ECE 421/521 Electric Energy Systems Power Systems Analysis I 3 Generators, Transformers and the Per-Unit System. Instructor: Kai Sun Fall 2013

Transcription

1 ECE 41/51 Electric Energy Systems Power Systems Analysis I 3 Generators, Transformers and the Per-Unit System Instructor: Kai Sun Fall 013 1

2 Outline Synchronous Generators Power Transformers The Per-Unit System

3 Synchronous Generators Salient-pole rotor Cylindrical/round rotor Field winding Armature winding Stator Field current 3

4 Types of Rotors Salient pole rotors Concentrated windings on poles and nonuniform air gap Short axial length and large diameter Hydraulic turbines operated at low speeds (large number of poles) Have damper windings to help damp out speed oscillations Round rotors 70% of large synchronous generators (150~1500MA) Distributed winding and uniform air gap Large axial length and small diameter to limit the centrifugal forces Steam and gas turbines, operated at high speeds, typically 3600 or 1800rpm ( or 4-pole) Eddy in the solid steal rotor gives damping effects 4 16 poles salient-pole rotor (1 MW) Round rotor generator under construction (Source:

5 d Generator Model n Flux linkage with coil a (leading the axis of aa by ωt) λa = Nφcosωt Induced voltage: dλa ea = = ωnφsinωt = Emax sinωt dt π = Emax cos( ωt ) m N S ψ e a q Axis of coil a (reference) E = ωnφ = π fnφ max (occurring when ωt=π/) E 4.44 fnφ = (rms value) P n f = (n: synchronous speed in rpm; P: the number of poles) 60 Assume: i a is lagging e a by ψ (i a reaches the maximum when mn aligns with aa ) 4 ia = Imax sin( ωt ψ) ib = Imax sin( ωt ψ π) ic = Imax sin( ωt ψ π) 3 3 5

6 Magnetic motive forces (mmf s) of three phases: F = Ki = KI sin( ωt ψ) = F sin( ωt ψ) a a max m Fb = Kib = KI max sin( ωt ψ π) = Fmsin( ωt ψ π) Fc = Kic = KI max sin( ωt ψ π) = Fmsin( ωt ψ π) 3 3 Total mmf F s due to three phases of the stator Component along mn: F1 = Fm sin( ωt ψ)cos( ωt ψ) + Fm sin( ωt ψ π)cos( ωt ψ π) Fm sin( ωt ψ π)cos( ωt ψ π) 3 3 F m π 4π F = [sin ( ωt ψ) + sin ( ωt ψ ) + sin ( ωt ψ )] = Component orthogonal to mn 6 m N S d F r ψ ψ e a π π 4π 4π F = Fm sin( ωt ϕ) sin( ωt ψ) + Fm sin( ωt ψ ) sin( ωt ψ ) + Fm sin( ωt ψ ) sin( ωt ψ ) F 4 3 F m [3 cos ( t ) cos ( t π π = ω ψ + ω ψ ) + cos ( ωt ψ )] = F sin α = ( 1 cos α) / 3 3 m F s = 3 F m F s, the resultant total armature/stator mmf has constant amplitude, and is orthogonal to mn and revolving synchronously with F r F s q n Axis of coil a (reference)

7 d Phasor Diagram F r F sr n No-load conditions: No phase currents and F s =0 In each phase, field mmf F r produces no-load E (excitation voltage) Loading conditions: Armature carries three-phase currents m N ψ ψ e a F S s q Axis of coil a (reference) F s 0 is orthogonal to mn and induces emf E ar The sum of F r and F s gives mmf F sr in air gap The resultant air gap flux φ sr induces emf E sr =E+E ar Define the reactance of the armature reaction X ar = -E ar /(ji a ) E = Esr + jx aria F r F sr Terminal voltage, resistance R a and leakage reactance X l E = + [ R + j( X + X )] I = + ( R + jx ) I a ar a a s a F s E sr E ar E X s =X l +X ar is known as the synchronous reactance (Here, we ignore the difference between d and 7 q axes, so it is good for round-rotor generators)

