
\newcommand{\thistitle}{Lecture 2: Measuring \& Converting Energy}

\input{title.tex}


\section{Measuring energy}


\begin{frame}
  \frametitle{Units of Energy: Joule and tonne of oil equivalent}

  \alert{Joule} (J) is the SI unit of energy.

  Conventional primary energy sources are often measured in units corresponding to their natural form: volume, mass etc.

  We can convert from measurements of mass [kg] and volume [m$^3$] to energy units using the \alert{calorific value} [J/kg, J/m$^3$], which measures the heat from combustion.

  Example: the unit \alert{tonne of oil equivalent} (toe) is the energy generated by burning one metric ton of oil. Since the calorific value of oil is 41.88 MJ/kg, we have
  \begin{equation*}
    1 \textrm{ toe} = 41.88 \textrm{ GJ}
  \end{equation*}

  \vspace{.4cm}
  [Reminder: k = kilo = 1e3, M = Mega = 1e6, G = Giga = 1e9, T = Tera = 1e12, P = Peta = 1e15, E = Exa = 1e18.]
\end{frame}
\begin{frame}
  \frametitle{Lower Heating Values of Energy Fuels}

  \centering
  \includegraphics[width=12cm]{fuel_lhv.png}

\end{frame}

\begin{frame}
  \frametitle{Higher and Lower Heating Values}


  \begin{itemize}
  \item \alert{Lower Heating Value (LHV)} is the maximum amount of usable heat from combustion without counting the condensation enthalpy of water vapor contained in the exhaust gas.
  \item \alert{Higher Heating Value (HHV)} includes the condensation enthalpy of water vapor contained in the exhaust gas. It is always higher than the LHV (e.g. 11\% higher for methane).
  \end{itemize}

  \vspace{.35cm}

  \centering
  \includegraphics[width=8cm]{Higher-and-lower-heating-values-for-various-fuels-22.png}

  \vspace{.35cm}


  \raggedright
 LHV is most commonly used in European statistics.
  HHV becomes relevant in e.g. condensing combined heat and power plants (CHP) where vapor is condensed.

\end{frame}


\begin{frame}
  \frametitle{Power: Flow of energy}

  \alert{Power} is the rate of consumption of energy.

  It is measured in \alert{Watts}:
  \begin{equation*}
     1 \textrm{ Watt } = 1 \textrm{ Joule per second }
  \end{equation*}
  The symbol for Watt is W, 1 W = 1 J/s.

  \centering
  1 kilo-Watt = 1 kW = 1,000 W

  1 mega-Watt = 1 MW = 1,000,000 W

  1 giga-Watt = 1 GW = 1,000,000,000 W

  1 tera-Watt = 1 TW = 1,000,000,000,000 W


\end{frame}



\begin{frame}
  \frametitle{Power: Examples of consumption}

  At full power, the following items consume:

  \ra{1.1}
  \begin{table}[!t]
    \begin{tabular}{lr}
      \toprule
      Item & Power\\
      \midrule
      New efficient lightbulb & 10 W \\
      Old-fashioned lightbulb & 70 W \\
      Single room air-conditioning & 1.5 kW \\
      Kettle & 2 kW \\
      Factory & $\sim$1-500 MW \\
      CERN & 200 MW \\
            Germany total demand & 35-80 GW \\
      \bottomrule
    \end{tabular}
  \end{table}

\end{frame}

\begin{frame}
  \frametitle{Power: Supplying world's energy with wind and solar}

  If all energy is electrified in 2050 and energy consumption
  equalises between nations, the average electricity consumption of
  the world would be around 12~TW.

  Suppose half is met with wind (capacity factor 33.3\%) and half is
  met with solar PV (capacity factor 16.6\%). [Capacity factor $=$ average
  generation $/$ capacity.] How much wind and solar capacity does the world need (assuming perfect lossless storage)?

  \pause

  \vspace{1cm}

  Wind: 6~TW / 0.333 $=$ 18~TW (around 743~GW was installed by 2020)

  Solar: 6~TW / 0.166 $=$ 36~TW (around 626~GW was installed by 2019)

\end{frame}

\begin{frame}
  \frametitle{Power: Supplying world's energy with wind and solar}

  If installed wind density on average is 10~MW/km$^2$ and solar is 72~MW/km$^2$, what percentage of world land (510 million km$^2$) is taken with each?
  \pause

  \vspace{.5cm}

  Wind: 18~TW/(10~MW/km$^2$) = 1.8 million km$^2$ (around 0.35\% of total land $=$ area of Indonesia)

  Solar: 36~TW/(72~MW/km$^2$) = 0.5 million km$^2$ (around 0.1\% of total land $=$ area of Spain)

  \vspace{.5cm}

  Nota Bene:
  \begin{itemize}
    \item Wind doesn't interfere with other land uses like agriculture
    \item 10~MW/km$^2$ is a \alert{local} maximum installation density for wind, but to allow wind replenishment over large areas 2~MW/km$^2$ is suitable as a \alert{wide-area} limit
      \item Solar can be rooftop or combined with agriculture = agrivoltaics
  \end{itemize}

\end{frame}


\begin{frame}
  \frametitle{Units of energy: Watt-hour}

  In the electricity sector, energy is usually measured in
  `Watt-hours', Wh.

