Solving the Reactor Kinetics Equations numericallyenables to reproduce the initial phase of the

Chernobyl Accident

Frigyes ReischNuclear Power Safety

KTH, Royal Institute of TechnologyStockholm, Sweden

Brookhaven National Laboratory3 November 2005

The classical reactor kinetic equationswith six groups of delayed neutrons

are not solved analyticallyHere they are solved numerically

with MATLAB and presented graphically

At the Chernobyl experimentdue to the abrupt decrease of the

speed of the main circulation pumpsat low reactor power

and heavy Xenon poisoningand also due to several other reasons

the void (steam) contentin the coolant channels

increased suddenly (~50%)Thus the positive void coefficient

(~30 pcm/%)caused a large reactivity insertion.

The neutron flux and thereby the reactor powerincreased very fast

Due to the thermal inertia of the fueland the limited amount of thefuel temperature coefficient

the Doppler effectcould not break the power excursion.

Thereforeto characterize the process

at the initial phaseto use only the

reactor kinetics equationsis sufficient.

The simplified neutron kinetics equations

6

1

i i

i

dN kN c

dt l

i ii i

dcN c

dt l

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 fractionsl 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

0dN

dt 0k N(0)=1

6

1

i i

i

cl

β1=0.000215 λ1=0.0124 β1/ λ1=0.0173400 1/ λ1=80.6451 sec

Constants and some related values for U235 thermal fission

β2=0.001424 λ2=0.0305 β2/ λ2=0.0466900 1/ λ2=32.7869 sec

β3=0.001274 λ3=0.1110 β3/ λ3=0.0114700 1/ λ2=32.7869 sec

β4=0.002568 λ4=0.3010 β4/ λ4=0.0085300 1/ λ4=3.32226 sec

β5=0.000748 λ5=1.1400 β5/ λ5=0.0006561 1/ λ5=0.87719 sec

β6=0.000273 λ6=3.0100 β6/ λ6=0.0000907 1/ λ6=0.33223 sec

6

1

i

i

0.0065

0.001l sec

At steady state

0idc

dt /

(0) (0)i i

ic Nl

cn1(0)=17.34 cn2(0)=46.69 cn3(0)=11.47cn4(0)=8.53 cn5(0)=0.6561 cn6(0)=0.0907

Initial values

The normalized value Nn(0)= 1

The one plus six differential equations

N’=(DeltaK/0.001-6.5)*N+0.0124*c1+ 0.0305*c2+ 0.111*c3+ 0.301*c4+1.14* c5+ 3.01*c6

c1’=0.2150*N-0.0124* c1

c2’=1.1424*N-0.0305* c2

c3’=1.2740*N-0.1110* c3

c4’=2.5680*N-0.3010* c4

c5’=0.7480*N-1.1400* c5

c6’=0.2730*N-3.0100* 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’

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/0.001-6.5)*x(1)+0.0124*x(2)…+0.0305*x(3)+0.111*x(4)+0.301*x(5)+1.14*x(6)+3.01*x(7);0.2150*x(1)-0.0124*x(2);1.1424*x(1)-0.0305*x(3);1.2740*x(1)-0.1110*x(4);2.5680*x(1)-0.3010*x(5);0.7480*x(1)-1.1400* x(6);0.2730*x(1)-3.0100* x(7)];

The instruction to plot the graphic

%Save as NeutronKin.m figurehold onfor i=-3:6:3[t,x]=ode45(@xprim7,[0 0.2],[1; 17.34;46.69;11.47;8.53;0.6561;0.0907],[] ,i); plot(t,x(:,1))endhold off

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.

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

1

2

3

4

5

6

7

8

9

10

Time, sec

Neu

tron

flu

x

Transients with insertion of 50% steam into the channels of Chernobyl type of reactor

void coefficient +30 pcm/%

void coefficient -30 pcm/%

1 pcm=10exp-5

A parameter study results in this graph:

With zero void coefficient there is neither power increase nor power decreaseWhile negative void coefficient would lead to the decline of the power i.e. to shut down

A study of the delayed neutrons separately with +30 pcm/%

0 0.1 0.2 0.3 0.4 0.5 0.6 0.710

-2

10-1

100

101

102

103

Time, sec

Del

ayed

neu

tron

s

Production of the delayed neutrons during the transient

1st 2nd 3rd 4th 5th6th

The 1st group has the largest time constant (1/ λ1=80.6451 sec)Therefore the time delay is the longest there.The 6th group has the shortest time constant (1/ λ6=0.33223 sec)Therefore the time delay is the least there.The 2nd group has the largest βi/ λi (=0.0466900)The 6th group has the smallest βi/ λi (=0.0000907)