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

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In summary, the conversation discusses the various factors and equations involved in calculating the Four Factor Formula and Six Factor Formula for a reactor. These include the reproduction factor, thermal utilization factor, resonance escape probability, fast fission factor, and probabilities of fast and thermal non-leakage. The values of these factors are influenced by the absorption of neutrons in different materials and the energy levels of the neutrons. The conversation also mentions the possibility of a low overall value for k, the multiplication factor.
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Homework Statement



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.

Homework Equations



[itex] k_i=ηfpε[/itex] where k_i is the Four Factor Formula; the six facor formula is
[itex] k=k_i*(P_FNL)*(P_TNL)[/itex]

the variables are...

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

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) [itex] f=(ƩF_a)/(Ʃ_a)[/itex]

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 [itex] ε=(total # of fission neutrons)/(# of fission neutrons from thermal fissions only) [/itex]

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

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 [itex] η= 10/9 =1.111...[/itex]

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* [itex] (total-absorbed)/total = (1000-900)/1000 = 1/10[/itex] ?

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 [itex]1000/450 = 2.222...[/itex] 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 [itex] (1000-125)/1000[/itex] = 0.875

If everything above is correct (which I doubt) my answer would be a pitifully low [itex]k≈0.108[/itex]
 
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. I'm mostly just wondering if I'm on the right track with the factors and how I might evaluate p since that's the one I'm most uncertain about.
 

1. What is the six-factor formula for calculating a given set of information?

The six-factor formula is a mathematical equation that is used to calculate the value of a given set of information. It is also known as the six-factor model or the six-factor regression equation. The formula takes into account six different factors, including sample size, mean, standard deviation, standard error, confidence level, and confidence interval, to estimate the true value of a population.

2. How do I determine the sample size for the six-factor formula?

The sample size is the number of individuals or data points included in a study. To determine the sample size for the six-factor formula, you need to consider the margin of error, the desired confidence level, and the variability of the data. A larger sample size will result in a more accurate estimation of the population value.

3. What is the significance of the confidence level in the six-factor formula?

The confidence level is the probability that the calculated interval will contain the true population value. It is typically expressed as a percentage, such as 95% or 99%. The higher the confidence level, the wider the interval will be, but there is a higher chance that the true population value falls within that interval. The confidence level is an important factor in determining the accuracy of the six-factor formula.

4. How does the mean and standard deviation affect the six-factor formula?

The mean, also known as the average, is the sum of all values divided by the number of values. The standard deviation measures the spread of the data around the mean. These two values are used in the six-factor formula to calculate the standard error, which is the estimated standard deviation of the sample mean. The standard error is then used to determine the confidence interval, which is a range of values that is likely to contain the true population value.

5. Can the six-factor formula be used for all types of data?

The six-factor formula is commonly used in scientific studies and research to estimate the true value of a population. It can be used for all types of data, including numerical, categorical, and ordinal data. However, the accuracy of the formula may vary depending on the type of data and the assumptions made during the calculation process. It is important to carefully consider the data and make appropriate adjustments when using the six-factor formula.

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