# How could a submarine-like vehicle steer and propel itself in a superfluid?

In a large body of liquid helium below the lambda point, what would be the best way(s) for submarine like vehicles to achieve acceleration and directional control?

I'm willing to accept an answer that invokes a way round the problem, such as intentionally heating helium over the lambda point, as long as it is reasonably justified as a practical solution.

• I have edited the title of question to be a better reflection of what you were asking. Navigating superfluids would be different question, Yours is about propulsion and manuoevrability of submarines in superfluids. If this isn't right, you can always re-edit. It's a great question. Just make sure your sub is leak-proof. – a4android Feb 10 '17 at 1:14
• @a4android: Can you imagine a liquid helium leak in a submarine? Things would start freezing for no reason, everyone's voice would get high pitched, then everyone would die. – Joe Bloggs Feb 10 '17 at 7:59
• @JoeBloggs It's a chilling thought. – a4android Feb 10 '17 at 8:09
• See this post on Physics. – JDługosz Feb 10 '17 at 9:55
• Mandatory XKCD: what-if.xkcd.com/50 – Goufalite Feb 10 '17 at 12:00

## Peristaltic pumps.

Superfluid liquids aren't any more compressible than other liquids. So you can use peristaltic pumps to move it around. For instance, from the front of you to the back of you. Much like a propeller or jet.

A single pump will produce thrust in discrete packets, but having an array of them can provide smooth sailing. Point them in whichever direction you need to thrust away from, like maneuvering jets on a spaceship, and you're in business.

You can also use heating elements like NASA, but they just don't purr the same.

• I'm not 100% sure but I think that impeller pumps would also work. Which raises the awesome possibility of liquid helium jet skis. – Joe Bloggs Feb 10 '17 at 7:55
• OK: Found someone that manufactures liquid helium impeller pumps. Now I just need to build a hull that isn't brittle at -269 C and I'm laughing. – Joe Bloggs Feb 10 '17 at 7:57
• Used in the ATLAS detector and with a couple of properties that make it perfect for this submarine (notably it's designed to prevent heat transfer from the motors to the pump). – Joe Bloggs Feb 10 '17 at 8:04
• @JoeBloggs post the hull material as a new question… I have ideas! – JDługosz Feb 10 '17 at 9:52
• There is a "Hunt for Red October" joke in here somewhere, but I keep loosing it in the sound of magma flows :) – Jason K Feb 10 '17 at 14:27

A propeller will still function, just without frictional losses.

Turning the propeller means the fluid mass must be accelerated backwards, pushing the blade forwards. Zero viscosity does not mean zero mass.

Propellers better in water than molasses because water has much lower frictional (and power-draining) losses. They don't work as well in air because of the much lower air density, but prop. planes prove they work well in a fluid with quite low viscosity one you speed them up to compensate for the low mass of air.

Propellers do generate thrust in superfluids, see google books - Superfluidity and Superconductivity

The nature of the flow within the wind-tunnel is indicated by the propeller which consists of two thin mica vanes ... passage goes on the explain how these applies specifically in the case of superfluids. It also notes that this effect depends up turbulence (so that the disturbed flow does not re-impact the back side of the black to make a net momentum transfer of 0 -- In fact, if slow enough, the net-thrust would be zero, i.e., once you reach the laminar flow region)

They also include a graph of experimental results the plots the net thrust against velocity.

An enclosed screw drive would be effective at very slow speeds even in a superfluid

For a less technical proof that propeller would work, how about a picture of a helium pump. Notice the use of an Impeller to push the helium.

• I'm not sure about that. If an object moves through superfluid with no resistance, it means that no momentum is transferred from the object to the liquid. If case of propeller it means that it should not create ant thrust. – Alexander Feb 10 '17 at 0:28
• Exactly like how a wheel still turns on a frictionless surface without frictional losses. – Samuel Feb 10 '17 at 0:30
• Can you demonstrate equivalence between flowing a superfluid past a propeller to generate torque on the propeller and spinning a propeller in stationary superfluid to generate thrust? It's not clear to me that it will work the same way. The source doesn't seem to actually say propellers generate thrust. – Samuel Feb 10 '17 at 1:05
• @Alexander: It makes perfect sense if you think about it purely in terms of particles: Assume a very simple propellor made of nothing but angled plates: The helium particles are being hit by the angled plate in such a way that they are deflected behind the ship. That's where the momentum transfer is coming from: not the frictional forces of the propellor 'grabbing' the water. – Joe Bloggs Feb 10 '17 at 7:51
• @JoeBloggs That only works if the particles stick long enough to gain the energy. Without viscosity, they won't do that. These things are not intuitive. The question was asked on physics.se, so we'll get some more informed people to add to the discussion. – Samuel Feb 10 '17 at 17:36

The vehicle could carry reaction mass to be ejected, and thereby navigate by Newton's Third.

