Estimating doses from criticality accidents

["Minnema, Douglas" <Douglas.Minnema@ns.doe.gov>]

Drat, got caught by the dreaded "from" bug.  Sorry about that, here is the
full text again.

RADSAFErs,

With the recent experience in Japan and questions raised on this listserver,
I dusted off some old material I had worked up for a class I did about 6
months ago on criticality safety.  Thought you might find it interesting.

While each criticality accident tends to be very unique, and expert advice
should always be sought, one can estimate the doses that might be received
from such an accident based on first principles.  To me it helps in
visualizing the levels of concern from such an accident. 

The key to evaluating the doses is to know the number of fissions involved
in the event.  This is probably where the greatest difficulty lies in
analyzing a hypothetical accident, although after the fact it is somewhat
easier since it can be determined from fission product yield measurements.
Therefore, I will assume for this discussion that the yield is 1E18
fissions.  Note that the results will scale directly with the number of
fissions.

Neutron Dose:

At a distance from the source, neutron flux generally follows 1/r^2.  Assume
that there are 2.5 neutrons per fission, and 50% leakage (leakage is
typically very high in such systems,  this is a guesstimate, but probably
not too far off).  Also assume that the energy of the leaking neutrons is
500 KeV (again, a guesstimate, but not too bad since many such accidents are
either unmoderated or undermoderated, and thermal neutrons get captured
close to the system), which results in a dose conversion factor of about
2.34E-9 rad/(n/cm^2).  The neutron dose would be:

		D(neutrons) = 2.34E-9 x (0.5 x 2.5 x 1E18)/(4 x Pi x r^2)
[rads]

At 10 meters (r = 1000 cm) this would be 232 rads.

Prompt Fission Gammas:

In Schaeffer, "Reactor Shielding for Nuclear Engineers" (USAEC 1973), I
found that there are about 3.75 photons per fission, with an average energy
of 0.324 MeV per photon.  At 0.3 MeV, the dose conversion factor that I have
is 2.22E-10 rad/(photons/cm^2).  The prompt fission dose would be:

		D(prompt gamma) = 2.22E-10 x (3.75 x 1E18)/(4 x Pi x r^2)
[rads]

At 10 meters this would be about 66 rads.

Early Fission Product Gammas:

Again in Schaeffer, I found that for the period of between 0.2 and 45
seconds after fission, for U-235, there are 3.31 photons per fission at an
average energy of 0.961 MeV per photon.  At that energy, the dose conversion
factor is about 5.28E-10 rad/(photons/cm^2).  The early fission product
gamma dose would be:

		D(early FP gammas) = 5.28E-10 x (3.31 x 1E18)/(4 x Pi x r^2)
[rads]

At 10 meters, this would be about 139 rads.

Note that we have ignored any potential shielding, self-absorption in the
system (except for neutrons), and attenuation in the air for all of the
above estimates.

While this is only a "back of the envelope calculation", I find that it
helps me in gaining an appreciation for such an accident.  How does this
compare to more refined calculations?  Detailed evaluations for various
types of critical systems can be found in "An Updated Nuclear Criticality
Slide Rule", NUREG/CR-6504, April 1997.  Below is a simple comparison of
results:

(Dose in rads)		Here         Metal system		U(5%)O2
U3(93.2%)O8

neutrons		232		300		    120
200

prompt gammas(1)	  66		 20		      90
80

Total			298		 320		     210
280

(1)  The Slide Rule did not have a comparison value for my "early FP gammas"
dose.

I certainly don't mean to imply that a few hand calcs are equivalent to the
NUREG, because there is much more to all of this than a few educated
guesses.  However, I feel it does help to understand where the doses come
from, and how significant they can be.  Recall that we are talking 300 rads
at a distance of 10 meters away, delivered almost instantaneously, before
you even include the fission product gammas!  That is why there was some
concern regarding the fact that there were public residences only 80 meters
away.

Now, if somebody wanted to, they could go to DOE's or NRC's release
fractions handbooks and estimate the amount of material released to the
environment for the accident.  However, we can save that story for another
day, although it would help to further put all of this in perspective.

Hope I didn't consume too much precious bandwidth, but thought you might
find this as interesting as I did.  If there are any questions, please feel
free to contact me personally at the address below.  If I made any mistakes
in the math, please forgive me.

Doug Minnema, Ph.D., CHP
Defense Programs, DOE
<Douglas.Minnema@ns.doe.gov>

"what few thoughts i have are truly my own"

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