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Niobium Volcano in a nutshell


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33 minutes ago, Yunru said:

You say that like it's a problem, rather than a joke at the learning experience we all go through. 

It's a learning experience that could cost you at least an entire asteroid and access to niobium for a very long time (possibly forever in future updates) and you will never learn from your mistakes without an external guide or sandbox testing or have any solid method to fix the issue in worst case scenario because insulation (still not available anyway) could melt and even though refined carbon could withstand the temperatures, you can't make tiles out it. That's not reasonable.

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17 hours ago, ZombieDupe said:

It's a learning experience that could cost you at least an entire asteroid and access to niobium for a very long time (possibly forever in future updates) and you will never learn from your mistakes without an external guide or sandbox testing or have any solid method to fix the issue in worst case scenario because insulation (still not available anyway) could melt and even though refined carbon could withstand the temperatures, you can't make tiles out it. That's not reasonable.

Two things.
1. Reload previous save
2. You'll find out if this is an "I told you so" moment once I have built something to deal with the heat output. The question won't be "if" I can deal with the heat. The question will be  "How many steam turbine do I need" and "How much thermal mass and therefore steam/water will I need in order to act as a heat sink and prevent damage to my autosweeper"

Here. I'll even get technical with some calculations. 

Specific heat capacity of water is 4.179, niobium is 0.265.
With the same amount of heat energy and for every degree change for 1kg of water,  you will need 15.769kg of niobium to achieve the same 1 degree change.


Total heat from my volcano

Niobium is spewed from the volcano at 3726.9 and for mine, a total mass of 267.2kg/s over 63s.
Total mass m
= 267.2kg/s * 63s
= 16,839.9 kg


Change in heat energy must be the difference between volcano output temp and max thermium operating temp so 3726.9 and 875.
ΔT = 2851.9
Total heat energy 

q = mc ΔT
16,839.9 * 0.265 * 2851.9
= 12726813.36 Joules


How much water for my heat sink?

Assume a resting temp of 125 degree for the steam and a need to keep the thermium autosweeper below 875 degrees.
Therefore our margin for heat energy  ΔT  is 750. I will need a larger margin due to the time it takes for heat to disperse but this is a good minimum bar for our heat sink calculation.

So going back and finding the mass of water we need, rearrange q = mc ΔT to     m = q/ (cΔT)
m = 12726813.36 /  (4.179* 750)
= 4061 kg


Closing notes 
So 4061 kg or 4 tiles worth of water is the MINIMUM mass required to act as a heat sink for my volcano. This is a trivial amount of water.
I have bolded and underlined the variables that will change for your niobium volcano if you want to carry out a calculation for the amount of water you need in your heat sink.

p.s. I told you so ;-)
 

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5 hours ago, Smithe37 said:

Two things.
1. Reload previous save
2. You'll find out if this is an "I told you so" moment once I have built something to deal with the heat output. The question won't be "if" I can deal with the heat. The question will be  "How many steam turbine do I need" and "How much thermal mass and therefore steam/water will I need in order to act as a heat sink and prevent damage to my autosweeper"

Here. I'll even get technical with some calculations. 

Specific heat capacity of water is 4.179, niobium is 0.265.
For every degree change for 1kg of water,  you will need 15.769kg of niobium to achieve the same 1 degree change.


Total heat from my volcano

Niobium is spewed from the volcano at 3726.9 and for mine, a total mass of 267.2kg/s over 63s.
Total mass m
= 267.2kg/s * 63s
= 16,839.9 kg


Change in heat energy must be the difference between volcano output temp and max thermium operating temp so 3726.9 and 875.
ΔT = 2851.9
Total heat energy 

q = mc ΔT
16,839.9 * 0.265 * 2851.9
= 12726813.36 Joules


How much water for my heat sink?

Assume a resting temp of 125 degree for the steam and a need to keep the thermium autosweeper below 875 degrees.
Therefore our margin for heat energy  ΔT  is 750. I will need a larger margin due to the time it takes for heat to disperse but this is a good minimum bar for our heat sink calculation.

So going back and finding the mass of water we need, rearrange q = mc ΔT to     m = q/ (cΔT)
m = 12726813.36 /  (4.179* 750)
= 4061 kg


Closing notes 
So 4061 kg or 4 tiles worth of water is the MINIMUM mass required to act as a heat sink for my volcano. This is a trivial amount of water.
I have bolded and underlined the variables that will change for your niobium volcano if you want to carry out a calculation for the amount of water you need in your heat sink.

p.s. I told you so ;-)
 

This is a real ONI fan answer as I like. If something looks impossible, it just means you need to engineer more !

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1 minute ago, nakomaru said:

That won't do! We must recover everything.

Unfortunately, niobium must be 24kg or below in order to form debris instead tiles, so this is a job for pumps.

