some typo correction

exam/summer2024/tex/task01.tex

@@ -5,9 +5,7 @@
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \taskGerman{Hubmagnet}
-% Lukas
-% EMA Uebungen PDF Binder 3a -> leicht veränderte Zahlenwerte 
-% Ergänzungsfrage: Bestimmung des Flusses sowie der Flussverkettung durch eine Spule 
+
 A lifting magnet made of steel with the dimensions shown in \autoref{fig:LiftingMagnet} is to carry an iron load with a mass of 700 kg. The average field line length in the load is $l_{\mathrm{load}} = \SI{250}{\milli\metre}$, the flux-carrying cross-section $A_{\mathrm{load}}$ is $\SI{0.0095}{\metre^2}$. The winding of both coils have ${N = 300}$ turns. The magnet has a circular cross-section with a diameter of $\SI{70}{\milli\metre}$. The roughness of the surfaces of the magnet and load results in an average air gap length of $\delta = \SI{0.5}{\milli\metre}$. Neglect all leakage fluxes.
 
 
@@ -100,7 +98,7 @@
     Since the leakage flux is neglected, the fluxes $\phi_{\updelta}$ in the air gap and $\phi_{\mathrm{load}}$ in the iron are equal. This results into
     $$\phi_{\updelta} = \phi_{\mathrm{load}} = A_{\updelta} B_{\updelta} = A_{\mathrm{load}} B_{\mathrm{load}} $$
 
-    thus, the magentic flux density in the load is determined as
+    thus, the magnetic flux density in the load is determined as
     $$ B_{\mathrm{load}} = \frac{A_{\updelta}}{A_{\mathrm{load}}} B_{\updelta} = \frac{\SI{0.0038}{\metre^2}}{\SI{0.0095}{\metre^2}} \cdot \SI{1.51}{\tesla} = \SI{0.60}{\tesla},$$
 
     and the corresponding magnetic field is $H_{\mathrm{load}} = \SI{2000}{\ampere\per\metre}$ derived from \autoref{fig:magCurve}. 

exam/summer2024/tex/task02.tex

@@ -5,9 +5,6 @@
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \taskGerman{Fremderregte Gleichstrommaschine}
-% Oliver
-% Fremderregte DC-Maschine in Anlehnung an Aufgabe 4.16 Aufgabensammlung EMA Binder (Teil2)
-% ggf. ergänzt um transiente Reaktion
 
 \begin{table}[htb]
     \caption{DC machine parameters.}

exam/summer2024/tex/task03.tex

@@ -5,7 +5,6 @@
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \taskGerman{Transformator}
-% Frage 5: Welche idealisierte sekundäre Ausgangsspannung ergibt sich, wenn der Transformator zum Spartransformator umgewickelt wird? Welcher primäre Kurzschlusstrom ergibt sich nun, falls sekundär kurzgeschlossen wird? 
 
 
 \subtask{Draw and label the T-type equivalent circuit diagram of a single phase transformer.}{1}

exam/summer2024/tex/task04.tex

@@ -2,16 +2,7 @@
 %%% Task 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \task{Synchronous generator}
-% Lukas 
-% Entsprechend Binder EMA Aufgabensammlung Teil 1, Aufgabe 1.8 (Windgenerator)
-% Fragen grundsätzlich identisch, ggf. mit einfachen weiteren Fragen vorab
-% Ergänzungsfrage 1: Bestimmung Nenndrehmoment, sofern angneommen wird das Pn = 1.5 MW die elektrische Abgabeleistung im Nennbetrieb ist und der Wirkungsgrad 0.95 beträgt (-> mech. Leistung im Generatorbetrieb ist dann 1.5/0.95 MW)
-% Ergänzungsfrage 2: Kann der Synchrongenerator der Windkraftanlage direkt am Netz betrieben werden? Nein -> weil die Drehzahl der WKA variert, während das Netz eine konstante Frequenz hat  
 
-
-
-
-% extern%%%%ally ecx%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \taskGerman{Synchrongenerator}
 