• This is a clever solution, could you elaborate? What mass would work well, and how would you choose to store it? etc. – Zxyrra Feb 10 '17 at 6:03
• A rocket-propelled submarine! Now that's something worth conjuring with it. But, yes, more information and explanation will improve your answer. – a4android Feb 10 '17 at 6:47
• If you take in the superliquid Helium and warm it above liquid temperature, you will have a fast expanding gas, which, sent through a nozzle, will generate a good amount of thrust. So, you can take the medium as reaction mass and don't need to carry it with you. – M.Herzkamp Feb 10 '17 at 11:09
• @M.Herzkamp That is a particularly good extension on this answer, If the OP of this answer doesn't update, I suggest you post it as an answer of your own. – Dubber Rucky Feb 11 '17 at 0:06

Preamble

I've really appreciated all the answers on this, There has been lots of excellent discussion, I'll be answering by drawing from my two favorites, peristaltic pumps, and heating helium to a gaseous state and exhausting it to generate thrust.

I will try and produce some maths for this. Where I am able I'll provide general case formulae for anyone who wants to use this idea. If commenters can produce or correct any of my formulae I will update accordingly. With that preamble over, I'll begin.

The Basic Engine design

Peristaltic pumps supply fluid to a chamber with a central helium fusion core and a series of metal meshes.

The helium is directed out of this chamber via multiple directional nozzles.

To travel at significantly higher speeds, the mesh is heated via direct heating from a helium fusion core within the chamber and a surge of electrical current, bringing the helium (violently) to it's vaporization point.

Importantly we can and possibly should use a material that is a superconductor at cryogenic temperatures [1]

(That source is cool, check it out)

Calculations of efficiency and thrust for both modes will be supplied in the main body of this answer.

(If anybody wants to provide an engineering drawing of this concept I would be very grateful)

Assumptions for my simple model

• I will assume $G$ (gravitational acceleration constant) is $10 \, \mathrm{m\ s^{-2}}$ I will assume the temperature of the helium is $1 \,\mathrm{k}$.
• I am assuming standard atmospheric pressure at surface (This assumption Requires the body of helium is sealed as I can think of no plausible element that would remain gaseous to provide pressure at $0.5 \, k$, perhaps we are in some kind of artificially created helium submarine arena at a lavish alien tournament?
• I will take the density of liquid helium as $125 \, \mathrm{Kg\ m^{-3}}$ [0]
• I will assume the submarine operates at a maximum of 100m depth (due to brittle hull materials at low temperatures).
• Given the above I have implicitly assumed the hull can withstand $1.25\times 10^5 \, \mathrm{Kg\ m^{-3}}$, using the formula Pressure $= R \times G \times H =$ Density $\times$ Gravitational constant $\times$ height of liquid above vessel.

Importantly the general Equation for the pressure is $P = R \times G \times H$.

Efficiency and thrust in pump powered mode

I was not able to locate a resource providing the efficiency of peristaltic pumps for liquid helium. However I am willing to make the assumption that they are very efficient, losing negligible heat to the helium. This is implied without any figures given here. [2]

I am going to be lazy and provide 90% efficiency. Given operation in a practically friction less environment, we will assume that acceleration characteristics for a general case are $$a = \frac{0.9p}{m V}$$ Where:

• $p$ is power
• $m$ is mass
• $V$ is velocity
• $0.9$ is the factor of efficiency

Because these figure may vary vastly depending on how much you want to wave your arms and say space-future-materials I have left the figures up to the reader. The primary point is that the efficiency is very high.

