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I agree. I enjoy anything she produces that is physics-related but the social commentary and science topics outside her area of expertise are hit-or-miss. I als
by timmytokyo 3y ago
I agree. I enjoy anything she produces that is physics-related but the social commentary and science topics outside her area of expertise are hit-or-miss. I also think she opens herself up to audience capture, especially now that she's being funded directly by viewers, and that concerns me.
- Retric 3y agoShe also made some significant errors in physics videos. One example is she calculated the power output of fusion reactions as energy created by fusion * heat to electricity conversion efficiency. However, energy that’s heating the plasma should also recoverable at the same conversion efficiency. Largely irrelevant a Q=1,000 but it plays a major role in what Q is breakeven and what Q is economically useful. Not on the surface such a big deal, but a sign she should reach out more to actual experts in these fields.
- jiggawatts 3y agoSo… she’s correct in the industrial regime that matters, but technically incorrect in the experimental regime where the energy is not recovered?
- Retric 3y agoIf she was talking about very high Q then it wouldn’t be a major issue, but she was specifically referring to the regimes where it was important. Aka energy breakeven and minimum economic viability. It also explains why Q is formulated in terms of energy added to the plasma vs energy gained from fusion which she didn’t understand. This formula for Q ends up being a term in the equation for economic viability, but she attacked it as if such a formulation was misleading. Basically the total energy spent on the facility including cryogenic cooling, pumps, etc matters + Q as defined by fusion experts + fusion output energy + conversion energy lets you find the total net energy gain which you then compare with the cost of the facility to find economic viability. Roughly, [ (Fusion energy) * (1 + 1/Q) * (conversion efficiency) - (Total energy use by facility) ] / (cost per hour of operation) = energy per dollar. There are other ways you can calculate that same number, but they require a lot more details.