5 ms·
The simplest answer that's part of the plan for most DT reactors is "as neutrons". The fraction of the energy that is retained as velocity of particles with a
by Tuna-Fish 4y ago
The simplest answer that's part of the plan for most DT reactors is "as neutrons".
The fraction of the energy that is retained as velocity of particles with a charge vs that is lost as the velocity of neutrons turns out to be conveniently just about where you'd want it to be. So, to capture the energy you need to surround your reactor with something that effectively converts fast neutrons to heat, such as a blanket of molten lithium, which you then use as a heat source. Lithium is proposed because it would also breed the necessary tritium.
- aziaziazi 4y agoDoes that mean you need to “feed” with new lithium regularly ? Any idea on the frequency, like lithium gram*watt produced ?
- dmurray 4y agoAssuming you need enough lithium to replace the tritium, and 100% energy efficiency, one atom of lithium makes one atom of tritium which makes 17.6 MeV of energy [0]. That's an energy output of 7 x 10^7 Wh per gram [1], or alternatively, a 70MW reactor needs one gram per hour, or alternatively, a very large 3.5GW plant needs 500 grams per hour, roughly the equivalent of the water drunk by a single thirsty person. [0] https://en.wikipedia.org/wiki/Deuterium%E2%80%93tritium_fusion https://en.wikipedia.org/wiki/Deuterium%E2%80%93tritium_fusi... [1] https://www.wolframalpha.com/input?i=17.6+MeV+divided+by+atomic+mass+of+lithium+in+watt+hours+per+gram https://www.wolframalpha.com/input?i=17.6+MeV+divided+by+ato...
- aziaziazi 4y agoThanks ! Following up on the napkin : The world final energy consumption (2018) [0] is around 9717Toe or 9,717*11.63=113,009TWh 1g for 70MW = 1T for 70TWh 113,009 / 70 = 1614 1614 ton of lithium every hour seems huge to me, does my calculations are wrong ? [0] https://en.m.wikipedia.org/wiki/World_energy_supply_and_consumption https://en.m.wikipedia.org/wiki/World_energy_supply_and_cons...
- willis936 4y agoMore than likely a plant will have a lifetime supply of lithium and other blanket materials on site before it is turned on. The fuel costs really do drop to effectively zero on MCF devices. Cost of cryogens, cost of steel, cost of working nitronic, cost of Beryllium, and cost of superconductors are the dominant sources to track.
- AtlasBarfed 4y agoAre you gradually irradiating the reactor? I mean that's what happens with fast neutrons and... Anything Your fuel costs are replaced by the reactor replacement cost. Spent nuclear rod fuel waste is replaced by the irradiated materials in the reactor.
- willis936 4y agoIndeed reactors are not free. The irradiated materials situation isn't as bad as fission though since they are low level activation materials that become safe within a few hundred years, as opposed to trans-uranics which are radioactive for thousands of years and will always be toxic. Since it is a complicated engineering problem there is space to turn engineering effort into reduced cost.
- jjk166 4y agoYou need slightly over 1 lithium atom per fusion event. So for ~40% heat to electricity efficiency, you'd need about 7 grams of lithium for 150 MWh of electricity. A 1 GWe plant would consume about 403 kg of lithium per year, about as much as in 5 Tesla Model 3 Battery packs.