\label{app.tide2}
\section{Power per unit area of tidal pools}
\label{pTidePool}
% \section{Tide pool power}
% \subsection{Production on ebb and flood}
% {\sc The power of an artificial tide pool in more detail.}
To estimate the power of an artificial \tidepool, imagine that it's filled
\marginfig{
\begin{center}
\mbox{\epsfbox{metapost/tide.3}}
\end{center}
\caption[a]{
A \tidepool\ in cross section.
The pool was filled at high tide, and now it's low tide.
We let the water out through the electricity generator
to turn the water's potential energy into electricity.
}\label{tidepool2}
}%
rapidly at high tide, and emptied rapidly at low tide.
Power is generated in both directions, on the ebb\index{tide!two-way generation}
and on the flood. (This is called \ind{two-way generation} or \ind{double-effect generation}.)
The change in potential energy of the water, each six hours, is
$mgh$, where $h$ is the change in height of the centre of
mass of the water, which is half the range. (The range is the
difference in height between low and high tide;
\figref{tidepool2}.)\label{pagetidepool2}
The mass per unit area covered by tide-pool
is $\rho \times (2 h)$, where $\rho$ is the density
of water ($1000\,\kg/\m^3$).
So the power per unit area generated by a {\tidepool} is
\[
\frac{ 2 \rho h g h}{ \mbox{6\,hours} },
\]
assuming perfectly efficient generators.
Plugging in $h = 2\,\m$ (\ie, range 4\,m), we find the
power per unit area of tide-pool
is $3.6\,\Wmm$.
Allowing for an efficiency of 90\% for conversion of this power to
electricity,\nlabel{Turbine90} we get
\[
\mbox{power per unit area of tide-pool}
% \frac{ 2 \rho h g h}{ \mbox{6\,hours} }
% = 0.9 * 2000 kg/m^3 * 4*9.81 m^2 m/s/s / (6*3600)s
% = 3.27 W/m^2
\: \simeq \: \OliveGreen{3\,\Wmm}.
\]
So to generate 1\,GW of power (on average),
we need a {\tidepool} with an area of about 300\,km$^2$.
A circular pool with diameter 20\,km would do the trick.
(For comparison, the area of the Severn estuary behind
the proposed barrage is about 550\,km$^2$,
and the area of the Wash is more than 400\,km$^2$.
If a {\tidepool} produces electricity in one direction only, the power per
unit area is halved.
The average power per unit area of the tidal barrage at La Rance,
where the mean tidal range is 10.9\,m, has been
% where the tidal range is at most 13.5\,m, has been
\OliveGreen{2.7\,\Wmm} for decades (\pref{pRanceFacts}).
% mean range from novak page 98 WilsonBallsTide
\section{The raw tidal resource}
% \subsection{Tides as tidal waves}
The tides around Britain are genuine
tidal waves.\index{tide!as waves}\index{wave!tides as waves}
(Tsunamis\index{tsunami}, which are called ``tidal waves,'' have nothing to do with tides:
they are caused by underwater landslides and earthquakes.)
The location of the high tide (the crest of the tidal wave)
moves much faster than the tidal
flow -- 100 miles per hour, say, while the water itself
moves at just 1 mile per hour.
% \subsection{Sanity-check using total arriving power}
\begin{figure}[thbp]
% \figuremargin{
\figuredangle{
\begin{center}
\mbox{\epsfbox{metapost/tide.1}}
\end{center}
}{
\caption[a]{A shallow-water wave.\index{wave!shallow-water}
Just like a deep-water wave, the wave has energy in two forms:
potential energy associated with raising water out of the
light-shaded troughs into the heavy-shaded crests;
and kinetic energy of all the water moving around
as indicated by the small arrows.
The speed of the wave, travelling from left to right, is indicated by
the much bigger arrow at the top.
For tidal waves,
a typical depth might be 100\,m,
%% v = sqrt(g d) = sqrt( 1000 )
the crest velocity 30\,m/s,
the vertical amplitude at the surface 1 or 2\,m,
and the water velocity amplitude
%% U = v h/d = 30 * 2/100 = 0.6
0.3 or 0.6\,m/s.
}
\label{fig.shallowwave}
}
\end{figure}
The energy we can extract from tides, using tidal pools
or {\tidefarm}s, can never
be more than the energy of these tidal waves
from the Atlantic. We can estimate the total power
of these great Atlantic tidal waves in the same
way that we estimate the power of
% their smaller cousins,
ordinary wind-generated waves.
The next section describes a
standard model for the power arriving in travelling waves
in water of depth $d$ that is shallow compared to the
wavelength of the waves (\figref{fig.shallowwave}).
The power per unit length of wavecrest of shallow-water tidal waves is
\beq
\rho g^{3/2} \sqrt{ d } h^2/2.
\label{eqRawPo}
\eeq
% From this formula, a ballpark estimate for the power coming
\Tabref{tab.pfluxtid} shows the power per unit length of wave crest
for some plausible figures.
If $d = 100\,\m$,
and $h = 1$ or $2\,\m$, the power per unit length of wave crest is
% in kW/m: p(h) = 9.81**1.5 * 10 * h**2
% in kW/m: p(h,d) = 9.81**1.5 * sqrt(d) * h**2
$150$\,\kWm\ or $600\,$\kWm\ respectively.\index{power per unit length!tide}
These figures are impressive compared with the raw power per unit
length of ordinary Atlantic deep-water waves, 40\,\kWm\ (\chref{ch.waves2}).
Atlantic waves and the Atlantic tide have similar
vertical amplitudes (about 1\,m), but the raw power in tides
is roughly 10 times bigger than that of ordinary wind-driven waves.
% Fortunately we don't have to depend on my rough model of incoming tidal
% power.
\citet{Taylor1920} worked out a more detailed model of tidal power that includes
important details such as the Coriolis effect (the effect produced
by the earth's daily rotation), the existence of
% counter-propagating
tidal waves travelling in the opposite direction, and the
direct effect of the moon on the energy flow in the Irish Sea. Since then,
experimental measurements and computer models have verified and
extended Taylor's analysis.
