1. Solving the Reactor Kinetics Equations numerically
enables to reproduce the initial phase of the
Chernobyl Accident
Frigyes Reisch
Nuclear Power Safety
KTH, Royal Institute of Technology
Stockholm, Sweden
Brookhaven National Laboratory
3 November 2005

2. The classical reactor kinetic equations
with six groups of delayed neutrons
are not solved analytically
Here they are solved numerically
with MATLAB and presented graphically

3. At the Chernobyl experiment
due to the abrupt decrease of the speed of the main circulation pumps
at low reactor power
and heavy Xenon poisoning
and also due to several other reasons
the void (steam) content
in the coolant channels
increased suddenly (~50%)
Thus the positive void coefficient
(~30 pcm/%)
caused a large reactivity insertion.

4. The neutron flux and thereby the reactor power increased very fast
Due to the thermal inertia of the fuel
and the limited amount of the
fuel temperature coefficient
the Doppler effect
could not break the power excursion.

5. to characterize the process at the initial phase to use only the
Therefore
to characterize the process
at the initial phase
to use only the
reactor kinetics equations
is sufficient.

6. The simplified neutron kinetics equations

7. t time (sec)
N neutron flux (proportional to the reactor power)
δk change of the neutron multiplication factor (k)
β sum of the delayed neutron fractions
neutron mean lifetime (sec)
λi i:th decay constant (sec-1)
ci concentration of the i:th fraction of the delayed neutrons
at steady state
N(0)=1

8. Constants and some related values for U235 thermal fission
β1=
λ1=0.0124
β1/ λ1=
1/ λ1= sec
1/ λ2= sec
β2=
λ2=0.0305
β2/ λ2=
1/ λ2= sec
β3=
λ3=0.1110
β3/ λ3=
λ4=0.3010
1/ λ4= sec
β4=
β4/ λ4=
λ5=1.1400
β5/ λ5=
β5=
1/ λ5= sec
λ6=3.0100
β6=
β6/ λ6=
1/ λ6= sec
0.0065
sec

9. The normalized value Nn(0)= 1
At steady state
Initial values
The normalized value Nn(0)= 1
cn1(0)= cn2(0)= cn3(0)=11.47
cn4(0)= cn5(0)= cn6(0)=0.0907

10. The one plus six differential equations
N’=(DeltaK/ )*N *c *c *c *c4+1.14* c *c6
c1’=0.2150*N * c1
c2’=1.1424*N * c2
c3’=1.2740*N * c3
c4’=2.5680*N * c4
c5’=0.7480*N * c5
c6’=0.2730*N * c5
The MATLAB notation
x(1)=N x(2)=c1’ x(3)=c2’ x(4)=c3’ x(5)=c4’ x(6)=c5’ x(7)=c6’

11. The MATLAB code %Save as xprim7.m function xprim = xprim7(t,x,i)
DeltaK=i*0.010*0.50;
%voidcoef=i*0.010pcm/percent void change, void increase 50percent
xprim=[(DeltaK/ )*x(1) *x(2)…
*x(3)+0.111*x(4)+0.301*x(5)+1.14*x(6)+3.01*x(7);
0.2150*x(1) *x(2);
1.1424*x(1) *x(3);
1.2740*x(1) *x(4);
2.5680*x(1) *x(5);
0.7480*x(1) * x(6);
0.2730*x(1) * x(7)];

12. The instruction to plot the graphic
%Save as NeutronKin.m
figure
hold on
for i=-3:6:3
0.2],[1; 17.34;46.69;11.47;8.53;0.6561;0.0907],[] ,i);
plot(t,x(:,1))
end
hold off

13. At the time of the Chernobyl accident the void coefficient was about +30 pcm/%
Now a days it is diminished (due to higher enrichment but still positive)
If the void coefficient was rather about -30 pcm/%
the transient would have terminated itself.

14. A parameter study results in this graph:
With zero void coefficient there is neither power increase nor power decrease
While negative void coefficient would lead to the decline of the power i.e. to shut down

15. A study of the delayed neutrons separately with +30 pcm/%
The 1st group has the largest time constant (1/ λ1= sec)
Therefore the time delay is the longest there.
The 6th group has the shortest time constant (1/ λ6= sec)
Therefore the time delay is the least there.
The 2nd group has the largest βi/ λi (= )
The 6th group has the smallest βi/ λi (= )