Wind energy potential in Portugal

Joao Pedro Pereirapereirajpp@gmail.com

January 17, 2012

Contents

1 Introduction 2

2 Physical limits to wind energy 22.1 Kinetic energy of wind . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Wind speed variation . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2.1 Across space . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.2 Through time . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Technological limits to wind generation 53.1 Betz limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.2 Specifications of some representative turbines . . . . . . . . . . . . 7

3.2.1 Sample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.2 Power curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.3 Power coefficient . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3 Wind farm optimization . . . . . . . . . . . . . . . . . . . . . . . . 103.3.1 Windmills spacing . . . . . . . . . . . . . . . . . . . . . . . 103.3.2 Theoretical expected total output . . . . . . . . . . . . . . 103.3.3 Feasible expected total output . . . . . . . . . . . . . . . . 12

4 The effect of wind variability in Portugal 134.1 Load factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Wind intermittency . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Estimates of wind generation in Portugal 17

1

1 Introduction

Wind is one of the renewable energy resources with greatest potential tosupply our energy needs. However, it is not clear how big is this potential.Different people mention estimates that seem to differ by orders of magnitude:

“The power in the winds [...] is some 100 times the total globalpower usage.” — Andrews and Jelley (2007).

“The earth’s wind resource is so large that it could technicallyprovide five times the total energy consumed by the entire worldfrom all sources.” — Gore (2009)

This paper describes the physical and technological limits to the produc-tion of energy from wind. Wind technology is mature so we do not expectsignificant efficiency improvements in the foreseeable future.

We then apply these concepts to Portugal. We estimate that wind farmsin Portugal will be able to output an average of 2.8 W per m2 of land area.This means that with wind farms covering around 2% of the territory, windcould supply all our electricity needs of 2010. This would require about 5times the wind hardware installed in 2010. With an area around 9%, windwould be able to generate the total amount of energy consumed in Portugalin 2010 from all sources (renewables and fossil fuels).

Wind does indeed have a big potential in Portugal. However, these av-erage values mask a huge variability in wind power through time. Windintermittency is probably the main challenge to the growth of the wind in-dustry.

2 Physical limits to wind energy

2.1 Kinetic energy of wind

A mass m or air moving at speed v generates a kinetic energy of

KE =1

2mv2 (1)

This has units kg.m2/s2 = (kg.m/s2).m = N.m = J . Consider that themass of air is passing through a vertical area A (which will be a circle for a

2

Figure 1: Mass of air passing through a circle (Source: MacKay (2009))

vt

A v

typical horizontal axis wind turbine) and that the length of the mass of airis L, as in figure 1.

Since mass equals volume times density, we have m = A×L×ρ, where ρis the density of air. At velocity v, during time t, we have L = v × t. Thus,

m = A× vt× ρ (2)

Replacing in (1),

KE =1

2(A× vt× ρ)v2 =

1

2A× t× ρ× v3 (3)

Since power equals energy per unit of time, the power of wind is

P = KE/t =1

2ρAv3 (4)

This has units kgm3 .m

2.m3

s3= (kg.m2/s2)/s = J/s = W . Interestingly, power

depends on the cube of wind velocity.

2.2 Wind speed variation

Wind is second-hand solar energy. The Sun radiation heats the Earth’s sur-face, which in turn heats the surrounding air. Different areas (sea, mountains,deserts, forests) absorb radiation differently, so differences in temperaturearise. The hotter air expands and rises, creating a pressure vacuum intowhich cooler air rushes, which results in winds. The Earth’s movement thengenerates variability in these relations through time. The final result is thatthe wind speed varies across space and over time.

3

2.2.1 Across space

Figure 2 shows the average speed across Portugal, estimated for a height of60 m. Most of the territory shows an average speed around 5 or 6 m/s.

Figure 2: Wind speed in Portugal at height of 60m (Source: Esteves (2004))

The wind speed increases with height (h). A typical model is

v(h) = v0

(h

h0

)α

where v0 is the measured speed at the reference height h0, typically h0 =10 m. The exponent α is strongly dependent on the terrain and even on thetime of day. A typical value is α = 1/7 (MacKay, 2009).

For example, if α = 1/7 and v0 = 4.645 m/s at h0 = 10 m, we getv(60 m) = 6 m/s, an optimistic average of the speed across Portugal. If weincrease the height to 80 m (closer to the average height of modern 2 MW

4

turbines, as described below), the speed increases to v(80 m) = 6.25 m/s,which represents a 13% increase in the power of the wind.

