All the different ways to increase the efficiency of a nuclear power plant

[foamy]

Also, in your case, of heatpipes as capacitor, you dont have a tool too measure amount of heat stored in plant, so you dont know when insert new cell, so , as result, in case of irregular usage, probably with combination of solars and some burners, or simple bug attack repelled by lasers you will feed nuclear cells nonstop. Which is, as you know, punishable by law.

Not having a way to measure heat stored directly is an issue, yes, but it's not insurmountable: Steam consumption can be measured quite precisely, and how much a nuclear cell generates is an exactly known quantity (if your plant has no other bottlenecks or desync). Ergo you can feed in a new cell precisely when the last cell was wholly consumed, and therefore respond with the absolute maximum possible precision to demand alterations.

[gGeorg]

dont think, you understnad what wasteless design means. It is fully automated system which can handle any spike or power ned drop for any period of time.
Example : You start to make a base redisign so power consumption drops from 500 - 600MW to 4MW. Where you have an solar field which gives 10MW. Meanwhile you redesign the base, bugs come and lasers make energy spike at 400MW. Power plant need to designed so no waste fuel happens. and everything get power when needed.

No, what you describe is self-regulating. Meaning the power plant will adjust its consumption and production itself. Thanks to reading steamflow, steam level, side fuel consumption and so on. This is the feedback that allow self-regulating.

Wasteless/wastelessness, means you don't insert fuel when the reactor is 1000°, because this energy is then wasted. It's less complex behavior than self-regulating.

No. Power plant can not adjust consumption. Also, You could waste fuel even if you insert cell on 700C. Waste is not linked to core temp, but thermal capacity of the plant.

[gGeorg]

Also, in your case, of heatpipes as capacitor, you dont have a tool too measure amount of heat stored in plant, so you dont know when insert new cell, so , as result, in case of irregular usage, probably with combination of solars and some burners, or simple bug attack repelled by lasers you will feed nuclear cells nonstop. Which is, as you know, punishable by law.

Not having a way to measure heat stored directly is an issue, yes, but it's not insurmountable: Steam consumption can be measured quite precisely, and how much a nuclear cell generates is an exactly known quantity (if your plant has no other bottlenecks or desync). Ergo you can feed in a new cell precisely when the last cell was wholly consumed, and therefore respond with the absolute maximum possible precision to demand alterations.

I made a test bed for offset timer for steam level controlled plant. I hope I can finalize and update my Perfect cloverleaf until Friday. It will be nice if you can test_ride/evaluate the resullt.

[mmmPI]

No. Power plant can not adjust consumption. Also, You could waste fuel even if you insert cell on 700C. Waste is not linked to core temp, but thermal capacity of the plant.

Does that count as automated way to nuclear power plant ?

If no automation counts, then this does indeed.
I'll add this in my sheets

I described the clock-mechanism and specifically asked OP if it was to be included in the topic. To which he said yes. That's a bit why i'm insisting.

I would say :It doesn't matter if the plant can not adjust consumption for "wasteless". If you set up to 50% and consume 50, it is wasteless, if you set to 75% and consume 75% also you have wastelessness.

You are right when you say you could insert at 700° and still waste. It has nothing to do with automation.

If the plant is able to auto-adjust power production/fuel consumption, it's another functionnality. I called it self-regulated, because the other would be player regulated.


If you want your power plant to function at 80% load while not wasting fuel, the sef-regulating design is not practical. The self-regulated design will adapt power production to your consumption meaning if your consumption drops at 50% the plant will function at 50% and if consumption increase to 95% the plant will adapt. You have no control over your fuel consumption. Sure most of the time you want that behavior, the automatic adaptation, but it's not the exact same as wasteless.

Here is an old picture of a modded game where the clock-system makes total sense :

nuclearfurnace.jpg (583.73 KiB) Viewed 5941 times

The 3 nuclear reactor burn modded fuel to generate heat that is required for the furnaces to process the silicon from the 2 giant miners.

Fuel and used fuel were delivered by train.

There is no steam to measure, you have to find a way to quantify the heat. It could be by calculating the amount of silicon processed, and you refuel every 200K for example.
Or you could have a timer that you set-up manually, and have a lot of thermal inertia to reduce the frequency at which you need to adapt and the precision required for the setting.
or something like a timer that runs only when the belt is full/empty, to stop fuel burning when the buffer are filled to prevent waste of fuel.

You cannot read electricity consumption and let the power plant cool down automatically when the buffer are full, because the fuel is not burned for electricity. The electricity could be solar for example, and auto adapt, you still need to control the fuel on the reactors that only heat furnaces because otherwise they will heat up the furnaces to 1000° and continue, you risk being punished by law no ?

Yes it's very particular, but if we are going to list and discuss "All the different ways", then we're not choosing the best one aren't we ?

Maybe someone sometimes will need something very particular and it will be helpful to have it. The method and the calculations, the explanations you did can be helpful for other things too. It doesn't help that my personnal version of it is very bad, to show when it can be useful , i admit

[mrvn]

dont think, you understnad what wasteless design means. It is fully automated system which can handle any spike or power ned drop for any period of time.
Example : You start to make a base redisign so power consumption drops from 500 - 600MW to 4MW. Where you have an solar field which gives 10MW. Meanwhile you redesign the base, bugs come and lasers make energy spike at 400MW. Power plant need to designed so no waste fuel happens. and everything get power when needed.

No, what you describe is self-regulating. Meaning the power plant will adjust its consumption and production itself. Thanks to reading steamflow, steam level, side fuel consumption and so on. This is the feedback that allow self-regulating.

Wasteless/wastelessness, means you don't insert fuel when the reactor is 1000°, because this energy is then wasted. It's less complex behavior than self-regulating.

But doing that by having the player set the desired output in MW is just stupid. Power consumption is never a constant.

