02 1 synchronous-machines

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02 1 synchronous-machines

  1. 1. SYNCHRONOUS MACHINESCopyright © P. KundurThis material should not be used without the authors consent 1539pk
  2. 2. Synchronous Machines Outline1. Physical Description2. Mathematical Model3. Parks "dqo" transportation4. Steady-state Analysis  phasor representation in d-q coordinates  link with network equations1. Definition of "rotor angle"2. Representation of Synchronous Machines in Stability Studies  neglect of stator transients  magnetic saturation1. Simplified Models2. Synchronous Machine Parameters3. Reactive Capability Limits SM - 2 1539pk
  3. 3. Physical Description of a Synchronous Machine Consists of two sets of windings:  3 phase armature winding on the stator distributed with centres 120° apart in space  field winding on the rotor supplied by DC Two basic rotor structures used:  salient or projecting pole structure for hydraulic units (low speed)  round rotor structure for thermal units (high speed) Salient poles have concentrated field windings; usually also carry damper windings on the pole face. Round rotors have solid steel rotors with distributed windings Nearly sinusoidal space distribution of flux wave shape obtained by:  distributing stator windings and field windings in many slots (round rotor);  shaping pole faces (salient pole) SM - 3 1539pk
  4. 4. Rotors of Steam Turbine Generators Traditionally, North American manufacturers normally did not provide special “damper windings”  solid steel rotors offer paths for eddy currents, which have effects equivalent to that of amortisseur currents European manufacturers tended to provide for additional damping effects and negative-sequence current capability  wedges in the slots of field windings interconnected to form a damper case, or  separate copper rods provided underneath the wedges Figure 3.3: Solid round rotor construction SM - 4 1539pk
  5. 5. Rotors of Hydraulic Units Normally have damper windings or amortisseurs  non-magnetic material (usually copper) rods embedded in pole face  connected to end rings to form short-circuited windings Damper windings may be either continuous or non- continuous Space harmonics of the armature mmf contribute to surface eddy current  therefore, pole faces are usually laminated Figure 3.2: Salient pole rotor construction SM - 5 1539pk
  6. 6. Balanced Steady State Operation Net mmf wave due to the three phase stator windings:  travels at synchronous speed  appears stationary with respect to the rotor; and  has a sinusoidal space distribution mmf wave due to one phase: Figure 3.7: Spatial mmf wave of phase a SM - 6 1539pk
  7. 7. Balanced Steady State Operation The mmf wave due to the three phases are: MMFa = Kia cos γ ia = Im cos( ωs t )  2π   2π  MMFb = Kib cos γ −  ib = Im cos ωs t −   3   3   2π   2π  MMFc = Kic cos γ +  ia = lm cos ωs t +   3   3  MMFtotal = MMFa + MMFb + MMFc 3 = KIm cos( γ − ωs t ) 2 SM - 7 1539pk
  8. 8. Balanced Steady State Operation Magnitude of stator mmf wave and its relative angular position with respect to rotor mmf wave depend on machine output  for generator action, rotor field leads stator field due to forward torque of prime mover;  for motor action rotor field lags stator field due to retarding torque of shaft load Figure 3.8: Stator and rotor mmf wave shapes SM - 8 1539pk
  9. 9. Transient Operation Stator and rotor fields may:  vary in magnitude with respect to time  have different speed Currents flow not only in the field and stator windings, but also in:  damper windings (if present); and  solid rotor surface and slot walls of round rotor machines Figure 3.4: Current paths in a round rotor SM - 9 1539pk
  10. 10. Direct and Quadrature Axes The rotor has two axes of symmetry For the purpose of describing synchronous machine characteristics, two axes are defined:  the direct (d) axis, centered magnetically in the centre of the north pole  The quadrature (q) axis, 90 electrical degrees ahead of the d-axis Figure 3.1: Schematic diagram of a 3-phase synchronous machine SM - 10 1539pk