8 Circuit Model (per Phase) F r F sr When F r is ahead F sr by δ r, the machine is operating as a generator When F r falls behind of F sr by δ r, (I a changes the direction) the machine acts as a motor Synchronous condenser (δ r =0), used to supply or absorb ARs to regulate line voltages. Since X l 0, δ r δ Generated real power: E F s jx s I a E P3 φ 3 sinδ X s Load δ r or δ is known as the power angle 8

9 E = + ( R + jx ) I a s a E Load I a = E δ 0 Z s γ Consider different power factors 9

10 Real and Reactive Power Outputs Consider a generator connected to an infinite bus, which is an equivalent bus of a large power network and has constant voltage (magnitude, angle and frequency). E = + jx I s a E δ 0 S = * 3ϕ 3I a E P3 φ = 3 sin δ X P max(3 φ ) = 3 s E X s T T P 3φ = = ω max = 3 E 3 sin δ ωx E ωx s s Q3 φ = 3 ( E cosδ ) 3 ( E ) X X s s 10

11 Consider constant real power output: P I I const * 3φ =R 3 a = 3 a cos θ =. I cos θ = ab = const. a E δ 0 cd = E sinδ = X I cosθ = 1 1 s a1 1 const. θ 3 θ 3 a b Q3 φ 3 ( E ) X s θ 1 >0, θ =0, θ 3 <0 Reactive power output can be controlled by means of the rotor excitation (adjusting E) while maintaining a constant real power output If E >, the generator delivers reactive power to the bus, and the generator is said to be overexcited. If E <, the generator is absorbing reactive power from the bus Generators are main source of reactive power, so they are normally operated in the overexcited mode δ E Minimum excitation: when δ=δ max =90 o, i.e. the stability limit 11

12 Examples 3.1 E δ 0 1

13 E δ 0 P 1 3φ X s δ = sin 3 E 13

14 Example 3. E δ 0 14

15 E δ 0 15

16 Comparison E δ I a θ 3.6k 10.6 o 807.5A o 18.7k 1.8 o 769.8A k 9.7 o 96.3A o 6.9K (minimum) 90 o 073A 68. o When excitation decreases, armature current first decreases and then increases (-shape function) power factor changes from lagging to leading 16

17 Curve of a synchronous machine: the curve of the armature current I a as the function of the field current I f Example 3.3: for the generator of Example 3.1, construct the curve for the rated power of 40MW with varying field excitation from 0.4 P.F. leading to 0.4 P.F. lagging. Assume the open-circuit characteristics in the operating region is given by E =000 I f () P = 40(MW) = 30 (k) 3 I a P3 φ 40 = = (ka) 3 cosθ 30 3 cosθ θ=cos -1 (0.4)~ -cos -1 (0.4) 30 E = + jx sia = + j9 Ia θ (k) 3 I = E 1000 / 000(A) f 17

18 Salient-Pole Synchronous Generators The model developed in previous slides assumes uniform magnetic reluctance in the air gap (round rotors). d n N E = + [ R + j( X + X )] I = + ( R + jx ) I a ar a a s a ψ e a q Axis of coil a (reference) For a salient-pole rotor, the reluctance in the rotor direct axis (d axis) is less than that in the quadrature axis (q axis) Represent the machine by two circuits respectively for d and q axes Define d-axis and q-axis reactances (X d >X q ) to replace a single synchronous reactance X s Resolve I a into d-axis and q-axis components: I d and I q m S 18

19 E = cosδ + XdId P3 φ = 3 I a cosθ I cosθ = ab + de = I cosδ + I sin δ a q d sin δ = XqIq I q = sin δ X q I d = E cosδ X d E Xd Xq P3 3 ( I cos δ q Id sin δ ) 3 sin 3 sin X δ = + = + φ X X δ d d q Approximately, the second item can be ignored: 19 E P3 φ 3 sinδ X d

20 Power Transformer Ideal transformers Winding resistance is negligible No leakage flux Permeability of the core is infinite (net mmf to establish flux in core is zero.) No core loss Real transformers: Windings have resistance Windings do not link the same flux Permeability of the core is not infinite Core losses (hysteresis losses and eddy current losses due to time varying flux) 0