[mmmPI]

But doing that by having the player set the desired output in MW is just stupid. Power consumption is never a constant.

it doesn't have to be constant, it has to average in the amount of cycle of fuel that you can buffer under the form of heat.


if you use enough heat pipes to store 3 cycle of fuel, the consumption can be 10% for 5 minutes then 90% for five minutes. If you set it at 50% it's fine, still no fuel loss. wasteless.

[gGeorg]

We are mixing 2 things then dont understand each other. Include OP.

Regarding usage of fuel, (efficiency of fuel), There are 2 different design challenges in Nuclear power plants.

1. When insert another cell ?
- cell always burn to the end in 200s (it doesnt pause when not needed like smelting oven)
- designer can use default principle - load as many as possible - which creates fuel waste
- designer can use some smart solution to insert fuel cell only when conditions are filled - fuel waste reduction problem
- ideal solution is wasteless design

2. How create and detect adequate thermal capacity to prevent void heat ?
- game mechanic makes void not used heat from burned cell
- player designer challenge is, use some smart solution to prevent thermal loss - thermal loss reduction problem
- thermal design solution creates such plant design which can absorb all the heat from all the Nuclear cell inserted at once. then prevent another cell load to allow consume stored heat first
- ideal solution is lossless design

For some reason, OP mixed both these problem to one question which is confusing.

There are mentioned few "systems" whithout noticing what is what, so confusion is seed is there.

Steam Level 2. How create and detect adequate thermal capacity to prevent void heat ?
Steam Production 1. When insert another cell ?
Steam Flow 1. When insert another cell ?
Accumulator Level Monitoring 1. When insert another cell ?
Side Channel Power Production Monitoring 1. When insert another cell ?
------------
I can add some other ideas for (1. When insert another cell ?) :
detect used cell unload event
detect fresh cell load event
measure time to burn the cell

it doesn't have to be constant, it has to average in the amount of cycle of fuel that you can buffer under the form of heat.

if you use enough heat pipes to store 3 cycle of fuel, the consumption can be 10% for 5 minutes then 90% for five minutes. If you set it at 50% it's fine, still no fuel loss. wasteless.

Well, by the rules I have just highlighted, you talk about partial lossless design. Thermal buffer capacity. e.g. such design which can absorb (buffer) an certain portion of heat.

[xaetral]

I don't think I get what you mean here...
You can just ask the power plant (using circuit-wired combinators) to add 8GJ per reactor to the "heat system" over 200s (though you might consider it to be instant because you need to be able to store that energy if you don't consume any power).
And the other point is that you need to wait for that heat system to loose enough energy (through power consumption) before adding a new batch of heat energy (tbh it's just the same point but said differently).
Here I described the most simple case but you may also use heat together with steam and electricity to store that energy, the point is that at the end you just have a black box that is only an energy storage that you can fill and empty.


Also yeah, as I said in the first post, I may break down the exact working principle of all these setups if you ask so.

[mmmPI]

We are mixing 2 things then dont understand each other.

yes maybe more than 2 haha.

1. When insert another cell ?
- cell always burn to the end in 200s (it doesnt pause when not needed like smelting oven)
- designer can use default principle - load as many as possible - which creates fuel waste
- designer can use some smart solution to insert fuel cell only when conditions are filled - fuel waste reduction problem
- ideal solution is wasteless design

This is close to what i called wasteless. the difference between default, load as many as possible, or having a way to control fuel consumption to reduce it when not needed is done to achieve wastelessness.

I think though that it is not a specificity that is proper to a design, but i would say it more the result of how you use designs.

If you use the default method, load as many as possible, but ingame you also use 100% power all the time, you will not waste fuel. The design itself is not smart. The situation that wil happen if you always consume 100% power will still be a situation where the power plant doesn't waste anything.

What op proposed as initial design, your clover leaf power plant, those have dynamic mechanism to seek wasteless situations. Getting information, and reacting to be wasteless no matter what electricity consumption. This means reducing refuel frequency when not needed but also having enough buffer for the energy.

We could say those design function ONLY wasteless. or try to.

While the default system can be wasteless sometimes if the consumption is exactly the same as the maximum production, but the rest of the time it waste fuel. the design is not waste-free waste-impossible, you need the efficiency police to control only when everything is fine, not random control haha.

2. How create and detect adequate thermal capacity to prevent void heat ?
- game mechanic makes void not used heat from burned cell
- player designer challenge is, use some smart solution to prevent thermal loss - thermal loss reduction problem
- thermal design solution creates such plant design which can absorb all the heat from all the Nuclear cell inserted at once. then prevent another cell load to allow consume stored heat first
- ideal solution is lossless design

This i think is more characteristic to the design itself versus the other part that is more the result of a situation. I don't really see the difference between wasteless and lossless. things are not organised the same way in my mind for me the keywords here are "detect" and "prevent void heat".

What i read here include the relation between the energy buffer capacity, in steam, or heat, and the fuel cycle.

Taking the example of your clover leaf, the fuel cycle worth 4(core)x3(adjacency)x8GJ= 96 GJ or 96 000MJ, which is 480MWx200sec. If consumption stops just after refuel, you have to buffer 96000 MJ in the form of a heat in few heatpipes, and/or steam in some tanks and iron pipes.

This is because the all the fuel will burn and the heat will be lost if it's not buffered.

This is the MINIMUM energy buffer you need if you want a power plant to be wasteless when the situation is no electricity consumption just after refuel.
This one is not one you can configure yourself, it's how it is in the game and you have to play with that. This is the "no consumption" extreme.

The other extreme is 100% consumption, max capacity, 200fuel timer, they all mean the same for me. In this case no need for buffer.

