Brown Hornet, on 12 July 2013 - 03:42 AM, said:
IO Oh no you didn't! That's a pretty naive assumption. How heat sinks work is at the heart of the
discussion. Conduction, convection, radiation and evaporation. They are the possible heat loss
mechanisms. If you're assuming that heat sinks work passively in the year 3050 then we've just gone
back to the stone ages.
The laws of physics aren't assumptions, silly
I've already covered most of this, and regardless of how heatsinks work, it's ultimately because of
the Second Law of Thermodynamics (and possibly the Stephan-Boltzmann law), unless of course the mech is in an environment approaching its reactor temperature, which I doubt said mech could function in.
Your problem is that you're still stuck on the reactor-heatsink part of the system, so you ignore the
much more important component of this thermodynamic system. Regardless of how a heatsink acquires
heat, the way in which it rids itself of that heat increases in effectiveness as it heats up.
Conduction and convection are both going to follow the heat transfer equation, so the heatsink rids
itself of heat with linearly increasing wattage as it heats up. Using water's heat of vaporization is
a good heat transport mechanism (vaporize near the reactor, condense near the heatsink), but the
heatsink itself still needs to bleed it into the surrounding air via conduction of some sort, so it's
still bound by the heat transfer equation. Radiation? It increases exponentially as it gets hotter.
There is no mechanism, none, whatsoever, that a heatsink will work by such that its interface to the
environment (the heatsink part of the heatsink, not its interface to the reactor) does not bleed heat
into the environment faster as it gets hotter. It will radiate exponentially more wattage, and conduct
linearly increasing wattage, with linearly increasing temperature.
Because you don't consider the environment, you don't consider the problem your claim creates. If the
heatsink is emitting more heat wattage as it gets hotter, then it has to be getting that heat from
somewhere to stay hot. Because of that, it has to be getting more wattage from the reactor, which
means the reactor has to be losing more wattage.
For your claim that heatsinks necessarily become less effective when they get hotter to be true, one
of two things would have to happen. Either your heatsink would have to lose higher wattage without
pulling more wattage from the reactor, or your heatsink would have to stay hot without the increase in
delta v between it and the environment pulling heat from the cooling surfaces faster.
The first violates conservation of energy, the second defies both the Stephan-Boltzmann equation and
the heat transfer equation, if not flat out defying the Second Law of Thermodynamics.
So either way, you're not only making a claim about how heatsinks must behave that has no evidence,
but you're claiming that they defy the laws of physics by getting hot without pulling more energy from the reactor, an effective thermodynamic impossibility. I say effective because if someone wanted to get really creative about sabotaging basic heatsink design, maybe they could pull something off, but I have no idea how such this hypothetical counter-productive heatsink would even work. If you think you can propose such a mechanism, then by all means.
Edited by Catamount, 13 July 2013 - 06:51 PM.