Originally posted by XL5
As for converting radiation into electricity. If they can make a chip or a sheet that is compact enough for the amount of energy a car needs (about
75HP AVG.), then we don't really need anything else. All that would be needed at that point is a thick lead box, lead terminals and a big lead bolt
to hold the radioactive rod. Hopefully this does get developed and they use thorium as the source.
in your consideration are some major flaws:
first of all: spontaneous decay of radioactive material has a very high energy density but a very low power density
second: proper to use isotopes (short half life time) have to be created by artificial means and are increadable expensive
(to sum those points up in an example.. the us government bought all the pu 238 needed für new horizons from russia for some million dollars...
unsatisfied with this necessity they developed a program to build a new reactor to create pu 238, this reactor will coast approx 200 mio $ and produce
5 kg pu 238 per year..
source)
now to your carpowering idea: the 75 hp you proposed are equal to 55162 watt, 1 kg pu 238 can only provide 450 watt (and this as radiation energy so
you will have losses) but even if you could get all power from the pu into the cars motor you would need
120 kg fuel (today about 40 mio $
worth material)
of course you stated an other material:
thorium (since we are talking on radioaktive isotopes this information is quite useless, but i will nevertheless give it a try)
first guess: you mean the most abundand thorium isotope 232 (100% of all thorium in nature)
this has a power density of some microwatt (in my guess 1.9) per kilo (meaning you would need 28807632120 kg thorium 232 to power 1 car with 75 hp
(for imagination this is a massive cube of thorium with edge length of 134 meters))
calculation:
(6.022*10^26)/232 = 2.596*10^24 th 232 atoms in 1 kg
(2.596*10^24)/2 = 1.298*10^24 th 232 atoms decayed in 1kg after its half life
(1.298*10^24*10^24)*(4.083*10^6)eV *1.602*10^-19 J/eV = 8.49*10^11 J of energy came out of 1 kg after the half life time of the th232
8.49*10^11 J / (1.405*10^10 * 365.25 * 24 * 60 * 60) = 1.91 µW as the average power of a 1 kg th 232 source
i did not consider further products in the thorium decay chain since it needs 102 mio years until 0.5% of the original thorium decayed to ra228
second guess:
you mean a classical reactor concept producing neutron radiation which can be absorbed by th 232 to get even more unstable th 233 decaying via
protactinium to uranium 233 --> thorium 229 and so on
problems: hughe efforts to seal and cool such a reactor.. very dangerous and complicated fuel replacement activities, big effort to clean and maintain
such a highly involved system
so until you come accross some totally different material: no chance