Nooblet Breeder Questions

    It seems to me that breeding takes a while. I am trying out THIS breeder currently. I read that 5 lava buckets instantly heats the hull to 10k. I have 20 blocks of water touching it. I put in the lavato get it hot, and yet it still seems to take a long time.


    So my question is, what can I do to breed faster? And my other question is, is there a design that will breed multiple cells quickly?


    Also, what do you guys think of THIS design?


    Thanks


    EDIT: I guess I should also ask a clarifying question. In the reactor planner app -let's say we're looking at the one I posted- in the ring of uranium/ isotope cells, is the outer ring of four the isotopes, or the inner cell the isotope?

    Your reactor doesn't heat up enough from the five lava buckets. You didn't remember or know that the coolant cells and IHDs absorb part of the heat - 10k each - so that you would actually require 210 lava buckets to heat it up, so you should rather use a heating setup to do that. Using less IHDs(for example, this ) than in that design is also highly recommended as they just increase the reactor's cost and time to heat it up, and the reactor plating isn't really necessary either for a six chamber design. The ice CASUC breeder seems alright, but I'd recommend using buckets, if possible. For it the lava bucket heating works as it doesn't have any cooling components expect the ice.
    In the first design you posted the outer ring is made of the isotope cells, otherwise it would blow up.

    Check out this awesome mod !

    Edited once, last by jppk1 ().

    Ok, thanks for reply. I have a couple more questions now.


    In the first design I posted, I had a ring of four uranium cells where you said there should be isotope cells, and it recharged 3 of the central isotope cells in a row, without overheating, let alone exploding. Is there a reason it didn't explode like you suggested? I see now that doing it this way isn't very efficient, but it seemed safe. So If I switched it out, and had one uranium cell in the middle, that one cell would charge 4 isotope cells around it? Would it take very long? Or would it be around the 2-hour mark?


    And also, if I understand the diagrams correctly now, in the second design, (ice casuc breeder) would the cells on the top row be uranium cells?


    Thanks

    I believe that your first design(if it is this , please link to the reactor planner whenever possible) could just about recharge 3 cells without exploding if you first put the five lava buckets in it, and that's without otherwise preheating the components. If you switched it out it would take half a cycle(=5000 seconds, about 1,5 hours) to charge the four isotope cells surrounding the uranium cell. It would be much more uranium efficient and safe then a isotope cell surrounded with uranium, and you wouldn't have to visit the reactor as often.
    THe ice casuc breeder you linked did have a row of uranium at the top, and that was to balance the heat to match the cooling from four ice blocks per second so that it doesn't cool down.

    Okay, thanks once again. And just two more questions.
    - The ice CASUC below the first row is isotopes alternating with uranium cells, correct?
    - Which one is more efficient? (exclude cost to make ice, I've figured out how to do that without using uranium)

    Here is a tweak of the CASUC design that uses buckets instead of ice. It also has the same amount of passive cooling as there are recharging cells, which means it should maintain a stable heat level when the reactor and cooling system are off (makes it easier & safer to pre-heat, since you don't have to be in a hurry).

    Hello,




    i'm using your thread because i'm going to precisely ask a noob breeder-related question;


    it is written here :
    Re-Enriched Uranium Cell

    Quote

    The fully charged state of an isotope cell, it will continue to produce only 1 heat and no EU but it will no longer react with adjacent Uranium Cells. Combined with another coal dust, it will become a brand new Uranium Cell.


    Does that 1 heat per tick happens when there is NO strictly-named "uranium cell"?? Because it seems it isn't happening.


    Slowly going to the conclusion than a reactor cannot generate heat without any strictly-named "uranium cell".




    (i did a breeder with automated pistons pushing blocs and liberating lava to maintain the heat when it stopped, and... it's still losing heat. I counted the 4 Re-Enriched Uranium Cells as "1 heat / tick", but if it's not the case, i must permit lava to be near the reactor 1 block more...)

    Quote from NewMilenium

    Hello, ...


    Yes, you probably did not notice that because cooling works even when there are no uranium cells around. Basically, all heat from isotope cells goes straight to hull, where it is terminated (with no mercy) by external cooling (internal too, when there is any).

