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Currently, we have good methods of sustained, slow dissipation of energy through radiators. They take quite large area (and mass) per watt dissipated though.

I know sublimators were used at least in the Apollo missions, as exhaustible but long-term means of heat dissipation. I'm not sure how efficient they were though.

Still, I wonder what means we have or have plans for, that could be used for rapid dissipation of huge amounts of power in a very short time - possibly in exhaustible, non reusable way. Ablators come to mind, but all I know about ablators is that they are used in the atmosphere, on reentry, and they strongly depend on convection, most of heat dissipated into the air, not into the ablator. I have no clue how one would work in vacuum.

This is related to my question on Harpoon Propulsion - in particular, brakes of the spool. Their energy output would be of orders of a megawatt, over a period of a couple minutes.

They are allowed to burn, melt, evaporate or explode afterwards, but they must keep on braking at constant force while they work. It may be a friction brake, or electrodynamic or any other, applied to linear motion (along the ribbon) or rotary (of the axis).

So, how can we sink that amount of energy?

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  • $\begingroup$ It may soon be possible to use the waste heat instead of just throwing it away. $\endgroup$ – Joe L. Feb 15 '16 at 18:11
  • $\begingroup$ @JoeL.: That assumes a convenient drain, which is suspiciously absent in space. $\endgroup$ – MSalters Jun 8 '16 at 11:33
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In a vacuum, there are only two ways to get rid of the heat, radiators and sublimators.

You can not simply get the energy in the heat back to useful forms like electricity, due to the laws of thermodynamics.

The efficiency of both is dependant on the temperature they operate at.

For radiators, we have the Stefan-Boltzmann law, that says that the power radiated is proportional to the fourth power of the temperature. That means that if you want to get rid of large amounts of heat, you can let the radiators run hotter.

The power level of a sublimator is proportional to the temperature and the heat capacity of the evaporated material. The best heat capacity possible is that of hydrogen at $14.267 J/gK$. Water is good too. Generally, go for light molecules.

Because the systems scale differently, sublimators are good for small, short duration missions with a low operational temperature, and radiators for larger, long duration missions with a higher operational temperature. The borders are fuzzy.

If you are planning to use heat to produce useful energy, you can do so if you have a temperature difference.

The overall efficiency is determined by the Chambadal-Novikov efficiency:

$$\eta = 1 - \sqrt{\frac{T_L}{T_H}}$$

Where $T_L$ is the lower temperature and $T_H$ the higher.

For a constant high-end operation temperature, you can combine this efficiency with the Stefan-Boltzmann law to find the ideal radiator temperature for the highest energy production to radiator mass ratio.

It occurs at:

$$T_L=\frac{64}{81}T_H$$

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  • $\begingroup$ I'm not trying to produce energy, I'm trying to discard it - I can use e.g. superconducting electrodynamic brakes which will hardly produce any heat, but enormous amounts of electricity... and how do I get rid of it then? $\endgroup$ – SF. Feb 15 '16 at 17:41
  • $\begingroup$ @SF the first half of the answer is about getting rid of the heat. The second part is how you can use it for something a little more valuable than throwing it away. $\endgroup$ – Hohmannfan♦ Feb 15 '16 at 17:43
  • $\begingroup$ What am I going to do with a gigajoule of electrical energy which I receive within less than 5 minutes? Some 300kWh of electricity? Coming in at some 3 megawatt over superconducting wires? The energy doesn't need to be thermal. $\endgroup$ – SF. Feb 15 '16 at 17:55
  • $\begingroup$ @SF A giglajoule? If you radiate that heat over an hour, and let the radiators run hot at a 1000 K, that is only going to require 5 m² of radiators. $\endgroup$ – Hohmannfan♦ Feb 15 '16 at 18:06
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    $\begingroup$ @SF. Actually that is precisely what you are asking for. More generally, Hoffmannfan is reiterating 2nd Law of Thermodynamics. The heat must go somewhere, either some mass you carry with you, or into the ether. Therefore the only alternative would be to break the 2nd Law, which is equivalent to turning the heat back into work. $\endgroup$ – Aron Feb 15 '16 at 18:37
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There are space radiators, and then there are space radiators: https://en.wikipedia.org/wiki/Liquid_droplet_radiator

Spraying droplets of your thermal fluid through space before collecting and recirculating them effectively creates a radiator of large surface area without the mass.

Plans were drawn up to fly a Liquid Droplet radiator demonstration unit on an STS mission but never reached fruition.

