Tutorial:Applied power math

Bu sayfada "Bir Fabrikayı çalıştırmak için ne kadar kömür gerekiyor?" sorusunu cevaplayacağız.

Her şeyden önce, fabrikamızın ne kadar enerji tükettiğini bilmeliyiz. Bunu görmek için elektrik teline tıklayıp electricity sekmesini incelemeniz kafidir.

Yukarıdaki görselde görülebileceği üzere radar 300kW'lık güç kullanıyor (299kW yazmasına aldırış etmeyin. Yuvarlama hatasından kaynaklı). Factorio gerçek bilimi kullanır. Watt(W) gücün birimidir. 1 Kilowatt(kW), 1000W ve 1 megawatt(MW) 1000kW'a eşittir. Eğer fabrikanızı yeterince büyütebilirseniz bu bilgi de işinize yarayacaktır: 1 gigawatt(GW) da 1000MW'a eşittir.

Fabrikamızın ne kadar güç harcadığını anlamış olduk. Eğer fabrikamızı tam gaz işletmek istiyorsak 300kW güç üretimini sürdürmeliyiz. Daha kolay işlem yapmak için, 300kW'ı 0.3MW'a dönüştüreceğiz.

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.

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.

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

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...

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.