21 Ideal Transformer Assuming sinusoidal flux φ = Φ max cos 1max ωt dφ e1 = N1 = ωn1φmax sin ωt dt = E cos + 90 E E ( ωt ) = π fn Φ 1max 1 max = 4.44 fn Φ 1 1 max i Note: Flux is lagging the induced voltage by 90 o For an ideal transformer No flux leakage: E = 4.44 fn Φ max Zero reluctance in core (exact mmf balance between the primary and secondary) IN = IN 1 E I N = = E I N 1 1 1

22 Real Transformers Modeling the current under no-load conditions (due to finite core permeability and core losses) I =0 I 1 =I 0 0 I 0 =I m +I c Ideal Transformer I m is the magnetizing current to set up the core flux In phase with flux and lagging E 1 by 90 o, modeled by E 1 /(jx m1 ) I c supplies the eddy-current and hysteresis losses in the core A power component, so it is in phase with E 1, modeled by E 1 /(R c1 ) Modeling flux leakages Primary and secondary flux leakage reactances: X 1 and X Modeling winding resistances Primary and secondary winding resistances: R 1 and R

23 Exact Equivalent Circuit E = + ZI E = N E 1 1 N N N = + N 1 1 N 1 1 N ZI N N = + N = + ZI ZI Z = R + jx N N = R + j X N 1 1 N Exact equivalent circuit referred to the primary side 3

24 Approximate Equivalent Circuits For no-load conditions: 1 E 1 R 1 and X 1 can be combined with R and X to obtain equivalent R e1 and X e1 ( ) = + R + jx I 1 e1 e1 N R = R + R 1 e1 1 N N X = X + X 1 e1 1 N I = 3 * S L Approximate equivalent circuit referred to the primary side ( ) = + R + jx I 1 e e Approximate equivalent circuit referred to the secondary side 4

25 Simplified Circuits Referred to One Side Power transformers are generally designed with very high permeability core and very small core loss. If ignoring the shunt branch: 5

26 Determination of Equivalent Circuit Parameters Open-circuit (no-load) test neglect (R 1 +jx 1 )I 0 Measure input voltage 1, current I 0, power P 0 (core/iron loss) P 0 R c1 = 1 I 1 c P = I 0 R m = I0 Ic c1 X m1 = I 1 m 6

27 Short-circuit test Apply a low voltage SC to create rated current I SC Neglect the shunt branch due to the low core flux I SC P SC SC Z e1 = I sc sc R P = sc e1 X ( I sc ) e1 = Ze 1 Re 1 7

28 Transformer Performance Efficiencies vary from 95% to 99% output power η = = input power Pout P + P + P out var const n S PF S PF = = n S PF + n P + P S PF + n P + P / n ( ) ( ) cu c cu c S =3 I full-load rated A n: fraction of the full load power S PF P cu =3R e I full-load copper loss (current dependent) P c : core/iron loss at rated voltage (mainly voltage dependent, almost constant) Maximum efficiency (constant PF) occurs when dη P = c 0 n = di P cu Learn Example 3.4 8

29 oltage Regulation % change in terminal voltage from no-load to full load (rated) Generator Transformer nl rated R= 100 rated E rated R= 100 rated nl rated R= 100 rated Utilizing the equivalent circuit referred to the primary/secondary side ( ) = + R + jx I 1 e1 e1 1 R= 100 ( ) = + R + jx I 1 e e 1 R= 100 9

30 Three-Phase Transformer Connections Four possible combinations: Y-Y, -, Y- and -Y Y-connection: lower insulation costs, with neutral for grounding, 3 rd harmonics problem (3 rd harmonic voltages/currents are all in phase, i.e. v an3 =v bn3 =v cn3 = m cos3ωt) -connection: more insulation costs, no neutral, providing a path for 3 rd harmonics (all triple harmonics are trapped in the loop), able to operate with only two phases (-connection) In Y-Y or - connection, H/L ratio is same for line and phase voltages Y- and -Y are commonly used as voltage step-down and step-up transformers, respectively 30

