The EON site in orkney2.gif is 10km long.
Raw power 400MW. Average power if 20\% efficient is 80MW.
Capacity of site is 100MW.
Costa Head site is about 6 km long.
6x40 is 240 MW.
Capacity 200 MW. OK.
In 500 kilometres of
\ind{wave-farm}, delivering {{1.2\,kWh/d per
person}}, there would be 3 million \tonnes\
of steel. That's roughly 40 times
the mass of the Magnus oil platform shown on page \pageref{Magnus} and
page \pageref{pTerryC2}.
(Which delivers 5\,GW; let's turn this into mass per MW
shall we? I should include an engine too --
the Wartsila-Sulzer weighs 2300 \tons, is 52\% efficient and delivers 80\,MW.
% The output of Magnus could feed 32 such engines.
The Magnus--engine oil-chain
has a weight-to-delivered-power ratio of 56\,\tonnes\ per MW;
56\,kg per kW. So the prototype wave machine has roughly ten times
the weight-to-power ratio of the evolved fossil fuel solution.)
%% 71000/(0.52* 5000) t/MW
%% 2300/80 t/MW
How do this chapter's estimates compare with those of
other bodies?
The IEE's 2002 estimate of the `technical potential' of wave -- ``an upper
limit that is unlikely ever to be exceeded even with quite dramatic changes in the
structure of our society and economy'' -- was 2.3\,kWh/d/p.
The Tyndall Centre estimated the `theoretical potential' of offshore wave power
to be 32\,kWh/d/p,
and the `practicable resource' to be 2.3\,kWh/d/p.
The Interdepartmental Analysts Group estimated that wave power could
contribute an average of 1.5\,kWh/d/p at a cost of 7p/kWh.
All three of these estimates are similar to my figure of 4\,kWh/d/p.
% The Scottish Executive in 2001 estimated that Scotland alone has a potential
% 14\,GW of wave energy (5.6\,kWh/d each when shared across the UK).
%% My estimate was that 1.6\,kWh/d each is a limit that the
%% whole UK was very unlikely to breach.
% I feel that the Scottish Executive
% have overestimated -- or perhaps their estimate has been misrepresented.
I speculate that overestimates may have been made by other bodies
who have got their wave-power estimate from the \ind{Atlas of Marine Resources},
because
some people don't realise that
{\em{waves are a resource per unit length of \ind{coastline}}}.
You can't have your cake and eat it. You can't collect wave energy
two miles off-shore {\em{and}\/} one mile off-shore.
% from the same bit of coastline.
Or rather, you can try, but the two-mile
facility will absorb energy that would have gone to the one-mile
facility, and it won't be replaced. The \ind{fetch} required for wind
to stroke up big waves is thousands of miles.
%% see atlastechnicalreportpart2.pdf
%% page 18
\begin{figure}
% \figuremargin{
\figuredangle{
\begin{center}
%\mbox{\epsfxsize=3in\epsfbox{../../images/wave.eps}}
\mbox{\epsfbox{metapost/wave.2}}
\end{center}
}{
\caption[a]{A 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 moving around
as indicated by the small arrows.
The speed of the wave, travelling from left to right, is indicated by
the bigger arrow at the top.
}
}
\end{figure}
%%% CUTTABLE
%%% long wavelength, smallest frequency, are most destructive.
\section{Other countries}
What about \ind{Africa}?
The length of Africa's exposed coastline is about 12\,000\,km.
At {40\,\kW$\!$/\m}, waves' maximum conceivable contribution
would be about $1\times 10^9\,\kWh$/day, or about 1\,\kWh\
per day
per person. (But in fact, the average power of waves is
smaller in equatorial waters, so this is an overestimate.)
\Figref{fig.coastline} shows the
coastline-to-population ratio, in metres per person,
for a few regions of the world.
% South Africa: length of coastline $\simeq$ 2400\,km.
Owing to
\ind{Ireland}'s habit of exporting its population to other countries,
\ind{Ireland}'s residents are relatively
% %and Portugal are
well-endowed with coastline.