That leaves 1 thing open : There is no MAXIMUM buffer, you can have 12 millions tanks, and activate the nuclear plant at max power only on mondays if the consumption is only 15% of max capacity. Then next week you know if you activate during 20hours on monday or 19hours or 24 hours, you can adapt ( you count the steam tank left on sunday haha).

Most self regulating/adaptive/automated design ( like your clover leaf) adapt in 1 cycle of fuel, because it reduce the amount of buffer required to store the energy/steam/heat. 12millions tanks is way too many = adapting every week is not often enough.

Those parameter are both embedded in the design, if you include "this" system of regulation, you need "that" amount of buffer.

This comes in second, the first question really is : "what situations can happens ?".

Most self regulating design (like your clover leaf) are designed to handle extreme situations while being wasteless : (A) 0% for long time, (B) 100% for long time, (C) 0% and then 100%, (D)100% and then 0%, (E) 100% then 0% then 100% then 0%, (D) any% (E) x% (F) y% and the list could be just summed up with "every situations".

First one say: " i want my power plant to handle EVERY situation" , then you need to know how many MW maximum, it means how many core, from that you can calculate the minimum buffer of energy, it's the 96000 MJ for a 2X2 power plant, 1 cycle of fuel worth. Only at this point you have multiple choices, you can use more buffer if you want but to reduce the building footprint and material cost the minimum buffer is the best. Then you need a mechanism to adapt the refuel timing to the electric consumption given the buffer you have choosen superior or equal to the minimum.

But if you say " my power plant willl be ON/OFF only" then you can design differently.
Or if you say " my uranium production is x per minutes" i want to only use 1/2 for power", you need to design differently.

Maybe the other uranium patch is very far away in biters land or in unknown location, the only patch you have is very small, you need nuclear power because low oil and no coal but you don't have the tech for enrichment yet.
Maybe the nuclear plant is only for the science bus and you research only 1 tech, with the starter base and outpost uses solar.
Maybe the nuclear plant is only for laser turret drain, and they use solar and accumulators for spikes.
Maybe the nuclear plant is on another planet or an asteroid and you send uranium using rockets which cost a lot.

that's how things are in my mind x)

it doesn't have to be constant, it has to average in the amount of cycle of fuel that you can buffer under the form of heat.

if you use enough heat pipes to store 3 cycle of fuel, the consumption can be 10% for 5 minutes then 90% for five minutes. If you set it at 50% it's fine, still no fuel loss. wasteless.

Well, by the rules I have just highlighted, you talk about partial lossless design. Thermal buffer capacity. e.g. such design which can absorb (buffer) an certain portion of heat.

Yes i understand i think what you call partial lossless design, it's partial because it's only happening if the player set the timer correctly and the consumption is somewhat predictable. Otherwise it's not lossless.

the other designs like the clover leaf would be "fully lossless" to compare. Or adapting to be lossless in every situation, in my mind . What called "self-regulated".

I can add some other ideas for (1. When insert another cell ?) :
detect used cell unload event
detect fresh cell load event
measure time to burn the cell

i don't understand 1 and 2 it seems to me that those are just mechanism that triggers, it's not where you read information to adapt the timers contrary to the other in OP's list.

3 i think is basically using time/a timer/clock to trigger refuel, a bit like what i started to do? except mine is unfinished you cannot change timer easily. Same as not reading any value in game right ? because a cell is always burnt in 200 sec.

All the other way are indirect means to measure electric consumption to adapt refuel timers.

[gGeorg]

I don't really see the difference between wasteless and lossless.

Wasteless is about : "do not feed new cell if you dont need power". (typicaly 2N design could save lot fuel becouse scalability of power output )
Looseless is more delicate, it is about make sure not even part ( %) of a cell is lost. e.g. make thermal buffer large enough for the whole cell energy.

Surprisngly, make plant loseless is quite easy. Factorio cheat sheed did a kiss of death by publish table with bunch of steam tanks. For cloverleaf you actualy need only 4 steam tanks!
Surprisngly a 2N design whic is designed as wasteless, has a logic to use only part of cores, could get a lossles property. Simply becouse, those unsed cores are great heat buffer.

[mrvn]

I don't really see the difference between wasteless and lossless.

Wasteless is about : "do not feed new cell if you dont need power". (typicaly 2N design could save lot fuel becouse scalability of power output )
Looseless is more delicate, it is about make sure not even part ( %) of a cell is lost. e.g. make thermal buffer large enough for the whole cell energy.

Surprisngly, make plant loseless is quite easy. Factorio cheat sheed did a kiss of death by publish table with bunch of steam tanks. For cloverleaf you actualy need only 4 steam tanks!
Surprisngly a 2N design whic is designed as wasteless, has a logic to use only part of cores, could get a lossles property. Simply becouse, those unsed cores are great heat buffer.

If you consider a 2N reactor that only uses part of its reactors wasteless then that is a bad definition of wasteless because you are wasting neighbor bonuses and therefore using more fuel cells than optimal. It's the opposite of efficient. So I would always want a wastefull but looseless reactor. Never a wasteless one.

[mmmPI]

If you consider a 2N reactor that only uses part of its reactors wasteless then that is a bad definition of wasteless because you are wasting neighbor bonuses and therefore using more fuel cells than optimal. It's the opposite of efficient.

I agree with this a 2N reactor should use all its core and have a bigger buffer for the energy, this way it maximize neighbour bonus, we can put numbers on this, let's start from the formula from the wiki:

The most efficient practical layout is an aligned double row of arbitrary length (number of reactors as needed). For even numbers of reactors, the total output of the array is 160n − 160 MW (where n = total number of reactors, and assuming all are fueled). Splitting the row, while possibly logistically beneficial, reduces total power output by 160 MW per split.

