P
Po 2t /DT
keff
1
keff
1.0025
1
1.0025
0.00249
k/k
¯
eff
eff
0.0070
0.00249
0.10 sec1 0.00249
18.1 sec
REACTOR KINETICS
DOE-HDBK-1019/2-93
Reactor Theory (Reactor Operations)
NP-04
Rev. 0
Page 18
Doubling time (DT) = (ln 2)
where:
= stable reactor period
ln 2
= natural logarithm of 2
When the doubling time is known, the power level change from P is given by the following
o
equation.
(4-12)
where:
t
= time interval of transient
DT
= doubling time
The following example problems reinforce the concepts of period and startup rate.
Example 1:
A reactor has a
of 0.10 sec and an effective delayed neutron fraction of 0.0070. If
eff
-1
k is equal to 1.0025, what is the stable reactor period and the SUR?
eff
Solution:
Step 1:
First solve for reactivity using Equation (3-5).
Step 2:
Use this value of reactivity in Equation (4-9) to calculate reactor period.
