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Minimum critical mass with natural uranium

  1. Feb 21, 2010 #1
    I wanted to know what is the minimum critical mass of natural uranium (99.28% U-238) when moderated by beryllium and light water, when in a optimumally designed lattice.
    As can be seen in this article (google Plutonium Production Using Natural Uranium) criticality is attained at about 10 tons of U on graphite and 800Kg of U on heavy water. Maybe those are not optimally yet but seem close by the articles text.
    With beryllium it should be an amount inbetween heavy water and graphite.
    Now here (google Volume 1, Issue 2, 2007 Burn-Up Profiles for a New Beryllium) there is a reactor design that uses both beryllium and light water but it takes 300 tons of natural uranium. It is not the minimum design for critical mass but it is used for power production.
    The article also says that a mixture of uranium, beryllium and water has a smaller critical mass then uranium with either beryllium or water alone. So that must be part of the solution too.
    If you can think of a technically and economically easier way to achieve critical mass with natural uranium please tell us. Particle aceleration driven systems are too expensive and technical. Heavy water seems much more technical to manufacture and manage then beryllium, although you would use less uranium it doesnt offset the costs. Also beryllium seems a better neutron reflector then heavy water.

    One way would be to make a subcritical pile that would breed plutonium fast enough to increase reactivity to become critical, but how long does that take? Maybe it would need to be too close to critical size to do such breeding in a reasonable time (5 years maybe) so it would be utterly futile to have to wait that long to spare 100kg of uranium.

    By recent prices, you would need 3300t of graphite, maybe $39.6 mil plus $11mil for the uranium , but beryllium in the natural uranium reactor (that I think is not optimized) would be $51mil (204t) so with $28mil (303t) from uranium you would potentially be $28mil more expensive.

    Can someone help us to figure how small can the beryllium reactor be? I guestimate 55.5t of Uranium and 144t of Be. Anyone with reactor physics knowlegde would please help?

    Here goes some info for who knows the formulas:
    "infinite reapeating slab geometries, Be and NU(Natural Uranium) had a maximum reactivity (kinf=1.08) at about 16 cm Be and 0.6 cm NU"
    M is the migration lenght, the square root of the migration area.
    Beryllium migration area=582cm2
    Beryllium density=1.85g/cm3
    Uranium density=19.05g/cm3

    Notice that adding some light water and less beryllium, the minimum mass could be smaller still. Anyone that can help?

    This articles cited on Hayes (Burn-Up Profiles for a New Beryllium Moderated Water Cooled Natural Uranium Reactor) seem to have the answer but I couldnt get a hold on them:

    Hayes R.B. A light-water-cooled nuclear reactor of natural uranium and beryllium. Journal of
    ASTM International, Vol. 3, No.8 Published on line, Paper ID JAI00274. DOI:

    Green R. Nuclear criticality safety concerns related to transport of U-BeO waste. Trans. Amer.
    Nucl. Soc 76:256, 1997.

    Miller B. A., Busch R. D. Minimum enrichments of uranium for criticality in homogenous and
    heterogeneous systems. Proceedings of ICNC ’99, 6th International Conference on Nuclear
    Criticality Safety.​
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2
    and who are you df001?
  4. Feb 21, 2010 #3
    [tex]\begin{align*}&M^{2}=582\left.cm^{2}\rightarrow M=24.12\left.cm

    Radius solves as:
    [tex]\begin{align*}&R_{crit}\cong\frac{\pi M}{\sqrt{k-1}}=\frac{3.14\times 24.12\left.cm}{\sqrt{1.08-1}}=\frac{75.77}{\sqrt{0.08}}\left.cm=\frac{75.77}{0.2828}\left.cm=267.92\left.cm
    \\&R_{crit}\cong 2.68\left.m

    And the volume:

    [tex]\begin{align*}&V_{sphere}=\frac{4}{3}\pi R^3=1.33\times 3.14\times (2.68\left.m)^{3}=4.19\times 19.23\left.m^{3}

    This k=1.08 is for a slab arrangement of 0.6 cm uranium slabs between 16 cm beryllium slabs. So calculating the volume fractions and the masses:
    Beryllium and Uranium densities​
    [tex]\begin{align*}&\mu_{Be}=1.85\left. g/cm^{3}
    \\&\mu_{U}=19.05\left. g/cm^{3}


