# Homework Help: Calculating Six-(Or Four? I think 6) Factor Formula from given inform

1. Mar 12, 2013

### whynotzoidberg

1. The problem statement, all variables and given/known data

In a reactor at the beginning of its life for every 1000 neutrons,

-450 neutrons are absorbed in U-235
-350 neutrons are absorbed in U-238
-75 neutrons are absorbed in coolant and cladding, and
-125 neutrons leak out of the reactor
nu=2.43

Note: In order to not cheat, the values of the first four quantities have been changed from the original problem. nu is constant (or nearly so) for U-235 so changing it doesn't make sense (to me anyhow). This also means that my values for the six factors (and the final answer) could be wildly off base.

2. Relevant equations

$k_i=ηfpε$ where k_i is the Four Factor Formula; the six facor formula is
$k=k_i*(P_FNL)*(P_TNL)$

the variables are...

η: reproduction factor (# fission neutrons produced per absorbtion in fuel) $η=nu*((σF_f)/(σF_a))$

where nu is the average # of neutrons per fission, σF_f and F_a are "the microscopic fission cross sections for fuel, accordingly" (I think F_f is U-235 and F_a is U-238)

f: thermal utilization factor (probability that a neutron that gets absorbed does so in the fuel material) $f=(ƩF_a)/(Ʃ_a)$

where ƩF_a is the macroscopic fission cross section of a (U-238?) and Ʃ_a is the total macroscopic cross section of a.

p: resonance escape probability (Fraction of fission neutrons that slow down from fission to thermal energies without being absorbed.) No equation given, can (maybe) figure out.

ε: Fast fission factor $ε=(total # of fission neutrons)/(# of fission neutrons from thermal fissions only)$

P_FNL: probability of fast non-leakage
P_TNL: probability of thermal non-leakage

3. The attempt at a solution

η is dependent on cross sections which are dependent upon energy (I wasn't given either) so I can only surmise there must be something in the given information that will suffice.

Perhaps 1000 are produced and 450+350 (900) are absorbed in fuel? So $η= 10/9 =1.111...$

f: Mathematically speaking, ratio of macroscopic x-sections is no more relevant than ratio of microscopic since macro=micro*N where N is neutron density so N would cancel.

absorbed: 450+350 (900); fuel is U_235 (right?) so f = 450/900 = 0.5

p: Fissioned neutrons start of fast (high energy), go through the resonance peak (hig absorbtion) energy region, then if they survive all that go to thermal (lower energy) state. So...

I don't know. Conceptually I get it (I wrote the above line about slowing down myself) but not sure how to evaluate this.
Seems like *maybe* $(total-absorbed)/total = (1000-900)/1000 = 1/10$ ?

ε: total # fission neutrons is the 1000, right? And just from thermal is... I don't know. I assume it has something to do with the 450 absorbed in U-235, but that's absorbed and not fissioned. So it *could* be $1000/450 = 2.222....$ but that seems wrong somehow.

P_FNL *and* P_TNL: Seems like the 125 given is total leakage and so the product of the two should be $(1000-125)/1000$ = 0.875

If everything above is correct (which I doubt) my answer would be a pitifully low $k≈0.108$