Using Statistical Distributions to Approximate Available Energy

The distribution of wind speeds in Table 12-1 provides a solid basis for calculating available energy at this site. However, suppose we only knew U = 5.57 m/s, and did not have the hourly bin data for the year. Observations have shown that in many locations, if the average wind speed is known, the probability of the wind speed being in a given range can be predicted using a statistical “Rayleigh” distribution.2 The prob­ability density function (PDF) for this distribution has the following form:

exp[-(x/a)[51]]

a2

Подпись: f (x) = 2x • Подпись: for x > 0 Подпись: (12-2a)

f (x) = 0, for x < 0

where a is a shape parameter for the Rayleigh function, and x is the independent vari­able for which the probability is to be evaluated. The cumulative distribution function (CDF) for the Rayleigh is then

Подпись: (12-2b)F(x) = 0, for x < 0

F(x) = 1 – exp[-(x/a)2], for x > 0

For analysis of the distribution of wind speed, it is convenient to rewrite the CDF of the Rayleigh function in terms of U, which for nonnegative values of x gives

average

F(x) = 1 – exp[(-tf/4Xx/Uawrage)2] (12-3)

The CDF is used to calculate the probability that the wind speed will be at or below a given value U, given a known value of U :

average

p(windspeed < U) = 1 – exp[(-n/4)(U/Uaverage)2] (12-4)

So, for example, substituting the value U = 4 m/s into Eq. (12-4), the probability with U = 5.57 m/s that the wind speed is less than or equal to this value is 32%. The Rayleigh distribution is actually a special case of the Weibull distribution with shape parameter к = 2 and scale parameter c = U x (4/p)1/2. It is possible to use the Weibull

average

distribution to more closely fit the observed wind distribution in some locations than is possible with the Rayleigh, but we will not consider this possibility here. The applica­tion of the Rayleigh distribution to working with wind speed bins can be understood in Example 12-1.

Example 12-1 Using the Rayleigh distribution and Uaverage = 5.57 m/s, calculate the probability that the wind is in bin 6.

Solution The probability that the wind is in the bin is the difference between the probability of wind at the maximum value for the bin, and the probability of the minimum value. Bin 6 is between 4 and 5 m/s. From Eq. (12-4) above

P(windspeed < 4) = 1 – exp[(-n/4)(4/5.57)2] = 0.333 = 33.3%

P(windspeed < 5) = 1 – exp[(-n/4)(5/5.57)2] = 0.469 = 46.9%

Therefore, the bin probability is 46.9% – 33.3% = 13.6%. This is the value obtained using the Rayleigh distribution; the observed value is 1254 h/8760 h/year = 14.3%.

Подпись: —■—Observed ♦ Rayleigh
Using Statistical Distributions to Approximate Available Energy

To show how well the estimate using the Rayleigh distribution fits the above data, we can plot them next to each other, as shown in Fig. 12-9. From visual inspection,

Bin number

Figure 12-9 Comparison of observed and Rayleigh estimated probabilities of wind speeds in a given bin for wind speeds up to 14 m/s.

the Rayleigh estimated curve fits the observed data fairly well. Note that the esti­mating technique must be used with caution: although it happened to fit this data set well, there is no guarantee that for another location, the fit might not be quite poor, for example, in a case where the observed distribution has more than one peak.

In order to understand the impact of using an estimated versus observed distribu­tion for wind speed, we will calculate the estimate of energy available in the wind using both methods. For a given wind speed U, the power P in watts available in the wind per square meter of cross-sectional area is calculated as follows:

P = 0.5 pU3 (12-5)

Here p is the density of air in kg/m3, which often has values on the order of 1 kg/m3, depending on elevation above sea level and current weather conditions. From Eq. (12-5), the amount of power available in the wind grows with the cube of the wind speed, so there is much more power available in the wind at high wind speeds. For example, at Uaverage and p = 1.15 kg/m3, the available power is 99 W/m2, but at the upper range of the bins at 14 m/s, the power available is 1578 W/m2.

Based on the actual bin data, we can calculate the annual power available at the site by calculating the power available in each bin, based on the bin average speed, and multiplying by the number of hours per year to obtain energy in kWh. Example 12-2 tests the accuracy of the Rayleigh function compared to the observed data.

Example 12-2 Suppose a wind analyst calculates the wind energy available at the site given in Table 12-1, knowing only Uaverage for the site and using a Rayleigh function to calculate probabilities for the given wind bins. By what percent will the estimated energy differ from the value obtained if the bin data are known? Use an air density value of p = 1.15 kg/m3.

Solution Using Eq. (12-5) to calculate the power available in each bin and the percentage of the year in each bin to calculate the number of hours, it is possible to generate a table of observed and estimated power by bin. We will use bin 6, for which we calculated statistical probability in Example 12-1, as an example. The average speed in this bin is 4.5 m/s, therefore, the average power available is

P = (0.5)(1.15)(4.5)3 = 52.4 W/m2

According to the Rayleigh estimate, the wind is in the bin for (0.136)(8760 h/year) = 1191 h/year. Therefore, the values of the observed and estimated output in the bin are, respectively

Подпись:Estimated =(1191y;ar)(52.45-,

Repeating this process for the each bin gives the following results:

Bin

Power (W/m2)

Annual Output

Observed (kWh/m2)

Estimated (kWh/m2)

1

0.00

0.0

0.0

2

0.07

0.0

0.0

3

1.94

1.0

1.2

4

8.98

7.2

8.5

5

24.65

29.9

27.9

6

52.40

65.7

62.4

7

95.67

119.2

108.2

8

157.91

162.2

155.9

9

242.58

172.0

194.2

10

353.12

193.9

214.1

11

492.99

218.4

212.2

12

665.63

218.3

191.2

13

874.50

193.3

158.0

14

1123.05

139.3

120.5

15

1414.72

84.9

85.1

Total

1605.1

1539.3

From the observed bin data, this value is 1605 kWh/m2, while for the Rayleigh estimate the value is 1539 kWh/m2, excluding any wind above bin 15. So the Rayleigh underestimates the energy available at the site by a factor of 4.1%.

" N

X (*,3)/N

Подпись: 1=1 Using Statistical Distributions to Approximate Available Energy Подпись: 6/n Подпись: (12-6)

Example 12-2 shows that the Rayleigh approximation gives an energy estimate quite close to the observed value, in this case; for example, the difference is less than the year-to-year variability value of +10% discussed earlier. Alternatively, it is possible to make a rough approximation of the available power by using an adjusted value of the average speed that takes into account the effect of cubing the wind speed in the power equation. For any group of N numbers nv n2,. . . nN, it can be shown that the relationship between the average of the cube and the cube of the averages is the following:

On the basis of Eq. (12-6), it is possible to adjust the average wind speed by multi­plying by (6/p)1/3 and using the resulting value in the power equation to quickly estimate available power. For example, for the data from Table 12-1, the adjusted speed is 6.91 m/s, and the energy available at this wind speed if it is supplied continuously year round is 1662 kWh/m2, or 3.6% more than the observed value. On the other hand, if we simply use the unadjusted wind speed of 5.57 m/s for all 8760 h of the year, the estimated energy
available would be only 870 kWh/m2. This comparison shows the important contribution of winds at higher than the average wind speed to overall wind energy output.

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