I am wondering if boiling water that is trapped inside of a semi-enclosed pipe would be a low-cost method of creating an underwater thruster that would function as a silent marine propulsion system.

Please reference the following conceptual drawing:

enter image description here

This drawing is showing a cross-sectional side view of a conceptual underwater thruster. There would be a copper pipe that would be partially positioned within a metal cylinder and within this metal cylinder there would be an AC induction heating element that would partially surround the closed end of the copper pipe. The other end of the copper pipe would be located outside of the metal cylinder and it would be open-ended. This heating element should create a rising water temperature gradient within the closed end of the copper pipe.

The AC induction heating element would be turned on and would remain on so that it would continually boil the water. This will cause the water pressure (psi) inside the pipe to be higher that the ambient water pressure outside of the pipe/thruster and this imbalance should cause the thruster to move forward as indicated on the drawing. This thruster could be mounted externally to a ship or submarine's hull or it could be installed internally in the stern area of either vessel.

The main premise of this idea is that the boiling water will not be able to exit the copper pipe so there should be continual high water pressure inside the pipe and thus continual thrust. Also, there should always be water inside of the pipe because there should be a continual circulating flow of warm water exiting the copper pipe while colder outside water enters it.

Could boiling water inside a semi-enclosed pipe be a low-cost method for a silent marine propulsion system?


Based on the answers and comments I have received to original design above, I have redesigned this 'Boiling Water Thruster', shown in the new drawing below:

enter image description here

I believe this new underwater thruster design should provide silent propulsion for the ship/submarine, although the thrust would be low compared to propeller driven craft.

  • 4
    $\begingroup$ 1. If the pressure is higher inside the pipe, why would cool water enter? 2. Why would this be silent? $\endgroup$
    – The Photon
    Commented Jul 16, 2019 at 22:10
  • $\begingroup$ Won't the momentum of the water going in the water going out cancel for no net thrust? $\endgroup$
    – zeta-band
    Commented Jul 16, 2019 at 22:10
  • $\begingroup$ @The Photon, well, it wouldn't be completely silent, yet there would be no mechanically moving parts. Perhaps the AC heating element could get its electricity from a nuclear reactor helping to keep the overall sound down. $\endgroup$
    – user18610
    Commented Jul 16, 2019 at 22:14
  • $\begingroup$ @zeta-band, the way I see it is that if the AC heating element can boil the water fast enough to keep the water boiling along the interior surface of the capped section of the pipe, and thus keep the water pressure up along this interior surface, then the water going in & out at the open end of the pipe shouldn't cancel anything out and net thrust should be maintained. $\endgroup$
    – user18610
    Commented Jul 16, 2019 at 22:19
  • $\begingroup$ Might Engineering be better suited for this question? $\endgroup$
    – Kyle Kanos
    Commented Jul 17, 2019 at 11:07

1 Answer 1


It can kinda work. Ish.

Your design is based, at its core, on the idea that warm water rises. Warm water only rises in cold water, so I am sceptical of the efficacy of simple thermal gradients. Nevertheless, bubble pumps have been in commercial use for many years, so if you get it just right I suppose it's possible.

However, you have some serious mass flow problems in your design. If I understand correctly, in your design you intend the flow path to be something like this: naive pipe pump

Now, this system is pretty hard to analyze simply. As the essential components are the pumping action and the flows, I think it will be easier to see on a simpler system that is effectively the same (the symbol is a generic pump): equivalent mass flow system

(let right be positive) I'm going to play a little loosy-goosy with mass and mass flow rate so as to not complicate the analysis with a whole lot of derivatives.

