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