Simple(st) Reactor Fuel control (without combinators)

[Halmatrix]

When I first started to play around with nuclear power, the day 0.15 came out, I was using steam from the Heat Exchangers directly. Soon I switched to storing steam in tanks. The principle is of course that turbines use steam from tanks, and when that steam runs out, all your reactors get exactly one fuel each at the same time, so they all get their neighbourhood bonuses and create fresh steam with maximum efficiency for the 200 seconds burn time of the Fuel Cell.

After some failed attempts I came up with the method outlined below, which has been running flawlessly ever since. No combinators or complex timers are used (yes, figuring out how to set these up is fun. But I wanted something simple and reliable in place fast, ‘cause I was wasting fuel cells by continuously inserting them into my reactors).

The setup:
Have a belt load fuel cells in chests next to the Reactors (limited to a single slot, no need to fill the entire chest). Wire one of your (connected) Steam Tanks to the inserters feeding your reactors and the inserters between chests in the control setup, as in this schematic:


How it works:
There is a single Fish in chest 1. When the steam level in the tank drops below 2000, Inserter A is enabled, moving the Fish to chest 2. Inserter B generates a pulse when it moves the fish to chest 3. This pulse is the trigger for the inserters feeding your reactors to load 1 fuel cell. Once the steam level is high enough, inserter C moves the fish to chest 4. This prevents small fluctuations of the steam to trigger multiple refuels. Inserter D moves the fish to the first chest, where it will stay till the next refuel is at hand.

in-game:
I made the sytem described above, and it hasn't malfunctioned in the 50 hours I've spend on the map since. Everything is tested 'in the wild', I don't use creative mode
My factory is quite small (haven’t even launched a rocket yet in this game), but the large island/continent it is on has been cleansed of Biters, so I’m not in a rush. I’ve invested resources in building 8 reactors (which count as 28 'effective' reactors with the bonuses), 28*4=112 Heat Exchangers and 28*2=56 28*3.5=98 Tanks for steam.
This would support 28*7=196 Turbines, but I need just 9 now. I've attached a speaker to the Fish pulse, so every two hours I get a *ding* that indicates my reactors have been fuelled up.

Notes:
-when placing the first Reactors, or adding new Reactors or Heat Exchangers, manually load fuel for the initial heating.

-make sure you have enough tanks to capture all the steam. Two Four tanks per 'effective' reactor does the trick. See below.

-the image above shows the control setup, not an actual layout for Heat Exchangers etc. In my game I stack rows like the image below for every 4 ‘effective’ Reactors (so when you extend your 2x8 setup with 2 extra Reactors to make it 2x10, you'd add two of these rows, as you're adding 8 'effective' Reactors). Every row is fed by two pumps, and has 16 Heat Exchangers,8 Tanks for steam should be 14 tanks for steam , and can be extended to a total of 28 turbines when your factory becomes power hungry.


- first time poster, hope I didn't re-invent a wheel, but couldn't find a setup like this here. Simplest is probably vanatteveldt's "Very simple controlled reactor setup" (to which I can't link 'cause URL's are not allowed in a first post, it appears :-/) . Wish that had been available when I was searching Hope it is of some use!

*edit: typooos :-/
*edit: doubled amount of tanks

[Hoeloe]

Your Setup is quite good, it should work as intended.
There is only one case where it may will fail: When your powerusage is close or plus 100% couldt'n it be possible that the steamtanks don't fill enough to give the Inserter C the signale to "reset" the fish?
I did not realy messed arround with nuclear yet, but I understand your idea (using 100% maximum output of 1 fuelcell and don't waste any) and made my own thougts about that. Beside your solution, to achieve this woudn't it also be important to store the surplus of energy? So I did a bit of math:
Fuelcell runtime: 200s
Turbine steam consumption: 60steam/s
200s x 60steam/s = 12'000steam/turbine @ 100%
= 1 Storage Tank (25k) per 2 Turbines (12k)
So what I actually try to tell is: You need 1 tank per 2 turbines to have a 100% buffer.
Going back to your "in-game" setup you should consider that your 16 heatexchangers actually produce steam for 28 turbines, so you would need 14 sorage tanks to store 100% of 1 fuelcell. Since you also have energy consumption in that time you can decrease your storage by that value. You are working with 8 tanks, so you can store 57% (100%/14x8) of 1 fuelcell which means you need to use at least 43% of your energy production to have not an energy loss.

Enough math for me, just tell me if I'm wrong.

[MBas]

viewtopic.php?f=208&t=48988

I belive that this one is much more simplier .

[Halmatrix]

Your Setup is quite good, it should work as intended.
There is only one case where it may will fail: When your powerusage is close or plus 100% couldt'n it be possible that the steamtanks don't fill enough to give the Inserter C the signale to "reset" the fish?

Thanks
And you are correct on the point of failure... But that's a long way off. I'm now at 1 refuel per 2 hours gameplay... Before I get to the point where the refuels get close to 200 seconds, I'll have expanded my power supply. Or, seeing that I'm close to 100% usage, remove the fish-control entirely and let the reactors run full time, of course

Beside your solution, to achieve this woudn't it also be important to store the surplus of energy? So I did a bit of math:
Fuelcell runtime: 200s
Turbine steam consumption: 60steam/s
200s x 60steam/s = 12'000steam/turbine @ 100%
= 1 Storage Tank (25k) per 2 Turbines (12k)
So what I actually try to tell is: You need 1 tank per 2 turbines to have a 100% buffer.

