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### lecture16

Course: EEL 4213, Fall 2011
School: FSU
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Word Count: 1283

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to three-phase single-line ground double-line to ground line-to-line faults unbalanced faults balanced faults 60-75% 15-25% 5-15% &lt;5% Percentage of total faults symmetrical components augmented component models Unbalance fault analysis requires new tools F a u l t ty p e s : Power Systems I Fault Analysis Ib1 120 120 abc sequence positive sequence 120 Ic1 Ia1 120 120...

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to three-phase single-line ground double-line to ground line-to-line faults unbalanced faults balanced faults 60-75% 15-25% 5-15% <5% Percentage of total faults symmetrical components augmented component models Unbalance fault analysis requires new tools F a u l t ty p e s : Power Systems I Fault Analysis Ib1 120 120 abc sequence positive sequence 120 Ic1 Ia1 120 120 Ia2 acb sequence negative sequence Ic2 120 Ib2 zero sequence 0 Ic0 Ib0 Ia0 applicable to current and voltages permits modeling of unbalanced systems and networks Representative symmetrical components Allow unbalanced three-phase phasor quantities to be replaced by the sum of three separate but balanced symmetrical components Power Systems I Symmetrical Components = I a1 1+ a + a2 = 0 a 3 = 10 = 1 + j 0 a 2 = 1240 = 0.5 j 0.866 a = 1120 = 0.5 + j 0.866 Operator a identities I c1 = I a1 ( + 120) = a I a1 I b1 = I a1 ( + 240) = a 2 I a1 I a1 = I a1 ( + 0) Positive sequence phasors Power Systems I Symmetrical Components I c 0 = I a 0 ( + 0) = I a 0 I b 0 = I a 0 ( + 0) = I a 0 I a 0 = I a 0 ( + 0) = I a 0 Zero sequence phasors I c 2 = I a 2 ( + 240) = a 2 I a 2 I b 2 = I a 2 ( + 120) = a I a 2 I a 2 = I a 2 ( + 0) = I a 2 Negative sequence phasors Power Systems I Symmetrical Components = I a 0 + a I a1 + a 2 I a 2 I c = I c 0 + I c1 + I c 2 I a 1 1 I = 1 a 2 b I c 1 a 1 I a 0 a I a1 2 a I a 2 = I a 0 + a 2 I a1 + a I a 2 I b = I b 0 + I b1 + I b 2 In matrix notation = I a 0 + I a1 + I a 2 I a = I a 0 + I a1 + I a 2 Relating unbalanced phasors to symmetrical components Power Systems I Symmetrical Components 1 a a 2 1 1 1 1 A = 1 a 3 1 a 2 I 012 = A 1 I abc 1 1* 2 a = A 3 a Solving for the symmetrical components leads to I abc = A I 012 1 1 1 a 2 A = 1 a [A] is known as the symmetrical components transformation matrix Power Systems I Symmetrical Components ) + aI ) Ia2 = 1 3 + a 2 Ib a c I a1 = 1 I a + aI b + a 2 I c 3 ( (I I a 0 = 1 (I a + I b + I c ) 3 In component form, the calculation for symmetrical components are Power Systems I Symmetrical Components * A T A* = 3 T * * * S 3 = 3V012 I * = 3 Va 0 I a 0 + 3 Va1 I a1 + 3 Va 2 I a 2 012 T S 3 = V012 A T A *I * 012 S 3 = (AV012 ) (AI 012 ) T T S 3 = Vabc I * abc The apparent power may also be expressed in terms of symmetrical components V012 = A 1 Vabc Vabc = A V012 Similar expressions exist for voltages Power Systems I Symmetrical Components I a0 = Solution (1.625) + (1.0180) + (0.9132) = 0.4596.5 I c = 0.9132 I b = 1.0180 I a = 1.625 Obtain the symmetrical components of a set of unbalanced currents 3 (1.625) + a(1.0180) + a 2 (0.9132) = 0.94 0.1 I a1 = 3 (1.625) + a 2 (1.0180) + a(0.9132) = 0.6022.3 I a2 = 3 Power Systems I Example Ib1 Ic1 abc set Ia Ia1 positive sequence