 ## Q. 6.15

Annual Energy Delivered Using a Spreadsheet. Suppose that a NEG Micon 60-m diameter wind turbine having a rated power of 1000 kW is installed at a site having Rayleigh wind statistics with an average windspeed of 7 m/s at the hub height.
a. Find the annual energy generated.
b. From the result, find the overall average efficiency of this turbine in these winds.
c. Find the productivity in terms of kWh/yr delivered per m² of swept area.

 TABLE 6.7 Examples of Wind Turbine Power Specifications Manufacturer: NEG Micon NEG Micon NEG Micon Vestas Whisper Wind World Nordex Bonus Rated Power (kW): 1000 1000 1500 600 0.9 250 1300 300 Diameter (m): 60 54 64 42 2.13 29.2 60 33.4 Avg. Windspeed v (m/s) v(mph) kW kW kW kW kW kW kW kW 0 0 0 0 0 0 0.00 0 0 0 1 2.2 0 0 0 0 0.00 0 0 0 2 4.5 0 0 0 0 0.00 0 0 0 3 6.7 0 0 0 0 0.03 0 0 4 4 8.9 33 10 9 0 0.08 0 25 15 5 11.2 86 51 63 22 0.17 12 78 32 6 13.4 150 104 159 65 0.25 33 150 52 7 15.7 248 186 285 120 0.35 60 234 87 8 17.9 385 291 438 188 0.45 92 381 129 9 20.1 535 412 615 268 0.62 124 557 172 10 22.4 670 529 812 356 0.78 153 752 212 11 24.6 780 655 1012 440 0.90 180 926 251 12 26.8 864 794 1197 510 1.02 205 1050 281 13 29.1 924 911 1340 556 1.05 224 1159 297 14 31.3 964 986 1437 582 1.08 238 1249 305 15 33.6 989 1006 1490 594 1.04 247 1301 300 16 35.8 1000 998 1497 598 1.01 253 1306 281 17 38.0 998 984 1491 600 1.00 258 1292 271 18 40.3 987 971 1449 600 0.99 260 1283 259 19 42.5 968 960 1413 600 0.97 259 1282 255 20 44.7 944 962 1389 600 0.95 256 1288 253 21 47.0 917 967 1359 600 0.00 250 1292 254 22 49.2 889 974 1329 600 0.00 243 1300 255 23 51.5 863 980 1307 600 0.00 236 1313 256 24 53.7 840 985 1288 600 0.00 230 1328 257 25 55.9 822 991 1271 600 0.00 224 1344 258 26 58.2 0 0 0 0 0.00 0 0 0

Source: Mostly based on data in www.windpower.dk.

## Step-by-Step

The 'Blue Check Mark' means that this solution was answered by an expert.

a. To find the annual energy delivered, a spreadsheet solution is called for. Let’s do a sample calculation for a 6-m/s windspeed to see how it goes, and then present the spreadsheet results.
From Table 6.7, at 6 m/s the NEG Micon 1000/60 generates 150 kW.
From (6.45), the Rayleigh p.d.f. at 6 m/s in a regime with 7-m/s average windspeed is

$f\left(v\right) \ = \ \frac{\pi \ v}{2 \bar{v}^{2}} \ \exp \ \left[-\frac{\pi }{4} \ \left(\frac{v}{\bar{v}}\right)^{2} \right] \ = \ \frac{\pi \ \cdot \ 6}{2 \ \cdot \ 7^{2}} \ \exp \ \left[- \frac{\pi}{4} \ \left(\frac{6}{7}\right)^{2}\right] \ = \ 0.10801$

In a year with 8760 h, our estimate of the hours the wind blows at 6 m/s is

$\text{Hours} \ @ 6 \ {m}/{s} \ = \ 8760 \ {h}/{yr} \ \times \ 0.10801 \ = \ 946 \ {h}/{yr}$

So the energy delivered by 6-m/s winds is

$\text{Energy} \ \left(@ 6 \ {m}/{s}\right) \ = \ 150 \ kW \ \times \ 946 \ {h}/{yr} \ = \ 141, \ 929 \ {kWh}/{yr}$

The rest of the spreadsheet is given below. The total energy produced is 2.85 × $10^{6}$ kWh/yr.

b. The average efficiency is the fraction of the wind’s energy that is actually converted into electrical energy. Since Rayleigh statistics are assumed, we can use (6.48) to find average power in the wind for a 60-m rotor diameter (assuming the standard value of air density equal to 1.225 kg/m³):

$\begin{matrix} \bar{P} \ = \ \frac{6}{\pi} \ \cdot \ \frac{1}{2} \rho A \bar{v}^{3} & = \ \frac{6}{\pi} \ \times \ 0.5 \ \times \ 1.225 \ \times \ \frac{\pi}{4} \left(60\right)^{2} \ \times \ \left(7\right)^{3} \\ & = \ 1.134 \ \times \ 10^{6} \ W \ = \ 1134 \ kW \quad \quad \quad \ \end{matrix}$

In a year with 8760 h, the energy in the wind is

$\text{Energy in wind} \ = \ 8760 \ {h}/{yr} \ \times \ 1134 \ kW \ = \ 9.938 \ \times \ 10^{6} \ kWh$

So the average efficiency of this machine in these winds is

$\text{Average efficiency} \ = \ \frac{2.85 \ \times \ 10^{6} \ {kWh}/{yr}}{9.938 \ \times \ 10^{6} \ {kWh}/{yr}} \ = \ 0.29 \ = \ 29 \%$

c. The productivity (annual energy per swept area) of this machine is

$\text{Productivity} \ = \ \frac{2.85 \ \times \ 10^{6} \ {kWh}/{yr}}{\left({\pi}/{4}\right) \ \cdot \ 60^{2} \ m^{2}} \ = \ 1008 \ {kWh}/{m^{2}} \ \cdot \ yr$
 Windspeed (m/s) Power (kW) Probability f (v) Hrs/yr at v Energy (kWh/yr) 0 0 0.000 0 0 1 0 0.032 276 0 2 0 0.060 527 0 3 0 0.083 729 0 4 33 0.099 869 28,683 5 86 0.107 941 80,885 6 150 0.108 946 141,929 7 248 0.102 896 222,271 8 385 0.092 805 310,076 9 535 0.079 690 369,126 10 670 0.065 565 378,785 11 780 0.051 444 346,435 12 864 0.038 335 289,551 13 924 0.028 243 224,707 14 964 0.019 170 163,779 15 989 0.013 114 113,101 16 1000 0.008 74 74,218 17 998 0.005 46 46,371 18 987 0.003 28 27,709 19 968 0.002 16 15,853 20 944 0.001 9 8,709 21 917 0.001 5 4,604 22 889 0.000 3 2,347 23 863 0.000 1 1,158 24 840 0.000 1 554 25 822 0.000 0 257 26 0 0.000 0 0 Total: 2,851,109

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