tutorial-3: minor corrections in solution

tutorial-3/worksheet/sheet03.tex

@@ -123,8 +123,8 @@
   \end{axis}
  \end{tikzpicture}
  \caption{Diurnal and synoptic variations of wind and solar power generation
-  \(G^{N}_{w}(t)\)
-  \autoref{figref:w} and \(G^{S}_{s}(t)\)
+  \(g^{N}_{w}(t)\)
+  \autoref{figref:w} and \(g^{S}_{s}(t)\)
   \autoref{figref:s}, and a constant load (all in per-unit) \(L(t)\)
   \autoref{figref:l}.}
  \label{fig:variations}

tutorial-3/worksheet/solution03.tex

@@ -122,8 +122,8 @@
   \end{axis}
  \end{tikzpicture}
  \caption{Diurnal and synoptic variations of wind and solar power generation
-	\(G^{N}_{w}(t)\)
-	\autoref{figref:w} and \(G^{S}_{s}(t)\)
+	\(g^{N}_{w}(t)\)
+	\autoref{figref:w} and \(g^{S}_{s}(t)\)
 	\autoref{figref:s}, and a constant load (all in per-unit) \(L(t)\)
 	\autoref{figref:l}.}
 \label{fig:variations}
@@ -180,15 +180,15 @@ For now, assume no power exchange between the regions and that the stores are lo
   G_{s,storage,dispatch}^N & = \max ( \pm \Delta^N(t))                                                              \\
                            & = \max (\pm [L^N(t) - G^N_w \cdot g^N_w(t)])                                    \\
                            & = \max (\pm [L^N(t) - \frac{A^N_l}{c_w}\cdot c_w\cdot (1+A_w \sin \omega_w t)]) \\
-                           & = \max (\pm [L^N(t) - A^N_l + A^N_l A_w \sin \omega_w t)])                      \\
-                           & = \max (\pm [A^N_l A_w \sin \omega_w t)])                                         \\
+                           & = \max (\pm [L^N(t) - A^N_l - A^N_l A_w \sin \omega_w t)])                      \\
+                           & = \max (\pm [-A^N_l A_w \sin \omega_w t)])                                         \\
                            & = A^N_l A_w = 0.9 \cdot 20 \si{\giga\watt} = 18 \si{\giga\watt}
  \end{align*}
 
  In the South:
 
  \begin{align*}
-  G_{s,storage,dispatch}^S & = \max ( \pm g_s^S(t))                                                              \\
+  G_{s,storage,dispatch}^S & = \max ( \pm \Delta^S(t))                                                              \\
                            & = \max (\pm [L^S(t) - G^S_s \cdot g^S_s(t)])                             \\
                            & = \max (\pm [L^S(t) - \frac{A^S_l}{c_s}\cdot c_s\cdot (1+A_s \sin \omega_s t)]) \\
                            & = \max (\pm [L^S(t) - A^S_l + A^S_l A_s \sin \omega_s t)])                      \\
@@ -219,7 +219,7 @@ For now, assume no power exchange between the regions and that the stores are lo
  In the South:
 
  \begin{align*}
-  e_{st}^S(t) & = \int_{0}^{t} -g_s^S(t') \;\mathrm{d}t' = \int_{0}^{t} A^S_l A_s \sin \omega_s t' \;\mathrm{d}t'           \\
+  e_{st}^S(t) & = \int_{0}^{t} -\Delta^S(t') \;\mathrm{d}t' = \int_{0}^{t} A^S_l A_s \sin \omega_s t' \;\mathrm{d}t'           \\
            & = A^S_l A_s \frac{-\cos(\omega_s t')}{\omega_s}\Big|_0^t = A^S_l A_s \frac{1-\cos(\omega_s t')}{\omega_s}
  \end{align*}
 
@@ -271,7 +271,7 @@ For now, assume no power exchange between the regions and that the stores are lo
  The minimal (lossless) corresponding cost to supply constant demand by using hydrogen as storage technology are
 
  \begin{align*}
-  P_h^S & = 1200 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{storage,dispatch}^S + 10 \text{\EUR{}/kWh} \cdot E_{st}^S \\
+  P_h^S & = 750 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{storage,dispatch}^S + 10 \text{\EUR{}/kWh} \cdot E_{st}^S \\
         & = 22.5 \cdot 10^9 \eur + 2.3 \cdot 10^9 \eur                                                                                  \\
         & = 24.8 \cdot 10^9 \eur
  \end{align*}