\amargintab{t}{
\begin{center}
\begin{tabular}{cc}\toprule
$h$ & $ \rho g^{3/2} \sqrt{ d } h^2/2$ \\
(m) & (\kWm) \\ \midrule
0.9& 125 \\
1.0& 155 \\
1.2& 220 \\
1.5& 345 \\
1.75& 470 \\
2.0 & 600 \\ % 1230
2.25& 780 \\ % 1555
\bottomrule
\end{tabular}
\end{center}
\caption[a]{Power fluxes (power per unit length of wave crest)
for depth $d=100\,\m$. }
\label{tab.pfluxtid}
}
%
\citet{Flather1976} built a detailed numerical model of the
lunar tide, chopping the continental shelf around the British Isles into
roughly 1000 square cells.
Flather estimated that the total average power entering this region
is 215\,GW\@.
According to his model,
180\,GW enters the gap between France and Ireland.
From Northern Ireland round to Shetland, the incoming power is 49\,GW\@.
Between Shetland and Norway there is a net loss of 5\,GW\@.
%
% [Compare with 215--250\,GW estimated in the literature; of which 64\,GW
% estimated to enter the Irish Sea. \cite{Blunden}.
As shown in \figref{fig.TideLine2}, \citet{Cartwright1980}
found experimentally that the
average
% M2
power transmission was
60\,GW between \ind{Malin Head} (Ireland) and \ind{Flor{\o}} (\ind{Norway})
and 190\,GW between \ind{Valentia}
(\ind{Ireland}) and the Brittany coast near Ouessant.
The power entering the Irish Sea was found to be
45\,GW, and entering the North Sea
via the Dover Straits, 16.7\,GW\@.
% Near the Orkneys the incoming powers
% are 14\,GW and 12\,GW\@.
\subsection{The power of tidal waves}
\label{sec.powerT}
This section, which can safely be skipped,
provides more details behind the
formula for tidal power used in the previous section.
I'm going to go into this model of tidal power in some detail
because most of the official estimates of the UK tidal resource
have been based on a model that I believe is incorrect.
\Figref{fig.shallowwave} shows a model for a tidal wave travelling
across relatively shallow water.
This model is intended as a cartoon, for example, of tidal
crests moving up the English channel or down the North Sea.
It's important to distinguish the speed $U$ at which the water itself moves
(which might be about 1 mile per hour)
from the speed $v$ at which the high tide moves,
which is typically 100 or 200 miles per hour.
The water has depth $d$.
Crests and troughs of water are injected from the left hand side
by the 12-hourly ocean tides.
The crests and troughs move with velocity
\beq
v= \sqrt{ g d }.
\label{eqv}
\eeq
We assume that the wavelength is much bigger than the depth, and
we neglect details such as Coriolis forces and density variations in the water.
Call the vertical amplitude of the tide $h$.
For the standard assumption of nearly-vorticity-free flow,
the horizontal velocity of the water is near-constant with depth.
The horizontal velocity $U$ is proportional to the surface displacement
and can be found by conservation of mass:
\beq
U = v h/d.
\label{eqU}
\eeq
If the depth decreases, the wave velocity $v$ reduces
%\begin{figure}
\marginfig{
\begin{center}
{\mbox{\epsfxsize=50mm\epsfbox{../../images/PUBLICDOMAIN/maps/northseaTID.eps}}}\\
\end{center}
\caption[a]{Average tidal powers measured by \citet{Cartwright1980}.
}
\label{fig.TideLine2}
}%
%\end{figure}
(\eqref{eqv}).
% and the amplitude of tidal wave increases (as $1/d^{1/4}$, as we'll see).
For the present discussion we'll assume the depth is constant.
Energy flows from left to right at some rate.
How should this total tidal power be estimated?
And what's the {\em{maximum}\/} power that could be extracted?
One suggestion is to choose a cross-section and estimate the average
{\em{flux of kinetic energy}\/} across that plane, then
assert that this quantity represents the power that could be extracted.
This kinetic-energy-flux
method was used by consultants
Black and Veatch\nocite{BlackVeatch} to estimate the UK
resource. In our cartoon model, we can
compute the total power by
other means. We'll see that the kinetic-energy-flux answer is
too small by a significant factor.
The peak kinetic-energy flux at any section
is
\beq
K_{\rm BV} = \frac{1}{2} \rho A U^3 ,
\eeq
where $A$ is the cross-sectional area. (This is the
formula for kinetic energy flux, which
we encountered in \chref{ch.wind2}.)
The true total incident power is not equal to this kinetic-energy flux.
The true total incident power in a shallow-water wave
is a standard textbook calculation;
one way to get it is to find the total energy present in one wavelength
and divide by the period.
% ; another option is to imagine replacing a vertical
% section by an appropriately compliant
% piston and computing the average work done on the
% piston. I'll do the calculation both ways.
%
The total energy per wavelength is the sum of the potential energy and the kinetic energy.
The kinetic energy happens to be identical to the potential energy.
(This is a standard feature of almost all things that wobble, be they
masses on springs or children on swings.)
So to compute the total energy all we need to do is compute one
of the two -- the potential energy per wavelength, or the
kinetic energy per wavelength -- then double it.
% Let's go for the potential energy.
%
The potential energy of a wave (per wavelength and per unit width
of wavefront) is found by integration to be
\beq
\frac{1}{4} \rho g h^2 \lambda .
\eeq
So, doubling and dividing by the period,
the true power of this model shallow-water tidal wave is
\beq
\mbox{power} = \frac{1}{2}( \rho g h^2 \lambda) \times w / T
= \frac{1}{2} \rho g h^2 v \times w ,
\eeq
where $w$ is the width of the wavefront.
Substituting
% $A = w d$ and
$v=\sqrt{ g d }$,
\beq
\mbox{power}
= \rho g h^2 \sqrt{ g d } \times w /2
= \rho g^{3/2} \sqrt{ d } h^2 \times w /2 .
