Tourist submersible visiting the Titanic is missing Part 2 69

[Tomfh]

18,590 lbs x 12,000 ft depth = 223,080,000 ft lbs of energy in the collapse.

It's close. The volume is slightly less though. I have 6.2m3.

So the energy in SI is 3600m * 6.2m3 * 1000kg * 9.8 = 218736000 J (161333530 ft lbs)

1 gram of TNT is 4184 J, i.e. the energy is 218736000/4184 = 52279 gram of TNT. That's a figure being bandied about in the first thread. 50kg of TNT

Even if you fixed the units, you're calculating gravitational potential energy, which is not relevant for this problem

The energy is the gravitational potential of the volume of displaced water

 

[SwinnyGG]

I know it's semi semantic and I'm being a pain in the ass but.. The hull was 2.54m long from flange to flange, 1.68m diameter, with hemispherical ends.

V_H = πr²L + (4πr³)/3
V_H = 3.14159 * 0.84² * 2.54 + (4*3.14159*0.84³)/3
V_H = 5.6 + 2.5 = 8.1m³

It makes a roughly 30% difference in the amount of energy we're talking about. Which is a lot.

 

[Tomfh]

The hull was 2.54m long from flange to flange, 1.68m diameter, with hemispherical ends.

Yes. But that’s the outer dimensions. It’s the inner dimension available for implosion.

 

[SwinnyGG]

The outer dimensions are the amount of displaced water, which is what defines the energy available. If the pressure hull was filled with solid steel, it would be unable to implode but would have the same submerged displacement. You'd really need to know the exact volume of the entire, fully assembled sub to get a close number - but since a lot of the structures outside the hull are not pressure tight it's hard to know what the true submerged displacement is.

 

[IRstuff]

The outer dimensions are the amount of displaced water, which is what defines the energy available.

True, but the noncompressible parts don't give up their energy, so all the titanium and 5 inches of CF displace water, but the water can't fill that back in when the structure collapses. Only the compressible volume can give back the potential energy.

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[Denial]

I do not believe that the energy released in the implosion can come from the gravitational potential energy of the 3600m high column of water above the vessel.[ ] The implosion is said to have happened in about 20[ ]milliseconds.[ ] The velocity of sound in seawater at the low temperature we are talking about is ~1400m/sec, so the column of water involved can be no more than about 30[ ]metres high.

As I said in my posts in the first thread on this catastrophe, I believe the implosion's energy comes from the strain energy released when the water in the vessel's immediate vicinity is suddenly decompressed.


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[Tomfh]

Denial it’s the same thing. All that intense pressure and strain is the gravitational energy of the water above. Calculating the gravitational potential is simply a way of neatly quantifying it rather than all the nonsense of calculating strained columns of water.

 

[1503-44]

Correct. The strain energy stored in the Hull, the energy stored in the compressed water, everything, was all put there by the weight of the water column above. Its far easier to that one calculation.

The velocity of sound is the velocity of the pressure wave in static water. We are not counting on any pressure wave to move through the water to distribute pressure to adjacent water particles. All of the energy is in the exact spot it needs to be in to be released upon implosion. All the water involved in the implosion was initially pressurised to the full 5200 something psi. No pressure wave traveling at sonic velocity had to be added. The simple removal of the CF shell counteracting the existing 5200 psi, unbalanced the pressure on the adjacent water particles, causing the 5200 psi on those next particles to accelerate into the void left by the shell, accelerating the collapsing shell further inward as well. That happened faster than sonic velocity, thereby no pressure wave traveled outward to even slightly resist the 5200 psi on any adjacent particle, causing them also to accelerate, with their full 5200 psi now unbalanced pressure, inward. No outbound pressure wave is started until water from both sides of the CF tube impact at the centerline, causing velocity to zero. That KE is then converted to pressure wave that can now travel outward, since the imploding water has come to a stop, no inward velocity now allows the pressure wave to begin its outward journey and to start pushing water particles in the opposite direction.

