advertisement

ELECTRICAL PROTECTION IV Dr. Oriedi Akumu Building 20-G01M [email protected] Consultation Hours: Monday: 7:00-10:00 Wednesday:7:00-12:00 Thursday:7:00-10:00 ELECTRICAL PROTECTION IV LEARNING OUTCOMES Understand the basic concept, definitions and be able to: Calculate symmetrical fault currents in power systems and determine their distribution taking into account transient and subtransient component currents Calculate unsymmetrical fault currents in power systems and determine their distribution taking into account transformer connections Select current transformers, circuit breakers, fuses, over-current relays and reclosers based on their characteristics, and evaluate their performance Implement protective relaying schemes and techniques for unit and line protection ELECTRICAL PROTECTION IV Text Book: Power System Analysis & Design Authors: J. D. Glover, M. S. Sarma and T. J. Overbye Assessment Type: Semester Tests Class Tests Assignments Design Project using Power World ELECTRICAL PROTECTION IV Electrical Power System ELECTRICAL PROTECTION IV Electrical Power System All plant & equipment necessary to: Generate Transmit Distribute Utilize electric power The equipment includes: Generators, transformers, transmission lines, loads, switching devices, controls, protection, measuring instruments and so on The requirements on such a system are: Highest reliability standards Minimum environmental impacts Lowest operation cost ELECTRICAL PROTECTION IV Protection is the science, skill, and art of applying and setting relays or fuses, or both, to provide maximum sensitivity to faults and undesirable conditions, but to avoid their operation under all permissible or tolerable conditions. ELECTRICAL PROTECTION IV Power System Protection Protection schemes are control systems that monitor the operation of the power system and initiate appropriate corrective measures. Protection schemes do not prevent the occurrence of faults and abnormal conditions. Protection design philosophy: Any equipment threatened with damage by a sustained fault is to be automatically taken out of service. ELECTRICAL PROTECTION IV Design Objectives The basic design objectives of any protective scheme are to: Maintain dynamic stability. Prevent or minimize equipment damage. Minimize the equipment outage time. Minimize the system outage area. Minimize system voltage disturbances. Allow the continuous flow of power within the emergency ratings of equipment on the system. ELECTRICAL PROTECTION IV Design Criteria of Relaying System Speed: for power system stability, reduction of damage & outage time, and to prevent development of other faults Selectivity: determination of fault point & tripping only the nearest CB Sensitivity: capability to reliably operate under actual conditions that produce least operating tendency Reliability: trip at all times when required – dependability; not to trip falsely – security Simplicity: use the minimum protective equipment necessary to achieve the protection objectives Economy: about 5% cost of protected equipment ELECTRICAL PROTECTION IV Protection Application Concept ELECTRICAL PROTECTION IV Types of Faults and Abnormalities Short-circuit faults (3, 2, 2g, 1g) Open-circuit faults (open conductor) Complex faults (inter-circuit, cross-country, broken conductor) Inter-turn faults in windings Abnormalities Overload and excessive operating temperature Real power deficit (underfrequency) Power Swings Power frequency overvoltage and undervoltage Underexcitation of synchronous machines Asynchronous operation of synchronous machines Overfrequency Mechanical defects (oil leaks, tap changer mechanism fault) ELECTRICAL PROTECTION IV Short-circuit type faults ELECTRICAL PROTECTION IV Short-circuit type faults with fault impedance ELECTRICAL PROTECTION IV Some causes of fault impedance Arc resistance Pole, tower or structure footing resistance to earth Resistance of things that may come into contact with a line e.g. tree, crane etc Contact resistance where a conductor falls to the ground ***Fault impedance can significantly reduce the magnitude of fault currents and should be a design consideration*** ELECTRICAL PROTECTION IV Complex type faults ELECTRICAL PROTECTION IV Faults in windings ELECTRICAL PROTECTION IV Protective Relaying System Components ELECTRICAL PROTECTION IV Relaying System Design Steps Power flow studies – to pick all system components, including protective devices, based on rated system quantities (voltages, currents, power …) Short circuit analysis – to pick circuit breaking devices based on maximum fault values Stability studies – to determine stability limits/critical clearing times for system faults (for time setting of relays) ELECTRICAL PROTECTION IV Typical Power Circuit Breakers CBs are located such that each component of the power system can be completely disconnected from the rest of the system They must have sufficient capacity to carry momentarily