Keep this in mind :

There is no MAXIMUM buffer, you can have 12 millions tanks, and activate the nuclear plant at max power only on mondays if the consumption is only 15% of max capacity. Then next week you know if you activate during 20hours on monday or 19hours or 24 hours, you can adapt ( you count the steam tank left on sunday haha).

This is super efficient, if you have a 51x2 core central, that's a 16 160 MW central at full capacity. We can also say it consume 102 fuel every 200 second. This means one fuel cycle is worth 16 160x200 =3 232 000 MJ. Or we can say 1 fuel is worth 3 232 000 MJ/102= 31,686 GJ, instead of the regular 8GJ, 396% efficiency.

Now take a 2X2 core central, that's 480 MW at full capacity. We can also say it consume 4 fuel every 200 second. This means one fuel cycle is worth 480x200= 96 000 MJ, Or we can say 1 fuel is worth 96 000 MJ /4= 24 GJ, instead of the regular 8GJ, 300% efficiency.

In other word you get, 1.3205 more energy per fuel cell or you need 0.75729... cells to produce the same amount of energy. [Edit: if you use the 51x2 it's more efficient than the 2X2]

(typicaly 2N design could save lot fuel becouse scalability of power output )

is that what you meant ?

In order to achieve this noticeable efficiency gain the question now becomes : how much buffer do you need ? For this too we can put numbers, let's use this intermediate result from OP

In theory, when you put a fuel cell in a reactor, it will produce a very accurate amount of steam (without reactor bonuses it's around 82474 steam per fuel cell).Well, in practice this is also the case.

A simple rule of three applied knowing 1 fuel cell is 8GJ would give a little less than 10 310 steam per GJ. (10 309,25 steam really but we are talking about having enough buffer so we'll round up anyway). Each tank being capable of storing 25 000 steam then, with 3 232 GJ of energy per fuel cycle that you get with a 51x2 central, you would be required 3 232x10 309,25/25 000= 1332,78 tanks.

This means if you have a 51x2 core central and 1333 tanks as buffer for steam you can stop it entirely just after refuel and nothing would be wasted.

But what you want is just compare with the 2X2 plant efficiency right ?

The 2X2 plant produce at max power 480MW, while the 51X2 produce 16 160 MW, this means the 51X2 is [16 160/480]=[101/3] around 33 times more powerfull. In order to make only 480MW of energy you'd need to activate only [3/101] % of time since you cannot really produce [3/101] of max production right ?

this is around 2,97% of time.

The refuel timers would then need to be extended so that the 200 second of fuel burning would only represent 2.97% or the total amount of time.

this means one cycle should last : 200x[101/3] = 6733 seconds. or almost 2 hours.

Conclusion: a 2x51 power plant require 1333 tanks to fully buffer energy, and with cycle of roughly 2 hours , it would be able to produce 480MW of energy requiring only a bit less than 76% the amount of fuel that a 2X2 plant woud consume.

[Edit: It doesn't STRICTLY REQUIRE 1333, you could be good with less, since some energy will be stored under the form of heat, 1333 is guaranteed to be enough even in the worst case if the heat is not wasted due to poor heat pipe layout]

Interestingly enough, this totally validate clocking inserter on time based on your estimated consumption and building huge buffers. Because no matter how much energy you need, you'd be more efficient burning more fuel at the same time, and then stopping for some time.



The bigger buffer you can have, the more efficiency you can have !


feel free to tell me if i did the math wrong

[gGeorg]

The bigger buffer you can have, the more efficiency you can have !
feel free to tell me if i did the math wrong

Well, mmmPI your math is wrong. Now, I dont have time explain why, but if no one else does, I would later. Meanwhile would you test a new version of my Perfect plant ?

I was finally successful to create Perfect cloverleaf 2.0
testing was a challenge, few hours to debug the small beast. So the main features:
- cold interval reduced to zero (fresh cell is inserted at the same tick as old is released )

- used mechanic for fuel control is not described in the list by OP : - ]
- side effect of new mechanic of control - power-plant auto-starts ( well, in case you plop it down at night, wait till morning)
- based on discussion here, I also added a predictable fuel cell buffer with Speaker to announce shortage
- based on discussion here, if you belive that start with certain minimal amount of cores is useful, you can modify one value of a combinator from 1 to whatever you like, for example 4 or 2
- some other minor improvements

Here to download & discuss :
viewtopic.php?p=554437#p554437

[xaetral]

I prefer to talk about stuff per reactor, it eases 2xN setups.
A fuel cell gives off 32GJ due to the neighbor bonus (at least it tends toward that), which is 32/2.425 ~= 13.2 storage tanks full of steam.

For each reactor you have around 27.5 steam turbines, already holding 5500 steam, a bit less than half a GJ.
The nuclear reactor itself holds a few GJ of heat energy (10MJ per °C).
The heat exchangers are holding heat as well (1MJ per °C) and since you need 16 of them that's less than 8GJ total (they also hold 3.2k steam but whatever).

Now on my setup I use around 50 heat pipes, 50 pipes, 3 pumps and that totals to around 6k steam storage (which is a bit more than half a GJ) and 25GJ in the heat pipes.

I know I assume a full range of 500°C to 1000°C but I also assume no power consumption (which virtually reduces the storage from 100% to 0% with a power draw from 0% to 100%)
Anyway I've got 0.5 + 5 + 8 + 0.5 + 25 = 39GJ per reactor, which is more than the 32GJ that are shoved in each time.

Now I can compute my working temperature range (the maximum temperature difference between the coldest and hottest point in the system):
I've got 7GJ to spare on the 5+8+25=38GJ that depends on temperature, meaning if we use only 31/38 ~= 80% of the temperature range we would still have enough energy to hold all the cell power assuming no power consumption.
That means we can work with a 400°C temperature change, leaving us 100°C difference between the coldest and hottest point without using any steam tank, which is perfectly fine.

