Calculating amount of fissionable material needed for reactor

[engineer23]

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!

 

[Andronicus1717]

The energy released from one fission of U-235 is ~200 MeV (Total energy - Neutrinos)

http://www.tpub.com/content/doe/h1019v1/css/h1019v1_85.htm

Plus there are efficiencies you might want to include in the calc?

 

[Mech_Engineer]

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.

 

[engineer23]

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.

 

[Mech_Engineer]

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.

 

[Andronicus1717]

Re: 1) The reactor in question could be using weapons grade Uranium. Such is a political concern and not an engineering requirement.

 

[Danger]

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.

 

[mgb_phys]

get a CANDU. They burn U-238 and are incapable of a meltdown.

Although they do create maple syrup as a waste product.

 

[Danger]

One man's waste product is another man's nectar.
Ahhh... maple syrup on bacon... mmmmmmm...

 

[Astronuc]

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).

 

[Danger]

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.
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.