  1 kWh = power consumption of 1 kW for one hour

  E.g. a 10 W lightbulb left on for two hours will consume

  10 W * 2 h = 20 Wh

  It is easy to convert this back to the SI unit for energy, Joules:

  1 kWh = (1000 W) * (1 h) = (1000 J/s)*(3600 s) = 3.6 MJ
\end{frame}





\begin{frame}
  \frametitle{Yearly energy to power}

  Germany consumes around 600 TWh per year, written 600 TWh/a.

  What is the \emph{average} power consumption in GW?

  \pause
  \begin{align*}
    600\textrm{ TWh/a} & = \frac{(600\textrm{ TW}) * (1\textrm{ h})}{(365*24 \textrm{ h})} \\
    & =  \frac{600}{8760} \textrm{ TW} \\
    & = 68.5 \textrm{ GW}
  \end{align*}

\end{frame}

\begin{frame}
  \frametitle{Tables for converting units}

  \centering
  \includegraphics[width=8cm]{conversion.jpg}

  \raggedright Units used in the United States:
  \begin{itemize}
  \item British thermal unit (Btu), 1 million Btu = MBtu (often written MMBtu) = 0.293 MWh
  \item Quad = 1e15 Btu
  \end{itemize}
\end{frame}


\section{Energy conversion}



\begin{frame}
  \frametitle{Energy conversion/transformation processes}
  \centering
  \includegraphics[width=14cm]{conversion_processes.png}

  \source{Zweifel, Praktiknjo \& Erdmann (2017)}

\end{frame}


\begin{frame}
  \frametitle{Energy conversion efficiency}

  \alert{Efficiency} of an energy conversion device (e.g. power plant, vehicle engine):
  \begin{equation*}
    \textrm{Efficiency, } \eta = \frac{\textrm{Useful energy output}}{\textrm{Energy input}}
  \end{equation*}


  Example: How much much natural gas is required for generating 100 MWh of
  electricity in a gas power plant with an efficiency of 50\%?

\end{frame}



\begin{frame}
  \frametitle{Efficiency}

  When fuel is consumed, much/most of the energy of the fuel is lost
  as waste heat rather than being converted to electricity.

  The thermal energy, or calorific value, of the fuel is given in
  terms of MWh${}_{\textrm{th}}$, to distinguish it from the
  electrical energy MWh${}_{\textrm{el}}$.

  The ratio of input thermal energy to output electrical energy is the
  \alert{efficiency}.

  \ra{1.1}
  \begin{table}[!t]
    \begin{tabular}{lrrrrr}
      \toprule
      Fuel & Calorific energy & Per unit efficiency & Electrical energy \\
       & MWh\(_{\text{th}}\)/tonne & MWh\(_{\text{el}}\)/MWh\(_{\text{th}}\) & MWh\(_{\text{el}}\)/tonne \\
\midrule
Lignite & 2.5 & 0.4 & 1.0 \\
Hard Coal & 6.7 & 0.45 & 2.7\\
Gas (CCGT) & 15.4 & 0.58 & 8.9\\
Uranium (unenriched) & 150000 & 0.33 & 50000 \\
      \bottomrule
    \end{tabular}
  \end{table}




\end{frame}


\begin{frame}
  \frametitle{Fuel costs to marginal costs}


  The cost of a fuel is often given in \euro/kg or \euro/MWh\(_{\text{th}}\).

  Using the efficiency, we can convert this to
  \euro/MWh\(_{\text{el}}\).

  For the full marginal cost, we have to
  also add the CO$_2$ price and the variable operation and maintenance
  (VOM) costs.


  \ra{1.1}
  \begin{table}[!t]
    \begin{tabular}{lrrrrr}
      \toprule
      Fuel &  Per unit efficiency & Cost per thermal & Cost per elec. \\
      &  MWh\(_{\text{el}}\)/MWh\(_{\text{th}}\)  & \euro/MWh\(_{\text{th}}\) & \euro/MWh\(_{\text{el}}\)\\
\midrule
Lignite  & 0.4  & 4.5 & 11\\
Hard Coal  & 0.45   & 11 & 24\\
Gas (CCGT) & 0.58  & 19 & 33\\
Uranium & 0.33 & 3.3 & 10\\
      \bottomrule
    \end{tabular}
  \end{table}



\end{frame}




\begin{frame}
  \frametitle{CO2 emissions per MWh}

  The \co2 emissions of the fuel.