[mmmPI]

Well, mmmPI your math is wrong. Now, I dont have time explain why, but if no one else does, I would later.

I'll be waiting

maybe it's wrong on several places. like mathing and/or reasonning.

I prefer to talk about stuff per reactor, it eases 2xN setups.
A fuel cell gives off 32GJ due to the neighbor bonus (at least it tends toward that), which is 32/2.425 ~= 13.2 storage tanks full of steam.

I understand that rounding for 2N reactor, yet i calculated without rounding the adjacency thanks to using the wiki formula for computing maximum power (n-1)160 where n is number of reactor. the 160MW per core is with full adjacency, the "-160" account for the lesser adjacency for the exterior reactor.

Of course the higher the numbers the closer to 32 GJ, but for a 2X2 it's only 24GJ which is noticeable difference as it's still few cores.

For each reactor you have around 27.5 steam turbines, already holding 5500 steam, a bit less than half a GJ.
The nuclear reactor itself holds a few GJ of heat energy (10MJ per °C).
The heat exchangers are holding heat as well (1MJ per °C) and since you need 16 of them that's less than 8GJ total (they also hold 3.2k steam but whatever)
Now on my setup I use around 50 heat pipes, 50 pipes, 3 pumps and that totals to around 6k steam storage (which is a bit more than half a GJ) and 25GJ in the heat pipes.
[...]
Anyway I've got 0.5 + 5 + 8 + 0.5 + 25 = 39GJ per reactor, which is more than the 32GJ that are shoved in each time.
.

make sense to count all kind of buffer including the reactor themselves i computed the higher bound, if you discard storage that is not a storage tank. Neglecting heat stored in heat pipes and reactor themselves.

You do more precise calculations to make sure you have enough energy buffer on a number of GJ rounded up earlier where i just calculated the amount of steam buffer neglecting heat storage that is not steam in tanks, but on a more precise number of GJ.

The most accurate math would join both the real number of GJ per fuel cell factoring in adjacency bonus, and would calculate the energy buffer to count the number of tanks required while also taking in account the heat stored in reactor and exchanger and steam stored in turbine and exchanger. Otherwise the roundings we used makes the minimum buffer look too big. Which is not too bad as it's something we need to be in excess or just the right amount.

Now I can compute my working temperature range (the maximum temperature difference between the coldest and hottest point in the system):
I've got 7GJ to spare on the 5+8+25=38GJ that depends on temperature, meaning if we use only 31/38 ~= 80% of the temperature range we would still have enough energy to hold all the cell power assuming no power consumption.
That means we can work with a 400°C temperature change, leaving us 100°C difference between the coldest and hottest point without using any steam tank, which is perfectly fine.

i understand that you remove the 1GJ of potential energy storage under the form of steam stored in turbines or heat exchanger and pipes to the 39 GJ of your total energy storage capacity. This makes it 38GJ you can store under the form of heat, temperature rising after the steam is backed up.
Out of this 38GJ storage capacity you will use 32GJ, or 80%, so you calculate the temperature will rise to 80% of what it can reach. Since the range of temperature used to calculate the 38GJ is from 500° to 1000°, it gives 400° if we take 80% of that range away you're left with 100°.

at this point tough i'm not sure how/why you conclude that the 100° left are the difference between the coldest and hottest point.

I would say your math meant that 100° is what you have left on average on each heat storing component before it reaches the upper bound of the range which is 1000°. Meaning that your system will be around 900° on average after it absord a full fuel cycle while not producing anypower starting from a situation where it was at 500° on average.

When you say "coldest and hottest point" you mean over time right ? not at a particular moment the "coldest and hottest point of the system" right ?

Here a point of detail that bother me for the conclusion:

I know I assume a full range of 500°C to 1000°C but I also assume no power consumption (which virtually reduces the storage from 100% to 0% with a power draw from 0% to 100%)

This could be troublesome to your mathing due to how you conclude. ( i think)

I would say the hottest and coldest point of the system at any time depends on the layout used for heat pipes. (that's why i asked earlier)

If instead of being 500°C on average your system has reactor at 750° when the coldest point of the system ( the furthest exchanger most likely)is at 500°. This is around the value i have for the one i showed when running full power, but if my heat pipes were even more streched out, my reactors may need to reach 800° or 900° for the heat flow to start sustaining 500° at the furthest point.

I avoided this by just counting tanks. It allowed me to count an equivalent GJ=Tanks, that is overestimated as i counted all heat storage to 0. ( if you have "this" amount of tank, you are sure that you have at least the minimum because there is also some storage of energy under the form of heat. that's what i meant in conclusion i will add clarifications.) It stays out of the imprecisions of the heat conveying dynamics and instead just tells you "if you manage to convey heat properly and have this amount of tanks you are fine". it maybe less tanks, but the amout i said is (should) be guarantee 100% to be enough.

On the other hand, you assumed a capacity of storage ( like 25 GJ in heat pipes and some in reactor ) that may factually not be achieved due to a slow-to-propagate-heat layout. If when running to 100% power, the average temperature of your heatpipes is 750°, due to having a massive cluster near the reactor and then very long and thing branch to reach heat exchanger. Then average energy storage per heat pipe is not 500 MJ but 250 MJ if you were to refuel and suddendly just after refuel completly stop consumption that's what would be the effective storage capacity. ( situation 100% for long time then 0% for long time).

(i think) one of your assumption can work for theoric reasonning but your conclusion is telling something that needs to be validated on each heat pipe layout because it could be false in practice.

while my conclusion work for theory, as long as you make it work in practice. hahaha.

i'm not sure i want to look again at the fluid mechanic of the game and try to be more precise with theory on how to layout pipes to predict heat flow precisely enough to be able to have a conclusion as precise as the one you tried x).

