WIND ENERGY SYSTEMS - Cd3wd

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Chapter 5—Electrical Network 5–18 of wind causes the input power to exceed the pullout power from the generator, the rotor will accelerate above rated speed. Large generator currents will flow and the generator will have to be switched off the power line. Then the rotor will have to be slowed down and the generator resynchronized with the grid. Rapid pitch control of the rotor can prevent this, but the control system will have to be well designed. Figure 12: Power flow from an ac generator as a function of power angle. The other feature illustrated by this power plot is that the power becomes negative for negative δ. This means the generator is now acting as a motor. Power is being taken from the electric utility to operate a giant fan and speed up the air passing through the turbine. This is not the purpose of the system, so when the wind speed drops below some critical value, the generator must be disconnected from the utility line to prevent motoring. Before working an example, we need to discuss generator rating. Generators are often rated in terms of apparent power rather than real power. The reason for this is the fact that generator losses and the need for generator cooling are not directly proportional to the real power. The generator will have hysteresis and eddy current losses which are determined by the voltage, and ohmic losses which are determined by the current. The generator can be operated at rated voltage and rated current, and therefore with rated losses, even when the real power is zero because θ = 90 degrees. A generator may be operated at power factors between 1.0 and 0.7 or even lower depending on the requirements of the grid, so the product of rated voltage and rated current (the rated apparent power) is a better measure of generator capability than real power. The same argument is true for transformers, which always have their ratings specified in kVA or MVA rather than kW or MW. A generator may also have a real power rating which is determined by the allowable torque in the generator shaft. A rating of 2500 kVA and 2000 kW, or 2500 kVA at 0.8 power factor, would imply that the machine is designed for continuous operation at 2500 kVA output, with 2000 kW plus losses being delivered to the generator through its shaft. There are always safety Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 5—Electrical Network 5–19 factors built into the design for short term overloads, but one should not plan to operate a generator above its rated apparent power or above its rated real power for long periods of time. We should also note that generators are rarely operated at exactly rated values. A generator rated at 220 V and 30 A may be operated at 240 V and 20 A, for example. The power in the wind is continuously varying, so a generator rated at 2500 kVA and 2000 kW may be delivering 300 kW to the grid one minute and 600 kW the next minute. Even when the source is controllable, as in a coal-fired generating plant, a 700-MW generator may be operated at 400 MW because of low demand. It is therefore important to distinguish between rated conditions and operating conditions in any calculations. Rated conditions may not be completely specified on the equipment nameplate, in which case some computation is required. If a generator has a per phase rated apparent power S R and a rated line to neutral voltage V R , the rated current is I R = S R V R (34) Example The MOD-0 wind turbine has an 1800 r/min synchronous generator rated at 125 kVA at 0.8 pf and 480 volts line to line[8]. The generator parameters are R s = 0.033 Ω/phase and X s = 4.073 Ω/phase. The generator is delivering 75 kW to the grid at rated voltage and 0.85 power factor lagging. Find the rated current, the phasor operating current, the total reactive power, the line to neutral phasor generated voltage E, the power angle delta, the three-phase ohmic losses in the stator, and the pullout power. The first step in the solution is to determine the per phase value of terminal voltage, which is The rated apparent power per phase is S R = 125 3 The rated current is then |V | = 480 √ 3 = 277 V/phase =41.67 kVA/phase = 41, 670 VA/phase I R = S R 41, 670 = = 150.4 A V R 277 The real power being supplied to the grid per phase is P = 75 3 =25 kW/phase = 25 × 103 W/phase From Eq. 14 we can find the magnitude of the phasor operating current to be Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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