the max short-circuit current, and interrupt this current In cases where auto-reclosure is permitted they must be able to withstand closing in on, and interrupting, a short circuit Types of Circuit Breakers Air Oil SF6 Vacuum ELECTRICAL PROTECTION IV Relay Types: Construction Electromagnetic attraction: An armature being attracted to the poles of an electromagnet or a plunger being drawn into a solenoid Electromagnetic induction: a pivoted aluminium disc placed in two alternating magnetic fields of the same frequency but displaced in time and space. Torque produced by interaction of one mag. field with current induced in the disc by the other ELECTRICAL PROTECTION IV Solid state relays Use low power components Require independent power supplies Flexible, smaller, more accurate, settings more repeatable Characteristics can be shaped by adjusting logic elements Require less mounting space Not affected by vibration and dust Higher cost ELECTRICAL PROTECTION IV Block diagram of microprocessor based relays Much higher precision and more reliable and durable Improve the reliability and power quality of electrical power systems before, during and after faults occur Capable of both digital and analog I/O Higher cost but with more functionality ELECTRICAL PROTECTION IV Relay Types: Operation Principle Magnitude relays: respond to the magnitude of input quantities (e.g overcurrent relays) Directional relays: respond to the phase angle difference between two quantities Ratio relays: respond to the ratio of two phasor quantities: ratio can be a complex number or its magnitude (e.g impedance relays) Differential relays: respond to the magnitude of the algebraic sum of two or more inputs Pilot relays: use communicated information from remote sites as input signals ELECTRICAL PROTECTION IV MODULE 1: SYMETRICAL FAULTS Learning Outcome: Students will be able to explain the concepts and methods for calculating symmetrical short circuit currents. Assessment Criteria: Effects of three-phase short circuits in series RL circuits and unloaded synchronous generator is explained Calculation in p.u. and kA values of sub-transient fault currents at the point of fault and other points of the power system is explained Construction and use of bus impedance matrix are explained Selection procedure demonstrated of circuit breakers and ELECTRICAL PROTECTION IV fuses is 1 1: Short Circuit Conditions and Transients in Series R-L Circuits ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV Example 1 ELECTRICAL PROTECTION IV Solution ELECTRICAL PROTECTION IV 1 2: Synchronous machine response to short-circuit conditions ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV Example 2 ELECTRICAL PROTECTION IV Solution ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV Power System Three-phase Short Circuit Calculations The following assumptions are made to compute subtransient fault current for three-phase short circuits in power systems: Transformers are represented by their leakage reactances. Winding resistances, shunt admittances, and delta/star phase shifts are neglected. Transmission lines are represented by their equivalent series reactances. Series resistance and shunt admittances are neglected. Synchronous machines are represented by constant voltage sources behind subtransient reactances. Armature resistance, saliency and saturation are neglected Non-rotating impedance loads are neglected Induction motors are either neglected (especially for small motors rated less than 37.3kW), or represented in the same way as synchronous machines. ELECTRICAL PROTECTION IV Consider the following single-line diagram Assuming a 3-phase short circuit occurs at bus 1, the positive sequence equivalent circuit becomes: ELECTRICAL PROTECTION IV On closing the switch, the fault may be represented by two opposing voltage sources; Using superposition, the above circuit results in: Pre-fault condition ELECTRICAL PROTECTION IV If VF is the pre-fault voltage at the fault, then the 2nd circuit represents the system before the fault occurs, which implies I”F2 = 0. VF has no effect and can be removed from 2nd circuit Hence the subtransient fault current from the 1st circuit: I”F = I”F1 The contribution to the fault from generator and motor respectively are: I”G = I”g1 +I”g2 =I”g1 +IL & I”M = I”m1 +I”m2 =I”m1 -IL Where IL is the pre-fault generator current ELECTRICAL PROTECTION IV Example 3 The generator in the previous case is operating at rated MVA, 0.95 p.f. lagging and at 5% above rated voltage when a bolted 3-phase short circuit occurs at bus 1. Calculate the per unit values of: a) subtransient fault current b) subtransient generator and motor currents, neglecting prefault current, and c) subtransient generator and motor currents, including prefault current Solution ELECTRICAL PROTECTION IV 1 ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV Bus Impedance Matrix The bus impedance, Zbus, is the inverse of the bus