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Bu sayfada "Bir Fabrikayı çalıştırmak için ne kadar kömür gerekiyor?" sorusunu cevaplayacağız.
In this tutorial we'll be answering the question: '''how much coal is needed to power a factory?'''


Her şeyden önce, fabrikamızın ne kadar enerji tükettiğini bilmeliyiz. Bunu görmek için elektrik teline tıklayıp electricity sekmesini incelemeniz kafidir.
First off, we need to know: how much power does our factory use? That's easy - you can check the electricity tab by clicking on a power pole.


[[File:power-details.png|800px]]
[[File:power-details.png|800px]]


Yukarıdaki görselde görülebileceği üzere fabrikamız 300kW'lık güç tüketiyor. (299kW yazmasına aldırış etmeyin. Yuvarlama hatasından oluşmuş). Factorio bilimi kullanır. Watt(W) gücün birimidir. 1 Kilowatt(kW), 1000W'a ve 1 megawatt(MW) da 1000kW'a eşittir. Eğer fabrikanızı yeterince büyütebilirseniz bu bilgi de işinize yarayacaktır: 1 gigawatt(GW) da 1000MW'a eşittir.
Here we see one radar using 300kW (the 299kW is a rounding error). Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.


Fabrikamızın ne kadar güç harcadığını gördük. Şimdi, fabrikamızı tam gaz işletmek istiyorsak 300kW güç üretimini sürdürmeliyiz. Daha kolay işlem yapmak için - birim dönüştüreceğiz - 300kW'ı 0.3MW olarak alacağız.
So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.


Akabinde yeni bir soru! Kömürden ne kadar güç elde edilir? Bu soruyu yanıtlamak çok kolay, yapmamız gereken imleci üzerine getirip beklemek akabinde 8MJ güç elde edilebildiğini görmüş olacağız.
Next question! How much power is stored in coal? That's also easy, because it tells us when we hover over it: 8 MJ.


[[File:coal-joules.png|300px]]
[[File:coal-joules.png|300px]]


Joule, enerjinin birimlerinden birisidir. 1 kilojoule(kJ), 1000J'e eşittir.
A Joule is another standard measure of stored energy. As with Watts a kilojoule (kJ) is 1000 J, and so on. There is a fixed relationship between Joules and Watts. '''1 Joule can provide 1 Watt of power for 1 second.''' As a formula:


Enerji = Güç * Zaman
    J = W × s
J = W * s


Eğer fark ettiyseniz; Zaman'ı her zaman 1 saniye olarak alırsak. Enerji ve Güç birbirlerinin aynısı oluyor. Yani fabrikamızı idame ettirmek için 0.3MW güce ihtiyacımız var ise bir saniyede 0.3MJ enerji tüketmemiz gerekecektir.
So to run our factory at 0.3 MW, we need to consume 0.3 MJ every second.


== Quiz ==
== Quiz ==


Mükemmel verimle çalıştığımızı düşünürsek, bir parça kömür (8MJ); bir radarı (0.3MW) kaç saniye çalıştırır?
Assuming perfect efficiency, how long could one piece of coal (8 MJ) run
our single radar (0.3 MW)?


<div class="toccolours mw-collapsible mw-collapsed">
<div class="toccolours mw-collapsible mw-collapsed">
'''Cevabı görmek için tıklayınız'''
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
<div class="mw-collapsible-content">
     8 MJ = 0.3 MW * s
     8 MJ = 0.3 MW × s
       s = 8 MJ / 0.3 MW
       s = 8 MJ ÷ 0.3 MW
       s = 26.67
       s = 26.67


Yaklaşık 27 saniye çalıştırır.
Nearly 27 seconds.
</div>
</div>
</div>
</div>


Kusursuz verime sahip olduğumuzu varsayarsak, bir tane kömür, 5 saniye içinde, azami güç tüketimde kaç watt güç üretir?
Assuming perfect efficiency, what is the maximum size factory (in watts)
a single piece of coal could run for 5 seconds?


<div class="toccolours mw-collapsible mw-collapsed">
<div class="toccolours mw-collapsible mw-collapsed">
'''Cevabı görmek için tıklayınız'''
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
<div class="mw-collapsible-content">
     8 MJ = MW * 5
     8 MJ = MW × 5
       MW = 8 MJ / 5
       MW = 8 MJ ÷ 5
       MW = 1.6
       MW = 1.6


1.6 megawatt üretir.
1.6 megawatts.
</div>
</div>
</div>
</div>


'''İlave Soru!''' Gerçek hayatta, bir ton kömür [http://hypertextbook.com/facts/2006/LunChen.shtml yaklaşık 21 GJ] enerji içerir. Bu durumda bir tane Factorio kömürü kaç kg'dır?
'''BONUS!''' A ton of real world coal [http://hypertextbook.com/facts/2006/LunChen.shtml contains about 21 GJ]. How much does a
piece of Factorio coal weigh?