% Ireland has a
Their
coastline-to-population ratio
is 0.12 metres per person, about 3 times that of the
UK. So perhaps, with enormous deployment
of hardware, wave power could deliver 10\,kWh/d per person there.
% see ../data/gnuc
\begin{figure}[tbp]
\figuremargin{
\begin{center}
\mbox{\includegraphics[width=2in,angle=-90]{../data/coastline.eps}}
\end{center}
}{
\caption[a]{Coastline, in metres per person, of various countries and continents.
These figures are rough estimates of {\em exposed\/} coastline. }
\label{fig.coastline}
}
\end{figure}
\begin{eqnarray*}
\mbox{Power per length of wavefront} &=& 2 \,
(\mbox{potential energy per period}) \times \mbox{frequency}
\\
&\simeq & 2 \left\{ \rho \left[ \frac{1}{2} \frac{\lambda}{2} h \right]
\right\} g h \frac{\omega}{2 \pi}
\\&=& \frac{1}{4\pi} \rho \lambda g h^2 {\omega}
\end{eqnarray*}
Use dimensional analysis to get frequency/wavelength relationship
deep water
\def\W{{\rm{W}}}
\def\kW{{\rm{kW}}}
\def\kWh{{\rm{kWh}}}
\def\m{{\rm{m}}}
\def\km{{\rm{km}}}
\def\s{{\rm{s}}}
\def\kg{{\rm{kg}}}
\begin{tabular}{cc} \hline\hline
{$\omega$} &{$T^{-1}$}\\
{$\lambda$} &{$L$}\\
{$\rho$} &{$M/L^3$}\\
{$g$} &{$L/T^2$}\\ \hline\hline
\end{tabular}
$\displaystyle
\:\:\Rightarrow\:\:
{ \frac{\lambda \omega^2}{g} = \kappa}$ (a dimensionless constant)\\
$\displaystyle
\mbox{Power} = \frac{1}{4\pi} \rho \lambda g h^2{\omega}$
$\displaystyle
= \frac{1}{4\pi} \rho \kappa (g/\omega^2) g h^2 {\omega}$
$\displaystyle
= \frac{1}{4\pi} \rho \kappa g^2 h^2 \frac{1}{2 \pi f}$
$\displaystyle
= \frac{1}{4\pi}
10^3\,\kg/\m^3 (10 \,\m/\s^2)^2 (0.7\,\m)^2 \frac{1}{2 \pi 0.1\,\s^{-1} }$
$\displaystyle
= \frac{1}{4 \times 40} 10^6\, \W/\m $
One way to think
$\displaystyle
= 6 \,\kW/\m $ \ \ {(I've read open sea waves have {40\,\kW/\m})}
$\displaystyle
= 150 \,\kWh/\m/\mbox{day} $
%% from quotes
The total power of waves breaking on the world's coastlines is put at
2 to 3 million megawatts, according to the US energy department. Areas
considered rich in tidal and wave power include: the western coasts of
Scotland, northern Canada, southern Africa, Australia, and the
north-east and north-west coasts of the US.
Divide by world population, 6e9,
get
0.5kW each, or 12 kWh per day. (If it were perfectly harvested.)
%%%%%%%%%%%%%%%%%%%%
%
% more notes nov 2006
%
% p178: once the amplitude of the waves exceeds c/omega = lambda/2 pi,
% the waves become more cusped, and white horses form.
% the mean energy per unit area of surface is 2 * 1/4 rho k a**2
% where a is amplitude.
% horizontal momentum os rho k a**2 / (2 c)
% geberal dispersion relation including ripples is
% c = sqrt( g/k + sigma k / rho )
% or omega**2 = gk + sigma k**3/rho
% minimum phase velocity is at 23 cm/s.
% group velocity for ripples is (3/2) times phase vel.
`Wave climate and the wave power resource'
[Mollison, Denis (1986), in Hydrodynamics of Ocean Wave-Energy Utilization
(eds. D Evans & A F de O Falcao), Springer, 133-156.]