 An externally excited synchronous generator is directly fed from a wind turbine without utilizing a gear.

lecture/tex/Lecture01.tex

@@ -272,7 +272,7 @@ \section{An initial overview of electrical machines and drives}
 		\begin{subfigure}[b]{0.49\textwidth}
 			\centering
 			\includegraphics[width=0.5\textwidth]{fig/lec01/Electric_Airbus_A320.jpg}
-			\caption{Electric aircrafts (source: \href{https://commons.wikimedia.org/wiki/File:Electric_Airbus_A320.jpg}{Wikimedia Commons}, M.~Weinold, \href{https://creativecommons.org/licenses/by-sa/4.0/deed.en}{CC BY-SA 4.0})}
+			\caption{Electric aircraft (source: \href{https://commons.wikimedia.org/wiki/File:Electric_Airbus_A320.jpg}{Wikimedia Commons}, M.~Weinold, \href{https://creativecommons.org/licenses/by-sa/4.0/deed.en}{CC BY-SA 4.0})}
 		\end{subfigure}
 		\\
 		\begin{subfigure}[b]{0.49\textwidth}
@@ -311,7 +311,7 @@ \section{An initial overview of electrical machines and drives}
 \begin{frame}
 	\frametitle{Why is knowledge about electric machines and drives important?}
 	\begin{varblock}{Electric machines and drives are an essential pillar of the modern society}
-		Without electric machines and drives, our todays' society would not be possible. Starting from providing electricity via electrical generators to powering electric vehicles, tools and entire factory production lines, electric machines and drives are everywhere, that is, they enable our today's living standard. 
+		Without electric machines and drives, our todays society would not be possible. Starting from providing electricity via electrical generators to powering electric vehicles, tools and entire factory production lines, electric machines and drives are everywhere, that is, they enable our today's living standard. 
 	\end{varblock}
 	\begin{varblock}{Energy efficiency and sustainability is key}<2->
 		Electric machines and drives utilize  approx. 50\,\% of the global electricity with about 8 billion electric motors in use in the EU (source: \href{https://commission.europa.eu/energy-climate-change-environment/standards-tools-and-labels/products-labelling-rules-and-requirements/energy-label-and-ecodesign/energy-efficient-products/electric-motors-and-variable-speed-drives_en}{European Commission} and \href{https://iea.blob.core.windows.net/assets/d69b2a76-feb9-4a74-a921-2490a8fefcdf/EE_for_ElectricSystems.pdf}{International Energy Agency}). Therefore, improving their efficiency is an essential factor to reduce the global energy consumption and the associated CO$_2$ emissions.

lecture/tex/Lecture02.tex

@@ -599,7 +599,7 @@ \section{Fundamental electromagnetic principles and magnetic materials}
 %% Hyteresis curve of permanent magnets %%
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
-	\frametitle{Hyteresis curve of permanent magnets}
+	\frametitle{Hysteresis curve of permanent magnets}
 	\begin{columns}
 		\begin{column}{0.45\textwidth}
             \begin{itemize}
@@ -629,7 +629,7 @@ \section{Fundamental electromagnetic principles and magnetic materials}
 %% Hyteresis curve of permanent magnets (temperature dependence) %%
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
-	\frametitle{Hyteresis curve of permanent magnets (temperature dependence)}
+	\frametitle{Hysteresis curve of permanent magnets (temperature dependence)}
 	\begin{columns}
 		\begin{column}{0.4\textwidth}
             \begin{itemize}