<div class="toccolours mw-collapsible mw-collapsed">
<div class="toccolours mw-collapsible mw-collapsed">
'''Cevabı görmek için tıklayınız'''
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
<div class="mw-collapsible-content">
Enerji oranları ile kütle oranları birbirlerini eşler, yani:
The energy ratios will match the weight ratios, so:


     8MJ / 21000MJ = kütle / 1000kG
     8 MJ / 21000 MJ = weight ÷ 1000 kG
             kütle = (8MJ / 21000MJ) * 1000kG
             weight = (8 MJ ÷ 21000 MJ) × 1000 kG
             kütle = 0.38kG
             weight = 0.38 kG


Yaklaşık 400g! Acaba bizim karakter bu kadar ağır kömürü nasıl taşıyor? Cevap yok...
About 400g! Still doesn't explain how our character can carry so much of it...
</div>
</div>
</div>
</div>


== Verim ==  
== Efficiency ==  


Hadi bu bilgileri test edelim! Bir radarı 27 saniye boyunca çalıştırabilmeliyiz, değil mi? Bir kazan, bir buhar makinesi ve bir radar ile sistemi kurduk. Şimdi, kazanın içine bir parça kömür koyacağız ve radarın kaç saniye çalışacağını göreceğiz.
Let's test it! We should be able to run a radar for 27 seconds, right? Using a setup with a single boiler, steam engine, and radar, we can drop a single piece of coal in the boiler and see how long the radar runs.


[[File:radar-example.png|800px]]
[[File:radar-example.png|800px]]


Zar zor 20 saniye çalıştı! Ne oldu? Beklediğimizden daha az enerji ürettik. Bir de akü kullanarak ölçelim:
We barely got 20s! What's wrong? We are generating less power than expected.
Let's measure using an accumulator.


[[File:Accumulator-example.png|800px]]
[[File:Accumulator-example.png|800px]]

Revision as of 07:07, 6 October 2018

In this tutorial we'll be answering the question: how much coal is needed to power a factory?

First off, we need to know: how much power does our factory use? That's easy - you can check the electricity tab by clicking on a power pole.

Power-details.png

Here we see one radar using 300kW (the 299kW is a rounding error). Factorio uses real science here. A Watt (W) is a measure of energy transfer. A kilowatt (kW) is 1000 W, a megawatt (MW) is 1000 kW, and if you're lucky enough to ever make a factory big enough, a gigawatt (GW) is 1000 MW.

So that answers the question of how much power our factory uses. To keep our factory running at full speed we need to maintain 300 kW. For ease of comparison, we'll convert that to 0.3 MW.

Next question! How much power is stored in coal? That's also easy, because it tells us when we hover over it: 8 MJ.

Coal-joules.png

A Joule is another standard measure of stored energy. As with Watts a kilojoule (kJ) is 1000 J, and so on. There is a fixed relationship between Joules and Watts. 1 Joule can provide 1 Watt of power for 1 second. As a formula: 

   J = W × s

So to run our factory at 0.3 MW, we need to consume 0.3 MJ every second.

Quiz

Assuming perfect efficiency, how long could one piece of coal (8 MJ) run our single radar (0.3 MW)?

Expand to reveal answer 

   8 MJ = 0.3 MW × s
      s = 8 MJ ÷ 0.3 MW
      s = 26.67

Nearly 27 seconds.

Assuming perfect efficiency, what is the maximum size factory (in watts) a single piece of coal could run for 5 seconds?

Expand to reveal answer 

   8 MJ = MW × 5
     MW = 8 MJ ÷ 5
     MW = 1.6

1.6 megawatts.

BONUS! A ton of real world coal contains about 21 GJ. How much does a piece of Factorio coal weigh?

Expand to reveal answer

The energy ratios will match the weight ratios, so: 

   8 MJ / 21000 MJ = weight ÷ 1000 kG
            weight = (8 MJ ÷ 21000 MJ) × 1000 kG
            weight = 0.38 kG

About 400g! Still doesn't explain how our character can carry so much of it...

Efficiency

Let's test it! We should be able to run a radar for 27 seconds, right? Using a setup with a single boiler, steam engine, and radar, we can drop a single piece of coal in the boiler and see how long the radar runs.

Radar-example.png

We barely got 20s! What's wrong? We are generating less power than expected. Let's measure using an accumulator.

Accumulator-example.png

We started with 8 MJ in the coal, but only 4 MJ ended up in the accumulator. This is because, like in the real world, some energy is lost along the way as waste. In this case, only 50% is making it into our accumulator. In Factorio, this loss happens in the boiler, and it actually tells us this in the item details! You can think of the waste 50% as going into the pollution it generates...

Boiler-details.png

Now we have everything we need to answer our initial question: how much coal do we need to power our factory? Well, the question actually needs to be a bit more precise: how much coal do we need per second to power our factory?

Aside: why did the radar run for 20s and not 13.5s (50% of 27s)?

Because it doesn't turn on and off instantly. As you can see in the graph, it ramps up then ramps down. Technically, if you measured the area under the curve it would give you the expected 4 MJ.

Quiz

How much coal per second is needed to power a 20 MW factory?

Expand to reveal answer

Since J = W × s, a 20 MW factory will consume 20 MJ/s. We'll need 40 MJ/s input due to the 50% boiler efficiency loss. Since coal contains 8 MJ, we'll need 5 coal per second (5 × 8 MJ = 40 MJ).

BONUS! How many mining drills are needed to produce that much coal?

Expand to reveal answer

Per the linked page, an electric mining drill produces ~0.5/s, so we'll need 10 electric mining drills (10 × 0.5 = 5).

SECOND BONUS! The average American household uses 3.2 GJ in a month (typically expressed as 888 kilowatt hours). How much Factorio coal would be needed to power one of these homes for that period, and how many homes could a steam engine support (900 kW), assuming the power is used at a constant rate? (Assume 30 days in a month.)

Expand to reveal answer 

   3.2 GJ = 3,200 MJ
   3200 MJ ÷ (8 MJ × 50%) = 800

   3200 MJ ÷ 30 days ÷ 24 hours ÷ 60 minutes ÷ 60 seconds = 1,235 W
   900,000 ÷ 1235 = 728

800 pieces of coal per household, and a single steam engine could power 728 households by approximate.

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