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Could you elaborate on that? What do you mean that the lasers could require more energy? Is it that in a specific volume they got X EM energy coming in from th

Could you elaborate on that? What do you mean that the lasers could require more energy? Is it that in a specific volume they got X EM energy coming in from the laser and Y thermal energy coming out, with Y>X BUT the electricity consumption of the lasers is Z>Y>X? If so that's sort of misleading, like the plethora of claims from ITER. I hoped this was different.

foxyv 4y ago
So the laser energy that went into the reaction in the form of light is less than what came out of the reaction. However, the energy needed to produce that laser energy may be orders of magnitude more depending on the laser. AKA: the Wall Plug Efficiency. Tabletop rigs can be as efficient as 50%, however high power such as we see here tends to come with drastically reduced efficiency.
ansible 4y ago
Not /u/acidburnNSA, but what was meant is that no laser is 100% efficient. Not only do they not convert 100% of their electrical input into laser energy, but they also require other support systems, notably cooling. So we need to consider the total energy costs of the building the fusion experiment is conducted in, not just the physically small area where the fusion reaction is happening inside the reactor. Still, this is an important step in the development of fusion energy reactors.
derefr 4y ago
Presumably they mean that there are efficiency losses in charging the supercapacitor banks used to fire the lasers; so that if you consider the system over multiple duty cycles rather than over a single cycle, it's no longer energy-positive. (I.e. if the system were capturing its emitted energy — and that emitted energy needed to be enough to act as a grid power source feeding input power to the supercapacitors, rather than merely being the equivalent of the direct output power of the lasers per shot — then it wouldn't be enough to sustain the reaction.) But personally, I don't know whether that's actually important. Power plants usually consume a nontrivial fraction of their own produced power to power themselves, and in fact consume more than 100% of produced power when starting from a full stop — meaning that in initial few-shot conditions, even when feeding back their own produced power into themselves, they still need (huge amounts of) external power input to get going, like a car engine needing a battery + starter motor. Only a rare few kinds of power plant can be used to "black start" a power grid. Most types of generator need to overcome initial higher resistances, e.g. inertia (and thereby back-EMF resistance at the transformer) in getting heavy turbines spinning from a stop. It wouldn't be at all strange if a practical fusion power plant turned out to be energy-negative over a few-shot run (i.e. required "bootstrapping"), but then became energy positive over a theoretical 24/7 run at whatever its optimal duty cycle is. And a single-shot run becoming net-positive would be a good point to start to consider those more practical calculations, since they'd have been useless to consider until then—a power plant can't possibly be net-positive over any kind of runtime + duty cycle, if its core reaction can't be net-energy-positive when considered in isolation. Which is, to me, why it probably does make sense for ITER to be excited. They've reached the point where they can stop using a lab-bench model of power efficiency, and start trying to come up with another, more full-scale model of power efficiency to replace it with.
danbruc 4y ago
Exactly. Looking at the Wikipedia article [1] suggest that they start out with 422 MJ stored in capacitors, turn this into 4 MJ IR laser light, convert it into 1.8 MJ UV laser light, this into x-rays of which 0.15 MJ heat the target of which finally 0.015 MJ heat the fuel. Depending on what in this chain you consider the input energy, you can get orders of magnitude different numbers - 15 kJ of energy produced could either be a gain of 1 or a gain of ‭0.0000036 or anything in between.‬ And this is before trying to capture the released energy and converting it into electricity, this will come with another sizable loss. [1] https://en.wikipedia.org/wiki/National_Ignition_Facility https://en.wikipedia.org/wiki/National_Ignition_Facility
- est31 4y ago
  From https://www.ft.com/content/4b6f0fab-66ef-4e33-adec-cfc345589dc7 https://www.ft.com/content/4b6f0fab-66ef-4e33-adec-cfc345589... > The fusion reaction at the US government facility produced about 2.5 megajoules of energy, which was about 120 per cent of the 2.1 megajoules of energy in the lasers, the people with knowledge of the results said, adding that the data was still being analysed. They probably upgraded the rig since the Wikipedia article was written, so most likely the 2.1 MJ refers to the UV light numbers.
  - danbruc 4y ago
    If this is assumption is true, they only produced 0.6 % of the energy they spent. Another question would then be, how relevant this is, i.e. could the UV light be produced much more efficiently than the experiment does? Maybe some constraints forces them to use a very inefficient process? In that case it might be reasonable to use the UV laser power as the reference for the gain.
    - kelnos 4y ago
      Sure, and if they upgrade the lasers themselves to current laser tech (as I understand it, the NIF's hardware is around 25 years out of date on that front), then that 0.6% number probably jumps to 20% or so. Which still isn't enough, but is way closer than 0.6%. Add to that the fact that improvements in laser efficiency is a hot research area (as lasers are used commercially in a lot of places, and cost-cutting is always a concern), and this is starting to feel a little more attainable.

