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### lecture20

Course: EEL 4213, Fall 2011
School: FSU
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State Steady Stability l l l l The ability of the power system to remain in synchronism when subject to small disturbances Stability is assured if the system returns to its original operating state (voltage magnitude and angle profile) The behavior can be determined with a linear system model Assumption: u u u the automatic controls are not active the power shift is not large the voltage angles changes are...

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State Steady Stability l l l l The ability of the power system to remain in synchronism when subject to small disturbances Stability is assured if the system returns to its original operating state (voltage magnitude and angle profile) The behavior can be determined with a linear system model Assumption: u u u the automatic controls are not active the power shift is not large the voltage angles changes are small Power Systems I Steady State Stability l Swing Equation H i d 2d i = Pmi - Pmax sin d 2 p f 0 dt l Small disturbance modeling d = d 0 + Dd Consider a small deviation H d 2 (d 0 + Dd ) = Pm - Pmax sin(d 0 + Dd ) 2 dt p f0 H d 2d 0 H d 2 Dd + = Pm - Pmax [sin d 0 cos Dd + cos d 0 sin Dd ] 2 2 p f 0 dt p f 0 dt Power Systems I Steady State Stability l Simplification of the swing equation H d 2d 0 H d 2 Dd + = Pm - Pmax [sin d 0 cos Dd + cos d 0 sin Dd ] 2 2 p f 0 dt p f 0 dt Substitute the following approximations Dd << d cos Dd 1 sin Dd Dd H d 2d 0 H d 2 Dd + = Pm - Pmax sin d 0 - Pmax cos d 0 Dd 2 2 p f 0 dt p f 0 dt Group steady state and transient terms H d 2d 0 H d 2 Dd - Pm + Pmax sin d 0 = - Pmax cos d 0 Dd 2 2 p f 0 dt p f 0 dt Power Systems I Steady State Stability l Simplification of the swing equation H d 2 Dd H d 2d 0 - Pm + Pmax sin d 0 = - Pmax cos d 0 Dd 2 2 p f 0 dt p f 0 dt H d 2 Dd 0= + Pmax cos d 0 Dd 2 p f 0 dt Steady state term is equal to zero dPe dd d0 d = Pmax sin d dd = Pmax cos d 0 = Ps d0 H d 2 Dd + Ps Dd = 0 Second order equation. 2 p f 0 dt The solution depends on the roots of the Power Systems I characteristic equation Stability l Stability Assessment u u When Ps is negative, one root is in the right-half s-plane, and the response is exponentially increasing and stability is lost When Ps is positive, both roots are on the jw axis, and the motion is oscillatory and undamped, the natural frequency is: p =H 2 p f0 PS s f0 wn = PS H Power Systems I Root locus jw a S-plane Damping Torque dd Damping force is due to air-gap interaction PD = D dt H d 2 Dd dDd +D + PS Dd = 0 2 p f 0 dt dt d 2 Dd p f 0 dDd p f 0 + + D PS Dd = 0 2 dt H dt H d 2 Dd d Dd 2 + 2z wn + w n Dd = 0 dt 2 dt D p f0 z= 2 H PS Power Systems I Characteristic Equation 2 2 n s + 2zw n s + w = 0 D z= 2 p f0 <1 H PS for normal operation conditions s1 , s2 = -zw n j w n 1 - z wd = wn 1 - z 2 Power Systems I 2 complex roots the damped frequency of oscillation Laplace Transform Analysis x1 = Dd , x2 = dDd dt 1 x1 & x = x = Ax - 2zw n 2 & x1 0 x = - w 2 &2 n & L {x = Ax} sX( s ) - x(0) = AX( s ) -1 X( s ) = (sI - A ) x(0) s ( sI - A ) = 2 w n Power Systems I -1 s + 2zw n s + 2zw n 1 -w 2 s n x ( 0) X( s ) = 2 s 2 + 2zw n s + w n Laplace Transform Analysis Dd ( s ) = (s + 2zw n )Dd 0 2 s 2 + 2zw n s + w n 2 w n Dd 0 Dw ( s ) = 2 2 s + 2zw n s + w n Dd (t ) = Dd 0 1-z 2 Dw (t ) = Power Systems I e -zw nt sin (w d t + q ), q = cos -1 z w n Dd 0 1-z 2 e -zw nt sin (w d t ) d (t ) = d 0 + Dd (t ), w (t ) = w 0 + Dw (t ) Example l A 60 Hz synchronous generator having inertia constant H = 9.94 MJ/MVA and a transient reactance Xd = 0.3 pu is connected to an infinite bus through the following network. The generator is delivering 0.6 pu real power at 0.8 power factor lagging to the infinite bus at a voltage of 1 pu. Assume the damping power coefficient is D = 0.138 pu. Consider a small disturbance of 10 or 0.1745 radians. Obtain equations of rotor angle and generator frequency motion. Power Systems I Example X12 = 0.3 X'd = 0.3 G inf Xt = 0.2 Power Systems I X12 = 0.3 V = 1.0 Example Delta, degree 30 25 20 15 10 0 0.5 1 1.5 t, s ec 2 2.5 3 0 0.5 1 1.5 t, s ec 2 2.5 3 60.1 f, Hz 60.05 60 59.95 59.9 59.85 Power Systems I
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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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