admittance, Ybus, i.e., Since we know that then and Zbus relates nodal current injections to the nodal voltages ELECTRICAL PROTECTION IV In power flow studies, we work with Ybus. The reason being that the power flow problem requires an iterative solution that can be made very efficient when we use Ybus, due to the sparsity (lots of zeros) in the matrix used in performing the iteration (the Jacobian matrix) In fault analysis, we use Zbus. The main reason is that Zbus quantities characterize conditions when all current injections are zero except one, corresponding to the faulted bus Once we have that one current injection, all bus voltages in the network can be obtained using (4), from which all currents are then calculated The Zbus is not sparse, but fault analysis does not require iterative solutions ELECTRICAL PROTECTION IV Zbus Elements Assuming a 3 node system, we can write: 1 11 2 = 21 3 31 12 22 32 13 23 33 1 2 3 (5) Here the independent sources are all current sources, which are the equivalent representation of the generator voltage sources For example 2 = 21 1 + 22 2 +23 3 (6) Hence driving point impedance becomes 22 = 2 − 21 1 − 23 3 2 (7) ELECTRICAL PROTECTION IV If we set I1 and I3 to 0, that is, open any current sources at nodes 1 and 3 so that there are no sources there (but there may be load impedances), (7) becomes: 22 = 2 2 1 =3 =0 (8) Z22 is the ratio of bus 2 voltage to the bus 2 current injection when all sources are idled - this is the Thevenin impedance looking into the network at bus 2 Therefore, the diagonal elements of the Zbus are the Thevenin equivalent impedances seen looking into the network at that bus An important attribute to building the Zbus for fault analysis is that the generator subtransient reactances should be included ELECTRICAL PROTECTION IV Zbus Modifications It is easier to develop the Ybus. From that, one can invert it to obtain the Zbus. This is only possible with small networks, due to matrix inversion difficulties for large matrices For large networks, we must turn to the so-called Zbus building algorithm, based on modification to an already existing Zbus Let’s assume that we have a 3-bus system with Ybus given by: (9) ELECTRICAL PROTECTION IV Assume the branch between buses 1 and 3 is numbered as branch 3. Then let’s modify branch 3 by adding another circuit between bus 1 and bus 3 having an admittance of Δy3. Add Δy3 to diagonal elements in positions (1,1) and (3,3) Subtract Δy3 from off-diagonal elements in positions (1,3) and (3,1) The resulting matrix is (10) ELECTRICAL PROTECTION IV So the new Ybus is just the old Ybus with the addition of Δy3 in four positions, as indicated by equation (10): (11) Define a vector corresponding to the modification of branch 3 (connected between bus 1 and bus 3) as: (12) ELECTRICAL PROTECTION IV Then (11) becomes: (13) In general, anytime we modify Ybus, then (14) where ak is constructed according to: 1 at position i and -1 at position j if we add or remove a branch between buses i and j. 1 at bus i if we add or remove a shunt at bus i ELECTRICAL PROTECTION IV The inverse of (14) is found to be: (15) where bk is an n×1 vector given by: (16) and γ is a scalar given by (17) where μ is a scalar ( = Δy ) and ak is an n×1 vector. ELECTRICAL PROTECTION IV Example 4 ELECTRICAL PROTECTION IV Fault calculations using Zbus Recall that (18) Since (18) represents a set of linear equations, superposition holds, and we may write: (19) This says that the change in voltage at all buses V may be computed if the change in injections at all buses I are known. We can write eq. (19) in expanded form as: ELECTRICAL PROTECTION IV (20) Now consider a bolted short-circuit at bus k, where the prefault voltage at bus k is Vf. Let the fault current be I’’f Since the fault is a short circuit, then ΔVk = -Vf. Also, since the fault current is out of bus k, then Δik = - I’’f. Substituting these into eq. (20) results in: ELECTRICAL PROTECTION IV (21) Therefore (22) ELECTRICAL PROTECTION IV We observe from row k that: (23) Solving (23) for I’’f results in: (24) Substituting (24) into (22), we get: (25) ELECTRICAL PROTECTION IV Consider any bus, say bus j, with a pre-fault voltage of Vj. Then we can compute the bus j voltage under the faulted condition as (26) We know that (27) Substitution of (27) into (26) results in: (28) ELECTRICAL PROTECTION IV Consider this simple system We can write that We can compute the subtransient current flowing from bus i to bus j under the fault condition as: ELECTRICAL PROTECTION IV This results in: ELECTRICAL PROTECTION IV ELECTRICAL PROTECTION IV Example 5 Solve problem in Example 3 using impedance matrix method Per-unit admittance values ELECTRICAL PROTECTION IV = − 9.9454 −3.2787 −3.2787 8.2787 = + 0.11565 0.04580 0.04580 0.13893 1.05 0° = = = −9.079 . . 11 0.11565 " 1 = 1 − 2 11 0.11565 = 1.05 − ∙ 1.05 = 0 11 0.11565 21 0.04580 = 2 − = 1.05 − ∙ 1.05 = 0.63 11 0.11565 ELECTRICAL PROTECTION IV