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The latest ARC claims from Whyte's last presentation was 3MW/m^3 density, and that's with currently achievable magnets, not those people believe are ultimately

The latest ARC claims from Whyte's last presentation was 3MW/m^3 density, and that's with currently achievable magnets, not those people believe are ultimately achievable. Counting just the PWR reactor vessel, and not the entire containment facility, seems a huge cheat. We're talking economics of energy. PWR Fission reactors have enormous footprints, averaging 1 square mile. An ARC reactor wouldn't need anywhere need that size. The very nature of using pressurized water means your containment facility has to be a factor of 1000 larger than the reactor vessel to handle a flash boil over. Current costs to build a 1GW fission plant are $6-9 billion. Predicted costs to build a 200MW ARC reactor are $4-5 billion, so the ARC reactor costs ~5x more at the high end, and 77% more at the low end, for the same output (e.g. build 5 ARC reactors to equal one fission plant). The economics are still viable, and your costs for storage, disposal, and de-comissioning are far lower, as is your land footprint. Note that these ARC costs are for the first reactor, include the sunk development costs, and not the economies of scale that would occur from dropping HTS costs. > It's all about the limits on areal power density at the reactor walls. No, he makes lots of assumptions about how the magnets are shielded, how breeding works, and how heat transfer works. All of these are addressed by FLiBe, there's no "structure" to the heat blanket to damage, it acts simultaneously as radioactive shielding for the magnets, heat transfer, and a breeder. The "blanket" in ARC isn't a surface sandwiched with a super structure, it's a volume that's continuously pumped and recirculated. He also makes the implicit assumption that reactor wall damage is a huge problem. Instead of seeing it as a problem, you should see it as a solution, like an ablative heat shield on a rocket or capsule. As long as the cost of maintenance or replacement is low, and reactor downtime is limited, it is not a problem. Current estimates for ARC, which is designed to be modular, so it can be easily maintained, is that the reactor walls will last about 1 year. He doesn't even mention the biggest unsolved engineering problem, removing the "ash" (Helium) that accumulates in the plasma. His article has all the trappings of reusable rocket naysayers in the 80s and 90s, who claimed reusables could never work, because carrying wings, landing legs, heat shielding, and structural reinforcement to make rapid reusability would cut down on your usable payload, and thus doom reusable rockets to always be substantially inferior. And this is because of an obsession with a single variable, like power density, or payload fraction, and ignoring the end to end economics of the entire system, including safety margins, maintenance, footprint, etc. And that's why I believe ARC will prove Lidsky wrong.

pfdietz 6y ago
> The latest ARC claims from Whyte's last presentation was 3MW/m^3 density, That's per volume of PLASMA, not volume of REACTOR. Since we're talking about the size of the things we have to pay for (the reactor) the latter is appropriate. > Counting just the PWR reactor vessel, and not the entire containment facility, seems a huge cheat. Fusion reactors will also require containment vessels, to prevent escape of tritium in accidents (unless you're ok with an accident rendering ground water in a large area around the reactor undrinkable for decades.) The containment vessel will have to be large enough to contain all the cryogens heated to accident temperature. Unlike in a PWR like the AP1000, these volatiles cannot be condensed by water cooling. In ITER, for example, complete volatilization of the cryogens would requires several times the volume of the AP1000 containment building. The ARC design adds to accident fun by having a large volume of titanium hydride shielding near the molten salt. TiH2 decomposes to Ti and hydrogen gas at 450 C, well below the temperature of the molten salt blanket. Fully decomposed, this TiH2 would release 15 tonnes of hydrogen, occupying 180,000 cubic meters at STP. Beyond that, the containment building around a fusion reactor will have to mitigate tritium leakage by many orders of magnitude. This will be expensive (in particular, polymer seals cannot be used on penetrations as they permeate tritium.) ARC will also require a large volume for disassembly of the reactor for vacuum vessel replacement, and for remote control demolition of activated vacuum vessels for disposal. All that volume will become contaminated and off limits to human access. > PWR Fission reactors have enormous footprints, averaging 1 square mile. An ARC reactor wouldn't need anywhere need that size. Really good farmland in the US is less than $10K/acre, or about $6M/square mile. Cheap range land is an order of magnitude less; really useless land another order of magnitude cheaper. The cost of land there is a very minor part (0.1%) of the cost of the power plant. https://www.howmuchisit.org/acre-of-land-cost/ https://www.howmuchisit.org/acre-of-land-cost/ > No, he makes lots of assumptions about how the magnets are shielded, how breeding works, and how heat transfer works. Again, you have not understood his argument. His argument would apply even if you had 100T magnets made of unobtainium and a plasma configuration with beta = 1. All that he requires is an upper limit on how much energy per area can be transmitted through the surface of the reactor. This is a function of the materials of the first wall, and has absolutely nothing to do with magnets or breeding. > Predicted costs to build a 200MW ARC reactor are $4-5 billion If you are referring to the cost estimate from the arxiv paper, that is just the estimated cost of the reactor itself, not the power plant containing the reactor. I will admit I found the estimation methodology in that paper very cursory, so it's possible they were overestimating the cost (80% of the cost was the stainless steel magnet support structure.) A fission reactor core is much cheaper than that estimate, per MW. > He doesn't even mention the biggest unsolved engineering problem, removing the "ash" (Helium) that accumulates in the plasma. And why should he? I mean, why does he have to kill the corpse a third time (after the power density and reliability arguments)? > And that's why I believe ARC will prove Lidsky wrong. As I said earlier, the power density figures from ARC show he was right.

