\amarginfignocaption{b}{
\begin{center}
\begin{tabular}{@{}c@{}}
\mbox{\epsfxsize=53mm\epsfbox{../../images/SimpleTurbineD.jpg.eps}} \\
\end{tabular}
\end{center}
% \caption[a]{ }
}
\section{The physics of wind power}
To estimate the energy in wind, let's imagine holding up
a hoop with area $A$, facing the wind
whose speed is $v$. Consider
the mass of air that passes through that hoop in one second.
Here's a picture of that mass of air just before it passes through the
\ind{hoop}:
\medskip\\
\begin{center}
{\mbox{\epsfbox{crosspad/newwind.1}}}
\medskip\\
\end{center}
\noindent
And here's a picture of the same mass of air one second later:
\medskip\\
\begin{center}
{\mbox{\epsfbox{crosspad/newwind.3}}}
\medskip\\
\end{center}
\noindent
The mass of this piece of air is the product of
its \ind{density} $\rho$, its area $A$, and its length, which is $v$ times $t$,
where $t$ is one second.%
\marginpar{
\fbox{\small
\begin{minipage}{50mm}
I'm using this formula again:
\[
\mbox{mass} = \mbox{density} \times \mbox{volume}
\]
\end{minipage}
}
\label{miniformulaGain}
}
\medskip\\
\begin{center}
{\mbox{\epsfbox{crosspad/newwind.2}}}
\medskip\\
\end{center}
\noindent
The \ind{kinetic energy} of this piece of air\index{cartoon!wind power} is
\beq
\frac{1}{2} m v^2 = \frac{1}{2} \rho A v t \, v^2 = \frac{1}{2} \rho A t v^3 .
\label{eq.windpower1}
\eeq
So the power of the wind, for an area $A$ -- that is, the kinetic
energy passing across that area per unit time -- is
\beq
\frac{ \frac{1}{2} m v^2}{t}
= \frac{1}{2} \rho A v^3 .
\label{eq.windpower}
\eeq
This formula may look familiar -- we derived an identical expression
on \pref{pageAvcubed} when we were discussing the power\index{power density!wind}
requirement of a moving car.
What's a typical wind speed? On a windy day, a cyclist really notices
the wind direction; if the wind is behind you,
\amarginfig{b}{\small
\begin{center}
\begin{tabular}{cccc} \toprule
miles/ & km/h & m/s & Beaufort \\
hour & & & scale \\ \midrule
% & 1 & 0.28 & \\
% 1 & 1.6 & & \\
2.2 & 3.6 & 1 & force 1 \\
7 & 11 &3 & force 2 \\
11 & 18 &5 & force 3 \\
13 & 21 &6 & \\
16 & 25 &7 & \raisebox{7pt}[0in]{force 4} \\
22 & 36 &10 & force 5 \\
29 & 47 &13 & force 6 \\
36 & 58 &16 & force 7 \\ % error: was 31!
%%
% [The definition of a moderate breeze (Beaufort force 4)
% is 13--18 miles per hour. A gale (force 8) is 39--46 miles per hour.]
% in met office records a day of gale is a day on which
% speed at 10m (averaged over 10 mins) was
% 34 knots / 39mph / 17.2m/s or more
% 15 & 24 & & Force 4 \\
% 20
%(19-24)
% & & 10 & 5 \\
% 25--31 & & 14 & 6 \\
% 32--38 & & 18 & 7 \\
%% & & & {\small{(moderate breeze)}} \\
%% 39-46
42 & 68 & 19 & force 8 \\
%% & & & {\small{(gale)}} \\
% & 100 & 28 & \\
% 70 & 110 & 31 & \\
% miles/ & km/h & m/s & Beaufort \\
% hour & & & scale \\ \midrule
49 & 79 & 22 & force 9\\
% 27 &97.2 &52.5 &60.4 &Force 10\\% Force 10
% 31 &111.6 &60.3 &69.3 &Force 11\\% Force 11
% 35 &126.0 &68.0 &78.3 &Force 12\\% Force 12
60 & 97 & 27 & force 10 \\
69 & 112 & 31 & force 11 \\
78 & 126 & 35 & force 12 \\
\bottomrule
\end{tabular}
\end{center}
\caption[a]{Speeds.
% 7 knots = 8\,mph = 13\,km/h = 3.6\,m/s.