  \ra{1.1}
  \begin{table}[!t]
    \begin{tabular}{lrrr}
      \toprule
Fuel & t\(_{\text{CO2}}\)/t & t\(_{\text{C02}}\)/MWh\(_{\text{th}}\) & t\(_{\text{CO2}}\)/MWh\(_{\text{el}}\)\\
\midrule
Lignite &  0.9 & 0.36 & 0.9 \\
Hard Coal  & 2.4 & 0.36 & 0.8 \\
Gas (CCGT) & 3.1 & 0.2 & 0.35\\
Uranium & 0 & 0 & 0 \\
      \bottomrule
    \end{tabular}
  \end{table}

  Current \co2 price in EU Emissions Trading Scheme (ETS) is around \euro 50-100/t${}_{\textrm{CO2}}$

\end{frame}


\begin{frame}
  \frametitle{You calculate: What CO$_2$ price to switch gas and lignite?}

  What CO$_2$ price, i.e. x \euro/t${}_{\textrm{CO2}}$, is required so
  that the marginal cost of gas (CCGT) is lower than lignite?

  NB: It helps to track units.

  \pause

  We need to solve for the switch point by adding the CO$_2$ price to the fuel cost.  Left is lignite, right is gas:
  \begin{equation*}
    11 \textrm{ \euro/MWh}_{\textrm{el}} + (0.9 \textrm{ tCO}_2/\textrm{MWh}_{\textrm{el}}) \cdot (x \textrm{ \euro/tCO}_2)
    =    33 \textrm{ \euro/MWh}_{\textrm{el}} + (0.35 \textrm{ tCO}_2/\textrm{MWh}_{\textrm{el}}) \cdot (x \textrm{ \euro/tCO}_2)
  \end{equation*}

  Solve:
  \begin{equation*}
    x = \frac{33 - 11}{0.9 - 0.35} = 40
  \end{equation*}

\end{frame}


\begin{frame}
  \frametitle{CO2 and import costs change over time...}

  \centering
  \includegraphics[width=14cm]{import_costs-2019.png}

 \source{\hrefc{https://www.agora-energiewende.de/fileadmin2/Projekte/2019/Jahresauswertung_2019/AGORA_State_of_Affairs_German_Power_Sector_2019_Webinar_28012020.pdf}{Agora Energiewende, 2019}}
\end{frame}

\begin{frame}
  \frametitle{...which affects the marginal costs of generation}

    \centering
  \includegraphics[width=14cm]{marginal_costs-2019.png}

\source{\hrefc{https://www.agora-energiewende.de/fileadmin2/Projekte/2019/Jahresauswertung_2019/AGORA_State_of_Affairs_German_Power_Sector_2019_Webinar_28012020.pdf}{Agora Energiewende, 2019}}
\end{frame}


\begin{frame}
  \frametitle{CO2 emissions from electricity sector}

  \co2
  emissions in electricity generation stagnated for years because of coal, which is slowly being pushed out by the \co2 price and in the longer term by the Kohleausstieg.

  \centering
  \includegraphics[width=11cm]{co2-strom-2022.png}
  \source{\hrefc{https://static.agora-energiewende.de/fileadmin/Projekte/2022/2022-10_DE_JAW2022/A-EW_283_JAW2022_WEB.pdf}{Agora Energiewende Jahresauswertung 2022}}
\end{frame}



\begin{frame}
  \frametitle{Hydrogen in REPowerEU by 2030}

  The European Commission's REPowerEU plan, published in March 2022, aims for 10~Mt/a of clean hydrogen to be produced domestically in the European Union by 2030, with another 10~Mt imported.

  If electrolysis of water to hydrogen is 70\% efficient (LHV) and there is 33~MWh/tH$_2$ (LHV), what will be the electricity consumption from electrolysis for hydrogen in the EU in 2030?

  \pause

  Consumption will be
  \begin{equation*}
   \frac{ 10 \textrm{ MtH}_2/\textrm{a} * 33 \textrm { MWh}_{\textrm{H}_2}/\textrm{tH}_2}{ 0.7 \textrm{ MWh}_{\textrm{H}_2}/\textrm{MWh}_{\textrm{el}}} = 471 \textrm{ TWh}_{\textrm{el}}/\textrm{a}
  \end{equation*}

  Compare to the current electricity consumption in Europe of around 3200~TWh\el/a.
\end{frame}


\begin{frame}
  \frametitle{Capacity Factors and Full Load Hours}


  A generator's \alert{capacity factor} is the average power generation divided by the power capacity.

  For variable renewable generators it depends on weather, generator model and
  curtailment; for dispatchable generators it depends on market
  conditions and maintenance schedules.

  A generator's \alert{full load hours} are the equivalent number of hours at full capacity the generator required to produce its yearly energy yield.  The two quantities are related:
  \begin{equation*}
    \textrm{full load hours} = \textrm{per unit capacity factor} \cdot 365 \cdot 24 = \textrm{per unit capacity factor} \cdot 8760
  \end{equation*}


  Typical values for Germany:
  \begin{table}[!t]
    \begin{tabular}{lrr}
      \toprule
Fuel & capacity factor [\%] & full load hours \\
\midrule
wind & 20-35 & 1600-3000 \\
solar & 10-12 & 800-1000 \\
nuclear & 70-90 & 6000-8000 \\
open-cycle gas & 20 & 1500 \\
      \bottomrule
    \end{tabular}
  \end{table}


\end{frame}


\end{document}
