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Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
Ece4762011 lect22
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Ece4762011 lect22

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  • 1. ECE 476 POWER SYSTEM ANALYSIS Lecture 22Power System Protection, Transient Stability Professor Tom Overbye Department of Electrical and Computer Engineering
  • 2. Announcements Be reading Chapters 9 and 10 After exam read Chapter 11 HW 9 is 8.4, 8.12, 9.1,9.2 (bus 2), 9.14; do by Nov 10 but does not need to be turned in. Start working on Design Project. Firm due date has been extended to Dec 1 in class Second exam is on Nov 15 in class. Same format as first exam, except you can bring two note sheets (e.g., the one from the first exam and another)  Exam/solution from 2008 will be posted on website shortly  Exam covers through Chapter 10 2
  • 3. In the News: Boulder municipalization• Last week Boulder, CO narrowly voted to move forward with municipalization of their electric grid • Currently Boulder is in the Xcel Energy electric service territory (Xcel is a large Investor Owned Utility)• Xcel has recently decided not to continue funding the Boulder “SmartGridCity” initiative, which has cost $45 million, triple its original cost.• Xcel does not wish to sell its electric grid in Boulder, saying it would be extremely expensive for Boulder to go on their own.Source: NY Times 11/3/11; Thanks to Margaret for pointing out this story 3
  • 4. Power System Protection Main idea is to remove faults as quickly as possible while leaving as much of the system intact as possible Fault sequence of events 1. Fault occurs somewhere on the system, changing the system currents and voltages 2. Current transformers (CTs) and potential transformers (PTs) sensors detect the change in currents/voltages 3. Relays use sensor input to determine whether a fault has occurred 4. If fault occurs relays open circuit breakers to isolate fault 4
  • 5. Power System Protection Protection systems must be designed with both primary protection and backup protection in case primary protection devices fail In designing power system protection systems there are two main types of systems that need to be considered:3. Radial: there is a single source of power, so power always flows in a single direction; this is the easiest from a protection point of view4. Network: power can flow in either direction: protection is much more involved 5
  • 6. Radial Power System Protection Radial systems are primarily used in the lower voltage distribution systems. Protection actions usually result in loss of customer load, but the outages are usually quite local. The figure shows potential protection schemes for a radial system. The bottom scheme is preferred since it results in less lost load 6
  • 7. Radial Power System Protection In radial power systems the amount of fault current is limited by the fault distance from the power source: faults further done the feeder have less fault current since the current is limited by feeder impedance Radial power system protection systems usually use inverse-time overcurrent relays. Coordination of relay current settings is needed to open the correct breakers 7
  • 8. Inverse Time Overcurrent Relays Inverse time overcurrent relays respond instan- taneously to a current above their maximum setting They respond slower to currents below this value but above the pickup current value 8
  • 9. Inverse Time Relays, contd The inverse time characteristic provides backup protection since relays further upstream (closer to power source) should eventually trip if relays closer to the fault fail Challenge is to make sure the minimum pickup current is set low enough to pick up all likely faults, but high enough not to trip on load current When outaged feeders are returned to service there can be a large in-rush current as all the motors try to simultaneously start; this in-rush current may re-trip the feeder 9
  • 10. Inverse Time Overcurrent Relays Current and time settings are ad- justed using dials on the relay Relays have traditionally been electromechanical devices, but are gradually being replaced by digital relays 10
  • 11. Protection of Network Systems In a networked system there are a number of difference sources of power. Power flows are bidirectional Networked system offer greater reliability, since the failure of a single device does not result in a loss of load Networked systems are usually used with the transmission system, and are sometimes used with the distribution systems, particularly in urban areas 11