\index{Beaufort scale}%
\index{speed}\index{conversion table, speed}\index{units!speed}
% A moderate breeze (Beaufort force 4)
% is 13--18 miles per hour. A gale (force 8) is 39--46 miles per hour
% \tinyurl{yfgfe7}{http://www.answers.com/topic/beaufort-scale}.
}
}%
% (You might enjoy noticing how similar this is
% to the formula for the power required by a high-speed car,
% on \pref{pageAvcubed}.)
you can go much faster than
normal;\index{cyclist}\index{estimation}\index{bicycle}
the speed of such a wind is therefore
comparable to the typical
speed of the cyclist, which is, let's say, 21\,\km\ per hour (13 miles per hour, or
6 metres per second). In Cambridge, the wind is only occasionally this big.
Nevertheless, let's use this as a typical British figure (and bear in mind that we may need
to revise our estimates).
%% In Europe, the wind is only rarely this big.
%% today's forecast: 10mph, tomorrow 21mph, then 12, 16, 23 (november gales)
%% http://www.answers.com/topic/beaufort-scale
The density of air
%% ykf4j9
%% http://hypertextbook.com/facts/2000/RachelChu.shtml
is about 1.3\,\kg\ per $\m^3$.
% \tinyurl{ykf4j9}{http://hypertextbook.com/facts/2000/RachelChu.shtml}.
(I usually round this to 1\,\kg\ per $\m^3$,
which is easier to remember, although
I haven't done so here.)
Then the typical power of the wind per square metre of hoop is
\beq
\frac{1}{2} \rho v^3 = \frac{1}{2} 1.3\,\kg/ \m^3 \times (6 \,\m/\s)^3
= 140 \,\Wmm .
\label{eq.windpower2}
\eeq
Not all of this energy can be extracted by a windmill. The windmill
slows the air down quite a lot, but it has to leave the air with {\em{some}\/}
kinetic energy, otherwise that slowed-down air would get in the way.
\Figref{fig.flowair} is
a cartoon of the actual flow past a windmill.
\begin{figure}[tbp]
\figuremargin{
\begin{center}
{\mbox{\epsfbox{crosspad/newwind.4}}}
%{\mbox{\epsfbox{crosspad/wind4.ps}}}
\end{center}
}{
\caption[a]{Flow of air past a windmill.\index{cartoon!windmill}
The air is slowed down and splayed out by the windmill. }
\label{fig.flowair}
}
\end{figure}
The maximum fraction of the incoming energy that can be extracted
by a disc-like windmill was worked out by a German physicist
called Albert \index{Betz, Albert}{Betz}\nlabel{betz} in 1919.
If the departing wind speed is one third of the arriving wind speed,
the power extracted is 16/27 of the total power in the wind.
16/27 is 0.59.
% Real windmills are not optimized to maximize this fraction, however
In practice let's guess that a windmill might be 50\% efficient.
In fact, real windmills are designed with particular
wind speeds in mind; if the wind speed is significantly greater than
the turbine's ideal speed, it has to be switched off.
As an example, let's assume a diameter of $d=25\,\m$, and a hub height of $32\,\m$,
which is roughly the
size of the lone windmill above the city of \ind{Wellington}, \ind{New Zealand}
(\figref{fig.wellingtonB}).
%% http://www.khantazi.org/Events/WindMill/WindMill.html
%% from the photo I think it is 32m high and has 24m diameter.
The power of a single windmill is
\beqan
& & \mbox{efficiency factor} \times \mbox{power per unit area} \times
\mbox{area} \nonumber \\
&=& 50\% \times \frac{1}{2} \rho v^3 \times \frac{\pi}{4} d^2
\label{eq.windpower3}
\\
&=& 50\% \times 140 \,{\Wmm} \times \frac{{\ensuremath{\pi}}}{4} (25\,\m)^2
% \times 25\,\m
\\
&=& 34\,\kW.
\eeqan
Indeed, when I visited this windmill on a very breezy day,
its meter showed it was generating 60\,\kW\@.
To estimate how much power we can get
% per person
from wind, we need to decide
how big our windmills are going to be, and how close together
we can pack them.
% How the figures work out will depend
% strongly on our assumptions about population density,
% a topic to which we will return in another chapter.