31 Y- and -Y Connections Y- and Y- connections will result in a phase shift of 30 o between the primary and secondary line-to-line voltages According to the American Standards Association (ASA), the windings are arranged such that the H side line voltage leads the L side line voltage by 30 o E.g. Y- Connection: YP An NY = = = N P ab a YL YP AB = = An 3 = = L P ab YL L AB = = ab 3a 31

32 Per-Phase Model for Y- or -Y Connection Neglect the shunt branch Replace the connection by an equivalent Y connection (Z Y =Z /3) Work with only one phase (equivalent impedances are line-to-neutral values) 3

33 Autotransformers The primary and secondary coils are electrically connected A conventional two-winding transformer can be changed into an autotransformers by connecting its two windings in series Secondary is either manually or automatically variable. (Source: EPRI Power System Dynamic Tutorial) 33 0/5 MA (69/40k) McGraw-Edison Substation Auto-Transformer (Y-Y) (Source:

34 Autotransformer Model I L 1 I N1 = = = a I N 1 N S = + I = (1 + ) I auto ( ) N1 1 = (1 + ) S = S + S a w w conducted N N = + = + = + = (1 + a ) 1 1 H 1 L L L N N 1 IL = I1 + I = I1 + I1 = (1 + a) IH N N Power rating advantage 34 Equivalent Circuit (if the equivalent impedance is referred to the N 1 -turn side)

35 Advantages and Disadvantages of Autotransformers Advantages Smaller and more efficient Lower leakage reactance Lower loss Lower exciting current Higher ka rating ariable output voltage when sliding contact is used. Disadvantage No isolation between primary and secondary. 35

36 Example 3.5 =1440x50x

37 Three-Winding Transformers Primary (P), secondary (S) and tertiary (T) windings Typical applications: The same source supplying two independent loads at different voltages Interconnection of two transmission systems of different voltages In a substation, the tertiary winding is used to provide voltage for auxiliary power purposes or to supply a local distribution system Reactive power compensation by switched reactors or capacitors connected to the tertiary winding 37

38 Three-Winding Transformer Model Estimate impedances by short-circuit tests: Z ps = measured P impedance, with S short-circuited and T open Z pt = measured P impedance, with T short-circuited and S open Z st = measured S impedance, with T short-circuited and P open Referring Z st to the P side: Z st N p = N s Z st Z p, Z s and Z t are impedances of three windings referred to the P side: Z = Z + Z ps p s Z = Z + Z pt p t Z = Z + Z st s t ( ) ( ) ( ) Z = Z + Z Z p ps pt st Z = Z + Z Z s ps st pt Z = Z + Z Z t pt st ps / / / 38

39 oltage Control of Transformers oltage magnitude Reactive power flow oltage phase angle Real power flow Two commonly used methods Tap changing Transformers Regulating transformers 39

40 Tap Changing Transformers Practically all power transformers and many distribution transformers have taps in one or more windings for changing the turns ratio Off-load tap changing transformers Can only be adjusted when the transformer current flow has been completely interrupted Typically, 4 taps with.5% of rated voltage each tap to allow an operating range of -5% ~ +5% of rated voltage Under-load tap changing (ULTC) transformers More flexible when the transformer is in-service May have built-in voltage sensing circuitry for automatic tap changing to keep voltage constant Typically, off-load tap changers at the H side for -5% ~ +5% ULTC at the L side with 3 taps of 0.65% each for an operating range of -10% ~ +10% 40

41 ULTC Model I P+jQ How to adjust t S and t R to maintain 1 and at desired levels? R S ( ) = + R + jx I Since phase shift δ is small S S cosδ = + ab + de R = + IRcosθ + IXsinθ R R X = + I cosθ + I R R R R R = + R RP φ + R XQ φ sinθ t = t + S t s 1 R RP φ t R + XQ 1 RPφ + XQ = tr + 1 tr φ φ 41