  11. 11. Mathematical Descriptions of a Synchronous Machine For purposes of analysis, the induced currents in the solid rotor and/or damper windings may be assumed to flow in two sets of closed circuits  one set whose flux is in line with the d-axis; and  the other set whose flux is along the q-axis The following figure shows the circuits involved Figure 3.9: Stator and rotor circuits SM - 11 1539pk
  12. 12. Review of Magnetic Circuit Equations (Single Excited Circuit) Consider the elementary circuit of Figure 3.10 dΨ ei = dt dΨ e1 = + ri dt Ψ = Li The inductance, by definition, is equal to flux linkage per unit current φ L=N = N2P i where P = permeance of magnetic path  = flux = (mmf) P = NiP Figure 3.10: Single excited magnetic circuit SM - 12 1539pk
  13. 13. Review of Magnetic Circuit Equations (Coupled Circuits) Consider the circuit shown in Figure 3.11 dΨ1 e1 = + r1i1 dt dΨ2 e2 = + r2i2 dt Ψ1 = L11i1 + L21i2 Ψ2 = L21i1 + L22i2 with L11 = self inductance of winding 1 L22 = self inductance of winding 2 L21 = mutual inductance between winding 1 and 2 Figure 3.11: Magnetically coupled circuit SM - 13 1539pk
  14. 14. Basic Equations of a Synchronous Machine The equations are complicated by the fact that the inductances are functions of rotor position and hence vary with time The self and mutual inductances of stator circuits vary with rotor position since the permeance to flux paths vary Iaa = L al + Igaa = L aa 0 + L aa 2 cos 2θ  2π  Iab = Iba = −L ab 0 + L ab 2 cos 2θ −   3   π = −L ab 0 − L ab 2 cos 2θ +   3 The mutual inductances between stator and rotor circuits vary due to relative motion between the windings Iafd = L afd cos θ Iakd = L akd cos θ  π Iakq = L akq cos θ +  = −L akq sin θ  2 SM - 14 1539pk
  15. 15. Basic Equations of a Synchronous Machine Dynamics of a synchronous machine is given by the equations of the coupled stator and rotor circuits Stator voltage and flux linkage equations for phase a (similar equations apply to phase b and phase c) dΨa ea = − R aia = pΨa − R aia dt Ψa = −laaia −labib − lacic + lafdifd + lakdikd + lakqikq Rotor circuit voltage and flux linkage equations e fd = pΨfd + R fdifd 0 = pΨkd + R kdikd 0 = pΨkq + R kqikq ψ fd = L ffdifd + L fkdikd   2π   2π   − L afd ia cos θ + ib cos θ −  + ic cos θ +    3   3  ψ kd = L fkdifd + Lkkdikd   2π   2π   − L afd ia cos θ + ib cos θ −  + ic cos θ +    3   3  ψ kq = Lkkdikq   2π   2π   + L akq ia sin θ + ib sin θ −  + ic sin θ +    3   3  SM - 15 1539pk
  16. 16. The dqo Transformation The dqo transformation, also called Parks transformation, transforms stator phase quantities from the stationary abc reference frame to the dqo reference frame which rotates with the rotor   2π   2π    cos θ cos θ −  cos θ +    3   3  id  ia  i  = 2  2π   2π   ib    3 q  − sin θ − sin θ − 3  − sin θ + 3       i  i0   1 1 1  c    2  2 2  The above transformation also applies to stator flux linkages and voltages With the stator quantities expressed in the dqo reference frame  all inductances are independent of rotor position (except for the effects of magnetic saturation)  under balanced steady state operation, the stator quantities appear as dc quantities  during electromechanical transient conditions, stator quantities vary slowly with frequencies in the range of 1.0 to 3.0 Hz The above simplify computation and analysis of results. SM - 16 1539pk
  17. 17. Physical Interpretation of dqo Transformation The dqo transformation may be viewed as a means of referring the stator quantities to the rotor side In effect, the stator circuits are represented by two fictitious armature windings which rotate at the same speed as the rotor; such that:  the axis of one winding coincides with the d-axis and that of the other winding with the q-axis  The currents id and iq flowing in these circuits result in the same mmfs on the d- and q-axis as do the actual phase currents The mmf due to id and iq are stationary with respect to the rotor, and hence:  act on paths of constant permeance, resulting in constant self inductances (Ld, Lq) of stator windings  maintain fixed orientation with rotor circuits, resulting in constant mutual inductances SM - 17 1539pk