  • 12. Network System Protection Removing networked elements require the opening of circuit breakers at both ends of the device There are several common protection schemes; multiple overlapping schemes are usually used3. Directional relays with communication between the device terminals4. Impedance (distance) relays.5. Differential protection 12
  • 13. Directional Relays Directional relays are commonly used to protect high voltage transmission lines Voltage and current measurements are used to determine direction of current flow (into or out of line) Relays on both ends of line communicate and will only trip the line if excessive current is flowing into the line from both ends – line carrier communication is popular in which a high frequency signal (30 kHz to 300 kHz) is used – microwave communication is sometimes used 13
  • 14. Impedance Relays Impedance (distance) relays measure ratio of voltage to current to determine if a fault exists on a particular lineAssume Z is the line impedance and x is thenormalized fault location (x = 0 at bus 1, x = 1 at bus 2) V1 V1Normally is high; during fault ≈ xZ I12 I12 14
  • 15. Impedance Relays Protection Zones To avoid inadvertent tripping for faults on other transmission lines, impedance relays usually have several zones of protection: – zone 1 may be 80% of line for a 3φ fault; trip is instantaneous – zone 2 may cover 120% of line but with a delay to prevent tripping for faults on adjacent lines – zone 3 went further; most removed due to 8/14/03 events The key problem is that different fault types will present the relays with different apparent impedances; adequate protection for a 3φ fault gives very limited protection for LL faults 15
  • 16. Impedance Relay Trip Characteristics Source: August 14th 2003 Blackout Final Report, p. 78 16
  • 17. Differential Relays Main idea behind differential protection is that during normal operation the net current into a device should sum to zero for each phase – transformer turns ratios must, of course, be considered Differential protection is used with geographically local devices – buses – transformers – generators I1 + I 2 + I 3 = 0 for each phase except during a fault 17
  • 18. Other Types of Relays In addition to providing fault protection, relays are used to protect the system against operational problems as well Being automatic devices, relays can respond much quicker than a human operator and therefore have an advantage when time is of the essence Other common types of relays include – under-frequency for load: e.g., 10% of system load must be shed if system frequency falls to 59.3 Hz – over-frequency on generators – under-voltage on loads (less common) 18
  • 19. Sequence of Events Recording During major system disturbances numerous relays at a number of substations may operate Event reconstruction requires time synchronization between substations to figure out the sequence of events Most utilities now have sequence of events recording that provide time synchronization of at least 1 microsecond 19
  • 20. Use of GPS for Fault Location Since power system lines may span hundreds of miles, a key difficulty in power system restoration is determining the location of the fault One newer technique is the use of the global positioning system (GPS). GPS can provide time synchronization of about 1 microsecond Since the traveling electromagnetic waves propagate at about the speed of light (300m per microsecond), the fault location can be found by comparing arrival times of the waves at each substation 20
  • 21. Power System Transient Stability In order to operate as an interconnected system all of the generators (and other synchronous machines) must remain in synchronism with one another – synchronism requires that (for two pole machines) the rotors turn at exactly the same speed Loss of synchronism results in a condition in which no net power can be transferred between the machines A system is said to be transiently unstable if following a disturbance one or more of the generators lose synchronism 21
  • 22. Generator Transient Stability Models In order to study the transient response of a power system we need to develop models for the generator valid during the transient time frame of several seconds following a system disturbance We need to develop both electrical and mechanical models for the generators 22
