While it is true that you do not need a target to transmit heat TO for vacuum radiation, it is also true that your hull is only going to emit ~230W per square meter (0.23kW or 2.3 * 10^-4 MW) of heat on a continual basis. This assumes a hull temp at surface of 300K (very generous in terms of how warm the skin would be, and therefore “conservative” for heat loss estimates), a background temp of 4K (real world background temp for deep space), and a hull emissivity of 0.5. If you start with a cold / shut down your reactor and all systems, your ship will eventually get very cold. The effects of the chill will be felt in a few hours, things will become hazardously cold in a few days.
Which means a 100 ton sphere would be able to self-radiate approximately 0.139MW. Not bad if you need to cool down with no power plant at all. But your power-plant, which if its anything akin to a modern nuclear plant will have an efficiency on the order of 35-40% MAX) is producing several megawatts, you still have to get rid of a load of heat. For the 100 ton example, lets say 4MW total produced for usable power. At 40% (which is actually a VERY high efficiency for a power plant), you are making TEN megawatts of heat. You still have 6 MW to get rid of, and only 0.139 MW bleeding out of your hull. And no, Traveller has never accounted for reactor efficiencies and even FF&S never rated the radiators according to how much heat generation is present, only actual power plant output.
A blackbody radiator is defined as a surface with an emissivity of 1.0, which is another one of those theoretical maximums. There are no such materials in the real world. Your tungsten filament is a very good example of a real world radiator, and it has a well documented emissivity of approximately 0.6. The closest we can approach to this in nature is the emissive spectra from a star, and even that never quite gets there.
As to the background temperature, please refer to the equation I posted several posts ago. That formula can be found in nearly any heat transfer text book. Background temp is the term Tsurr^4. It is VERY relevant, as it determines what your NET transfer rate is. When that temperature is a mere 4K, it’s not that big a deal. But when your environment is on the order of several thousand K, it’s effect is quite pronounced.
Waste heat can be used? To a very small degree. In boilers: economizers, superheaters, and the like make an effort to use up the residual heat of the system, but they do not use the WASTE HEAT, which is the heat rejected to the cold sink for your Rankine cycle. And note your cycle efficiency is determined by the ability to reject that heat at the end of the process. So, yeah, cooking and maybe sauna heating would be nice uses for it, but not enough to make a difference.
As to the refrigerator/air conditioner, yes you are correct they use convection to dissipate the accumulated heat, but that is because convection is available. The same devices if used in a vacuum would be forced to rely on a radiator if they wanted the heat outside the system, unless they transferred to another system somewhere else onboard. But at some point, you have to get that heat overboard or you will start to build it up somewhere. An internal thermal-battery for the 100 ton example above is going to have to be able to store 6 megawatt-weeks worth of energy. Or roughly 3.63 * 10^12 joules of waste heat.
I don’t know of any medium, even at twenty tons of displacement mass that has that kind of thermal capacitance. The temperature rise in such a medium btw is
dT = E / m * Cp where
dT = change in temperature
E = applied energy
M = mass in kg
Cp = thermal capacitance
20 tons of liquid water will go from a barely liquid 4°C to boiling with the application of only 1.92 * 10^9 joules of energy (assuming a constant Cp of 1000 joules/kg*°C). Phase change from ice to liquid, and again to steam will account for another few billion joules, but still not much of a dent. There goes the ballast if water was used, and you haven’t even made a dent in the heat that was generated in that time frame. With what you have left over, you can melt twenty more tons of rock into lava, and still power your sauna.
There is a reason that real world power plants have such HUGE heat rejection systems. Take a look at a pic of a nuclear reactor sometime. Those HUGE towers are cooling arrays. They take up 90% of the land footprint of any given system. And yes, there are hundred plus megawatt reactors in use.
Now, if you go with cold fusion as a viable option (which it might be someday, but for now it doesn’t seem to be the case)… then all of this thermodynamic mumbo jumbo goes right out the airlock. No heat production means no heat rejection to deal with.
Which is all a very long winded way of getting around to saying, yeah, I hope the j-space bubble ISN’T reflective. If the bubble is reflective, our poor ships are going to get cooked.
Pass the marshmallows.