\marginfig{
%\begin{figure}
%\figuremargin{
\begin{center}
\mbox{\epsfxsize=53mm\epsfbox{../../images/windmill.ps}}
\end{center}
% }{
\caption[a]{The
\ind{Brooklyn windmill} above \ind{Wellington}, \ind{New Zealand},
% Brooklyn windmill above Wellington, N.Z.,
with people providing a scale at the base.
On a breezy day, this windmill was producing 60\,kW,
(1400\,kWh per day).
Photo by {Philip Banks}.}
\label{fig.wellingtonB}
}
%\end{figure}
% For the moment, let's take the population density of
% England: 380 people per square kilometre.
% The taller a windmill is, the bigger the wind speed it encounters.
How densely could such windmills be packed?
Too close and the upwind ones will cast wind-shadows on the downwind
ones.
Experts say that windmills can't be spaced closer than
5 times their diameter without losing significant power.
%\begin{figure}\figuremargin{
\marginfig{
\begin{center}
{\mbox{\epsfbox{metapost/wind2.55}}}
%{\mbox{\epsfbox{crosspad/wind5a.ps}}}
\end{center}
% }{
\caption[a]{{\Windfarm} layout.}
}%
% \end{figure}
At this spacing, the power that windmills can generate per unit land area is
\beqan
\frac{ \mbox{power per windmill (\ref{eq.windpower3})}}
{ \mbox{land area per windmill} }
& =& \frac{ \frac{1}{2} \rho v^3 \frac{\pi}{8} d^2 }
{ (5 d)^2 }
\label{eq.windpower4}
\\
& = & \frac{\pi}{200} \frac{1}{2} \rho v^3
\label{eq.windpower5}
\\
& = & 0.016 \times 140 \,\Wmm \\
& = & 2.2 \,\Wmm .
\eeqan
This number is worth remembering:
a {\windfarm} with a wind speed of
6\,m/s produces a power of 2\,W per m$^2$ of land area.
Notice that our answer does not depend on the diameter of the windmill.
The $d$s cancelled because bigger windmills have to be spaced
further apart.
\margintab{
% \begin{figure}
\begin{center}
\begin{tabular}{cc} \toprule
\multicolumn{2}{l}{\sc Power per unit area }\\ \midrule
{\windfarm} & 2\,\Wmm\\
(speed 6\,m/s) \\
\bottomrule
\end{tabular}
\end{center}
% }{
\caption[a]{Facts worth remembering: {\windfarm}s.
%% , number 1
}
}%
% \end{figure}
Bigger windmills might be a good idea in order to
catch bigger windspeeds that exist higher up (the taller a windmill is,
the bigger the wind speed it encounters),
or because of economies of scale, but
those are the only reasons for preferring big windmills.
This calculation depended sensitively on our estimate of
the windspeed.
Is 6\,m/s plausible as a long-term typical windspeed
in windy parts of Britain?
Figures \ref{fig.camb.wind}
and \ref{fig.cairngorm}
showed windspeeds in Cambridge and Cairngorm.
\Figref{Mawgan} shows the mean winter and summer
windspeeds in eight more locations around Britain.%
%\marginfig{
\begin{figure}[!bp]\figuremargin{
\begin{center}
\begin{tabular}{@{}c@{\hspace*{-3mm}}c}
\mbox{\epsfbox{metapost/wind.2}}
&
\raisebox{47mm}{\epsfig{file=../../images/PUBLICDOMAIN/maps/wind.eps,angle=270}}
\\
\end{tabular}
\end{center}
}{
\caption[a]{Average summer windspeed (dark bar)
and average winter windspeed (light bar)\index{wind!data}\index{windspeed!data}
in eight locations around Britain.
Speeds were measured at the standard weatherman's height of
10 metres.
Averages are over the period 1971--2000.
}
\label{Mawgan}
}
\end{figure}
% (These are the windspeeds at the standard weatherman's height of
% 10\,\m; a)
%
I fear 6\,m/s was an overestimate of the typical speed
in most of Britain!
%% \,per\,second
If we replace 6\,m/s by Bedford's 4\,m/s as our estimated
windspeed, we must scale our estimate down,
multiplying it by $(4/6)^3 \simeq 0.3$.
(Remember, wind power scales as wind-speed cubed.)