2.2.2 Through time

At each location, the wind speed varies through time. According to Andrewsand Jelley (2007), the wind speed distribution is often well described by theRayleigh distribution, which has density

f(v) =2v

c2exp

[−(v/c)2

](5)

where c = 2v/√π and v is the average wind speed. Figure 3 shows this

function for v = 6 m/s.

Figure 3: Rayleigh pdf for a mean wind speed of 6 m/s

0 5 10 15 20 25 300

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Den

sity

, f(v

)

Wind speed (m/s)

3 Technological limits to wind generation

3.1 Betz limit

A windmill cannot absorb all kinetic energy in the wind. To do that, it wouldhave to fully stop the wind coming through, but then this slowed-down air

5

would get in the way. The windmill has to leave the air with some kineticenergy. Figure 4 shows how the air slows down and sprays out as it passesthrough a windmill.

Figure 4: Mass of air passing through a windmill (Source: MacKay (2009))

Define the power coefficient, Cp, as

Cp :=Power extracted by rotor

Power of incoming wind(6)

The physicist Albert Betz showed in 1919 that the maximum fraction ofthe incoming energy that can be extracted by a windmill is 59%, that is,maxCp = 0.59.

Proof. (Betz limit — intuitive proof) Let v1 and v2 denote, respectively, thewind velocity before and after the rotor. From equation (4), the full powerof the incoming wind is

P1 =1

2ρAv31

Let KE1 and KE2 denote, respectively, the kinetic energy of the wind beforeand after the rotor. By conservation of energy, the energy extracted by therotor, Er, must equal

Er = KE1 −KE2 =1

2m(v21 − v22)

where we used (1). From (2), m = A× vrt× ρ, where vr is the wind velocitythrough the rotor. Assume that this speed is an average of the speeds beforeand after the turbine, vr = v1+v2

2(in the more formal proof of Betz this a

result, not an assumption). The power extracted by the rotor is thus

Pr := Er/t =1

2

(A× v1 + v2

2× ρ

)(v21 − v22) =

1

4Aρ(v1 + v2)(v

21 − v22)

6

The ratio between the power extracted by rotor and full incoming power is

Cp =Pr

P1

=1

2

(v1 + v2)(v21 − v22)

v31=

1

2

(1−

(v2v1

)2)(

1 +v2v1

)Define x := v2/v1 and write the efficiency as a function of x,

Cp =1

2

(1− x2

)(1 + x)

By maximizing this function, we find argmaxx Cp = 1/3, that is, the max-imum efficiency is achieved when the departing wind speed is 1/3 of thearriving wind speed, v2 = (1/3)v1, which means that the speed at the rotoris vr = (2/3)v1. Replacing the optimum x = 1/3 above, we get the maximumpower coefficient:

maxx

Cp = 16/27 = 0.59

This value is called the Betz or Lanchester-Betz limit.

3.2 Specifications of some representative turbines

3.2.1 Sample

Table 1 describes the characteristics of some representative turbines availabletoday. According to DGEG, the average turbine size installed in Portugalup to 2010 is 1.9 MW, so we focus on this range.

3.2.2 Power curve

Interestingly, the power output of a turbine does not increase monotoni-cally with wind speed. Instead, the actual power output increases from theminimum cut-in wind speed until it reaches the rated power at the ratedwind speed. After this rated wind speed, the power output stays constant atthe rated value and eventually drops to zero if the wind speed exceeds themaximum safe operating speed (cut-out speed).

Figure 5 shows the power curve for the Vestas V112 turbine. Note thatthis is a variable-speed variable-pitch turbine, so the power curve alreadyreflects any possible optimal adjustments for each wind speed.

However, looking only at the power curve may be misleading. For exam-ple, one might conclude that the Vestas V112 turbine is only indicated for

7

Table 1: Characteristics of some representative turbinesSource: manufacturers’ websites.

RatedPower

Cut-inwindspeed

Ratedwindspeed

Cut-outwindspeed

Rotordiam-eter

Hubheight

Turbine (kW) (m/s) (m/s) (m/s) (m) (m)

Enercon E53 800 2 13.0 28 52.9 60Vestas V90 2 000 4 13.5 25 90 80Enercon E82 2 000 2 13.0 28 82 78Siemens SWT 113 2 300 3 12.5 25 113 100Vestas V112 3 075 3 13.0 25 112 84Siemens SWT 120 3 600 4 12.5 25 120 90Enercon E126 7 500 3 16.0 28 127 135

Figure 5: Power curve for Vestas V112 turbine

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sites with wind speeds close to 13m/s. The same applies to all other turbinesin table 1. This is odd as there are very few sites with average speeds as highas 13 m/s... In reality, these turbines are actually optimized for much loweraverage wind speeds, as discussed in the following section.

