Question

135Xe will be shown in future discussions to be a major concern in nuclear reactors. Assume there are 10,000 particles of 135Xe initially in a volume. The production rate is 20,000 particles per hour. Plot the number of 135Xe particles over one week in one hour increments(don't forget that 135Xe also decays, so that is a loss rate) using ODE45 and the analytic solution. Is the system close to equilibrium(within 5 % of actual equilibrium value)?

Answer 1

You should have given the value of decay constant of Xe, but anyway I'll use the data I found on the internet.

so that

The differential equation is

where R is the constant production rate given along with the initial condition

Analytical solution is easily obtained.

Equilibrium no.of particles correspond to

N_0=10000;
lam=0.0753;
R=20000;
DE=@(t,N)R-lam*N(1);
[t,n]=ode45(DE,[0 168],N_0); %using ode45 solver
plot(t,n)

A=zeros(1,168);
A(1)=N_0;
for i=1:167 %Increments of 1 hour
A(i+1)=A(i)+R-lam*A(i);
end
hold on
plot(A);
legend('By ode45','In increments of 1 hour')
xlabel('Time, t in hours')
ylabel('Number of particles, N(t)');
title('N vs t')

The system reaches equilibrium very early before 1 week. Even mid-week values are quite close to equilibrium.