TO: Clifford K. Beck
FROM: Raymond L. Murray
CC: Reactor Committee
SUBJECT: Equilibrium of Radioactive Product of Intermittent Reactor Operation.

Problem: An element or radioactive half-life [lambda]1, is generated at a rate g
during the operating time [tau] of a reactor; no production takes place during an
idle time (k-1)[tau] the cycle is repeated with period k[tau]. A daughter product of
half-life [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 (1)
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 , so that Several features of this equation are of interest.

1. The solution is applicable to time after the termination of reactor
operation from o to (k-l)[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 = (k-l)[tau], and is thus a factor e-[lambda]1(k-1)[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 = (K-1)[tau]
and going to t = k[tau]
---(3)
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.

Note; as shown in
Appendix 4, the
predecessor of
I133 can be assumed
to decay instantly. Operating time [tau] = 6 hours = 0.25d.
thusk = 4, k[tau] = 1d

Half-lives:	parent I133 22 hr = 0.917d daughter Xe133 5.3d
Decay constants:	[lambda]1= 0.693/0.917 = 0.756 d-1 [lambda]2 = 0.693/5.3 = 0.131 d-1

Xe133 activity from equation (3) obtained by calculating different
constants and terms.

	[lambda]1[tau] = 0.189, e-[lambda]1[tau] = 0.828 K[lambda]1[tau] - 0.756, e-k[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, e-k[lambda]2[tau] = 0.877

A2/g	1/-0.625[.131x0.324(0.8280.470-0.877)+0.131(1-0.828)-0.756(1-0.968)]
	(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(1-e-[lambda]1t)
	t=0 to t=k[tau]	t=(k-1)[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. when t=0, N2=0,

1st N2
N2

when t=0, N2=0,

2nd N2
N2

or simplifying
N2
For t=0 to t=(K-1)[tau], use only N1. (off-time)
for t=(K-1)[tau] to t=K[tau], use N1 + N2 (on-time)

Appendix 4.
Justification of neglect of predecessors of I133.
Actual chain:

consider the 60 m decay from continuously generated Te.
No = g/[lambda]o(1-e-[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: Clifford K. Beck
FROM: Raymond L. Murray
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.