3.2.3 Power coefficient

What really determines the performance of a given turbine is its power co-efficient (Cp, as defined in equation (6)). Figure 6 shows both the powercurve and the power coefficient (as a function of wind speed, Cp(v)) for theEnercon E82 turbine.1

Figure 6: Power coefficient function for Enercon E82 turbine

The E82 is rated at 2 MW, which is only achieved at wind speeds of13 m/s or higher. However, Cp(v) reaches a maximum of 0.5 at v = 9 m/sand is above 0.4 from v = 4 m/s to v = 11 m/s. Hence, this turbine isoptimized for wind speeds in this lower range, much more in line with whatis actually achievable in most onshore sites.

1Surprisingly, some manufacturers, like Vestas and Siemens, do not show the Cp(v)curve.

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Importantly, the maximum value of 0.5 is very close to the Betz limit of0.59. Hence, we conclude that wind turbine technology is mature, with verylittle room for efficiency improvements. Any substantial improvement in thefraction of energy that we are able to extract from wind will have to comefrom a radically different technology.

3.3 Wind farm optimization

3.3.1 Windmills spacing

If windmills are packed too close to each other, the upwind ones will castwind-shadows on the downstream ones. According to MacKay (2009), thedistance between turbines should be at least five times their diameter (d), asin figure 7. Andrews and Jelley (2007) suggest a higher spacing. For example,they state that a spacing of 8d downwind by 5d crosswind decreases the farmoutput by around 10% when compared to the output of the turbines sitedseparately. We follow the more optimistic 5d by 5d assumption and ignorelosses.

Figure 7: Wind farm layout (source: MacKay (2009))

d

5d

3.3.2 Theoretical expected total output

At a constant speed, the power of wind is given by (4). Since speed variesthrough time, we can compute the expected power of the wind as

E[Pwin] =

∫1

2ρAv3f(v)dv

10

where f(v) denotes the density of the wind speed distribution. For a turbinewith power coefficient function Cp(v), the expected output power is

E[Pout] =

∫Cp(v)

1

2ρAv3f(v)dv

Define the expected output power density as

E[Pden] :=E[Pout]

Land area(7)

Assuming that the turbines are spaced at 5 diameters (land area = (5d)2)and replacing A = πd2/4, we get

E[Pden] =

∫Cp(v)

12ρ(πd2/4)v3f(v)dv

(5d)2(8)

=π

200ρ

∫Cp(v)v

3f(v)dv (9)

Interestingly, the power density does not depend on the diameter (d) of therotor. The ds canceled because bigger windmills have to be spaced furtherapart. This means that wider turbines will not give us more electricity(though higher turbines will because the wind speed increases with height).However, larger turbines typically have some advantages, like lower mainte-nance and grid infrastructure costs. Also, many ridges only allow a singleline of turbines, so wider rotors will produce more electricity.

To illustrate (9), we compute the expected output power density for theEnercon E82 turbine. The Cp(v) function is given by the manufacturer forspeeds up to 25 m/s (see figure 6, though the turbine brochure gives theactual numbers in a table). We assume that the wind speed at the sitefollows the Rayleigh distribution in (5) with a mean of 6 m/s. We use thestandard air density of ρ = 1.225 kg/m3. Discretizing the integral in (9) overthe 25 wind-speed points given by the manufacturer (∆v = 1), we get

E[PE82den ] ≈ π

2001.225

25∑v=1

Cp(v)v3f(v) = 3.15 W/m2 (10)

Having an estimate for the power density of the average representativeturbine, we could then easily extrapolate to any land size covered by wind-mills. The expected total output power, E[P T

out], would be

E[P Tout] = E[Pden]× Land area (11)

11

However, the expected output power estimate depends critically on twopieces of information. First, it depends on the distribution of wind speeds,f(v). Even if the Rayleigh distribution is a good approximation to the truedistribution, the resulting estimate is very sensitive to the assumed aver-age speed. For example, for an average speed of 4 m/s, the power den-sity decreases to E[PE82

den ] = 1.05 W/m2, while for 8 m/s it increases toE[PE82

den ] = 5.33 W/m2. Unfortunately, we don’t have a good model to de-scribe the wind speed distribution over different locations in Portugal. Sec-ond, the expected output power depends on the power coefficient functionCp(v). This is a property of the turbine itself and unfortunately is not readilyavailable for many turbines. Hence, we use the alternative approach in thefollowing section.

