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# Electrical power drivers et3026 wb lecture 7-1

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### Electrical power drivers et3026 wb lecture 7-1

1. 1. March 5, 2009 1 Elektrische Aandrijvingen WTB Lokatie/evenement P.BAUER
2. 2. March 5, 2009 2 Figure 16.1 Schematic iagram and cross-section view of a typical 500 MW synchronous generator and its 2400 kW dc exciter. The dc exciting current Ix (6000 A) flows through the commutator and two slip-rings. The dc control current Ic from the pilot exciter permits variable field control of the main exciter, which, in turn, controls Ix.
3. 3. March 5, 2009 3 Figure 16.13a Generator operating at no-load.
4. 4. March 5, 2009 4 Synchronous generator
5. 5. March 5, 2009 5 Figure 16.17 Equivalent circuit of a 3-phase generator, showing only one phase. Determining the value of xs
6. 6. March 5, 2009 6 Example 16.2 • 3 phase, SG produces an open circuit line voltage of 6928 V when the dc exc. current is 50 A. The AC terminals are short circuited and line currents are 800A. • Calculate synchronous reactance • Terminal voltage if three R 12 ohm are connected • Eo = EL / Sqrt(3)= 4000 V • Xs = Eo / I= 4000/800 • Z = Sqrt (R2 + Xs 2)
7. 7. March 5, 2009 7 SG under load
8. 8. March 5, 2009 8 Example 16.4 • 36 MVA, 20,8 kV 3phase, synchr. Reactance 9 ohm, the no load saturation curve is given.Terminal voltage remain 21kV., calculate the exciting current and draw the phasor diagram at: • No load • Resistive load of 36 MW • Capacitive load of 12 MVAr • E = 20,8 / Sqrt(3)= 12kV • Eo = E • Is = 100
9. 9. March 5, 2009 9 Example 16.4 • 36 MVA, 20,8 kV 3phase, synchr. Reactance 9 ohm, the no load saturation curve is given.Terminal voltage remain 21kV., calculate the exciting current and draw the phasor diagram at: • No load • Resistive load of 36 MW • Capacitive load of 12 MVAr • P = 36 / 3 = 12MW • I = P/E = 1000 A • Ex = j I Xs • Eo= • Is = 100
10. 10. March 5, 2009 10 Example 16.4 • 36 MVA, 20,8 kV 3phase, synchr. Reactance 9 ohm, the no load saturation curve is given.Terminal voltage remain 21kV., calculate the exciting current and draw the phasor diagram at: • No load • Resistive load of 36 MW • Capacitive load of 12 MVAr • Q = 12 / 3 = 12MVAr • I = Q/E = 333A • Ex = j I Xs • Eo= • Is = 70A
11. 11. March 5, 2009 11 Figure 16.23 Regulation curves of a synchronous generator at three different load power factors.
12. 12. March 5, 2009 12 Synchronization • The generator frequency is equal to system frequency • Voltage equal • Voltage in phase • Phase sequence the same
13. 13. March 5, 2009 13 Figure 16.26a Generator floating on an infinite bus. Synchronous generator on an infinite bus • The exciting current • Mechanical torque • Float on the line • Imposes voltage and frequency
14. 14. March 5, 2009 14 Figure 16.26b Over-excited generator on an infinite bus. • Ex = E0 - E Synchronous generator on an infinite bus • I = (E0 – E) / Xs • Increase the exciting current
15. 15. March 5, 2009 15 Figure 16.26c Under-excited generator on an infinite bus. • Ex = E0 - E • I = (E0 – E) / Xs Synchronous generator on an infinite bus • Decrease the exciting current
16. 16. March 5, 2009 16 Figure 16.27 a. Turbine driving the generator. b. Phasor diagram showing the torque angle d. Effect of varying the mechanical torque • Ex = E0 - E • Rotor accelerates, angle σ increases, electrical power too • Electrical power=Mechanical power
17. 17. March 5, 2009 17 Figure 16.27 a. Turbine driving the generator. b. Phasor diagram showing the torque angle d. Effect of varying the mechanical torque • Ex = E0 - E • Difference of potential is created when 2 equal voltages are out of the phase • Increased phase angle=increased I=incr. active power
18. 18. March 5, 2009 18 Figure 16.28a The N poles of the rotor are lined up with the S poles of the stator. Physical interpretation δ = p alpha /2
19. 19. March 5, 2009 19 Example 16.6 • The rotor poles of an 8 pole SG shift by 10 mechanical degrees from no load full load • Calculated the torque angle between Eo and E • Which voltage E or Eo is leading ? δ = p alpha /2 = 8 . 10 /2 = 40
20. 20. March 5, 2009 20 Figure 16.29 Graph showing the relationship between the active power delivered by a synchronous generator and the torque angle. Active power • P = E0 E sin δ / XS
21. 21. March 5, 2009 21 Example 16.7 • 36 MVA, 21 kV 1800 r/min 3 phase generator , Xs= 9 ohm per phase, Eo=12kV, E= 17,3kV • Calculated the active power when δ=30 • The peak power ? • P = E0 E sin δ / XS
22. 22. March 5, 2009 22 Figure 16.30 Variation of generator reactance following a short-circuit. Transient reactance
23. 23. March 5, 2009 23 Figure 16.32 Change in current when a short-circuit occurs across the terminals of a generator. See Example 16-8.
24. 24. March 5, 2009 24 Figure 16.33 Power flow between two voltage sources. Power transfer • E1 = E2 + jIX • P = E2 I cos θ • P = E1E2 sin δ / XS
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