how many reactors takes for full consuption of half conveyor of uranium?

[napouser]

how many reactors takes for full consuption of half conveyor of uranium?



hello

assuming a megabase of 10000spm that requires 50gw of power lets say 400 reactors in complex of 4. 1 reactor is 40mw 4 is 160 but with bonus is 120x4 so 480. lets say 0.5gw x100 of these complex



apparently it seems that 1/2 conveyor of uranium is plenty to feed them + trains + some left over uranium



how many reactors u need to build to exceed the 22.5 rate?



well i ve done some calculations but iam not sure they are correct so i am doing some live tests by turning on and off some reactors to see the balance around 646 reactors



if those calculations end up producing a number lets say 646 then if u build 647 eventually ur uranium buffers will be depleted

if u have 645 then ur buffers eventually will be filled

naturally it will take years for the buffers to stabilize so its better to predict it safely with mathematics. so the question is if the following is correct





ok so if have nothing to do with ur life and u want to waste some time lets dive in into reactor uranium cost

(btw if u making a 10000 megabase its really bad idea to go uranium since solar is ups friendly)



assuming half conveyor 22.5 ore is = A ores per second





A/10 x120% equals B the output uranium/s (lets assume u get 100% bad uranium and 0% good uranium for now)



2) B/20 x140% x10 equals to C cells /s (assuming the receipe requires 20 bad 0 good to make 10 cells)



3) C equals to D used sells /s (after 300secs)



4) D/5 x3 x120% equals to E uraniums/s from recycling



5) repeat steps 2-4 infinite times to get the total value of all uranium produced by constant recycling



6) assuming the repetitive process gives 50% of cells each circle then after infinite repetitions u will end up with ~2x the original amount D

so D + 1/2D + 1/4D + 1/8D + ..... = ~ 2D = E (assuming 60% u can end up with 3D assuming 30% u can end up with 1.5D etc)



7) assuming E amount of cells/s u can feed E reactors /sec

the first sec u feed E the 2nd second u feed +E more then each sec +E



8) the reactor can survive for 200 secs without refelling so u can feed 200xE equals G reactors total . after that u stop feeding new reactors and start from scratch. so G is ur limit. thats the number of reactors we are looking for. GREAT SUCCESS. wait no. still needto account the good uranium



9) back to step 2 the cells are being made by 19 bad 1 good. so now we have to count good. back to step 1

10) good are made by kovarex 40good + 5 bad = 41good +2 bad = > 3bad make 1 good . +120% => 3b = 1.2g => 2.5b = 1g



11) so now step 2 becomes B/21.5 x140% x10 equals to C cells /s



12) also need to count the uranium lost on trains. assuming 300 trains the consumption should be around 1 fuel per 5 secs so 2.5 uranium /s for trains needs to be substructed from reactor recources





here is a excel with some calculations



https://docs.google.com/spreadsheets/d/ ... sp=sharing





tl:dr

number should be 662 with 0 trains or 613 with around 300 trains

some1 confirm/deny?

[mrvn]

It's not the numbers of reactors that is relevant but the number of GW of power you consume. Running twice the number of reactors but only fueling it every 400s is more efficient. Although the difference becomes rather small at 640 reactors vs 1280 reactors.

But you should calculate this for an infinitely large reactor too and then show how many GW one belt of ore will produce.

[napouser]

It's not the numbers of reactors that is relevant but the number of GW of power you consume. Running twice the number of reactors but only fueling it every 400s is more efficient. Although the difference becomes rather small at 640 reactors vs 1280 reactors.

But you should calculate this for an infinitely large reactor too and then show how many GW one belt of ore will produce.

Ur consumption is irrelevant since reactors burn fuel regardless of using the power on not.

And i dont get ur idea of feeding every 400? How are u gonna do that?

[astroshak]

22.5 fuel cells per second, times 200 seconds per cell per reactor, yields 4500 reactors.

How you choose to array those reactors, how much power you withdraw from that power plant, is up to you.

FWIW, 2xN is the optimal (vanilla) reactor layout. Every time you start a new 2xN instead of extending it by those reactors costs you 160 MW of potential power.

[FuryoftheStars]

It's not the numbers of reactors that is relevant but the number of GW of power you consume. Running twice the number of reactors but only fueling it every 400s is more efficient. Although the difference becomes rather small at 640 reactors vs 1280 reactors.

But you should calculate this for an infinitely large reactor too and then show how many GW one belt of ore will produce.

