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Chapter 6—Asynchronous Generators 6–27 The base power for this machine is √ 3(230 V)(96 A) = 38,240 W. The rated power is √ 3(230)(96) cos θ, which will always be smaller than the base power. Because of this feature of the per unit system, the mechanical input power should not exceed 1.0 pu except for very short periods because the machine is already overloaded at P m = 1.0 pu. The base impedance is 132.8/96 = 1.383 Ω. The base capacitance is 1/(Z base ω base ) = 1/[(1.383)(377)] = 1918 µF line to neutral. A line to line capacitance of 385 µF, for example, would be represented in our analysis by a line to neutral capacitance of 3(385) = 1155 µF, which has a per unit value of 1155/1918 = 0.602 pu. A good starting point for the capacitance on experimental induction generators in the 5-50 hp range seems to be about 0.6 pu. Changing capacitance will change performance, but oscillation should occur with this value of capacitance. Returning to our discussion of Fig. 14, we see that for curve 1, representing a load of 2.42 pu and a capacitance of 0.602 pu, the voltage varies from 0.68 pu to 1.13 pu as P m varies from 0.22 pu to 0.59 pu. The variation is nearly linear, as would be expected. When the resistance is decreased to 1.39 pu with the same capacitance, we get curve 2. At the same P m of 0.59 pu, the new voltage will be about 0.81 pu. The electrical power out, V 2 L /R L, will remain the same if losses do not change. We see that the voltage is determined by the resistance and not by the capacitance. Curves 2 and 3 and curves 4 and 5 show that changing the capacitance while keeping resistance essentially constant does not cause the voltage to change significantly. Changing the capacitance will cause the frequency of oscillation to change and therefore the machine speed. We see how the speed varies with P m in Fig. 15. A decrease in capacitance causes the speed to increase, for the same P m . The change will be greater for heavy loads (small R L ) than for light loads. The speed will also increase with P m for a given R L and C. The increase will be rather rapid for light loads, such as curves 4 and 5. The increase becomes less rapid as the load is increased. We even have the situation shown in curve 7 where power is changing from 0.4 to 0.6 pu with almost no change in speed. The frequency will change to maintain resonance even if the speed does not change so we tend to have high slip where the speed curves are nearly horizontal. For this particular machine the efficiency stayed at about 90 percent even with this high slip and no other operational problems were noted. However, small increases in load would cause significant increases in speed, as seen by comparing curves 6 and 7. It would seem therefore, that this constant speed-high slip region should be avoided by adding more capacitance. Curves 7, 3, and 2 show that speed variation becomes more pronounced as capacitance is increased from 0.446 pu to 0.602 pu. We could conclude from this argument that a capacitance of 0.524 pu is the minimum safe value for this machine even though a value of 0.446 pu will allow operation. We now want to consider the proper strategy for changing the load to maintain operation under changing wind conditions. The mechanical power output P m from the wind turbine is assumed to vary from 0 to 1.0 pu. A capacitance value of 0.524 is assumed for discussion purposes. At P m = 1.0 pu the voltage is 1.09 pu and the speed is 1.01 pu for R L = 1.38 pu. These are good maximum values, which indicate that good choices have been made for R L and C. As input power decreases to 0.44 pu the speed decreases to 0.944 pu. If input power is decreased still more, the induction generator gets out of the nonlinear saturation region and Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 6—Asynchronous Generators 6–28 oscillation will cease. We therefore need to decrease the load (increase R L ). Note that there is a gap between curves 3 and 4, so we may have a problem if we change from R L =1.38pu to 4.66 pu. The voltage will be excessive on the larger resistance and we may lose oscillation with the smaller resistance, while trying to operate in the gap area. We need an intermediate value of R L such that the curve for the larger R L will intersect the curve for the smaller R L , as is the case for curves 1 and 2. Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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