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Suppose you open a portal between Earth and space with a diameter of 1 ft. The portal is an opening in space that immediately connects both sides, i.e., it’s like a Portal portal.

How strong would the air be rushing in, and how much would it effect the area around it? Let's say the portal is 1 foot in diameter.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – L.Dutch ♦
    Commented Jul 1, 2021 at 2:27
  • $\begingroup$ could you define portal $\endgroup$
    – Scrooge McDuck
    Commented Jul 3, 2021 at 15:42
  • $\begingroup$ This question is missing details asked for in comments shunted to chat. VTC. $\endgroup$
    – rek
    Commented Jul 4, 2021 at 5:47

6 Answers 6

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Using https://www.fujikin.co.jp/en/support/calculator/,

I get a flow rate of $85,000 m^3/h$. This corresponds to $327 m/s$ through the hole, or slightly less than the speed of sound.

However, at a distance of $1 m$ ($3ft$), the air would move at a speed of around $4 m/s$ ($10 mph$), and as you get further away, the air's speed will quickly decrease.

All in all, anything within around $20 cm$ ($8 in$) will experience hurricane strength winds, but anything much further than that will remain relatively unaffected.

My inputs were "flow rate" for calculation type, "gas" for fluid type, 1 for specific gravity (compared to air), 3500 for $c_v$, inlet pressure of 101kpa and outlet pressure of 0. $c_v$ is a number that describes the ease at which fluid flows through the valve. It is dependent on the diameter of the hole, the shape of the hole, and any other properties of the hole that influence flow rate.

Also, here is another source that gives the approximately same answer.

It makes reasonable physical sense that air will exit the portal at about the speed of sound. Here are two sources that say that air will enter a vacuum at approximately the speed of sound.

Note: This differs from Willk's answer because that calculator assumes the fluid is water.

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    $\begingroup$ Just to make the comparision with the other answer easier. This answer comes to around 1400m^3/ min, Willk's answer is around 50m^3/min. I have no idea which one is more realistic. $\endgroup$
    – quarague
    Commented Jun 30, 2021 at 9:31
  • $\begingroup$ I don’t think this tool is any trustworthy. It doesn’t include the diameter at all, yet gives a result in m³/h. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 11:47
  • $\begingroup$ @Wrzlprmft it actually does. The Cv value depends on diameter. $\endgroup$
    – Robert
    Commented Jun 30, 2021 at 12:24
  • $\begingroup$ @Robert: Sure, but where do you enter the diameter into that calculator? If your solution doesn’t use a parameter that clearly affects the outcome, that’s a strong indicator it’s wrong. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 12:36
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    $\begingroup$ Just so I get this right: You treat an arbitrary opening as a valve, then use some equation to determine the flow through that opening, use that to estimate a valve coefficient ($c_\text{v}$) for a liquid flowing through that valve, translate that to a gas flowing through the same valve, and finally translate that back to a case without a valve, different geometry (no pipes), pressure and probably scale? First, your answer should document that. Second, there are almost certainly assumptions broken on the way. I would guess that all the tools assume laminarity, while the problem is turbulent. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 13:19
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I used this tool.

https://www.copely.com/tools/flow-rate-calculator/

Starting values 304 mm (~1 foot) pipe, 1 bar pressure (atmospheric pressure), 0.1 meter hose.

I got 50246 liters/minute. Velocity was for some reason in imperial and was 37.8 feet / second which I converted to 25.7 miles per hour.

A hair dryer makes wind at 40 mph. But this portal is 12 inches across so it would be easier to dry your whole head at once in the wind. The only problem is that it sucks instead of blows and if you are not paying attention after your shower you could get sucked up against the 12 mm hole and if you occlude the entire hole with your wet body you will get a serious hickey.

Which is OK too! I am not going to judge. It is a new era and each of us needs to live our own truths.