3.3.3 Feasible expected total output

Define the rated, or nominal, power density as

P ratedden :=

rated power per windmill

land area per windmill

Table 2 shows an estimate of the power density for the sample of turbines intable 1, assuming that each turbine occupies a square of land of size (5d)2.Excluding the biggest 7.5 MW turbine that seems to be an outlier (probablyneeds a bigger area of land), the average rated power density for all turbinesis

P ratedden ≈ 10 W/m2

Define the load factor as the ratio of actual generated power over theinstalled capacity:

Load Factor :=output power

rated power(12)

The expected output power density defined in (7) can also be written as

E[Pden] :=E[Pout]

land area=

E[Pout]

rated power× rated power

land area

Therefore, the expected total output power can also be estimated by

E[P Tout] = E[Load Factor]× P rated

den × Land area (13)

12

Table 2: Rated power density at 5d sittingUsing the data from table 1, we compute the rated power density as the ratio of rated

power to the land area required for each turbine, assumed to equal (5d)2.

RatedPower

Rotordiameter

Area perturbine

Rated powerdensity

Turbine (kW) (m) (m2) (W/m2)

Enercon E53 800 52.9 69 960 11.4Vestas V90 2 000 90 202 500 9.9Enercon E82 2 000 82 168 100 11.9Siemens SWT 113 2 300 113 319 225 7.2Vestas V112 3 075 112 313 600 9.8Siemens SWT 120 3 600 120 360 000 10.0Enercon E126 7 500 127 403 225 18.6

The advantage of this approach is that we can use the load factor actuallyobserved in a given region. This will capture the interaction between theactual varying wind speed and the turbines’ efficiencies over all wind farmsinstalled in that region. Hopefully, this will be a reasonable estimate forother future wind farms. The following section describes the load factor inPortugal.

4 The effect of wind variability in Portugal

4.1 Load factor

To compute the load factor defined in (12), we use the hourly wind produc-tion for all wind parks in Portugal during 2010 (data available from REN).According to DGEG, the installed wind capacity at the beginning of 2010was 3507 MW and at the end was 3865 MW. We assume that the capacityincreased linearly throughout the year. This results in a time series with8760 hourly load factors.

Figure 8 shows the histogram of the load factor.2 Across the full year of

2Note that since our data series is the aggregate output of all wind parks in Portugal,we never observe a load factor of 1, even though it is possible that individual parks mayoperate at full capacity for some hours.

13

Figure 8: Load factor histogram for Portugal, 2010Load factor (production over installed capacity) histogram using hourly values for all wind

parks in Portugal during 2010. We assume that the installed capacity increases linearly

throughout the year from 3507 MW up to 3865 MW. Hourly production is from REN.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90

100

200

300

400

500

600

Load factor

Num

ber

of h

ours

2010, the mean load factor was 0.28 and the median was 0.22. Hence, we usethe following reference value:

E[Load Factor] ≈ 0.28 (14)

Note that this may be a somewhat optimistic estimate because the existingwind parks very likely already took many of the best locations.

For example, if we apply this average load factor to the Enercon E82turbine, we get an expected output power density

E[PE82den ] = E[Load Factor]× P rated

den = 0.28× 11.9 = 3.33 W/m2 (15)

which is very close to our previous estimate of 3.15 W/m2 in (10).Using our reference average value of P rated

den ≈ 10 W/m2, we get

E[P Tout] = E[Load Factor]× P rated

den × Land area

≈ 2.8 (W/m2)× Land area (m2) (16)

14

This will be the main model to estimate the wind energy potential in Portu-gal.

4.2 Wind intermittency

Figure 8 already shows a lot of variability in the load factor, which is a directresult of wind speed variability through time. In fact, wind speed intermit-tency is commonly pointed out as the main challenge to the integration ofsignificant wind production in a national electricity grid. Hence, we take abrief detour to examine the distribution of the load factor in more detail.

