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Next: Tests involving the inverse Up: Preliminary tests power method Previous: Preliminary tests power method   Contents   Index

Tests involving the power method

The test is divided up into two parts one part which deals with a super-critical ($\lambda_0 > 1$) and another part which deals with a sub-critical ($\lambda_0 < 1$) reactor.

For the super-critical test example we use the elevation data and cross sections data given in the introductory section 4.2 together with the absorber tip position given by (4.3).

In order to achieve a sub-critical assembly we increase the fast absorption cross section of the two fuel slabs to the value given by (4.4).

In both tests we use a nearly uniform computational grid with steplengths of $10 \mbox{cm}$.

For the super-critical reactor assembly we obtain the eigenvalue sequence plotted in Figure 4.6. As we can see the exact eigenvalue is approached from above--this is of major importance if we would want to use the power method to generate a good estimate of the eigenvalue for use in the fractional iteration method.

\begin{figure}
% latex2html id marker 16754\rule{\textwidth}{0.2mm}
\rule{0cm}...
...$}) represent the exact numerical eigenvalue, $\lambda^{(\infty)}$.}\end{figure}

In Figure 4.7 the estimated eigenvector error corresponding to Figure 4.6 at the kth iterate, Ek, defined as

\begin{displaymath}
E_k \;\hbox{$=$\kern-0.68em\raise1.1ex
\hbox{$\scriptscrip...
...{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(k-1)} \Vert _\infty
\end{displaymath} (4.3)

is plotted as a function of the iteration number, k. It is this error estimate which is used in the convergence criterion (2.4) described in section 2.2 (see p. [*]).

\begin{figure}
% latex2html id marker 16803\rule{\textwidth}{0.2mm}
\rule{0cm}...
...{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(0)} = [1,1,\ldots,1]^T$.}\end{figure}

We have furthermore calculated the exact eigenvector error at the kth iterate, $E_{k,{\mbox{\protect\scriptsize exact}}}$, defined as

\begin{displaymath}
E_{k,{\mbox{\protect\scriptsize exact}}} \;\hbox{$=$\kern-0...
... \hspace{0.2ex}\underline{\phi}{}\hspace{0.15ex} \Vert _\infty
\end{displaymath} (4.4)

where $\hspace{0.2ex}\underline{\phi}{}\hspace{0.15ex}$ represents the exact eigenvector (in our case the one calculated numerically by Matlab).

The exact error on the eigenvector for the super-critical reactor assembly is depicted in Figure 4.8. In section 2.2 we saw that the asymptotic behaviour of the exact error follows

\begin{displaymath}
\frac{E_{k+1,{\mbox{\protect\scriptsize exact}}}}{E_{k,{\mb...
...}}}} = \left( \frac{\vert \lambda_1
\vert}{\lambda_0} \right)
\end{displaymath} (4.5)

where $\lambda_1$ is the eigenvalue with the next largest modulo and $\lambda_0$ is the largest eigenvalue. Now drawing a straight line through the last 40 iteration points in Figure 4.8 we calculate the ratio of the two largest eigenvalues as follows

\begin{displaymath}
\frac{E_{k+1,{\mbox{\protect\scriptsize exact}}}}{E_{k,{\mb...
...t( \frac{1.75 \cdot
10^{-5}}{0.5} \right)^{1/70} \approx 0.86
\end{displaymath} (4.6)

The exact ratio calculated by values in Table 4.1 is

\begin{displaymath}
\left( \frac{\vert \lambda_1 \vert}{\lambda_0} \right) = \frac{1.0739}{0.9277}
\approx 0.86386
\end{displaymath} (4.7)

These two value are so close together that we with confidence can state that the used implementation of the power method is working correctly.

\begin{figure}
% latex2html id marker 16850\rule{\textwidth}{0.2mm}
\rule{0cm}...
...{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(0)} = [1,1,\ldots,1]^T$.}\end{figure}

Finally, we have plotted the points of iterations for the exact eigenvalue error, $E_{\lambda,k}$, defined as

\begin{displaymath}
E_{\lambda,k} \;\hbox{$=$\kern-0.68em\raise1.1ex
\hbox{$\s...
...}}\;\lambda^{(k)} - \lambda_{\mbox{\protect\scriptsize exact}}
\end{displaymath} (4.8)

where $\lambda_{\mbox{\protect\scriptsize exact}}$ is the exact eigenvalue (in our case the one calculated numerically by Matlab).

The resulting plot is depicted in Figure 4.9. We note, as postulated in section 2.2, that the error on the eigenvalue is significantly less that the error on the eigenvector (see Figure 4.8 for comparison).

\begin{figure}
% latex2html id marker 16900\rule{\textwidth}{0.2mm}
\rule{0cm}...
...{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(0)} = [1,1,\ldots,1]^T$.}\end{figure}




For the sub-critical reactor we obtain the results plotted in Figures 4.10 and 4.11. As we can see there is no difference in the behaviour for a sub-critical and super-critical reactor assembly. We note, as previously stated, that the error in the eigenvector is larger than the error on the eigenvalue. As in the super-critical case the exact eigenvalue $\lambda_0$ is approached from above -- an important feature in regard to the feasibility of the power method eigenvalue estimate serving as input to the inverse power method.

\begin{figure}
% latex2html id marker 16948\rule{\textwidth}{0.2mm}
\rule{0cm}...
...{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(0)} = [1,1,\ldots,1]^T$.}\end{figure}

\begin{figure}
% latex2html id marker 16997\rule{\textwidth}{0.2mm}
\rule{0cm}...
...{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(0)} = [1,1,\ldots,1]^T$.}\end{figure}

Finally, a test was made to investigate if a start guess equal to a sinus works better, ie we investigate the iteration scenario when both the group 1 (fast) and group 2 (thermal) part of the eigenvector start out as the first half period of a sine. In mathematical terms (cf. (1.91) and (1.77)) we may state this as

\begin{displaymath}
\hspace{0.2ex}\underline{\phi}{}\hspace{0.15ex}^{(0)} = \le...
...-1} / z_{\mbox{\protect\scriptsize top}})
\end{array} \right]
\end{displaymath} (4.9)

From a qualitative point of view a sinus curve has a obviously resemblance to the sought eigensolution. As we observe by looking at the results plotted in Figures 4.12 and 4.13 it however turns out that the sinus start guess in fact results in larger errors and the sequence of eigenvalue errors iterates shows a highly unwanted oscillating course. Therefore, in the preceding we discard the sinus start guess altogether and we will use the unity start guess exclusively.

\begin{figure}
% latex2html id marker 17047\rule{\textwidth}{0.2mm}
\rule{0cm}...
... equal to
the sinus curve given by (\protect\ref{Sin_start_guess}).}\end{figure}

\begin{figure}
% latex2html id marker 17093\rule{\textwidth}{0.2mm}
\rule{0cm}...
...$k=2$\ through 7 are negative and that these value are not plotted.}\end{figure}


next up previous contents index
Next: Tests involving the inverse Up: Preliminary tests power method Previous: Preliminary tests power method   Contents   Index  
 
 
 
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