%\section{Compressed air energy storage}
% removed to _storage.tex
\section{Pumped storage}
\label{pstorage2}
The working volume required, $V$, depends on the height drop
we assume. If $\epsilon$ is the efficiency of
potential energy to electricity conversion,
%
% $ \epsilon \rho V g h = 50\, \GWh $
%
% was 100
\beq
V = 50\, \GWh / (\rho g h \epsilon) .
\label{eq.Vstore}
\eeq
%
Assuming the generators have an efficiency
of $\epsilon = 0.9$, \tabref{tab.pumpedstorage} (\pref{tab.pumpedstorage}) shows
a few ways of storing 50\,GWh, for a range of height drops.
\section{Tidal pumped storage}
\label{pTESS}
TBC
% This calculation goes with the discussion on \pref{pTESS0}. (cut to _storage.tex)
\begin{figure}
\figuremargin{\small
\begin{center}
\begin{tabular}{cc}
{\mbox{\epsfysize=0.851in\epsfbox{figs/TideStore.eps}}} &
{\mbox{\epsfysize=1in\epsfbox{figs/TideStore2.eps}}} \\
(a) & (b) \\
\end{tabular}
\end{center}
}{
\caption[a]{Two combined pumped-storage and tidal generators.
}
\label{tidestore}
}
\end{figure}
Let's denote the tidal range by $2h$
% (for example 4\,m)
and the extra head by $b$.
We start a daily cycle one evening, with high and low lagoons both
at the mid-tide level.
Overnight we pump the high lagoon up and the low lagoon down at the optimal times
near to high and low tide.
The total energy cost of the pumping, per unit area of facility, is
\[
\half \rho g b^2 / \epsilon_{\rm{p}}
\]
where $\epsilon_{\rm{p}}$ is the pump's efficiency.
In the design of \figref{tidestore}(a),
the energy returned at the time of peak demand, per unit area, is
\[
\epsilon_{\rm{g}} \half \rho g (h+b)^2.
\]
The alternative design, \figref{tidestore}(b),
delivers slightly greater power by picking which pool to
exploit depending on the sea level. At a randomly chosen time,
% the time when electricity is most profitable
there is a 50\% chance that
the tide is in quite an extreme state -- more than
71\% of the way to high tide or more than 71\% of the
way to low tide;
and
there is a 66\% chance that
the tide is either beyond half-high or beyond half-low.
\marginfig{\begin{center}
{\mbox{\epsfxsize=50mm\epsfbox{figs/sin.eps}}} \\
{\mbox{\epsfxsize=50mm\epsfbox{figs/sin5.eps}}}
\end{center}
\caption[a]{Half the time,
the tide is more than
71\% of the way to high tide or more than 71\% of the
way to low tide;
two thirds of the time,
the tide is above half-high or below half-low.
}
}
So, as long as the profitable period lasts more than a couple of hours,
both of the pools will have a good chance of serving high-value electricity
and the energy returned, per unit area, is about
\[
\epsilon_{\rm{g}} \half \rho g \left(\frac{3}{2}h+b\right)^2.
\]
% Additionally,
%\[
% \epsilon_{\rm{g}} \half \rho g \left(\frac{3}{2}h+b\right)^2.
%\]
%
%%% see also tide2.tex
% It is as if all water goes from high to mid-tide level.
Let's find the size required for a system with a stored energy of 10\,GWh (about
the same as Dinorwig), assuming $2h=4$\,m and $b=10$\,m.
% see figs/TS.gnu
That's about 57\,km$^2$, which would fit inside an 8\,km square,
and would need about 40\,km of wall.
Equipped with generators like Dinorwig's,
it would deliver 2\,GW of power during 5 hours of peak demand each day.
And the energy cost of delivering that 10\,GWh would be just
6.7\,GWh.
% *** explain this
For comparison with other renewable sources, the power density
of the useful power delivered at peak demand would
be 10\,W/m$^2$, averaged over 24 hours.
The average net power delivered would be 3.3\,W/m$^2$.
The economics for such a scheme
favour large facilities: the cost
of the lagoons' walls would be proportional to the walls' length,
but the area enclosed, and thus the delivered power, would increase
as the {\em{square}\/} of the length.