Puzzle: propagation of momentum in water

[HS-experiment]

Hello again Physics Forums! I have a question about fluid dynamics. Perhaps someone here can help me out.

I am trying to understand how a plume of water moving at some speed carries momentum within a body of water. For instance the ‘exhaust’ from a submarine propeller. I am having a hard time understanding how momentum is conserved in such a situation. I do not care about surface effects like waves.

A traveling plume of water always displaces a volume of water equal to its own, and that volume of water effectively fills the space left behind by the plume. So the net momentum of a moving plume of water and the remainder of the system is always zero.

The submarine also displaces a volume of water equal to its own as it travels. If the submarine has the same density as the water, the system also has a momentum of zero. But if the density of the submarine is different than water, that system has a non-zero momentum.

To conclude, the ( submarine + system ) have a non-zero momentum, but the ( plume of water it generates + system ) cannot have any momentum. How is momentum conserved?

I hope I have written it so that it is comprehensible. If someone could please comment I would really appreciate it. :)

 

[jbriggs444]

To conclude, the ( submarine + system ) have a non-zero momentum, but the ( plume of water it generates + system ) cannot have any momentum. How is momentum conserved?

What makes you say that the water has no momentum when it contains a moving submarine? That water has a moving hole in it and, accordingly, a net movement in the opposite direction.

 

[.Scott]

For simplicity, let's push the sub by pumping water from its bow to its stern. And we will also assume zero viscosity.
If the sub is at zero buoyancy, there is no issue because the net momentum will always be zero.
But if the sub is heavier, then accelerating the sub will require that it ultimately push back on something to accelerate. In this case, pumping the water will necessarily create a lower pressure at the bow and higher water pressure at the stern and a lower pressure. The fluid will be, to a minor extent, compressible. In the extreme case, relativistic mechanics would guarantee this. So a pressure wave will develop that carries the momentum to the surface or to some other object that will feel the push back.

Also, you don't need a submarine to demonstrate this problem. What if I simply had a piston/cylinder arrangement that I placed under water? If I pull the piston back, it displaces water. How would that water move out of the way?

 

[HS-experiment]

What makes you say that the water has no momentum when it contains a moving submarine? That water has a moving hole in it and, accordingly, a net movement in the opposite direction.

Hi JBriggs. I'm not sure if we are on the same page. I try to decompose the system into two parts, one ( submarine + water ), the other ( plume + water ). The ( submarine + water ) has net momentum only if the submarine has a density different from water. If the density is the same, the volume of water that the submarine displaces as it moves has the same mass as the submarine, which means their combined momentum is zero.

For simplicity, let's push the sub by pumping water from its bow to its stern. And we will also assume zero viscosity.
If the sub is at zero buoyancy, there is no issue because the net momentum will always be zero.
But if the sub is heavier, then accelerating the sub will require that it ultimately push back on something to accelerate. In this case, pumping the water will necessarily create a lower pressure at the bow and higher water pressure at the stern and a lower pressure. The fluid will be, to a minor extent, compressible. In the extreme case, relativistic mechanics would guarantee this. So a pressure wave will develop that carries the momentum to the surface or to some other object that will feel the push back.

Also, you don't need a submarine to demonstrate this problem. What if I simply had a piston/cylinder arrangement that I placed under water? If I pull the piston back, it displaces water. How would that water move out of the way?

Hi Scott. I am considering this simplified case where the submarine is pumping water from bow to stern, and zero viscosity. In this case we can ignore factors like hydrodynamics, drag, and the streamlines of water flowing past the hull. The water is just pumped from the bow and comes out the stern without disturbing the surrounding water. Is this right?

I think what happens is that as soon as there is pumping in the sub, it pushes on water behind the sub, which pushes on water in its surroundings, and this pressure wave then reaches something to push on. I take a guess when I say this pressure wave travels at the speed of sound.

In that case, the pressure wave travels very quickly compared to the submarine. Momentum is defined in SI units as kg.m/s, so the kg component can be greatly reduced. Without doing the calculations, it seems plausible.

