- In This Task A Doubly Fed Induction Generator Will Be Studied Grid F P Gen Pmech Too Asg Compensation 1 Figure 2 Fixe 1 (71.09 KiB) Viewed 63 times
- In This Task A Doubly Fed Induction Generator Will Be Studied Grid F P Gen Pmech Too Asg Compensation 1 Figure 2 Fixe 2 (107.01 KiB) Viewed 63 times
- In This Task A Doubly Fed Induction Generator Will Be Studied Grid F P Gen Pmech Too Asg Compensation 1 Figure 2 Fixe 3 (58.38 KiB) Viewed 63 times
- In This Task A Doubly Fed Induction Generator Will Be Studied Grid F P Gen Pmech Too Asg Compensation 1 Figure 2 Fixe 4 (127.01 KiB) Viewed 63 times
- In This Task A Doubly Fed Induction Generator Will Be Studied Grid F P Gen Pmech Too Asg Compensation 1 Figure 2 Fixe 5 (65.54 KiB) Viewed 63 times
1.2 Pel 1.0 0.8 Electrical Power P/PN 0.6 0.4 0.2 Wind Speed 0 0 0.2 0.4 1.4 1.6 1.8 0.6 0.8 1.0 1.2 Turbine Speed n/nn 2 Figure 3: Electrical power vs. wind speed.
Grid th 40- -ziz 3- SG th = 3- Box Filter 4 Figure 4: Full converter-fed wind turbine system.
a) Explain why variable speed operation is needed in order to harvest maximum energy from the wind, Figure 3. b) Explain the drawbacks of the old fixed speed Danish concept. c) What are the advantages of the DFM (Figure 5) compared with the SG (Figure 4)? = = d) Consider a 70kW, 1600rpm, 50Hz, 4 pole doubly fed induction generator. Find the slip at rated speed, the frequency of the converter into the rotor, and the power flowand direction at the stator-side and the rotor side. Neglect losses in the machine. [Answer: s= -0.0667, fr = 3.33Hz, Pstator = 65.62kW, Protor = 4.375kW] e) Consider subtask d) with 1250rpm. The maximum stator power flow is 100kW. What is the maximum mechanical power at this speed? One needs to find the slip and the power flows. Neglect losses in the machine. (Answer: s = 0.1667, Pmech = 83.33kW, Protor = –16.67kW] =
Grid foto Pgen 1 DEM s*P GenT s*PGen 3- 3- Filter Converter 5 Figure 5: Doubly fed induction generator system.