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Is it possible to use thermal byproduct of daylight operation to power night time operation of volatile computer storage?

In my setting, a moon-like moon with almost no atmosphere has one of its sides (the far side) populated with nodes powered with sunlight (solar flux of said planet is around $1247 W/m^2$), and day-night cycle is 27 days long. The computer storage on this node is volatile, meaning it needs continuous power supply to keep it from losing its memory. The node's storage requires constant power at $4\times10^6 W$.

Each node operates at $2\times10^7 W$ of power, easily powered by sunlight during 13.5 days long daytime. About $4\times10^6 W$ of it is deposited in flywheel energy storage during daylight, that is 13.5 days. Naturally, each node radiates $2\times10^7 W$ as thermal energy, so I equip them with radiators that operate at 600K temperature, with coolants exiting at 400K, radiating at a rate of $7656 W/m^2$ according to equation on this answer.

As mentioned above, at daylight the radiator outputs coolant at 400K temperature, and this heat is stored on water reservoirs, for the sake of this analysis assume that the reservoir is near-perfectly insulated. This reservoir stores $3.52\times10^8$kg of water, that stores about $1.87\times10^14J$ in the form of heat.

Now, at night the node is allowed to operate at lower temperature, let's say it could be as low as 273K, and with hot reservoir at 400K, this is 127K of temperature difference. Again as in linked answer, the temperatures of the hot and cold reservoirs of your power generation system define Carnot efficiency, and in this case it is ideally at 37.5% efficient. But let our heat engine works at 35%, this means from the stored energy only ~$5.6\times10^6W$ of usable work could be extracted for the entire night time (13.5 days).

Flywheel energy storage charged from 13 days of continuous operation at $4\times10^6W$ during daylight is now extracted, and due to limited conversion efficiency at around 80%, only $3.2\times10^6W$ of power could be extracted.

In total, energy from heat engines and flywheel energy storage provide us with $8.8\times10^6W$ of usable energy during night time, enough to power our computer storage, and some energy is left for node maintenance.

Now, this system also radiates energy at around ~$2\times10^7W$ (heat from reservoirs and from energy generated by extracting the flywheels). The radiator at night time took 400K coolants and outputs it at temperature of 250K, according to linked question the radiator would radiate heat at a rate of ~$1400W/m^2$.

My concern is that although the configuration appears realistic enough to be workable for me, this results in total of constantly radiated heat of $2\times10^7W$ during day and night, despite the fact that each node is powered by $2\times10^7W$ of sunlight only in daylight. At first I thought it is possible because of different working temperature, 400K during daytime and down to 273K during night time, but I can't be sure.

Therefore, I came up with this question: Is this configuration of utilising thermal byproduct to power night time operation possible? If it is not, then where is my error?

[This question was on sandbox]