  18. 18. Per Unit Representation The per unit system is chosen so as to further simplify the model The stator base quantities are chosen equal to the rated values The rotor base quantities are chosen so that:  the mutual inductances between different circuits are reciprocal (e.g. Lafd = Lfda)  the mutual inductances between the rotor and stator circuits in each axis are equal (e.g., Lafd = Lakd) The P.U. system is referred to as the "Lad base reciprocal P.U. system" One of the advantages of having a P.U. system with reciprocal mutual inductances is that it allows the use of equivalent circuits to represent the synchronous machine characteristics SM - 18 1539pk
  19. 19. P.U. Machine Equations in dqo reference frame The equations are written with the following assumptions and notations:  t is time in radians  p = d/dt  positive direction of stator current is out of the machine  each axis has 2 rotor circuits Stator voltage equations e d = pψ d − ψ qωr − R aid e q = pψ q + ψ dωr − R aiq e 0 = pψ 0 − R a i 0 Rotor voltage equations e fd = pψ fd + R fdifd 0 = pψ 1d + R 1di1d 0 = pψ 1q + R 1qi1q 0 = pψ 2q + R 2qi2q SM - 19 1539pk
  20. 20. P.U. Machine Equations in dqo Reference Frame (contd) Stator flux linkage equations ψ d = −( Lad + Ll ) id + Lad ifd + Lad i1d ψ q = − ( Laq + Ll ) iq + Laq i1q + Laq i 2 q ψ 0 = −L0 i0 Rotor flux linkage equations ψ fd = L ffdifd + L f 1di1d − L adid ψ 1d = L f 1difd + L11di1d − L adid ψ 1q = L11qi1q + L aqi2q − L aqiq ψ 1q = L aqi1q +L 22qL 2 q − L aqiq Air-gap torque T e = ψ diq − ψ qid SM - 20 1539pk
  21. 21. Steady State Analysis Phasor RepresentationFor balanced, steady state operation, the stator voltages maybe written as: e a = Em cos( ωt + α ) eb = Em cos( ωt − 2π 3 + α ) e c = Em cos( ωt + 2π 3 + α )with ω = angular velocity = 2πf α = phase angle of ea at t=0Applying the d,q transformation, e d = Em cos( ωt + α − θ ) e q = Em sin( ωt + α − θ )At synchronous speed, the angle θ is given by θ = ωt + θ0with θ = value of θ at t = 0Substituting for θ in the expressions for ed and eq, e d = Em cos( α − θ 0 ) e q = Em sin( α − θ 0 ) SM - 21 1539pk
  22. 22. Steady State Analysis Phasor Representation (contd) The components ed and eq are not a function of t because rotor speed ω is the same as the angular frequency ω of the stator voltage. Therefore, ed and eq are constant under steady state. In p.u. peak value Em is equal to the RMS value of terminal voltage Et. Hence, e d = E t cos( α −θ0 ) e q = E t sin( α −θ0 ) The above quantities can be represented as phasors with d-axis as real axis and q-axis as imaginary axis Denoting δi, as the angle by which q-axis leads E e d = E t sin δi e q = E t cos δi SM - 22 1539pk
  23. 23. Steady State Analysis Phasor Representation (contd) The phasor terminal voltage is given by ~ in the d-q coordinates E t = ed + je q in the R-I coordinates = E R + jE l This provides the link between d,q components in a reference frame rotating with the rotor and R, I components associated with the a.c. circuit theory Under balanced, steady state conditions, the d,q,o transformation is equivalent to  the use of phasors for analyzing alternating quantities, varying sinusoidally with respect to time The same transformation with θ = ωt applies to both  in the case of machines, ω = rotor speed  in the case of a.c. circuits, ω = angular frequency SM - 23 1539pk
  24. 24. Internal Rotor Angle Under steady state e d = −ωψq − idR a = ωL qiq − idR a = X qiq −idR a Similarly e q = ωψd − iqR a = − X did + X adifd − iqR a Under no load, id=iq=0. Therefore, ψ q = −L qiq = 0 ψ d = L adifd ed = 0 e q = L adifd ~ and E t = e d + jeq = jL adifd Under no load, Et has only the q-axis component and δi=0. As the machine is loaded, δi increases. Therefore, δi is referred to as the load angle or internal rotor angle. It is the angle by which q-axis leads the phasor Et SM - 24 1539pk
  25. 25. Electrical Transient Performance To understand the nature of electrical transients, let us first consider the RL circuit shown in Figure 3.24 with e = Emsin (ωt+α). If switch "S" is closed at t=0, the current is given by di e=L + iR dt solving ( L) t Em sin( ωt + α − φ) −R i = Ke + Z The first term is the dc component. The presence of the dc component ensures that the current does not change instantaneously. The dc component decays to zero with a time constant of L/R Figure 3.24: RL Circuit SM - 25 1539pk
