Question 6.11: Annual Energy Delivered by a Wind Turbine. Suppose that a NE......
Annual Energy Delivered by a Wind Turbine. Suppose that a NEG Micon 750/48 (750-kW generator, 48-m rotor) wind turbine is mounted on a 50-m tower in an area with 5-m/s average winds at 10-m height. Assuming standard air density, Rayleigh statistics, Class 1 surface roughness, and an overall efficiency of 30%, estimate the annual energy (kWh/yr) delivered.
We need to find the average power in the wind at 50 m. Since “surface roughness class” is given rather than the friction coefficient α, we need to use (6.16) to estimate wind speed at 50 m. From Table 6.4, we find the roughness length z for Class 1 to be 0.03 m. The average windspeed at 50 m is thus
\left(\frac{v}{v_{0}} \right) \ = \ \frac{\ln \left({H}/{z}\right)}{\ln \left({H_{0}}/{z}\right)} (6.16)
v_{50} \ = \ v_{10} \frac{\ln \left({H_{50}}/{z}\right) }{\ln \left({H_{10}}/{z}\right) } \ = \ 5 \ {m}/{s} \ \cdot \ \frac{\ln \left({50}/{0.03}\right) }{\ln \left({10}/{0.03}\right) } \ = \ 6.39 \ {m}/{s}Average power in the wind at 50 m is therefore (6.48)
\bar{P} \ = \ \frac{6}{\pi} \ \cdot \ \frac{1}{2} \rho A \bar{v} ^{3} \quad \left(\text{Rayleigh assumptions}\right) (6.48)
Since this 48-m machine collects 30% of that, then, in a year with 8760 hours, the energy delivered would be
| TABLE 6.4 Roughness Classifications for Use in (6.16) | |||||
| Roughness Class |
Description | Roughness Length z(m) |
|||
| 0 | Water surface | 0.0002 | |||
| 1 | Open areas with a few windbreaks | 0.03 | |||
| 2 | Farm land with some windbreaks more than 1 km apart | 0.1 | |||
| 3 | Urban districts and farm land with many windbreaks | 0.4 | |||
| 4 | Dense urban or forest | 1.6 | |||