TO:
FROM:
CC: Reactor Committee
SUBJECT: Equilibrium of Radioactive Product of Intermittent Reactor Operation.
Problem: An element or radioactive halflife [lambda]1, is generated at a rate g
during the operating time [tau] of a reactor; no production takes place during an
idle time (k1)[tau] the cycle is repeated with period k[tau]. A daughter product of
halflife [lambda]2 is liberated continuously from the accumulation of parent. The
equilibrium level of the latter and the amount of daughter activity at various
times in the cycle is to be determined.
Analysis: The solution of the differential equation for the parent during
reactor operation
is
At the end of a time [tau], the accumulation is:
During the period thereafter, simple decay takes place, according to
The variable t inserted in this equation is taken as zero at time [tau].
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 + k[tau], t + 2k[tau],....., t + ik[tau].
inserted, as may be seen from Figure 1.
The sum of the form Several features of this equation are of interest. 
1. The solution is applicable to time after the termination of reactor
operation from o to (kl)[tau], 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/[lambda], where B is the expression in parentheses.
Figure 2 indicates the trend schematically.
2. When one cycle has elapsed, t = k[tau], the peak is again reached, as may
be demonstrated by substitution. (Appendix 1)
3. The minimum level occurs at t = (kl)[tau], and is thus a factor e[lambda]1(k1)[tau]
lower than the peak.
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
at a level 1/k times the actual generation rate. (Appendix 2)
Accumulated activity of daughter over a period k .
Assume that the daughter product grows during the interval between successive
times of shutting down the reactor, ie. between times t = o and t = k[tau].
The amount collected is the solution of the equation
As shown in Appendix 3, the result is the sum of two terms, the first
applicable to times in the range t = otot = k[tau], the second starting at t = (K1)[tau]
and going to t = k[tau]
At the end of the next reactor operation t k[tau], the activity, A2 = N2[lambda]2, is
Application of theory to the case of the mass 133 cycle.
Appendix 4, the
predecessor of
I133 can be assumed
to decay instantly.
thusk = 4, k[tau] = 1d
Halflives:  parent I133 22 hr = 0.917d daughter Xe133 5.3d 
Decay constants:  [lambda]1= 0.693/0.917 = 0.756 d1 [lambda]2 = 0.693/5.3 = 0.131 d1 
[lambda]1[tau] = 0.189, e[lambda]1[tau] = 0.828 K[lambda]1[tau]  0.756, ek[lambda]1[tau] = 0.470 

[therefore] B = 0.324  [lambda]2[lambda]1 = 0.525 .625 [lambda]2[tau] = 0.0327, e[lambda]2[tau] = 0.968 K[lambda]2[tau] = 0.131, ek[lambda]2[tau] = 0.877 
A2/g  1/0.625[.131x0.324(0.8280.4700.877)+0.131(10.828)0.756(10.968)] 
(0.0424)(0.407) (0.131)(0.172) (0.756)(.032)  
0.0172 0.0225 0.0241 = 0.0188  
= 0.0333  
A2/g  0.0301 
Evaluate g, using the fact that mass 133 appears with 4.5%
yield in fission, and assuming a 10kw reactor power level.
thus the accumulated Xe activity is (0.03330.0301)(365) = 12.211.0 curies
Appendix 1
Proof that solution is periodic.
N1)total  
origin of time at [tau] substitute t=k[tau] 
origin of time at 0 substitute [tau] 

the starting peak value, at t=0. 
Appendix 2
Proof that average number of parent atoms is 1/k times
continuous equilibrium case
= g/[lambda]1 1/k
Appendix 3
Solution of Daughter Equation
where N1  = g/[lambda]1Be[lambda]1t +  g/[lambda]1(1e[lambda]1t) 
t=0 to t=k[tau]  t=(k1)[tau] to k[tau] using range t=0 to t=[tau] 
Separate solution into two parts, superpose contribution of
1st and 2nd terms in N1.
1st N2  
N2 
2nd N2  
N2 
Appendix 4.
Justification of neglect of predecessors of I133.
Actual chain:
consider the 60 m decay from continuously generated Te.
No = g/[lambda]o(1e[lambda]ot)
By the previous theory the peak equilibrium level of this element
occurs at the end of the operating period [tau]
[lambda]o = 0.693/1/24 = 16.6, [lambda]o[tau] = 4.15, k[lambda]o[tau] = 16.6
[therefore]Bo
thus the activity is within 1.6% of being at its equilibrium value
Ao = [lambda]oNo = g.
TO:
FROM:
CC: Reactor Committee
Errata: Equilibrium of Radioactive Product of Intermittent Reactor Operation.
The following typographical errors in the original note of the
above title have been found:
Page 2.  Equation for N2. Delete the [lambda]2 in the denominator of the first term. Equation for A2)K[tau]. Multiply B by [lambda]2. Third line from bottom should read [lambda]2  [lambda]1 = 0.625. 
Page 3.  A2/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. 