lecture/tex/Lecture03.tex

@@ -95,7 +95,7 @@ \section{DC machines}
     \begin{columns}
 		\begin{column}{0.42\textwidth}
             \begin{itemize}
-				\item To ensure a quasi-continous torque, the current through the conductor loop(s) in the rotor must have a constant direction.
+				\item To ensure a quasi-continuous torque, the current through the conductor loop(s) in the rotor must have a constant direction.
 				\item<2-> This is achieved by using a commutator (brushes).
 				\item<3-> Compared to homopolar machines, DC machines require a mechanical rectification of the current.
 			\end{itemize}
@@ -860,7 +860,7 @@ \section{DC machines}
     \begin{figure}
         \centering
         \includegraphics[width=0.75\textwidth]{fig/lec03/Armature_reaction.pdf}
-        \caption{Superpostion of the field and armature magnetic excitation and the resulting air gap field normal components (adapted from W.~Novender, \textit{Elektrische Maschinen}, Technische Hochschule Mittelhessen, 2023)} 
+        \caption{Superposition of the field and armature magnetic excitation and the resulting air gap field normal components (adapted from W.~Novender, \textit{Elektrische Maschinen}, Technische Hochschule Mittelhessen, 2023)} 
 		\label{fig:Armature_reaction}
     \end{figure}
 \end{frame}
@@ -1326,7 +1326,7 @@ \section{DC machines}
 	\begin{figure}
 		\centering
 		\includegraphics[scale=1.05]{fig/lec03/Universal_machine_time_signals.pdf}
-		\caption{Qualitative voltage, current and torque signals for an universal motor}
+		\caption{Qualitative voltage, current and torque signals for a universal motor}
 	\end{figure}
 \end{column}
 \end{columns}

lecture/tex/Lecture04.tex

@@ -381,7 +381,7 @@ \section{Transformers}
 %% Core loss model (hyteresis and eddy current losses) %%
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
-	\frametitle{Core loss model (hyteresis and eddy current losses)}
+	\frametitle{Core loss model (hysteresis and eddy current losses)}
 	To also consider the iron losses inside the transformer core, a first-order model with the additional core loss resistance $R_{\mathrm{c}}$ can be introduced:
 	\begin{equation}
 		P_{\mathrm{l,c}} \approx R_{\mathrm{c}} I_{\mathrm{c}}^2 \approx \frac{U_1^2}{R_{\mathrm{c}}}. 
@@ -491,7 +491,7 @@ \section{Transformers}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
 	\frametitle{Voltage transformer application: measuring high AC voltages}
-	If the voltage to be measured is too high for direct measurement, a voltage transformer can be used to step-down the voltage to a suitable level:
+	If the voltage to be measured is too high for direct measurement, a voltage transformer can be used to step down the voltage to a suitable level:
 	$$u_2(t) = \frac{1}{\ddot{u}} u_1(t).$$
 	\pause
 	Hence, we choose $\ddot{u} > 1$. \pause Moreover, the voltage sensor on the secondary side comes with a high internal resistance $R_\mathrm{i}$ to avoid a significant current and, therefore, power flow. Neglecting the leakage inductance, we can model the voltage transformer as shown in \figref{fig:Voltage_transformer_meas_ECD} with
@@ -518,7 +518,7 @@ \section{Transformers}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
 	\frametitle{Current transformer application: measuring high AC currents}
-	If the current to be measured is too high for direct measurement, a current transformer can be used to step-down the current to a suitable level:
+	If the current to be measured is too high for direct measurement, a current transformer can be used to step down the current to a suitable level:
 	$$i_2(t) = \ddot{u} i_1(t).$$ \pause
 	Hence, we choose $\ddot{u} < 1$. \pause Moreover, the current sensor on the secondary side comes with a minimal internal resistance $R_\mathrm{i}$ to avoid a significant ohmic power losses. Likewise, the transformer should be designed for low $R_1$ and $R_2$ (e.g., $N_1=1$ on the primary and sufficiently large cable cross-sections).\\ \pause
 	\vspace{1em}
@@ -747,7 +747,7 @@ \section{Transformers}
            \begin{itemize}
 				\item If the flux zero component $\phi_0$ can be avoided, a three-leg design as shown in \figref{fig:Three_phase_transformer_3_legs_star} can be used. \pause
 				\item However, if $\phi_0 \neq 0$ due to an asymmetric design, magnetic saturation or non-ideal symmetrical operation, the zero component will act as a stray field leaving the core. \pause
-				\item This can lead to increased losses in auxillary components (e.g., housing) and electromagnetic interference issues.
+				\item This can lead to increased losses in auxiliary components (e.g., housing) and electromagnetic interference issues.
 			\end{itemize}
 		\end{column}
         \hfill