Figure 9 shows the distribution of the load factor for all wind parks inPortugal during 2010, conditional on the the season of the year and the timeof day. First, we note that the load factor depends on the season of the year.Winter shows the highest values with a median load around 0.4, while Sum-mer shows the lowest values with median loads always below 0.2. The higheravailability in Winter is good as it matches the higher demand for heating.Second, the load varies throughout the day: the median load factor decreasesin the early afternoon in all seasons, except Winter. This is unfortunate asthese are hours of high demand. If the standards of living increase and airconditioning becomes more common in Portugal, the demand may peak inthe middle of hot summer days, exactly when wind is less available. Finally,we note that the distribution of the load factor is always very wide. There isa consistently large range between the 25th and 75h percentiles and the tailsof the distribution extend a lot beyond those points, reaching values veryclose to zero in almost every hour of every season (Summer is an exception,but because the distribution is packed in lower values).

In summary, wind energy is very intermittent. Currently, since wind hasdispatch priority in Portugal, the intermittency of wind is compensated bynatural gas power plants, which have fast online/offline times. In the future,the intermittency of wind may also be compensated by better electricitystorage techniques (for example, batteries in electric vehicles), by a supergrid able to integrate many countries, or by other renewable technologies.In particular, solar energy is probably a good match to wind: solar will bestronger exactly during the periods where wind is weaker (the middle of theday and the Summer months).

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Figure 9: Conditional load factor distribution for Portugal, 2010Each figure describes the daily evolution the load factor, defined as production over in-

stalled capacity, for all wind parks in Portugal during 2010. We assume that the installed

capacity increases linearly throughout the year from 3507 MW up to 3865 MW. Hourly

production is from REN. On each box, the central mark is the median, the edges of the box

are the 25th and 75th percentiles, the whiskers extend to the most extreme data points

not considered outliers, and outliers are represented individually as dots.

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Hour of day

Winter

Load

fact

or

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Hour of day

SpringLo

ad fa

ctor

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Hour of day

Summer

Load

fact

or

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Hour of day

Autumn

Load

fact

or

16

5 Estimates of wind generation in Portugal

Table 3 shows the energy demand in Portugal during 2010 (source DGEG).The amount of electricity consumed was near 50 TWh and the total primaryenergy was 206 TWh.

Table 3: Annual energy demand in Portugal, 2010.(Source: DGEG)

Demand TWh

1 Total Electricity 49.901.1 Electr. from Wind 9.181.2 Electr. from other Renewables 16.951.3 Electr. from fossil fuels and imports 23.772 Primary Energy for other usages 156.293=1+2 Total Energy Demand 206.18

To estimate how much energy we can generate from wind, we use equation(16) for different areas of Portugal covered by windmills. More precisely, wefind the required area for wind to supply each of the demand items describedin the previous table.

Table 4 shows the results. We estimate that we need 374 km2 of land togenerate the 9.18 TWh of wind energy in 2010. According to DGEG, therewere 208 wind farms in Portugal, so it seems that our model is reasonablywell calibrated (it would be interesting to know the actual area of the existingwind farms, but this information is not available).

With 2.2% of the area of Portugal, we would be able to generate all ofthe electricity demanded in 2010. This would require less than 5 times thecurrently installed wind hardware. With 9.1% of the area of Portugal, wewould be able to generate enough wind energy to cover all our primary energyneeds from all sources!

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Table 4: Annual potential wind energy supply in PortugalThe “expected total output power” represents the amount of power that is generated from

the land area in the first column, assuming a ratio of 2.8 W/m2, as defined in equation

(16). The “expected total energy” is this total power times 8760 hours.

Land area Expectedtotal

outputpower

Expectedtotal

energy

Corresponding in 2010 to

(km2) (%) (MW) (TWh)

2 034 2.2% 5 696 49.90 Total Electricity374 0.4% 1 048 9.18 Electr. from Wind691 0.7% 1 935 16.95 Electr. from other Renewables969 1.0% 2 713 23.77 Electr. from fossil fuels and im-

ports6 372 6.9% 17 841 156.29 Primary Energy for other us-

ages8 406 9.1% 23 537 206.18 Total Energy Demand92 300 100.0% 258 440 2 263.93 Total area of Portugal

References

Andrews, J., and N. Jelley, 2007, Energy Science — principles, technologies,and impacts. Oxford University Press.

Esteves, T. M. V., 2004, “Base de dados do potencial energetico do vento emPortugal,” Master Thesis, Universidade de Lisboa.

Gore, A., 2009, Our choice - a plan to solve the climate crisis. Bloomsbury.

MacKay, D. J., 2009, Sustainable Energy — without the hot air. UIT Cam-bridge Ltd.

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