Talk:Power production: Difference between revisions
No edit summary |
|||
(6 intermediate revisions by 3 users not shown) | |||
Line 16: | Line 16: | ||
:The differnece is rather clear: One is about producing power, one is about transporting power and the mechanics of it. Another example of that is [[oil processing]] and [[fluid system]], one is about production of certain fluids, one about transportation of fluids. Due to this being a common pattern, the page being popular and long, the pages will not be merged. -- [[User:Bilka|Bilka]] ([[User talk:Bilka|talk]]) - <span style="color:#FF0000">Admin</span> 14:59, 10 May 2018 (UTC) | :The differnece is rather clear: One is about producing power, one is about transporting power and the mechanics of it. Another example of that is [[oil processing]] and [[fluid system]], one is about production of certain fluids, one about transportation of fluids. Due to this being a common pattern, the page being popular and long, the pages will not be merged. -- [[User:Bilka|Bilka]] ([[User talk:Bilka|talk]]) - <span style="color:#FF0000">Admin</span> 14:59, 10 May 2018 (UTC) | ||
== Power Production - Feedback loop == | |||
The situation discussed is a feedback loop. In a limited boiler, steam engine, mining drill (or pumpjack/refinery/chemical plant) system it would be a positive feedback loop due to the fact the fuel production cycle in and of itself produces sufficient, indeed more, fuel than is needed to produce that fuel. However, in a real system that is overloaded, the overloading elements (the factory) change the amplification factor in the fuel production cycle so that it requires more energy production than the fuel cycle makes available. That is, even thought he mining drill is a net gain in energy, it is not producing enough net gain to overcome the drain on the power grid from the rest of the factory, leading to a situation where there is less and less power available to the mining drill over time. This reduction of the amplification ratio is what makes it a negative feedback system not positive. The simplified model of only the fuel cycle and the power conversion cycle starts with a positive feedback ratio. That is mining drill mines coal at a base rate of 0.525/s Yielding 8 MJ at a cost of 47250 J or an amplification factor of (8MJ / 47250 J) = 169.312. however, the conversion in a boiler is only 50% efficient meaning the 8 MJ of coal only produces 4MJ of power reducing the amplification to (4 MJ / 47250 J) = 84.656. If we further consider the use of at least 1 additional iron mine and one copper mine, the amplification is now 4 MJ / (3 * 47250 ) = 28.218. As more systems are added to the power grid the power consumption begins to exceed the net power yield of this system leading to a negative feedback situation or more specifically a gain of < 1.0. [[User:AnthonyQBachler|AnthonyQBachler]] ([[User talk:AnthonyQBachler|talk]]) 11:48, 17 May 2018 (UTC) | |||
Sorry, just woke up. The number 47250 above should be 171428.571, so the amplification ratios are ~ 3.6 times worse. Nothing else is off. [[User:AnthonyQBachler|AnthonyQBachler]] ([[User talk:AnthonyQBachler|talk]]) 12:20, 17 May 2018 (UTC) | |||
:''See [[User talk:AnthonyQBachler]]'' -- [[User:Bilka|Bilka]] ([[User talk:Bilka|talk]]) - <span style="color:#FF0000">Admin</span> 12:23, 17 May 2018 (UTC) | |||
== Optimal ratio == | |||
Correct me if I'm wrong, in the section it says the optimal ratio is 0.84, and in the example it gives 200 accumulators vs. 240 solar panels and says it's differs with 20 extra solar panels. | |||
But as the optimal ratio is 0.84, for 200 accumulators you need 238 panels (200 / 0.84 = 238.09), isn't it needs only 2 extra solar panels rather than 20 or is it my misunderstanding? | |||
Ok I got it, I missed the 10MW, which should be multiplied, my bad. [[User:Kurax|Kurax]] ([[User talk:Kurax|talk]]) 02:07, 11 July 2018 (UTC) | |||
: I think you were right. A solar panel produces 42kW of power on average so 238 of them already produce 10MW (42kW * 238 = 9996kW). --[[User:Lk|Lk]] ([[User talk:Lk|talk]]) 10:14, 12 May 2019 (UTC) |
