diff --git a/tutorial-3/sheet03.pdf b/tutorial-3/sheet03.pdf index 287cd41714ff619cdc615ef87712c1eaa0beb721..9f86d68c72afbf0106312926f92794b28a373b76 100644 Binary files a/tutorial-3/sheet03.pdf and b/tutorial-3/sheet03.pdf differ diff --git a/tutorial-3/sheet03.tex b/tutorial-3/sheet03.tex index 1eb011c6fb46a3df58d145ce0f39409d88791a79..189c554787af25bcc9933d7303164a136d68918d 100644 --- a/tutorial-3/sheet03.tex +++ b/tutorial-3/sheet03.tex @@ -112,58 +112,59 @@ \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} \end{wrapfigure} -Imagine a two-node Germany. The South can install solar panels with a capacity factor $Cf_s$ to cover its load $L_S$, while the North uses wind turbines that have a capacity factor $Cf_w$ -to feed their load $L_N$. Figure \ref{fig:variations} shows approximations to the daily and synoptic variations of per-unit wind and solar power generation $$G_{N,w}(t)$$ and $$G_{S,s}(t)$$ and a constant load $$L_{N/S}(t)$$: +Imagine a two-node Germany. The South can install solar panels with a capacity factor $Cf_s$ to cover its load $L^S$, while the North uses wind turbines that have a capacity factor $Cf_w$ +to feed their load $L^N$. Figure \ref{fig:variations} shows approximations to the daily and synoptic variations of per-unit wind and solar power generation $$G^{N}_{w}(t)$$ and $$G^{S}_{s}(t)$$ and a constant load $$L^{N/S}(t)$$: \vspace{-0.5em} \begin{align*} - G_{N,w}(t) & = Cf_w(1+A_w \sin \omega_w t), \\ - G_{S,s}(t) & = Cf_s(1+A_s \sin \omega_s t), \\ - L_{N/S}(t) & = A_{l,N/S}. + G_{w}^N(t) & = Cf_w(1+A_w \sin \omega_w t), \\ + G_{s}^S(t) & = Cf_s(1+A_s \sin \omega_s t), \\ + L^{N/S}(t) & = A_{l}^{N/S}. \end{align*} The capacity factors and constants are \vspace{-0.25em} \begin{align*} - A_{l,N} & = 20 \si{\giga\watt}, & A_{l,S} & = 30 \si{\giga\watt}, \\ + A_{l}^{N} & = 20 \si{\giga\watt}, & A_{l}^{S} & = 30 \si{\giga\watt}, \\ Cf_w & = 0.3, & A_w & = 0.9, & \omega_w & = \frac{2\pi}{7 \text{d}}, \\ Cf_s & = 0.12, & A_s & = 1.0, & \omega_s & = \frac{2\pi}{1 \text{d}}. \\ \end{align*} \vspace{-0.3em} -For now, assume the stores are lossless. Losses will be considered in III.2. +For now, assume the stores are lossless. Losses will be considered in Problem III.2. \begin{enumerate}[(a)] % (a) - \item How much wind capacity $G_{N,w}$ must be installed in the North and solar capacity $G_{S,s}$ in the South? + \item How much wind capacity $G^{N}_{w}$ must be installed in the North and solar capacity $G_s^S$ in the South? % (b) - \item What is the amount of store and dispatch power capacity $G_{s,store}=\max(-g_s(t))$ and $G_{s,dispatch}= \max g_s(t)$ the storage units must have in the North and in the South? + \item What is the amount of store and dispatch power capacity $G_{st,store}=\max(-\Delta(t))$ and $G_{st,dispatch} = \max \Delta(t)$ the storage units must have in the North and in the South to account for the mismatch $\Delta(t)=L(t)-G_{w/s}\cdot G_{w/s}(t)$? % (c) \item What is the amount of energy capacity one needs in the North and in the South? \begin{equation*} - E_s = \max_t e_s(t) = \max_t \int_{0}^{t} (-g_s(t')) \;\mathrm{d}t' + E_{st} = \max_t e_{st}(t) = \max_t \int_{0}^{t} -\Delta(t') \;\mathrm{d}t' \end{equation*} % (d) \item Should they choose hydrogen or battery storage? And how much would