\section{The physics of deep-water waves}
% \index{cheese@gouda}
% Would place gouda in the list right next to cheese, as if gouda was actually spelled cheese
Waves contain energy\index{wave}\index{energy!wave}
in two forms: \ind{potential energy}, and \ind{kinetic energy}.
The potential energy is the energy required to move all the water
from the troughs to the crests.
The kinetic energy is associated with the water moving around.
People sometimes assume that when the crest of a wave moves across
an ocean at 30\,miles per hour, the water in that crest must also
be moving at 30\,miles per hour in the same direction. But this isn't so.
It's just like a \ind{Mexican wave}\index{wave!Mexican}.
When the wave rushes round the \ind{stadium},
the humans who are making the wave aren't themselves moving round the stadium:
they just bob up and down a little. The motion of a piece of water
in the ocean is similar: if you focused on a bit of seaweed floating
in the water as waves go by, you'd see that the seaweed moves up and down,
and also a little to and fro in the direction
of travel of the wave -- the exact effect could
be recreated in a Mexican wave if people moved like window-cleaners,
polishing a big piece of glass in a circular motion.
The wave has potential energy because of the elevation of the crests
above the troughs. And it has kinetic energy because of
the small circular bobbing motion of the water.
\marginfig{
\begin{center}
\hspace*{3mm}\begin{tabular}{l}
%\mbox{\ \ \epsfxsize=45mm\epsfbox{../wave/wPowerDen.eps}} \\[0.16in]
\mbox{\epsfxsize=49mm\epsfbox{../wave/wWindSpeed.eps}} \\[0.16in]
\mbox{\epsfxsize=49mm\epsfbox{../wave/wPeriod.eps}} \\[0.16in]
\mbox{\epsfxsize=49mm\epsfbox{../wave/wWavelength.eps}} \\[0.16in]
\mbox{\epsfxsize=49mm\epsfbox{../wave/wPower.eps}} \\[0.16in]
\end{tabular}
\end{center}
\caption[a]{
Facts about deep-water waves.\index{wave!deep-water}
In all four figures the horizontal axis is the wave speed in m/s.
From top to bottom the graphs show:
wind speed (in m/s) required to make a wave with this wave speed;
period (in seconds) of a wave;
wavelength (in m) of a wave;
and power density (in kW$\!$/m) of a wave with amplitude 1\,m.
}
\label{fig.waveTheory}
}
%% power density per unit area is 1000 W/m**2 for a wave of
%% speed 7.5 m/s (ie period 5s) made by a force 7 wind.
\begin{figure}
% \figuremargin{
\figuredangle{
\begin{center}
%\mbox{\epsfxsize=3in\epsfbox{../../images/wave.eps}}
\mbox{\epsfbox{metapost/wave.1}}
\end{center}
}{
\caption[a]{A \ind{wave}
% , yesterday. 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 within a few wavelengths
of the surface -- the speed of the water is 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.
}\label{waveYesterday}
}
\end{figure}
Our rough calculation of the power in ocean waves will require
three ingredients:
an estimate of the period $T$ of the waves (the time between crests),
an estimate of the height $h$ of the waves,
and a physics formula\index{formula!wave}
that tells us how to work out the speed $v$ of the
wave from its period.
The \ind{wavelength} $\lambda$
and \ind{period} of the \ind{wave}s (the distance and time
respectively between
two adjacent crests) depend on the \ind{speed} of the \ind{wind}
that creates the waves, as shown in \figref{fig.waveTheory}.
The height of the waves doesn't depend on the windspeed; rather, it
depends on how long the wind has been caressing the water surface.
You can estimate the period of ocean waves
by recalling the time between
waves arriving on an ocean \ind{beach}. Is 10 seconds reasonable?
For the height of ocean waves, let's assume an amplitude of 1\,m, which
means 2\,m from trough to crest. In waves this high, a man in a dinghy can't
see beyond the nearest crest when he's in a trough;
I think this height is bigger than average, but we can revisit this estimate if
we decide it's important.
