Metal Refinery heat data?

[SchlauFuchs]

good to know. thanks

[Lilalaunekuh]

Ok you shouldn´t do math while drunk but where is my mistake here:

 

If I use molten glass as coolant, it will heat up by ~10K for each shown Kj of energy (1170K if I refine steel which shows 117Kj / tested with other ingredients too).

Shouldn´t it take roughly 80Kj to heat up 400Kg molten glass ((0,2j/g)/K) by just one degree ?

(So 80Kj/K * 1170K = 93600Kj equal to 800times the shown energy output)

 

[SchlauFuchs]

maybe grams to kilograms conversion wrong here?

x K = g / 0.2 j = kg/ 0.2 kj 

x K = 400kg / 0.2 * 117Kj

x = 17K 

At least if I got my brains together today.

[0xFADE]

19 minutes ago, Lilalaunekuh said:

Ok you shouldn´t do math while drunk but where is my mistake here:

 

If I use molten glass as coolant, it will heat up by ~10K for each shown Kj of energy (1170K if I refine steel which shows 117Kj / tested with other ingredients too).

Shouldn´t it take roughly 80Kj to heat up 400Kg molten glass ((0,2j/g)/K) by just one degree ?

(So 80Kj/K * 1170K = 93600Kj equal to 800times the shown energy output)

 

From that screenshot the numbers worked out.  the 117 heat * 2 / the heat capacity which is .2 equals 1170

[SchlauFuchs]

why times 2?

[0xFADE]

I don't know why the *2 is there but it is working in each of the calculations.

[Lilalaunekuh]

3 minutes ago, 0xFADE said:

* 2 / the heat capacity which is .2

*2/0,2 are the 10K for each Kj I saw in my tests

(in the linked thread there was something strange:

LIQUID_COOLED_HEAT_PORTION = 0.8f;

)

But how much heat energy does it take to heat up 1kg of molten glass by 1K ?

(Shouldn´t it be 0,2Kj ? / 80Kj for 400Kg molten glass)

 

[SchlauFuchs]

((0,2j/g)/K) -> it requires 0.2 joules per gram to heat molten glass by 1K. Or 0.2KJ for a Kilogram, or 80kJ for 400kg - for one degree Celsius or Kelvin. Not sure where that factor 2 came in here.

[0xFADE]

Could be that.

400kg / 0.8(liquid_cooled_heat_portion) = 500.  500 * 2 is 1000kg

So the 2 could be from that 400kg being 500 kg after that heat portion value, and that is only half of 1000kg so you multiply it by 2

 

[Lilalaunekuh]

117Kj result in ~1170K temperature change of a liquid which should take 80Kj/K

(117Kj should heat the 400Kg molten galss liquid by not even 2 degree K)

 

=> There is something more than just a factor 2 missing

[SchlauFuchs]

1 minute ago, Lilalaunekuh said:

 

=> There is something more than just a factor 2 missing

Probably the dev's skills to properly apply physical equations?

[Lilalaunekuh]

Right now it´s our ability to derive an equation for the stuff happening behind the curtain.

=> Take the shown Kj and multiply them with 800 and you have the Kj added to your coolant.

 

[SchlauFuchs]

I will mention this in the wiki page.

[0xFADE]

The 2 is there because if you were heating up 1000kg of liquid with that 117 heat with a LIQUID_COOLED_HEAT_PORTION of 1.0 you wouldn't need to adjust it.  But you are only heating up half as much because of the .8 turning 400kg in to what counts as 500kg

[SchlauFuchs]

I think I might found a way the numbers make sense.  The value of 117kJ is not added per recipe execution, but per kg of coolant. and the factor 0.8 is just some devs idea of gaming physics.

[Lilalaunekuh]

Now I know why there is a factor 2:

800times the shown energy will be distributed between 400Kg of coolant resulting resulting in 2 times the shown heat energy FOR EACH Kg.

[SchlauFuchs]

and twice applied because why?

 

[Lilalaunekuh]

Thats the question

But the 800 times the total energy is a tested value. (or call it 2times the shown energy for each Kg of coolant)

 

 

[0xFADE]

117000j * 0.8 = 93600j

93600j / 400kg = 234

234 / .2(heat capacity) is 1170c increase

You don't need all that extra stuff when all you need to do is take the number * 2 and divide by heat capacity like I said early on but everyone has to show their work I guess.

[SchlauFuchs]

2 minutes ago, 0xFADE said:

117000j * 0.8 = 93600j

93600j / 400kg = 234

234 / .2(heat capacity) is 1170c increase

You don't need all that extra stuff when all you need to do is take the number * 2 and divide by heat capacity like I said early on but everyone has to show their work I guess.

just that the formula is J/g or kJ/kg - you show us here J/kg which swallows a factor 1000. 

[Lilalaunekuh]

15 minutes ago, 0xFADE said:

117000j * 0.8 = 93600j

93600j / 400kg = 234

234 / .2(heat capacity) is 1170c increase

You don't need all that extra stuff when all you need to do is take the number * 2 and divide by heat capacity like I said early on but everyone has to show their work I guess.

You get more like 93,6Mj that´s the reason we are talking here ^^

[SchlauFuchs]

I have added the knowledge about all of this into https://oxygennotincluded.gamepedia.com/Metal_Refinery. Please review and comment

[Lilalaunekuh]

1.Ok the kj for each item you can refine is not really usefull:

A) show the added Kj for each Kg of liquid (2x times the shown value)

B)show the total added Heat energy (800x times the shown value)

2.Maybe it would be nice to know how much energy each refining process needs (Kj / not W)

 

Everything else looks really nice and I like your tables

 

[SchlauFuchs]

re 2), I understand that the Refinery consumes 1200W power, but is the energy release of a recipe defining how much energy is consumed, i.e. does it take longer for Steel than for Gold to complete?

B) I was doing some Excel calculations, if you consider the released heat value as per kg, not per batch of coolant, it is just a factor of 2 then.

 

[Argelle]

Nice work, I have one only question:

Quote

The temperature change to coolant depends on metal that is being refined and decreases linearly with coolant's specific heat.

Is this heat capacity? (on first approximation as the formula is complex as you state in the wiki page)