% = \rho g h^2 \sqrt{ g } \times A / \sqrt{d} ,
\label{eqP}
\eeq
Let's compare this power with
the kinetic-energy flux $K_{\rm BV}$.
Strikingly, the two expressions scale differently with the amplitude $h$.
Using the amplitude conversion relation (\ref{eqU}),
the crest velocity (\ref{eqv}), and $A = w d$,
we can re-express
the kinetic-energy flux
as
\beq
K_{\rm BV} = \frac{1}{2} \rho A U^3
= \frac{1}{2} \rho w d (v h/d)^3
% = \frac{1}{2} \rho w (\sqrt{ g d })^3 h^3 / d^2 .
= \rho \left( g^{3/2} / \sqrt{ d } \right) h^3 \times w/2 .
\label{eq.cubed}
\eeq
% *** OHB here ***
So the kinetic-energy-flux method suggests that the total
power of a shallow-water wave scales as amplitude {\em{cubed}\/} (\eqref{eq.cubed});
but the correct formula shows that the power scales as amplitude {\em{squared}\/}
(\eqref{eqP}).
The ratio is
\beq
\frac{ K_{\rm BV} }
{ \mbox{power} }
= \frac{ \rho w \left( g^{3/2} / \sqrt{ d } \right) h^3
}{ \rho g^{3/2} h^2 \sqrt{ d } w }
= \frac{ h
}{ d} .
\eeq
Because $h$ is usually much smaller than $d$ ($h$ is about 1\,m or 2\,m,
while $d$ is 100\,m or 10\,m),
estimates of tidal power resources that are
based on the kinetic-energy-flux method may be
{\em{much too small}}, at least in cases where this
shallow-water cartoon of tidal waves is appropriate.
Moreover, estimates based on the kinetic-energy-flux method
incorrectly assert that the total available power at springs (the biggest
tides) is eight times greater
than at neaps (the smallest tides),
assuming an amplitude ratio, springs to neaps, of two to one;
but the correct answer is that the total available power of a travelling wave
scales as its amplitude squared, so the springs-to-neaps
ratio of total-incoming-power is four to one.
% \section{Back to the shallow-water tidal wave model}
\subsection{Effect of shelving of sea bed, and Coriolis force}
If the depth $d$ decreases gradually and the width remains constant
such that there is minimal reflection or absorption of the incoming power,
then the power of the wave will remain constant.
This means
$\sqrt{d} h^2$ is a constant, so we
deduce that the height of the tide scales with depth
as
$h \sim 1/d^{1/4}$.
%% st kilda: springs = 2.8m, neaps = 1.3
%% shillay springs = 3.5m, neaps >= 1.5, maybe 1.8
%% /home/mackay/sustainable/refs/DTIAtlas> xv Z100.gif M100.gif
%\subsection{Application to the UK}
This is a crude model.
One neglected detail is the Coriolis effect.
The Coriolis force causes tidal crests and troughs to
tend to drive on the right -- for example,
going up the English Channel, the high tides are higher
and the low tides are lower on the French side of the
channel. By neglecting this effect I may have introduced
some error into the estimates.
% tide is like wind
\section{Power per unit area of tidal stream farms}
Imagine sticking underwater windmills on the sea-bed.
The flow of water will turn the windmills.
Because the density of water is roughly 1000 times that of
air, the power of water flow is 1000 times
greater than the power of wind at the same speed.
% (see \pref{eq.windpower1} for the equations).
% (equations \ref{eq.windpower1}--\ref{eq.windpower3}).
\begin{figure}[tp]\small
\figuredangle{
\begin{center}
\begin{tabular}{@{}c@{\,\,\,\,\,\,\,\,\,\,}c}
% \fullmoon \leftmoon \newmoon \rightmoon \fullmoon \\
\raisebox{1.162in}{\makebox[0in][l]{\footnotesize speed (m/s)\raisebox{-2.3mm}{%
\newmoon \hspace{24.13mm} \rightmoon \hspace{24.13mm} \fullmoon }}}
\ \ \ \ \ {\mbox{\epsfxsize=3.124in%
\mono{\epsfbox{../tide/mono/current.eps}}
{\epsfbox{../tide/current.eps}}%
}}%
&
\raisebox{1.162in}{\makebox[0in][l]{\footnotesize power (\Wmm)\raisebox{-2.3mm}{%
\hspace*{-7.6mm}\newmoon \hspace{54.3mm} \rightmoon }}}
\ \ \ \ \ {\mbox{\epsfxsize=3.124in\mono%
{\epsfbox{../tide/mono/power.eps}}%
{\epsfbox{../tide/power.eps}}%
}} \\[0.12in]
\multicolumn{1}{r}{\footnotesize time (days)} &
\multicolumn{1}{r}{\footnotesize time (days)} \\[-0.12in]
{\footnotesize (a)} & {\footnotesize (b)}
\\
\end{tabular}
\end{center}
}{
\caption[a]{(a) Tidal current over a 21-day period
% as a function of time
at a
% North Sea
location where the maximum current
at spring tide is 2.9\,knots (1.5\,m/s)
and the maximum current at neap tide is 1.8\,knots (0.9\,m/s).
(b) The power per unit sea-floor area over a nine-day period
extending from spring tides to neap tides.
The power peaks four times per day, and has a maximum of about
27\,\Wmm.
The average power of the tide farm
% oct.m 25.5W/m^2
is 6.4\,\Wmm.
}
\label{tidepoolNS}
}
\end{figure}
%
What power could tidal stream farms extract?
It depends crucially on whether or not we
can add up the power contributions
of tidefarms on {\em{adjacent}\/} pieces of sea-floor.
% densities per unit area
% sea-floor for many adjacent pieces of sea-floor.
For wind, this additivity assumption is
believed to work fine:
as long as the wind turbines are spaced a standard distance
apart from each other, the total power delivered by
10 adjacent {\windfarm}s is the sum of the
powers that each would deliver if it were alone.
{% begin troublesomepage hack
% this should be *before* the start of troublesome page
\renewcommand{\floatpagefraction}{0.8}
Does the same go for {\tidefarm}s?
