August 30, 1951

The question was raised in a recent meeting on the amount of depression of
flux produced by a control rod in a reactor. In this note, calculations of flux
distributions for a simple reactor rod geometry are reviewed.

Assumptions.
• 1. The reactor is a bare, infinitely long circular cylinder, containing
  a water solution of NH/N235 = 400.
• 2. A control rod, black to thermal neutrons, with arbitrarily chosen diameter
  of 1" is located along the central axis of the reactor.
• 3. One group diffusion theory is applicable to the estimate of critical
  size and flux distribution.

Theory.
• (a) Reactor without control rod.
  The solution of the reactor equation in cylindrical coordinates r,
  l/r d/dr (r d[phi]/dr) + K² [phi] = 0 ----(1)
  
  is [phi] = [phi]o Jo(Kr)
  
  where [phi]o is the central flux and Jo(Kr) is the 1st order Bessel function
  of the argument Kr. In order for [phi] to go to zero at the critical radius
  R, K = 2.405/R.
  
  The constant K depends on the composition of the reactor solution ac-
  cording to K² = lnk/L²+[tau]
  
  where k is the infinite multiplication constant, L is the thermal dif-
  fusion length, and [tau] is the "age." For the case NH/N235 = 400, K is
  approximately 0.128, so that R = 2.405/0.128 = 18.8 cm.
•  
• (b) Reactor with control rod.
  Since the thermal flux goes to zero at the surface of the control rod,
  of radius taken as a, the solution of equation (l) must be written as a
  linear combination of the bessel functions Jo, and No (using the nota-
  tion of Johnke and Ernde). Thus assume

• [phi]¹ = A Jo (Kr) + B No (Kr)
  
  where the prime distinguishes the case with a rod, and apply the boundary
  conditions [phi]¹ = o at r = a and r = R¹.
  This yields the condition
  
  
  
  from which the critical radius R¹ is computed.
  Since one of the constants A and B is arbitrary, we take B =+Jo (Ka).
  Thus A = -No (Ka) and [phi]¹ = - No (Ka) Jo (Kr) + Jo (Ka) No (kr)
  For the case of K 0.l28, Ka 0.163, No (Ka) = -1.215, Jo (Ka) = 0.994.
  Thus, if we let KR¹ = x
  No(x)/Jo(x) = -1.215/0.994 = -1.222
  
  It x = 3.08, this equation is satisfied.
  The central radius is thus R¹ = 3.08/0.128 = 24.1 cm, larger, as expected,
  than that for, the reactor without a rod.
  The flux is given by
  
  [phi]¹ = 1.215 Jo (0.128r) + 0.994 No (0.128r) ----(3)

Since the method of calculation involves two separate reactors, with only a
common active solution, there is no preferable mode of normalization for compari-
son. In order to illustrate the rod effect, a simple Bessel function and the
distribution with a rod are plotted in the attached graph so that the ordinates
agree near the walls. The flux is affected strongly for about a third of the way
out.