% http://www.ma.hw.ac.uk/~denis/wave.html
% http://www.ma.hw.ac.uk/~denis/wave.html
`Wave power availability in the NE Atlantic' [490 Kb pdf]
[Mollison, Buneman & Salter (1976), Nature 263, 223-226.]
% same URL
Deep offshore sites on the
west coast of the UK offer 40\,kW per metre {\em{net power}}.
Net power is the amount of power per unit
length crossing a line facing in the best
direction for a site.
Source:
D.\ Mollison {The {UK} wave power resource}.
% http://www.ma.hw.ac.uk/~denis/wave.html
% http://www.ma.hw.ac.uk/~denis/wave.html
Winter wave power is roughly five times the Summer
wave power.
Sea-to-duck efficiency using the biggest (16\,m diameter) ducks
suggested: about 85\% at high wave frequencies;
less at lower frequencies where most of the power is
(my estimate of average
efficiency, from his graphs, averaging
over all frequencies, is 50\%)
(source: Nature paper).
Duck-to-London
(`hydrodynamically underprivileged area')
efficiency (conversion to electricity,
and transmission): about 60\%.
So if we
completely fill one third of the 1000\,km exposed Atlantic deep-sea-coastline
with 50\% efficient ducks or Pelamises, followed by 60\%
conversion and transmission efficiency,
we get
$40\,\kW/\m \times (1/3) \times 0.5 \times 0.6 \times 10^6 \m
= 4\,GW$
or 1.6 kWh/d.
Typical fetch to make a fully developed
sea is 700\,km or so.
(page 9 of the 24-page doc)
Typically 70\% of energy is lost
through bottom-friction
between 100\,m and 15\,m depth.
The estimated mean output for Salter Ducks
at a deep water site (where
the raw net power is 45\,MW/km) off
South Uist is 19\,MW/km, electricity,
including transmission to central Scotland.
(page 17 of the 24-page doc)
Losses because
a) highest power levels
are not economical to extract;
b) extraction efficiency about 75\%;
c) conversion and transmission about 80\%.
d) 5\% non-availability.
`Difficult to imagine a device that
extracts more than about 2/3 of the mean
resource'.
Assuming 15-20\,MW/km,
total of 50\,GW for Europe's
Atlantic coast, of which UK
and Ireland might get 12\,GW
each.
The DTI Atlas estimate of total
available wave power is wrong.
Page 18 of atlastechnicalreportpart2.pdf
They say
'The conversion from these values to annual output in TWh/annum can be achieved by
integrating in space and time (e.g. multiply mean power by approx. 12000 metres per
Atlas grid box, and 365 days x 24 hours x 60 minutes x 60 seconds). It is important to
note that values determined in this manner do not directly provide a realistic
quantification of practical output production as the scale of any energy converter and
its efficiency rating and consequential downstream energy losses would all need to be
considered.'
Indeed!
They said this also on page 4
'Calculations of energy yield do not consider the effects of downstream
losses as a consequence of the installation of a technology.'
What size is an atlas grid box?
So where do they get 12000 m per atlas grid box from???
I think they are using 12km squares.????!!!!
Here are the URLs for the rubbish report.
http://www.dti.gov.uk/energy/sources/renewables/renewables-explained/wind-energy/page27741.html
http://www.dti.gov.uk/files/file27746.pdf
http://www.dti.gov.uk/files/file27747.pdf
The project team was led by ABP Marine Environmental Research
page 23 gives examples of them botching the 'Total MW available (annual)' too.
%% see also
%% http://www.engineering.lancs.ac.uk/REGROUPS/LUREG/REPORTS/thorpe.pdf
Pelamis facts
%% http://www.oceanpd.com/PDFS/Pel%20conf%20history%20and%20status.pdf
%The OPD Pelamis WEC: Current Status and Onward Programme
%R W Yemm, R M Henderson, C A E Taylor
%Ocean Power Delivery Ltd., Edinburgh, UK.
