The synchronous machine (SM) in the power system (2) (Where does the electricity come from)?

Transcription

1 1 The synchronous machine (SM) in the power system (2) (Where does the electricity come from)?

2 2 Lecture overview Synchronous machines with more than 2 magnetic poles The relation between the number of poles and rotation speed The magnetic field in the SM Circuit model for a SM and power when we have a salient pole rotor Introduction to the dynamics of machines Photos of practical SM s

3 3 3 phase, 2 pole synchronous machine Rotor A round rotor Stator

4 4 A 3 phase, 2 pole generator Rotor windings (or field windings) for Direct current with direction into the figure The stator windings for phase a with direction into the figure Rotor windings (or field windings) for Direct current with direction out of the figure A salient pole rotor The rotor The stator Magnetic poles, N and S The stator windings for phase -a with direction out of the figure

5 5 A 3 phase, 4 pole generator Rotor windings (or field windings) for Direct current with direction into the figure A salient pole rotor -a c b -b -c N S Rotor Stator The stator windings for phase a with direction into the figure a -c b S -a N c -b a Magnetic poles, N and S Rotor windings (or field windings) for Direct current with direction out of the figure The stator windings for phase a with direction out of the figure

6 6 A 3 phase, 6 pole machine We have and increased number of poles both on the rotor and stator

7 7 f e Speed of rotation and number of poles p p n p = f n m = = ω m = 2 ωe p f e = 50 Hz f e = 60 Hz Legend: ω e = Electrical radians ω m = Mechanical radians f e = Frequency of the AC voltage (50 or 60 Hz) f m = Revolutions per second p = Number of poles (2,4,6...) n = Revolutions per minute (rpm) 6000 = p n n= 7800 = p n n= p p 7800 p An example for the number of poles: n (f = 50 Hz)

8 8 The Magnetic Field in the SM No load condition δ = 0 A phase angle α between the rotor and stator current (stator field) This angle is also called δ

9 9 The Magnetic Field in the SM (2)

10 Reactive Power from a 10 Synchronous Machine Assume the machine is connected to an infinite bus with constant voltage V. φ I δ V E jxi Locus for vector I with constant real power and variable exitation Locus for vector E with constant real power and variable exitation and reactive power The length of this line is proportional to E cosδ V and hence the reactive power generation and exitation

11 Phasor diagram with Power a Engineering constant - Egill Benedikt Hreinsson 11 magnetization E = V + jx I I f t s a E V E V f t f t a = = + j jx s jx s jx s X s V jx t s = j V X t s δ 1 E jx f 2 f 2 s = j E jx E X s f 1 f 1 s = j E X s E f 2 δ 2 E f 1 The locus for a vector of the constant excitation voltage δ 1 ji X δ 2 a1 s A locus for the vector of armature current in the case of constant excitation voltage φ 1 I a1 φ 2 I a2 V t ji a2 X s

12 12 Salient pole rotor q-axis q-axis d - and q - axis are different d-ás V d I d I q φ I d-axis δ V V q E q jx I d d jx I q q q-ás d-axis d - axis is directged along the pole direction. q axis is perpendicular to the pole directions in electrical degrees Geometry Flux Inductance Currents and voltages E = V + jx I + jx I q d d q q

13 13 P and Q for salient pole rotor P + jq = V I = ( V + jv ) ( I + ji ) * * e e d q d q P + jq = ( V + jv ) ( I ji ) e e d q d q Pe = VdId + VqIq V = Vsinδ V = V cosδ d q V d I d d-axis I q φ I δ V Vq E q jx I d d q-axis jx I q q P = I Vsinδ + I V e d q We substitute: I d = E q V X d q I q cosδ = V X d q Eq Vq Vd Pe = Vsinδ + V X X d cosδ Eq V cosδ V sinδ Pe = Vsinδ + V X X d q q cosδ