kingledion
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Hendrik Lie
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    What's wrong with batteries? This complicated setup is only useful if there is something fundamentally wrong with batteries. As to the question, the dubious step is where you make the strange assumption that the water in the reservoirs will be heated to the exact same temperature as the coolant in the black body radiators; I cannot think of any specific reason of why this would be the case. And anyway, since you want to trap all the black body radiation and store it as heat, then why do you bother with black body radiation at all and don't use a plain counter-current heat exchanger? – AlexP Dec 08 '17 at 15:37
  • @AlexP this should be a good answer. The reason I don't use battery is that I don't want the battery to decay over time, as any rechargeable chemical battery do, so I opted for flywheels, but then I add this thermal engine. Apparently I overlooked the option you described. Can you please expand it to a full answer? Perhaps I am more interested in the last sentence of your comment. – Hendrik Lie Dec 08 '17 at 15:47
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    @HendrikLie The bearings of your flyweels will wear out, memeory will suffer bitrot, processors will burn out. Everything breaks with time. Given the scale of the mega-structure it's probably been designed with enough redundancy to handle selective shutdown for maintenance of all of the systems. – sphennings Dec 08 '17 at 16:06
  • @sphennings you're right. To counter that, it is not the only node, there are literally millions more spread across half the surface of the moon. However i opted for flywheel with magnetic bearings over battery power, as somehow I'm convinced that with similar capacity, flywheels are long lived. – Hendrik Lie Dec 08 '17 at 16:12
  • @sphennings is correct. You're storing the energy as kinetic energy. Anything that moves has problems. But to be honest, your real problem is a lack of atmosphere. You're not addressing thermal conductivity in the structure, which will cause stress over time. I've watched ICs vaporize because the thermal conductivity of the package coudln't handle the heat generated by the device. Out of curiosity, how efficient is your system? Theoretically, a superconducting transistor generates no heat. Only resistance generates heat (well... resistance and magnetic reactance, but let's not go there.) – JBH Dec 08 '17 at 16:56
  • So because batteries have a limited number of cycles you use mechanical systems with moving parts? That's a brave engineering decision. And you are also a very good friend of thermodynamics: "Each node operates at 2E7 W of power, easily powered by sunlight during 13.5 days long daytime. About 4E6 W of it is deposited in flywheel energy storage during daylight, that is 13.5 days. Naturally, each node radiates 2E7 W as thermal energy, so I equip them with radiators that operate at 600 K". Ordinary people would have to add some 1E7 W to the power consumption of the system for the heat pump. – AlexP Dec 08 '17 at 19:55
  • @AlexP can you please expand it into an answer? I would really like to get more details on your proposed design. I didn't know that we must have some 1E7 W to power the heat pump. – Hendrik Lie Dec 09 '17 at 02:18
  • @JBH originally i assume they have radiators and coolants to deliver heat away from the computer, but that's as far as I know. I don't know how efficient the system should be, i didn't even know that i should have considered that in the first place. Can you please expand it into an answer? – Hendrik Lie Dec 09 '17 at 02:22
  • No, I cannot. You are using the hard-science tag. That tag literally means, "if you can't back it up with full mathematics and authority, or links to same, it's an invalid answer." It's the most difficult tag on WB:SE, and you've applied it to the future. I doubt anyone can answer this as it stands. – JBH Dec 09 '17 at 02:54
  • @JBH hard-science tag removed. I may have overestimated the quality of my question when i applied that tag. What I want is just mere mathematically and physically supported, not that it has to be true down to atomic level analysis – Hendrik Lie Dec 09 '17 at 03:02
  • I appreciate your decision, it will help with answers. However, please do not expect much from the mathematics. A future technology with unknown material science means only the most basic of physics can be applied. But, at least now you'll begin to receive insight into what you can do with the question. Pay attention to the answers. Together with your question they may lead to a more specific question that achieves the goals you're looking for. – JBH Dec 09 '17 at 03:09
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    Heat does not move by itself from cooler objects to warmer objects. If you want to move heat from the CPUs (which I assume will work at about 100° C or 400 K) to the radiators at 600 K then you must use a heat pump; heat pumps consume energy to work, just like your refrigerator; using 1E7 W to move 2E7 W of heat is a decent assumption. I won't comment further; I would have used static RAM and avoided the entire issue. While static RAM is more expensive than dynamic RAM, eternal flywheels, motors, generators etc. are expensive too. – AlexP Dec 09 '17 at 06:17
  • @AlexP can you please expand it into an answer? Your suggestion seems to be within the scope of my question – Hendrik Lie Dec 09 '17 at 07:31

1 Answers1

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  1. Why at night the system dissipates 2E7 W when it is only powered by 2E7 W during the day?

    Because during the day it doesn't dissipate any heat. During the day, the OP says, the waste heat of the data processing subsystem is used to heat up water.

  2. Does it work?

    No it doesn't, not as such. The bewildering array of numbers in the question fails to take into account the energy necessary to operate the radiators (coolant pumps, heat pumps, etc.) Otherwise it's fine, except that it's overly complicated. Why radiate heat as electromagnetic radiation just to capture it and heat some water? Instead of this strange detour, the system should heat the water directly using a countercurrent a heat exchanger.

    If the pilot installation finds out that the energy stored as hot water is not enough to power the data processing subsystem at night then some more solar panels should be added to heat the water directly. Thermal energy storage is quite well understood and it's one of the proposed solutions to the necessity to have solar power plants work at night.

On the other hand, static RAM is a thing, and static CPUs are not unheard of. The data processing subsystem can simply halt in place at dusk and resume operation at dawn with no power requirements during the night, or at least with minimal power requirements to maintain air circulation and avoid deep freezing.

AlexP
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  • Perhaps a bit too late, but soon after I realize that my implementation is wrong, and ever since revised it. However your answer do answer my question: this doesn't work. So, accepting the answer :) – Hendrik Lie Feb 26 '18 at 16:34
  • Static circuits use minimal power to keep their state during night - which a stored heat source could possibly be used for - though a battery seems more normal. It is the electrical equivalent of keeping pressure in a pipe, but without any water flowing except for very small leaks. – user253751 Mar 22 '24 at 05:33