Calculating amount of fissionable material needed for reactor

In summary, the amount of fissionable material (U-235) needed for a reactor to operate at 300 kW thermal power for twenty years is more complicated than simply calculating the mass of U-235 consumed. It also needs to take into account the U-235 required to maintain a critical system throughout the production cycle, the accumulation of fission products which compete for neutrons, and the transformation of U-238 into Pu-239 and higher isotopes. Current LWRs in the US utilize fuel with burnups of about 60 GWd/tU, while fast reactors can reach levels of 200 GWd/tU. CANDUs burn U-235 at natural enrichments and have lower max discharge burnups than LWR
  • #1
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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  • #3
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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  • #4
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.
 
  • #5
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.
 
  • #6
Re: 1) The reactor in question could be using weapons grade Uranium. Such is a political concern and not an engineering requirement.
 
  • #7
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.
 
  • #8
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. :tongue:
 
  • #9
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. :tongue:
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.
 

1. How do you calculate the amount of fissionable material needed for a reactor?

To calculate the amount of fissionable material needed for a reactor, you need to consider several factors, including the type of reactor, its power output, and its fuel efficiency. Generally, you can use the following equation: Amount of Fissionable Material = (Power Output x Burnup x Operation Time) / (Energy per Fission x Fuel Utilization x Fission Cross Section). This equation takes into account the energy requirements of the reactor, the efficiency of the fuel, and the amount of time the reactor will be in operation.

2. What is burnup and how does it affect the calculation of fissionable material?

Burnup is a measure of how much energy a fuel has produced in a reactor. It is typically measured in gigawatt days per metric ton of initial heavy metal (GWd/MTIHM). The higher the burnup, the more energy has been produced by the fuel, and thus, less fissionable material will be needed for the reactor.

3. What is fuel utilization and why is it important in calculating the amount of fissionable material?

Fuel utilization is a measure of how much of the fissionable material is consumed in the reactor. It is typically expressed as a percentage. A higher fuel utilization means that more of the fissionable material is being used to produce energy, thus reducing the amount of material needed for the reactor.

4. How does the type of reactor affect the calculation of fissionable material?

The type of reactor can greatly affect the calculation of fissionable material needed. Different reactor designs have different fuel requirements and efficiencies. For example, a pressurized water reactor (PWR) will have different fuel requirements compared to a boiling water reactor (BWR). It is important to consider the specific design of the reactor when calculating the amount of fissionable material needed.

5. Are there any safety considerations when calculating the amount of fissionable material needed for a reactor?

Yes, safety is a crucial factor to consider when calculating the amount of fissionable material needed for a reactor. It is important to ensure that the reactor has enough fuel to operate efficiently, but not so much that it poses a safety risk. Additionally, the handling and storage of fissionable material must be carefully planned to avoid any potential accidents or hazards.

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