Latest revision as of 10:14, 12 May 2019
http://www.factorioforums.com/forum/viewtopic.php?f=6&t=2015&p=14783#p14783
Posting For Discussion http://www.factorioforums.com/forum/viewtopic.php?f=8&t=5585
Move proposal
In order to be consistent with other wikis, I recommend moving "Power production" to "Power Production". Most wikis capitalize each word in the title. It's not really necessary, but it's nice to be consistent. Also, it seems that page "Power Production" has a redirect to Power production, so it's not like links are going to fail, but as I said, it's best to be consistent.--Twisted Code (talk) 11:25, 7 June 2015 (CEST)
- The discussion for this decision is here: http://www.factorioforums.com/forum/viewtopic.php?f=50&t=6111
- I added a posting there! Ssilk (talk) 15:14, 8 June 2015 (CEST)
Remove/move/rewrite/rename/merge proposal
It is not clear what makes this different from Electric system. Perhaps these should be merged, or perhaps the difference should be clarified. -- SafwatHalaby (talk) 13:49, 10 May 2018 (UTC)
- The differnece is rather clear: One is about producing power, one is about transporting power and the mechanics of it. Another example of that is oil processing and fluid system, one is about production of certain fluids, one about transportation of fluids. Due to this being a common pattern, the page being popular and long, the pages will not be merged. -- Bilka (talk) - Admin 14:59, 10 May 2018 (UTC)
Power Production - Feedback loop
The situation discussed is a feedback loop. In a limited boiler, steam engine, mining drill (or pumpjack/refinery/chemical plant) system it would be a positive feedback loop due to the fact the fuel production cycle in and of itself produces sufficient, indeed more, fuel than is needed to produce that fuel. However, in a real system that is overloaded, the overloading elements (the factory) change the amplification factor in the fuel production cycle so that it requires more energy production than the fuel cycle makes available. That is, even thought he mining drill is a net gain in energy, it is not producing enough net gain to overcome the drain on the power grid from the rest of the factory, leading to a situation where there is less and less power available to the mining drill over time. This reduction of the amplification ratio is what makes it a negative feedback system not positive. The simplified model of only the fuel cycle and the power conversion cycle starts with a positive feedback ratio. That is mining drill mines coal at a base rate of 0.525/s Yielding 8 MJ at a cost of 47250 J or an amplification factor of (8MJ / 47250 J) = 169.312. however, the conversion in a boiler is only 50% efficient meaning the 8 MJ of coal only produces 4MJ of power reducing the amplification to (4 MJ / 47250 J) = 84.656. If we further consider the use of at least 1 additional iron mine and one copper mine, the amplification is now 4 MJ / (3 * 47250 ) = 28.218. As more systems are added to the power grid the power consumption begins to exceed the net power yield of this system leading to a negative feedback situation or more specifically a gain of < 1.0. AnthonyQBachler (talk) 11:48, 17 May 2018 (UTC)
Sorry, just woke up. The number 47250 above should be 171428.571, so the amplification ratios are ~ 3.6 times worse. Nothing else is off. AnthonyQBachler (talk) 12:20, 17 May 2018 (UTC)
- See User talk:AnthonyQBachler -- Bilka (talk) - Admin 12:23, 17 May 2018 (UTC)
Optimal ratio
Correct me if I'm wrong, in the section it says the optimal ratio is 0.84, and in the example it gives 200 accumulators vs. 240 solar panels and says it's differs with 20 extra solar panels.
But as the optimal ratio is 0.84, for 200 accumulators you need 238 panels (200 / 0.84 = 238.09), isn't it needs only 2 extra solar panels rather than 20 or is it my misunderstanding?
Ok I got it, I missed the 10MW, which should be multiplied, my bad. Kurax (talk) 02:07, 11 July 2018 (UTC)