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In theory, MOX fuel:
- has a lifetime of 5,000 seconds, half as much as uranium
- has exactly the same base EU output, heat output and efficiency scaling values as uranium
- generates more EU the closer to 100% heat the reactor is, up to x5 at 100%
- generates more EU the higher the reactor's actual heat level is (scaling factor unknown, but smaller than heat%) (was a bug, got fixed)
The takeaway:
- 100% reactor heat is, shall we say, "slightly impractical". A more realistic number is x4.4 at 85% heat, or x3.8 at 70% heat (no hurt)
- Neutron reflectors last two times as many cycles and yield much higher total EU bonuses, potentially making them cost effective for once
- The reactor must maintain its heat level exactly in order to give good results; this can be harder than you think
Why can it be harder than you think? Because even designs that on the surface have cooling power equal to heat output will actually be unable to run at heat levels greater than 0. That is because you also need to look at a number the planner won't tell you anything about, and that is core transfer rate. Many reactor designs involving overclocked vents can actually pull more heat from the hull per tick than they are capable of dissipating. This works fine so long as common uranium reactors run at zero heat, but start the system off with several thousand degrees and the cooling system will melt itself even though it seemed stable in the planner.
The following components have core transfer rates:
- Heat exchanger, 4 core transfer
- Advanced heat exchanger, 8 core transfer
- Reactor heat exchanger, 72 core transfer
- Reactor heat vent, 5 core transfer
- Overclocked heat vent, 36 core transfer
The reactor vent has 5 cooling, and thus will keep itself perfectly balanced on its own. The three exchangers will only utilize their transfer rates in order to make their own temperature (and that of adjacent components) match the reactor hull. The overclocked vent, however, will always pull the full 36 if it can, and it only has 20 cooling of its own. It will melt itself given the chance. However, so long as we know these transfer rates, we can use these parts build a reactor that satisfies the requirements for running MOX fuel.
The simplest reactors, of course, get away without using core transfer at all: sample 1, sample 2 from another thread
They simply have the fuel rods transfer heat directly into an adjacent component that can accept heat. If such a component is available, the fuel rods will never transfer heat to the core. Thus the reactor will maintain its heat level perfectly, and all we have to worry about is to provide enough cooling. The downside, of course, is that you're either running fairly low efficiency numbers (every fuel rod side occupied by a cooling component is one you can't use for another fuel rod or a reflector), or you'll invariably need a CRCS system to refresh spent coolant cells. Also, advanced vents are very expensive.
The other option is to deal with core transfer rate: sample 1, sample 2 that I just made to illustrate the point
This results in much higher efficiency reactors, with the downside being that efficiency means heat, and heat means you need to spend a lot of room on components. Which means less room for plating, which means less total heat capacity, which means less of an output multiplier. The trick question here is whether or not there is a point at which you lose more output by removing heat plating than you get from building a higher efficiency reactor. Absolute heat level seems to be a much smaller multiplier than total heat percentage, but it needs to be examined.
Another downside of this approach is the fact that if you switch the reactor off (or the fuel rods run out), then the reactor will cool itself down to 0 unless you remove all cooling components or swap the fuel rods quickly. If you're slow, you might end up having to re-heat the components, or even the whole reactor, for the next cycle.
Design Q&A:
- why does sample 1 work, the core transfer is higher than both heat output and available cooling! Answer: because the reactor heat exchanger only draws as much as it needs to keep itself at the same temperature as the hull. That means it will draw exactly as much as the cooling components steal from it each tick, which is exactly as much as the fuel rod pumps into the hull each tick.
- why does sample 2 work, it has more cooling than heat output! Answer: because it only has enough core transfer to pull 168 heat per tick from the hull, even though the cooling system could handle 172. If it bothers you, you can move one of the component vents so it only touches one component instead of two... but then you lose THE SYMMETRY!
Please share your thoughts on the subject, and most importantly, your design ideas. Let's come up with a few that are worthy of a new section in the Official Reactor Design thread!