Research topic about energy

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EditorRUS
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Research topic about energy

Post by EditorRUS »

This might be made already by someone, but I don't care.
Testing lab:
Image
It's just a water tank.

1. Heat capacity of water.
This one is pretty easy to begin with.
Q = C * M * (T1-T0) // C
Q/C = M * T1-T0 // Q
C = (M * (T1-T0)) / Q
Heat capacity formula: C = M * (T1-T0) / Q * N
C is heat capacity; T1, T0 is end temp and start temp and Q is amount of heat (energy used). N is efficiency (50% for boiler).

Image
I will use 1 coal as fuel (8 MJ of energy so Q is known) until it burns out.

10 units are heated up on 85 degrees...
10 units are heated up on 85 degrees...
10 units are heated up on 85 degrees...
10 units are heated up on 85 degrees...
10 units are heated up on 60 degrees...
Using 8 MJ of energy

So we've got:
40 * 85 + 10 * 60
C = --------------------------------
8 000 * .5

C = 1 kJ per 1 degree by 1 unit.
P.S. Real world water takes 4.2 kJ per 1 degree per 1 kg. That's one of the biggest heat capacities.

2. How much energy yields a steam engine when consuming X units of water of Y degrees?
Pretty tough, but no.
Let's use system from previous research and pump 10 units of 100 degrees water into a steam engine and see how much energy it will yield.

Image
Image
It yielded 850 kJ of energy for 10 units of 100 degrees water. So steam engines' efficiency is (850 / (10 * 85 * 1)) * 100% = 100%. So we can assume it will yield 85 kJ of energy for 1 unit of 100 degree water.
Image
Image
Yep.

So formula is X * (Y-15). In kJ.

3. Steam engines' consumption of water.
That's a bit tricky, but can be calculated.
Maximum output of steam engine is 6*85 + 1 = 510 kW (not 511). Why 6*85? There's a formula: max(6x, 0) where X is cur. temp of water minus 15. You can find it yourself, but I did that by just having two water tanks with heated water and approximating the function.
It's really obvious: max amount of water is 10, steam engine will yield 850 kJ in case of 100 degrees water. It will be eaten in 1 + 2/3 seconds, so consumption of steam engine of water depends on, well, consumption of energy. The fastest is ~6 units of water per sec, the slowest is 0 (obviously)

4. Offshore pump. How many units of water it yields per sec?
That's hard. I am not looking in source code or wiki because I want to find it out hard way.
It's time for that. THE STOPWATCH.
...
~16.5 seconds for filling one water tank up to 1000.

1000 / ~16.5 = ~60 water units per sec. But pipes currently can hold only 10 units of water so one pipe is filled in 0.17 seconds? No need to say that's really fast.

5. Ideal boiler/pump ratio.
So now this is time for something useful in game.
One boiler uses 390 kJ per seconds for heating water. If we make straight pump to boiler chain, we'll get next:
60 units of water requires 60 kJ to be heated on 1 degree.
So 390 kJ / 60 = +6.5 degrees.

In order to fully heat up the water, we need to heat it up to 100 degrees (+85)
So 390X / 60 = +85
X = 13,076923076923076923076923076923 (about ~13-14 boilers)

1 : 13-14 pump : boilers

Results:
Maximum output:
Pump: 60 units of water per sec. One unit of 100 degrees water holds 85 kJ of usable energy therefore one pump should be used for 5.1 MW
Steam engine: 510 kW per sec.
Boiler: 390 kW per one unit.

Max. capacity if we use water tank with hot water as accumulator: 2 500 units of 100 degrees water: 212 500 kJ. That's WAY MORE than usual accumulator (42.5 times better)

Ideal ratio of pump : steam engine is 5 100 / 510 = 10
So, we get: 1 pump, 13-14 boilers and 10 steam engine to get 5.1 MW.

I hope you enjoyed reading that. Or not.
Last edited by EditorRUS on Thu Sep 25, 2014 4:53 pm, edited 1 time in total.

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DerivePi
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Re: Research topic about energy

Post by DerivePi »

I believe the boiler output energy is 390 kW (the input energy is therefore 780kW).

Instead of using a stopwatch, I think you can set up a timed system using an assemblers production time and chest. Link an inline pump on the same electric grid as the assembler and let it run for a minute and then unplug it. Count the number of copper cables say and then compare it to the volume of water pumped in (you will need 2 liquid storage containers).. That should give you a more accurate rate.

The established ratio is 1 pump to 14 boilers to 10 steam engines.

EditorRUS
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Re: Research topic about energy

Post by EditorRUS »

Instead of using a stopwatch, I think you can set up a timed system using an assemblers production time and chest. Link an inline pump on the same electric grid as the assembler and let it run for a minute and then unplug it. Count the number of copper cables say and then compare it to the volume of water pumped in (you will need 2 liquid storage containers).. That should give you a more accurate rate.
Looks fun. But I meant Offshore pump, not small one. Offshore pump is trickier.
I believe the boiler output energy is 390 kW (the input energy is therefore 780kW).
Easy to check.
/me starts his stopwatch
~10 seconds for one coal. In this case you are right. 8 MJ were wasted in ~10 seconds and that means it's really 780. Gonna rewrite something.


I decided to make stopwatch test again.
Image
And you know what? ~1000 units of water in ~16.5 seconds (I started at 10). In this case it's about 60-61.

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ssilk
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Re: Research topic about energy

Post by ssilk »

Filling the storage tank is not a save measurement, cause the filling speed is logarithmic (fast in the beginning, slower at the end).

Besides that, the used energy is explained here https://forums.factorio.com/wiki/index.php?title=Fuel
And the effectivity here https://forums.factorio.com/wiki/inde ... tle=Boiler
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Re: Research topic about energy

Post by DaveMcW »

ssilk wrote:Filling the storage tank is not a save measurement, cause the filling speed is logarithmic (fast in the beginning, slower at the end).
It is linear until 2000 units.

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Re: Research topic about energy

Post by ssilk »

But can store only 1000. 8-)
Edit: Perhaps I'm missing something here. All my experiments with pipes showed me, that the filling speed depends on the pressure and that depends on the pump, which fills it.
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Re: Research topic about energy

Post by EditorRUS »

Edit: Perhaps I'm missing something here. All my experiments with pipes showed me, that the filling speed depends on the pressure and that depends on the pump, which fills it.
Fluid tank is just a pipe with huge volume.
Filling the storage tank is not a save measurement, cause the filling speed is logarithmic (fast in the beginning, slower at the end).
In this case that means common pipes aren't filled linearly too.

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Re: Research topic about energy

Post by Tartaros »

i cant see pictures... help

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