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Chapter 9—Wind Power Plants 9–27 while for the double circuit case it is C s =4($0.363) + $1 = $2.452/ft C d =8($0.178) + $2 = $3.42/ft Note that both cases includes a neutral of the same ampacity as the three phase conductors. We now proceed to Table 2 of Appendix C to find the resistance of our conductors. We find that 2 AWG has a resistance of 0.2561 Ω per 1000 ft while 3/0 has a resistance of 0.1013 Ω per 1000 ft. The resistances R s and R d of Fig. 5 are therefore R s = (300/1000)(0.1013) = 0.03039 Ω R d = (300/1000)(0.2561) = 0.07683 Ω The power losses at rated current are P s = 3(220) 2 (0.03039) = 4413 W P d =1.5(220) 2 (0.07683) = 5578 W In this particular case, the double circuit is both more expensive and more lossy than the single circuit. If a sufficiently large ampacity is required, then the double circuit will be less expensive. The losses for a double circuit will always be greater than the losses of a single circuit of the same ampacity. We now need the yearly energy loss in the low voltage conductors before we can complete our economic analysis. We cannot just multiply the loss at rated current by 8760 hours per year because the turbines are operating at rated current only a small fraction of the year. From the wind speed duration curves and the curve of turbine power versus wind speed we can calculate the fraction of time that the turbine is at each power level. However, this does not give the full picture since the power factor of the generator decreases as power production decreases. This means that if a rated current of 220 A, for example, occurs at rated power, the current at half power will be greater than 220/2 = 110 A. A detailed analysis will require a histogram of current versus time for one year, which may be more trouble than it is worth. A crude estimate of low voltage loss can be obtained by starting from the capacity factor CF for turbines at this site. A capacity factor of 0.2, for example, means that the yearly energy production of a turbine can be calculated by assuming the turbine is producing full Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 9—Wind Power Plants 9–28 power for 20% of the time and is off the remaining time, or is producing 20% power all the time. In the first case, the yearly energy loss would be W s1 =(0.2)(8760)(3)(I) 2 R s (12) and in the second case, if we assume the current drops to 0.3I for 20% power, W s2 =(1.0)(8760)(3)(0.3I) 2 R s =0.45W s1 (13) W s1 is an upper bound for conductor losses. Depending on the variation of current with power, W s2 is a reasonable estimate for the lower bound of losses. An assumption of 0.6W s1 or 0.7W s1 should be adequate for most purposes. A similar argument can be made for the load losses of transformer T 1 . The current will be the same as in the low voltage conductors, and the resistance will just be that of the transformer windings. An estimate of (0.6)(CF)(8760)P LL should be quite acceptable. The no load losses of T 1 can be found by multiplying the no load power from Table 1 by the number of hours per year that the transformer is energized. Significant amounts of energy can be saved by opening the circuit breakers CB2 during long periods of low winds. These circuit breakers are not intended for frequent cycling, but a few times each week should be acceptable. The losses of the distribution voltage conductors can be determined in a manner similar to that of the low voltage conductors. Actually, the losses in these conductors will be much lower than the low voltage conductor losses and can be ignored without significant error. The reason for this is that the distribution voltage conductors must be sized for the worst case condition of one circuit breaker open and all the loop current flowing through the other one. In this case, the loop current increases from zero next to the open circuit breaker to rated at the operating circuit breaker. The average current in the loop would be approximately half the rated current, with losses on the order of one fourth the losses we would expect if the entire loop carried the same current. In normal operation, however, both circuit breakers CB2 will be operating, so the maximum current in the loop will be half the rated current. At some point around the midpoint of the loop, the current will actually be zero. The losses in this case will be on the order of one tenth the expected losses for a uniform current throughout the loop. This will usually be less than 0.5% of the energy produced by the windfarm, hence is not very significant. 8 PROTECTIVE RELAYS The circuit breakers are operated by a variety of protective relays which sense various operating conditions that may be harmful to the utility, the windfarm, or to operating personnel. Some functions, such as overcurrent, would be common to all the circuit breakers. Others, such as Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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