pfdietz 6y ago

> The latest ARC claims from Whyte's last presentation was 3MW/m^3 density, That's per volume of PLASMA, not volume of REACTOR. Since we're talking about the size of the things we have to pay for (the reactor) the latter is appropriate. > Counting just the PWR reactor vessel, and not the entire containment facility, seems a huge cheat. Fusion reactors will also require containment vessels, to prevent escape of tritium in accidents (unless you're ok with an accident rendering ground water in a large area around the reactor undrinkable for decades.) The containment vessel will have to be large enough to contain all the cryogens heated to accident temperature. Unlike in a PWR like the AP1000, these volatiles cannot be condensed by water cooling. In ITER, for example, complete volatilization of the cryogens would requires several times the volume of the AP1000 containment building. The ARC design adds to accident fun by having a large volume of titanium hydride shielding near the molten salt. TiH2 decomposes to Ti and hydrogen gas at 450 C, well below the temperature of the molten salt blanket. Fully decomposed, this TiH2 would release 15 tonnes of hydrogen, occupying 180,000 cubic meters at STP. Beyond that, the containment building around a fusion reactor will have to mitigate tritium leakage by many orders of magnitude. This will be expensive (in particular, polymer seals cannot be used on penetrations as they permeate tritium.) ARC will also require a large volume for disassembly of the reactor for vacuum vessel replacement, and for remote control demolition of activated vacuum vessels for disposal. All that volume will become contaminated and off limits to human access. > PWR Fission reactors have enormous footprints, averaging 1 square mile. An ARC reactor wouldn't need anywhere need that size. Really good farmland in the US is less than $10K/acre, or about $6M/square mile. Cheap range land is an order of magnitude less; really useless land another order of magnitude cheaper. The cost of land there is a very minor part (0.1%) of the cost of the power plant. https://www.howmuchisit.org/acre-of-land-cost/ https://www.howmuchisit.org/acre-of-land-cost/ > No, he makes lots of assumptions about how the magnets are shielded, how breeding works, and how heat transfer works. Again, you have not understood his argument. His argument would apply even if you had 100T magnets made of unobtainium and a plasma configuration with beta = 1. All that he requires is an upper limit on how much energy per area can be transmitted through the surface of the reactor. This is a function of the materials of the first wall, and has absolutely nothing to do with magnets or breeding. > Predicted costs to build a 200MW ARC reactor are $4-5 billion If you are referring to the cost estimate from the arxiv paper, that is just the estimated cost of the reactor itself, not the power plant containing the reactor. I will admit I found the estimation methodology in that paper very cursory, so it's possible they were overestimating the cost (80% of the cost was the stainless steel magnet support structure.) A fission reactor core is much cheaper than that estimate, per MW. > He doesn't even mention the biggest unsolved engineering problem, removing the "ash" (Helium) that accumulates in the plasma. And why should he? I mean, why does he have to kill the corpse a third time (after the power density and reliability arguments)? > And that's why I believe ARC will prove Lidsky wrong. As I said earlier, the power density figures from ARC show he was right.