The speed of deep-water waves is related to
the time $T$ between crests by the physics formula
(see \citeasnoun{Faber}, p170):
\[
v = \frac{g T}{2 \pi} ,
\]
where $g$\index{g@$g$} is the acceleration of \ind{gravity} (9.8\,m/s$^2$).
For example, if $T = 10$ seconds, then
$v = 16\,\m/\s$.
The \ind{wavelength} of such a wave -- the distance between crests --
\index{$\lambda$}\index{l@$\lambda$}is $\lambda = vT = g T^2 / 2\pi = 160\,\m$.
For a wave of wavelength $\lambda$
and period $T$,
% = 10$ seconds,
if the height of
each crest and depth of each trough
is $h = 1$\,m, the potential energy passing per unit time, per unit length,
is
\beq
{P}_{\rm potential} \simeq m^* g \bar{h} / T,
\eeq
where $m^*$ is the mass per unit length,
which is roughly $\half \rho h (\lambda/2)$ (approximating the
area of the shaded crest in \figref{waveYesterday}
by the area of a triangle),
and $\bar{h}$ is the change in height of the centre-of-mass
of the chunk of elevated water, which is roughly $h$.
So
\beq
{P}_{\rm potential} \simeq \frac{1}{2} \rho h \frac{\lambda}{2} g h / T .
\eeq
(To find the potential energy properly, we should have done an integral
here; it would have given the same answer.)
Now $\lambda/T$ is simply the speed at which the wave travels, $v$, so:
\beq
{P}_{\rm potential} \simeq \frac{1}{4} \rho g h^2 v .
\eeq
Waves
have kinetic energy as well as potential energy,
and, remarkably, these are exactly equal, although I don't
show that calculation here;
so the total power of the waves is double the power calculated
from potential energy.
\beq
{P}_{\rm total} \simeq \frac{1}{2} \rho g h^2 v .
\label{eqPwrong}
\eeq
%% in page 17 of the dti atlas 'see appendix D1' they use
%% the formula
%% 0.0623 rho g H^2 c_g
%% where H is the ``significant wave height'' which is defined to be the height of the highest
%% one third of waves in the sea.
%% My height is probably H_rms ?? = H_s / 1.416
%% c_f = 0.5c (1 + 2 k h / sinh 2 k h )
%% (from appendix D) where h is vater depth and k is wavenumber and c = sigma/k.
%% c=gT/ 2 pi in deep water.
%% and c_g is the wave group speed, which is HALF of the phase velocity
%% My formula says
%% (1/2) rho g h^2 * (2 c_g )
%% which is 16 times bigger than their figure. WHAT????
%% I guess their definition of H is whacky
%% they start from (1/8) rho g h^2 * c_g
There's only one thing wrong with this answer: it's too big, because we've
neglected a strange property of dispersive waves: the energy in the wave doesn't actually
travel at the same speed as the crests; it travels at a speed called the group
velocity, which for deep-water waves is {\em{half}\/} of the speed $v$.
You can see that the energy travels slower than the
crests by chucking a pebble in a pond and watching the
expanding waves carefully.
% -- you'll see that the crests move faster then the disturbance
% itself. Rather paradoxical, but true.
What this means is that \eqref{eqPwrong}
is wrong: we need to halve it. The correct power per unit length of wave-front is
\beq
{P}_{\rm total} = \frac{1}{4} \rho g h^2 v .
\label{eqPwrongright}
\eeq
Plugging in $v=16\,\m/\s$ and $h=1\,\m$, we find
\beq
{P}_{\rm total} = \frac{1}{4} \rho g h^2 v
% = 0.5 \times 1000\, \times 10 x 1 x 1 \times 16.
= 40 \, \kW\!/\m .
\eeq
This rough estimate agrees with real measurements in the Atlantic
\citep{Mollison85}. (See \pref{pMolli}.)
% Deep water power in Australia averages 18\,kW/m
The losses from \ind{viscosity} are minimal:
a wave of 9\,seconds period would have to go three times round the
world to lose 10\% of its amplitude.
