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How Much do Reactors Weigh?(2)

SparWeb (Aerospace) (OP) 
4 Oct 04 15:10 
In another forum ( Thread2599737), I have talked myself into a corner regarding nuclear electric propulsion of spacecraft. When it came to numbercrunching time, it turns out my propulsion textbooks don't have any data regarding mass/power of the nuclear power plants they describe, not even enough data to work it out approximately. Harumph. I poked around at nuclear submarine numbers, but that research didn't yeild much, probably due to the fact that nobody wants to widely publicize such numbers about military vessels. Nuclear RTG's used on probes like Cassini and Galileo don't scale up very well. They only generate a kilowatt, and we're looking for MW, here. Does anybody have approximate numbers handy on powertoweight ratios for a 1 MW power plant, assuming it could be optimized for weight, needing shielding only in one direction? I know a bit about heavywater systems, and I think there are other variations used in nuclear subs, which leads me to wonder if different moderators exist that would be more weightefficient, too. Since Xenon is used in Ion Electric thrusters, could the same Xenon be used as a moderator? (Now I'm really exposing my ignorance on the finer points of capturing neutrons). Hope somebody's interested. Thanks in advance, for any suggestions. Steven Fahey, CET "Simplicate, and add more lightness"  Bill Stout 

byrdj (Mechanical) 
15 Oct 04 22:03 

SparWeb,
Here is some information on reactor weights. I hope this is helpful.
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 halflife. So, if a space craft is on a decadeslong mission, A Pu power source should be able to function continuously. The downside of plutonium is that it is extremely heavy. Considering plutonium238 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 crosssection. Let’s assume the flux is 1E14 neutron/cm2/sec, the cross section for fast fission of Pu239 is about 2 barns (2E24 cm2), the energy release per fission is 204 MeV, and the Pu239 number density is 4.939E22 atoms/cm3. Then the power is
P = flux * number density * cross section * Mev per fission * 1.602E13 Watt/MeV P = 1E14 * 4.939E22 * 2E24 * 204 * 1.602E13 = 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 Pu239, 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 * 1.602E13 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 Pu239 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 Pu239 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.
TREMOLO


SparWeb (Aerospace) (OP) 
5 Nov 04 14:59 
Tremolo, Your explanation is highly appreciated! From it, I should be able to bring together other information into a more complete picture  now that I know where to look. Could you please elaborate on the different meanings between the specific power your first calculated 89 W/kg, and the subsequent 16.5 kW/kg? Is this the difference between RTG's that convert heat into electricity with thermocouples, versus the theoretical limit? Do you also care to comment on the use of different working fluids in the heattransfer cycle? There are proposals to use potassium in a 2phase system, or Xenon, in a purely gas state. It sounds like NASA won't use water. Steven Fahey, CET "Simplicate, and add more lightness"  Bill Stout 

SpareWeb,
The difference in the power densities is due to the use of natural radioactive decay to generate electricity in the first case vs nuclear fission in the second case. It’s hard to beat the power density of a fission reaction.
As far as the working fluid, in space applications water is less attractive than gas because of it relatively higher density. Also, you can get better conversion efficiencies with a gas cycle (40%) than with a water cycle (33 to 35%). Finally, I provided calculations for a plutonium fueled fast reactor. Traditionally, this type of reactor has used gas as the working fluid.
As a side note, I reviewed my text books and they report a power density for a gas cooled fast reactor of 90 kW/kg as opposed to the 16.5 kW/kg that I calculated, so maybe there’s something off in my calculation.
TREMOLO. 

SparWeb (Aerospace) (OP) 
10 Nov 04 10:34 
Are they accounting for the cooling and power generation system when they quote 90 kW/kg, or are they also referring only to the core? I am only going by the textbook that I have at hand, but steam cycles can also reach 40% efficiency, according to them (Applied Thermodynamics, Eastop & McConkey). Is the working fluid relevant to the efficiency of the cycle? Peak temperatures in any system are limited by structural materials, not fluid properties, aren't they? When mass of the whole system is a concern, there may also be room to reduce weight when liquids are being handled instead of gases. Steven Fahey, CET "Simplicate, and add more lightness"  Bill Stout 

SparWeb,
The 90 kW/kg is based on fuel weight, only.
The efficiencies are based on the thermodynamic cycle and safety margin.
TREMOLO. 

First thing allow me to say that only nuclear power can provide enough power to take us to planetary missions (like Jupiter), and back. These missions require 7 to 12 years of continues operation. Solar and chemical simply can not do it. Now to your question of powertoweight ratios for a 1 MW power plant? For a space reactor this will produce approximately 100 kWe, depending on the reactor design and the power conversion. For such a system the power to weight ratio is 55 w/kg as oppose to 22 w/kg for a solar system. Here are some numbers for a very interesting design (Heat Pipe Reactor), for 1 MWt Reactor Fuel region 157 kg Reflector 154 kg Heat pipes 117 kg Reactor control 33 kg Other support 32 kg Total Reactor mass 493 kg [Ref LOS ALAMOS SCIENTIFIC LABORATORY] Ned Xoubi Nuclear Engineer 

Ned: Your statement regarding "only nuclear" seems a bit arbitrary. How about a (obviously huge) solar sailcraft powered by a (also huge) pair of reflectors which orbit as close to the sun (or other star) as possible and concentrate a beam onto the travelling craft? Course I agree, you'd probably want to use a reactor to send the orbiting reflectors required for the deceleration half and for the return journey across to the target star first. Otherwise... Pechez les vaches. 