Ur consumption is irrelevant since reactors burn fuel regardless of using the power on not.

And i dont get ur idea of feeding every 400? How are u gonna do that?

Circuits. You should check out some of the nuclear reactor threads.

[mmmPI]

22.5 fuel cells per second, times 200 seconds per cell per reactor, yields 4500 reactors.

I think it's not 22.5 fuel cells per second, but 22.5 uranium ore per second, i was trying to look for kovarex process, and nuclear fuel reprocessing with prod modules to yield as many fuel cell as possible from those 22.5 uranium ore. I'm not sure i can follow and compare the OP's reasonning though.

It would explain why (A/10)*1.2

assuming half conveyor 22.5 ore is = A ores per second
A/10 x120% equals B the output uranium/s (lets assume u get 100% bad uranium and 0% good uranium for now)

this is the https://wiki.factorio.com/Uranium_processing steps. And later the Kovarex and recycling are mathed out.

[Tertius]

According to my calculation it's slightly higher than 661.75 [reactors].

Long calculation goes like this.
I included ore refining, kovarex for U-235, reprocessing, all with max amount of prod modules (2 resp. 4). Only flaw: In the calculation, I throw away the U-235 from ore refining and only use Kovarex for U-235.

Reactor consumption:
Per 1 reactor per 200 seconds: [1]
1 fuel cell -> 1 used cell
or per 1 reactor per 1 second:
1/200 fuel cell -> 1/200 used cell

Fuel cell manufacturing (assembling machine, 4 prod modules) [2]
per 10 * 1.4 fuel cells:
(19 U-238 + 1 U-235) / 1.4 -> 10 fuel cells
per 1 fuel cell:
(1.9 U-238 + 0.1 U-235) / 1.4 -> 1 fuel cell
19/14 U-238 + 1/14 U-235 -> 1 fuel cell

U-238 manufacturing (centrifuge, 2 prod modules): [3]
10000 U-Ore -> (993 U-238 + 7 U-235) * 1.2
10000 U-Ore -> 993 * 1.2 U-238 + 7 * 1.2 U-235
10000/1.2 U-Ore -> 993 U-238 + 7 U-235
or per different U:
10*0,993/1.2 U-Ore -> 1 U-238
10*0.007/1.2 U-Ore -> 1 U-235

9.93/1.2 U-Ore -> 1 U-238 [3.1]
0.07/1.2 U-Ore -> 1 U-235 [3.2]

Kovarex enrichment: (centrifuge, 2 prod modules): [4]
3 U-238 -> 1 U-235 * 1.2
3 / 1.2 U-238 -> 1 U-235

Reprocessing (centrifuge, 2 prod modules): [5]
5 used cells -> 3 U-238 * 1.2
5 / 1.2 used cells -> 3 U-238
5/3.6 used cells -> 1 U-238


consumption c per second:
c = 1/200 fuel cell / s
insert [2]:
c = 1/200 * (19/14 U-238 + 1/14 U-235) / s
c = 19/(14*200) U-238 + 1/(14*200) U-235 / s
consumption U-238:
c_U-238 = 19/(14*200) U-238 / s [20]
consumption U-235:
c_U-235 = 1/(14*200) U-235 / s

U-238 in ore according to [3.1]
c_U-238 = 19/(14*200) * (9.93/1.2 U-Ore) / s
c_U-238 = (19 * 9.93) / (14 * 200 * 1.2) U-Ore / s
c_U-238 = 0.0561 U-Ore / s

U-235 in ore according to [3.2]
c_U-235 = 1/(14*200) * (0.07/1.2 U-Ore) / s
c_U-235 = (1 * 0.07) / (14*200 * 1.2) U-Ore / s
c_U-235 = 2.08e-5 U-Ore / s

U-235 in Kovarex U-238 according to [4]
c_U-235 = 1/(14*200) U-235 / s
c_U-235 = 1/(14*200) * (3 / 1.2) U-238 / s
c_U-235 = 3 / (14*200 * 1.2) U-238 / s [40]

We have a production of U-238 according to [5]:
5/3.6 used cells -> 1 U-238
1 used cell -> 3.6/5 U-238
with 1 (used) cell per 1/200 s, this is:
1 used cell per 200 s -> 3.6/5 U-238 per 200 s
p_U-238 = 3.6/(5*200) U-238 per s
p_U-238 = 0.0036 U-238 per s [50]

