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sam

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Fusion Power Plant Simulator

fusionenergybase.com
193 points·by sam·3 bulan yang lalu·129 comments

Progress toward fusion energy gain as measured against the Lawson criteria

fusionenergybase.com
232 points·by sam·tahun lalu·142 comments

comments

sam
·3 bulan yang lalu·discuss
Well said about the “lossy crankshaft”. As for what determines Q, that’s up next, stay tuned :)
sam
·3 bulan yang lalu·discuss
Power to magnets (at least those not contributing to heating) are assumed to be included in the house load.
sam
·3 bulan yang lalu·discuss
Thanks - what browser?
sam
·3 bulan yang lalu·discuss
Yeah, it’s slightly buried - the publication with all the details, including the math is linked to in footnote 3 of the explainer article,

https://pubs.aip.org/aip/pop/article/29/6/062103/2847827/Pro...

It’s open access and you can download the PDF directly from there.
sam
·5 bulan yang lalu·discuss
This is mistaken. In space a radiator can radiate to cold (2.7K) deep space. A thermos on earth cannot. The temperature difference between the inner and outer walls of the thermos is much lower and it’s the temperature difference which determines the rate of cooling.
sam
·tahun lalu·discuss
In the context implied above it is the ratio of fusion energy released to laser energy on target or the laser energy crossing the vacuum vessel boundary (they are the same in this case). So it would have been more precise to say "target gain" or "scientific gain".
sam
·tahun lalu·discuss
We are careful to always specify what kind of “breakeven” or “gain” is being referred to on all graphs and statements about the performance of specific experiments in this paper.

Energy gain (in the general sense) is the ratio of fusion energy released to the incoming heating energy crossing some closed boundary.

The right question to ask is then: “what is the closed boundary across which the heating energy is being measured?” For scientific gain, this boundary is the vacuum vessel wall. For facility gain, it is the facility boundary.
sam
·tahun lalu·discuss
It’s the ratio of fusion energy released to heating energy crossing the vacuum vessel boundary.
sam
·tahun lalu·discuss
Companies like Commonwealth Fusion Systems are an example of those utilizing high-temperature superconductors which did not exist commercially when ITER was being designed.
sam
·tahun lalu·discuss
Author here - some other posters have touched on the reasons. Much of the focus on high performing tokamaks shifted to ITER in recent decades, though this is now changing as fusion companies are utilizing new enabling technologies like high-temperature superconductors.

Additionally the final plot of scientific gain (Qsci) vs time effectively requires the use of deuterium-tritium fuel to generate the amounts of fusion energy needed for an appreciable level of Qsci. The number of tokamak experiments utilizing deuterium tritium is small.