However this is where we run into trouble, peristaltic pumps are fairly slow, one industrial machines I sourced boasts rates of $40\, \mathrm{Lmin^{-1}}$, equivalent to $0.667 \times 10^{-3}\, \mathrm{m^3\ s^{-1}}$ Weighing in at $40\, \mathrm{Kg}$. [3]

Assuming the material can be passed through a very narrow aperture due to being a superfluid, we can impart a fairly high velocity to it.

Let us take the case of an aperture of area $0.001\, \mathrm{m^2}$.

The equation for velocity is $V = \frac{Q}{A}$ Where:

• $V$ is velocity
• $Q$ is volumetric flow rate
• $A$ is Area (that the flow passes through)

So $V = \frac{0.667 \times 10^{-3}}{0.001} = 0.667\, \mathrm{ms^{-1}}$

At that exit velocity the craft experiences a force of $0.667 \times 0.667 = 0.445\, \mathrm{N}$

Thats pretty bad for a $40\, \mathrm{Kg}$ motor. Even exploiting that our fluid is 0 viscosity.

That is to say that $1\, \mathrm{Kg}$ of motor will produce $11.1 \times 10^{-3}$ newtons of thrust :( Pathetic.

(If my calculations or sources are wrong here, please let me know)

Efficiency and thrust in helium vaporization jet powered mode

It's pretty clear from the above that unless we find large hunks of metal very slowly drifting around cool then we better flip the big red switch labeled ACTIVATE ROCKETS

But before we do that, let's figure out what we are going to loose in efficiency, what our power draw is, and what kind of acceleration we might expect.

I didn't bother with hull or equipment weights earlier, because it was quite clear that performance was going to be lackluster no matter which way you cut it.

Let's apportion ourselves one (metric) ton to work with and divide it among hull, pumps, heating, power and control systems. Just to give us some figures to work with, your mileage may vary and I in no way take responsibility for anyone who's submarine collapses at depth in a helium lake.

I will stick $200\, \mathrm{Kg}$ in my super strong magic future material tough at at $0.5\, \mathrm{k}$ hull.

With my remaining $800\, \mathrm{kg}$, I will put $300\, \mathrm{Kg}$ into a super duper all the power you can eat space tech triple alpha helium reactor (I mean we have a lot of helium so why not use it), especially as this produces carbon, that's right, any excuse for a sooty exhaust. [4]

I will take $240\, \mathrm{kg}$ of pump, giving me an easy to calculate $\frac{240}{60} = 4\, \mathrm{L}$ or $0.5\, \mathrm{Kg}$ of helium to work with per second.

with that done, we'll add a $50\, \mathrm{kg}$ AI to control it because we aren't so irresponsible as to put organics on this deathtrap.

$160\, \mathrm{kg}$ will be allotted to physical control systems, thrust vectoring etc.

the final $50\, \mathrm{Kg}$ will be our heat dispersal and heating element.

Whilst most of that was frivolous the important factor is this, we have $0.5\, \mathrm{Kg}$ of helium per second to convert into gas and into energy. (To go slower in this mode we would lower the speed of the pumps, reducing our fuel.)

The reactions for this type of fusion are as follows (Wikipedia verbatim quote):

He + He → Be (−91.8 keV)

Be + He → C (+7.367 MeV)

The net energy release of the process is 7.273 MeV (1.166 pJ).

As a side effect of the process, some carbon nuclei fuse with additional helium to produce a stable isotope of oxygen and energy:

C + He → O (+7.162 MeV)

Unfortunately it's not clear to me what quantity becomes oxygen, regardless it might be different for the case of this implausible wave-your-arms-and-hope-tech reactor.

In terms of joules,making a blind assumption of $10\%$ of carbon molecules becoming oxygen molecules, then:

31 He → 9C + O $\left( 7.237\times 10 + 7.162 = 79.532 \mathrm{MeV}\right)$

Converting to Joules yielded per gram $1.2742436502359998\times 10^{-11} \approx 1.27\times 10^{-11}$ joules per 31 molecules (online calculator [5])

The molecular weight of helium is $4.0026$ [0] so roughly speaking one gram of helium contains Avogadros constant/4

so $\frac{1}{4}\left( 6.02214085\times 10^{23}\right) = 1.505535213×10^{23}$ atoms of helium per gram.

Multiply that by the yield per atom $1.505535213\times 10^{23} \times \left(1.2742436502359998\times 10^{-11} \times \frac{1}{31}\right)$ gives $61884473701$ Joules per gram. Let's round to $6.19\times 10^{10}$ joules.

