Commit 5e6102b5 authored by sp2668's avatar sp2668

clean tex files

parent 591cf4f9
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...@@ -71,11 +71,11 @@ ...@@ -71,11 +71,11 @@
\begin{document} \begin{document}
\begin{center} \begin{center}
\textbf{\Large Energy System Modelling }\\ \textbf{\Large Energy System Modelling }\\
{SS 2018, Karlsruhe Institute of Technology}\\ {SS 2018, Karlsruhe Institute of Technology}\\
{Institute of Automation and Applied Informatics}\\ [1em] {Institute of Automation and Applied Informatics}\\ [1em]
\textbf{\textsc{\Large Tutorial II: Network Theory and Power Flow}}\\ \textbf{\textsc{\Large Tutorial II: Network Theory and Power Flow}}\\
\small Will be worked on in the exercise session on Friday, 13 July 2018.\\[1.5em] \small Will be worked on in the exercise session on Friday, 13 July 2018.\\[1.5em]
\end{center} \end{center}
...@@ -86,34 +86,34 @@ ...@@ -86,34 +86,34 @@
%===================================================================== %=====================================================================
\begin{wrapfigure}[10]{r}{0pt} \begin{wrapfigure}[10]{r}{0pt}
\centering \centering
\begin{tikzpicture} \begin{tikzpicture}
[scale=.81,auto=left,every node/.style={circle,fill=gray!20}] [scale=.81,auto=left,every node/.style={circle,fill=gray!20}]
\node (n0) at (2,14) {0}; \node (n0) at (2,14) {0};
\node (n1) at (2,8) {1}; \node (n1) at (2,8) {1};
\node (n2) at (0,4) {2}; \node (n2) at (0,4) {2};
\node (n3) at (3,0) {3}; \node (n3) at (3,0) {3};
\node (n4) at (4,4) {4}; \node (n4) at (4,4) {4};
\node (n5) at (5,7) {5}; \node (n5) at (5,7) {5};
\foreach \from/\to in {n0/n1,n1/n2,n1/n4,n1/n5,n2/n4,n2/n3,n3/n4,n4/n5} \foreach \from/\to in {n0/n1,n1/n2,n1/n4,n1/n5,n2/n4,n2/n3,n3/n4,n4/n5}
\draw (\from) -- (\to); \draw (\from) -- (\to);
\end{tikzpicture} \end{tikzpicture}
\caption{Simple Network} \caption{Simple Network}
\label{fig:network} \label{fig:network}
\end{wrapfigure} \end{wrapfigure}
Consider the simple network shown ins Figure \ref{fig:network}. Calculate in Python or by hand: Consider the simple network shown ins Figure \ref{fig:network}. Calculate in Python or by hand:
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item Compile the \textit{nodes list} and the \textit{edge list}.\\~\\ \item Compile the \textit{nodes list} and the \textit{edge list}.\\~\\
\textbf{Remark:} While graph-theoretically both lists are unordered sets, let's agree on an ordering now which can serve as basis for the matrices in the following exercises: we sort everything in ascending numerical order, i.e.\ node 1 before node 2 and edge (1,2) before (1,4) before (2,3). \textbf{Remark:} While graph-theoretically both lists are unordered sets, let's agree on an ordering now which can serve as basis for the matrices in the following exercises: we sort everything in ascending numerical order, i.e.\ node 1 before node 2 and edge (1,2) before (1,4) before (2,3).
\item Determine the \textit{order} and the \textit{size} of the network. \item Determine the \textit{order} and the \textit{size} of the network.
\item Compute the\textit{ adjacency matrix} $A$ and check that it is symmetric. \item Compute the\textit{ adjacency matrix} $A$ and check that it is symmetric.
\item Find the \textit{degree} $k_n$ of each node $n$ and compute the \textit{average degree} of the network. \item Find the \textit{degree} $k_n$ of each node $n$ and compute the \textit{average degree} of the network.
\item Determine the \textit{incidence matrix} $K$ by assuming the links are always directed from smaller-numbered node to larger-numbered node, i.e.\ from node 2 to node 3, instead of from 3 to 2. \item Determine the \textit{incidence matrix} $K$ by assuming the links are always directed from smaller-numbered node to larger-numbered node, i.e.\ from node 2 to node 3, instead of from 3 to 2.
