Nuclear Fuel Cycle

by muink

Efficient use of Nuclear Fuel.

Content
5 years ago
0.16 - 0.17
14
Power

g 1 atomic bomb = 1 u235 + 15 u238 ?

6 years ago
(updated 6 years ago)

this seems a bit imbalanced.. the vanilla atomic bomb needs 30 u235.
i suggest 10 breeder fuel cells producing like 15 plutonium instead of 38

6 years ago

The 13 pu-239 in atomic bomb will be adjust to 20

6 years ago

it needs to be at least 26

6 years ago
(updated 6 years ago)

the critical mass based on u235 is 15kg and the critical mass of pu239 is 10kg
so 20 is a reasonable value without considering power
and breeder recipes i will reconsider

6 years ago
(updated 6 years ago)

I have adjusted breeder recipes in v0.1.1
🍴

6 years ago
(updated 6 years ago)

now 1 atomic bomb is 20 u238...
i'm not really sold on this balance yet.
i guess the actual price is the time it takes? (at 3.5 gj per fuel cell still not a whole lot)

6 years ago
(updated 6 years ago)

yes
produce one U bomb: 50s * 30 + 50s = 1550s

kovarex enrichment process: 1 U235/50s
atomic bomb energy required: 50s
atomic bomb U235 required: 30

produce one PU bomb: 2s * 10 + 10 * 3500MJ / 40MW/s + 80s + 50s = 1025s

produce 10 breeder fuel: 1 fuel/2s * 10
time to burn breeder fuel: 3.5GJ / 40MW/s
process 10 used breeder fuel: 20 PU239/80s
atomic bomb energy required: 50s
atomic bomb PU239 required: 20

but considering that the centrifuge can use speed module, and the manufacturing cost is cheaper than the reactor.
U-bomb can be manufactured faster.

and future will also add startup settings to customize ingredient and prevent the reactor no-load work

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