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Course: EEL 4213, Fall 2011
School: FSU
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Word Count: 1418

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8: Chapter Transient Analysis of Synchronous Machines FAMU-FSU College of Engineering Synchronous Machines Steady state modeling rotor mmf and stator mmf are stationary with respect to each other flux linkage with the rotor are invariant with time no voltages are induced on the rotor circuits Transient modeling flux linkage changes with time differential equations have time-varying coefficients...

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8: Chapter Transient Analysis of Synchronous Machines FAMU-FSU College of Engineering Synchronous Machines Steady state modeling rotor mmf and stator mmf are stationary with respect to each other flux linkage with the rotor are invariant with time no voltages are induced on the rotor circuits Transient modeling flux linkage changes with time differential equations have time-varying coefficients Parks transformation dynamic behavior (c) Feb 2004 IG XS emf RA VT sub-transient period, transient period, and steady-state period Power Systems I 2 Transient Analysis Transient analysis will be applied in the dynamic study of generators Generators experience dynamic behavior during switching load faults Consider the transient behavior of an RL circuit with a switched voltage source R t=0 L i(t) V(t) (c) Feb 2004 Power Systems I 3 Transient Analysis The voltage source is sinusoidal: v (t ) = Vm sin ( t + ) The KVL equation are: d i (t ) 0 = Vm sin ( t + ) R i (t ) L dt i (t ) = I m sin ( t + ) I m e t sin ( ) R Vm L where : I m = , = , Z R 1 L = tan , V(t) R t=0 L i(t) and Z = R 2 + 2 L2 (c) Feb 2004 Power Systems I 4 Example Solve for the time-domain solution of the current for a faulted generator having the following characteristics R = 0.125 L = 10 mH v(t) = 151 sin (377 t + ) which will give (a) zero dc offset current, (b) maximum dc offset current (c) Feb 2004 Power Systems I 5 Example Z = 0.125 + j (377 )(0.01) = 0.125 + j 3.77 = 3.77288.1 151 0.01 = Im = = 40 A = 0.08 s 3.772 0.125 i (t ) = 40 sin(377 t + 88.1) 40 e t 0.08 sin( 88.1) (a) Let = 88.1 i (t ) = 40 sin (377 t ) (b) Let = (88.1 90) = 1.9 i (t ) = 40 sin (377 t 90) 40 e t 0.08 sin( 90) = 40 cos(377 t ) + 40 e t 0.08 (c) Feb 2004 Power Systems I 6 Example 40 (a) zero dc offset current i(t) 20 0 -20 -40 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.2 0.25 0.3 0.35 t, sec 80 60 40 i(t) (b) maximum dc offset current 20 0 -20 -40 0 0.05 0.1 0.15 t, sec (c) Feb 2004 Power Systems I 7 Transient Analysis Synchronous Machines Models and analysis were previously developed for steady state behaviors rotor and stator magnetic fields are stationary with respect to each other the flux linkage in the rotor circuit are constant in time the per phase equivalent circuit becomes a constant generated emf in series with a simple impedance Under transient conditions (time varying) the above assumptions are no longer valid (c) Feb 2004 changing stator current are reflected in a dynamic flux linkage changing flux linkage induces transient currents in the rotor transient rotor currents in turn react with the stator and the induced voltages Power Systems I 8 Synchronous Machine Model The synchronous machine consist of: three ac stator windings mounted on the stator one field winding mounted on the rotor two fictitious windings which model short-circuited paths of the damper windings When modeling, the following are assumed: a synchronously rotating reference frame with a speed of the reference frame is along the axis of phase a at time zero For transient analysis of an ideal synchronous machine The machine is represented as a group of magnetically coupled rotating circuits with inductances which depend on the angular position of the rotor (c) Feb 2004 Power Systems I 9 Synchronous Machine Model Reference Axis m Direct Axis Quadrature Axis b Q a c m = t + + 2 D F Q a F D c b Physical Layout (c) Feb 2004 Power Systems I 10 Synchronous