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_fusi...
[1] https://www.wolframalpha.com/input?i=17.6+MeV+divided+by+ato...
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_cons...
The worldwide primary energy consumption is ~600EJ. 1g/70MWh * 600EJ ~= 2400t per year.
Cost of cryogens, cost of steel, cost of working nitronic, cost of Beryllium, and cost of superconductors are the dominant sources to track.
Your fuel costs are replaced by the reactor replacement cost. Spent nuclear rod fuel waste is replaced by the irradiated materials in the reactor.
Since it is a complicated engineering problem there is space to turn engineering effort into reduced cost.