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Metal Refinery Coolant Heating Formula


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Quick question for those of you that like playing with the back end numbers.

I'm trying to calculate (via Excel formula) exactly how much heat change any particular coolant will take when run through a Metal Refinery with particular materials.

I thought I had the formula figured out but my result calculations aren't lining up with what is actually happening. We pretty much all know that cooling Iron with Crude Oil raises the oil by just about 79.5 degrees C (91.8 to 171.3 in my most recent check). My problem is figuring out the formula that gives this number.

What am I either missing or getting wrong? Refining 100kg of Iron generates 67.1kJ (kDTU?) of heat according to the tool tip. Because there's 4 times the amount of coolant as refined metal, we can just divide the 67.1 by 4 and then apply the 1.69 Specific Heat of the oil to it, yes? Obviously I have this wrong since I'm calculating an increase of only 28.4 degrees C...

Working backwards I get a different result that is off by almost exactly 8 times. 79.5 degree change is 134.4 DTU/g, thus 53.742 million DTUs. This is basically 8 times the 6.71 million DTU that 67.1kDTU applied to 100kg of Iron equates to.

Normally math doesn't annoy me like this but I don't understand why both methods are "wrong" according to what I actually see in the game.

Anyone have ideas?

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27 minutes ago, Coolthulhu said:

Make sure that you have at least 400kg of spare coolant before starting the job and that the output is completely empty.

There is a bug causing jobs to start with <400kg of coolant, which causes the output to heat up more.

In this case, I can guarantee that all aspects functioned properly. This was a brand new refinery with oil straight from the bottom of the map and only a single order on the newest preview build. 800kg of Crude Oil along with the 100kg of Iron Ore was loaded in properly. 400kg of that coolant was moved into a separate section within the refinery as soon as the work errand was assigned. (The visible fill level on the refinery itself dropped to half and there were 2 separate lines in the "Contents" with 400kg each)

Then the "main tank" was filled back up to 800kg while the dupe was running the refinery, resulting in 1,200kg being held in the refinery. Once the job was finished, the 400kg part that was originally set aside changed temperature and was output.

As an aside, I don't remember the 400kg being set aside and another 400kg being loaded into the refinery while it is running. In fact, at one point after the initial order, I had oil flowing both into and out of the refinery while a dupe was using it. So they may have finally gotten a handle on that bug. >Crosses his fingers<

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Correct formula you can find in this post:

To calculate what you want you need to use iron melting temperature and crude oil initial temperature. Take a look at my calculator with numbers for this case:

http://www.oni-heat-calc.cba.pl/?mt1=solids&mt2=liquids&m1=Iron&m2=Crude Oil&t1=1495&t2=92&mm1=100&mm2=400&st=0

Strange thing is about that 67.1kDTU in tooltip. This would mean that metal refinery melts iron at 1495oC instead of 1535oC.

Edit: Looks like I confused some things here. I don't understand why the tooltip shows 67.1kDTU. Such amount of heat will increase 100kg iron temperatue by less than 1oC. So why it shows only 67.1kDTU?

Final temperature in the calculator calculations is not exactly 171oC as probably some heat is spread to refinery surroundings, but is still pretty close. I guess that also this 400kg of heated crude oil interacts with the other 400kg of still not heated oil so it can lower the final temperature.

EDIT2: actually the calculator has no use here. Better to just count heat generated by melting a given ore and take 80% of that value and then calculate by how much such heat will change temperature of a coolant.

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The total heat seems to be correct in the wiki (https://oxygennotincluded.gamepedia.com/Metal_Refinery), for iron 53 696 808 DTU (pretty much what your backward calculations showed.

The crude oil temperature change thus is:

  Amount Specific Heat Start Temp   End temp Heating/Cooling
  g (DTU/g)/°C °C   °C DTU
Crude Oil 400 000 1,690 0   79,4 53 696 808
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26 minutes ago, thejams said:

The total heat seems to be correct in the wiki (https://oxygennotincluded.gamepedia.com/Metal_Refinery), for iron 53 696 808 DTU (pretty much what your backward calculations showed.

