2. The table below shows a portion of a discretized estimate of the energy delivered by a Siemens 2300-kW, 101-m diameter wind turbine (Table 7.5) exposed to Rayleigh winds with average speed 6 m/s. For example, for winds blowing around 4 m/s (3.5 ≤ v ≤ 4.5), the turbine produces 100 kW of power and in a year's time, a Rayleigh pdf predicts it delivers 107,841 kWh. 2300 6 m/s P(MW) v avg v (m/s) kW kWh/yr 0 0 1 0 2 0 3 0 4 100 5 107.841 2 etc a. How many kWh/yr would be generated for winds blowing at 5 m/s winds? Do this by hand. Hint: first, use equation (7.46) to find the probability for wind in the 5 m/s window (v-5), then from table (7.5), find the power (kW) delivered by this turbine at 5 m/s wind speed window. The total energy produced with the 5 m/s wind per year is calculated by Energy (power at 5 m/s window) x (hours per year) x (probability of 5 m/s wind) = TABLE 7.5 Examples of Wind Turbine Power Specifications Brand: D (m): Wind Vestas Vestas Suzlon Suzlon GE GE GE Siemens Siemens Vergnet PR (kW): 7000 3075 2100 2100 2500 1600 1500 3000 2300 275 164 112 97 88 103 100 77 Power Power Power Power Power Power Power 101 101 32 Power Power Power (m/s) (kW) (kW) (kW) (kW) (kW) (kW) (kW) (kW) (kW) (kW) 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 3 0 20 20 0 10 4 4 60 0 0 10 4567890 120 130 80 10 85 60 40 130 100 3 480 300 220 130 205 190 120 280 230 18 950 550 420 305 400 460 250 480 420 36 1630 900 678 540 695 750 420 765 720 58 2550 1350 1020 830 1130 1080 640 1175 1100 98 3750 1920 1433 1180 1630 1390 920 1650 1530 141 5000 2500 1830 1523 2050 1540 1200 2200 2000 189 11 5950 2950 2050 1845 2340 1595 1360 2700 2240 243 12 6695 3060 2090 2040 2480 1620 1450 2900 2300 272 13 6960 3072 2100 2080 2500 1620 1490 2970 2300 275 14 6995 3075 2100 2100 2500 1620 1510 2990 2300 275 15 7000 3075 2100 2100 2500 1620 1510 3000 2300 275 16 7000 3075 2100 2100 2500 1620 1510 3000 2300 275 17 7000 3075 2100 2100 2500 1620 1510 3000 2300 275 20 21 23 622222 18 7000 3075 2100 2100 2500 1620 1510 3000 2300 275 19 7000 3075 2100 2100 2500 1620 1510 3000 2300 275 7000 3075 2100 2100 2500 1620 1510 3000 2300 275 7000 3075 0 0 2500 1620 1510 3000 2300 0 7000 3075 0 0 2500 1620 1510 3000 2300 0 7000 3075 0 0 2500 1620 1510 3000 2300 0 24 7000 3075 0 0 2500 1620 1510 3000 2300 0 25 7000 3075 0 0 2500 1620 1510 3000 2300 0 Hint: first, use equation (7.46) to find the probability for wind in the 5 m/s window (v-5), then from table (7.5), find the power (kW) delivered by this turbine at 5 m/s wind speed window. The total energy produced with the 5 m/s wind per year is calculated by Energy (power at 5 m/s window) x (hours per year) x (probability of 5 m/s wind) Solution: • Wind Probability density functions Rayleigh probability density function πυ f(v) = Probability density f(v) 0.20 0.16- 0.12 0.08 exp[-1] (7.46) 252 exp 4 Wind speed (mph) 30 10 20 V=4 m/s (8.9 mph) V=6 m/s (13.4 mph) 40 50 V-8 m/s (17.9 mph) 0.04 0.00- 0 5 10 15 20 25 Wind speed (m/s) FIGURE 7.27 The Rayleigh probability density function with varying average wind speeds.