14 14 P and Q for salient pole rotor P e 2 2 EV q V V = sinδ sinδ cosδ + sinδ cosδ X X X d d q Power P e 2 EV q V 1 1 Pe = sinδ + sin 2δ = Pfield + P Xd 2 Xq X d Q e We use: sin 2δ sinδ cosδ = q 2 = cos + reluctance EV sin δ cos δ δ V X d Xq X d Note when: δ P field P reluctance X d = X q

15 15 The swing equation P e is the power delivered from one or more machines into the power system J is the moment of inertia (one or more machines) P m is the shaft mechanical power from the turbine p is the number of poles dw m = P P m e The energy balance in rotation, or the swing dt equation is as follows: ω m f e Wm = Jωm = W0 = W0 Kinetic energy in rotation 2 ωm0 f e0 as a function of frequency: The relation between rotation angle and the electrical angle of the stator current: ω m = 2 ωe p ω ω e m = 2π f e ω = 2π f e0 e0 = 2 π f ω = m m0 2 π f m0 W m is the kinetic rotation energy of one or more machines w e, f e is the electrical angle and frequency for the electric oscillation w m, f m is rotation angle and frequency of the rotor mechanical rotation. (The index 0, means that the quantity is a constant at the nominal frequency, 50 or 60 Hz)

16 16 The swing equation (2) We differentiate this equation... 1 ω 2 2 ω f 2 m e Wm = J m = W0 = W0 2 ωm0 f e0 dw dω 2W ω dω = ω J = dt dt dt...and then we get: 0 m m m m m 2 ωm0 We substitute into the swing equation: dw m = P P m e dt 2W0 dωm Pm Pe = 2 m0 dt m m ω ω ω 2W0 dωm 2 m0 dt ω = T T m e

17 17 The inertia constant, H H = We define the following quantity: The kinetic energy of the rotating mass Rated power for the generators in this mass 1 J 2 W ωm m0 H 2 S S = = rating rating The unit for H is: Ws/VA = s We examine now the former version of the swing equation... 2W0ωm dωm 2 ωm0 dt and make the m m0 m m = = 1 2 following + approximation where the deviation form the nominal frequency, Δω is small relative to ω = P P ω 1 ω +Δω 1 Δω 1 ω ω ω ω ω ω m0 m0 m0 m0 m0 m0 m e

18 18 The swing equation (3) We then get the following equation: 2W0 dωm ω m0 dt = P m P e we divide by the rating S rating and get: ω 2W0 dωm Pm Pe = S dt S S m0 rating rating rating By defining the power from the turbine (or system) as a fraction of the rated power, we get: H ω 2 m m0 dω dt = P P mn ( ) en ( )

19 19 Two equilibrium points Synchronizing torque from δ deviation P e P m dp e /dδ > 0 for δ < 90 - stable equilibrium dp e /dδ < 0 for δ > 90 - unstable equilibrium δ

20 H on different MVA bases 20 Machine base Steam turbines 4-9 s Gas turbines 3-4 s Hydro turbines 2-4 s Synchronous compensator s Common base H ~generator size Infinite bus has infinite H Narrow range!

21 21 Hydroeelectric rotor The multiple poles are shown clearly on the rotor

22 22 Inside the Búrfell power house

23 23 Rotor in a hydro station

24 A rotor from Laxá hydro station, 24 North Iceland

25 A stator from Laxá hydro station, 25 North Iceland

26 A stator from Laxá hydro station, 26 North Iceland

27 27 The stator in a hydroelectric generator

28 MVA, 83.3 rpm stator

29 References 29 P.C. Sen: "Principles of Electric Machines and Power Electronics"; 2. edition: John Wiley & Sons, 1997 O.I. Elgerd: Electric Energy Systems Theory, McGraw-Hill, 1983 A.E. Fitzgerald, C. Kingsley Jr., S.D. Umans: Electric Machinery, McGraw-Hill 2003 P. Kundur: Power System Stability and Control, McGraw-Hill 1994 D.P.Kothari, I.J. Nagrath: Modern Power System Analysis, McGraw-Hill 2003

30 30 Example 7

31 31 Example 8

32 32 Example 9

33 33 Example 7 - solution

34 34 Example 8 - solution

35 35 Example 9 solution