Machine Model r rF VF r ia ic LF Laa rD Lcc LD LQ Lbb r in Schematic of mutually coupled circuits ib (c) Feb 2004 Power Systems I 11 Synchronous Machine Model The KVL equations for the model r 0 va 0 r v b 0 0 vc = 0 0 vF 0 0 0 0 0 0 v abc R abc v = 0 FDQ (c) Feb 2004 0 0 0 0 0 0 r 0 0 0 rF 00 0 rD 0 0 0 0 ia a 0 ib b 0 ic d c 0 iF dt F D 0 iD rQ iQ Q 0 i abc d R FDQ i FDQ dt Power Systems I FDQ abc 12 Synchronous Machine Model The magnetic inductance equations a Laa Lab Lac LaF L Lbb Lbc LbF b ba c Lca Lcb Lcc LcF = F LFa LFb LFc LFF D LDa LDb LDc LDF Q LQa LQb LQc LQF abc L SS L SR i abc = L L RR i FDQ FDQ RS (c) Feb 2004 LaD LbD LcD LFD LDD LQD LaQ ia LbQ ib LcQ ic LFQ iF LDQ iD LQQ iQ Power Systems I 13 Salient Pole Machines Rotors have two types of construction Cylindrical Salient The cylindrical rotor has an evenly spaced air gap and a constant self-inductance The Salient has a non-uniform air gap and a self-inductance that varies periodically Cylindrical maximum when inductance the direct axis coincides with the stator coil magnetic axis minimum inductance when the quadrature axis coincides with the stator coil magnetic axis (c) Feb 2004 Power Systems I Salient 14 Salient Pole Machines The salient pole machine can be represented by cosines of second harmonics Stator quantities Laa = LS + Lm cos 2 Lbb = LS + Lm cos 2( 2 ) 3 Lcc = LS + Lm cos 2( + 2 ) 3 Lab = Lba = M S Lm cos 2( + 1 ) 6 Lbc = Lcb = M S Lm cos 2( 1 ) 2 Lca = Lac = M S Lm cos 2( + 5 ) 6 (c) Feb 2004 Power Systems I 15 Salient Pole Machines Rotor quantities - All the rotor self inductances are constant since the effects of the stator slots are negligible LFF = LF LDD = LD LQQ = LQ LFD = LDF = M R LFQ = LQF = 0 LDQ = LQD = 0 (c) Feb 2004 Power Systems I 16 Salient Pole Machines Mutual inductance between the stator and rotor circuits LaF = M F cos 2 LbF = M F cos 2( 2 ) 3 LcF = M F cos 2( + 2 ) 3 LaD = M D cos 2 LaQ = M Q sin 2 LbD = M D cos 2( 2 ) LbQ = M Q sin 2( 2 ) 3 3 LcD = M D cos 2( + 2 ) LcQ = M Q sin 2( + 2 ) 3 3 (c) Feb 2004 Power Systems I 17 Park Transformation Changes the abc frame of reference to the dq0 frame of reference Voltages and currents on the stator are changed to equivalent values based on the rotors frame of reference The transformation is based on the two-axis theory the electrical quantities are projections onto three new axes: direct axis - along the direct axis of the rotor field winding quadrature axis - tangent to the direct axis of the rotor field winding zero axis - a stationary axis 1 2 2 P= cos 3 sin (c) Feb 2004 1 2 cos( 2 ) 3 sin ( 2 ) 3 2 cos( + 3 ) sin ( + 2 ) 3 Power Systems I 1 2 18 Park Transformation The Park transformation for current ia 2 2 cos( 3 ) cos( + 3 ) ib 2 2 sin ( 3 ) sin ( + 3 ) ic Similarly applied to all electrical quantities i 0 dq = P i abc i0 id = iq 1 2 2 cos 3 sin v 0 dq = P v abc 0 dq (c) Feb 2004 =P 1 2 1 2 in matrix notation abc Power Systems I 19 Park Transformation The Park transformation matix is orthogonal: 1 2 1 T P =P = 1 3 1 2 2 2 cos sin 2 2 cos( 3 ) sin ( 3 ) cos( + 2 ) sin ( + 2 ) 3 3 Applying the Park transformation to the generator (c) Feb 2004 Power Systems I 20 Park Transformation Transforming the time-varying inductance to obtain a rotor frame of reference P 0 = 0 I FDQ 33 0 dq FDQ abc P 1 0 0 dq or = 0 I 33 FDQ FDQ P 1 0 0 dq L SS L SR P 1 0 i 0 dq = L 0 I 33 FDQ RS L RR 0 I 33 i FDQ 0 dq P 0 L SS L SR P 1 0 i 0 dq = 0 I L L RR 0 I 33 i FDQ FDQ 33 RS (c) Feb 2004 abc Power Systems I 21 Park Transformation Resulting inductance matrix 0 0 Ls 2 M s 3 0 Ls + M s + 2 Lm 0 3 0 0 Ls + M s 2 Lm 3 0 0 2 MF 3 0 0 2 MD 3 0 0 2 MQ (c) Feb 2004 Power Systems I 0 3 2 MF 0 0 3 2 MD 0 LF MR MR LD 0 0 0 3 2 MQ 0 0 LQ 0 22 Park Transformation Applying the transformation to the machine model KVL vabc Rabc 0 iabc d abc v = 0 R i dt FDQ FDQ FDQ FDQ P1 0 v0dq Rabc 0 P1 0 i0dq d P1 0 v = 0 R i 0 I33 FDQ 0 I33 FDQ dt 0 I33 FDQ FDQ 0dq v0dq P 0 Rabc 0 P1 0 i0dq P 0 d P1 0 i 0 I v = 0 I 0 R 0 I33 FDQ 33 dt 0 I33 FDQ FDQ 33 d v0dq Rabc 0 i0dq P dt P1 0 0dq d 0dq v = 0 R i 0 I33 FDQ dt FDQ FDQ FDQ FDQ (c) Feb 2004 