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submitted 5 months ago by DadBear@slrpnk.net to c/technology@slrpnk.net

I can't remember if I saw the argument here or on Reddit, but this is my preferred platform so it's going here.

Summary of argument: a user should have been using water for their thermal battery, not sand, because water has better heat capacity (4.18 joules per unit of mass person unit heat - 4.18/gK). Sand's thermal capacity is significantly lower (0.835J/gK).

Looking at these numbers alone in the post I understood why someone would say that; it also made me question why so much research is being done on sand batteries. The user who argued against sand batteries missed a crucial factor: material density. Water has a density of 1000kg per m^3. Dry sand (regular not pure quartz sand) has a density of 1730 kg per m^3. I found no satisfactry response to the argument in that thread, but that thread is now lost to me. I have also been curious about how much better regular sand is for heat batteries than water.

When designing large batteries, the goal is usually energy per volume. Let's compare 1m^3 of each (roughly 3.3ft cube) and how much heat it can hold before the next state change (which matters a lot when managing the pressure from steam).

Total stored energy = mass (g) * thermal capacity (J/gK) * heat (kelvin).

Water: 1,000,000 * 4.18 * 373.15 = 1,559,767,000J Sand: 1,730,000 * 0.835 * 1996.15 = 2,883,538,482.5J

Over 1 billion more joules per m^3. I hope this makes it clearer why sand batteries are such an area of interest lately. It certainly did to me.

Disclaimer: I am not an expert, so there may be mistakes. All the numbers and relevant equations were found on the internet.

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[-] ji59@kbin.social 5 points 5 months ago

But I would argue its harder to put extra heat into sand that has already 2000K, and also bigger heat loss. But I don't know anything about how it works, just some common sense.

[-] magiccupcake@lemmy.world 5 points 5 months ago

You can fight a lot of that with bigger batteries. Surface area goes up by r^2, but volume goes up by r^3.

More expensive batteries will also warrant better insulation.

Combine those and sand batteries make decent sense.

[-] DadBear@slrpnk.net 2 points 5 months ago

From the math I looked at, that doesn't seem to be the case. What we're actually doing is fighting radiative and convective heat loss, basically requiring more energy per second to compensate for increasing heat losses per second. An adequately insulted sand battery would negate a lot of that.

this post was submitted on 26 May 2024
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