foxyv 4y ago

So the laser energy that went into the reaction in the form of light is less than what came out of the reaction. However, the energy needed to produce that laser energy may be orders of magnitude more depending on the laser. AKA: the Wall Plug Efficiency. Tabletop rigs can be as efficient as 50%, however high power such as we see here tends to come with drastically reduced efficiency.

ansible 4y ago

Not /u/acidburnNSA, but what was meant is that no laser is 100% efficient. Not only do they not convert 100% of their electrical input into laser energy, but they also require other support systems, notably cooling. So we need to consider the total energy costs of the building the fusion experiment is conducted in, not just the physically small area where the fusion reaction is happening inside the reactor. Still, this is an important step in the development of fusion energy reactors.

derefr 4y ago

Presumably they mean that there are efficiency losses in charging the supercapacitor banks used to fire the lasers; so that if you consider the system over multiple duty cycles rather than over a single cycle, it's no longer energy-positive. (I.e. if the system were capturing its emitted energy — and that emitted energy needed to be enough to act as a grid power source feeding input power to the supercapacitors, rather than merely being the equivalent of the direct output power of the lasers per shot — then it wouldn't be enough to sustain the reaction.) But personally, I don't know whether that's actually important. Power plants usually consume a nontrivial fraction of their own produced power to power themselves, and in fact consume more than 100% of produced power when starting from a full stop — meaning that in initial few-shot conditions, even when feeding back their own produced power into themselves, they still need (huge amounts of) external power input to get going, like a car engine needing a battery + starter motor. Only a rare few kinds of power plant can be used to "black start" a power grid. Most types of generator need to overcome initial higher resistances, e.g. inertia (and thereby back-EMF resistance at the transformer) in getting heavy turbines spinning from a stop. It wouldn't be at all strange if a practical fusion power plant turned out to be energy-negative over a few-shot run (i.e. required "bootstrapping"), but then became energy positive over a theoretical 24/7 run at whatever its optimal duty cycle is. And a single-shot run becoming net-positive would be a good point to start to consider those more practical calculations, since they'd have been useless to consider until then—a power plant can't possibly be net-positive over any kind of runtime + duty cycle, if its core reaction can't be net-energy-positive when considered in isolation. Which is, to me, why it probably does make sense for ITER to be excited. They've reached the point where they can stop using a lab-bench model of power efficiency, and start trying to come up with another, more full-scale model of power efficiency to replace it with.

danbruc 4y ago

Exactly. Looking at the Wikipedia article [1] suggest that they start out with 422 MJ stored in capacitors, turn this into 4 MJ IR laser light, convert it into 1.8 MJ UV laser light, this into x-rays of which 0.15 MJ heat the target of which finally 0.015 MJ heat the fuel. Depending on what in this chain you consider the input energy, you can get orders of magnitude different numbers - 15 kJ of energy produced could either be a gain of 1 or a gain of ‭0.0000036 or anything in between.‬ And this is before trying to capture the released energy and converting it into electricity, this will come with another sizable loss. [1] https://en.wikipedia.org/wiki/National_Ignition_Facility https://en.wikipedia.org/wiki/National_Ignition_Facility

est31 4y ago
From https://www.ft.com/content/4b6f0fab-66ef-4e33-adec-cfc345589dc7 https://www.ft.com/content/4b6f0fab-66ef-4e33-adec-cfc345589... > The fusion reaction at the US government facility produced about 2.5 megajoules of energy, which was about 120 per cent of the 2.1 megajoules of energy in the lasers, the people with knowledge of the results said, adding that the data was still being analysed. They probably upgraded the rig since the Wikipedia article was written, so most likely the 2.1 MJ refers to the UV light numbers.
- danbruc 4y ago
  If this is assumption is true, they only produced 0.6 % of the energy they spent. Another question would then be, how relevant this is, i.e. could the UV light be produced much more efficiently than the experiment does? Maybe some constraints forces them to use a very inefficient process? In that case it might be reasonable to use the UV laser power as the reference for the gain.
  - kelnos 4y ago
    Sure, and if they upgrade the lasers themselves to current laser tech (as I understand it, the NIF's hardware is around 25 years out of date on that front), then that 0.6% number probably jumps to 20% or so. Which still isn't enough, but is way closer than 0.6%. Add to that the fact that improvements in laser efficiency is a hot research area (as lasers are used commercially in a lot of places, and cost-cutting is always a concern), and this is starting to feel a little more attainable.