% 130 m long, 3.5 m diameter, five segments.
% deisgned to be moored in depth about 50m
% A 30MW wavefarm would occupy a square km of ocean. (what length?)
% I assume that would be about 250m x 4km.
%5 actually the plan view of the wave farm shows THREE rows.
%% http://www.oceanpd.com/Pelamis/theFuture.html
%% So the width of the farm is a bit more than 400m
%% So 2.5km long.
% So they are saying 30MW/2.5km or 10MW/km.. 10kW/m delivered.
% so they are saying 25% of incoming energy is delivered to the
% grid?!
The effective cross section of a Pelamis is 7\,m (\ie, for
good waves, it extracts 100\% of the energy that would cross 7\,m).
This is roughly consistent with putting in say 65\,kW/m and getting 10\,kW/m
from three rows with effective spacing between things of 50\,m.
% http://www.oceanpd.com/Anims/pelamis_V4.html
% ETSU changed its name to Future Energy Solutions which got absorbed.
% part of AEA
% http://www.aea-energy-and-environment.co.uk/
From the 151-page document by Thorpe,
which includes thorough review and economic costing of
deep water and shallow water wave systems,
% aka ETSU R120?
Osprey would cost 8p per kWh
10 devices each of size 24m wide would
deliver 5.4MW average.
Wave: 46GWh per year from 10 devices,
that's 5.2MW. (allowing for a bit of down time)
Same documents says 5p--8p per kWh for the duck too.
ETSU-R120
A Brief Review of Wave Energy
A report produced for The UK Department of Trade and
Industry
T W Thorpe
May 1999
address: AEA Technology plc
`Westwave' --
Pelamises in Cornwall: DTI grant of 4.5 million to E.ON and ocean power
delivery.
`5 MW' from 7 pelamises (3000 homes, i.e. 1.5\,MW average).
%% http://www.oceanpd.com/LatestNews/default.html
this is not needed:::
\section{Other ways of grabbing wave energy}
The oyster: a 12.5\,m high buoyant slab that flops to and fro
in 12\,m deep water.
Economically, this may be a good way to collect wave energy.
But shallow-water wave power is less than deep-water wave power,
so the oyster option has no effect on my estimate of the total
conceivable power from waves.
%% http://www.westcornwallpasty.co.uk/
\section{Other estimates}
%% http://www.carbontrust.co.uk/Publications/publicationdetail.htm?productid=CTC601
The carbon trust claims that wave and tide
could supply one fifth of UK electricity;
they think offshore wave is bigger than tidal stream
and near-shore wave.
``Building Options for UK
Renewable Energy indicated a practical worldwide wave
energy resource of between 2000 and 4000\,TWh/year,
[The practical resource allows for practical
and economic factors that combine to
make developments commercially attractive. The technical resource describes
the amount of energy technically available
before such constraints are applied.]
while the UK practical offshore wave energy resource has
been estimated at 50 TWh/year [ETSU], (about one seventh of
current UK electricity consumption). New findings during
the MEC suggest that the technical UK tidal stream
resource is 18 TWh/year [Black and Veatch],
which is about 10-15\% of the
known worldwide tidal stream resource. The UK practical
near-shore and shoreline wave energy resources have been
re-estimated at 7.8 TWh/year and 0.2 TWh/year,
respectively.''
From Atlas of Marine Resources:
``The conversion from these values to annual output in TWh/annum can be achieved by
integrating in space and time (e.g.\ multiply mean power by approx. 12000 metres per
Atlas grid box, and 365 days $\times$ 24 hours $\times$ 60 minutes $\times$ 60 seconds).
{\em It is important to
note that values determined in this manner do not directly provide a realistic
quantification of practical output production\/} as the scale of any energy converter and
its efficiency rating {\em{and consequential downstream energy losses}\/} would all need to be
considered.''
[Emphasis added.]
\qa{
Will periods of low wave power coincide with periods of
low wind power?
