Simulation of point kinetics reactor equations
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Hello!Those two equations are needed to be solved (the attached picture)
The initial conditions n(0)=0.1, c(0)=0
The required: find the required time to increase n from 0.1 to 1
I got errors regarding syms functions. I am not sure if I do that right.I attached to this a matlab file which contains all the parameters and what I tried to do.
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Accepted Answer
Alan Stevens
on 7 Oct 2021
There are seven equations if you are using all six delayed neutron groups. You don't give your reactivity, nor the individual beta values. The program below uses arbitrary data for rho and the timespan, and Glasstone and Sesonske values for beta_i. If you are only interested in the case of a single group of delayed neutrons you should be able to modify the following appropriately:
% Point reactor kinetics 6 groups of delayed neutrons
beta = [0.00021; 0.00141; 0.00127; 0.00255; 0.00074; 0.00027]; % Taken from Glasstone and Sesonske
betasum = sum(beta);
rho = 1.1*betasum; % reactivity
% Initial consitions
c0 = zeros(6,1);
n0 = 0.1;
tspan = [0, 1];
nc0 = [n0; c0];
[t, nc] = ode45(@(t,nc) kinetics(t,nc,rho,beta,betasum), tspan, nc0);
n = nc(:,1);
c = nc(:, 2:7);
plot(t,n),grid,
xlabel('time'), ylabel('n')
% figure
% plot(t,c),grid
% xlabel('time'), ylabel('c')
function dncdt = kinetics(~,nc,rho,beta,betasum)
L = 0.0001;
lam = [0.0126; 0.0337; 0.111; 0.301; 1.14; 3.01];
n = nc(1);
c = nc(2:7);
dndt = (rho - betasum)/L + sum(lam.*c);
dcdt = beta*n/L - lam.*c;
dncdt = [dndt; dcdt];
end
6 Comments
Cody James
on 31 Oct 2021
Is it possible to simultaniously solve for multiple values of rho and plot them? For example: rho = [0.01, 0.02, 0.03]
Alan Stevens
on 1 Nov 2021
Something like this?
% Point reactor kinetics 6 groups of delayed neutrons
beta = [0.00021; 0.00141; 0.00127; 0.00255; 0.00074; 0.00027]; % Taken from Glasstone and Sesonske
betasum = sum(beta);
rho = [0.01 0.02 0.03];
% Initial consitions
c0 = zeros(6,1);
n0 = 0.1;
tspan = [0, 1];
nc0 = [n0; c0];
for i = 1:numel(rho)
[t, nc] = ode45(@(t,nc) kinetics(t,nc,beta,betasum,rho(i)), tspan, nc0);
n = nc(:,1);
c = nc(:, 2:7);
figure
plot(t,n),grid,
xlabel('time'), ylabel('n')
legend(['rho = ' num2str(rho(i))])
end
function dncdt = kinetics(~,nc,beta,betasum,rho)
L = 0.0001;
lam = [0.0126; 0.0337; 0.111; 0.301; 1.14; 3.01];
n = nc(1);
c = nc(2:7);
dndt = (rho - betasum)/L + sum(lam.*c);
dcdt = beta*n/L - lam.*c;
dncdt = [dndt; dcdt];
end
More Answers (1)
Swu
on 7 Mar 2023
in "function dncdt = kinetics(~,nc,rho,beta,betasum)"
"dndt = (rho - betasum)/L + sum(lam.*c);"
should be ""dndt = (rho - betasum)* n/L + sum(lam.*c);
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