Power Systems I FDQ 0dq 23 Park Transformation Evaluate the expression d P dt P 1 d P dt P 1 1 2 12 1 2 0 sin cos = 2 cos cos( 23 ) cos( + 23 ) 0 sin ( 23 ) cos( 23 ) 3 sin sin ( 23 ) sin ( + 23 ) 0 sin ( + 23 ) cos( + 23 ) 0 0 0 = 0 0 1 0 1 0 (c) Feb 2004 Power Systems I 24 Park Transformation Substituting the original terms into the transformation 0 0 r 0 v0 0 r Lq 0 vd 3 v q 0 Ld r 2 M F = 0 rF 0 0 vF 0 0 0 0 0 0 0 0 0 0 0 0 0 L0 0 3 Ld 0 2 MF 0 Lq 0 0 3 0 2MF 0 LF 3 0 MR 0 2 MD 3 0 0 0 2 MQ (c) Feb 2004 0 0 3 2 MD 0 rD 0 0 3 2MD 0 MR LD 0 Power Systems I i0 3 2 M Q id 0 iq 0 iF 0 iD rQ iQ 0 0 i 0 i0 d 3 i 2 MQ d q i 0 dt F 0 iD iQ LQ 25 Park Transformation Observations The transformation has constant coefficients provided that the speed is assumed to be constant The first equation (the zero sequence) is not coupled to the other equations, and it can be treated separately While the transformation technique is a mathematical process, it gives insight into the internal phenomena of the rotor and the effects of transients (c) Feb 2004 Power Systems I 26
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FSU - EEL - 4213
Balanced 3-Phase Short CircuitConsider a synchronous generator operating at 60 Hzwith constant excitationExamine the impact on the stator currents when a threephase short circuit is applied to the generator terminalsThe initial currentsParkia (0+ )
FSU - EEL - 4213
power flow - evaluate normal operating conditionsfault analysis - evaluate abnormal operating conditionsthree-phasesingle-line to ground and double-line to groundline-to-line faultsunbalanced faultsbalanced faultsspecifying ratings for circuit brea
FSU - EEL - 4213
inversion of the bus admittance matrix is a n3 effortfor small and medium size networks, direct building of the matrixis less effortfor large size networks, sparse matrix programming withgaussian elimination technique is preferredDirect formation of
FSU - EEL - 4213
three-phasesingle-line to grounddouble-line to groundline-to-line faultsunbalanced faultsbalanced faults60-75%15-25%5-15%&lt;5%Percentage of total faultssymmetrical componentsaugmented component modelsUnbalance fault analysis requires new tools
FSU - EEL - 4213
TransformersEquivalent Series Impedance:Transformer bank of three single-phase transformersZ 0 = Z1 = Z 2 = Z Three-phase transformer with a three-leg coreZ1 = Z 2 = Z Z0 &gt; ZWye-Delta Wound TransformersWiring connection will always cause a phase s
FSU - EEL - 4213
Common Unbalanced Network FaultsSingle-line-to-ground faultsDouble-line-to-ground faultsLine-to-line faults March 2004Power Systems I1Single Line to Ground FaultVa = 0IaVaI f = IaIbVbIb = Ic = 0IcVcI a0 1 1 1 I a I = 1 1 a a 2 0 a1 3
FSU - EEL - 4213
StabilitylThe ability of the power system to remain in synchronismand maintain the state of equilibrium following adisturbing forceuSteady-state stability: analysis of small and slow disturbancesnugradual power changesTransient stability: analys
FSU - EEL - 4213
Steady State StabilityllllThe ability of the power system to remain in synchronismwhen subject to small disturbancesStability is assured if the system returns to its originaloperating state (voltage magnitude and angle profile)The behavior can be
FSU - EEL - 4213
Transient StabilityThe ability of the power system to remain in synchronismwhen subject to large disturbancesLyapunov energy functionsLarge power and voltage angle oscillations do not permitlinearization of the generator swing equationssimplified en
FSU - EEL - 4213
Solving Non-linear ODEObjectiveTime domain solution of a system of differential equations Given a function or a system of functions: f(x) or F(x) Seek a time domain solution x(t) or x(t) which satisfy f(x) or F(x)Integration of the differential equat
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Multi-machine SystemsEach synchronous machine is represented by a constantvoltage source behind the direct axis transient reactanceThe input powers are assumed to remain constantUsing the pre-fault voltage, all loads are converted toequivalent admitt