Overall U-238 consumption:
c_all_U-238 = c_U-238 + c_U_235 - p_U-238 ([20] + [40] - [50])
c_all_U-238 = 19/(14*200) U-238 / s + 3 / (14*200 * 1.2) U-238 / s - 3.6/(5*200) U-238 / s
c_all_U-238 = (1/200 * (19/14 U-238 + 3 / (14 * 1.2) U-238 - 3.6/5 U-238)) / s
c_all_U-238 = 1/200 * 0.8157 U-238 / s
c_all_U-238 = 0.00408 U-238 / s

or in Uranium ore:
consumption in ore = 9.93 / 1.2 * 0.00408 U-Ore / s
consumption in ore = 0,03375 U-Ore / s

Actually it's 10 / 1.2 uranium ore, because we ignored the 0.7% U-235 from ore, so:
consumption in ore = 10 / 1.2 * 0.00408 U-Ore / s = 0.034 U-Ore /s

If you want 22.5 ore per second, this is a factor of:
22.5 U-Ore / s / 0,034 U-Ore / s = 661.75

This calculation is correct, if we throw away the tiny amounts of U-235 we get from refining and only use Kovarex.
So the real factor is slightly higher.

[napouser]

According to my calculation it's slightly higher than 661.75 [reactors].

Long calculation goes like this.
I included ore refining, kovarex for U-235, reprocessing, all with max amount of prod modules (2 resp. 4). Only flaw: In the calculation, I throw away the U-235 from ore refining and only use Kovarex for U-235.

Reactor consumption:
Per 1 reactor per 200 seconds: [1]
1 fuel cell -> 1 used cell
or per 1 reactor per 1 second:
1/200 fuel cell -> 1/200 used cell

Fuel cell manufacturing (assembling machine, 4 prod modules) [2]
per 10 * 1.4 fuel cells:
(19 U-238 + 1 U-235) / 1.4 -> 10 fuel cells
per 1 fuel cell:
(1.9 U-238 + 0.1 U-235) / 1.4 -> 1 fuel cell
19/14 U-238 + 1/14 U-235 -> 1 fuel cell

U-238 manufacturing (centrifuge, 2 prod modules): [3]
10000 U-Ore -> (993 U-238 + 7 U-235) * 1.2
10000 U-Ore -> 993 * 1.2 U-238 + 7 * 1.2 U-235
10000/1.2 U-Ore -> 993 U-238 + 7 U-235
or per different U:
10*0,993/1.2 U-Ore -> 1 U-238
10*0.007/1.2 U-Ore -> 1 U-235

9.93/1.2 U-Ore -> 1 U-238 [3.1]
0.07/1.2 U-Ore -> 1 U-235 [3.2]

Kovarex enrichment: (centrifuge, 2 prod modules): [4]
3 U-238 -> 1 U-235 * 1.2
3 / 1.2 U-238 -> 1 U-235

Reprocessing (centrifuge, 2 prod modules): [5]
5 used cells -> 3 U-238 * 1.2
5 / 1.2 used cells -> 3 U-238
5/3.6 used cells -> 1 U-238


consumption c per second:
c = 1/200 fuel cell / s
insert [2]:
c = 1/200 * (19/14 U-238 + 1/14 U-235) / s
c = 19/(14*200) U-238 + 1/(14*200) U-235 / s
consumption U-238:
c_U-238 = 19/(14*200) U-238 / s [20]
consumption U-235:
c_U-235 = 1/(14*200) U-235 / s

U-238 in ore according to [3.1]
c_U-238 = 19/(14*200) * (9.93/1.2 U-Ore) / s
c_U-238 = (19 * 9.93) / (14 * 200 * 1.2) U-Ore / s
c_U-238 = 0.0561 U-Ore / s

U-235 in ore according to [3.2]
c_U-235 = 1/(14*200) * (0.07/1.2 U-Ore) / s
c_U-235 = (1 * 0.07) / (14*200 * 1.2) U-Ore / s
c_U-235 = 2.08e-5 U-Ore / s

U-235 in Kovarex U-238 according to [4]
c_U-235 = 1/(14*200) U-235 / s
c_U-235 = 1/(14*200) * (3 / 1.2) U-238 / s
c_U-235 = 3 / (14*200 * 1.2) U-238 / s [40]

