The problem with (d,t) fusion:17.6 MeV/fusion (3.5 alpha, 14.1 n), 5 amu (D, T)
## \frac{17.6 MeV}{5 amu(d+t)} * (1.6021 E-13 J/MeV) / (1.66054 E-27 kg/amu) ## =
3.396 E14 J/kg(D+T) - That's a lot of energy.
https://physics.nist.gov/cgi-bin/cuu/Value?ukg
1.660 539 066 60(50) x 10-27 kgWant to produce 3000 MJ per second for 1 year
8.8336 E-6 kg for 3000 MJ (3.0E9 J) thermal which could be used to produce 1000 MJ electricity with a 33.4% conversion efficiency, or more if one could separate the charges (alpha particle and electrons)
Seconds in one year = 3.15576 E7
So 8.8336 E-6 kg/s * 3.15576 E7 s/yr = 278.77 kg/yr * 0.6 T/(D+T) = 167.26 kgT (that's a lot of T). It means capturing the neutrons from the fusion reaction in Li, usually Li-6, if the neutrons are thermalized, in order to take advantage of the high thermal cross section for the 6Li(n,alpha)T. And these numbers assume 100% conversion of (D,T). If the efficiency is 50%, then the D,T fuel must be doubled, and the unused D,T has to be captured/recovered and recycled. If the efficiency is only 0.10, then that requires a lot of D,T and a lot of recovery/recycling.The recent IFE shot produced 3.15 MJ using input of 2.05 MJ, but the system had to use around 500-600 MJ (still waiting to learn the actual number) just to get 2.05 MJ into the hohlraum - and that is one shot. Doing this for 3.15576 E7 s doesn't appear to be on the horizon any time soon, and probably not ever.
Note also, that T has a half=life of 12.3 years, so some T will decay to He3 while waiting to be used, and the He3. A d+3He reaction (18.3 MeV) would be ideal, but it requires a higher temperature/compression.
It's both a physics and engineering problem.