%\beq
% \mbox{Maximum conceivable wind power (assuming 4\,\m/s)}
% = 40\,\kWh\,\mbox{per person per day}.
%\eeq
On the other hand, to estimate the typical power, we shouldn't take the
mean wind speed and cube it; rather, we should find the mean cube of the windspeed.
The average of the cube is bigger than the cube of the average.
But if we start getting into these details, things get even more complicated,
because real wind turbines don't actually deliver a power proportional to wind-speed cubed.
Rather, they typically have just a range of wind-speeds within which they deliver the ideal
power;
at higher or lower speeds real wind turbines deliver less than
the ideal power.
\subsection{Variation of wind speed with height}
Taller windmills see higher wind speeds.
The way that wind speed increases with height
is complicated and depends on the roughness of the
surrounding terrain and on the time of day.
As a ballpark figure, doubling the height
typically increases wind-speed by 10\%
and thus increases the power of the wind by 30\%.
Some standard formulae for speed $v$ as
a function of height $z$ are:
\begin{enumerate}
\item According to the wind shear formula from
NREL {\tinyurl{ydt7uk}{http://www.nrel.gov/business_opportunities/pdfs/31235sow.pdf}}%
%(Oct 2001)
%NREL
, the speed varies as a power of the height:
\[
v(z) = v_{10} \left( \frac{z}{10\,\m} \right)^{\alpha} ,
\]
\amarginfig{c}{\small
\begin{center}
Wind speed versus height\\[0.05in]
\mbox{\ \ \ \epsfxsize=45mm\epsfbox{figs/vversush.eps}} \\[0.2916in]
Power density of wind v.\ height\\[0.05in]
\mbox{\ \ \ \epsfxsize=45mm\epsfbox{figs/powerd.eps}} \\[0.16in]
\end{center}
\label{fig.WindZTheory}
% COLON
\caption[a]{
Top: Two models of wind speed and wind power as a function of
height.
DWIA = Danish Wind Industry Association;
NREL = National Renewable Energy Laboratory.
For each model the speed at 10\,m has been fixed to 6\,m/s.
For the Danish Wind model, the roughness length is set to
$z_0=0.1$\,m.
Bottom: The power density (the power per unit of upright area) according
to each of these models.
}
}%
where $v_{10}$ is the speed at 10\,m,
and a typical value of the exponent $\alpha$ is 0.143 or $1/7$.
% Thus
%\[
% v(z) \propto z^{1/7} .
%\]
The one-seventh law ($v(z)$ is proportional to $z^{1/7}$)
is used by \citet{ElliottWindy}, for example.
\item
The wind shear formula from the Danish Wind Industry Association
\tinyurl{yaoonz}{http://www.windpower.org/en/tour/wres/shear.htm}
is
\[
v(z) = v_{\rref} \frac{ \log ( z/z_0 ) }{ \log ( z_{\rref} / z_0 ) },
\]
where $z_0$ is a parameter called the roughness length, and $v_{\rref}$
is the speed at a reference height $z_{\rref}$ such as 10\,m.
The roughness length for typical countryside
(agricultural land with
some houses and sheltering hedgerows with some 500-m intervals -- ``roughness class 2'')
is $z_0 = 0.1\,\m$.
\end{enumerate}
In practice, these two wind shear formulae give
similar numerical answers.
That's not to say that they are accurate at all
times however. \citet{vdBerg2004} suggests that different
wind profiles often hold at night.
\subsection{Standard windmill properties}
The typical \ind{windmill} of
today has a \ind{rotor} diameter of around 54 metres
centred at a height of 80 metres; such a machine has
a ``capacity'' of 1\,MW\@. The ``\ind{capacity}'' or ``\ind{peak} power''
is the {\em maximum\/} power
the windmill can generate in optimal conditions.
\begin{figure}[tbp]
\figuredangle{
{\epsfxsize=153mm\epsfbox{../../images/qr6kW.eps}}%
}{
\caption[a]{The qr5 from \myurl{quietrevolution.co.uk}. Not a typical windmill.}
}
\end{figure}
Usually, wind turbines are designed to start running at wind
speeds somewhere around 3 to 5\,m/s
and to stop if the wind speed reaches gale speeds of
25\,m/s.\label{pWindFacts}
% \tinyurl{ymfbsn}{http://www.windpower.org/en/tour/wres/powdensi.htm}.
