Lucian Solutions 04
Lucian Solutions Number 04
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The Question
By Baxter – If dropped into Earth’s atmosphere, would a kernel of popcorn, pop, before being incinerated by re-entry?
The Answer
It would not be able to pop because the water would evaporate and crack the shell. The vacuum of space would cause the water inside the kernel to begin to evaporate and off-gas. The gaseous water will cool and expand the inside surface of the shell (the weakest strength side of the shell) causing irregular internal pressure which will crack the shell. This would remove both of the necessary elements for popping a kernel of corn. Both are likely to happen, but if either the water escapes or the shell cracks, then the kernel cannot pop. Therefore, since at least one of these would occur over the time from tossing the kernel overboard to its eventual re-entry, it would incinerate and never pop.
Decision Tree
Decision tree for popcorn in space
Immediately upon tossing into space
1 – Shell cracks from internal off-gassing pressure
2 – Shell remains intact
A – Internal moisture slowly freezes and cracks shell
B – Internal moisture slowly freezes but shell remains intact
This will cause some kernels never to pop.
i – Kernel will not pop due to being frozen
ii – Frozen kernel still viable
a – Aerodynamically stable free-fall of frozen intact kernel
1 – Heat applied quickly to one side of shell
This will cause the frozen moisture to act as an insulator which will result in the shell heating unevenly. Vapor pressure from the melting moisture under the front of the shell combined with temperature differentials between the front and the back of the kernel (which is still frozen) will crack the shell.
2 – Heat applied slowly to one side of shell,
This will cause more moisture to begin to melt. The reduced exterior pressure and the temperature difference between the front and the back of the shell will cause irregular internal pressure and the shell will crack. (This is why you must cook them in oil – to spread the heat over the outside of the kernel and prevent scorching on one side.)
b – Aerodynamically unstable free-fall of frozen intact kernel
1 – Heat applied quickly to all sides of shell
This will cause the shell to heat before the inside moisture can melt. The temperature over time differential for the heating of the shell and the warming of the moisture inside the kernel do not match. In order to pop, the shell must be heated (but remain intact) until the inside moisture reaches the vapor pressure and temperature to explode. On the planet, the atmospheric pressure will assist in holding the shell together until the moisture pressure is heated past boiling. The reduced pressure would allow the slowly melting moisture to go straight to vapor, however. As soon as it surpassed the internal pressure capacity of the kernel it would burst the shell. It would do this before the interior had reached the temperature necessary to do this on earth, because the pressure is lower. Therefore the kernel would not pop.
2 – Heat applied slowly to all sides of shell
This will cause more of the internal moisture to begin to heat. The reduced pressure will cause the moisture to evaporate at a lower temperature. The internal pressure will reach the bursting fatigue point before the internal temperature reaches the same point necessary to cause popping on the planet. This means the bursting fatigue point / internal temperature profiles will not be the same for a kernel in space and a kernel on the planet surface. This profile is unique to each kernel but has a range within which the kernel will pop. Only this profile range will allow for popping.
a – The internal temperature / bursting pressure profile does not match the parameters needed for popping to occur
b – The internal temperature /bursting pressure profile does match the parameters needed for popping to occur
***********
IT POPS
***********
For this last phenomenon to occur, the kernel would have to be heated at the exact rate necessary to heat and melt the interior moisture, taking into account the lower vapor pressure needed to break the shell.
calculated probability = 0.000000164 = 0.0000164 %
0.000000164 = 1.6 out of 10,000,000 chance
to make it even = 8.0 out of 50,000,000 chance
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