  26. 26. Short Circuit Currents of a Synchronous Machine If a bolted three-phase fault is suddenly applied to a synchronous machine, the three phase currents are shown in Figure 3.25. Figure 3.25: Three-phase short-circuit currents SM - 26 1539pk
  27. 27. Short Circuit Currents of a Synchronous Machine (contd) In general, fault current has two distinct components: a) a fundamental frequency component which decays initially very rapidly (a few cycles) and then relatively slowly (several seconds) to a steady state value b) a dc component which decays exponentially in several cycles This is similar to the short circuit current in the case of the simple RL circuit. However, the amplitude of the ac component is not constant  internal voltage, which is a function of rotor flux linkages, is not constant  the initial rapid decay is due to the decay of flux linking the subtransient circuits (high resistance)  the slowly decaying part of the ac component is due to the transient circuit (low resistance) The dc components have different magnitudes in the three phases SM - 27 1539pk
  28. 28. Elimination of dc Component by Neglecting Stator Transients For many classes of problems, considerable computational simplicity results if the effects of ac and dc components are treated separately Consider the stator voltage equations e d = pψ d − ω ψ q − idR a e q = pψ q + ω ψ d − iqR a transformer voltage terms: pψd, pψq speed voltage terms: ω ψ q , ω ψ d The transformer voltage terms represent stator transients:  stator flux linkages (ψd, ψq) cannot change instantaneously  result in dc offset in stator phasor current If only fundamental frequency stator currents are of interest, stator transients (pψd, pψq) may be neglected. SM - 28 1539pk
  29. 29. Short Circuit Currents with Stator Transients Neglected The resulting stator phase currents following a disturbance has the wave shape shown in Figure 3.27 The short circuit has only the ac component whose amplitude decays Regions of subtransient, transient and steady state periods can be readily identified from the wave shape of phase current Figure 3.27: Fundamental frequency component of short circuit armature current SM - 29 1539pk
  30. 30. Synchronous Machine Representation in System Stability Studies Stator Transients (pψd, pψq) are usually neglected  accounts for only fundamental frequency components of stator quantities  dc offset either neglected or treated separately  allows the use of steady-state relationships for representing the transmission network Another simplifying assumption normally made is setting ω = 1 in the stator voltage equations  counter balances the effect of neglecting stator transients so far as the low-frequency rotor oscillations are concerned  with this assumption, in per unit air-gap power is equal to air-gap torque (See section 5.1 of book for details) SM - 30 1539pk
  31. 31. Equation of Motion (Swing Equation) The combined inertia of the generator and prime- mover is accelerated by the accelerating torque: dωm J = Ta = Tm − Te dt where Tm = mechanical torque in N-M Te = electromagnetic torque in N-m J = combined moment of inertia of generator and turbine, kg•m2 m = angular velocity of the rotor in mech. rad/s t = time in seconds SM - 31 1539pk
  32. 32. Equation of Motion (contd) The above equation can be normalized in terms of per unit inertia constant H 1 Jω2 m H= 0 2 VA base where  0m = rated angular velocity of the rotor in mechanical radians per second Equation of motion in per unit form is d ωr 2H = Tm − Te dt where ωmωr = = per unit rotor angular velocity ω0m Tmω0mTm = VA base = per unit mechanical torque Te ω0mTe = VAbase = per unit electromechanical torque Often inertia constant M = 2H used SM - 32 1539pk
  33. 33. Magnetic Saturation Basic equations of synchronous machines developed so far ignored effects of saturation  analysis simple and manageable  rigorous treat a futile exercise Practical approach must be based on semi- heuristic reasoning and judiciously chosen approximations  consideration to simplicity, data availability, and accuracy of results Magnetic circuit data essential to treatment of saturation given by the open-circuit characteristic (OCC) SM - 33 1539pk
