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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, 2016 at 18:11
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    $\begingroup$ @JoeL.: That assumes a convenient drain, which is suspiciously absent in space. $\endgroup$
    – MSalters
    Jun 8, 2016 at 11:33
  • $\begingroup$ Current US EVA suits still use sublimators. $\endgroup$
    – Organic Marble
    Apr 18, 2019 at 14:17
  • $\begingroup$ In the situations where your brakes do generate electricity, then send it to an efficient radio transmitter. I can't turn this into an answer because finding the frequency of maximum efficiency is a bit of a challenge. It's probably going to be somewhere between 10 MHz and 1 GHz, but there are different problems in each frequency range. You'll have to radiate the losses thermally or soak it up with water. $\endgroup$
    – uhoh
    Jul 14, 2019 at 13:40
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    $\begingroup$ @uhoh: It would certainly be an option if there was a transmitter of a very good efficiency and power of the order described. And preferably a good power:mass ratio too. Are there? $\endgroup$
    – SF.
    Jul 14, 2019 at 15:23

8 Answers 8

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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, 2016 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, 2016 at 17:02
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    $\begingroup$ 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, 2016 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, 2016 at 21:27
  • $\begingroup$ @SF 500kJ/kg is the wrong number to use for flywheels. Flywheel energy density should scale linearly with the specific strength of the material it is made out of. Since the scenario this is based on already assumes buckytube rope, you can use the same buckytubes for the flywheel. $\endgroup$
    – Lex
    Jul 16, 2019 at 3:54
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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 dependent 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, 2016 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. $\endgroup$
    – SE - stop firing the good guys
    Feb 15, 2016 at 17:43
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    $\begingroup$ 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, 2016 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. $\endgroup$
    – SE - stop firing the good guys
    Feb 15, 2016 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, 2016 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, 2016 at 7:36
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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, 2016 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, 2016 at 10:29
  • $\begingroup$ @SF. Which is a sublimator. $\endgroup$
    – Aron
    Feb 15, 2016 at 18:40
  • $\begingroup$ Bad news, output is in the GW range, not MW. See your other question. $\endgroup$
    – Hobbes
    Feb 15, 2016 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, 2016 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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  • 1
    $\begingroup$ 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, 2016 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, 2016 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, 2016 at 9:41
  • $\begingroup$ The problem is waste heat, not energy excess! Both methods will increase the onboard heat! $\endgroup$
    – SE - stop firing the good guys
    Feb 15, 2016 at 16:30
  • $\begingroup$ Ahhhh - my misunderstanding. Thanks @Hohmannfan $\endgroup$
    – Rory Alsop ♦
    Feb 15, 2016 at 16:35
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It depends on the entropy. In the case of the 'reverse winch launch', the energy is in a potentially useful form. In this case, in theory with a sufficiently efficient generator, electrical circuitry and (say) antenna, in theory you can approach near perfect efficiency in creating EM radiation from this energy, with out an obvious lower bound on how much the system would need to weigh.

If you have to deal with (or allow) entropy to accumulate, then you start to hit thermodynamics problems that it is hard to see a good way around. Namely having a negative Gibb's energy (being thermodynamically favorable) see https://en.wikipedia.org/wiki/Exergonic_process and https://en.wikipedia.org/wiki/Gibbs_free_energy.

Without stretching the limits of credulity too much this is going to involve heat dissipation, and as discussed this is going to involve either refining either of the 2 methods already discussed (venting something hot or black body radiation. Both methods are going to be difficult to make game changing improvements to. However increasing the effective surface area, conductivity, and survivable temperature of a radiator, (for a given mass) seems the least infeasible way.

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  • $\begingroup$ Yeah, so... any calculations or at least estimates for how big/heavy either the antenna or radiator strategies would need to be, for the winch application that the OP focuses on? $\endgroup$
    – leftaroundabout
    Jul 15, 2019 at 10:17
  • $\begingroup$ @leftaroundabout That's not easy to answer. In theory a tungsten radiator near its melting point would be into Mw territory with just a square meter, the antenna even smaller. But how thin could you make this and still circulate a coolant effectively? How hot is it okay to let the heating part part of the craft get? The answer could range from grams if we are will to hand wave away everything that's 'just' an engineering problem, to: the whole thing is impossible if we are limited to today's technology. $\endgroup$
    – ANone
    Jul 15, 2019 at 10:40
  • $\begingroup$ There's certainly no reason to go into “impossible” territory (a simple example of a usable tungsten radiator would be a light bulb, perhaps without the bulb). As for the antenna, I guess the critical consideration would be how efficient the amplifier/resonator circuit can be. There should be reasonably hard established boundaries for that in electronics. $\endgroup$
    – leftaroundabout
    Jul 15, 2019 at 10:45
  • $\begingroup$ @leftaroundabout right. The radiation component of heat dissipation of a tungsten filament is extraordinary. Easily enough to achieve OP's requirements with a tiny mass. BUT its not the limiting factor. it wouldn't be anywhere near that easy with a single heat source. A filament generates its own heat almost exactly where it is radiated. its hard to imagine a mechanism that could do this for OP's situation. Maybe not impossible to get close, but how close is going to depend on how much hand waving. $\endgroup$
    – ANone
    Jul 15, 2019 at 10:57
  • $\begingroup$ @leftaroundabout. Also by impossible I meant that tether braking, even if we had a magic box that absorbed heat, would still be beyond our current means. $\endgroup$
    – ANone
    Jul 15, 2019 at 10:57
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If it isn't heat yet but some kind of kinetic energy, you could convert it to electricity and power a laser or some other transmitter of some kind.

Or if you're talking heat, dump it into a big chunk of wax and let the phase transition soak it up, then either jettison the wax or let it radiate and resolidify over time depending on your need. There's more about "space wax" in this question.

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  • $\begingroup$ nice to see you again!! there are a few unanswered or insufficiently answered space law questions listed here if you are interested: law.meta.stackexchange.com/q/814/24570 $\endgroup$
    – uhoh
    Jul 15, 2019 at 23:49
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“Orders of a megawatt” is not that hard to handle via radiation from a disk brake.

Carbon disk brakes on earth can operate around 800C with peaks to 1000C.

At even 1100K, each square meter would radiate about 1/2 megawatt into space (one side).

A 4m diameter disk, not so big for the end of a spool carrying kms of cable, completely open to space would happily radiate 5MW once it came up to temperature.

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  • $\begingroup$ A concrete-numbers no-nonsense answer, I like it. I wonder how carbon brakes would behave at extremely fast velocities. (But if that's a problem, the velocity could be scaled – maybe instead of a single 4 m disk we'd need a whole bunch of much smaller ones, and a sufficiently large spool, then the brake pads will move at more normal speed over the disk rotors.) $\endgroup$
    – leftaroundabout
    Jul 16, 2019 at 8:22

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