We have a production of U-238 according to [5]:
5/3.6 used cells -> 1 U-238
1 used cell -> 3.6/5 U-238
with 1 (used) cell per 1/200 s, this is:
1 used cell per 200 s -> 3.6/5 U-238 per 200 s
p_U-238 = 3.6/(5*200) U-238 per s
p_U-238 = 0.0036 U-238 per s [50]

Overall U-238 consumption:
c_all_U-238 = c_U-238 + c_U_235 - p_U-238 ([20] + [40] - [50])
c_all_U-238 = 19/(14*200) U-238 / s + 3 / (14*200 * 1.2) U-238 / s - 3.6/(5*200) U-238 / s
c_all_U-238 = (1/200 * (19/14 U-238 + 3 / (14 * 1.2) U-238 - 3.6/5 U-238)) / s
c_all_U-238 = 1/200 * 0.8157 U-238 / s
c_all_U-238 = 0.00408 U-238 / s

or in Uranium ore:
consumption in ore = 9.93 / 1.2 * 0.00408 U-Ore / s
consumption in ore = 0,03375 U-Ore / s

Actually it's 10 / 1.2 uranium ore, because we ignored the 0.7% U-235 from ore, so:
consumption in ore = 10 / 1.2 * 0.00408 U-Ore / s = 0.034 U-Ore /s

If you want 22.5 ore per second, this is a factor of:
22.5 U-Ore / s / 0,034 U-Ore / s = 661.75

This calculation is correct, if we throw away the tiny amounts of U-235 we get from refining and only use Kovarex.
So the real factor is slightly higher.

My calcs get ~661.98
Whats that [20] [50] etc?

Also can u bother do an addition of assuming 1 train fuel consumed every 10 and every 20 secs?

[mrvn]

It's not the numbers of reactors that is relevant but the number of GW of power you consume. Running twice the number of reactors but only fueling it every 400s is more efficient. Although the difference becomes rather small at 640 reactors vs 1280 reactors.

But you should calculate this for an infinitely large reactor too and then show how many GW one belt of ore will produce.

Ur consumption is irrelevant since reactors burn fuel regardless of using the power on not.

And i dont get ur idea of feeding every 400? How are u gonna do that?

If you know you consume exactly 50% of the reactor output then you can use a simple timer to activate inserters every 400 seconds. But you never know your consumption that exactly.

So you need some feedback mechanism and that usually means adding a tank to the steam output and measuring it's contents. As long as the reactor is hot enough steam is produced and the tank remains full. As soon as it starts to drop you know the reactor has cooled too much so you throw in 1 fuel cell in every reactor and wait at least 200s before repeating that. When you throw in fuel faster than you consume electricity the steam tank will fill up again and stop you from throwing in more fuel. So the system will throw in fuel as fast as you consume power and no faster over the long run.

[napouser]

It's not the numbers of reactors that is relevant but the number of GW of power you consume. Running twice the number of reactors but only fueling it every 400s is more efficient. Although the difference becomes rather small at 640 reactors vs 1280 reactors.

But you should calculate this for an infinitely large reactor too and then show how many GW one belt of ore will produce.

Ur consumption is irrelevant since reactors burn fuel regardless of using the power on not.

And i dont get ur idea of feeding every 400? How are u gonna do that?

If you know you consume exactly 50% of the reactor output then you can use a simple timer to activate inserters every 400 seconds. But you never know your consumption that exactly.

So you need some feedback mechanism and that usually means adding a tank to the steam output and measuring it's contents. As long as the reactor is hot enough steam is produced and the tank remains full. As soon as it starts to drop you know the reactor has cooled too much so you throw in 1 fuel cell in every reactor and wait at least 200s before repeating that. When you throw in fuel faster than you consume electricity the steam tank will fill up again and stop you from throwing in more fuel. So the system will throw in fuel as fast as you consume power and no faster over the long run.

Aha i see
Yea interesting consept altho it doesnt apply to me since i am only using reactors to burn uranium hoping to get rid off it from the base. Obviusly its so much that u need millions of reactors to consume the whole map but for now i am satisfied if i find the requirement to go from half conveyor to full

[Tertius]

Whats that [20] [50] etc?

Also can u bother do an addition of assuming 1 train fuel consumed every 10 and every 20 secs?

[20], [50] etc. are references to the corresponding numbered intermediate formulas, so you know from where equations come.

I don't understand what you mean with the train question.