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    $\begingroup$ Can you give me some numbers - e.g. energy dissipation for a kilogram of working liquid? $\endgroup$ – SF. Jun 6 '16 at 7:36
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Don't convert it into heat in the first place, and then don't dissipate it fast.

MSL used an electromagnetic brake to slow the descent of the rover from the skycrane. It dissipated the energy into a resistor. Here, instead store the energy in a superconducting magnet. It can remain there as long as you can keep the magnet cold, and you can then either use the energy, or dissipate it over a long period of time through small radiators.

The conversion of mechanical energy into electrical energy won't be 100% efficient, so you will inevitably be left with some heat energy to dissipate. However such conversion efficiencies can be 98% or better, so you have reduced your immediate problem by a factor of 50.

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  • $\begingroup$ 12.5 kJ/kg energy density. With zero losses the magnetic storage will saturate upon converting $v= \sqrt{E_k \over m}$ = 111m/s of own speed to charge. page 3 $\endgroup$ – SF. Jun 6 '16 at 7:25
  • $\begingroup$ Good point. The specific energy of these things aren't so great. You could find other means of energy storage. A flywheel perhaps? $\endgroup$ – Mark Adler Jun 6 '16 at 17:02
  • $\begingroup$ 500kJ/kg sounds better. (never mind that we can push the wheel right to the limit and beyond it; if it falls apart and explodes, so be it, just don't let the shards hit the craft.) Still, that's only 1km/s to the break-even point. We still need at least an order of magnitude more. $\endgroup$ – SF. Jun 6 '16 at 21:21
  • $\begingroup$ (correcting previous calculation; $v= \sqrt{ 2E_k \over m } = 158 m/s$. Forgot the 1/2 in the $E_k$ equation.) $\endgroup$ – SF. Jun 6 '16 at 21:27
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Water has one of the highest heat capacities available, so this would be the most efficient material to use as dissipation mass (the least mass necessary per MW of heat dissipated).
You could use the steam in various ways: build a steam engine to drive a generator, heat the spacecraft and/or or use the steam exhaust as propulsion.

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  • $\begingroup$ Using ice at -60C and letting it evaporate into steam of 2200C (shortly before it starts breaking up into oxygen and hydrogen, and 500C below decomposition of nanotubes) we're getting usage of 129kg/s to keep up with 1 megawatt output. Almost 26 tons for 200s of operation. $\endgroup$ – SF. Feb 15 '16 at 10:17
  • $\begingroup$ ...that's 4km/s for a craft of 25 tons, at 2g. (not counting the heat sinking or rope mass.) With 400km of rope. $\endgroup$ – SF. Feb 15 '16 at 10:29
  • $\begingroup$ @SF. Which is a sublimator. $\endgroup$ – Aron Feb 15 '16 at 18:40
  • $\begingroup$ Bad news, output is in the GW range, not MW. See your other question. $\endgroup$ – Hobbes Feb 15 '16 at 18:59
  • $\begingroup$ I think the Space Shuttle had water tanks and a heat exchanger system it used immediately after launch to extract waste heat from the SSMEs and blow it overboard as steam before they could heat-soak the whole structure. I just lost fifteen minutes trying to track down a reference and failed. $\endgroup$ – Kengineer Jun 5 '16 at 22:03
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Obviously we cannot use ordinary radiator panels to increase the rate of energy dissipation (other than adding more of them)

There is always the option of using this excess energy:

  • accelerate reaction wheels
  • power a laser or some other energy transmission device

etc

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  • $\begingroup$ I don't think that amount of energy could be meaningfully stored in any reaction wheels of space-faring size. And lasers have an efficiency below 30%, meaning 70% of the energy would need to be dissipated from the laser construction by other means. $\endgroup$ – SF. Feb 15 '16 at 9:29
  • $\begingroup$ Oh, I totally agree - it's just an extra. If we can't dissipate, then let's see what we can use it for :-) $\endgroup$ – Rory Alsop Feb 15 '16 at 9:33
  • $\begingroup$ Personally, I thought if an engine similar in nature to NTR (except non-nuclear) could be powered, using the propellant for cooling and reducing strain by providing rocket propulsion... but I'm afraid the weight (of the engine and propellant) would increase braking mass -> energy output more than its dissipation capability would provide. $\endgroup$ – SF. Feb 15 '16 at 9:41
  • $\begingroup$ The problem is waste heat, not energy excess! Both methods will increase the onboard heat! $\endgroup$ – Hohmannfan♦ Feb 15 '16 at 16:30
  • $\begingroup$ Ahhhh - my misunderstanding. Thanks @Hohmannfan $\endgroup$ – Rory Alsop Feb 15 '16 at 16:35

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