\newcommand{\thistitle}{Lecture 4: Networks Introduction}
\input{title.tex}
\section{Single Location Versus Country Versus Continent}
\begin{frame}
\frametitle{Variability: Single wind site in Berlin}
Looking at the wind output of a single wind plant over two weeks, it is highly
variable, frequently dropping close to zero and fluctuating strongly.
\centering
\includegraphics[width=12cm]{variability-berlin}
\end{frame}
\begin{frame}
\frametitle{Variability: Single country: Germany}
For a whole country like Germany this results in valleys and peaks that are somewhat smoother, but the profile still frequently
drops close to zero.
\centering
\includegraphics[width=12cm]{variability-de}
\end{frame}
\begin{frame}
\frametitle{Variability: A continent: Europe}
If we can integrate the feed-in of wind turbines across the European continent, the
feed-in is considerably smoother: we've eliminated most valleys and
peaks.
\centering
\includegraphics[width=12cm]{variability-eu}
\end{frame}
\begin{frame}
\frametitle{Duration curve: Berlin}
A \alert{duration curve} shows the feed-in for the whole year, re-ordered by from highest to lowest value. For a single location there are many hours with no feed-in.
\centering
\includegraphics[width=12cm]{duration-berlin}
\end{frame}
\begin{frame}
\frametitle{Duration curve: Germany}
For a whole country there are fewer peaks and fewer hours with no feed-in.
\centering
\includegraphics[width=12cm]{duration-de}
\end{frame}
\begin{frame}
\frametitle{Duration curve: Europe}
For the whole of Europe there are no times with zero feed-in.
\centering
\includegraphics[width=12cm]{duration-eu}
\end{frame}
\begin{frame}
\frametitle{Statistical comparison}
\ra{1.1}
\begin{table}[!t]
\begin{tabular}{lrr}
\toprule
Area & Mean & Standard deviation\\
\midrule
Berlin & 0.21 & 0.26 \\
Germany & 0.26 & 0.24 \\
Europe (including offshore) & 0.36 & 0.19 \\
\bottomrule
\end{tabular}
\end{table}
\vspace{1cm}
\alert{Conclusion}: Wind generation has much lower variability if you integrate it over a continent-sized area.
\end{frame}
\begin{frame}
\frametitle{Fourier spectrum: weekly variations suppressed}
% from ~/energy/playground/pypsa-eur/notebooks/fourier_of_synoptic.ipynb
% (tom3) tom@darth:~/energy/playground/pypsa-eur/notebooks$ cp german_wind.pdf ../../../courses/ei2-2019/graphics-ei2/fourier_german_wind.pdf
% (tom3) tom@darth:~/energy/playground/pypsa-eur/notebooks$ cp europe_wind.pdf ../../../courses/ei2-2019/graphics-ei2/fourier_europe_wind.pdf
The \alert{synoptic} (2-3 weeks) variations in the Fourier spectrum are
also suppressed between Germany (left) and the Europe
profile (right), however the seasonal variations remain.
\vspace{.7cm}
\includegraphics[width=6.5cm]{fourier_german_wind.pdf}
\includegraphics[width=6.5cm]{fourier_europe_wind.pdf}
\end{frame}
\begin{frame}
\frametitle{Why does this work? Consider the correlation length of wind}
\begin{columns}[T]
\begin{column}{5.5cm}
\includegraphics[height=7.5cm]{simeon-correlation}
\end{column}
\begin{column}{8.5cm}
The Pearson correlation coefficient of wind time series with a point in
northern Germany decays exponentially with distance.
Determine the \alert{correlation length} $L$ by fitting the
function:
\begin{equation*}
\rho \sim e^{-\frac{x}{L}}
\end{equation*}
to the radial decay with distance $x$.