}{
Yes, some correlation is expected, because\index{wind}\index{wave!source of}
{\em{wind makes waves}}. Waves travel a long
way, so Atlantic waves
arriving at the British Isles on Friday
usually were made by wind blowing on
a piece of Atlantic roughly 2000 miles away the preceding Monday.
So the correlation between waves and wind power will be the same as the correlation
between (wind in mid-atlantic 4 days ago) and (wind here now).
}
% from waves2.tex
The relationship between the frequency of a deep-water wave
and its wavelength can be derived in a couple of ways.
One neat trick is to ask what {\em{possible}\/} form the relationship
could take. Relationships are {\em{possible}\/} if they
don't involve adding apples to oranges (or equating them to each other).
(Two examples of `possible'
relationships are `area of triangle = half base times height'
and `area of triangle = base times base'; both are `possible' because
they equate an area to a product of two lengths.
An illegal relationship, in contrast, is `area of triangle = base + ${\mbox{height}}$'; this relationship can be ruled out
instantly by `dimensional analysis', because it equates an area to a
length, and areas are not lengths, they are squared-lengths.)
If we find that only one form of relationship
is possible, then, like Sherlock Holmes, we can deduce that that
relationship must be true.
[More details of dimensional analysis here.]
\begin{tabular}{cc} \toprule
{$\omega$} &{$T^{-1}$}\\
{$\lambda$} &{$L$}\\
{$\rho$} &{$M/L^3$}\\
{$g$} &{$L/T^2$}\\ \bottomrule
\end{tabular}
$\displaystyle
\:\:\Rightarrow\:\:
{ \frac{\lambda \omega^2}{g} = \kappa}$ (a dimensionless constant)
%% omega^2 = g k
%% omega/k = g/omega
%% 2 omega domega/dk = g
%% domega/dk (group vel) = (1/2) g/omega
\[
v = \frac{g T}{2 \pi} .
\]
Or if you want speed in terms of wavelength,
\[
v = \sqrt{ \frac{g \lambda }{2 \pi} } .
\]
\section{CAT report}
The CAT report (zerocarbonbritain)
assumes that you could get 250\,TWh/y out of a technical
potential of 600--750\,TWh.
Where did they get this figure?
Answer:
BWEA. (2007) Marine Renewable Energy. British Wind
Energy Association, www.bwea.com/marine/resource.html
% [Live June 2007]
This gives a throw-away figure of `as much as 700\,TWh/y',
but then says that ETSU reckoned 50\,TWh/y.
% a (Capacity credit for wind is approximately the square root of the installed capacity, i.e. for 81.5 GW of wind,
% capacity credit = 9 GW = 11%.)
% (Derived from Laughton, Renewable Energy and the Grid, Earthscan, 2007.)
% Laughton, Renewable Energy and the Grid, Earthscan, 2007.
(This formula is explained in this chapter's end-notes.)
So if we completely fill half
% one third
of the 1000\,km of exposed Atlantic deep-sea-coastline
with 50\% efficient ducks, followed by 60\%
conversion and transmission efficiency,
we get
$40\,\kW/\m \times (1/2) \times 0.5 \times 0.6 \times 10^6 \m = 6\,\GW$
or 2.4\,kWh/d per person.
% $40\,\kW/\m \times (1/3) \times 0.5 \times 0.6 \times 10^6 \m = 4\,\GW$
% or 1.6\,kWh/d each.
Memorable figures:
%% page 18
From \cite{Mollison85}:
``The large scale resource of the NE Atlantic,
from Iceland to North Portugal, has a net resource of
40--50\,MW/km, of which 20--30\,MW/km is potentially
economically extractable.''
Using a conservative mean output estimate of 15--20\,MW/km, the
potential power delivered is
UK: 12\,GW (100\,TWh/y, or 5\,kWh/d per person);
Ireland: 12\,GW; Faeroes, France, Spain, Portugal:
5--8\,GW each.
If 50\% of the exposed Atlantic deep-sea coastline (500\,km) were occupied by
Pelamis wave-farms (a total of 7500 devices in 200 wavefarms) then the power
delivered would be 5\,GW, or 2\,kWh/d per person in the UK.
