Titanic tourist sub goes missing sparking search

[O-ring]

Is anybody an oral surgeon or work in a dentist's office? Sounds like we need to secure a supply of teeth so we can do some hands on testing.

 

[Wookie]

Maybe, but there is a bunch of energy released and I'm betting on high temperatures.

I’ve seen a calculation that as the air compressed, temps as high as 6000C were calculated. I don’t have the inclination to check the math.

 

[O-ring]

I’ve seen a calculation that as the air compressed, temps as high as 6000C were calculated. I don’t have the inclination to check the math.

That sounds way less fun than pounding on some teeth with a hammer.

 

[Dan]

Another thought comes to mind is some of the implosion energy are being absorbed by the crushing of the carbon-fiber cylinder, so the impact on the objects inside the submersible would be less.

 

[tridacna]

I’ve seen a calculation that as the air compressed, temps as high as 6000C were calculated. I don’t have the inclination to check the math.

(This is not my work)

Modeling the compression of the air in the sub like adiabatic compression:

dU + δW = δQ=0


Treat the air inside the sub as an ideal gas:

δW = PdV


U = αPV = αnRT

dU = d(αPV) = αVdP+αPdV



Substituting into conservation:

dU = -δW

αVdp+αPdV = -PdV


Integrate:

∫-(α+1)pdV = ∫αVdP

ln(P/Po) = -((α+1)/α) ln(V/Vo)

We will write ɣ = ((α+1)/α) then:

(P/Po) = (Vo/V)^(ɣ)


Finally substitute V = nRT/P and Vo = nRTo/Po


Simplify to get:

T = To(P/Po)^((ɣ-1)/ɣ)


ɣ = (5DOF + 2)/5DOF = 1.4 for air

Pressure inside the sub Po ~1atm

Pressure at 13000ft P ~ 400atm

Initial temperature inside the sub To ~273K (Its cold down there)

Plugging in you get... 1500K almost exactly. That's approximately how hot the air inside that sub got for a moment as it collapsed to that pressure.

 

[Wookie]

(This is not my work)

Modeling the compression of the air in the sub like adiabatic compression:

dU + δW = δQ=0


Treat the air inside the sub as an ideal gas:

δW = PdV


U = αPV = αnRT

dU = d(αPV) = αVdP+αPdV



Substituting into conservation:

dU = -δW

αVdp+αPdV = -PdV


Integrate:

∫-(α+1)pdV = ∫αVdP

ln(P/Po) = -((α+1)/α) ln(V/Vo)

We will write ɣ = ((α+1)/α) then:

(P/Po) = (Vo/V)^(ɣ)


Finally substitute V = nRT/P and Vo = nRTo/Po


Simplify to get:

T = To(P/Po)^((ɣ-1)/ɣ)


ɣ = (5DOF + 2)/5DOF = 1.4 for air

Pressure inside the sub Po ~1atm

Pressure at 13000ft P ~ 400atm

Initial temperature inside the sub To ~273K (Its cold down there)

Plugging in you get... 1500K almost exactly. That's approximately how hot the air inside that sub got for a moment as it collapsed to that pressure.

1226 C. Uncomfortably hot, but not incineration in .02 seconds hot.

 

[tridacna]

1226 C. Uncomfortably hot, but not incineration in .02 seconds hot.

Not even close. You would need a few minutes even at that temp to incinerate bones. Contrary to popular belief, incinerators at crematoria do not produce ash. Huge magnets pull out any ferrous metal then the remains are crushed mechanically into fine ash. Who knows what they found down there.

 

[Pumpkin King]

Cremulator

 

[HalcyonDaze]

Loring Chien, former principal engineer at Fortune 1000 company (2002-2016) answer this question in Quora Quest:

“Were the corpses of Titan Submarine found?”

His answer:
“It is very likely that the bodies were severely mangled by the implosion of the hull. The contents of the hull would have been crushed to about 1/400th the volume by the 6000 PSI of pressure. The inrushing water would be like a huge piston causing both enormous pressure rise and probable very high flash temperatures as the existing air was compressed in an instant.

They have said they recovered some human remains. What exactly they were and how small and what condition was thankfully not described.”

One YouTuber put it this way - "You stop being biology and become physics."

 

[chillyinCanada]

And a number of the latter posts are why I very nearly didn't make my previous post.