it cost them with the prices in Table 1? Is the South or the North paying more for their energy? +% (e) \item What do you imagine would change if you considered the storage losses given in Table 1 in your results (a)-(d)? Support your statement with a graphical illustration. \end{enumerate} Now we lift the restriction against transmission and allow them to bridge their 500 km separation with a transmission line. -\begin{enumerate}[(e)] - % (e) +\begin{enumerate}[(f)] + % (f) \item Estimate the cost-optimal technology mix by assuming wind energy in the North is only stored in the North and solar energy in the South is likewise only stored in the South! What would happen if you dropped that assumption? \end{enumerate} @@ -189,7 +190,7 @@ Now we lift the restriction against transmission and allow them to bridge their Python for Power System Analysis (PyPSA) is a free software toolbox for optimising modern power systems that include features such as variable wind and solar generation, storage units, etc\.. Use the toolbox to extend on your findings in Problem III.1. \begin{enumerate}[(a)] - \item Build a network in PyPSA with the two buses North and South and attach the load at each bus and attach the wind and solar generators with availability according to $G_{N,w}(t) = Cf_w(1+A_w\sin \omega_w t)$ and $G_{S,s}(t) = Cf_s(1+A_s\sin \omega_s t)$ for a year (you have to call \texttt{set\_snapshots} for the year) and with \texttt{p\_nom\_extendable} set to True. As help you should have a look at the minimal LOPF example\footnote{\url{https://www.pypsa.org/examples/minimal_example_lopf.html}}, understand what the components documentation\footnote{\url{https://pypsa.org/doc/components.html}} of PyPSA gives you and that you can find the underlying objective function and constraints in the LOPF documentation\footnote{\url{https://pypsa.org/doc/optimal_power_flow.html\#linear-optimal-power-flow}}. + \item Build a network in PyPSA with the two buses North and South and attach the load at each bus and attach the wind and solar generators with availability according to $G^{N}_{w}(t) = Cf_w(1+A_w\sin \omega_w t)$ and $G^{S}_{s}(t) = Cf_s(1+A_s\sin \omega_s t)$ for a year (you have to call \texttt{set\_snapshots} for the year) and with \texttt{p\_nom\_extendable} set to True. As help you should have a look at the minimal LOPF example\footnote{\url{https://www.pypsa.org/examples/minimal_example_lopf.html}}, understand what the components documentation\footnote{\url{https://pypsa.org/doc/components.html}} of PyPSA gives you and that you can find the underlying objective function and constraints in the LOPF documentation\footnote{\url{https://pypsa.org/doc/optimal_power_flow.html\#linear-optimal-power-flow}}. \item Attach extendable storage units at the North and the South! The storage units have to be modelled as a \texttt{H2-bus} (a bus with \texttt{carrier='H2'}) linked to the \texttt{AC-bus} North with a \texttt{Link} where \texttt{p\_nom\_extendable=True} with the \texttt{capital\_cost} of the power capacity and an also extendable \texttt{Store} with the \texttt{capital\_cost} of the energy capacity, for instance. The losses can be set on the links as \texttt{efficiency}. \item Run an investment optimization by calling the \texttt{lopf} function. \item