Next we need to figure out how much heat it's going to take to vaporize our helium. I suspect if my calculations are correct we can probably manage a plasma without breaking a sweat.

$2326\, \mathrm{J\ kg^{-1}}$ is given here to vaporize helium [0]

Given that figure, 1 gram of fused helium can vaporize $2.66\times 10^7 \mathrm{kg}$ of helium.

With those orders of magnitude, we have the breathing room to pick a temperature for our exhaust gas to fit our liking. let us say we want to exhaust helium at room temperature to make for slightly easier calculations.

We are heating $0.5\, \mathrm{kg}$ of helium (more or less) which is equivalent to $\frac{500}{4} = 125\, \mathrm{mol}$

Using a scientific calculator [6] we find that the exhaust helium has a volume of of about 6200L for the pressure of $125000\, \mathrm{kg\ m^{-2}}$ at lowest operating depth.

As I understand it ( and I may need correcting ), thrust is calculated by momentum transferred to fuel. So for a given volume and mass per second, momentum transferred is affected by exhaust aperture.

For easy calculation I will assume an aperture area of $0.5\, \mathrm{m^2}$

We have room as previously discussed to produce a near arbitrary output heat.

To this end I chose the gas output temperature as $1500 \,\mathrm{C}$. This yields . Circa $14743 \,\mathrm{L}$

Plugging this new value into the thrust equation we get:

$14743\, \mathrm{l\ s^{-1}}$ must travel at $29.5\, \mathrm{m\ s^{-1}}$ through an aperture of this size $\left( 1000\, \mathrm{l} = 1\, \mathrm{m^3}\right)$. imparting $29.5\, \mathrm{m\ s^{-1}}$ to $0.5\, \mathrm{kg}$ per second produces a thrust of $14.75$ newtons for a $1000\, \mathrm{kg}$ craft. So we can expect an acceleration of $0.0148\, \mathrm{m\ s^{-2}}$

Still fairly low. Some optimisations required. Who knew jet powered superfluid submarines could be so sedate.

Since this figure is appalling I propose that as nice as the idea of peristaltic pumps and jets is, anybody sensible will heat up the fluid to a non super-fluid temperature, or perhaps if really wise, won't venture into a lake of liquid helium at all.

Finally

I'm really sorry I rained so thoroughly on the parade of peristaltic pumps and jets, I didn't set out to the maths just worked out that way. They were a great starting point and key to this answer.

I'm sorry this answer is currently a shambles at the end, it took a long time to write and unfortunately rocket science is where my understanding of physics wears thin.

Sources

(yes I added this one later and couldn't be bothered to renumber, I was 3 hours in dammnit)

You wouldn't be able to use propeller or any kind of wheel in superfluid - but you still can use jets. The problem would be how to suck up the liquid in and then expel it in desired direction, but as long as you able to implement leakproof valves, it's not going to be a real problem.

P.S. After looking into superfluidity a little more, I found that there is a way to make propellers work, even in superfluid. Superfluidity seems to be limited by a speed of sound in superfluid liquid (https://en.wikipedia.org/wiki/Superfluid_helium-4) That means that an object that moves faster than speed of sound will find itself in an ordinary liquid instead of superfluid. So, if your propeller rotates faster that speed of sound (220-240 m/s for superfluid He), you should be good.

P.P.S. In the reference shared by Gary Walker, propeller experiment was working with the speeds far lower that theoretically predicted (generating thrust at speeds as low as 6 mm/s). This fact was reflected in the paper, however, scientific explanation to this phenomenon is still pending, as I understand.

• Propeller-like pumps do exist, see here - this proves your first sentence to be false. – Mołot Feb 10 '17 at 11:19
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Frostfyre Feb 10 '17 at 13:07
• @Frostfyre actually this is an answer. "use jets" would answer the question. It is a bad answer, I admit, and at least in part false one, but still an answer. – Mołot Feb 10 '17 at 14:26
• @Mołot That's a liquid helium pump, not a superfluid liquid helium pump. There is a very big difference. – Samuel Feb 10 '17 at 15:50
• @Mołot The OP implies it wouldn't, cursory research implies it wouldn't, and this was the first answer, so it doesn't need to address any of the other answers. – Samuel Feb 10 '17 at 16:06