Cavitation at/in Fast Water Stream

[A.T.]

In this video at 7:15 you see a rifle fired underwater:

https://www.youtube.com/watch?v=cp5gdUHFGIQ

Apparently even before the bullet leaves the barrel, you see a cavitation trace coming out. The author explains it with the water being pushed out of the barrel by the bullet, forming a fast water stream. The fast stream supposedly creates an under-pressure (Bernoulli) which leads to cavitation.

How exactly does this work? Is the fast water in the stream evaporating? Already in the barrel or when it gets outside? Or is it the slow surrounding water that gets sucked inwards that evaporates. Are there possible alternative explanations for this first bubble trace (e.g. shock wave)?

I don't quite understand the Bernoulli explanation, in a situation with just two fluid parts in relative motion. Since movement is relative, how does one know which fluid part is actually moving, and therefore must have the lower pressure?

 

[A.T.]

To be more specific in my questions:

1) Does Bernoulli predict, that a stream of moving water has a smaller pressure than the surrounding stationary water?

2) While it is being pushed out by the bullet, does the water inside the barrel have a smaller pressure than the water outside the barrel?

3) Since the barrel is rifled, wouldn't it be possible, that the explosion gasses/shock waves pass the bullet within the barrel, end exit it before the bullet?

 

[Naty1]

[I did not watch the video. ]

Bernoulli's relationships derives from a steady flow...does not seem that applies to a bullet pumping out water from a barrel under high pressure. Seems more like a high pressure cutting tool, or a pressure washer, or even a hose at household pressure. Does not appear anything really unusual is going on...high pressure water displaces low pressure water.


Now that I think about it so does a propeller on a boat...but there is a tiny area of low pressure cavitation at the tips...and bubbles induced there can wear high speed prop tips away. The overall effect is low pressure on the front side and high pressure thrust aft when in forward gear.

In 1958 North American Aviation developed a system using ultra high pressure liquid cutting to cut hard materials. This system used a 100,000 psi pump to deliver a hypersonic liquid jet that could cut high strength alloys such as PH15-7-MO stainless steel.

http://en.wikipedia.org/wiki/Water_jet_cutter
1) Does Bernoulli predict, that a stream of moving water has a smaller pressure than the surrounding stationary water?

I can't see that in this situation

2) While it is being pushed out by the bullet, does the water inside the barrel have a smaller pressure than the water outside the barrel?

higher

3) Since the barrel is rifled, wouldn't it be possible, that the explosion gasses/shock waves pass the bullet within the barrel, end exit it before the bullet?

I'm no gun expert, but that sure seems reasonable...there has to be some clearance between
the bullet and the barrel as well as the rifling. Seems like it would serve to change the shape of the shock wave relative to a perfectly sealed round tipped bullet.

 

[Integral]

Not clear to me that the Bernoulli principle is the route to a good solution to this.

Bernoulli assumes incomprehensibility, not clear that this is the case here. The pressure in the barrel shortly after firing MUST be higher then the outside pressure. It has the outside pressure at the open end of the barrel and very high pressure in front of the bullet. In fact the pressure and speed of the bulled may well compress the water in the barrel thus moving out of the Bernoulli realm. I would bet that the pressures due the shock wave force dissolved gases out of solution so now you are dealing with a gas water mixture. Again we are out of the Bernoulli realm.

 

[SteamKing]

Not clear to me that the Bernoulli principle is the route to a good solution to this.

Bernoulli assumes incomprehensibility, not clear that this is the case here. The pressure in the barrel shortly after firing MUST be higher then the outside pressure. It has the outside pressure at the open end of the barrel and very high pressure in front of the bullet. In fact the pressure and speed of the bulled may well compress the water in the barrel thus moving out of the Bernoulli realm. I would bet that the pressures due the shock wave force dissolved gases out of solution so now you are dealing with a gas water mixture. Again we are out of the Bernoulli realm.