Assuming the CF shell shattered and the titanium ends simply came off the shell intact, nothing was compressed, all of those bits retained their individual original volumes, none of that total material volume ever displaced any water, so none of it was part of the energy released, ie nothing vanished into thin water, so its the inside diameter that defines the space available for energy extraction.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[SWComposites]

re health monitoring, there is an ASTM guide for using acoustic emission for SHM:

https://www.astm.org/e2983-14r19.html

 

[Reverse_Bias]

I am still wondering if they heard any sort of a boom at the surface ship. Maybe it was too faint to get their attention. Clearly the energy traveled to the surface if the navy was able to pick it up. Whether it made it out of the water into the air and still be audible is a different story.

 

[IRstuff]

The standard is barely 3 and half pages and is more about the requirements for a SHM is supposed to do as opposed to how to do it.

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[SwinnyGG]

the noncompressible parts don't give up their energy

They don't contain any 'energy' - the energy in this equation comes from the displaced water. They contribute to that.

Anyway, we're onto semantics. Not trying to be a pain in the ass. I won't get all worked up about it. Either way its a gigantic reservoir of energy available.

 

[Tomfh]

They don't contain any 'energy' - the energy in this equation comes from the displaced water. They contribute to that.

They entire hull does displace water, that's correct. However only the space within the hull provides implosive energy. The hull material itself does not, because it cannot implode. It's not a matter of semantics.

 

[IRstuff]

They don't contain any 'energy' - the energy in this equation comes from the displaced water. They contribute to that.

Yes, but if they are incompressible, then no water displacement change takes place, so they do not contribute to the energy of the implosion. It's no different than the external components of the submersible, which displaced water as well, but they obviously contributed nothing to the energy of the implosion.

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[1503-44]

Inrushing water can only occupy the space created by changes in volume. Presumably the CF density did not change, the bits and pieces still occupy their total original material volume. The sardine can thickness can be ignored.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[SwinnyGG]

The quantity of water that has been displaced, ie the entire submerged displacement of the sub, has been 'lifted' - it has had its level of potential energy increased.

When the implosion happens, there are three energy sinks which are able to absorb portions of this available potential energy:

1) The gas bubble, which absorbs potential energy from the water column, by rapidly collapsing under high pressure

2) The hard components of the sub, which absorb potential energy as they are bent/broken/shattered/scattered by the implosion

3) The water itself, which absorbs some energy as heat, and whatever energy is absorbed in order to dissolve the gas bubble back into the water (although I'm not 100% sure of the physics on that, maybe it's an energy neutral interaction)

The energy it takes to break all the broken stuff, bend all the bent stuff, and scatter all the stuff that's lost also comes from the water column and is completely separate from the energy consumed to collapse the bubble.

 

[1503-44]

I'd rather not try to calculate that strain energy bit. It requires too much data for one.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[SwinnyGG]

I'd rather not try to calculate that strain energy bit

I don't intend to try - even if we had all the required information, it would be a complicated thing to try to calculate. And we have very little of the information we would need. But if you wanted to figure out the actual amount of energy absorbed in the implosion to the Joule, you'd need to calculate it.

Some of the available energy is going to go into heating the water that's involved, which would be some small percentage of the total. That would be a very complicated thing to try to compute as well.

Since we can't calculate every component of the whole, the best we can do is establish what the upper limit of absorbed energy is - and that is represented by the potential energy of the total submerged displacement of the entire vehicle at that depth.

The potential energy of the displacement of the air in the hull alone is still a useful number to have for the thought exercise we're all conducting - it represents the lower limit of absorbed energy. If the air bubble collapsed and all of the sub components were perfectly unharmed, and there is no temperature change in the water, the air volume potential energy would be the exact amount of energy absorbed.

So really the best we can say is that the total energy absorbed in the implosion is somewhere between my number and Tom's number. I would estimate it's closer to the high end of that window, but that's based on nothing other than intuition since we can't calculate the delta T of the water or the energy lost to breaking things.

 

[dik]

I'm not so sure... there is a lot of strain energy stored in the compressed CF material of the hull that would be released. I don't know what the magnitude of this is. The damage caused would be similar to an explosion compression wave, but with a non-compressible fluid, I suspect.

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[1503-44]

The videos, although small scale, show it pretty much just flattens out. Does flattening out a circle into a pancake take that much work?

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."