How do your results \texttt{objective} and \texttt{{generators,stores,links}.p\_nom\_opt} compare with the results of III.1(d)? diff --git a/tutorial-3/solution03.pdf b/tutorial-3/solution03.pdf index d04ac8a7b40b4612767975f5927c8f02f5ac650d..0f336868b276e9a0bb92067b6a7b52bb16926fab 100644 Binary files a/tutorial-3/solution03.pdf and b/tutorial-3/solution03.pdf differ diff --git a/tutorial-3/solution03.tex b/tutorial-3/solution03.tex index ce6bce971b3766c8d7175e2b29b2d7976586926c..0d38968a52a99e81a653ca5e59ac215b90459b8e 100644 --- a/tutorial-3/solution03.tex +++ b/tutorial-3/solution03.tex @@ -111,110 +111,112 @@ \end{axis} \end{tikzpicture} \caption{Diurnal and synoptic variations of wind and solar power generation - $$G_{N,w}(t)$$ - \autoref{figref:w} and $$G_{S,s}(t)$$ - \autoref{figref:s}, and a constant load (all in per-unit) $$L(t)$$ - \autoref{figref:l}.} - \label{fig:variations} + $$G^{N}_{w}(t)$$ + \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} \end{wrapfigure} -Imagine a two-node Germany. The South can install solar panels with a capacity factor $Cf_s$ to cover its load $L_S$, while the North uses wind turbines that have a capacity factor $Cf_w$ -to feed their load $L_N$. Figure \ref{fig:variations} shows approximations to the daily and synoptic variations of per-unit wind and solar power generation $$G_{N,w}(t)$$ and $$G_{S,s}(t)$$ and a constant load $$L_{N/S}(t)$$: +Imagine a two-node Germany. The South can install solar panels with a capacity factor $Cf_s$ to cover its load $L^S$, while the North uses wind turbines that have a capacity factor $Cf_w$ +to feed their load $L^N$. Figure \ref{fig:variations} shows approximations to the daily and synoptic variations of per-unit wind and solar power generation $$G^{N}_{w}(t)$$ and $$G^{S}_{s}(t)$$ and a constant load $$L^{N/S}(t)$$: + +\vspace{-0.5em} \begin{align*} - G_{N,w}(t) & = Cf_w(1+A_w \sin \omega_w t), \\ - G_{S,s}(t) & = Cf_s(1+A_s \sin \omega_s t), \\ - L_{N/S}(t) & = A_{l,N/S}. +G_{w}^N(t) & = Cf_w(1+A_w \sin \omega_w t), \\ +G_{s}^S(t) & = Cf_s(1+A_s \sin \omega_s t), \\ +L^{N/S}(t) & = A_{l}^{N/S}. \end{align*} The capacity factors and constants are - +\vspace{-0.25em} \begin{align*} - A_{l,N} & = 20 \si{\giga\watt}, & A_{l,S} & = 30 \si{\giga\watt}, \\ - Cf_w & = 0.3, & A_w & = 0.9, & \omega_w & = \frac{2\pi}{7 \text{d}}, \\ - Cf_s & = 0.12, & A_s & = 1.0, & \omega_s & = \frac{2\pi}{1 \text{d}}. \\ +A_{l}^{N} & = 20 \si{\giga\watt}, & A_{l}^{S} & = 30 \si{\giga\watt}, \\ +Cf_w & = 0.3, & A_w & = 0.9, & \omega_w & = \frac{2\pi}{7 \text{d}}, \\ +Cf_s & = 0.12, & A_s & = 1.0, & \omega_s & = \frac{2\pi}{1 \text{d}}. \\ \end{align*} - -For now, assume the stores are lossless. Losses will be considered in III.2. +\vspace{-0.3em} +For now, assume the stores are lossless. Losses will be considered in Problem III.2. \begin{enumerate}[(a)] % (a) - \begin{shaded}\item How much wind capacity $G_{N,w}$ must be installed in the North and solar capacity $G_{S,s}$ in the South?\end{shaded} + \begin{shaded}\item How much wind capacity $G^{N}_{w}$ must be installed in the North and solar capacity $G_s^S$ in the South?