\vspace{.2cm}
\centering
\includegraphics[height=4.5cm]{correlation_length.png}
\end{column}
\end{columns}
\source{\hrefc{https://doi.org/10.1016/j.apenergy.2011.10.039}{Hagspiel et al, 2012}}
\end{frame}
\begin{frame}
\frametitle{Reminder: Pearson correlation coefficient}
The \alert{Pearson correlation coefficient} measures the linear correlation between two random variables $X,Y$. It is defined as the \alert{covariance} of the two random variables normalised by their standard deviations, so that $\rho_{X,Y} \in [-1,1]$.
\begin{equation*}
\rho_{X,Y} = \frac{\textrm{cov}(X,Y)}{\sigma_X \cdot \sigma_Y} = \frac{\mathbb{E}\left[(X - \mu_X)(Y - \mu_Y)\right]}{\sigma_X \cdot \sigma_Y}
\end{equation*}
If the two random variables are perfectly proportional $Y = kX$ then we get
\begin{equation*}
\rho_{X,Y} = \frac{\mathbb{E}\left[(X - \mu_X)k(X - \mu_X)\right]}{\sigma_X \cdot | k| \sigma_X} = \frac{k\textrm{Var}(X)}{|k|\sigma_X^2} = \frac{k}{|k|} = \textrm{sign}(k)
\end{equation*}
so that if $k > 0$ they are perfectly correlated with $\rho = 1$ and if $k < 0$ they are perfectly anti-correlated with $\rho = -1$.
\end{frame}
\begin{frame}
\frametitle{Wind speed correlation lengths}
\begin{columns}[T]
\begin{column}{6cm}
% left bottom right top
\includegraphics[trim=0 0cm 25cm 0cm,width=6cm,clip=true]{cc-correlation.png}
\end{column}
\begin{column}{6cm}
\begin{itemize}
\item Typically correlation lengths for wind are around $400-600$~km.
\item Smoothing requires aggregating uncorrelated sources, so need a bigger area, i.e. a continent (Europe is about 3500~km tall and 3100~km wide).
\item Correlation lengths are longer in the North than
the South because of big weather systems that roll in from
the Atlantic to the North (in the South they get dissipated).
\end{itemize}
\end{column}
\end{columns}
\source{\hrefc{https://arxiv.org/abs/1805.11673}{Schlott et al, 2018}}
\end{frame}
\begin{frame}
\frametitle{What sets these lengths?}
\begin{columns}[T]
\begin{column}{6cm}
% left bottom right top
\includegraphics[trim=0 0cm 0cm 0cm,width=6cm,clip=true]{Evolution-of-the-southern-edge-of-the-Rossby-waves-as-the-cold-air-mass-in-blue-moves_W640.jpg}
\end{column}
\begin{column}{8cm}
\begin{itemize}
\item Reminder: atmospheric \alert{Rossby waves} are giant meanders in high-altitude winds that have a major influence on weather.
\item The wavelength of the Rossby waves sets the correlation length.
\end{itemize}
\vspace{0.5cm}
\includegraphics[trim=0 0cm 0cm 0cm,height=3.5cm,clip=true]{jupiter.jpg}
\includegraphics[trim=0 0cm 0cm 0cm,height=3.5cm,clip=true]{saturn.jpg}
\end{column}
\end{columns}
\source{\hrefc{http://dx.doi.org/10.3934/environsci.2019.1.14}{Stuart et al, 2019}}
\end{frame}
\begin{frame}
\frametitle{Mismatch between load and renewables}
How does the mismatch change as we integrate over larger areas?
If we have for each time $t$ a demand of $d_t$ and a `per unit'
availability $w_t$ for wind and $s_t$ for solar, then if we have $W$
MW of wind and $S$ MW of solar, the effective \alert{residual load}
or \alert{mismatch} is
\begin{equation*}
m_t = d_t - Ww_t - Ss_t
\end{equation*}
We choose $W$ and $S$ such that on \alert{average} we cover all the load
\begin{equation*}
\langle m_t \rangle = 0
\end{equation*}
and so that the 70\% of the energy comes from wind and 30\% from solar ($W = 147$ GW and $S = 135$ GW for Germany).
This means
\begin{equation*}
W\langle w_t \rangle = 0.7\langle d_t \rangle \hspace{2cm} S\langle s_t \rangle = 0.3\langle d_t \rangle
\end{equation*}
\end{frame}
\begin{frame}
\frametitle{Mismatch between load and renewables}
Let $p_t$ be the balance of power at each time. Because we cannot
create or destroy energy, we need $p_t = 0$ at all times.