Or do underwater windmills interfere with each other's
power extraction in a different way?
I don't think the answer to this question is known in general.
We can name two alternative assumptions, however,
and identify cartoon situations in which each assumption seems valid.
The
``\ind{tide is like wind}'' assumption says that you can put
tide-turbines
all over the sea-bed, spaced about 5 diameters apart from each other,
and they won't interfere with each other, no matter how much of the
sea-bed you cover with such {\tidefarm}s.
% This assumption seems to me
% to be valid if the heights of the turbines are small compared to the
% water depth. I think it might be valid even for tall turbines, but
% I'm not sure.
The ``\ind{you can have only one row}'' assumption, in contrast,
asserts that the maximum
power extractable in a region is the power that would be delivered by
a {\em{single}\/} row of turbines facing the flow.
A situation where this assumption is correct is the special case of
a hydroelectric dam: if the water from the dam passes through a single
well-designed turbine, there's no point putting any more turbines behind that one.
You can't get 100 times more power by putting 99 more
turbines downstream from the first.
The oomph gets extracted by the first one, and
there isn't any more oomph left for the others.
The ``\ind{you can have only one row}'' assumption is the right assumption
for estimating the extractable power in a place where
water flows through a narrow channel
from approximately stationary water at one height
into another body of water at a lower height.
(This case is analysed by \citet{Garrett05,Garrett07}.)
% -- from the Atlantic into the Mediterranean through the Strait of Gibraltar, for
% example.
I'm now going to nail my colours to a mast. I think that in many places
round the British Isles, the ``\ind{tide is like wind}'' assumption
is a good approximation.
Perhaps some spots have some of the character of a narrow channel.
% the Straits of Gibraltar.
In those spots, my estimates may be over-estimates.
Let's assume that
the rules for laying out a sensible {\tidefarm} will be similar to
those for {\windfarm}s, and that the efficiency of the tidemills
will be like that of the best windmills, about $1/2$.
We can then steal the formula for the power of a {\windfarm} (per unit land area)
from \pref{eq.windpower5}.
The power per unit sea-floor area is
\beqan
\frac{ \mbox{power per tidemill}}
{ \mbox{area per tidemill} }
%& =& \frac{ \frac{1}{2} \rho v^3 \frac{\pi}{8} d^2 }
% { (5 d)^2 }
%\label{eq.tidepower4}
%\\
& = & \frac{\pi}{200} \frac{1}{2} \rho U^3 .
\label{eq.tidepower5}
\eeqan
Using this formula, table \ref{tidetable} shows this {\tidefarm} power
for a few tidal currents.
\amargintab{c}{
\begin{center}
% 1 knot = 0.514444444444444444 m/s
\begin{tabular}{ccc} \toprule
\multicolumn{2}{c}{$U$} & {\tidefarm} \\
(m/s) & (knots) & power\\
& & (\Wmm) \\ \midrule
0.5 & 1 & 1 \\
1 & 2 & 8 \\
2 & 4 & 60 \\ % 62.8
3 & 6 & 200 \\ % 212
4 & 8 & 500 \\ % 502
5 & 10 % 9.72
& 1000 \\ % 981.7
\bottomrule
\end{tabular}
\end{center}
%}{
\caption[a]{{\Tidefarm} power per unit area
(in watts per square metre
of sea-floor) as a function of flow speed $U$.
(1 knot = 1 nautical mile per hour = 0.514\,m/s.)
The power per unit area is computed using
$\frac{\pi}{200} \frac{1}{2} \rho U^3$
(\protect\eqref{eq.tidepower5}).
}
\label{tidetable}
}%
Now, what are typical tidal currents?
Tidal charts usually give the currents
associated with the tides with the largest range (called spring tides)
and the tides with the smallest range (called neap tides).
Spring tides occur shortly after each full moon and each new moon.
Neap tides occur shortly after the first and third quarters of the moon.
The power of a tide farm would vary throughout the day
in a completely predictable manner.
\Figref{tidepoolNS} illustrates the variation
of power per unit area of a {\tidefarm} with a maximum current
of 1.5\,m/s. The average power per unit area of this {\tidefarm}
would be 6.4\,\Wmm.
There are many places around the British Isles where the power per unit
area of \tidefarm\ would be
% sea-floor is
$6\,\Wmm$ or more.
This power per unit area is similar to our estimates of
the powers per unit area of {\windfarm}s (2--3\,\Wmm)
and of photovoltaic solar farms (5--10\,\Wmm).
% And about the same as our estimate of the power density of
% an off-shore {\windfarm} ($5\W/\m^2$).
% Tide power is not to be sneezed at!
% How would it add up, if we assume that there are no economic
% obstacles to the exploitation of tidal power at all the
% hot spots around the UK?
We'll now use this ``tide farms are like {\windfarm}s'' theory
to estimate the extractable power from tidal streams in promising
regions around
the British Isles. As a sanity check, we'll also
work out the total tidal power crossing each of these regions,
using the ``power of tidal waves'' theory,
to check our {\tidefarm}'s estimated power isn't bigger than the total power
available.
\label{maintide}The main
locations around the British Isles where tidal currents are large are
shown in \figref{tideUK}.
\begin{figure}
\figuremarginwidecapab{
%\figuremargin{
\begin{center}\vspace*{-2.5mm}
\mbox{\epsfxsize=2.72in\epsfbox{../../images/PUBLICDOMAIN/Tides.eps}}
\end{center}
}{
\caption[a]{Regions around the British Isles where
peak tidal flows exceed 1\,m/s.
The six darkly-coloured regions are included
in table \protect\ref{tab.tide}:
\begin{enumerate}
\item {} \
the English channel (south of the Isle of Wight);
\item {} \
the Bristol channel;
\item {} \
to the north of Anglesey;
\item {} \
to the north of the Isle of Man;
\item {} \
between Northern Ireland, the Mull of Kintyre, and Islay; and
\item {} \ the
Pentland Firth (between Orkney and mainland Scotland),
and within the Orkneys.