[mmmPI]

Aha i see
Yea interesting consept altho it doesnt apply to me since i am only using reactors to burn uranium hoping to get rid off it from the base. Obviusly its so much that u need millions of reactors to consume the whole map but for now i am satisfied if i find the requirement to go from half conveyor to full

I have played with mods that added plutonium, obtainable only during recycling process, i made hundreds of reactor to increase plutonium production i was just wasting the heat to get used cell fast. It felt like the same math problem you are showing.

proof

20190922120415_1.jpg (886.91 KiB) Viewed 1903 times

But if your goal is to remove ore from the map faster, it would require less reactor if removing the productivity module that you are using in your calculations i think. is this cheating ?

The other thing is that for the refining process, you have some % or uncertainty, so it require a buffer for the "good uranium" if you want the system to stabilize. Maybe during one hour you will have 50 good uranium, maybe the next hour only 5 from the refining process, this will average over time but depending on how you build the system, it is possible that it never stabilize and keep oscillating semi randomly forever around a certain value. At least in some parts where you store the excess good uranium during lucky hour.

[napouser]

Aha i see
Yea interesting consept altho it doesnt apply to me since i am only using reactors to burn uranium hoping to get rid off it from the base. Obviusly its so much that u need millions of reactors to consume the whole map but for now i am satisfied if i find the requirement to go from half conveyor to full

I have played with mods that added plutonium, obtainable only during recycling process, i made hundreds of reactor to increase plutonium production i was just wasting the heat to get used cell fast. It felt like the same math problem you are showing.

proof

20190922120415_1.jpg

But if your goal is to remove ore from the map faster, it would require less reactor if removing the productivity module that you are using in your calculations i think. is this cheating ?

The other thing is that for the refining process, you have some % or uncertainty, so it require a buffer for the "good uranium" if you want the system to stabilize. Maybe during one hour you will have 50 good uranium, maybe the next hour only 5 from the refining process, this will average over time but depending on how you build the system, it is possible that it never stabilize and keep oscillating semi randomly forever around a certain value. At least in some parts where you store the excess good uranium during lucky hour.

best way to get rid of uranium fast is to set up a spidertron throwing nuclear bombs at a place non stop with circuits sending him back and forth
only that kind of consumption can get rid of uranium
other than that u need something close to 30 000 reactors running non stop

[napouser]

Whats that [20] [50] etc?

Also can u bother do an addition of assuming 1 train fuel consumed every 10 and every 20 secs?

[20], [50] etc. are references to the corresponding numbered intermediate formulas, so you know from where equations come.

I don't understand what you mean with the train question.

oh i get it now
so u count the u238 produced and substract the u238 thats coming from reactors with recycling


yea i used a different more complicated method by adding it and count a new amount of cells each time and do this a bunch of times until the extra is meaningless so i guess urs is cleaner


when i am talking about trains i mean can u add to ur calculations the condition if 1 train fuel is requested every 10 secs or 20secs
thats basically 3 bad uraniums turned into 1 good then 1 good into 1 fuel . + productivity

based on my calculations the reactors go from 662 down to 650 and 637.5

[Tertius]

I'm sorry, I will not calculate this. Too much work for a useless (in my opinion) result.

I once calculated how much rocket fuel a bigger base will consume, and I found out this is negligible in comparison to the consumption of nuclear fuel cells for this base.
I wanted to know how big my nuclear processing facility needs to be if I build a base that will use a nuclear power plant consisting of 36 reactors (2x18 = 5.6 GW).
One nuclear fuel allows a train running for 2016 seconds.
500 trains need one fuel every 4 seconds.
One assembling machine with 8 beacons (speed modules) and 4 prod modules is able to produce 1 in 9 seconds.
And trains don't run continuously. My busy lab trains run 1/6 of the time. If the factor is only 1/2, you can serve 500 trains still with a single (beaconed) assembling machine.
So I guess I can grow a small nuclear processing factory by growing with the factory, and not really need more than a small amount of centrifuges and assembling machines. A regular facility can start with 8 nuclear processing centrifuges, 2 recycling centrifuges and 1 every other machine for the other functions (Kovarex, fuel cells, etc.), everything with 2 resp. 4 prod modules. Amount of output can be increased by just adding beacons with speed modules as necessary.
And if I really should need more, I can very easily build a double or triple sized nuclear processing facility with a still tiny area footprint in comparison to the rest of the base. Not really big planning required.

So if there is big planning required for a huge factory, the nuclear processing facility is not a candidate. Only, perhaps, if you use uranium ammo for defense, but I chose to use laser turrets everywhere, so this is not a factor for me.