\end{shaded} In the North: - $$\expect{L_N} = \expect{G_{N,w} \cdot G_{N,w}(t)}$$ + $$\expect{L^N} = \expect{G^N_w \cdot G^N_w(t)}$$ - $$\Rightarrow \quad A_{l,N} = G_{N,w}\cdot Cf_w$$ + $$\Rightarrow \quad A^N_l = G^N_w\cdot Cf_w$$ \DIVIDE{20}{0.3}\res - $$\Rightarrow \quad G_{N,w} = \frac{A_{l,N}}{Cf_w} = \frac{20\si{\giga\watt}}{0.3} = \rd{\res}\si{\giga\watt}$$ + $$\Rightarrow \quad G^N_w = \frac{A^N_l}{Cf_w} = \frac{20\si{\giga\watt}}{0.3} = \rd{\res}\si{\giga\watt}$$ In the South: - $$\expect{L_S} = \expect{G_{S,s} \cdot G_{S,s}(t)}$$ + $$\expect{L^S} = \expect{G^S_s \cdot G^S_s(t)}$$ - $$\Rightarrow \quad A_{l,S} = G_{S,w}\cdot Cf_w$$ + $$\Rightarrow \quad A^S_l = G_{S,w}\cdot Cf_w$$ \DIVIDE{30}{0.12}\res - $$\Rightarrow \quad G_{S,s} = \frac{A_{l,S}}{Cf_s} = \frac{30\si{\giga\watt}}{0.12} = \rd{\res}\si{\giga\watt}$$ + $$\Rightarrow \quad G^S_s = \frac{A^S_l}{Cf_s} = \frac{30\si{\giga\watt}}{0.12} = \rd{\res}\si{\giga\watt}$$ % (b) - \begin{shaded}\item What is the amount of store and dispatch power capacity $G_{s,store}=\max(-g_s(t))$ and $G_{s,dispatch}= \max g_s(t)$ the storages must have in the North and in the South?\end{shaded} + \begin{shaded}\item What is the amount of store and dispatch power capacity $G_{st,store}=\max(-\Delta(t))$ and $G_{st,dispatch} = \max \Delta(t)$ the storage units must have in the North and in the South to account for the mismatch $\Delta(t)=L(t)-G_{w/s}\cdot G_{w/s}(t)$?\end{shaded} In the North: \begin{align*} - G_{s,storage,dispatch}^N & = \max ( \pm g_s^N(t)) \\ - & = \max (\pm [L_N(t) - G_{N,w} \cdot G_{N,w}(t)]) \\ - & = \max (\pm [L_N(t) - \frac{A_{l,N}}{Cf_w}\cdot Cf_w\cdot (1+A_w \sin \omega_w t)]) \\ - & = \max (\pm [L_N(t) - A_{l,N} + A_{l,N} A_w \sin \omega_w t)]) \\ - & = \max (\pm [A_{l,N} A_w \sin \omega_w t)]) \\ - & = A_{l,N} A_w = 0.9 \cdot 20 \si{\giga\watt} = 18 \si{\giga\watt} + 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}{Cf_w}\cdot Cf_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)]) \\ + & = 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)) \\ - & = \max (\pm [L_S(t) - G_{S,s} \cdot G_{S,s}(t)]) \\ - & = \max (\pm [L_S(t) - \frac{A_{l,S}}{Cf_s}\cdot Cf_s\cdot (1+A_s \sin \omega_s t)]) \\ - & = \max (\pm [L_S(t) - A_{l,S} + A_{l,S} A_s \sin \omega_s t)]) \\ - & = \max (\pm [A_{l,S} A_s \sin \omega_s t)]) \\ - & = A_{l,S} A_s = 1.0 \cdot 30 \si{\giga\watt} = 30 \si{\giga\watt} + & = \max (\pm [L^S(t) - G^S_s \cdot G^S_s(t)]) \\ + & = \max (\pm [L^S(t) - \frac{A^S_l}{Cf_s}\cdot Cf_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)]) \\ + & = \max (\pm [A^S_l A_s \sin \omega_s t)]) \\ + & = A^S_l A_s = 1.0 \cdot 30 \si{\giga\watt} = 30 \si{\giga\watt} \end{align*} % (c) \begin{shaded}\item What is the amount of energy capacity one needs in the North and in the South? - - $$- E_s = \max_t e_s(t) = \max_t \int_{0}^{t} (-g_s(t')) \;\mathrm{d}t' -$$ + + \begin{equation*} + E_{st} = \max_t e_{st}(t) = \max_t \int_{0}^{t} -\Delta(t') \;\mathrm{d}t' + \end{equation*} \end{shaded} In the North: \begin{align*} - e_s^N(t) & = \int_{0}^{t} -g_s^N(t') \;\mathrm{d}t' = \int_{0}^{t} A_{l,N} A_w \sin \omega_w t' \;\mathrm{d}t' \\ - & = A_{l,N} A_w \frac{-\cos(\omega_w t')}{\omega_w}\Big|_0^t = A_{l,N} A_w \frac{1-\cos(\omega_w t')}{\omega_w} + e_{st}^N(t) & = \int_{0}^{t} -\Delta^N(t') \;\mathrm{d}t' = \int_{0}^{t} A^N_l A_w \sin \omega_w t' \;\mathrm{d}t' \\ + & = A^N_l A_w \frac{-\cos(\omega_w