Ahhh. That makes sense. I suppose my attitude is just that there's going to be so much niobium anyway that there's no need to be maximally efficient.

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It comes out FAST. Trying to get a plinko to work to divide and bead the flow enough, but it is not easy. Good thing tungsten is heavier than niobium or I don't think even this would be possible.

image.thumb.png.fca2bc72dbb3bfea7daf9163bc13925d.png

With a staged system to bring the output down to ~2700C first, we can use doors to better control this, but I can't figure out how to get ~20kg precision.

Ooooooh I did it.

 :juggling:

But it's quite slow :(

qXhCo2XBuA.thumb.gif.f8dd4b657be2cb1753773788c9dfb92b.gif

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image.thumb.png.a4270b7074fd8c150ebc13659e759f8c.png

Top right would be the volcano. Niobium column has to be 10 cells high to build enough pressure to force it around the bend. Spits out nice small drops even with 17,000 kg of niobium painted in all at once at the volcano location. Add extra steam, turbines, rails, and such as needed.

Small problem is the amount of niobium in the column, takes some time to fill.

Major problem is you have to solidify ~20kg of 4000K niobium every other tick, 100% consistently on that specific tick, while it is dripping. Temp exchange is no big deal with enough plates, bridges, and whatnot. Game hiccups will screw you over on the other hand. Auto saves are the devil.

 

Edit:

Perhaps it would be more stable dripping the niobium off some molten tungsten so you get actual liquid cells falling instead of quantum droplets. Gradual cooling on the way down might seem beneficial, but you have to ensure the freezing niobium doesn't remelt on the next tick when the newly formed debris touches the next falling blob of liquid niobium. I guess you could explicitly separate the cooling into a precooler stage and then a freezing stage, but that sounds annoying.

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12 hours ago, nakomaru said:

so this is a job for pumps.

Agreed.

1709430907_NiobiumVolcanoTamer.thumb.png.6b2ce63b5b207e6cac3ba9966c726b56.png

 

1013989549_NiobiumVolcanoTamer-Plumbing.thumb.png.aa9e04a48400f49d204b9e23425d2de5.png

 

The trickiest part is finding a non-insulation pipe which won't melt.  I'm using obsidian insulated pipes here and running cold naphtha through the entire length.  The pipe temperature has stabilized between 1400C and 1500C.

 

(Please ignore the missing volcano.  I can't seem to find one to play with.  Do they not spawn anymore?)

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@nakomaruPutting in niobium masses from the video of 76.6kg above the gold and 153.1kg one cell leftwards, I too get a >25kg drop. The strange thing is I can't reproduce starting at those masses in the first place. I've tried a bunch of times starting from a column filled to various degrees and the worst drop I've seen coming out is 22.9kg (it quickly trends to 20ish kg drops). I wonder if there is a fundamental difference in the mechanics or I just got lucky and you unlucky.

Can you get the dropper to behave in a general test without going to the trouble of all the other cooling stuff? Maybe moving the gold over a cell to where the airflow is might help? It's safe there AFAICT.

image.thumb.png.bfdf8fdef6bda407349608b64620bc6f.png

 

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Had a stab at a self-powered niobium tamer. Built in debug mode, but should really be easier to build than usual tamers as the volcano can be isolated for most of the build.

Obsidian insulated pipes for niobium pump are key, thanks @ghkbrew! Tiles in direct contact with niobium should probably be airflow tiles, but ceramic or even igneous would last long enough.

Bottom heat injector is used to restart everything after dormancy period (use steel for the door and automation).

niobium_tamer.thumb.jpg.b51a69f00b7b93900225cb006d8e36f9.jpgniobium_tamer_plumb.thumb.jpg.dfcf2706771189df49077d740959b792.jpgniobium_tamer_auto.thumb.jpg.418ea23c05f5c58ba491a471c5e209a0.jpg

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9 hours ago, HeatEngine said:

Obsidian insulated pipes for niobium pump are key, thanks @ghkbrew! Tiles in direct contact with niobium should probably be airflow tiles, but ceramic or even igneous would last long enough.

 

The pipes will melt.
You need tungsten pipes, but then the refrigerant can boil.

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7 hours ago, mukca said:

The pipes will melt.

Assumptions:

  1. Heat the pipe exchanges with the content: q = klow * (Tpipe - Tcontent) * 10
  2. Niobium temp: 3727*C
  3. Naphtha temp: 200*C
  4. Equal amount of naphtha and niobium pass through the pipe

Equilibrium point is when q1 = - q2 (head added by niobium is taken away by naphtha), solving for Tpipe gives 1963.5*C, in my testing the the temp stabilized around 2000*C (assumption 4 is inaccurate?), well below 2727*C melting point for obsidian.

It's important to filter out niobium and not naphtha otherwise you end up with sections of the pipe that don't get any cooling.

 

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