%% http://www.windpower.org/en/tour/wres/powdensi.htm
The actual average power delivered is the ``capacity'' multiplied
by a factor that describes the fraction of the time that wind conditions
are near optimal. This factor, sometimes called the ``\ind{load factor}''
or ``\ind{capacity factor},''
depends on the site; a typical load
factor for a {\em{good}\/} site in the UK is $30\%$.\nlabel{pCapFac}
In the
Netherlands, the typical load factor is 22\%;
in Germany, it is 19\%.
% Source: page 28 of the dutch wind doc
\subsection{Other people's estimates of {\windfarm} power per unit area}
In the government's study [\myurlb{www.world-nuclear.org/policy/DTI-PIU.pdf}{http://www.world-nuclear.org/policy/DTI-PIU.pdf}]
the \UK\ onshore wind resource is estimated
using an assumed {\windfarm} power per unit area of
at most
% 9\,MW/km$^2$, which is
9\,\Wmm\ (capacity, not average production).
% Include a typical capacity factor here to deduce the average {\windfarm}
% power density.
If the capacity factor is 33\% then the average power
production would be 3\,\Wmm.
% , which is just 50\% different from our 2\,W/m$^2$ estimate.
% whiteless
% 129 MW, aka 200,000 homes
% corresponds to 0.644 kW per home. or 15.5kWh/day PER HOME
% Humph, what is actual per person electricity consumption?
% Elec = 380 TWh/y for UK (TOTAL, not domestic)
% which is 17 kWh/day each.
% How many people in a home, and what fraction is domestic?
% I guess 3 people per home, and 50% domestic, so
% we need 3*17 / 2 per home. Yep, I think they are mis-estimating
% what a ``home'' really is.
The London Array
is an offshore {\windfarm} planned for the outer Thames Estuary. With its 1\,GW
capacity, it is expected to become the world's largest offshore {\windfarm}.
% The site is seven miles off the North Foreland on the Kent coast in the area of Long Sand and Kentish Knock [2]. It will cover 90 square miles between Margate in Kent and Clacton in Essex.
The completed {\windfarm} will consist of 271 wind turbines in
245\,km$^2$
\tinyurl{6o86ec}{http://www.londonarray.com/london-array-project-introduction/offshore/}
%% 90 square miles,
and will deliver
an average power of 3100\,GWh per year (350\,MW). (Cost \pounds 1.5\,bn.)
That's a power per unit area of 350\,MW/245\,km$^2$ = 1.4\,\Wmm.
This is lower than other offshore farms because, I guess,
the site includes a big channel (Knock Deep) that's too deep (about 20\,m)
for economical planting of turbines.
\myquote{%
I'm more worried about what these plans [for
% an electricity substation for
the proposed London Array {\windfarm}]
will do to this landscape and our way of life than I ever was about a
\ind{Nazi invasion} on the beach.\nlabel{pNazi}
}{
Bill Boggia
% , whose family owns and runs several caravan parks around
of Graveney,
where the undersea cables \\ \hfill of the \windfarm\ will come ashore.
%% http://news.independent.co.uk/environment/article2086678.ece
}
%\end{quote}
\section{Queries}
\beforeqa
\qa{What about \ind{micro-generation}?
If you plop one of those \ind{mini-turbine}s on your roof, what
energy can you expect it to deliver?
}{
Assuming a windspeed of 6\,m/s, which, as I said before, is
{\em above\/}
the average for most parts of Britain; and assuming a diameter of 1\,m,
\amarginfig{c}{
%\begin{figure}
%\figuremargin{
% \begin{center}
\mbox{%
\epsfysize=41.53mm\epsfbox{../../images/Warw13C.eps}%
\,%
\epsfysize=41.53mm\epsfbox{../../images/Warw14.eps}%
}\par
% \end{center}
% }{
\caption[a]{An Ampair ``600\,W'' \ind{micro-turbine}.
The average power generated by this micro-turbine
in Leamington Spa is 0.037\,kWh per day (1.5\,W).
% 30th October 2007 to 9th May 2008
% Photos by {Robert MacKay}.}
}
\label{fig.warwickwind}
}%
the power delivered would be 50\,W\@.
%% pr 70*pi/4.0 * 24
That's 1.3\,kWh per day -- not very much.