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    $\begingroup$ 12 inches or 3-4 mm, not 12 mm. I'm sure it was just a typo. $\endgroup$
    – NomadMaker
    Commented Jun 29, 2021 at 19:32
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    $\begingroup$ To put the numbers in a wider context, 50,000 L/min is 50 m³/min. Spreading out the flow of air to the area of a house door, about 3 m², you get a current of air through that door flowing at about 16 meters/minute, or about 0.27 m/sec; which speed is about 1 km/h or 0.6 mph, about one quarter of the speed of a person walking -- in other words, barely any wind at all, a very gentle breeze. $\endgroup$
    – AlexP
    Commented Jun 29, 2021 at 19:36
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    $\begingroup$ @DWKraus Gases are fluids (as far as physics is concerned). Wikipedia: "In physics, a fluid is a liquid, gas, or other material that continually deforms (flows) under an applied shear stress, or external force." Though the tool linked in this answer does seem to be specific to water. $\endgroup$
    – David Z
    Commented Jun 30, 2021 at 4:24
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    $\begingroup$ That tool is made by a hose manufacturer and context makes it clear that it is about water. There certainly is no general answer that is valid for every fluid since it depends on the viscosity of the fluid. For a simple example, see the Hagen–Poiseuille equation (and no, you cannot apply that one here either, since it assumes a laminar fluid, which air under these conditions is almost certainly not). $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 11:32
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    $\begingroup$ Agree this value seems low. Windspeeds on earth are much higher than this, but are caused by pressure differentials that are only fractions of an atmosphere at most. Intuitively, I'd expect a pressure differential of 1 full atmosphere to generate much more wind. Does the limited portal size really slow things down that much? $\endgroup$
    – Nuclear Hoagie
    Commented Jun 30, 2021 at 14:01
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I'll go partially against the other answers, because I'm going to claim the portal has zero length , as opposed to a piece of pipe that's 0.1 m long in one of the answers. Next, I'm going to claim that the portal "material" is frictionless, so there's no drag (not that drag coefficient has any meaning for a 2-Dimensional orifice)

Now we are essentially in the known situation of a hole in the wall of a spaceship, or the ISS. This is discussed, among other places, in space.StackExchange
EDIT: using the calculator referenced at that page, I get a flow rate of about 4600 l/s for our 30.5 cm diameter orifice at 1 atmosphere. Velocity is roughly 63 m/s, well subsonic so other nasty effects don't come into play.

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    $\begingroup$ This is not an answer; it is a commenty, or, at best, a link to an article which might provide an answer. It would be much more helpful if you could summarize the content of the linked article and apply it to the actual question. $\endgroup$
    – AlexP
    Commented Jun 30, 2021 at 12:38
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    $\begingroup$ Now we are essentially in the known situation of a hole in the wall of a spaceship, or the ISS – That spaceship hole (and others) have a completely different size and flows do not scale simply. Even laminar flows scale with the fourth power of length, and here we (almost certainly) have a turbulent flow that is even more complicated. Finally, a hole in a wall is different from a hole in a portal, since air can flow in from behind the portal – which may matter considerably, since we have turbulence. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 12:47
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    $\begingroup$ Not to mention that this is closely related to a couple of what-if.xkcd columns. Go look them up :-) $\endgroup$
    – Carl Witthoft
    Commented Jun 30, 2021 at 13:28
  • $\begingroup$ I am skeptical that you can apply that calculator here: 1) At least it says something about accounting for turbulence, but it also says that the coefficients it uses only affect the result “marginally”, which can only mean that turbulence is asumed to be low (which I don’t expect to be the case in the scenario in question). 2) It says that it uses tabulated values, which certainly don’t exist for the scenario in this question. 3) It is all about pipes (like all the other answers), which have a considerably different geometry than the scenario in question. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 13:40
  • $\begingroup$ Not to mention that this is closely related to a couple of what-if.xkcd columns. – If you mean this one, that’s about water, doesn’t give any exact number, and only links an equation that assumes no viscosity or turbulence. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 13:48
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The typical speed of air molecules is (not coincidentally) approximately the speed of sound, and precisely the fact that the molecules moving at this speed no longer encounter any obstacles when passing through the portal is what causes air flow in this "experiment". Hence the speed of the air flow will be some 300 m/s (at typical ground conditions, but varying with altitude, pressure, temperature). Multiply with the area of the portal to obtain the volume flow.