  34. 34. Assumptions Normally Made in the Representation of Saturation Leakage inductances are independent of saturation Saturation under loaded conditions is the same as under no-load conditions Leakage fluxes do not contribute to iron saturation  degree of saturation determined by the air-gap flux For salient pole machines, there is no saturation in the q-axis  flux is largely in air For round rotor machines, q-axis saturation assumed to be given by OCC  reluctance of magnetic path assumed homogeneous around rotor periphery SM - 34 1539pk
  35. 35.  The effects of saturation is represented as L ad = K sdL adu (3.182) L aq = K sqL aqu (3.183) Ladu and Laqu are unsaturated values. The saturation factors Ksd and Ksq identify the degrees of saturation. As illustrated in Figure 3.29, the d-axis saturation is given by The OCC. Referring to Figure 3.29, ΨI = Ψat 0 − Ψat (3.186) Ψat (3.187) K sd = Ψat + ΨI For the nonlinear segment of OCC, can be ΨI expressed by a suitable mathematical function: ΨI = A sat eBsat ( Ψat − ΨTI ) (3.189) SM - 35 1539pk
  36. 36. Open-Circuit Characteristic (OCC) Under no load rated speed conditions id = iq = Ψq = e d = 0 E t = e q = Ψd = L adifd Hence, OCC relating to terminal voltage and field current gives saturation characteristic of the d-axis Figure 3.29: Open-circuit characteristic showing effects of saturation SM - 36 1539pk
  37. 37.  For salient pole machines, since q-axis flux is largely in air, Laq does not vary significantly with saturation  Ksq=1 for all loading conditions For round rotor machines, there is saturation in both axes  q-axis saturation characteristic not usually available  the general industry practice is to assume Ksq = Ksd For a more accurate representation, it may be desirable to better account for q-axis saturation of round rotor machines  q-axis saturates appreciably more than the d- axis, due to the presence of rotor teeth in the magnetic path Figure 3.32 shows the errors introduced by assuming q-axis saturation to be same as that of d-axis, based on actual measurements on a 500 MW unit at Lambton GS in Ontario  Figure shows differences between measured and computed values of rotor angle and field current  the error in rotor angle is as high as 10%, being higher in the underexcited region  the error in the field current is as high as 4%, being greater in the overexcited region SM - 37 1539pk
  38. 38.  The q-axis saturation characteristic is not readily available  It can, however, be fairly easily determined from steady-state measurements of field current and rotor angle at different values of terminal voltage, active and reactive power output  Such measurements also provide d-axis saturation characteristics under load  Figure 3.33 shows the d- and q-axis saturation characteristics derived from steady-state measurements on the 500 MW Lambton unit Figure 3.33: Lambton saturation curves derived from steady-state field current and rotor angle measurements SM - 38 1539pk
  39. 39. Example 3.3 Considers the 555 MVA unit at Lambton GS and examines  the effect of representing q-axis saturation characteristic distinct from that of d-axis  the effect of reactive power output on rotor angle Table E3.1 shows results with q-axis saturation assumed same as d-axis saturation Table E3.1 Pt Qt Ea (pu) Ksd δi (deg) ifd (pu) 0 0 1.0 0.889 0 0.678 0.4 0.2 1.033 0.868 25.3 1.016 0.9 0.436 1.076 0.835 39.1 1.565 0.9 0 1.012 0.882 54.6 1.206 0.9 -0.2 0.982 0.899 64.6 1.089 Table E3.2 shows results with distinct d- and q-axis saturation representation Table E3.2 Pt Qt Ksq Ksd δi (deg) ifd (pu) 0 0 0.667 0.889 0 0.678 0.4 0.2 0.648 0.868 21.0 1.013 0.9 0.436 0.623 0.835 34.6 1.559 0.9 0 0.660 0.882 47.5 1.194 0.9 -0.2 0.676 0.899 55.9 1.074 SM - 39 1539pk
  40. 40. Simplified Models for Synchronous Machines Neglect of Amortisseurs  first order of simplification  data often not readily available Classical Model (transient performance)  constant field flux linkage  neglect transient saliency (xd = xq) Et x′d E´ Steady-state Model  constant field current  neglect saliency (xd = xq = xs) Et xs Eq = Xadifd Eq SM - 40 1539pk