If the mismatch is positive $m_t > 0$, then we need \alert{backup power} $b_t = m_t$
to cover the load in the absence of renewables, so that
\begin{equation*}
p_t = b_t - m_t = b_t - d_t + Ww_t + Ss_t = 0
\end{equation*}
If the mismatch is negative $m_t < 0$ then we need \alert{curtailment} $c_t = -m_t$
to reduce the excess feed-in from renewables, so that
\begin{equation*}
p_t = - m_t - c_t = -c_t - d_t + Ww_t + Ss_t = 0
\end{equation*}
At any one time we have either backup or curtailment
\begin{equation*}
p_t = b_t - m_t - c_t = Ww_t + Ss_t + b_t - d_t - c_t = 0
\end{equation*}
\end{frame}
\begin{frame}
\frametitle{Mismatch for Germany}
Backup generation needed for 31\% of the total load.
Peak mismatch is 91\% of peak load (around 80~GW).
\centering
\includegraphics[width=12cm]{mismatch-duration-DE}
\end{frame}
\begin{frame}
\frametitle{Mismatch for Europe}
Requires 750~GW each of onshore wind and solar.
Backup generation needed for only 24\% of the total load.
Peak mismatch is 79\% of peak load (around 500~GW).
\centering
\includegraphics[width=12cm]{mismatch-duration-EU}
\end{frame}
\begin{frame}
\frametitle{Conclusions}
\begin{itemize}
\item Integration over a larger area smooths out the fluctuations of
renewables, particularly wind.
\item Wind backs up wind.
\item This means we need \alert{less backup energy}.
\item And \alert{less backup capacity}.
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Greiner papers}
\hrefc{http://www.sciencedirect.com/science/article/pii/S0360544215002212}
{`Cost-optimal design of a simplified, highly renewable pan-European
electricity system'} by
Rolando A. Rodriguez, Sarah Becker, Martin Greiner,
Energy 83 (2015) 658-668
{\footnotesize `average backup power' indicates backup energy; zero has no transmission between countries versus cooperative}
\centering
\includegraphics[width=6cm]{backup_energy.png}
\includegraphics[width=6cm]{backup_capacity.png}
\end{frame}
\begin{frame}
\frametitle{Flexibility Requirements}
\hrefc{http://www.sciencedirect.com/science/article/pii/S0360544214002680}{`Integration of wind and solar power in Europe: Assessment of flexibility requirements'} by Huber, Dimkova, Hamacher,
Energy 69 (2014) 236e246
1-hour net load ramp duration curves at the regional, country and European
spatial scales at 50\% share of renewables and 20\% PV in the wind/PV mix for the
meteorological year 2009.
\centering
\includegraphics[width=7cm]{huber-hamacher}
\end{frame}
\begin{frame}
\frametitle{Big Caveat}
There is a big caveat to this analysis.
We've assumed that we can move power around Europe without penalty.
However, in reality, we can only transport within restrictions of the power network.
In general we will have different power imbalances $p_{i,t}$ at each
location/node $i$ and instead of $p_t = 0$ we will have
\begin{equation*}
\sum_i p_{i,t} = 0
\end{equation*}
(neglecting power losses in the network).
Moving excess power to locations of consumption is the role of the network.
\end{frame}
\section{Electricity Networks}
\begin{frame}
\frametitle{Electricity Transport from Generators to Consumers}
%https://upload.wikimedia.org/wikipedia/commons/thumb/4/41/Electricity_grid_simple-_North_America.svg/640px-Electricity_grid_simple-_North_America.svg.png
Electricity can be transported over long distances with low losses using the
high voltage transmission grid (losses go like $I^2R$, power transmission like $VI$, so reduce $I$ by raising $V$):
\centering
\includegraphics[width=14cm]{Electricity_grid_simple-_North_America}
\raggedright
Usually in houses the voltage is $230~V$, but in the transmission
grid it is transformed up to hundreds of thousands of Volts.
