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?

  • 6
    Obligatory XKCD – Cort Ammon Nov 20 at 7:31
  • 4
    Bit of a bigger hole, but obligatory What If XKCD – CalvT Nov 20 at 10:10
  • There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated. – abukaj Nov 22 at 9:46
up vote 20 down vote accepted

Pressure from depth

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.

$$\begin{align}p &= 1030\text{ kg/m}^3\cdot9.81\text{ m/s}^2\cdot3000\text{ m} + 101000\text{ Pa}\\ &=30.4 \text{ MPa}\end{align}$$

Flow velocity from pressure

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 from flow velocity

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.

  • 4
    To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path. – Falco Nov 20 at 10:27
  • 4
    @Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball. – Martin Bonner Nov 20 at 13:17
  • @MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid. – kingledion Nov 20 at 13:37
  • @MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?) – Falco Nov 20 at 13:38
  • 2
    Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet. – MindS1 Nov 20 at 13:47

This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 \, \mathrm{km} .$ $$ V_{\text{exit}} = \sqrt{2 H g} = \sqrt{2 \times 3000 \times 9.81} = 242.61 \, \frac{\mathrm{m}}{\mathrm{s}} $$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.