\end{enumerate}
There are also enormous
currents around the Channel Islands, but they are not governed by
the UK\@. Runner-up regions include
the North Sea, from the Thames (London) to the Wash (Kings Lynn).
The contours show water depths greater than 100\,m.
Tidal data are from Reed's Nautical
Almanac and DTI\ Atlas of UK Marine Renewable
Energy Resources (2004).
}
\label{tideUK}
}{90mm}{80mm}
\end{figure}
\begin{table}
\figuremargin{\small
\begin{tabular}{@{}c@{\,}c@{}}
\begin{tabular}{cc@{\,\,\,}r@{\,\,\,}crr}\toprule
{Region} & \multicolumn{2}{c}{$U$}%{peak current}
&
power & area & average \\
& \multicolumn{2}{c}{(knots)} &
density & & power \\
& N & S &
(\Wmm) & (km$^2$) & (kWh/d/p) \\ \midrule
1
%English Channel (South of Isle of Wight)
%%% power-density area power
& 1.7 & 3.1 & 7 & 400 & \OliveGreen{1.1} \\ % was 2.82
2
% Bristol Channel
& 1.8 & 3.2 & 8 & 350 & \OliveGreen{1.1} \\ % 2.37
3
% Irish Sea (near Anglesey)
& 1.3 & 2.3 & 2.9 & 1000 & \OliveGreen{1.2} \\ % 1.178
4
% North of Isle of Man
& 1.7 & 3.4 & 9 & 400 & \OliveGreen{1.4} \\ % 1.40
% Irish Sea (Ireland) & 2.0 & 3.0 & 7.5 & 1000 & 3.0$^*$ \\ % 3.00
5
% Between Northern Ireland, Islay and Mull of Kintyre
&
1.7 & 3.1 & 7 & 300 & \OliveGreen{0.8} \\ % 2.82
%% the sea area of PF is definitely 100 km^2 but let's say 50km^2. was 1.4, now 7.0, now 3.5
6
% Pentland Firth
& 5.0 & 9.0 & 170 & 50 & \OliveGreen{3.5} \\ % 5.57 using oct.m Aka 8.5GW
% North Sea near Cromer
% % Kings Lynn 3.5+1.7= 5.2
% & 1.2 & 2.2 & 2.5 & 1000 & 1.0 \\ % 4.02 using oct.m
% Southern North Sea & 1.1 &1.9 & 1.7 & 1000 & 0.7 \\ % 0.676
\midrule
\multicolumn{3}{l}{Total} & & & \OliveGreen{{\bf 9}} \\ % 16.67 including Ireland part
\bottomrule
\end{tabular}
&
\begin{tabular}{cccc}\toprule
%\multicolumn{2}{c}{Region} &
& & \multicolumn{2}{c}{raw power} \\%& {\tidefarm}\\
%& &
$d$% depth
&
$w$ %width
& N & S \\%& estimate\\
%& &
(m) & (km) & \multicolumn{2}{c}{(kWh/d/p)} \\
\midrule
%% B+V: depth 30-40 or >40 Best bit has depth 30m. width 13km
%1&
%English Channel (South of Isle of Wight)
% &
30 & 30 & 2.3 & 7.8 \\%& 1.1 \\ %%% CAUTION: my est is 25% of TOTAL was 2.8,now 1.1
%% B+V: depth 25-40 / 30-40
%2&
%Bristol Channel
%&
30 & 17 & 1.5 & 4.7 \\%& 1.1 \\ %%% CAUTION: my est is 50% of TOTAL was 2.4 now 1.2
%% > 40m aka Carmel Head depth=35m width=7.5km
%3&
%Irish Sea (near Anglesey)
% &
50 & 30 & 3.0& 9.3 \\%& 1.2 \\ %%% looks fine
%% much of it is >40
%4&
% North of Isle of Man
%&
30 &20 & 1.5& 6.3 \\% & 1.4 \\ %%% looks fine
%% Mainly >40m page 27 says depth=29/30m width=7.3km BUT there is a height node at Islay so mistrust
%5&
% Between Northern Ireland, Islay and Mull of Kintyre
%&
40 & 10 & 1.2 & 4.0 \\%& 0.8 \\ %%% CAUTION: my est is 100% of TOTAL was 2.8, now 0.8
%6&
% Pentland Firth
%&
70 & 10 & 24 & 78 \\%& 3.5 \\ %%% OK
\midrule
\,
% \multicolumn{2}{l}{Total} & & & & & {\bf 9}
\\ %
\bottomrule
\end{tabular}
\\
(a)&(b)\\
\end{tabular}
}{
\caption[a]{(a) Tidal power estimates assuming that stream farms are like {\windfarm}s.
The power density is the average power per unit area of sea floor.
%% Total power is divided by the population of UK.
% (*) The estimated Irish Sea (Ireland) production
% would be in the territorial waters of Ireland.
The six regions are indicated in \protect\figref{tideUK}.
% This table lists them clockwise starting from London.
N = Neaps.
S = Springs.
(b) For comparison, this table
% compares the \tidefarm\ estimates
% of extractable power of \tabref{tab.tide}(a) with
shows the raw incoming power
estimated using \eqref{eqRawPo} (\pref{eqRawPo}).
% on \pref{sec.powerT}.
% The six regions are indicated in \protect\figref{tideUK}.
% This table lists them clockwise starting from London.
}
\label{tab.tide}\label{tab.tide2}
}
\end{table}
%% Of all the estimates in this book, this estimate of
% the tidal resource of Britain is the one I am least certain
% about.
I estimated\label{pSteal1} the typical peak currents at six locations
with large currents by looking at
tidal charts in {\tem{Reed's Nautical Almanac}}. (These estimates
could easily be off by 30\%.)
Have I over-estimated or under-estimated
the area of each region? I haven't surveyed the sea floor
so I don't know if some regions might be unsuitable in some
way -- too deep, or too shallow, or too tricky to build on.
% I've included only six small regions with large
% currents. Perhaps I should also have included an
% estimate for the power from the much larger regions
% with small currents.