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    $\begingroup$ I see two problems with this: 1) Once the first molecules have gone through the portal, the pressure around the portal will decrease, possibly considerably so. 2) The limiting factor is not the air molecules currently passing through the portal, but getting them there. And here they cannot easily move with the speed of sound anymore, because they are always interacting with each other. $\endgroup$
    – Wrzlprmft
    Commented Jun 30, 2021 at 20:37
  • $\begingroup$ @Wrzlprmft that's exactly how fast the rest of the gas in the container will spread out, for the same reason. The molecules adjacent to the low-pressure area will either be moving in the right direction already, or they will be going in the wrong direction and very quickly bounce off another molecule which sends it back in the right direction. The next layer of molecules follows suit, after a delay caused by a similar bounce time. That is, the pressure difference is a sound wave. $\endgroup$
    – JDługosz
    Commented Jul 2, 2021 at 14:04
  • $\begingroup$ @JDługosz: Which container? — they will be going in the wrong direction and very quickly bounce off another molecule which sends it back in the right direction. – But that will slow down that other molecule and so on. Also that still ignores that you have depressurisation (which in turn affects temperature and thus molecule speed) and so on. Most importantly, if your argument holds, it would apply to any pressure difference, and those are certainly not resolved at the speed of sound. $\endgroup$
    – Wrzlprmft
    Commented Jul 2, 2021 at 14:25
  • $\begingroup$ This is more of an arbitrary guess than a calculated answer. $\endgroup$
    – Monty Wild ♦
    Commented Jul 4, 2021 at 16:18
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This scenario is quite similar to that in To the World of Death?, though with significant differences in pressure. Choked Flow is assumed to occur in dry air if the ratio of upstream to downstream pressure is >1.893. Since the downstream pressure is 0, this ratio's value is infinite.

So, we are looking to calculate choked flow of air through a double-sided portal 1'/0.3m in diameter.

So, the formula we need is:

$\dot{m} = C_d A \sqrt{\gamma \rho_0 P_0 ({2 \over {\gamma + 1}})^{{\gamma + 1} \over {\gamma -1}}}$

  • $\dot{m}$ is the choked mass flow rate.
  • $C_d$ is the discharge coefficient, which we assume to be 1.0, with the actual discharge rate being equal to the theoretical discharge rate.
  • A is the area of the portal. A 1 foot/30cm diameter circular portal has an area of ~0.07 $m^2$
  • $\gamma$ is the heat capacity ratio of the gas, for air at ~20°C: 1.400
  • $\rho_0$ is the gas density: 1.225 $kg/m^3$
  • $P_0$ is the upstream pressure. 1.0 atmospheres, or 101.325 kilopascals (kPa), or 101325 $kg/{m/s^2}$.

So, if we plug the numbers into the formula, we get:

$\dot{m} = 1 × 0.07 m^2 × \sqrt{1.400 × 1.225 kg/m^3 × 101325 kg/{m/s^2} × ({2 \over {1.400 + 1}})^{{1.400 + 1} \over {1.400 -1}}}$

$\dot{m} = 17.6 kg/s$

Converting to volume, we get $17.6 kg/s /1.225 kg/m^3 = 14.4 m^3/s$ per side, for a total of $28.8 m^3/s$ for both sides of the double-sided portal.

This means that air would be rushing into each side of the portal at a speed of $14.4 m^3/s / 0.07 m^2 = 206m/s$... i.e. at about 741kph.

Now, like light, as the distance from the portal increases, the speed of the air would decrease proportional to the square of the distance. So, at a distance of 2× the portal's diameter (2'/60cm), the wind speed would be around 185 kph, at 3'/90cm, it would be around 82 kph, 46 kph at 4'/1.2m, 30 kph at 5'/1.5m, 21 kph at 6'/1.8m and so on.

