User:SolinNull/Applied power math/tr - Revision history

Bilka: Bilka moved page Tutorial:Applied power math/tr to User:SolinNull/Applied power math/tr without leaving a redirect: No WIP pages in the main namespace

2018-10-17T13:47:34Z

Bilka moved page Tutorial:Applied power math/tr to User:SolinNull/Applied power math/tr without leaving a redirect: No WIP pages in the main namespace

← Older revision		Revision as of 13:47, 17 October 2018
(No difference)

SolinNull at 07:19, 6 October 2018

2018-10-06T07:19:01Z

← Older revision		Revision as of 07:19, 6 October 2018
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	[[File:Accumulator-example.png\|800px]]		[[File:Accumulator-example.png\|800px]]

	~~We started with 8 MJ in the coal~~, ~~but only 4 MJ ended up in the accumulator~~.		Kömürün içinde potansiyel olarak 8MJ enerji vardı, lakin aküde 4MJ enerji oluştu.
	~~This is because, like in the real world, some energy is lost along the way as~~		Neden böyle oldu? Factorio'nun bilimi kullanmasından kaynaklanıyor. Gerçek hayatta olduğu gibi Factorio'da da enerji kaybı meydana gelir. Bu durumda akümüze kömürün enerjinin %50'sini verebildik. Factorio'da kayıp kazanda gerçekleşiyor. Kullandığınız eşyanın detaylarını incelerseniz
	~~waste~~. ~~In this case, only~~ 50~~% is making it into our accumulator~~. In Factorio,		verimini görebilirisiniz. %50'lik enerji kaybını, enerji üretilirken oluşan kirlilik olarak düşünebilirsiniz.
	~~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..~~.

	[[File:boiler-details.png]]		[[File: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~~		İlk soruyu "Bir Fabrikayı çalıştırmak için ne kadar kömür gerekiyor?" cevaplamak için gerekli tüm bilgilere sahibiz. Peki, gelin şu soruyu daha belirgin bir şekilde tekrar soralım: Fabrikamızı 'bir saniye' çalıştırmak için kaç tane kömüre ihtiyacımız var?
	~~more precise~~: ~~how much coal do we need ''per second~~'' ~~to power our factory~~?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">

SolinNull at 07:07, 6 October 2018

2018-10-06T07:07:41Z

← Older revision		Revision as of 07:07, 6 October 2018
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	{{Languages}}		{{Languages}}
	~~In this tutorial we'll be answering the question: '''how much coal is needed to power a factory~~?~~'''~~		Bu sayfada "Bir Fabrikayı çalıştırmak için ne kadar kömür gerekiyor?" sorusunu cevaplayacağız.

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

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

	~~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~~.		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.

	~~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~~.		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.

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

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

	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:~~		Joule, enerjinin birimlerinden birisidir. 1 kilojoule(kJ), 1000J'e eşittir.

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

	~~So to run our factory at~~ 0.~~3 MW, we need to consume~~ 0.~~3 MJ every second~~.		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.

	== Quiz ==		== Quiz ==

	~~Assuming perfect efficiency~~, ~~how long could one piece of coal~~ (~~8 MJ~~) ~~run~~		Mükemmel verimle çalıştığımızı düşünürsek, bir parça kömür (8MJ); bir radarı (0.3MW) kaç saniye çalıştırır?
	~~our single radar~~ (0.~~3 MW~~)?

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Expand to reveal answer~~'''		'''Cevabı görmek için tıklayınız'''
	<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

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

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

	<div class="toccolours mw-collapsible mw-collapsed">		<div class="toccolours mw-collapsible mw-collapsed">
	'''~~Expand to reveal answer~~'''		'''Cevabı görmek için tıklayınız'''
	<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 ~~megawatts~~.		1.6 megawatt üretir.
	</div>		</div>
	</div>		</div>

	'''~~BONUS~~!''' A ton ~~of real world coal~~ [http://hypertextbook.com/facts/2006/LunChen.shtml ~~contains about~~ 21 GJ]. ~~How much does a~~		'''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?
	~~piece of~~ Factorio ~~coal weigh~~?

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

	~~8 MJ~~ / ~~21000 MJ~~ = ~~weight ÷ 1000 kG~~		8MJ / 21000MJ = kütle / 1000kG
	~~weight~~ = (~~8 MJ ÷ 21000 MJ~~) ~~× 1000 kG~~		kütle = (8MJ / 21000MJ) * 1000kG
	~~weight~~ = 0.~~38 kG~~		kütle = 0.38kG

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

	== ~~Efficiency~~ ==		== Verim ==

	~~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~~.		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.

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

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

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

SolinNull: Created page with "{{Languages}} 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..."

2018-10-06T06:06:07Z

Created page with "{{Languages}} 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..."

New page

{{Languages}}
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.

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

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.

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

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)?

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

Nearly 27 seconds.
</div>
</div>

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">
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
8 MJ = MW × 5
MW = 8 MJ ÷ 5
MW = 1.6

1.6 megawatts.
</div>
</div>

'''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">
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
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...
</div>
</div>

== 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.

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

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]]

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...

[[File: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?

<div class="toccolours mw-collapsible mw-collapsed">
'''Aside: why did the radar run for 20s and not 13.5s (50% of 27s)?'''
<div class="mw-collapsible-content">
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.
</div>
</div>

== Quiz ==

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

<div class="toccolours mw-collapsible mw-collapsed">
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
Since <code>J = W × s</code>, 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''' (<code>5 × 8 MJ = 40 MJ</code>).
</div>
</div>

'''BONUS!''' How many [[Mining|mining drills]] are needed to produce that much coal?

<div class="toccolours mw-collapsible mw-collapsed">
'''Expand to reveal answer'''
<div class="mw-collapsible-content">
Per the linked page, an electric mining drill produces ~0.5/s, so we'll need '''10 electric mining drills''' (<code>10 × 0.5 = 5</code>).
</div>
</div>

'''SECOND BONUS!''' The average American household uses [https://www.eia.gov/tools/faqs/faq.php?id=97&t=3 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 ([[Steam_engine|900 kW]]), assuming the power is used at a constant rate? (Assume 30 days in a month.)

<div class="toccolours mw-collapsible mw-collapsed">
'''Expand to reveal answer'''
<div class="mw-collapsible-content">

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.
</div>
</div>