% according to my oct estimate of the backup region
% in the north sea of area 10,000 km^2 with currents (0.6kn, 1kn)
% we'd get an extra 1 from that.
% I can imagine my estimate might be off by a factor of four
% in either direction.
Admitting all these uncertainties, I arrive at an estimated
total power of \OliveGreen{9\,kWh/d per person} from tidal
stream-farms.
% ; plus another 2\,kWh/d per person from tidal lagoons and tidal barrages.
This corresponds to 9\% of the raw incoming power
mentioned on \pref{ptide100}, 100\,kWh per day per person.
% The total power crossing the lines in \figref{fig.TideLine}
% has been measured; on average it amounts to 100\,kWh per day per person.
(The extraction of 1.1\,kWh/d/p in the Bristol channel, region 2,
might conflict with power generation by the Severn barrage; it would depend on
whether the {\tidefarm} significantly {\em{adds}\/} to the existing natural
friction created by the channel,
or {\em{replaces}\/} it.)
%\subsection{Double-check}
% In table \ref{} I use
% the total power per unit length $\rho g^{3/2} \sqrt{ d } h^2$
%% U = sqrt(gd) h/d. U = sqrt(g) h/sqrt(d). h = U sqrt(d)/sqrt(g)
% $= \rho g^{3/2} \sqrt{ d } U^2 d/g$
% $= \rho g^{1/2} { d }^{3/2} U^2$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Power(d,U) = 1000 * sqrt(9.81) * d**(1.5) * (0.51444*U)**2
% power in W/m. w is in km, so mul by 1000 for m, then divide by 1000 for kW
% M(U,d,w) = w*Power( d,U ) * 24.0 / 60e6
% 1 knot = 0.514444444444444444 m/s
%%% almost all these large areas offer 1.5-2.5 m/s only.
% _tide.tex contains another version of this table
\section{Estimating the tidal resource via bottom friction}
Another way to estimate the power available from tide is
to compute how much power is already dissipated by friction on
the sea floor.
A coating of turbines placed just above the sea floor
could act as a substitute bottom, exerting roughly
the same drag
on the passing water as the sea floor used to exert,
and extracting roughly the same
amount of power as friction used to dissipate,
without significantly altering the
tidal flows.
So, what's the power dissipated by ``bottom friction''?
Unfortunately, there isn't a straightforward
model of bottom friction. It depends on the roughness of the
sea bed and the material that the bed is made from -- and even given
this information, the correct formula to use is not settled.
One widely used model says that
the magnitude of the stress (force per unit area) is
$R_1 \rho U^2$, where $U$ is the average flow velocity
and $R_1$ is a dimensionless quantity called the shear
friction coefficient.
We can estimate the power dissipated per unit area
by multiplying the
stress by the velocity.
Table \ref{tidefrictiontable} shows the power dissipated in friction,
$R_1\rho U^3$, assuming $R_1 = 0.01$ or $R_1 = 0.003$.
% (Add citations.)
% see _tide.tex
\begin{table}
\figuremargin{
\begin{center}
% 1 knot = 0.514444444444444444 m/s
\begin{tabular}{ccrrr} \toprule
$v$ & $v$ & \multicolumn{2}{c}{Friction power} & {\tidefarm} power \\
(m/s) & (knots) & \multicolumn{2}{c}{density (\Wmm)} & density \\
& & $R_1=0.01$ & $R_1=0.003$ & (\Wmm) \\ \midrule
0.5 & 1 & 1.25 & 0.4 & 1 \\
1 & 2 & 10 & 3 & 8 \\
2 & 4 & 80 & 24 & 60 \\ % 62.8
3 & 6 & 270 & 80 & 200 \\ % 212
4 & 8 & 640 & 190 & 500 \\ % 502
5 & 10 % 9.72
& 1250 & 375 & 1000 \\ % 981.7
\bottomrule
\end{tabular}
\end{center}
}{
\caption[a]{Friction power per unit area $R_1\rho U^3$ (in watts per square metre
of sea-floor) as a function of flow speed, assuming $R_1 = 0.01$
or $0.003$. \citet{Flather1976} uses $R_1=0.0025$--$0.003$;
\citet{Taylor1920} uses 0.002.
% (I don't have a factor of $1/2$ in the formula.)
(1 knot = 1 nautical mile per hour = 0.514\,m/s.)
The final column shows the {\tidefarm} power estimated in \tabref{tidetable}.
For further reading see \cite{Kowalik04}, \cite{Sleath}.
}
\label{tidefrictiontable}
}
\end{table}
For values of the shear
friction coefficient in this range, the
friction power is very similar to the estimated power that a tide farm
would deliver. This is good news, because it suggests that
planting a forest of underwater windmills on the sea-bottom,
spaced five diameters apart, won't radically alter the
flow.
% in the region.
The natural friction already has an
effect that is in the same ballpark.
%I disagree somewhat with the friction coefficient
%that Stephen uses in his calculations. I have a paper where he uses a
%"friction value" of 0.017, multiplied by an area of 23 x 10 km^2 and a
%velocity of 3 m/s to arrive at a power dissipation of 54 GW\@. In contrast
%to this I have an oceanography paper which gives C_d = 0.0026, and I am
%informed by an oceanographer colleague that 0.0026 is sensible.
}% end troublesoem page hack
\section{Tidal pools with pumping}
``The pumping trick''
artificially increases the amplitude of the tides
in a tidal pool so as to amplify the power obtained.
The energy cost of pumping {\em{in}\/} extra water at high tide is
repaid with interest when the same water is let {{out}} at low tide;
similarly, extra water can be pumped {\em{out}\/}
at low tide, then let back in at high tide.
The pumping trick is sometimes used at La Rance, boosting its
net power generation by about 10\%
% 494 GWh net gernated in 1984, when pumping was happening 22.6% of time
% and genreating was happening 73.4% + 4% of the time
% total generated was 609 GWh page 90 of Novak
\citep{WilsonBallsTide}.
Let's work out the theoretical limit for this technology.
