Upload
gellert-rendor
View
32
Download
3
Tags:
Embed Size (px)
DESCRIPTION
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. - PowerPoint PPT Presentation
Citation preview
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 k N cdt l
i ii i
dc N cdt 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
0dNdt
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
0idcdt
/(0) (0)i iic 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
flux
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
trons
Production of the delayed neutrons during the transient
1st 2nd 3rd 4th 5th 6th
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)