Calculating amount of fissionable material needed for reactor

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Discussion Overview

The discussion revolves around calculating the amount of fissionable material, specifically U-235, required for a reactor to operate at 300 kW thermal power for twenty years. Participants explore various aspects of this calculation, including energy requirements, reactor efficiencies, and the complexities involved in sustaining a critical system over time.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the total energy requirement as 1.89E14 J and derives a very small mass of U-235 needed, questioning if their approach is correct.
  • Another participant notes that the energy released from one fission of U-235 is approximately 200 MeV and suggests considering reactor efficiencies in the calculations.
  • A different viewpoint emphasizes the distinction between the total energy needed and the amount of material required to sustain a reactor's operation, suggesting that a simple division of energy by available energy per kg of U-235 may not suffice.
  • One participant provides a calculation indicating that about 1.64 kg of U-235 would be needed based on energy density, which is seen as a more reasonable estimate.
  • Concerns are raised about the actual mass of reactor-grade enriched uranium needed, factoring in that it contains only about 4% U-235, leading to a much larger required mass of approximately 40.98 kg.
  • Another participant mentions that the reactor could potentially use weapons-grade uranium, which introduces political considerations rather than engineering ones.
  • Some participants discuss alternative reactor designs, such as CANDU reactors, which utilize U-238 and have different safety profiles, although there is some disagreement about their fuel composition and burnup rates.
  • A later reply elaborates on the complexities of fuel consumption, including the effects of fission products and the need for void volume in the fuel system.
  • One participant corrects a previous claim about CANDUs, clarifying that they do burn U-235 at natural enrichment levels, which is lower than that of typical light water reactors (LWRs).

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions regarding the amount of U-235 needed for reactor operation. There is no consensus on the exact amount of fissionable material required, and multiple competing perspectives are presented throughout the discussion.

Contextual Notes

Participants highlight various factors that complicate the calculations, including reactor efficiencies, the distinction between energy needed for operation versus decay, and the impact of fission products on reactor dynamics. Additionally, the discussion touches on the differences in fuel composition and burnup rates between reactor types.

engineer23
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If I have a reactor that needs to operate at 300 kW thermal power for twenty years, how much fissionable material (U-235) do I need? I calculated that this will require 1.89E14 J, or 1.18E27 MeV. 931 MeV = 1u, so I have 1.27E24 u. This is .002 kg, which seems really small.

Am I doing something wrong? I'm really just looking for a simple, ballpark analysis here.

Thanks!
 
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I believe there is a significant difference between the amount of material you would need to release a total 1.89E14 J, and the amount of material you would need to sustain a 300kW load for 20 years. Just multiplying the power by the time only gives you how much material needs to DECAY, not how much you need in the reactor.

A straight division of the total required energy by U-235's available energy (77 TJ/kg according to Wikipedia) gives about 1.64 kg however.
 
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Thank you! That is a much more reasonable number. I saw another design of a reactor with comparable power required .8 kg of fuel to run for ten years.
 
Of course, that number doesn't really tell the whole story for a couple of reasons-

1) Reactor-grade enriched uranium is around 4% U-235, so in reality the "actual mass" you'll need is more like 40.98 kg, 96% of which will be U-238.

2) Efficiencies in the reactor will be less than ideal.
 
Re: 1) The reactor in question could be using weapons grade Uranium. Such is a political concern and not an engineering requirement.
 
Of course, you could always lose that stupid Yankee/Frence design and get a CANDU. They burn U-238 and are incapable of a meltdown.
 
Danger said:
get a CANDU. They burn U-238 and are incapable of a meltdown.
Although they do create maple syrup as a waste product. :-p
 
One man's waste product is another man's nectar.
Ahhh... maple syrup on bacon... mmmmmmm...
 
  • #10
engineer23 said:
If I have a reactor that needs to operate at 300 kW thermal power for twenty years, how much fissionable material (U-235) do I need? I calculated that this will require 1.89E14 J, or 1.18E27 MeV. 931 MeV = 1u, so I have 1.27E24 u. This is .002 kg, which seems really small.

Am I doing something wrong? I'm really just looking for a simple, ballpark analysis here.

Thanks!
The amount of fissile material is more complicated than simply calculating the mass of U-235 consumed, which is simply given by the total energy (power * time). In addition to U-235 consumed, one needs to include the U-235 required to maintain a critical system throughout the production cycle, and as burnup accumulates (energy/mass of fuel), fission products accumulate which compete for neutrons in the reactor. In addition, the fission products cause changes in the dimensional and physical properties of the fuel material. For each U-235 fission, two new atoms are produced, so for 1% of fuel volume used, there will be a 2% increase in volume. Gaseous fission products, e.g. Xe and Kr, cause gaseous swelling of the fuel, so a void volume is required in the fuel system to accommodate that gas.

Fuel consumption/utilization is usually measured in terms of GWd/tU (MWd/kgU) or related units of burnup (exposure), which is just thermal energy produced per unit mass of fuel used. One GWd/tU is roughly 1% of initial metal (U) atoms (1% FIMA).

Also, during operation of a uranium based reactor, U-238 absorbs neutrons and via beta decay transforms to Pu-239 and higher isotopes, as well as some Am and Cm isotopes, which increase with exposure. At high burnups, as much as half the fission occur in Pu-isotopes, so the core design must take this into account.

Currently LWRs in the US utilize fuel to about 60 GWd/tU peak rod, with local burnps approaching 75 GWd/tU (~7.5% FIMA). In fast reactors, burnups have reached levels of 200 GWd/tU (and possibly in special cases ~250-300 GWd/tU). Generally fast reactor fuel is clad in special steels in fuel rods with smaller diameter (volume) than commerical fuel, but they also have much larger internal void volumes for fission gases. Restrictions on thermo-mechanical behavior of LWR fuel are more stringent than fast reactor fuel, since fast reactors have generally been operated by government or reasearch insititions under government contract.

One correction on CANDUs: They do burn U-235, but at natural enrichments (~0.7% U-235) rather than 4-5% in LWRs. Max. discharge burnups (~10-15 GWd/tU) are much lower than LWRs (50-60 GWd/tU).
 
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  • #11
Astronuc said:
One correction on CANDUs: They do burn U-235, but at natural enrichments (~0.7% U-235) rather than 4-5% in LWRs. Max. discharge burnups (~10-15 GWd/tU) are much lower than LWRs (50-60 GWd/tU).

Party pooper. :-p
You're right of course, but you take a lot of the fun out of Yank-baiting. Still, the safety factor of a CANDU is a pretty good selling point. It can't even overheat, never mind melt down.
 

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