  41. 41. Reactive Capability Limits of Synchronous Machines In voltage stability and long-term stability studies, it is important to consider the reactive capability limits of synchronous machines Synchronous generators are rated in terms of maximum MVA output at a specified voltage and power factor which can be carried continuously without overheating The active power output is limited by the prime mover capability The continuous reactive power output capability is limited by three considerations  armature current limit  field current limit  end region heating limit SM - 41 1539pk
  42. 42. Armature Current Limit Armature current results in power loss, and the resulting heat imposes a limit on the output The per unit complex output power is ~ * S = P + jQ = E t ~t = E t It ( cos φ + j sin φ ) I where Φis the power factor angle In a P-Q plane the armature current limit, as shown in Fig. 5.12, appears as a circle with centre at the origin and radius equal to the MVA rating Fig 5.12: Armature current heating limit SM - 42 1539pk
  43. 43. Field Current Limit Because of the heating resulting from RfdI2fd power loss, the field current imposes the second limit The phasor diagram relating Et, It and Eq (with Ra neglected) is shown in Fig. 5.13 Equating the components along and perpendicular to the phasor Et X adifd sin δ i = X slt cos φ X adifd cos δ i = E t + X slt sin φ Therefore X ad P = E tlt cos φ = E tifd sin δi Xs X E2 Q = E tlt sin φ = ad E tifd cos δi − t Xs Xs The relationship between P and Q for a given field current is a circle centered at on the Q-axis and with as the radius. The effect of the maximum field current on the capability of the machine is shown in Fig. 5.14 In any balanced design, the thermal limits for the field and armature intersect at a point (A) which represents the machine name-plate MVA and power factor rating SM - 43 1539pk
  44. 44. Field Current LimitFig. 5.13: Steady state phasor diagramFig. 5.14: Field current heating limit SM - 44 1539pk
  45. 45. End Region Heating Limit The localized heating in the end region of the armature affects the capability of the machine in the underexcited condition The end-turn leakage flux, as shown in Fig. 5.15, enters and leaves in a direction perpendicular (axial) to the stator lamination. This causes eddy currents in the laminations resulting in localized heating in the end region The high field currents corresponding to the overexcited condition keep the retaining ring saturated, so that end leakage flux is small. However, in the underexcited region the field current is low and the retaining ring is not saturated; this permits an increase in armature end leakage flux Also, in the underexcited condition, the flux produced by the armature current adds to the flux produced by the field current. Therefore, the end-turn flux enhances the axial flux in the end region and the resulting heating effect may severely limit the generator output, particularly in the case of a round rotor machine Fig. 5.16 shows the locus of end region heating limit on a P-Q plane SM - 45 1539pk
  46. 46. End Region Heating LimitFig. 5.15: Sectional view end region of a generator Fig. 5.16: End region heating limit SM - 46 1539pk
  47. 47. Reactive Capability Limit of a 400 MVAHydrogen Cooled Steam Turbine Generator Fig. 5.18 shows the reactive capability curves of a 400 MVA hydrogen cooled steam turbine driven generator at rated armature voltage  the effectiveness of cooling and hence the allowable machine loading depends on hydrogen pressure  for each pressure, the segment AB represents the field heating limit, the segment BC armature heating limit, and the segment CD the end region heating limit Fig. 5.18: Reactive capability curves of a hydrogen cooled generator at rated voltage SM - 47 1539pk
  48. 48. Effect of Changes in Terminal Voltage Et Fig. 5.17: Effect of reducing the armature voltage on the generator capability curve SM - 48 1539pk

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