I have one problem remaining. Consider two subs of equal mass but different densities. One sub has a density same as water. The other sub is twice as dense. In a non-viscous environment these subs will behave exactly the same, right? An engine with the same power will accelerate them at the same rate, and produce the same plume of water and pressure wave. And yet, only the pressure wave produced by the denser sub is allowed to carry momentum. The buoyant sub displaces its weight in water, meaning the momentum of (sub + water) is always zero, meaning the pressure wave it generates also should have a momentum of zero. But isn’t it the same pressure wave generated by the same engine?

It’s pretty specific for a hypothetical scenario, but I think the two subs example can also be replicated in a viscous environment since hydrodynamics is affected by both volume and shape. So you can have two objects of different volume and density that have the same hydrodynamic drag, and would thus behave the same given the same engine. Although in that case one would produce a more turbulent flow, making it very difficult to analyze without a proper background in fluid dynamics.

 

[jbriggs444]

I try to decompose the system into two parts, one ( submarine + water ), the other ( plume + water ).

The plume + water has non-zero momentum. It has a submarine-sized hole in it.

The submarine + water (inclusive of plume) has zero momentum. It has no hole.

 

[.Scott]

Hi Scott. I am considering this simplified case where the submarine is pumping water from bow to stern, and zero viscosity. In this case we can ignore factors like hydrodynamics, drag, and the streamlines of water flowing past the hull. The water is just pumped from the bow and comes out the stern without disturbing the surrounding water. Is this right?

Right.

I think what happens is that as soon as there is pumping in the sub, it pushes on water behind the sub, which pushes on water in its surroundings, and this pressure wave then reaches something to push on. I take a guess when I say this pressure wave travels at the speed of sound.

In that case, the pressure wave travels very quickly compared to the submarine. Momentum is defined in SI units as kg.m/s, so the kg component can be greatly reduced. Without doing the calculations, it seems plausible.

I have one problem remaining. Consider two subs of equal mass but different densities. One sub has a density same as water. The other sub is twice as dense. In a non-viscous environment these subs will behave exactly the same, right?

Wrong. When the heavier sub first begins to accelerate, it will require greater pressure.

Of course, we are ignoring the fact that the heavier sub will also require positive lift.

 

[mfig]

I am trying to understand how a plume of water moving at some speed carries momentum within a body of water.

Because it has mass. Any mass that moves transports momentum. It doesn't matter if the mass is fluid or solid.

 

[HS-experiment]

The plume + water has non-zero momentum. It has a submarine-sized hole in it.

The submarine + water (inclusive of plume) has zero momentum. It has no hole.

I see what you mean now. If the submarine has the same density as water, water moving in the opposite direction as the hole has the same mass as the submarine. It cancels out. Even if the submarine is moving very rapidly, the entire system has no momentum. However if the submarine is denser than water, it no longer cancels out and the system has net momentum. But, conservation of momentum dictates that the ( plume + water + hole ) can only have net momentum if the plume transfers its momentum outside of of the system.

Please consider net momentum of a system consisting of a plume of water moving within a body of water. If you instead treat this plume as the hole, water displaced by the hole has an equivalent mass as the hole and moves in exactly the opposite direction. So the system has no net momentum. So how is momentum transferred, since it isn't the plume transferring the momentum? Scott suggests that it is a pressure wave, which seems like the best explanation right now.

Wrong. When the heavier sub first begins to accelerate, it will require greater pressure.

They both have the same mass, one is just denser. So they require the same power and pressure to push at the same speed , in a non-viscous fluid.

Of course, we are ignoring the fact that the heavier sub will also require positive lift.

For the sake of the scenario the submarine can be in micro-gravity ;)

 

[jbriggs444]

So how is momentum transferred, since it isn't the plume transferring the momentum?

Who says it isn't the plume?

The sub moves forward. The plume moves backward. The plume plus the water now has equal and opposite momentum to the sub, just as conservation of momentum demands.