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inserted by moving the rod from 10 inches withdrawn to 6 inches withdrawn.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
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