\section{Real wave power systems}
\subsection{Deep-water devices}
%% see _waves.tex
How effective are real systems at extracting power from waves?
\index{Salter, Stephen!duck}\index{duck, Salter}Stephen Salter's
``duck'' has been well characterized:
a row of 16-m diameter ducks, feeding off \ind{Atlantic} waves with
an average power of 45\,k\Wm, would
deliver 19\,k\Wm,
including transmission to central Scotland
\citep{Mollison85}.
% suck about 50\% of the power out
% of incoming \ind{Atlantic} waves, and the efficiency of the remaining
% steps (conversion to electricity, and
% transmission to a ``\ind{hydrodynamically underprivileged area}''
% such as London) would be about 60\%.
% Thus deep water Salter Ducks, feeding off 45\,k\Wm, would
% deliver 19\,k\Wm,
% including transmission to central Scotland
% \citep{Mollison85}.
The \ind{Pelamis}\index{sea snake} device, created by
\ind{Ocean Power Delivery},
has taken over the Salter duck's mantle as the
leading floating deep-water wave device.
Each snake-like device is 130\,m long and is made of
a chain of four segments, each
3.5\,m in diameter. It has a maximum power output of 750\,kW\@.
The Pelamises are designed to be moored in a depth of about 50\,m.
In a wavefarm, 39 devices in three rows would face the principal
wave direction, occupying an area of ocean
about 400\,m long and 2.5\,km wide (an area of 1\,km$^2$).
The effective cross-section of a single Pelamis is 7\,m (\ie, for
good waves, it extracts 100\% of the energy that would cross 7\,m).
The company says that such a wave-farm would deliver about 10\,k\Wm.
% (10\,MW/km).
% # width required for 400m at 55.8311, Longitude = -6.4561
% -- 55.8310, Longitude = -6.4479
% change in long of
\subsection{Shallow-water devices}
Typically 70\% of energy in ocean waves is lost
through \index{friction!waves}\index{bottom friction}bottom-friction
as the depth decreases from
100\,m to 15\,m.
% need a source for that
So the average wave-power per unit length of coastline
in shallow waters is reduced to about 12\,k\Wm.
% per metre.
The \ind{Oyster}, developed by \ind{Queen's University Belfast}
and \ind{Aquamarine Power Ltd} [\myurlb{www.aquamarinepower.com}{http://www.aquamarinepower.com/}],
is a bottom-mounted flap, about 12\,m high,
that is intended to be deployed in
waters about 12\,m deep,\nlabel{oyster}
%OysterÂ® modules are designed for deployment around 10m depth,
in areas where the average incident wave power is greater than 15\,k\Wm.
% \myurl{http://www.aquamarinepower.com/}
Its peak power is 600\,kW\@.
A single device would produce about 270\,kW in wave heights greater than 3.5\,m.
It's predicted that an Oyster would have
% Assume its width is about 20\,m? Then the ideal power for a device of
% cross-section 20\,m would be 240\,kW\@.
% A rough average figure is about 100\,kW\@.
% A smaller power than the Pelamis,
% but
a bigger power per unit mass of hardware than a Pelamis.\index{shallow water}\index{water!shallow}
% *** WHAT MASS oyster
% this no good:
% http://www.raeng.org.uk/policy/reports/pdf/energy_2100/Trevor_Whittaker.pdf
% *** Need to add cite here
Oysters could also be used to directly drive \index{reverse osmosis}{reverse-osmosis}
\ind{desalination} facilities.
``The peak freshwater output of an Oyster
desalinator is between 2000 and 6000\,m$^3$/day.''
That rate of
production has a value, going by the \ind{Jersey} facility (which uses
8\,kWh per m$^3$),
equivalent to
% 16\,000--48\,000\,kWh per day, that is,
600--2000\,kW of electricity.
%%% Need to add notes with citation. see sustainable/refs/wave/Oyster for papers, and accompanying email from jessica.owen for citations