% Two-way generation in CHina: there's only one big plant, at Jiangxia, which
% generates 12 GWh/y
% see also storage.tex where same calcns are done
I'll assume that generation has an efficiency of $\epsilon_{\rm{g}} = 0.9$
and that pumping has an efficiency of $\epsilon_{\rm{p}} = 0.85$.
% (These figures are based on the pumped storage system at Dinorwig.)
% , whose round-trip efficiency is about 75\%.)
Let the tidal range be $2h$.
I'll assume for simplicity that the prices of buying and selling electricity
are the same at all times,
% high tide and low tide,
so that the optimal height boost $b$ to which the pool is pumped above
high water is
given by (marginal cost of extra pumping = marginal return of extra water):
\[
b / \epsilon_{\rm{p}} = \epsilon_{\rm{g}} ( b + 2h ) .
\]
Defining the round-trip efficiency
$\epsilon = \epsilon_{\rm{g}}\epsilon_{\rm{p}}$, we have
\[
b = 2h \frac{\epsilon }{1- \epsilon } .
\]
For example, with a tidal range of $2h=4\,\m$, and a round-trip efficiency
of $\epsilon=76\%$, the optimal boost is $b=13\,\m$.
This is the maximum height to which pumping can be justified if the
price of electricity is constant.
Let's assume the complementary trick is used at low tide. (This requires
the basin to have a vertical range of 30\,m!)
The delivered power per unit area is then
\[
\left. \left( \frac{1}{2} \rho g \epsilon_{\rm{g}} ( b + 2h )^2
- \frac{1}{2} \rho g \frac{1}{\epsilon_{\rm{p}}} b^2 \right) \right/ T ,
\]
where $T$ is the time from high tide to low tide.
We can express this as the maximum possible
power per unit area without pumping,
$\epsilon_{\rm{g}} { 2 \rho g h^2}/{ T }$,
scaled up by a boost factor
\[
% \left( \frac{1}{1- \epsilon} \right)^2
%- \frac{1}{\epsilon} \left( \frac{\epsilon}{1- \epsilon} \right)^2
\left( \frac{1}{1- \epsilon} \right),
\]
which is roughly a factor of 4.
\begin{table}\figuremargin{
%% see figs/tide.gnu
\begin{center}
\begin{tabular}{cccc}\toprule
tidal amplitude & optimal boost & power & power \\
(half-range) $h$ & height $b$ & with pumping & without pumping \\
(m) & (m) & (\Wmm) & (\Wmm) \\
\midrule
% 0.5 & 3.3 & 0.9 & 0.2\\
1.0 & 6.5 & 3.5 & 0.8\\
2.0 & 13\phantom{.01} & 14 & 3.3\\
3.0 & 20\phantom{.02} & 31 & 7.4\\
4.0 & 26\phantom{.02} & 56 & 13 \\
%0.5 & 3.255 & 0.8696 & 0.2
%1.0 & 6.510 & 3.4787 & 0.8
%2.0 & 13.02 & 13.914 & 3.3
%3.0 & 19.53 & 31.308 & 7.4
%4.0 & 26.04 & 55.659 & 13 \\
\bottomrule
\end{tabular}
\end{center}
}{\caption[a]{Theoretical power per unit area from tidal
power using the pumping trick,
assuming no constraint on the height of the basin's walls.}\label{figTidePump1}
}
\end{table}
\Tabref{figTidePump1} shows the theoretical power per unit area
that pumping could deliver.
Unfortunately, this pumping trick will rarely be exploited
to the full because of the economics of basin construction:
full exploitation of pumping requires the total height of the pool
to be roughly 4 times the tidal range, and increases the delivered
power four-fold. But the amount of material in a sea-wall of height $H$ scales
as $H^2$, so the cost of constructing
a wall four times as high will be
more than four times as big. Extra cash would probably be better spent on
enlarging a tidal pool horizontally rather than vertically.
The pumping trick can nevertheless be used for free on any
day when
the range of natural tides is smaller than the maximum range:
the water level at high tide can be pumped up to the maximum.
Table \ref{tab.Pump2} gives the power delivered if the boost height
is set to $h$, that is, the range in the pool is just double the external range.
\begin{table}\figuremargin{
\begin{center}
\begin{tabular}{cccc}\toprule
tidal amplitude & boost height & power & power \\
(half-range) $h$ & $b$ & with pumping & without pumping \\
(m) & (m) & (\Wmm) & (\Wmm) \\ \midrule
% 0.5 &0.5 & 0.4 &0.2\\
1.0 &1.0 & 1.6 &0.8\\
2.0 &2.0 & 6.3 &3.3\\
3.0 &3.0 & 14\phantom{.01} &7.4\\
4.0 &4.0 & 25\phantom{.02} &13\phantom{.01} \\
%0.5 &0.5 & 0.39305 &0.2\\
%1.0 &1.0 & 1.57221 &0.8\\
%2.0 &2.0 & 6.28887 &3.3\\
%3.0 &3.0 & 14.1499 &7.4\\
%4.0 &4.0 & 25.1554 &13\\
\bottomrule
\end{tabular}
\end{center}
}{
\caption[a]{Power per unit area offered by the pumping trick,
assuming the boost height is constrained to be the same as the
tidal amplitude.
This assumption applies, for example, at neap tides, if the
pumping pushes the tidal range up to the springs range.
}\label{tab.Pump2}
}
\end{table}
A doubling of vertical range
is easy at neap tides, since neap tides are typically
about half as high as spring tides.
Pumping the pool at neaps so that the full springs range
is used thus allows neap tides to deliver roughly twice as much
power as they would offer without pumping. So a system with pumping
would show two-weekly variations in power of just a factor
of 2 instead of 4.
\subsection{Getting ``always-on'' tidal
power by using two basins}
Here's a neat idea:\label{pHaishan}
have \index{tidal power!two basins}two basins, one of which is the ``full'' basin
and one the ``empty'' basin; every high tide, the full basin is topped up;
every low tide, the empty basin is emptied.
These toppings-up and emptyings could be done
either passively through sluices, or
actively by pumps (using the trick mentioned
above).