\source{Wikipedia}
\end{frame}
\begin{frame}[fragile]
\frametitle{European transmission network}
Flows in the European transmission network must respect both
Kirchoff's laws for physical flow and the thermal and/or other limits of the power lines.
Taking account of network flows and constraints in the electricity market is a major and exciting topic at the moment.
\centering
\includegraphics[width=8cm]{europe-transmission}
\source{ENTSO-E}
\end{frame}
\begin{frame}
\frametitle{Network Bottlenecks and Loop Flows}
Electricity is traded in large market zones. Power trades between
zones (``scheduled flows'') do not always correspond to what flows
according to the network physics (``physical flows''). This
leads to political tension as wind from Northern Germany flows to Southern Germany via Poland and the Czech Republic.
\centering
\includegraphics[width=9cm]{germany_loop_flows}
\source{THEMA Consulting Group}
%https://ec.europa.eu/energy/sites/ener/files/documents/201310_loop-flows_study.pdf
\end{frame}
\begin{frame}
\frametitle{Solar resource distribution in Germany}
\begin{columns}[T]
\begin{column}{5.5cm}
\includegraphics[trim=0 0cm 0 0cm,width=5.5cm,clip=true]{SolarGIS-Solar-map-Germany-de}
\end{column}
\begin{column}{4.5cm}
\begin{itemize}
\item Solar insolation at top of atmosphere is on average 1361~W/m${}^2$
(orbit is elliptical).
\item In Germany average insolation on a horizonal surface is
around 1200~kWh/m${}^2$.
\item A 1~kW solar panel (around 7~m${}^2$) will generate around 1000~kWh/a.
% 1000~kWh/a/kW_p; 0.150 kW_p/m^2 means 150~kWh/a/m^2, efficiency of 0.125
\end{itemize}
\end{column}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Wind resource distribution in Germany}
\begin{columns}[T]
\begin{column}{8.5cm}
\includegraphics[trim=0 0cm 0 0cm,width=8.5cm,clip=true]{43_Mittlere_Windgeschwindigkeit_100_m_Deutschland}
\end{column}
\begin{column}{3.5cm}
\begin{itemize}
\item Best wind speeds in Germany in North and on hills.
\item In theory power output goes like cube $\propto v^3$ of wind
speed $v$.
\item In practice power-speed relationship is only partially
cubic.
\end{itemize}
\end{column}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{The Problem}
Renewables are not always located near demand centres, as in this example from Germany.
\begin{columns}[T]
\begin{column}{6cm}
\includegraphics[width=6cm]{scigrid-load}
\end{column}
\begin{column}{6cm}
% left bottom right top
\includegraphics[trim=0 0cm 0 1cm,width=6.3cm,clip=true]{scigrid-wind}
\end{column}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{The Problem}
\begin{columns}[T]
\begin{column}{5cm}
\includegraphics[width=6cm]{scigrid-loading}
\end{column}
\begin{column}{6cm}
\begin{itemize}
\item This leads to \alert{overloaded lines} in the middle of Germany, which
cannot transport all the wind energy from North Germany to the load
in South Germany
\item It also overloads lines in neighbouring countries due to
\alert{loop flows} (unplanned physical flows `according to least
resistance' which do not correspond to traded flows)
\item It also \alert{blocks imports and exports} with
neighbouring countries, e.g. Denmark
\end{itemize}
\end{column}
\end{columns}
\end{frame}
\begin{frame}{Different types of networks: radial networks}
In a \alert{radial} or \alert{tree-like} network there is only one path between any two nodes on the network.
The power flow is thus completely determined by the nodal power imbalances.
\centering
\includegraphics[width=7cm]{radial}
\source{Biggar \& Hesamzadeh}
\end{frame}
\begin{frame}{Different types of networks: meshed networks}
In a \alert{meshed} network there are at least two nodes with multiple paths between them.
The power flow is now not completely determined. We need new information: the impedances in the network.
\centering
\includegraphics[width=7cm]{meshed}
\source{Biggar \& Hesamzadeh}
\end{frame}
\end{document}