set Power Systems I Ib Ic Example Ib2 Ic2 negative sequence set Ia2 zero sequence set Ia0, Ib0, Ic0 Obtain the original unbalanced voltages: Va 2 = 0.8 30 Va1 = 1.030 Va 0 = 0.690 The symmetrical components of a set of unbalanced voltages are Power Systems I Vc = (0.690) + a (1.030) + a 2 (0.8 30) = 1.7088155.8 Vb = (0.690) + a 2 (1.030) + a (0.8 30) = 0.490 Va = (0.690) + (1.030) + (0.8 30) = 1.708824.2 Example Vc1 Va2 negative sequence set Power Systems I Vc2 Vb2 zero sequence set Va0, Vb0, Vc0 Example Vc Vb1 abc set Vb positive sequence set Va1 Va positive, negative, and zero sequence impedances wye-connected balanced loads transmission line 3-phase transformers generators Augmented network models The impedance offered to the flow of a sequence current creating sequence voltages Power Systems I Sequence Impedances Va Vc Ib Ic Power I Vb Ia Zs Zs In Zs Zn ZM ZM Systems ZM Balanced Loads Vabc = Z abc I abc Va Z S + Z n V = Z + Z n b M Vc Z M + Z n I n = I a + Ib + Ic ZS + Zn ZM + Zn ZM + Zn Z M + Z n I a Z M + Z n I b Z S + Z n I c Vc = Z M I a + Z M I b + Z S I c + Z n I n Vb = Z M I a + Z S I b + Z M I c + Z n I n Va = Z S I a + Z M I b + Z M I c + Z n I n M o de l a n d governing equations ] A] Z abc Power Systems I 1 Z S + Z n Z M + Z n Z M + Z n 1 1 a 2 Z M + Z n Z S + Z n Z M + Z n 1 a 2 a Z M + Z n Z M + Z n Z S + Z n 1 a 0 0 Z S + 3Z n + 2 Z M = 0 ZS ZM 0 0 0 Z S Z M 1 1 1 1 = 1 a 3 1 a 2 Z 012 V012 = A 1Z abc A I 012 V012 = Z 012 I 012 [ = [A Vabc = Z abc I abc (A V012 ) = Z abc (A I 012 ) Balanced Loads 1 a 2 a Ib Ic Power Systems I Vc1 In Vb1 Va1 Ia Zn Zs Zs Vn Vabc1 = Z abc I abc + Vabc 2 Zn ZS + Zn Zn Vc2 Vb2 Va2 Va1 Z S + Z n V = Z n b1 Vc1 Z n Zs Transmission Line I a Va 2 Z n I b + Vb 2 Z S + Z n I c Vc 2 Zn I n + I a + Ib + Ic = 0 Vn = 0 + Z n I n Vc1 = Z S I c Z n I n + Vc 2 Vb1 = Z S I b Z n I n + Vb 2 Va1 = Z S I a Z n I n + Va 2 Power Systems I 1 Z S + Z n a 2 Z n a Z n 0 Z S + 3Z n 0 0 Z S 0 = 0 0 Z S 1 1 1 = 1 a 3 1 a 2 Z 012 = A 1Z abc A ZS + Zn Zn Zn 1 1 Z n 1 a 2 Z S + Z n 1 a Zn V0121 = A 1Z abc A I 012 + V012 2 = Z 012 I 012 + V012 2 1 a 2 a Vabc1 = Z abc I abc + Vabc 2 A V0121 = Z abc A I 012 + A V012 2 Transmission Line X2 ~ Xd the zero sequence reactance is approximated to the leakage reactance X0 ~ XL zero sequence values are isolated from the airgap of the machine remember that the transient fault impedance is a function of time positive sequence values are the same as Xd, Xd, and Xd negative sequence values are affected by the rotation of the rotor Similar modeling of impedances to sequence impedances Typical values for common generators Power Systems I Generators Power Systems I Negative Sequence Positive Sequence Zero Sequence E1 Generator Model X2 X1 X0 VT2 VT1 VT0 Ec Power Systems I Zn Ea Eb Z S + 3Z n 0 = 0 ZS ZS Z 012 ZS Impedance Grounded Generators 0 0 ZS 0 0 Z S the magnetization current and core losses represented by the shunt branch are neglected (they represent only 1% of the total load current) the transformer is modeled with the equivalent series leakage impedance the series leakage impedance is the same for all the sequences the series leakage impedance is the same for the positive and negative sequence only Z1 = Z 2 = Z Z 0 = Z1 = Z 2 = Z Three-legged core three-phase units Three single-phase units & five-legged core three-phase u n it s Series Leakage Impedance Power Systems I