I assume you know that $F = ma$? In this system, there are two forces on the submarine: the force from sucking in the inlet water, which pushes to the left, and the force from pushing out the effluent water, which pushes to the right. Ok, so it is more useful to use the momentum forms of Newton's laws here: $$ F = \frac{dp}{dt}, \Delta p_{net} = 0$$

This means he change of the momentum of the craft is equal to the change of momentum of the water. At the inlet pipe, the water is moving into the sub at some velocity $+v_{in}$, and some mass $m_{inlet}$ is coming in. As $p = mv$ and the momentum imparted to the sub is exactly opposite to that imparted on the water, so $$\Delta p_{in, sub} = -\Delta p_{in, water} = -V_{in} \cdot m_{in}$$ It is similar at the outlet pipe, so $$\Delta p_{out, sub} = -\Delta p_{out, water} = +V_{out} \cdot m_{out}$$

Now, unless you are somehow generating water inside the craft, $m_{in}$ must equal $m_{out}$. Therefore, the total momentum imparted on the craft is: $$ \Delta p_{tot} = \Delta p_{in, sub} + \Delta p_{out, sub} = V_{out} \cdot m_{out} - V_{in} \cdot m_{in} = (V_{out} - V_{in})\cdot m_{water}$$

And there's the key problem with the design! If $V_{out} = V_{in}$, then $\Delta p = 0} and the craft cannot accelerate, no matter how fast you pump the water.

Ok, now for some fluid mechanics. How can we make $V_{out} \ne V_{in}$? Well, the mass $ \dot{m}$ flowing through an orifice is proportional to area, velocity, and density: $$ \dot{m} = A \cdot V \cdot \rho $$ because we know $\dot{m}_{in} = \dot{m}_{out}$, then $$ A_{in} \cdot V_{in} \cdot \rho_{in} = A_{out} \cdot V_{out} \cdot \rho_{out}$$ Now while the would change from inlet to outlet if they are different temperatures, water only has a density change of 3% over its entire liquid range, so I am going to disregard that, so: $$ A_{in} \cdot V_{in} = A_{out} \cdot V_{out} $$ Thus the only way that we can cause the craft to accelerate is to have a smaller outlet orifice than inlet (I'll let you work that one out).

If we move back to your design, I can't see how you could possibly enforce that in a simple tube. The inlet and outlet streams are almost completely uncontrolled (they touch! you're going to have Rayleigh waves out the wazoo!) and while you could maybe do something (put a mesh to restrict flow on the bottom?) I can't see an efficient one.

Also, boiling is loud. Have you ever boiled a pot of water? It's certainly not silent. You would also need to get power from somewhere. If it's from a nuclear reactor (as you mention in a comment) those are also really loud.

If you're dead set on this bubble pump thing, I feel it would be much more feasible to simply put the inlet at the front:

bubble pump with inlet at front

  • $\begingroup$ first, I did not know that a bubble pump existed, I'll have to read up on those. Second, I am only theorizing on the water circulation. The way I see it is that since hot water rises, the hot water should move towards the top of the copper pipe and then be pushed out of the pipe since it is trying to expand. As this hot water leaves, the cold outside water pushes in to replace it. $\endgroup$
    – user18610
    Commented Jul 16, 2019 at 23:57
  • 1
    $\begingroup$ @HRIATEXP okie dokie, while your design is super cool, it is a simple problem of momentum. The fact that it has no moving parts is really impressive, and it almost works with little modification. I edited my answer and did a much more rigorous treatment of it. $\endgroup$
    – Ash Pera
    Commented Jul 17, 2019 at 3:47
  • 1
    $\begingroup$ @HRIATEXP If the pressure is higher inside the machine, then a little bit of water comes out, and then the pressure is not higher inside the machine (because now it has a little bit less water, to make up for the temperature) so no more water comes out, and no water comes in either because the pressure is not lower inside the machine. $\endgroup$ Commented Jul 17, 2019 at 4:34
  • 1
    $\begingroup$ @Ash Pera, you've given me something to think about today. I enjoy challenges, hopefully I can come up with a simple solution because I would like to keep the overall design as simple as possible. $\endgroup$
    – user18610
    Commented Jul 17, 2019 at 10:16
  • 1
    $\begingroup$ There won't be any "internal water pressure." If the pipe is open at the end, everything in the pipe will be at the same pressure as the water the boat is floating in. $\endgroup$
    – alephzero
    Commented Jul 17, 2019 at 23:22

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