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EEL 3216 Introduction to Power SystemsHomework # 2CHAPTER 1.8 and 21. A three phase wye-connected unbalanced load is supplied by a balanced three-phasedelta connected source. The three source voltages and two of the line currents are:Vab = 13.2 kV 90
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 3CHAPTER 3(Problems 3-1, 3-2, 3-3, 3-4, 3-6, 3-7, 3-8, pp 154-157 in Textbook)1. The secondary winding of a transformer has a terminal voltage ofvs(t) = 282.8 sin 377t V. The turns ratio of the transfo
FSU - EEL - 3216
Life Long LearningProfessional development: seminars, workshops, conferences,summer courses, PE licenseLiterature resourcesReference books, i.e. McGraw Hill Standard Handbook for ElectricalEngineersMagazines, i.e. IEEE Power &amp; Energy, CIGRE Electra
FSU - EEL - 3216
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Induced Voltage in Stator Coil P-PolesFor a P-pole machine the flux per pole remains the same(integrating d =rlBmcos(P/2 )d over one pole pitch - /P to + /P )2rlBMThe electrical frequency becomesP2elmTherefore, the induced voltage per pole-pair (
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 1SOLUTIONS1. In the 60Hz circuit shown in Figure 1, VS = 240.0VR1 = 2.530 and I1 = 38.8A-36.2.L1 = 0.015 Ha) Determine the phasor currents, IS and I2, and the impedance Z2.b) Calculate the apparent
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 2SOLUTIONS1. A three phase wye-connected unbalanced load is supplied by a balanced three-phasedelta connected source. The source voltages and load impedances are:Vab = 13.2 kV 90Vbc = 13.2 kV 210Vca
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 3SolutionsCHAPTER 3(Problems 3-1, 3-2, 3-3, 3-4, 3-6, 3-7, 3-8, pp 154-157 in Textbook)1. The secondary winding of a transformer has a terminal voltage ofvs(t) = 282.8 sin 377t V. The turns ratio of t
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 4SOLUTIONS1. A Y-connected bank of three identical 100-kVA, 7967/480-V transformers is suppliedwith power directly from a large constant-voltage bus. In the short-circuit test, therecorded values on th
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 5, SOLUTIONS4-2. A three-phase four-pole winding is installed in 12 slots on a stator. There are 40turns of wire in each slot of the windings. All coils in each phase are connected inseries, and the thr
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 6SOLUTIONS4-6. The flux density distribution over the surface of a two-pole stator of radius r andlength l is given by)B BM cos( m tProve that the flux density under each pole face is2rlBMSolve fir
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EEL 3216 Introduction to Power SystemsHomework # 7 Solution1. A 3-phase, 4-pole synchronous machine rotates CCW with 1500 rpm and produces arated torque of 668.45 Nm at the shaft in CW direction. The total rated losses are5 kW and the rated stator (co
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EEL 3216 Introduction to Power SystemsHomework # 8A balanced 3-phase impedance type load is rated 3 MVA at 4.16 kV, with a laggingpower factor of 0.75. It is supplied by a generator via a 2 km long overhead transmissionline. The GMD of the conductors
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 8A balanced 3-phase impedance type load is rated 3 MVA at 4.16 kV, with a laggingpower factor of 0.75. It is supplied by a generator via a 2 km long overhead transmissionline. The GMD of the conductors
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 9Problems from book 9-6, 9-11, and 9-15 as shown below.Problem 9-6 refers to a single-phase, 8 kV, 50 Hz, 50 km long transmission line consisting of twoaluminum conductors with a 3 cm diameter separated
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 9Problems from book 9-6, 9-11, and 9-15 as shown below.Problem 9-6 refers to a single-phase, 8 kV, 50 Hz, 50 km long transmission line consisting of twoaluminum conductors with a 3 cm diameter separated
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 10Problem from book 5-29 as shown below.