t')}{\omega_w}\Big|_0^t = A^N_l A_w \frac{1-\cos(\omega_w t')}{\omega_w} \end{align*} \begin{equation*} - E_s^N = \max_t e_s(t) = \frac{2A_{l,N}A_w}{\omega_w} = \frac{2\cdot 20 \si{\giga\watt}\cdot 0.9}{2\pi} \cdot 7\cdot 24 \si{\hour} = 1 \si{\tera\watt\hour} + E_{st}^N = \max_t e_{st}(t) = \frac{2A^N_lA_w}{\omega_w} = \frac{2\cdot 20 \si{\giga\watt}\cdot 0.9}{2\pi} \cdot 7\cdot 24 \si{\hour} = 1 \si{\tera\watt\hour} \end{equation*} In the South: \begin{align*} - e_s^S(t) & = \int_{0}^{t} -g_s^S(t') \;\mathrm{d}t' = \int_{0}^{t} A_{l,S} A_s \sin \omega_s t' \;\mathrm{d}t' \\ - & = A_{l,S} A_s \frac{-\cos(\omega_s t')}{\omega_s}\Big|_0^t = A_{l,S} A_s \frac{1-\cos(\omega_s t')}{\omega_s} + 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' \\ + & = 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*} \begin{equation*} - E_s^S = \max_t e_s(t) = \frac{2A_{l,S}A_s}{\omega_s} = \frac{2\cdot 30 \si{\giga\watt}\cdot 1}{2\pi} \cdot 24 \si{\hour} = 230 \si{\giga\watt\hour} + E_{st}^S = \max_t e_{st}(t) = \frac{2A^S_lA_s}{\omega_s} = \frac{2\cdot 30 \si{\giga\watt}\cdot 1}{2\pi} \cdot 24 \si{\hour} = 230 \si{\giga\watt\hour} \end{equation*} % (d) @@ -224,12 +226,12 @@ For now, assume the stores are lossless. Losses will be considered in III.2. The cost of renewable generation in the North is - $$P_w^N=G_{N,w} \cdot 1200 \text{\EUR{}}\si{\per\kilo\watt}=80\cdot10^9\text{\EUR{}}$$ + $$P_w^N=G^N_w \cdot 1200 \text{\EUR{}}\si{\per\kilo\watt}=80\cdot10^9\text{\EUR{}}$$ The minimal (lossless) corresponding cost to supply constant demand by using hydrogen as storage technology are \begin{align*} - P_h^N & = 750 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{s,storage,dispatch}^N + 10 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_s^N \\ + P_h^N & = 750 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{storage,dispatch}^N + 10 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_s^N \\ & = 13.5 \cdot 10^9 \eur + 10 \cdot 10^9 \eur \\ & = 23.5 \cdot 10^9 \eur \end{align*} @@ -237,7 +239,7 @@ For now, assume the stores are lossless. Losses will be considered in III.2. The minimal (lossless) corresponding cost to supply constant demand by using batteries as storage technology are: \begin{align*} - P_b^N & = 300 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{s,storage,dispatch}^N + 200 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_s^N \\ + P_b^N & = 300 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{storage,dispatch}^N + 200 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_s^N \\ & = 5.4 \cdot 10^9 \eur + 200 \cdot 10^9 \eur \\ & = 205.4 \cdot 10^9 \eur \end{align*} @@ -256,12 +258,12 @@ For now, assume the stores are lossless. Losses will be considered in III.2. The cost of renewable generation in the South is - $$P_s^S=G_{S,s} \cdot 600 \text{\EUR{}}\si{\per\kilo\watt}=150\cdot10^9\text{\EUR{}}$$ + $$P_s^S=G^S_s \cdot 600 \text{\EUR{}}\si{\per\kilo\watt}=150\cdot10^9\text{\EUR{}}$$ 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_{s,storage,dispatch}^S + 10 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_s^S \\ + P_h^S & = 1200 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{storage,dispatch}^S + 10 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_{st}^S \\ & = 22.5 \cdot 10^9 \eur + 