[quyxkh]

Because of you I was playing around with trying to use 3 burner inserter, one for fuel injection one for fuel ejection, and another one to fuel the 2nd. If you put nuclear fuel, because that's what is done in nuke plant, you have 1210000kJ/70kJ=17285.7143 swing per nuclear fuel. Counting for fuel ejection and fuel injection every 200 second, that means 1 nuclear fuel would last 3457142.86 second in those 2, ( around 40 days ).

Which means the burner inserter that refuel the ejection inserter will do a swing every 40 days, which means it will need to confiscate one fuel to refuel itself every:

17 285.7143 x 3 457 142.86= 59 759 183 772.245 sec, 5.9E10 sec, that's 691 657 days, or 1894 years.

Is it bad that I feel no guilt?

If you keep your power consumption in a 1GW range you can have just one steerable 1GW reactor, everything else always-on. Gut check says very, very few bases have to worry about long-term power cycling in excess of 1GW, am I just too pedestrian here?

That means for almost all your power generation you can use stacked load/unload, one loading swing every 600s, with a loading inserter wired read-hold to the power switch on the unload circuit. Use fast inserters for unload, they can take 3 spent cells in the half-swing the power switch is closed, and you get ten minutes per swing, basically zero power consumption for the unload, 17285 swings is 120 *days* per fuel cell. One in the tank, one in the buffer, that's 240 days. Or I suppose it would be even cheaper to use just one burner for one reactor's reload, and use fast-inserters-powered-for-46-ticks-every-ten-minutes for all the rest.

At that point, you might as well just not bother with a refueling supply, your solstice festival can include a reactor-service-inserter-refueling celebratory ritual with lots of pomp and circumstance, maybe even an inserter dance!

[mrvn]

Interestingly enough, this totally validate clocking inserter on time based on your estimated consumption and building huge buffers. Because no matter how much energy you need, you'd be more efficient burning more fuel at the same time, and then stopping for some time.



The bigger buffer you can have, the more efficiency you can have !


feel free to tell me if i did the math wrong

I don't see that at all. A timed approach always assumes the time you set is right. Which it never is unless you go for 100% duty cycle. Any lower and consumption will have valleys and spikes and not match your timing.

You need a feedback mechanism to regulate fuel input to actual energy consumption. And that means measuring steam or accumulator levels. There is no way around that for an efficient plant.


As for bigger buffers: You are wrong.

Say you have a buffer of X tanks. You say X+1 tanks will be more efficient with the same amount of reactors and fuel insertion. But that would mean X tanks is wasting fuel, the reactors are overheating and energy is lost. Clearly if you have more empty tanks than one round of fuel can fill with steam (and your heat pipes aren't totally stupid) then no heat gets lost. And you can't be more efficient than 0 loss.

What the math shows is that more reactors in a 2xN setup is more efficient and even that doesn't hold if you look closely. The neighbor bonus approaches 3 per reactor while a 2x2 only has 2 per reactor. Each additional reactor pair gains some efficiency but that goes towards 0. At some point the loss due to more inserters eats up the gain. So for a 480MW reactor there actually is optimal size.

[mmmPI]

I don't see that at all. A timed approach always assumes the time you set is right. Which it never is unless you go for 100% duty cycle. Any lower and consumption will have valleys and spikes and not match your timing.

You need a feedback mechanism to regulate fuel input to actual energy consumption. And that means measuring steam or accumulator levels. There is no way around that for an efficient plant.

I see 3 differents points that need clarifications :

1) usual feedback mechanism to regulate fuel works with 1 cycle delay, you consume energy first then measure how much, then refuel accordingly.This is why you need to buffer the energy in the first place, in case the next cycle is no consumption. This means you have the information beforehand, you know how much you have consumed, so you can put the timer exact if you want, no need to guess so you can set the timer 100% of time right. Have you forgotten how/why your automatic machine functions ? Just look at the previous consumption averaged as you can read it when you click a power pole.
If you have a 1 hours buffer of energy it's very easy to read the last hour value, and set the timer to refill the exact quantity of heat that was consumed. Just a % of the max capacity.

2) it doesn't matter if the consumption have valleys and spike or if you don't need 100% energy, the only important thing is that you can refill the amount of energy consumed and that you can read the average consumption. If you can buffer 20 cycle of fuel, you just read the consumption for the last hour. Then you set to refill exactly what you have consumed. So if your consumption was 240 MW withs tons of valley and spike the last hour, it means the last hour the power plant shoudl have been 50% of the time burning fuel and 50% of the time idling with a used fuel cell for example. Now if you put the timer to 400 sec. The next hour what will happen is that the power plant will refuel and burn during 200 sec, then wait 200 sec , then refuel and burn during 200 sec then. At the end of this hour you will have refilled what was consumed the hour before. You read your new hourly consumption, and prepare to refill the buffer during the next hour.

If you need 100% of energy, it's just simpler it means the next hour you can't get caught not burning fast enough for the running hour in case it would more than the previous.
(Which again stress out the importance of buffer x) )


3) efficient doesn't mean self-regulated, it's efficient when you get more MW per fuel cell, you said it yourself earlier than this thread was about MW per fuel cell. If you consume a constant amount of energy you don't a feedback mechanism to be efficient. worse : you get more MW per fuel cell if you don't.

You say X+1 tanks will be more efficient with the same amount of reactors and fuel insertion

no i don't say that, you're trying to strawman me or what ?

ofc you need to adapt the number of reactor if you have a bigger buffer. Just make your power plant 10x the required size and use it 10% of the time. that's the meaning of the reasonning i'm sorry it wasn't explained clearly enough

I said : The bigger buffer you can have, the more efficiency you can have !