The crude oil temperature change thus is:

  Amount Specific Heat Start Temp   End temp Heating/Cooling
  g (DTU/g)/°C °C   °C DTU
Crude Oil 400 000 1,690 0   79,4 53 696 808

400 000 * 1,69 * 79,4 = 53 696 808 ????

The value 53 696 808 comes from heating to 1534,9oC a 100kg of iron at 40oC and taking 80% of the heat generated:

0.8*(1534,9-40)*100*0,449 = 53 696 808

But indeed such amount of heat will change crude oil temperature by ~79,4oC

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1 hour ago, Angpaur said:

400 000 * 1,69 * 79,4 = 53 696 808 ????

Well, 79,4331479289941°C, but yes :D

Bottom line is, considering you said tooltip gives kJ, it's seems like it hasn't been updated for quite a while.  The wiki values seem to be in line with the observed temperature change and you are correct it is 80% of the heat of cooling iron from melting temp to 40°C

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3 minutes ago, SamLogan said:

Note the formula has been change with QoL Upgrade.

I don't see a change here. Looks like it is still valid for water:

400 000 * 4,179 * 32,1 = 53 658 360

It is quite close for DTU value of smelting iron (53 696 808)

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Just now, Angpaur said:

I don't see a change here. Looks like it is still valid for water:

400 000 * 4,179 * 32,1 = 53 658 360

It is quite close for DTU value of smelting iron (53 696 808)

I didn't calculate but the fact is with Space Industry the polluted water was overheat and with QoL, I can use it many times to refine steel.

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3 minutes ago, SamLogan said:

I didn't calculate but the fact is with Space Industry the polluted water was overheat and with QoL, I can use it many times to refine steel.

So what are temperatures of polluted water before and after smelting?

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1 minute ago, Angpaur said:

I don't see a change here. Looks like it is still valid for water:

400 000 * 4,179 * 32,1 = 53 658 360

It is quite close for DTU value of smelting iron (53 696 808)

Well, it is quite likely that the actual temperature change is a little bit bigger than 32,1 but smaller than 32,2

1 minute ago, SamLogan said:

I didn't calculate but the fact is with Space Industry the polluted water was overheat

No, I have been using Polluted Water for all my Refinery needs in SI, which also included several tons of Steel and even in Rocketry, I might have been using clean water instead however they have the same SHC (since Expressive? Well, does not matter) but never did it overheat. The supply in SI has been 40°C at least.

3 minutes ago, SamLogan said:

From the official patch notes :

which sure is vague, quite possibly referring to a bug which may more or less frequently occur. It is still a matter of fact that one can see in SI that the added heat is the very same as it is now..Those are not new values but old values which the gamepedia had since long.

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Still all seems to be as it was before. In the post you linked you wrote that water temperature increased by 56oC

DTU from steel smelting is: 93 566 480

400 000 * 4,179 * 56 =  93 609 600

4 minutes ago, SakuraKoi said:

Well, it is quite likely that the actual temperature change is a little bit bigger than 32,1 but smaller than 32,2

No, it is not quile likely ;) it is exactly 32,12300072

But we can assume that game is rounding this to 32,1.

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5 minutes ago, Angpaur said:

But we can assume that game is rounding this to 32,1.

No, not quite if you are referring to the temperature change being 32,1, there is just a very high chance that the result will be rounded down instead of up especially for Iron (or mayhaps the game never rounds up so that one has to reach the threshold of 0 instead of 5). In the thread linked I actually tested myself and had Copper and Gold 0,1 hotter than Sam.

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So, to summarize:

1) We have no clue what the displayed 67.1kJ means. 

2) The proper calculation is: 80% of (((OreMeltingTemp - OreStartTemp) * OreSpecificHeat * 100kg) / (CoolantSpecificHeat) / 400)

3) The 80%, 100kg and 400kg can be pulled out of the formula and be simplified down to multiplying the rest of the formula by 0.2 (or 20%).

4) 0.1°C has feelings, too. 

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