Flow rate of water

Consider a 2-m diameter cylindrical. the tank is filled with water to a depth of 0.85 m above a 2.2 mm diameter hole is the sidewall of the tank. the entrance to the hole is well rounded. assuming frictionless flow estimate the velocity and the volume flow rate of the water escaping from the hole.

At the level of hole one can apply the Bernoulli law. Again, we consider the pressures inside and outside the cylinder at the hole level. There is no position pressure.
The static pressure of the water inside the cylinder is equal to the dynamic pressure of the water outside the cylinder.
$\rho g h = \rho*V^2/2$
$h$ is the filling level of the cylinder, $V$ is the speed of the water outside the hole
$V =\sqrt{(2 g h)} = \sqrt{(2*9.81*0.85)} =4.08 m/s$
The volume flow rate can be found by knowing the diameter of the hole. In one second the volume flow is
Volume flow = hole area$*V = \pi*r^2*V =\pi*(1.1*10^{-3})^2*4.08 =1.55*10^{-5} m^3/s$
$=1.55*10^{-2} liter/s$