Numerical Derivative Assignment Help
The differential equations for reactor power is modelled using point reactor kinetics equation as shown below in our numerical derivative assignment help. If the third variable is not present in the equation the system will be solved analytically. Euler’s method is performed to find the reactor period and reactor temperature. In numerical derivative Matlab homework help, we have solved the one-dimensional boundary value problems by estimating the finite difference Euler’s method. Finally we obtained the distribution of values by computation of eigenvectors.
- In the first problem we are dealing with a system of coupled differential equations for reactor power and precursor concentration. The equations are given by: The system would be analytically solvable if the reactivity wasn’t a third variable following the equation:Instead of solving the system of three numerical derivative matlab assignment help differential equations, we can apply simple finite difference forward Euler method. It can be derived from the Taylor’s expansion by ignoring the super linear terms or simply from the derivative definition by making the infinite limiting value finite:It should be noted the above equation is in the vector form so it works for a system as well as a single equation. The forward part originates from the fact that the previous value is used in the derivative calculation so the equation is explicit.The last formula required in this section is for numerical derivative assignment help the reactor period. It is defined as the time required for the reactor to increase the power by the factor of e. This can be formulated as:For the smaller insertion we have the following plot with marked powers of interest: By using the shown values we get the two respective periods as 19.46s and 74.48s. The end temperature is 320K. By inserting a larger amount of reactivity numerical derivative matlab homework help the nature of the curve changes drastically:
Peak power of 6.4GW occurs after only 0.055 seconds. Total energy produced is the area under the curve and is equal to 80.15MJ.
clear all;close all;clc %Define boundary points x=0:1:100; n=length(x); %Define widths delta=diff(x);
%iterate while either condition is active while abs(K_eff-K_eff_last)>1e-6 || d>1e-12 phi_last=phi; K_eff_last=K_eff; [A,b]=createTri(delta,D,Sa,Q,1); %Tridiagonal solving dg=spdiags(A); dg1=dg(1:end-1,1); dg2=dg(:,2); dg3=dg(2:end,3); phi=Tri(dg1,dg2,dg3,b)'; K_eff=sum(abs(phi)); phi=phi/K_eff; d=sum(abs(phi-phi_last)); Q=nSf.*phi(2:101); end figure plot(x,phi) xlabel('Cell') ylabel('End distribution')
- One-dimensional boundary value problems can be solved by using a similar approximation to the finite difference Euler method, but one that allows a non-uniform grid. If we assume that no parameter numerical derivative homework help changes over a single cell, we can use piecewise integrations to approximate the solution to the equation:It should be noted that with the constant numerical derivative assignment help values of diffusion constant, source strength and absorption the system has an easy analytical solution: Three cases are processed in this version.(a) When we have constant values: source strength of 5000 neutron/cm^3 per sec, D = 5 cm and Σa = 0.2 cm-1 in every cell the uniform numerical derivative assignment solution applies as calculated With changed values that change over the length, the solution cannot be analytically obtained as easily. Source is from 0.0 to 10 cm (the left-most 10 cells) of strength 5000 neutron/cm^3 per sec in each cell and no sources in the other 90 cells. The left quarter of the slab (cells 1 through 25) consists of a highly diffusive medium with low absorption, D = 60 cm and Σa = 0.07 cm-1. The right three quarters of the slab (cells 26 through 100) consists of a highly absorptive numerical derivative problem solution medium with low diffusion, D = 1cm and Σa = 0.2 cm-1. The plot is now: Type I cells (cells 1 – 10, 21 – 40, 51 – 70) simulate fuel with numerical derivative online experts sources of strength 5000 neutron/cm^3 per sec, D = 1cm and Σa = 0.2 cm-1. Type II cells (all the rest) simulate water with no source, D = 40 cm and Σa = 0.025 cm-1. In this case we have a distribution with three peaks: Left boundary flux is , and there is a conclusion to be made here. In each case the left boundary flux can be numerical derivative project help approximated by the uniform case. The smaller the diffusivity, the better the approximation is obviously the rule here.2.By following the instructions for eigenvector online numerical derivative tutor calculation, we obtain the following distribution of values:
plot(time,log(nt)) xlabel('Time(s)') ylabel('Log n(W)') title('Large insertion') grid on %Peak calculation [Pmax,tmax]=max(nt); %Display results fprintf('Maximum power value:%f W\n',Pmax) fprintf('Maximum power time in seconds:%f s\n',time(tmax)) fprintf('Total energy in the first second:%f J \n',E)