I think you mean 'incompressibility'. If Bernoulli assumed 'incomprehensibility', then we would not be able to analyze anything.

 

[cjl]

To be more specific in my questions:

1) Does Bernoulli predict, that a stream of moving water has a smaller pressure than the surrounding stationary water?

That's dependent on the situation. However, in this situation, no it does not. Bernoulli assumes incompressibility (not necessarily true in this case), steady flow (definitely not true in this case), and it assumes that you are comparing two points along the same streamline (definitely not true in this case).

A lot of people try to use bernoulli to state that faster moving fluids always have lower pressure than stationary ones that they are in contact with, and this is simply not true. Bernoulli is (at its core) a statement of energy conservation, and thus it only applies if the fluids started at the same state (or ended at the same state with no losses). Otherwise, it's absolutely possible to have a high velocity jet through a stationary fluid in which the jet is at a higher pressure.

2) While it is being pushed out by the bullet, does the water inside the barrel have a smaller pressure than the water outside the barrel?

Nope. In fact, it must have a higher pressure than the water outside the barrel, since it is accelerating out of the barrel. This doesn't contradict bernoulli at all, for the reasons I explained above.

3) Since the barrel is rifled, wouldn't it be possible, that the explosion gasses/shock waves pass the bullet within the barrel, end exit it before the bullet?

It's definitely possible that there is some gas leakage around the bullet, though I don't know if that's what's happening here. I can say that the explanation given in the video is wrong though.

 

[cjl]

I think you mean 'incompressibility'. If Bernoulli assumed 'incomprehensibility', then we would not be able to analyze anything.

It's definitely an amusing typo though...

 

[Integral]

What can I say, spell checkers gone wild!

 

[256bits]

I don't quite understand the Bernoulli explanation, in a situation with just two fluid parts in relative motion. Since movement is relative, how does one know which fluid part is actually moving, and therefore must have the lower pressure?

Well the guy did not give a most explicit explanation, did he.

But cavitation does result from the static pressure falling below the vaporization pressure of the liquid at the specific temperature, and due to the lowered pressure the liquid will turn into vapour. According to Bernouli if the fluid accelerates to a high enough velocity the local static pressure can drop enough to cause cavitation.

What you see is not the result of a high speed jet of water coming out of the gun barrel and just due to having a high velocity will thus have an inherent low pressure. That is incorrect.

If the water in the gun barrel was moving out slowly, then the water encountered in the pool would have enough time to move out of the way and both would mix easily.

My understanding is as follows:
Since the impluse given to the water in the barrel is large enough from explosive charge in the bullet casing, when the water in the barrel meets the water in pool, and since both masses being incompressible , they have no where to go except in the radial direction. The vast majority of momentum is transferred from horizontal to radial. The radial velecity is so great - you can consider it as a thin film moving outwards, that the static pressure in that interface is lowered below the vapourization pressure and you get the cavitation and bubbles seen in the video.

 

[A.T.]

Since the impluse given to the water in the barrel is large enough from explosive charge in the bullet casing, when the water in the barrel meets the water in pool, and since both masses being incompressible , they have no where to go except in the radial direction. The vast majority of momentum is transferred from horizontal to radial. The radial velecity is so great - you can consider it as a thin film moving outwards, that the static pressure in that interface is lowered below the vapourization pressure and you get the cavitation and bubbles seen in the video.

So the intial fast water stream acts just like the bullet itself acts later: It pushes the water radially out of the way, which due to the water's inertia leaves a low pressure area behind. Is that a correct summary?

 

[cjl]

So the intial fast water stream acts just like the bullet itself acts later: It pushes the water radially out of the way, which due to the water's inertia leaves a low pressure area behind. Is that a correct summary?

I would be more willing to believe that explanation than the one given in the video. It at least sounds somewhat plausible.

 

[AlephZero]

So the intial fast water stream acts just like the bullet itself acts later: It pushes the water radially out of the way, which due to the water's inertia leaves a low pressure area behind. Is that a correct summary?