If you'd fire up 8 reactors (still counting as 28) they'd produce 1.153.600 steam - you need 46.1 tanks to capture all the steam. That's 1.6 per reactor. I rounded it up to 2, for safety and convenience. The setup I use captures all the steam - I tested it without turbines connected, so see if the tanks would fill, as the amount of tanks felt low, inuitively. I could remove 1 of the 8 tanks and still catch all the steam, as only 6.5 tanks are needed.

I think the confusion comes from the difference between running full time without tanks, and storing steam. The 1 reactor : 4 HeatEx : 7 turbines ratio is based on running full time, where all the generated heat is continuously used, and the steam topped up all the time.

I must admit that I can't point to an error in your logic, nor in my tests, seeing that if you remove the tanks this setup would be a 'normal' 1:4:7 ratio, and the tanks don't overflow. If the tanks would get empty within 200 seconds, the reactor would still be running, and that ratio would be wrong. Hmmm.
But, as you say, that's besides the fuel inserting solution

[Halmatrix]

viewtopic.php?f=208&t=48988

I belive that this one is much more simplier .

Indeed, very elegant. Nice! Had I known of this, I'd have used it instead of tinkering myself

[Hoeloe]

I think the confusion comes from the difference between running full time without tanks, and storing steam. The 1 reactor : 4 HeatEx : 7 turbines ratio is based on running full time, where all the generated heat is continuously used, and the steam topped up all the time.

You missunderstand my point there BTW: I am working with these Ratios.
When you have 8 Reactors (28 effective) you are working with 112 Heatexchangers, like you said. With those you can 100% satisfy 193 Turbines. 1 Turbine will use 60steam/s etc. [insert my calculation form above here]
Or with your way of calculation:
8 Reactors = 28 real count Reactors x 4 Heatexchangers/Reactor = 112 Heatexchangers
112 Heatexchangers x ~1.75 Turbines/Heatexchanger = ~193 Turbines (see Ratios)
193 Turbines x 60steam/s x 200s/fuelcell = 2'316'000 Steam
2'316'000 Steam : 25'000 Steam/Storage Tank = ~93 Storage Tanks

Remember: Using Tanks/Reactor does not work since it's the Turbine which defines the amount of used steam.
Remember 2: With this calculation you will only fill the tanks when you have 0kW powerusage!

With 8 Reactors you can produce 1120MW, when you need about 560MW (50%) you will also only have 560MW (50%) overflow, meaning you can reduce your storagetanks to 50%.

1 Refuel per 2 hours means you are running at 2% so about 22.4MW...
But 9 Turbines running means (9x5.82MW) 52.4MW...
And assuming you have no steamoverflow using only 8 Tanks means you cannot store about 43% of your powerproduction = ~480MW powerusage

Tell me: how much power do you realy use?

With the given information I can't (prove it ) tell you, but I'm pretty sure: Eather you are able to store the Steam somewhere since not all reactos are running together or you actually have enough tanks together with your powerusage,
Or: You have alot of heatloss (= powerloss) eather in your heatpipingsystem or in form of steamoverflow. Maybe you just didn't realized it... Or maybe I'm still completly wrong

[Halmatrix]

I think the confusion comes from the difference between running full time without tanks, and storing steam. The 1 reactor : 4 HeatEx : 7 turbines ratio is based on running full time, where all the generated heat is continuously used, and the steam topped up all the time.

You missunderstand my point there BTW: I am working with these Ratios.
When you have 8 Reactors (28 effective) you are working with 112 Heatexchangers, like you said. With those you can 100% satisfy 193 Turbines. 1 Turbine will use 60steam/s etc. [insert my calculation form above here]
Or with your way of calculation:
8 Reactors = 28 real count Reactors x 4 Heatexchangers/Reactor = 112 Heatexchangers
112 Heatexchangers x ~1.75 Turbines/Heatexchanger = ~193 Turbines (see Ratios)
193 Turbines x 60steam/s x 200s/fuelcell = 2'316'000 Steam
2'316'000 Steam : 25'000 Steam/Storage Tank = ~93 Storage Tanks

Remember: Using Tanks/Reactor does not work since it's the Turbine which defines the amount of used steam.
Remember 2: With this calculation you will only fill the tanks when you have 0kW powerusage!

With 8 Reactors you can produce 1120MW, when you need about 560MW (50%) you will also only have 560MW (50%) overflow, meaning you can reduce your storagetanks to 50%.

1 Refuel per 2 hours means you are running at 2% so about 22.4MW...
But 9 Turbines running means (9x5.82MW) 52.4MW...
And assuming you have no steamoverflow using only 8 Tanks means you cannot store about 43% of your powerproduction = ~480MW powerusage

Tell me: how much power do you realy use?

With the given information I can't (prove it ) tell you, but I'm pretty sure: Eather you are able to store the Steam somewhere since not all reactos are running together or you actually have enough tanks together with your powerusage,
Or: You have alot of heatloss (= powerloss) eather in your heatpipingsystem or in form of steamoverflow. Maybe you just didn't realized it... Or maybe I'm still completly wrong

You are not wrong - you're 100% right! Somewhere in re-re-rebuilding that HeatExchanger/tank/turbine layout, I lost half the tanks Didn't notice in game, cause a field of tanks from previous tinkering was still connected by an underground pipe *phew*

Time to clean up this mess - thanks for your help!

[Hoeloe]

No problem and thank you for your reply
I would have been realy ashamed if I had been wrong

Now to get back to the initial post: I need to say the solution of MBas is also very nice and simple.

[milo christiansen]

Rather than using a fish in four chests, use a fish in four belt pieces placed in a loop. It's more compact, and requires no power.