I said it require a buffer because if you are going to use your maxi-oversized 100X power plant but just 1% of the time to get supergood efficency, the footprint of the nuclear plant would most likely be 90% heatpipes/steamtanks. hence i didn't say the higher number of core you have, because that's just a tiny bit of what would be a super-efficient power plant.the bigger buffer you can have the more efficiency you can have !i maintain

What the math shows is that more reactors in a 2xN setup is more efficient and even that doesn't hold if you look closely. The neighbor bonus approaches 3 per reactor while a 2x2 only has 2 per reactor. Each additional reactor pair gains some efficiency but that goes towards 0. At some point the loss due to more inserters eats up the gain. So for a 480MW reactor there actually is optimal size.

No way my math shows something like that.

if you take 1 reactor and 1 inserter. The ratio [cost of inserting the fuel/energy worth of 1 swing] is biggest.
if you take 2 reactor and 2 inserter. The [cost of inserting the fuel] has doubled, [the energy worth of 1 swing] a bit more than doubled.
if you take 4 reactor and 4 inserter. The [cost of inserting the fuel] has doubled, [the energy worth of 1 swing] a tiny bit more than doubled.

At some point it only look like [the energy worth of 1 swing] double without the tiny bits. But it never look like [the energy worth of 1 swing] would less than double right ?

What i've highlighted in your reasonning doesn't appear to be true i don't think there is a "loss due to more inserter eating up the gain" maybe i'm wrong i would be curious to know what would be the proposed method to calculate the optimal number of core to produce 480MW of constant energy because i can't think of one.

[mmmPI]

Is it bad that I feel no guilt?

why would you ask ME such things ? it was just a joke playing with inserter wasn't that much trouble

If you keep your power consumption in a 1GW range you can have just one steerable 1GW reactor, everything else always-on. Gut check says very, very few bases have to worry about long-term power cycling in excess of 1GW, am I just too pedestrian here?

That's again a difficult question, there's 2 things i'm not sure i understand correctly :

1) You mean if you consume 22GW or 21 GW roughly that would be in 1GW range ? and then you have a steerable 1 GW reactor always on ? i'm lost there, for 1GW you'd need between 6 and 8 cores depending on how many you have already. what does steerable means ? like what kind of control you'd want ?

2) It doesn't sound like "pedestrian" is used to describe your means of locomotion but i'm not sure which other meaning of the word in the list of synonym i should pick before answering the question, the list being " Lacking in distinction or imaginativeness; ordinary; commonplace; dull; insipid; prosaic" and not willing to be offensive i would just answer no no it's fine.

That means for almost all your power generation you can use stacked load/unload, one loading swing every 600s, with a loading inserter wired read-hold to the power switch on the unload circuit. Use fast inserters for unload, they can take 3 spent cells in the half-swing the power switch is closed, and you get ten minutes per swing, basically zero power consumption for the unload, 17285 swings is 120 *days* per fuel cell. One in the tank, one in the buffer, that's 240 days. Or I suppose it would be even cheaper to use just one burner for one reactor's reload, and use fast-inserters-powered-for-46-ticks-every-ten-minutes for all the rest.

At that point, you might as well just not bother with a refueling supply, your solstice festival can include a reactor-service-inserter-refueling celebratory ritual with lots of pomp and circumstance, maybe even an inserter dance!

oddly enough this i understand better i think haha,

1) cut on the energy drain from inserters doing the unload using power switch that activate thanks to the burner who itself has no drain and therefore can wait from the logic signal for free.

2)cut on the nuclear fuel consumption from the feeding inserter and energy from unload swings by stacking the cell 3 by 3 when doing the loading and unloading.

Doing the later you triple the life expectancy of the nuclear fuel in the feeding burner inserter from 40 days if a refuel every 200 second, to 120 days of ingame time. or twice that if you put 2 nuclear fuel in the burner by rotating it 4 time to make the first one disappear right away.

The condition for doing 2) is again having a bigger buffer for energy i keep saying it ! because one would then need to have a minimum buffer of energy able to handle 3 cells burnt in a row, with possibly no electric consumption for those 10 minutes.

I still can't find out the meaning of the previous sentence which sounds like conditions for using those mechanism ?

maybe even an inserter dance!

i'm considering it

[mrvn]

I don't see that at all. A timed approach always assumes the time you set is right. Which it never is unless you go for 100% duty cycle. Any lower and consumption will have valleys and spikes and not match your timing.

You need a feedback mechanism to regulate fuel input to actual energy consumption. And that means measuring steam or accumulator levels. There is no way around that for an efficient plant.

I see 3 differents points that need clarifications :

1) usual feedback mechanism to regulate fuel works with 1 cycle delay, you consume energy first then measure how much, then refuel accordingly.This is why you need to buffer the energy in the first place, in case the next cycle is no consumption. This means you have the information beforehand, you know how much you have consumed, so you can put the timer exact if you want, no need to guess so you can set the timer 100% of time right. Have you forgotten how/why your automatic machine functions ? Just look at the previous consumption averaged as you can read it when you click a power pole.
If you have a 1 hours buffer of energy it's very easy to read the last hour value, and set the timer to refill the exact quantity of heat that was consumed. Just a % of the max capacity.

Well, the previous comments about timing where about setting a delay. I read it as you set one in e.g. a constant combinator and then you get a fuel cell every 362s or whatever value you set.


There are 2 general ways (with a few variations, see at the start of the thread) to measure consumption: Throughput or buffered amount.

You can measure the throughput of the steam with some clever tank and pump tricks. That is the only practical way to actually measure how much energy you consume anywhere realtime.

Or you measure the amount of steam (tanks) or power (accumulator) buffered. This tells you when the reactors energy production is below the consumption plus gives you the amount of energy you can buffer even if nothing will be consumed in the future. You can time how soon the buffer decreases and if it does how fast to estimate the current consumption. The first has a rather long lag (200s for the fuel cells to be used up / fraction of power used). The second is usually falling very quickly and there is a long delay between inserting fuel and full power generation again. So the later needs huge buffers.

