# How much D2O does it take to power an interstellar generation ship O'Neill cylinder for 10,000 years?

Did I do the math right? I'm using full precision but write only 2 sig figs.

https://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_candidates_for_terrestrial_reactions

a D +   D  →   T  (1.01 MeV) + p (3.02 MeV)
b D +   D  → 3-He (0.82 MeV) + n (2.45 MeV)
c D + 3-He → 4-He (3.6 MeV)  + p (14.7 MeV)


We do reaction a and b, remove T from the plasma and let it decay to 3-He. So we have 2 3-He and do 2 c, which uses up another 2 D and end up with:

d 6 D → 2 He + 3 p + n + 43.9 MeV


I live in Austria and food grows fine here, so I picked the third shade of blue which is 130 W/m2 insolation. Since plants don't need IR I pick 100 W/m2 as LED based "insolation".

https://en.wikipedia.org/wiki/Luminous_efficacy#Examples_2

Assuming 33% efficient LEDs we need 300 W/m2 electric (there, plenty of IR :).

Two 32 km long and 8 km diameter O'Neill cylinders have 1.6E9 m2 mantle area, so we need 4.8E11 W or 1.5E23 J in 10,000 years.

https://en.wikipedia.org/wiki/Electronvolt

Now we need to make fusion reaction d

1.5E23/43.9 MeV=2.1E31 times

to get that energy. Assuming 75% electrical efficiency (we convert the charged fusion products directly to electricity, and only the waste heat goes to the steam turbine), so let's make that 2.1E31/0.75=2.9E31 times.

This gives us

2.9E31/6.022E23=4.8E7 moles of this reaction which needs 3 D2O per reaction and thus we need 4.8E7*3*(2*2+1*16)/1000=2.9E6 kg D2O.

https://en.wikipedia.org/wiki/Heavy_water

That is 2.9E6/1107 kg/m3=2596 m3 D2O.

If we cover the mantle of the O'Neill cylinders with D2O we'd need a 2596 m3/1.6E9 m2=1.6E-6 m=1.6 um thick film.

Bugger, last time I did the calculation (when I didn't factor in LED and fusion power plant efficiency) I got 0.4 mm.

Is 1.6 um right?