42 Limitations of ULTC in oltage Control Normally, when the turns ratio is adjusted, the Mvar flow across the transformer is also adjusted However since a transformer itself absorbs Mvar to build its internal magnetic field, when its secondary voltage is raised via a tap change, its Mvar usage increases and its primary voltage often drops. The greater the tap change and the weaker the primary side, the greater the primary voltage drop. If the primary side is weak, the tap change may not necessarily increase the secondary voltage. Therefore, spare Mvar must be available for a tap change to be successful. An example: ±10% / 33-position ULTC (5/16x10%= +3.15% ~ 14.3k) (Neutral point: 0% ~ 138k) (+10% ~ 151.8k) 4 (Source: EPRI Dynamic Tutorial)

43 Regulating Transformers oltage Magnitude Control Exciting transformer Series transformer an = an + Using an an to induce a voltage (//) added to an 43

44 Regulating Transformers Phase Angle Control Phase shifting transformer (PST), phase angle regulator (PAR) or Quadrature booster an = an + bc bc Using bc to induce a voltage ( ) added to an Series transformer Exciting transformer MA 0/155k phase-shifting transformer (source: wikipedia.org)

45 Application of PST 40MW 90MW (Source: EPRI Dynamic Tutorial) 45

46 46 (Source: EPRI Dynamic Tutorial)

47 The Per-Unit System Quantity in per-unit = Actual quantity / Base value of quantity Why per-unit system? Neglecting different voltage levels of transformers, lines and generators Powers, voltages, currents and impedances are expressed as decimal fractions of respective base quantities A minimum of four base quantities are required to completely define a per-unit system S pu S S = pu B = pu B I I I = pu B Z = Z Z B 47

48 Usually, two bases are selected: Three-phase base volt-ampere S B or MA B Line-to-line base voltage B or k B The other two are calculated accordingly I B = SB 3 B Z B ( ) ( k ) B / 3 B B = = = I S MA B B B The phase and line quantities expressed in per-unit are the same S = Z I pu = pui pu pu pu pu Relationship between the load power and per-unit impedance S L 3 3 P P ( φ) = I Z P P 3 P L = L P L(3 ) SL(3 ) = = I S φ φ Z pu Z P L L SB L L SB SB pu ZB SL(3 φ ) B B S S S L 3φ L 3φ L pu = = = = = pu ( ) ( ) ( ) ( ) 48

49 ULTC Per-Unit Model t s 1 RPφ + XQ = tr + 1 tr φ If all quantities are in per-unit, t S 1pu and t R 1pu Assuming t S t R =1 t S = 1 1 RP + XQ φ 1 φ t R =1/t S Example 3.6: t S =1.08pu and t R =0.96pu 49

50 Change of Base Z old pu Z = = Z Ω old B Z Ω S old B old ( ) B Z S = new old new old pu B pu Zpu old new SB B Z new pu Z = = Z Ω new B Z Ω S new B new ( ) B S old = Z (if B is the same) pu S new pu old B The nameplates of transformers and generators usually give impedance as x% based on its rated voltage and rated MA. Also, it is usually operated under the same (or very close to) rated voltage level while the base MA may be quite different. Then, the impedance under the new, system base can be calculated using the above simplified equation. 50

51 Advantages of the Per-Unit System Gives a clear idea of relative magnitudes of various quantities, e.g. voltage, current, power and impedance The per-unit impedance of equipment of the same general type based on their own ratings fall in a narrow range regardless of the rating of the equipment. Whereas their impedance in ohms vary greatly with the rating The per-unit values of impedance, voltage and current of a transformer are the same regardless of whether they are referred to the primary or the secondary side Idea for the computerized analysis and simulation of complex power system problems The circuit laws are valid in per-unit systems. For three-phase systems, factors of 3 and 3 are eliminated by properly selecting base quantities 51

52 Example 3.7 B1 B B3 B4 Z B B5 Z B5 B6 B1 Z B4 5

53 Example

54 Homework #4 Read through Saadat s Chapter 3 ECE51: 3., 3.4, 3.5, 3.6, 3.9, 3.10, 3.14, 3.15, 3.17 ECE41: 3., 3.5, 3.6, 3.9, 3.10, 3.14, 3.15 Due date: 10/4 (Friday) submitted in class or by 54