But really you don't want to measure consumption. At best it's a prediction of the future from past events. Maybe the next second aliens will attack and power consumption with increase 10 fold. You can't know.

So the better use of the measurement is to determine when it is safe to put in another fuel cell. Only use the measure of how much additional energy you can buffer right now. If that is less than 1 fuel cell generates then throw in one fuel cell. Most of the designs shown so far use that method. If you predict power usage you can potentially insert fuel cells earlier (or later) and use fewer steam tanks or accumulators. But if you guess wrong you either waste fuel or have a brownout. And steam tanks are cheap and you only need like 2 tanks to be on the safe side.

2) it doesn't matter if the consumption have valleys and spike or if you don't need 100% energy, the only important thing is that you can refill the amount of energy consumed and that you can read the average consumption. If you can buffer 20 cycle of fuel, you just read the consumption for the last hour. Then you set to refill exactly what you have consumed. So if your consumption was 240 MW withs tons of valley and spike the last hour, it means the last hour the power plant shoudl have been 50% of the time burning fuel and 50% of the time idling with a used fuel cell for example. Now if you put the timer to 400 sec. The next hour what will happen is that the power plant will refuel and burn during 200 sec, then wait 200 sec , then refuel and burn during 200 sec then. At the end of this hour you will have refilled what was consumed the hour before. You read your new hourly consumption, and prepare to refill the buffer during the next hour.

If you need 100% of energy, it's just simpler it means the next hour you can't get caught not burning fast enough for the running hour in case it would more than the previous.
(Which again stress out the importance of buffer x) )

Well, if you really do want to buffer 20 cycles of fuel.... Normal design shown so far buffer 1-2 cycles only. Mine (1 tank per reactor) doesn't even buffer 1 cycle because I assume a base load on the reactor that consumes the rest.


And there you go again saying "if you put the timer to 400 sec". Me? I don't want to put anything. The reactor should run itself. Or do you mean have some combinator construct that calculates the time to 400 sec? Please be precise.

Note: If you have 20 cycles buffer then you can just throw in a fuel cell whenever the buffer is below 15 cycles. At most 5 fuel cells get inserted so even if consumption stops the most you reach is full buffers. No need to measure times and estimate consumption. Just set the inserters to "steam < 15 cycles". No combinators needed, which safes energy and therefore is more efficient.

3) efficient doesn't mean self-regulated, it's efficient when you get more MW per fuel cell, you said it yourself earlier than this thread was about MW per fuel cell. If you consume a constant amount of energy you don't a feedback mechanism to be efficient. worse : you get more MW per fuel cell if you don't.

You say X+1 tanks will be more efficient with the same amount of reactors and fuel insertion

no i don't say that, you're trying to strawman me or what ?

ofc you need to adapt the number of reactor if you have a bigger buffer. Just make your power plant 10x the required size and use it 10% of the time. that's the meaning of the reasonning i'm sorry it wasn't explained clearly enough

I said : The bigger buffer you can have, the more efficiency you can have !

I said it require a buffer because if you are going to use your maxi-oversized 100X power plant but just 1% of the time to get supergood efficency, the footprint of the nuclear plant would most likely be 90% heatpipes/steamtanks. hence i didn't say the higher number of core you have, because that's just a tiny bit of what would be a super-efficient power plant.the bigger buffer you can have the more efficiency you can have !i maintain

What the math shows is that more reactors in a 2xN setup is more efficient and even that doesn't hold if you look closely. The neighbor bonus approaches 3 per reactor while a 2x2 only has 2 per reactor. Each additional reactor pair gains some efficiency but that goes towards 0. At some point the loss due to more inserters eats up the gain. So for a 480MW reactor there actually is optimal size.

No way my math shows something like that.

if you take 1 reactor and 1 inserter. The ratio [cost of inserting the fuel/energy worth of 1 swing] is biggest.
if you take 2 reactor and 2 inserter. The [cost of inserting the fuel] has doubled, [the energy worth of 1 swing] a bit more than doubled.
if you take 4 reactor and 4 inserter. The [cost of inserting the fuel] has doubled, [the energy worth of 1 swing] a tiny bit more than doubled.

At some point it only look like [the energy worth of 1 swing] double without the tiny bits. But it never look like [the energy worth of 1 swing] would less than double right ?

What i've highlighted in your reasonning doesn't appear to be true i don't think there is a "loss due to more inserter eating up the gain" maybe i'm wrong i would be curious to know what would be the proposed method to calculate the optimal number of core to produce 480MW of constant energy because i can't think of one.

You are forgetting that inserters drain energy while idle. If you have 1000 reactors or 2000 reactors the power per swing basically doubles. The cost of the swing doubles. But at the 480MW consumption rate the time between refuels basically doubles. So that balances out. The tiny bit more is near 0. It's there but near 0. The time of the swing compared to time between swings is, well, rounding error . So you have double the number of inserters and twice the time they are idle. That means the inserters consume 4 times as much power per cycle.

It's easier if you do the math with a fixed fuel cells per second value. Then 2x2N reactor will swing half as often as a 2xN reactor. Keeps the numbers simpler. Don't compare cycles as the fuel input per cycle isn't constant. Twice the number of inserters means double the power drain if you ignore the swing. The problem is that "a tiny bit more than doubled" has to make up for the double power drain of the inserters. That works going from 4 to 8 reactors. I don't know where it stops working but I suspect 1000 to 2000 won't make up the increased drain.

Earlier someone suggested using burner inserter and showed the math how much energy the inserters consume and how much burner inserter would safe. So my suggestion would be to scroll back to that and use that math.

Note: If you use burner inserter then there is no drain, only the swing. All the above is for electrical inserters. With burner inserters the bigger the better unless you have a combinator per reactor or other drain.