# How much electrical energy is required to dissipate heat?

I'm modeling the electrical energy consumption of a chemical reactor for the Moon base. The reaction taking place in this reactor is highly exothermic, and the device must be cooled.

My question is how do I estimate the electrical energy required to dissipate a certain amount of heat to space. Does anyone know publications on this topic. Like the Wattage of the thermal control pumps, and the relationship of Heat generated in some system to amount of pumping required.

There is an extensive post on heat sink here: Finned heatsinks in space

a nice image of ISS thermal control system:

• I don't think this has a good answer, as clearly the answer could be 'none at all'. For instance you could use a thermal siphon (won't work for the ISS!) to cool a system completely passively. But such a system will probably be much larger than a pumped system, say. You need to specify what the constraints are: without that there simply isn't a good answer, I think. – user21103 Apr 13 at 13:47
• a comparison might be made to the kilopower nuclear reactors for space. they dissipate up to tens of kW of exhaust heat from a Stirling engine and use heat pipes to move some if not all of that heat around. If you used something similar you could recover and reuse some of that waste heat. – uhoh Apr 13 at 14:10
• I had the same idea as uhoh. With the right setup, surely you could store that heat and recover some of it as electricity later, or do the conversion immediately. – SpaceLawyer Apr 13 at 19:05

As you used the ISS as an example, one may look up what kind of pump power is required for all of that: https://www.nasa.gov/pdf/473486main_iss_atcs_overview.pdf

2x PFCS ("Pump Flow Control Subassemblies"), each at 275W.

What costs energy in a cooling system is moving fluid around. In the case of the ISS, it's around 7.5tons per hour. (But if you really wanted to, moving heat from hot to cold is an opportunity to recover energy, at the complexity cost of a heat engine).

For a rough estimate, you could get the heat capacities of your preferred coolant, divide your wattage by that, and scale up the ISS pumps from above.

• Thanks for the link. – Organic Marble Apr 13 at 20:50
• – uhoh Apr 14 at 1:08
• As a preliminary estimate, do you think saying something like: on ISS it takes 0.55 kW of electricity to operate two coolant pumps, in order to radiatively dissipate ~14kW to space. Hence, a preliminary estimate is 0.04 kW of electricity required to dissipate 1 kW to space. ISS has large coolant piping system compared to mine, but since ammonia is very low viscosity, I think this does not add major pumping costs. – dlight Apr 30 at 14:09
• @dlight I think it would make sense as a order of magnitude estimate. – SE - stop firing the good guys Apr 30 at 15:23

It takes no energy to dissipate heat into space via black body radiation, which is what I assume you're talking about. A "heat pump" can also be a wide range of things, and the general category doesn't have any math that would be very useful to you. I recommend an edit for clarification, so your question can be answered better than a simple, "zero".

• It may not have not be crystal clear, but I meant the pumping energy required to dissipate heat through the ammonia coolant loop. – dlight Apr 30 at 14:02