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 10SOLUTION
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 11Problem from book 7-5, 7-7, 7-11 as shown below.7-57-77-11
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 11 SOLUTIONSProblem from book 7-5, 7-7, 7-11 as shown below.7-5Attention: this solution, as provided by the author of the book, is inconsistent with the text of thebook. The core losses are supplied by
FSU - EEL - 3216
EEL 3216 Fundamentals of Power SystemsHomework # 12You are encouraged to solve the two problems below in a group effort. Therefore, amaximum of 5 students may submit one assignment together. If you do so, clearly writethe names of the 5 students who h
FSU - EEL - 3216
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FSU - EEL - 3216
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FSU - EEL - 3216
Notes on Phasor Notation andDiagramsMathematically correctComplex quantity(underlined)VmMagnitudeV Vm e jvttVm cos( t )Re(V) Vm cosvtorjVm sin( t )tIm(V) Vm sintPractical notation and usageComplex quantity, butI = 10 Aunderlining omitt
FSU - EEL - 3216
Ideal 1~ Transformer ModelFlux linkages (winding flux)Ideali1i2N1N1 ,1N22N2SourceV2av1e1 , v2e2aV1N1N2e1e2ZLN1N2Loadv1v2i2i1Dot conventionCurrent into doted end produces positive MMF (NI) orampere-turnsTherefore, orientat
FSU - EEL - 3216
FSU - EEL - 3216
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FSU - EEL - 3216
Changing BasesThe impedances of generators, transformers,transmission lines, and loads are supplied by themanufacturers in per unit based on theequipments own rating.When placing the equipment into a system, theimpedance must be converted to the sy
FSU - EEL - 3216
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FSU - EEL - 3216
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FSU - EEL - 3216
AmpacityLecture 15Fundamentals of Power Systems1Geometric Mean Radius (GMR)'L102r4lnDr02Inductance of conductorup to a distance of 1 unit(e.g. 1 foot)'L102lnDGMRr4L'GMR1Dlnr110lnGMR2lnInductance of conductor in Hpe
FSU - EEL - 3216
ExampleM1TST = 1.5 MVAX = 3%NH/NL = 5.4586BusV = 4.16 kVM2M150 Hz, 6 pole, Y-connectedPR = 75 kW, Xm = 7.2R1 = 0.082 , R2 = 0.07X1 = 0.19 , X2 = 0.18Friction and windage losses 1.3 kWCore losses 1.4 kWMiscellaneous losses 150 WM2Rated 10
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 1CHAPTER 1.8 and 21. In the 60Hz circuit shown in Figure 1, VS = 240.0VR1 = 2.530 and IS = 35 A15.L1 = 0.015 Ha) Determine the phasor currents, I1 and I2, and the impedance Z2.b) Calculate the appa
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 71. A 3-phase, 4-pole synchronous machine rotates CCW with 1500 rpm and produces arated torque of 668.45 Nm at the shaft in CW direction. The total rated losses are5 kW and the rated stator (copper) los
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 3CHAPTER 31. (Problem 3-10 in Textbook)2. (Problem 3-22 in Textbook)3. In a certain single-phase system the per unit values of the impedance, current, voltage,and apparent power are:Z = 0.65 pu I = 1
FSU - EEL - 3216
Power Flow in a SM - GeneratorPout3V I A cos( )PoutReactive QoutpowerNote: is the phase anglebetween armature voltage V andarmature current IA. It is NOT theangle between VT and IL!Qout3 VT I L cos( )3V I A sin( )3 VT I L sin( )Lecture 13Fu
FSU - EEL - 3216
Magnetomotive Force and FluxDistributionThe magnetic fieldused to inducesinusoidal voltages(in the stator) has tobe distributedsinusoidal along thecircumference of thestator/rotor (air gap)Snapshot of oneinstantaneousvalue in timeThis spatial
FSU - EEL - 3216
T-Line Equivalent CircuitsThree general models for equivalent transmissionline circuitsChoice influenced by the line length, type (cable oroverhead line), and operating voltage levelChoice based on the analysis (e.g., short circuit orvoltage drop)M
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