Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
Lower ‘control rods’ into reactor to absorb excess neutrons<br />
Materials used: 113 Cd (σ c = 20,000 barns)<br />
10<br />
B (σ c = 4,000 barns)<br />
cross section is for thermal neutrons (~0.025 eV)<br />
[σ c (max) ~ π(λ/2π) 2 ∼ 2.6 10 7 barns]<br />
(withdraw control rods if reactivity gets too low)<br />
In practice control would be virtually impossible but for the existence<br />
<strong>of</strong> ‘delayed’ neutrons<br />
Delayed neutrons are released only after the β-decay <strong>of</strong> a fission product<br />
Typically, about 1% <strong>of</strong> neutrons produced by fission are delayed by<br />
10-20 seconds, which is enough time for small adjustments in the<br />
position <strong>of</strong> the control rods (automatically controlled)<br />
eg<br />
87<br />
Br β −<br />
t 1/2<br />
54.5 s<br />
n<br />
87<br />
Kr *<br />
86<br />
Kr<br />
137<br />
I β −<br />
t 1/2<br />
21.8 s<br />
n<br />
137<br />
Xe *<br />
136<br />
Xe<br />
Neutron Population Growth<br />
η is number <strong>of</strong> neutrons emitted per neutron absorbed<br />
Because <strong>of</strong> losses the mean number is k, where k < η<br />
k is the effective multiplication factor<br />
If all neutrons were prompt the neutron population would grow like<br />
dn/dt = n(k−1)/τ = nq/τ<br />
where q=(k−1) and τ is the average neutron lifetime in the reactor<br />
So<br />
n = n o exp{qt/τ}<br />
eg q = 0.001, τ = 0.001 second gives n = n o exp(t), so after 10 s<br />
n/n o increases by a factor <strong>of</strong> 22,000<br />
In 235 U the mean lifetime <strong>of</strong> the<br />
groups <strong>of</strong> delayed neutrons is<br />
about τ d = 9 s and they represent<br />
a fraction β = 0.65% <strong>of</strong> the total<br />
neutron emission<br />
235<br />
U delayed neutrons<br />
If q