In a setting where portals can be opened between one location and another,
Suppose you open a portal of area one square meter, with one end at the surface and the other end at the bottom of an ocean, three kilometers down, where the pressure is three hundred bar. Obviously a jet of water will come through the portal.
What will be the speed of the jet? Equivalently, how many cubic meters per second will come through?
Hydrostatic pressure is the pressure felt due to the weight of things above it. The pressure at any depth in the ocean can be calculated by the equation for hydrostatic pressure:
$$p - p_0 = \rho gh.$$
Here, $p_0$ is atmospheric pressure, and $p$ is pressure at the desired depth. The density of the fluid in the ocean is $\rho$, $g$ is the acceleration due to gravity, and $h$ is the height of the column of fluid.
The density of water changes slightly with temperature, and even more slightly with pressure. We will assume a standard value of 1030 kg/m$^3$, which is accurate enough to three significant digits.
$$\begin{align}p &= 1030\text{ kg/m}^3\cdot9.81\text{ m/s}^2\cdot3000\text{ m} + 101000\text{ Pa}\\ &=30.4 \text{ MPa}\end{align}$$
Bernoulli's equation governs incompressible flow of fluids at a low Mach number. Since our flow will be well below the speed of sound (as we'll see) and since we've already made an incompressibility assumption by using constant density in the last equation, we can use Bernoulli's equation here. Bernoulli's equation is
$$c = \frac{v^2}{2} + gz + \frac{p}{\rho}$$
where $c$ is a constant, $v$ is the velocity of fluid flow, $z$ is elevation above reference, and $g$, $p$, and $\rho$ are as before.
The constant can be factored out of the equation by rewriting the equation in terms of total head, which has units of meters. Before we re-write, we will cancel out $z$ as zero. You are asking for flow through a portal with no depth, there is no net elevation change between the pressure and low pressure zones. The energy head equation is
$$H = \frac{p}{\rho g}+\frac{v^2}{2g}.$$
For the zero-velocity situation, i.e. on the ocean side of the portal, we set
$$H = \frac{p_O}{\rho g}.$$
On the atmosphere side of the portal, we set
$$H = \frac{p_a}{\rho g}+\frac{v^2}{2g}.$$
Now we can set these two quantities equal to each other to solve for the velocity of the fluid flowing through the portal.
$$\begin{align}\frac{p_O}{\rho g} &= \frac{p_a}{\rho g} + \frac{v^2}{2g}\\ \frac{30400000\text{ Pa}}{1030\text{ kg/m}^3\cdot9.81\text{ m/s}^2} &= \frac{101000\text{ Pa}}{1030\text{ kg/m}^3\cdot9.81\text{ m/s}^2}+\frac{v^2}{2\cdot9.81\text{ m/s}^2}\\ v^2 &= 2\cdot9.81\cdot\left(\frac{304000000}{10100}-\frac{101000}{10100}\right)\text{ m}^2\text{/s}^2\\ v&= 243 \text{ m/s} \end{align}$$
Volumetric flow is expressd as $$\dot{V} = vA,$$ where $A$ is the area of the portal. Sinc $A = 1\text{ m}^2$, we have
$$\dot{V} = 243 \text{ m}^3\text{/s}.$$
This is equivalent to the flow of the Tiber river at Rome.
This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 \, \mathrm{km} .$ $$ V_{\text{exit}} = \sqrt{2 H g} = \sqrt{2 \times 3000 \times 9.81} = 242.61 \, \frac{\mathrm{m}}{\mathrm{s}} $$