    [tex]\begin{align*}& F_{Be}=16/16.6=0.9638
    \\& F_{U}=0.6/16.6=0.0361

    Volumes of Be and U​
    [tex]\begin{align*}& V_{Be}=V_{total}F_{Be}=80.56\left. m^{3} \times 0.9638=77.65\left.m^{3}
    \\& V_{U}=V_{total}F_{U}=80.56\left. m^{3} \times 0.0361=2.91\left.m^{3}

    Masses of Be and U (unfinished becouse needs converting the densities to cubic meters)​

    [tex]\footnotesize\begin{align*}& M_{Be}=\mu_{Be}V_{Be}=1.85\left. \frac{ g}{cm^{3}} \times 77.65\left.m^{3}=77.65\left.m^{3}
    \\& M_{U}=\mu_{U}V_{U}=80.56\left. m^{3} \times 2.91\left.m^{3}=2.91\left.m^{3}

    Of course this migration length is for pure beryllium and here we have 16/16.6 volume fraction of beryllium and 0.6/16.6 uranium by volume. How would that change the migration length? Perhaps it is inversely proportional to the density because it might be inversely proportional to the macroscopic cross section, like the diffusion length. And the cross section is proportional to the density.
  5. Feb 22, 2010 #4
    Why isn't anyone contributing? Uninteresting or don't have the knowledge?
  6. Feb 22, 2010 #5


    User Avatar

    Staff: Mentor

    There is a fair amount of detailed involved in core configuration and design, in addition to doing a criticality calculation. Give folks more time to digest the details.

    One complication of Be is its toxicity. Search on Berylliosis.

    http://chppm-www.apgea.army.mil/documents/FACT/64-003-0302.pdf [Broken]
    Last edited by a moderator: May 4, 2017
  7. Feb 22, 2010 #6
    Thank you Astronuc. I understand what you say and I don't want the answer right away, I just wanted someone to show some curiosity about the question, maybe some points about the answer. You are right, it is something to think about.

    About the toxicity, what I read about is that beryllium is toxic as much as uranium or asbestos, much less important concern then dealing with radiation. As a nuclear moderator there seems to be little risk of inhaling dust and much less physical contact.

    Either way I still think it is a reasonable question to know, as it seems to be the third most important neutron moderator economically and in scientific research, maybe fourth.

    Maybe this question is a way to understand how to find the minimal critical mass of natural uranium in different conditions.

    For example I read that the infinite neutron multiplication factor of water and natural uranium is 0.93 in a slab geometry. How much would it be in a optimized lattice using polyethylene instead of water, can you get to 1? Polyethylene is 20,80% hydrogen denser then water, so you could get a denser system with polyethylene. Would that be enough to reach critical conditions?
  8. Feb 22, 2010 #7


    User Avatar

    Staff: Mentor

    If k < 1, then there is no way to get k > 1 geometrically. One has to increase the enrichment of the fissile isotope, e.g., U-233, U-235, or Pu-239, or others.

    Before one minimizes the potential hazard from Be, one should look at the health costs associated with mining U or asbestos, particuarly asbestos. If the producing companies don't observe good practices and protect workers, then default on their responsibilities, then ultimately the rest of society (e.g., taxpayers) bear the burden of the health care.

    One is focused on the core/reactor design, but realize that fuel assemblies and components have to be manufactured by someone, somewhere. It is during manufacture that people may be exposed unnecessarily if proper practices are not observed.

    Ultimately one has to look at the parasitic absorption by hydogen or oxygen (if water) and structural materials. Then one has to consider the effects of heterogeneity and geometry, and then also, the thermal effects, e.g., Doppler resonances, and then the accumulation of fission products with exposure (burnup).

    CANDU reactors were designed to exploit the low absorption of neutrons by deuterium, as compared to protons in LWRs. CANDUs also use Zr-alloys, e.g., Zircaloy-4, Zircaloy-2 and/or Zr-2.5Nb. Zr has a relatively low absorption cross-section as compared to other structural materials, as well as having a lower oxidation/corrosion rate.

