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You mentioned “Fusion reactors also have strong diseconomies of scale” so that’s what I am referring to. A fusion reactor 2x larger with the same energy flow p

by Retric 2y ago

You mentioned “Fusion reactors also have strong diseconomies of scale” so that’s what I am referring to. A fusion reactor 2x larger with the same energy flow per m2 of wall could be cheaper per kWh than a more compact reactor. The exact details matter a great deal and not every design works at every scale. Anyway, a fission reactor’s complexity is mostly outside the device. Things like multiply redundant cooling systems, backups for your backup generators, fuel processing, etc that are completely unnecessary or simpler on a fusion reactor. So overall a large fusion power plant could easily be a simpler overall system.

pfdietz 2y ago
The diseconomy of scale follows thus: The magnetic pressure (and hence density, at given beta and temperature) scales as B^2. The fusion power density scales as density squared, or as B^4. Now, the mass of the support structure scales as volume x B^2. So, power per support structure mass (and hence, per cost of that structure) scales as B^2. However, fusion power density is limited by what the first wall can withstand. Therefore, a large reactor must reduce B (or, reduce beta) to stay within this limit. Reducing B reduces the power/support mass, and hence power/$ of support structure cost. Fusion reactors require equipment outside the reactor as well, probably much more complex than in a fission reactor. Online tritium recovery, robotics equipment for internal maintenance, and space for first wall replacement (since known materials do not survive the life of the reactor, unlike structures in a fission reactor core.) If you look at ITER, there's a large volume around the core where robotic maintenance devices can slide on a track. The ARC concept required volume and machinery for lifting the entire top of the reactor and replacing the vacuum vessel (then disassembling the old, activated vessel.)
- Retric 2y ago
  > Now, the mass of the support structure scales as volume X B ^2 No it does not, just as an example the fusion blanket is a constant thickness for a given level of neutron flux per m2 and sub linear scaling as flux increases. Same deal with your cooling loop etc. You’re making several other mistakes, but because at a fundamental level the costs relationships are more complex than what you’re modeling. Consider despite being a novel experimental design and having a huge research staff etc, Vogtle Electric Generating Plant still cost more than ITER to build. Sure the scale is an order of magnitude smaller, but fission has been around for a half century at this point is seemingly much simpler and you’d assume there was a great number of efficiency gains to be had. https://en.wikipedia.org/wiki/Vogtle_Electric_Generating_Plant https://en.wikipedia.org/wiki/Vogtle_Electric_Generating_Pla...
  - pfdietz 2y ago
    Yes, it does. This follows from the fact that structural mass is proportional to stored magnetic energy, just as the mass of a pressure vessel is proportional to volume x pressure (for materials of a given strength). Viewed another way: if we double B, we quadruple the forces on each segment of conductor, and therefore must quadruple the strength of the supports. I explicitly said "support structure", the structure that is resisting the force of the magnets. There are other parts that scale more slowly, but I don't need to show that all components scale poorly to show that overall there are diseconomies of scale. Note that this is NOT like in a fission reactor. In a fission reactor, everything scales just about linearly with power.
    - Retric 2y ago
      > I don't need to show that all components scale poorly to show that overall there are diseconomies of scale. Yes you do. If A represents 0.001% of the cost then increasing it by 20x doesn’t matter and the support structure is cheap. Same deal with a fission reactor there’s several diseconomies of scale for example around the length of control rods passive cooling after an accident etc, most of them just don’t matter much.
      - pfdietz 2y ago
        Very well. I will simply note that 2/3rds of the mass of ARC is the support structure for the magnets. And the rest of the mass includes the blanket.
        Retric 2y ago
        Now just look up how expensive steel is per kg vs superconductors or lithium and … ahh yea weight is a poor predictor of cost.
        pfdietz 2y ago
        You do realize that almost all the mass of a PWR reactor is steel, right? The steel pressure vessel, steel supports for the fuel bundles. So the argument you are making there is that the mass discrepancy between fusion reactors and PWRs grossly understates the difference in cost.

by Retric 2y ago

pfdietz 2y ago

The diseconomy of scale follows thus: The magnetic pressure (and hence density, at given beta and temperature) scales as B^2. The fusion power density scales as density squared, or as B^4. Now, the mass of the support structure scales as volume x B^2. So, power per support structure mass (and hence, per cost of that structure) scales as B^2. However, fusion power density is limited by what the first wall can withstand. Therefore, a large reactor must reduce B (or, reduce beta) to stay within this limit. Reducing B reduces the power/support mass, and hence power/$ of support structure cost. Fusion reactors require equipment outside the reactor as well, probably much more complex than in a fission reactor. Online tritium recovery, robotics equipment for internal maintenance, and space for first wall replacement (since known materials do not survive the life of the reactor, unlike structures in a fission reactor core.) If you look at ITER, there's a large volume around the core where robotic maintenance devices can slide on a track. The ARC concept required volume and machinery for lifting the entire top of the reactor and replacing the vacuum vessel (then disassembling the old, activated vessel.)

Retric 2y ago
> Now, the mass of the support structure scales as volume X B ^2 No it does not, just as an example the fusion blanket is a constant thickness for a given level of neutron flux per m2 and sub linear scaling as flux increases. Same deal with your cooling loop etc. You’re making several other mistakes, but because at a fundamental level the costs relationships are more complex than what you’re modeling. Consider despite being a novel experimental design and having a huge research staff etc, Vogtle Electric Generating Plant still cost more than ITER to build. Sure the scale is an order of magnitude smaller, but fission has been around for a half century at this point is seemingly much simpler and you’d assume there was a great number of efficiency gains to be had. https://en.wikipedia.org/wiki/Vogtle_Electric_Generating_Plant https://en.wikipedia.org/wiki/Vogtle_Electric_Generating_Pla...
- pfdietz 2y ago
  Yes, it does. This follows from the fact that structural mass is proportional to stored magnetic energy, just as the mass of a pressure vessel is proportional to volume x pressure (for materials of a given strength). Viewed another way: if we double B, we quadruple the forces on each segment of conductor, and therefore must quadruple the strength of the supports. I explicitly said "support structure", the structure that is resisting the force of the magnets. There are other parts that scale more slowly, but I don't need to show that all components scale poorly to show that overall there are diseconomies of scale. Note that this is NOT like in a fission reactor. In a fission reactor, everything scales just about linearly with power.
  - Retric 2y ago
    > I don't need to show that all components scale poorly to show that overall there are diseconomies of scale. Yes you do. If A represents 0.001% of the cost then increasing it by 20x doesn’t matter and the support structure is cheap. Same deal with a fission reactor there’s several diseconomies of scale for example around the length of control rods passive cooling after an accident etc, most of them just don’t matter much.
    - pfdietz 2y ago
      Very well. I will simply note that 2/3rds of the mass of ARC is the support structure for the magnets. And the rest of the mass includes the blanket.
      - Retric 2y ago
        Now just look up how expensive steel is per kg vs superconductors or lithium and … ahh yea weight is a poor predictor of cost.
        pfdietz 2y ago
        You do realize that almost all the mass of a PWR reactor is steel, right? The steel pressure vessel, steel supports for the fuel bundles. So the argument you are making there is that the mass discrepancy between fusion reactors and PWRs grossly understates the difference in cost.