Discharge velocity of deposit with hole

Answer

This is the typical question that can be solved by using Bernoulli equation between two points. So, let's start by stating that point 1 will be exactly at the water top surface (in the deposit), while point 2 will be at the discharge surface in the bottom.

$$P_1 + \frac{\rho}{2} \cdot V_1^2 + \rho \cdot g \cdot h_1 = P_2 + \frac{\rho}{2} \cdot V_2^2 + \rho \cdot g \cdot h_2$$

When we apply Bernoulli in this kind of systems, it is common that we can simplify some of the variables:

If the deposit is open to the atmosphere, then the pressure at both points is the atmospheric pressure: $P_1 = P_2$
If the deposit diameter is much bigger than the hole diameter ($D \gg d$), then $V_1 \approx 0$, so the term $\frac{\rho}{2} \cdot V_1^2$ can be neglected
We will choose that the origin in the vertical direction is right at the center of the hole, so $h_2 = 0$ and $h_1 = H$

Introducing these simplifications in the equation, we are left with only two members, with only one unknown, which is the discharge velocity:

$$\cancel{P_1} + \cancel{\frac{\rho}{2} \cdot V_1^2} + \rho \cdot g \cdot h_1 = \cancel{P_2} + \frac{\rho}{2} \cdot V_2^2 + \cancel{\rho \cdot g \cdot h_2}$$ $$\rho \cdot g \cdot H = \frac{\rho}{2} \cdot V_2^2$$

Solving the equation for $V_2$:

$$V_2 = \sqrt{2\cdot g \cdot H}$$