    Since the fission process generates heat, one must also have cooling system.

    A reactor design is a balance between nuclear physics, thermal-hydraulics (heat transfer), and structural integrity. It's got to be safe and economical.
  9. Feb 22, 2010 #8
    Maybe the question posed here is not supposed to encompass all the engineering considerations of a nuclear reactor, but only the nuclear physics aspects, at least for now. As you said yourself, all these considerations take a large amount of planning and analysis.
    About the polyethylene, of course you need to reach k>0. My question is if it is possible, given that in the sub-optimally geometric design of a slab geometry, you can reach k=0.97 with H20 (water). Polyethylene should be much more efficient then water as it is 20.80% denser in hydrogen. If such density increase is proportional to the possible reduction in moderator volume I calculate 11.88% reduction in the total system (slabs of 1 cm U with 1.6 cm of H20 substituted for 1.43 cm of polyethylene). If such systems follow the rule that the multiplication factor is inversely proportional to the square of the density, the multiplication factor should be 1.2142. But the inverse square proportionality works for the critical mass not the multiplication factor. If someone can find the formula relating density to neutron multiplication that would help.
  10. Feb 26, 2010 #9
    I think that the slow responses are indicative of the fact that there is no easy way of answering your question. Calculating the reactivity of a heterogeneous reactor is not done with a simple formula. Practically, it would be done with a monte-carlo or neutron diffusion theory.

    I think what Astronuc is saying with regards to beryllium is that because of it toxicity it is not suitable for many applications. Although the nuclear physic properties might meet requirements there are engineering concerns with the material.

    I'm not sure I like the way you preformed your volume conversion from water moderator to polyethylene. While might provide a crude estimate, this doesn't consider the effect of the carbon atoms (which are also significant moderators and present in significant quantity) nor does it consider the neutron absorption properties of the materials. Both of these factors would affect your optimal slab thicknesses.

    If you want to keep working with an equivalent layer approach then you want to consider how this decreases the volume which reduces the surface area. Assuming that the neutron leakage is proportional to surface area would give you an approximate answer. In reality it is more complicated because more compact cores have much higher neutron buckling and therefore higher leakage.
  11. Feb 28, 2010 #10
    I've redone much of my reasoning of the density change impact on neutron multiplication and critical mass. As much as it would be nice otherwise, changing polyethylene for water dont change the infinite neutron multiplication factor because although the migration area diminishes, the macroscopic cross section increases, the two balancing each other. So the polyethylene slab absorbs as much neutrons as a water slab.
    The best estimate way to rethink the problem would be to star with the optimum beryllium slab and substitute some of it by polyethylene and use the equations I have to estimate the migration area, infinite neutron multiplication factor and then the critical radius.
    Astruc and Hologram0110:
    I get all your criticisms and although I dont agree with some arguments I apreciate your input. Could you help me just answer my basic question, besides the other relevant points you made? What is the minimum critical mass of natural uranium with different moderators arrangements? I am starting with beryllium. I mostly got the answer for graphite and heavy water form one of the articles I cited.

    Hologram0110: Carbon can be mostly ignored in polyethylene because the migration area is orders of magnitude larger than hydrogen's. In other words, the amount of carbon in polyethylene does not moderate 1/1000th of the neutrons that go through it.
    Hydrogen is as light as a neutron and a scattering event has tens of times more chances of thermalizing it than carbon.
  12. Feb 28, 2010 #11
    Masses of Be and U​
    [tex]\begin{align*}& M_{Be}=\mu_{Be}V_{Be}=1.85\left. \frac{tonnes}{m^3}\left. \times 77.65\left.m^{3}=143.65\left. tonnes
    \\& M_{U}=\mu_{U}V_{U}=19.05\left. \frac{tonnes}{m^3} \times \times 2.91\left.m^{3}=55.44\left. tonnes
    \\ &1 \left. tonne = 1000 \left. kg

    Although the 16/16.6 beryllium density should increase migration length by 3.7%, and so volumes/masses should be 11.68% larger (61.91 tonnes U and 160.42 tonnes Be).
    Should a optimum lattice geometry be much better than this slab geometry?
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