2.3 \cdot 10^9 \eur \\ & = 24.8 \cdot 10^9 \eur \end{align*} @@ -269,7 +271,7 @@ For now, assume the stores are lossless. Losses will be considered in III.2. The minimal (lossless) corresponding cost to supply constant demand by using batteries as storage technology are: \begin{align*} - P_b^S & = 300 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{s,storage,dispatch}^S + 200 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_s^S \\ + P_b^S & = 300 \text{\EUR{}}\si{\per\kilo\watt} \cdot G_{storage,dispatch}^S + 200 \text{\EUR{}}\si{\per\kilo\watt\hour} \cdot E_{st}^S \\ & = 9 \cdot 10^9 \eur + 46 \cdot 10^9 \eur \\ & = 54 \cdot 10^9 \eur \end{align*} @@ -288,16 +290,16 @@ For now, assume the stores are lossless. Losses will be considered in III.2. Without taking losses into account, both regions should choose hydrogen storages. Overall, the North can provide electricity at a lower rate than the South: - $$P_{w+h}^N = \frac{104 \cdot 10^9 \eur}{20 \si{\giga\watt}} = 5 \cdot 10^9 \eur \si{\per\giga\watt}$$ + $$P_{w+h}^N \geq \frac{104 \cdot 10^9 \eur}{20 \si{\giga\watt}} = 5 \cdot 10^9 \eur \si{\per\giga\watt}$$ - $$P_{s+h}^S = \frac{175 \cdot 10^9 \eur}{30 \si{\giga\watt}} = 6 \cdot 10^9 \eur \si{\per\giga\watt}$$ + $$P_{s+h}^S \geq \frac{175 \cdot 10^9 \eur}{30 \si{\giga\watt}} = 6 \cdot 10^9 \eur \si{\per\giga\watt}$$ % (e) \begin{shaded} \item What do you imagine would change if you considered the storage losses given in Table 1 in your results (a)-(d)? Support your statement with a graphical illustration. \end{shaded} - To compensate for the energy losses wind and solar capacity $G_{N/S,w/s}$, store and dispatch power capacities $G_s$ and storage energy capacities $E_s$ have to increase. + To compensate for the energy losses wind and solar capacity $G_{w/s}$, store and dispatch power capacities $G_{storage,dispatch}$ and storage energy capacities $E_{st}$ have to increase. % (f) \begin{shaded}\item Now we lift the restriction against transmission and allow them to bridge their 500 km separation with a transmission line. Estimate the cost-optimal technology mix by assuming wind energy in the North is only stored in the North and solar energy in the South is likewise only stored in the South! What would happen if you dropped that assumption?\end{shaded} @@ -305,7 +307,9 @@ For now, assume the stores are lossless. Losses will be considered in III.2. Because $P_{w+h}^N < P_{w+h}^S$ there will be energy exports from North to South: $$E^N > E^S \quad \text{and} \quad E^N + E^S = 50 \si{\giga\watt}$$ - + + Note that we can consider energy as $\si{\giga\watt}$ as we have a constant load! Otherwise, we would have to pay more attention! + The total price of electricity is given by \begin{align*} @@ -320,7 +324,7 @@ For now, assume the stores are lossless. Losses will be considered in III.2. E^N & = 50 \si{\giga\watt} \end{align*} - such that all power would be produced from wind in the North and the total system cost is + such that all power would be produced from wind in the North. This is because the term $(P_{w+h}^N - P_{s+h}^S + 400 \eur \si{\per\kilo\watt})$ is negative. The resulting total system cost is \begin{align*} \min P_{tot} & = \frac{-0.6 \cdot 50 \cdot 10^9 \eur + 5.8 \cdot 50 \cdot 10^9 \eur}{50 \si{\giga\watt}} \\