lecture/tex/Lecture05.tex

@@ -262,7 +262,7 @@ \section{Rotating field theory}
                 \end{split}
             \end{equation}
             \onslide<2->
-            The index 's' indicates stator quantities, but is omitted in the following as we will only consider stator quantities until further notice, i.e.,, 
+            The index 's' indicates stator quantities, but is omitted in the following as we will only consider stator quantities until further notice, i.e., 
             $$i_\mathrm{s,a}(t) = i_\mathrm{a}(t), \quad i_\mathrm{s,b}(t) = i_\mathrm{b}(t), \quad i_\mathrm{s,c}(t) = i_\mathrm{c}(t)$$
             and $\hat{i}_{\mathrm{s}}=\hat{i}$. 
         \end{column}

lecture/tex/Lecture07.tex

@@ -1020,7 +1020,7 @@ \section{Synchronous machines}
 \begin{frame}
 	\frametitle{Steady-state torque (cont.)} 
 	\onslide<+->
-	Moreover, from \eqref{eq:SM_voltage_steady_state} we can express the field winding current (amplitude, DC quantitity) as
+	Moreover, from \eqref{eq:SM_voltage_steady_state} we can express the field winding current (amplitude, DC quantity) as
 	\begin{equation}
 		I_\mathrm{f} = \sqrt{2}\frac{\left|\underline{U}_\mathrm{i}\right|}{\omega_\mathrm{s}M}.
 	\end{equation}
@@ -1051,7 +1051,7 @@ \section{Synchronous machines}
 				\item<1-> From \eqref{eq:SM_torque_steady_state} we see that the torque depends on $\sin(\theta)$.
 				\item<2-> Beyond $\theta = \pm 90^\circ$, the absolute torque decreases again.
 				\item<3-> If the SM is operated with a fixed stator excitation, e.g., by a stiff grid voltage, the load angle $\theta$ is determined by the mechanical load.
-				\item<4-> If the absolute mechanical load is increased such that $|\theta| > 90^\circ$ applies, the SM will lose synchronism and stall.
+				\item<4-> If the absolute mechanical load is increased such that $|\theta| > 90^\circ$ applies, the SM will lose synchronicity and stall.
 				\item<5-> Hence, the stable operation range is limited to $|\theta| \leq 90^\circ$ (while in practice an additional safety margin is considered).
 			\end{itemize}
         \end{column}
@@ -1076,7 +1076,7 @@ \section{Synchronous machines}
 	\begin{equation}
 			\underline{S}= 3\underline{U}_\mathrm{s} \overline{\underline{I}}_\mathrm{s} =  3(P + \mathrm{j}Q) = 3S e^{\mathrm{j}\varphi} 
 	\end{equation}
-	with $\overline{\underline{X}}$ being the complex conjugate and the factor $3$ results from the representation of the three-phase machine in an orthogonal coordinate system (cf. Clarke transf.) plus the RMS phasor reprentation of currents and voltages. \pause Above, $S$ is the apparent power, $P$ and $Q$ are the active and reactive power, respectively. \pause The active power is
+	with $\overline{\underline{X}}$ being the complex conjugate and the factor $3$ results from the representation of the three-phase machine in an orthogonal coordinate system (cf. Clarke transf.) plus the RMS phasor representation of currents and voltages. \pause Above, $S$ is the apparent power, $P$ and $Q$ are the active and reactive power, respectively. \pause The active power is
 	\begin{equation}
 		P = 3\mathrm{Re}\left\{ \underline{U}_\mathrm{s} \overline{\underline{I}}_\mathrm{s}\right\} = 3 U_\mathrm{s} I_\mathrm{s} \cos(\varphi)
 	\end{equation}