CONTROL RODS DOE-HDBK-1019/2-93 Reactor Theory (Nuclear Parameters) Example 2: Using  the  differential  rod  worth  curve  provided  in  Figure  11,  calculate  the  reactivity inserted by moving the rod from 10 inches withdrawn to 6 inches withdrawn. Solution: The solution is basically given by the area under the curve for the interval.  The answers obtained in the following approximation may vary slightly depending upon the degree of approximation. Method 1. Treating  the  range  from  10  inches  to  6  inches  as  a  trapezoid,  that  is, taking  the  end  values  of  pcm/inch  and  multiplying  their  average  by  the 4 inches moved yields the following.                             8 pcm inch 3 pcm inch 2 4  inches         22  pcm This is negative because the rod was inserted. Method 2. Using the central value of rod position at 8 inches yields an average rod worth of 5.5 pcm/inch.   Multiplying by the 4 inches of rod travel yields the answer. (5.5 pcm/in.)(4 in.) = -22 pcm Method 3. Breaking  the  rod  travel  total  into  two  parts  (10  inches  to  8  inches  and 8 inches to 6 inches) yields:                                 8 pcm inch 5.5        pcm inch 2 2  inches         13.5  pcm                                 5.5        pcm inch 3        pcm inch 2 2  inches          8.5  pcm ( - 13.5 pcm) + ( - 8.5 pcm) = -22 pcm NP-03 Rev. 0 Page 54