Transformers the positive sequence quantities rotate by +30 degrees the negative sequence quantities rotate by -30 degrees the zero sequence quantities can not pass through the transformer independent of the winding order (-Y or Y- ) the positive sequence line voltage on the HV side leads the corresponding line voltage on the LV side by 30 consequently, for the negative sequence voltages the corresponding phase shift is -30 USA standard Wye-delta transformers create a phase shifting pattern for the various sequences Power Systems I Transformers primary winding - wye / wye-grounded / delta secondary winding - wye / wye-grounded / delta Zero-sequence network connections of the transformer d e p e n d s o n t h e w i n d i n g c o n n e c ti o n Power Systems I Transformers delta wye-grounded Power Systems I wye wye-grounded wye-grounded wye-grounded Transformers delta delta Power Systems I delta wye Transformers
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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
FSU - EEL - 4213
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
FSU - EEL - 3216
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
Transmission LinesOverhead ConductorOverhead Spacer CableUnderground CableThree-Conductor CableService CablesFundamentals of Power SystemsLecture 141Overhead ConductorsACSRAluminum Conductor withinner Steel ReinforcedstrandsACARAluminum Con
FSU - EEL - 3216
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
FSU - EEL - 3216
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
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 # 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
Electric MotorsThree basic types of motorsSynchronous, SMInduction (asynchronous), IMDC motorsIn the past: preferred for variable speed applicationsDecreasing in popularity; replaced by combination of IM and powerelectronic variable speed drive (VS
FSU - EEL - 3216
U.S. Electric Power Industry ExistingNet Summer Capacity by State, 2003Fundamentals of Power SystemsLecture 11U.S. Electric Power TransmissionSystem in 2000Fundamentals of Power SystemsLecture 1Page 12Power Systems ComponentsPrimary power sour
FSU - EEL - 3216
Electric Powersingle-phase (1~) systemsOne sourceTwo energy-carrying wiresPulsating instantaneous power (2f)three-phase (3~) systemsThree sources, 120 phase shiftedLecture 2Fundamentals of Power Systems1Single Phase Powerptv, i, pPVvtptDe
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
Three-phase transformersCore arrangementsTransformer bank =&gt; three 1~ transformers (core typeor shell type)3~ transformer (3, 4, 5 legged core-type)Winding arrangementsY-Y-YYWith tertiaryFundamentals of Power SystemsLecture 61Core Type vs. She
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
AC Machine FundamentalsGeneratorsMotors3-phase1-phaseSynchronousInductionFundamentals of Power SystemsLecture 91AC Machines SystematicFundamentalsPrinciple of generating a rotating magnetic fieldMagnetomotive force and induced voltagePrincip
FSU - EEL - 3216
SM No-Load CharacteristicRotor flux magnitudeInduced stator voltage magnitudeEAkCharacteristic of theiron (stator and rotor)Rotor field currentRotor field currentEffect of saturation: for large field currents the no-loadarmature voltage magnitud
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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