Whenever power is required, water is allowed to flow
from the full basin to the empty basin, or (better in power terms)
between one of the basins and the sea.
% The power per unit area is the same
%%%%%%%%%%%%%% as that of the single-basin tidal barrage;
The capital cost of a two-basin scheme may be bigger because
of the need for extra walls;
the big win is that power is available all the time,
so the facility can follow demand.
% http://www.fibrowatt.com/tidal-power.html
% http://cat.inist.fr/?aModele=afficheN&cpsidt=9408982
We can use power generated from the empty basin to pump
extra water into the full basin at high tide,
and similarly use power from the full basin to pump down the
empty basin at low tide. This self-pumping
would boost the total power delivered by the facility
without ever needing to buy energy from the grid.
It's a delightful feature of a two-pool solution that
the optimal time to {\em{pump}\/} water into the high pool is high tide, which
is also the optimal time to {\em{generate}\/} power from the low pool.
Similarly, low tide is the perfect time to pump down the low pool,
and it's the perfect time to generate power from the high pool.
In a simple simulation, I've found that a two-lagoon
system in a location with a natural tidal range of 4\,m
can, with an appropriate pumping schedule, deliver a {\em{steady}\/}
power of 4.5\,\Wmm\
\citep{MacKayLagoons}. One lagoon's water level is always kept
above mean sea-level; the other lagoon's level is always kept below
mean sea-level.
This power per unit area of 4.5\,\Wmm\ is 50\% bigger than
the maximum possible average
power per unit area of an ordinary tide-pool in the same location (3\,\Wmm).
The steady power of the lagoon system would be more valuable than
the intermittent and less-flexible power from the ordinary tide-pool.
A two-basin system could also function as a pumped-storage
facility.
% I'll come back to this in chapter \ref{ch.storage}.
% (\pref{pTESS})
%{\em (Perhaps move all this material to one place). }
\begin{figure}
\figuremargin{\small
\begin{center}
\begin{tabular}{cc}
{\mbox{\epsfxsize=52mm\epsfbox{figs/TideStore.eps}}} &
{\mbox{\epsfxsize=52mm\epsfbox{figs/TideStore2.eps}}} \\
(a)&(b)\\
\end{tabular}
\end{center}
}{
\caption[a]{Different ways to use the tidal pumping trick.
Two lagoons are located at sea-level.
(a) One simple way of using two lagoons
is to label one the high pool and the other the low pool; when
the surrounding sea level is near to high tide, let
water into the high pool, or actively pump it in (using electricity
from other sources);
and similarly, when the sea level is near to low tide, empty
the low pool, either passively or by active pumping;
then whenever power is sufficiently valuable, generate power on
demand by letting water from the high pool to the low pool.
(b)
Another arrangement that might deliver more power per unit area
has no flow of water between the two lagoons.
While one lagoon is being pumped full or pumped
empty, the other lagoon can deliver steady, demand-following
power to the grid.
Pumping may be powered by bursty sources such as wind, by spare power
from the grid (say, nuclear
power stations), or by the other half of the facility,
using one lagoon's power to pump the other lagoon up or down.
% to a greater height.
}\label{fig.TESS}
}
\end{figure}
\small
\section{Notes}
\beforenotelist
\begin{notelist}
\item[page no.]
\item[\npageref{Turbine90}]
{\nqs{Efficiency of 90\%\ldots}}
Turbines are about 90\%
efficient for heads of 3.7\,m or more.
\cite{BakerSwansea}.
% December
\item[\npageref{pHaishan}]
{\nqs{Getting ``always-on'' tidal
power by using two basins.}}
% There is a two-basin tidal power plant at \ind{Haishan}.
% \subsection{Haishan}
There is a two-basin tidal power plant
at \ind{Haishan}, Maoyan Island, \ind{China}. A single generator
located between the two basins, as shown in \figref{fig.TESS}(a),
% can generate up to 75\,kW,
delivers power continuously, and generates 39\,kW on average.
% average power =
\tinyurl{2bqapk}{http://wwwphys.murdoch.edu.au/rise/reslab/resfiles/tidal/text.html}.
% The Haishan TPP is noteworthy as it is the only linked-basins plant in existence in the world a plant featuring a high and a low-basin with the power plant in between these two basins, generating energy from water flowing from the high into the low-basin. The plant is located on Maoyan Island in Zhejiang province where it serves an isolated community of 760 families. The plant was designed for two 75kW units of which only one was installed and commissioned in 1975. This unit operated continuously. The energy was used partly to pump fresh water for domestic and irrigation use into the community reservoir. The plant has since been upgraded to an installed capacity of 0.25MW, producing 0.34GWh per year (Wilmington Media Ltd, 2004).
%% authors: Katrina O'Mara and Mark Rayner, with assistance from Philip Jennings (Murdoch University) in June 1999. Edited and Updated by Katrina Lyon in June 2004 and Mark McHenry in January 2006.
%% see also
%% http://www.waterpowermagazine.com/story.asp?storyCode=2022354
%% http://www.waterpowermagazine.com/story.asp?storyCode=2022354
%% recommends using a flow-through layout -- in at one end, out at other.
\item[Further reading:]
\cite{ShawWatsonTrading,Blunden,CharlierTideArticle,TidalStreamCharlier}.
For further reading on
bottom friction and variation of flow with depth, see \cite{Sleath}.
For more on the estimation of the UK tidal
resource,\index{tide!UK resource}
see \citet{MacKayUnderestimationTide}.
For more on \ind{tidal lagoon}s,
see \citet{MacKayLagoons}.
\end{notelist}
\normalsize
% \section{Notes}
% Area behind the Bristol barrage is 480\,km$^2$.
% So with average output of 2\,GW, it would indeed deliver 4\,\Wmm.
%% Actually 185 sq mi from the FOE site.
%% 4.536e+8 m² / 4.536e+4 hectares / 453.6 km² / 4.882e+9 ft² / 1.121e+5 acres / 175.1 mile²
%% measured by me
%% http://www.acme.com/planimeter/