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Nuclear Reactor Digitization Project

Raymond L. Murray Reactor Project Notebook

Prepared for the North Carolina State University Science and Technology Electronic Text Center

The lineation of the manuscript has been maintained and all end-of-line hyphens have been preserved.

Keywords in the header are a local Science and Technology Electronic Text Center scheme to aid in establishing analytical groupings.

Problem: An element or radioactive half-life λ1, is generated at a rate g
during the operating time τ of a reactor; no production takes place during an

idle time (k-1)τ the cycle is repeated with period kτ. A daughter product of

half-life λ

2 is liberated continuously from the accumulation of parent. The
equilibrium level of the latter and the amount of daughter activity at various

Analysis: The solution of the differential equation for the parent during
reactor operation

If the reactor has been operating for a long time, the number of parent atoms
is the sum of contributions from all previous cycles, ie, the sum of terms like

(2) with times t,

t + kτ, t + 2kτ,....., t + ikτ.
inserted, as may be seen from Figure 1.

The sum of the form

1. The solution is applicable to time after the termination of reactor
operation from o to (k-l)τ, i.e. until operation begins again. At that time a

term of the form of equation (1) must be added.

The peak level is, of course

g B/λ, where B is the expression in parentheses.
Figure 2 indicates the trend schematically.

2. When one cycle has elapsed, t = kτ, the peak is again reached, as may

3. The minimum level occurs at t = (k-l)τ, and is thus a factor e-λ 1(k-1)τ

4. The average number (or activity) of the parent atoms over a cycle is
exactly that expected if the generation rate were assumed to be continuous

Accumulated activity of daughter over a period k .

Assume that the daughter product grows during the interval between successive

The amount collected is the solution of the equation

Application of theory to the case of the mass 133 cycle.

Half-lives:parent I 133 22 hr = 0.917d
daughter Xe133 5.3d

A 2/g

Evaluate g, using the fact that mass 133 appears with 4.5%
yield in fission, and assuming a 10kw reactor power level.

Appendix 1
Proof that solution is periodic.

Appendix 2

Appendix 3

Separate solution into two parts, superpose contribution of
1st and 2nd terms in N

Appendix 4.

consider the 60 m decay from continuously generated Te.
N

The following typographical errors in the original note of the
above title have been found:

Page 2.2. Delete the λ2 in the denominator
of the first term.

Equation for A

2)Kτ. Multiply B by λ2.
Third line from bottom should read λ

2 - λ1 = -0.625.Page 3.2/g = 1/-0.625[(0.131)(0.324)(0.470 - 0.877) + (0.131)
(1 - 0.828) - 0.756(1 - 0.968)] = 0.030

-----------------

The accumulated activity is (0.030)(365) = 11 curies.