  • 23. Example of Transient Behavior 23
  • 24. Generator Electrical Model The simplest generator model, known as the classical model, treats the generator as a voltage source behind the direct-axis transient reactance; the voltage magnitude is fixed, but its angle changes according to the mechanical dynamics VT Ea Pe (δ ) = sin δ Xd 24
  • 25. Generator Mechanical ModelGenerator Mechanical Block DiagramTm = Jα m + TD + Te (δ )Tm = mechanical input torque (N-m)J = moment of inertia of turbine & rotorα m = angular acceleration of turbine & rotorTD = damping torqueTe (δ ) = equivalent electrical torque 25
  • 26. Generator Mechanical Model, cont’dIn general power = torque × angular speedHence when a generator is spinning at speed ωs Tm = Jα m + TD + Te (δ ) Tm ωs = ( Jα m + TD + Te (δ )) ωs @ Pm Pm = Jα mωs + TDωs + Pe (δ )Initially well assume no damping (i.e., TD = 0)Then Pm − Pe (δ ) = Jα mωsPm is the mechanical power input, which is assumedto be constant throughout the study time period 26
  • 27. Generator Mechanical Model, cont’dPm − Pe (δ ) = J α mωs θm = ωst + δ = rotor angle dθ m ωm = = θ m = ωs + δ & & dt α m = ωm = δ& &&Pm − Pe (δ ) = J ωsα m = J ωsδ && J ωs = inertia of machine at synchronous speedConvert to per unit by dividing by MVA rating, S B ,Pm Pe (δ ) J ωsδ 2ωs && − =SB SB S B 2ωs 27
  • 28. Generator Mechanical Model, cont’dPm Pe (δ ) J ωsδ 2ω s && − =SB SB S B 2ω s 2Pm − Pe (δ ) J ω s 1 && = δ (since ω s = 2π f s ) SB 2S B π f s 2 Jω sDefine @H = per unit inertia constant (sec) 2S BAll values are now converted to per unit H && HPm − Pe (δ ) = δ Define M = π fs π fsThen Pm − Pe (δ ) = M δ && 28
  • 29. Generator Swing EquationThis equation is known as the generator swing equation Pm − Pe (δ ) = M δ&&Adding damping we get Pm − Pe (δ ) = M δ + Dδ& &&This equation is analogous to a mass suspended bya spring k x − gM = Mx + Dx && & 29
  • 30. Single Machine Infinite Bus (SMIB) To understand the transient stability problem we’ll first consider the case of a single machine (generator) connected to a power system bus with a fixed voltage magnitude and angle (known as an infinite bus) through a transmission line with impedance jXL 30
  • 31. SMIB, cont’d EaPe (δ ) = sin δ Xd + XL EaM δ + Dδ& && = PM − sin δ Xd + XL 31
  • 32. SMIB Equilibrium PointsEquilibrium points are determined by setting theright-hand side to zeroMδ && + Dδ& = PM − Ea sin δ Xd + XL Ea PM − sin δ = 0 Xd + XL Define X th = Xd + XL −1  P X th  δ = sin  M  Ea   32
  • 33. Transient Stability Analysis For transient stability analysis we need to consider three systems2. Prefault - before the fault occurs the system is assumed to be at an equilibrium point3. Faulted - the fault changes the system equations, moving the system away from its equilibrium point4. Postfault - after fault is cleared the system hopefully returns to a new operating point 33
  • 34. Transient Stability Solution Methods There are two methods for solving the transient stability problem2. Numerical integration  this is by far the most common technique, particularly for large systems; during the fault and after the fault the power system differential equations are solved using numerical methods3. Direct or energy methods; for a two bus system this method is known as the equal area criteria  mostly used to provide an intuitive insight into the transient stability problem 34
  • 35. SMIB Example Assume a generator is supplying power to an infinite bus through two parallel transmission lines. Then a balanced three phase fault occurs at the terminal of one of the lines. The fault is cleared by the opening of this line’s circuit breakers. 35
  • 36. SMIB Example, cont’dSimplified prefault system The prefault system has two equilibrium points; the left one is stable, the right one unstable −1  P X th  δ = sin  M  Ea   36
  • 37. SMIB Example, Faulted SystemDuring the fault the system changes The equivalent system during the fault is then During this fault no power can be transferred from the generator to the system 37
  • 38. SMIB Example, Post Fault SystemAfter the fault the system again changes The equivalent system after the fault is then 38
  • 39. SMIB Example, DynamicsDuring the disturbance the form of Pe (δ ) changes,altering the power system dynamics: 1  EaVth δ =&&  PM − X sin δ  M  th  39

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