Here's something I've never noticed anyone asking when it comes to wormholes/portals: If they allowed air to pass through them, how fast would the wind be? Should atmospheric pressure differences be a concern?

Here's what I've got:

There's a round 2D wormhole with a diameter of 2m (imagine a green warp pipe with an inner diameter of 2m if that helps simplify things). On one end the barometer reads 1050mb (15.229psi), the other end reads 950mb (13.779psi); meaning there is a 100mb (1.45psi) difference in pressure between the two ends of the wormhole.

Assuming all other factors are static on both sides (68°F, same elevation, earth-like atmosphere, etc), can someone walk me through how to find the flow rate and velocity of the air traveling through this wormhole, please?

(Every online calculator I look at keeps asking me for more stuff like the flow rate in CFM, and I have no idea how to find that.)


3 Answers 3


Well, one thing, you can forget about it being a wormhole. Physically, it's simply a 2m pipe the same length as the distance between the entry and exit points of the wormhole.

This article would appear to answer your question if the wormhole is short -- no more than maybe six meters long. If it's longer, the calculation becomes more complicated and the velocity is probably somewhat less.

The far-end pressure is about 90% of the source pressure. The gas is air. D (the diameter) is about 80". If I do the arithmetic right, the flow is around 350,000 SCFM. Since the area of the hole is about 34 sq ft, the corresponding air speed is 170 ft/sec or about 115 mph. (And that seems physically quite plausible to me.)

Added per request in comments: The conversion CFM to MPH is done by dividing the CFM rate by the area of the hole in square feet, giving the speed (in feet/minute) at which the gas must be moving to get all those cubic feet through in a minute. It's easy to convert from that to MPH.

  • $\begingroup$ Diameter is 2m, not 2ft, though it shouldn't matter to the end result. $\endgroup$ Apr 9, 2019 at 21:03
  • $\begingroup$ @Starfish Prime Thanks for the correction. (I should read more carefully.) I've corrected my answer. $\endgroup$
    – Mark Olson
    Apr 9, 2019 at 21:15
  • $\begingroup$ @MarkOlson Thank you for the answer; this is just what I've been looking for. I went through the equations myself and pretty much got what you estimated. The only thing left I could ask for is the equation you used to convert CFM to MPH. (But thankfully that shouldn't be hard to find.) And thank you for going through the steps; it really helps me make sure I'm doing it right. $\endgroup$ Apr 10, 2019 at 19:23
  • $\begingroup$ @Cameron Farley: Paragraph added. I hope that helps. (I'll add that in S. M. Stirling's excellent In the Courts of the Crimson Kings, it ends with a 1 mile wormhole being opened from an Earth-like planet to Mars. It's fun to calculate the wind through the portal and the length of time it takes for the atmospheres to equilibrate. $\endgroup$
    – Mark Olson
    Apr 10, 2019 at 21:19

Reverse the following process (Source):

Speed to Pressure Conversion

Write this equation converting wind speed in meters per second (m/s) to pressure in Newton per square meter (N/m^2):

Pressure = 0.5 x C x D x V^2

C = Drag coefficient D = Density of air (kg/m^3) V = Speed of air (m/s) ^ = "to the power of"

Obtain the wind speed value you wish to convert to pressure. It needs to be in meters per second or the equation will not work.

Example: V = 11 m/s

Estimate the drag coefficient based on the shape of the surface of your object that faces the wind.

Example: C for one face of a cubic object = 1.05

Others include:

Sphere: 0.47 Half-Sphere: 0.42 Cone = 0.5 Corner of a Cube = 0.8 Long Cylinder = 0.82 Short Cylinder = 1.15 Streamlined body = 0.04 Streamlined half-body = 0.09

For additional information regarding these shapes, visit the link in the Resources section.

Plug the values into the equation and calculate your answer:

Pressure = 0.5 x 1.05 x 1.25 kg/m^3 x (11 m/s)^2 = 79.4 N/m^2

Unit Conversions

Perform any necessary conversions to the units you desire. The wind speed must be in meters per second for the equation to be accurate.

Convert mph to meters per second (m/s) by multiplying the speed in mph by 0.447. This value is obtained by dividing the number of meters in 1 mile, 1609, by the number of seconds in 1 hour, 3600.

Example: 23 mph x 0.447 = 10.3 m/s

Convert Newton per square meter (N/m^2) to psi by multiplying the pressure in N/m^2 by 0.000145. This number is based on the number of Newtons in a pound and the number of square inches in a square meter.

Example: 79.4 N/m^2 x 0.000145 = 0.012 psi

Your biggest problem should be determining a useful drag coefficient. You get to pick one, though I'd suggest using the value of a long cylinder (0.82). Since your wormhole/portal is a fictional object of your own creation, you can choose whatever coefficient you want.

Once you've done that, you can use this resource to move back and forth between CFM and MPH.

  • 1
    $\begingroup$ Drag is about the energy loss due to friction between the air and surface, or the friction within the air resulting from the effects of the air surface friction. Unless the wormhole has a surface that creates friction with the air on it somewhere (or is close to another surface), there will actually be no drag in this case. That is because the air will behave irrotationally. $\endgroup$
    – XRF
    Apr 10, 2019 at 4:21
  • $\begingroup$ @XRF Aaaaand we don't know squat about the interior of a wormhole. This is where the OP gets to invent his/her world, because whether or not the inner surface of a wormhole creates friction is a choice of fiction. $\endgroup$
    – JBH
    Apr 10, 2019 at 4:24
  • 1
    $\begingroup$ The OP states it is flat so my assumption was basically a pair of flat portals. My comment about friction creating surfaces was about the back of the wormhole, d'Alembert's paradox states that for form drag to exist, skin friction must first exist. $\endgroup$
    – XRF
    Apr 10, 2019 at 4:29
  • $\begingroup$ @XRF, I see where you're coming from. I assumed flat from the perspective of the flat ends of a cylinder, but the long walls of the cylinder itself are of unknown (read: "defined by the OP") composition. $\endgroup$
    – JBH
    Apr 10, 2019 at 4:32
  • $\begingroup$ If it is a long enough cylinder inside, then Poiseuille flow equations would be able to solve for drag without much guessing. I am pretty sure the cylinder drag value you gave is for outside flow. $\endgroup$
    – XRF
    Apr 10, 2019 at 4:41

You can just use the jet thrust equation for this. Force = Mass Flow Rate * change in velocity. If we assume a calm wind that reduces to Force = Area * Density * Velocity^2, which is equivalent to Pressure = Density * Velocity^2 or Velocity = sqrt(Pressure * Density). This equation works using kg/m^3, N/m^2, and m/s as units in metric, can't tell you how to pick the units in imperial though.


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