Iill buy that as well.

Look at how the bubbles move (or don't move). They have no velocity along the track of the bullet. There seems to be a cylinder (or an annulus) of bubbles corresponding to the low pressure region. Then I would guess there is pressure wave moving radially inwards which destroys the bubbles, but when this reaches the center line it is "reflected" outwards again to produce another pulse of low pressure and a second smaller region of bubbles.

There are analogies to this is the "shock waves" in solid structures caused by explosions or impacts. The initial wave of compression often does less damage later waves of tension - most materials are easier to break by stretching them rather than squashing them. The consequence is that things don't always break in the "obvious" place you might expect.

FWIW I assume the guy in the video knew what he was doing, but shooting a gun under water would seem like a good way to burst the barrel (or worse), considering the inertia of a gun barrel full of water is a lot more than a barrel full or air.

 

[256bits]

So the intial fast water stream acts just like the bullet itself acts later: It pushes the water radially out of the way, which due to the water's inertia leaves a low pressure area behind. Is that a correct summary?

The pool water AND the fast water stream are BOTH deflected radially at the interface. The interface moves forward until there is no more fast water stream remaining. For the interface to move forward some of the pool water also has to be deflected radially.

At the interface in the horizontal direction, the pressure would be quite high due to the inertia of the stream, while just off a radius sideways the water pressure is still that of the pool at that level, so where will the water go - radialy.

In essence, yes, you have a water bullet but the bullet just keeps disintegrating.

 

[256bits]

FWIW I assume the guy in the video knew what he was doing, but shooting a gun under water would seem like a good way to burst the barrel (or worse), considering the inertia of a gun barrel full of water is a lot more than a barrel full or air.

Apparantly quite a few guns can fire underwater. But in air, do not fire a gun with a chamber and barrel full of water. Either the gun does not fire, or the mechansim is destroyed. I was under the assumption as you or taught to keep your gun dry.

 

[256bits]

Here is an outstanding still of a pistol shot underwater. The rifling of the barrel is reflected in the twist of the cavitation bubble.
http://www.neatorama.com/2012/09/28/This-Is-What-It-Looks-Like-When-You-Fire-a-Gun-Underwater/

 

[A.T.]

Here is an outstanding still of a pistol shot underwater. The rifling of the barrel is reflected in the twist of the cavitation bubble.
http://www.neatorama.com/2012/09/28/This-Is-What-It-Looks-Like-When-You-Fire-a-Gun-Underwater/

The rifling of the barrel or the rotation of the bullet?

 

[256bits]

The rifling of the barrel or the rotation of the bullet?

Either/ or - the rifling is what makes the bullet rotate,

 

[A.T.]

Either/ or - the rifling is what makes the bullet rotate,

Ok, I though you meant that the rifling of the barrel directly creates that pattern. I guess it is the tip of the bullet (eventually deformed hollow point?). Or the bullet is already rolling over, like here:

https://www.youtube.com/watch?v=vQvqESdCk7U

 

[A.T.]

It's definitely possible that there is some gas leakage around the bullet, though I don't know if that's what's happening here.

Looks like this is actually happening in this video

https://www.youtube.com/watch?v=otpFNL3yem4

 

[256bits]

Ok, I though you meant that the rifling of the barrel directly creates that pattern. I guess it is the tip of the bullet (eventually deformed hollow point?). Or the bullet is already rolling over, like here:

https://www.youtube.com/watch?v=vQvqESdCk7U

In the still we do not know the location of the bullet so I won't even make a blind guess if the twisting is that of the cavitation caused by the rotating bullet or that of the rotating water from the barrel.

From this video the grooves in the bullet do twist the water which does make sense.

High speed slow motion is always interesting.

 

[A.T.]

From this video the grooves in the bullet do twist the water which does make sense.

I doubt that the grooves in the bullet from the barrel rifling are deep enough to cause the twist in the bubble surface. I think it happens when the bullet is already tumbling