Stealthflanker wrote:
This information is kinda rare.. I'm also looking for it.
Nonetheless, you can try estimating them based from following table :
Thanks Stealthflanker . I kinda did the math on this based on the numbers that you provided .
It seems Russian reactors for submarines are quite heavy compared to US ones . Not sure of the reason though .
First, the power generation you mentioned in the kilowatt range for actual unmanned spacecraft is probably from a thermoelectric generator as opposed to a fission reactor.
A thermoelectric generator converts the heat generated from natural radioactive decay into electric power. Due to design considerations it is not practical for thermoelectric generators to be large enough to produce more than about a kilowatt of electricity. Based on pure physics, however, here is the thermoelectric generator calculation.
Plutonium is often used in space applications because of its long half-life. So, if a space craft is on a decades-long mission, A Pu power source should be able to function continuously. The downside of plutonium is that it is extremely heavy. Considering plutonium-238 for the moment, its specific power is 0.56 Watts/gm due to natural radioactive decay. So, to generate 1 kW, we need 1000*0.56 gm = 560 gm or 0.56 kg. This is thermal energy due to radioactive decay. The thermoelectric generator which converts the thermal energy to electric energy has an efficiency of only a few percent. For our example, let's assume a 5% conversion efficiency. Then, the 0.56 kg will generate 0.05*1000 Watts = 50 Watts electric power, or, we can get 89 Watts per kilogram. For 1 Kw, our mass is 11 Kg. For 1 MW, our mass is 11,000 kg.
Now, onto a more practical means for generation 1 MW of power using a Plutonium fission reaction.
To calculate the mass required to obtain a certain power level, we have to know the neutron flux and the fission cross-section. Let’s assume the flux is 1E14 neutron/cm2/sec, the cross section for fast fission of Pu-239 is about 2 barns (2E-24 cm2), the energy release per fission is 204 MeV, and the Pu-239 number density is 4.939E22 atoms/cm3. Then the power is
P = flux * number density * cross section * Mev per fission * 1.602E-13 Watt/MeV
P = 1E14 * 4.939E22 * 2E-24 * 204 * 1.602E-13 = 323 W/cm3
So, for 1 MW, we need 1E6/323 = 3100 cm3. Given a density of 19.6 gm/cm3, this is 19.6*3100 = 60,760 gm or 60.76 kg.
The next question to ask is: how long do you want to sustain this reaction? In other words, what is the total energy output?
For example, a Watt is one Joule per second. So, to sustain a 1 MW reaction for 1 year, the total energy is 1E6 J/s * 3.15E7 s/year = 3.15E13 J
For Pu-239, we have 204 Mev per fission and we have 6.023E23./239 = 2.52E21 atoms/gm. So, the energy release per gram is 2.52E21 * 204 Mev/fission * 1.602E-13 J/Mev = 8.24E10 J/gm.
Therefore, to sustain 1 MW for 1 year, we will use 3.15E13 J / 8.24E10 J/gm = 382 gm of Pu-239 or 0.382 kg. This is only a small fraction of the total 60.76 kg needed for the fission reaction.
Finally, this is thermal energy. Our current light water reactors have about a 35% efficiency for conversion to electric power. So, you can take these numbers and essentially multiply by 3 to get a rough answer for the total Pu-239 needed: 3 x 60.76 = 182 kg. Rounding up, you would need roughly 200 kg for a long term sustained 1 MW fission reaction with a 35% conversion efficieny.
These calculations assume quite a bit and I wouldn’t use these numbers to design a real reactor, but they should give you a ballpark idea of the masses involved.