So, anyone or anything that got too close would be violently sucked in, and depending on its size, might variously find themselves in space, sliced into pieces on the edge of the portal (if it has one), or might block the portal and possibly suffer from a severe vacuum injury. However, the portal would be relatively harmless to humans at any range much over a metre, most birds would be safe at a distance of 1.5m, and really only a threat to insects out to a distance of a few metres.

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  • $\begingroup$ Can you comment on whether the equation you are using for a choked flow assumes a laminar or turbulent flow? At first glance, I cannot find anything explicit on this, though many of the materials I can find strongly suggest that laminarity is assumed. For example the Wikipedia articles on the Venturi effect and rocket-engine nozzles explicitly assume laminarity. Mind that in contrast to rocket engines or pipes, we here have no guiding structures to keep the flow laminar. $\endgroup$
    – Wrzlprmft
    Commented Jul 4, 2021 at 16:31
  • $\begingroup$ @Wrzlprmft Flow could be assumed to be largely laminar, depending upon the shape of the edge of the portal. In the case of turbulent flow,we could assume a lower air velocity through the portal due to an effectively smaller cross-section. $\endgroup$
    – Monty Wild ♦
    Commented Jul 4, 2021 at 16:42
  • $\begingroup$ Your calculation for the wind speed away from the portal is wrong. The correct formula should be $v=\frac{\dot{V}}{A_{portal} + 2\pi r^2}$. While your approximation works when you get far from the portal, because the portal is not a point, your approximation fails for short distances. $\endgroup$
    – Robert
    Commented Jul 4, 2021 at 18:41
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It wouldn't flow at all

Gravity is what keeps air on the surface of the planet. Opening a portal from a low gravity environment to a high gravity environment. The flow is from the low to the high not from high to low.

Air would not rush through into outer space any more than if you had a long pipe running into outer space.

If portals worked like such, you could create perpetual motion machines dropping water at the top of a mountain and generating energy as it run down hill until going through the portal again.

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    $\begingroup$ If portals worked like such, you could create perpetual motion machines dropping water at the top of a mountain and generating energy as it run down hill until going through the portal again. – If you have a world where portals exist, I don’t think the conservation of energy is something you would hold onto. The asker did not specify that the portal somehow conserves energy or (equivalently) that you have to work against the difference of gravitational potential between the two portal end points, so I see no reason to assume this. $\endgroup$
    – Wrzlprmft
    Commented Jul 4, 2021 at 11:29
  • $\begingroup$ As I said, it's no difference between having a long pipe to space. It's not going to rush out. $\endgroup$
    – Thorne
    Commented Jul 4, 2021 at 12:52
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    $\begingroup$ If you build a long pipe to space, do not evacuate it, then air would not flow in because it has to work against the pressure caused by the weight of a kilometre-high column of air. It’s for the same reason the atmosphere doesn’t escape. If you evacuate the same pipe and then open it, air would flow in quite a bit until we have reached the pressure profile of the previous case or the atmosphere, respectively. The air still wouldn’t reach space because each molecule has to travel kilometres working against the gravitational force pulling it down. […] $\endgroup$
    – Wrzlprmft
    Commented Jul 4, 2021 at 15:57
  • $\begingroup$ […] The latter is crucially different for our portal: The air does not have to travel a large distance and particularly it doesn’t have to work against any force. The only way, your answer makes sense would be if there were a force acting upon anything that travels through the portal to compensate for the difference in gravitational potential between both ends. Given that this force would be extremely strong, it is safe to assume that the asker would have mentioned it if they wanted it to exist. $\endgroup$
    – Wrzlprmft
    Commented Jul 4, 2021 at 15:57
  • $\begingroup$ You must not be familiar with the source material the OP mentions - the portals in Portal are exactly this type of non-energy conserving, physics-breaking magic. Travel through them requires no energy input regardless of potential energy differences between the source and destination. This isn't like running a pipe to space, it's like putting a hole in the wall of a space station. $\endgroup$
    – Nuclear Hoagie
    Commented Jul 14, 2021 at 14:21

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