\item Compute the \textit{Laplacian} $L$ of the network using $k_n$ and $A$. Remember that the Laplacian can also be computed as $L=KK^T$ and check that the two definitions agree. \item Compute the \textit{Laplacian} $L$ of the network using $k_n$ and $A$. Remember that the Laplacian can also be computed as $L=KK^T$ and check that the two definitions agree.
\item Find the \textit{diameter} of the network by looking at Figure \ref{fig:network}. \item Find the \textit{diameter} of the network by looking at Figure \ref{fig:network}.
\end{enumerate} \end{enumerate}
...@@ -128,32 +128,32 @@ If you map the nodes to countries like \texttt{0=DK, 1=DE, 2=CH, 3=IT, 4=AT,5=CZ ...@@ -128,32 +128,32 @@ If you map the nodes to countries like \texttt{0=DK, 1=DE, 2=CH, 3=IT, 4=AT,5=CZ
The linear power flow is given by The linear power flow is given by
\begin{equation} \begin{equation}
p_i = \sum_j \tilde{L}_{i,j}\theta_j \qquad \text{and} \qquad f_l = \frac{1}{x_l} \sum_i K_{i,l}\theta_i, \qquad \text{where} \qquad \tilde{L}_{i,j}= \sum_l = K_{i,l}\frac{1}{x_l} K_{j,l} p_i = \sum_j \tilde{L}_{i,j}\theta_j \qquad \text{and} \qquad f_l = \frac{1}{x_l} \sum_i K_{i,l}\theta_i, \qquad \text{where} \qquad \tilde{L}_{i,j}= \sum_l = K_{i,l}\frac{1}{x_l} K_{j,l}
\end{equation} \end{equation}
is the weighted Laplacian. For simplicity, we assume identity reactance on all links $x_l = 1$. is the weighted Laplacian. For simplicity, we assume identity reactance on all links $x_l = 1$.
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item Compute the \textit{voltage angles }$\theta_j$ and \textit{flows} $f_l$ for the first hour in the dataset with the convention of $\theta_0 = 0$; i.e.\ the slack bus is at node 0.\\~\\ \item Compute the \textit{voltage angles }$\theta_j$ and \textit{flows} $f_l$ for the first hour in the dataset with the convention of $\theta_0 = 0$; i.e.\ the slack bus is at node 0.\\~\\
\textbf{Remark:} Linear equation systems are solved efficiently using \texttt{numpy.linalg.solve}. \textbf{Remark:} Linear equation systems are solved efficiently using \texttt{numpy.linalg.solve}.
\item Determine the average flow on each link for 01-2017 and draw it as a directed network. \item Determine the average flow on each link for 01-2017 and draw it as a directed network.
\end{enumerate} \end{enumerate}
\begin{figure}[h] \begin{figure}[h]
\centering \centering
\begin{tikzpicture} \begin{tikzpicture}
[scale=.7,auto=left,every node/.style={circle,fill=gray!20}] [scale=.7,auto=left,every node/.style={circle,fill=gray!20}]
\node (n0) at (2,14) {DK}; \node (n0) at (2,14) {DK};
\node (n1) at (2,8) {DE}; \node (n1) at (2,8) {DE};
\node (n2) at (0,4) {CH}; \node (n2) at (0,4) {CH};
\node (n3) at (3,0) {IT}; \node (n3) at (3,0) {IT};
\node (n4) at (4,4) {AT}; \node (n4) at (4,4) {AT};
\node (n5) at (5,7) {CZ}; \node (n5) at (5,7) {CZ};
\foreach \from/\to in {n0/n1,n1/n2,n1/n4,n1/n5,n2/n4,n2/n3,n3/n4,n4/n5} \foreach \from/\to in {n0/n1,n1/n2,n1/n4,n1/n5,n2/n4,n2/n3,n3/n4,n4/n5}
\draw (\from) -- (\to); \draw (\from) -- (\to);
\end{tikzpicture} \end{tikzpicture}
\caption{Simple Network} \caption{Simple Network}
\label{fig:network} \label{fig:network}
\end{figure} \end{figure}
%\bibliographystyle{unsrt} %\bibliographystyle{unsrt}
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