y$ is the height of water calculated from the base.

And further simply solve$$-\int_H^0\frac{dy}{\sqrt{y}}=\sqrt{2g}\int_0^t dt$$