9.0 Wind Turbine Operation | 83 | |
The example below illustrates the above formula: |
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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)? |
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Solution: |
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Corrected Output = Standard Output *Ct *Ca |
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Ps = 50 kW | ||
519 | = .951 | |
Ct = |
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| ||
86 + 460 |
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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 |