And in reality,
in a typical urban location in England,
a micro-turbine delivers just 0.2\,kWh per day
-- see \pref{pmicrowind}.
Perhaps the worst windmills in the world are
a set in \ind{Tsukuba City}, \ind{Japan}, which actually
consume more power than they generate. Their installers were so
embarrassed by the stationary turbines that they imported power to make them
spin so that they looked like they were working!
\tinyurl{6bkvbn}{http://www.timesonline.co.uk/tol/news/world/asia/article687157.ece}
}
% WHAT FOR? ***
% See \citeasnoun{Faber}, p.\,63.
\small
\section{Notes and further reading}
\beforenotelist
\begin{notelist}
\item[page no.]
\item[\npageref{betz}] {\nqs{The
maximum fraction of the incoming energy that can be extracted
by a disc-like windmill\ldots}}%
\marginfig{
\begin{center}
\mbox{\epsfxsize=42mm\epsfbox{../../images/Iskra5kW.eps}}
\end{center}
% }{
\caption[a]{% AT5-1
A 5.5-m diameter Iskra 5\,kW turbine
[\myurlb{www.iskrawind.com}{http://www.iskrawind.com/}]
having its annual check-up.
This turbine, located in Hertfordshire (not the windiest of locations
in Britain), mounted at a height of 12\,m,
has an average output of
% 4000 kWh per year
11\,kWh per day.
A {\windfarm} of machines with this performance, one per
30\,m $\times$ 30\,m square, would have a power per unit area of
% 0.507
\pdcol{0.5\,\Wmm}.
}
\label{fig.Iskra}
}
There is a nice explanation
of this
on the Danish Wind Industry Association's
website.
\tinyurl{yekdaa}{http://www.windpower.org/en/stat/betzpro.htm}.
%% http://www.windpower.org/en/stat/betzpro.htm
\item[\npageref{pWindFacts}]
{\nqs{Usually, wind turbines are designed to start running at wind
speeds around 3 to 5\,m/s}}.
\tinyurl{ymfbsn}{http://www.windpower.org/en/tour/wres/powdensi.htm}.
%% http://www.windpower.org/en/tour/wres/powdensi.htm
\item[\npageref{pCapFac}] {\nqs{a typical
\index{capacity factor}\ind{load factor} for a {{good}} site is $30\%$}.}
In 2005, the average load factor of all major UK
{\windfarm}s was
28\%\
% 28.4\%\
\tinyurl{ypvbvd}{http://www.ref.org.uk/images/pdfs/UK_Wind_Phase_1_web.pdf}.
The load factor varied during the year, with a low of 17\%
in June and July.
The load factor for the best region in the country --
\ind{Caithness}, \ind{Orkney} and the \ind{Shetland}s -- was 33\%.
The load factors of the two \index{offshore wind!load factor}{offshore} {\windfarm}s operating in 2005
were 36\% for \ind{North Hoyle} (off North \ind{Wales})
and
29\% for \ind{Scroby Sands} (off \ind{Great Yarmouth}).
% http://www.clowd.org.uk/Downloads/clowdCarbonSavings.pdf
% is a good report on exaggerated claims for a bedford {\windfarm}
Average load factors in 2006 for ten regions
were:
Cornwall 25\%;
Mid-Wales 27\%;
Cambridgeshire and Norfolk 25\%;
Cumbria 25\%;
Durham 16\%;
Southern Scotland 28\%;
Orkney and Shetlands 35\%;
Northeast Scotland 26\%;
Northern Ireland 31\%;
offshore 29\%.
% Average of all: 27\%
\tinyurlb{wbd8o}{http://www.ref.org.uk/energydata.php}
\item[] \cite{WatsonEtAl}
say a minimum annual mean wind speed
% (AMWS)
of 7.0\,m/s is
currently thought to be necessary for commercial viability of
wind power. About 33\% of UK land area
has such speeds.
% The British Wind
%Energy Association estimates that the UK has 65\,252 km$^2$ of land area suitable for wind
%generation.
% At an average power density of 2\,\Wmm,
%% 2\,MW per km$^2$
% this area of {\windfarm}s would deliver 130\,GW, or 52\,kWh/d per person.
%% the report then reduces this number using constraints of network, visual impact, etc
\end{notelist}
\normalsize