%% The distance between Land's End and John o' Groats is 968.618 km (605 miles)
% this is copied into main wave chapter
What is the mass of each device? 700 tons, including 350 or 400 tons of
ballast.
%% source: e-mail from OPD.
Compare with the steel-requirements for offshore wind:
An offshore wind-turbine with a maximum power of 3\,MW weighs 500 tons, including
its foundation. So the total weight per kW of a wave-machine, half a ton per kW, is not
much bigger than that of a wind-turbine.
% check displacement:
% Depth in water: 1.5m; average width: 3m; length: 150m. -> 700 tons, grand.
If 400\,km of the shallow coastline were peppered with Oysters, three of them every
hundred metres, then the power delivered would be
1.2\,GW, or about 0.5\,kWh/d per person.
%% 00,000
The total from deep- and shallow-water devices would be at most
5.2\,GW (perhaps a little less, if they were installed in such a
way that the one shadowed the other).
(2.1\,kWh/d/p.)
\section{Notes}
Facts about deep-water waves.
% The group velocity is half of $v$.
% (Double check the constant in Faber.) done
% p178: once the amplitude of the waves exceeds c/omega = lambda/2 pi,
% the waves become more cusped, and white horses form.
% the mean energy per unit area of surface is 2 * 1/4 rho k a**2
% where a is amplitude.
% horizontal momentum os rho k a**2 / (2 c)
% geberal dispersion relation including ripples is
% c = sqrt( g/k + sigma k / rho )
% or omega**2 = gk + sigma k**3/rho
% minimum phase velocity is at 23 cm/s.
% group velocity for ripples is (3/2) times phase vel.
\section{Conventions}
Most of this is to be cut.
From \citet{Mollison85}, we learn the following facts and \ind{conventions}.
A wave-train of amplitude $a$ has root-mean-square wave-height $H$ given by
\[
H^2 = \frac{1}{2} a^2 .
\]
The power flux is
\[
P = \frac{\rho g^2}{4 \pi} H^2 T ,
\]
where $T$ is the period. The phase velocity is $U= gT/(2 \pi)$;
the wavelength is $L = gT^2/ (2 \pi)$.
The ``significant wave height'' is
\[
H_{\rm S} = 4 H_{\rm rms} .
\]
\subsection{Wave growth}
Wave \index{energy density!of waves}{energy density} per unit area, $E = \rho g H^2$, grows
roughly linearly with fetch $x$:
\[
H^2_{\rm rms} \simeq 1.6 \times 10^{-7} U_{10}^2 \frac{x}{g} ,
\]
where $U_{10}$ is the windspeed measured 10\,m
above the sea surface.
The greater the \ind{fetch}, the greater the mean period $T_{\rm e}$ of the
waves created. Empirically $T_{\rm e} \sim x^{1/3}$.
In fully developed seas,
$H_{\rm rms} = 0.0053 U_{19.5}^2$
(presumably SI units).
And the mean period is
$T_{\rm e} = 0.625 U_{19.5}$ -- again, presumably
SI units.
The associated phase velocity for this period of wave is
$U_{\rm e} = gT_{\rm e}/(2 \pi)
= \frac{g}{2 \pi} 0.625 U_{19.5} = 0.98 U_{19.5}$.
The power in such a \ind{fully developed sea}\index{sea!fully-developed}
is proportional to
$U^5$ and the fetch required to generate it is proportional to
$U^2$. This fetch is roughly 3000 wavelengths. (Using the
wavelength associated with the mean period.)
For example, if $U_{19.5} = 20$\,m/s, $H_{\rm rms} = 2.12$\,m,
$T_{\rm e} = 12.5$\,s, and the required fetch is about 700\,km.
The power density attained is 440\,kW$\!$/m, and the
mean rate of energy transfer from wind to sea is 0.6\,\Wmm.
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.
%% Mollison[33] estimate the bottom friction coefficient
%% for energy loss in depths 20-100 m.