    I'm sorry to have to say it, but you're not answering at all my question;
    i totally noticed cooling applies all the time (otherwise "cooldown" would have no meaning in nuclear engineering), my question is about this Re-Enriched Uranium Cell; does it still produce heat when NO uranium cell is in the reactor?
    Because my automated heating system for my breeder seems not to be enough to maintain the heat when the calculation takes in count the four Re-Enriched Uranium Cells.


    Having those 4 heat/tick going in the hull or not doesn't count much to me, as i am supposed to have exactly the same cooling value (4/tick). Here are the designs;
    the breeder,
    the breeder when uranium cell is gone (thanks to pistons - the 4 "unused blocks" are reinforced stones in my case).

    Ok, once again.

    Quote

    because cooling works even when there are no uranium cells around

    This means cooling works always. Even internal cooling. It is not shown in the planner but it works, so make a sum of internal cooling from the first design and external cooling from the second (because external changes there). Result is 40 (36int + 4ext). Thats because coolant cells can access hull through HDs and are cooling it all the time.

    Quote

    by external cooling (internal too, when there is any)

    If you remove internal cooling by simply removing HDs from your reactor, only external one is applied and reactor remains hot (4 heat - 4 ext. cooling = 0 overal gain). BUT you have not takem into account this scenario, where uranium cell may turn into near depleted uranium cell, which is also generating 1 heat per sec (i think). It may result in explosion if left unnoticed.

    [explained my problem on chat]
    [then : ]


    Big thanks to voker57 and Cassandra.


    So, raGan, this answer enlightens that even a near-depleted uranium cell created by an uranium cell that ended in the breeder won't produce heat, contrary to what you said.

    Oh my...
    So, i have no solution, as i cannot anticipate an uranium cell disappearing OR becoming a near-depleted cell...


    By the way, why are your 5 near-depleted cells only producing 4 heat per second in total instead of 5?



    edit : If i create a redstone input on a reactor chamber, and this input is put to "on" when the reactor isn't producing any EU, will stopping the reactor that way stop any heat production too?

    Quote from NewMilenium

    Oh my...
    So, i have no solution, as i cannot anticipate an uranium cell disappearing OR becoming a near-depleted cell...


    By the way, why are your 5 near-depleted cells only producing 4 heat per second in total instead of 5?


    Reactor itself has its own external cooling.
    Second thing - Always anticipate that, having 1 excess cooling is nothing horrible, it means 1 bucket of lava per 2000 seconds. (approx. 33 mins)


    Edit: NOPE, redstone only affects uranium cells

    Oh, you had 4 air around it? I see 1 on your image, at least.


    Do you know, by the way, if i was right to suppose that 1 opaque bloc = exactly 0 cooling? As i cannot trust the "planner" on everything.



    edited : you edited & answered.
    Damn, how am i gonna do... 1 lava bucket every 33 minutes means a lot of lava buckets in several days.

    Ooooh, right! Thank you.
    Got another question i can't find the answer on;
    external lava is diminishing external cooling to a minimum of 0, or ALL cooling (internal too)?

    Thanks a lot for all those answers.
    And I assume the answer to my question "Do you know, by the way, if i was right to suppose that 1 opaque bloc = exactly 0 cooling?" was "yes".



    So, i have absolutely no way to get around those 36 internal cooling...


    If i understand you correctly, the numbers are those (considering the new piston-mechanism i just made) :


    WITH Uranium : 20 air,
    18 external cooling (13 by reactor&chambers + 5 by air),
    36 internal,
    heat : 54
    TOTAL = 0


    WITHOUT uranium, with 4 depleted : 5 lava, 11 air, 4 blocs
    0 external cooling (13 by reactor&chambers + 2 by air - 5*3 by lava)
    36 internal,
    heat : 4
    TOTAL = 32 cooling / second


    WITHOUT uranium, with 5 depleted : 5 lava, 11 air, 5 blocs
    0 external cooling (13 by reactor&chambers + 2 by air - 5*3 by lava)
    36 internal,
    heat : 5
    TOTAL = 31 cooling / second

    You can set filter to the reactor that takes all HDs and depleted cells when the cycle is finished.
    Use detector cable to know when reactor stops generating power and start timer to remove all HDs AND depleted uranium cells. When you come to refill your reactor, it will still have same heat as before, but you have to manually refill it :(. This way you can be sure to not have any excess cooling.