9.0 Wind Turbine Operation  | 83  | |
The example below illustrates the above formula:  | 
  | |
If a 50 kW wind turbine generates 50 kW at 13.4 m/s (30 mph) wind speed under  | ||
standard conditions, what would the output be at an altitude of 1,980 m (6,500 feet) and  | ||
30°C (86°F)?  | 
  | |
Solution: | 
  | |
Corrected Output = Standard Output *Ct *Ca  | 
  | |
Ps = 50 kW  | ||
519  | = .951  | |
Ct =  | 
  | |
  | ||
86 + 460  | 
  | |
Ca = .9666500/1000 = .799 | ||
Corrected Output = (50kW) (.951) (.799) = 38.0 kW  | ||
This approach can also be used, albeit with less accuracy, to correct annual energy (kWh) output predictions.
9.4.5 Acoustics
The AOC 15/50 has been designed as a 
NOTICE: Use of the material contained in this document is subject to the warning on page Iv and the disclaimer on page v of this document.
DOC012R02 | AOC 15/50 User Manual  | Nov 2001  |