ELECTRICAL POWER SYSTEM – II [ ] ACTIVE LEARNING ASSIGNMENT: TOPIC : SYMMETRICAL THREE PHASE FAULTS. UNIVERSITY :GUJARAT TECHNOLOGICAL UNIVERSITY. COLLEGE : VADODARA INSTITUTE OF ENGINEERING. DEPARTMENT : ELECTRICAL ENGINEERING [E.E.– I]. SEMESTER : VI. PREPERED BY : [ BHARGAV M. JAYSWAL ] [ JESTY JOSE ] [ JOBIN ABRAHAM ] GUIDED BY : ASST. PROF. PUSHPA BHATIA. [ELECTRICAL DEPARTMENT] [ELECTRICAL DEPARTMENT] ACTIVE LEARNING ASSIGNMENT 1

 Contents:  Introduction  Transients on a transmission line  Selection of circuit breakers  Short circuit of a synchronous machine (on no load)  Short circuit of a loaded synchronous machine  Algorithm for short circuit studies  References ACTIVE LEARNING ASSIGNMENT 2

 Introduction:  Technical definition:  The fault on the power system which gives rise to the symmetrical fault currents i.e. equal fault currents in the line with phase displacement is called a symmetrical fault.  Due to balance nature of fault, for the analysis of the fault only one phase is to be considered as faults in other two cases will be identical.  The following points should be particularly noted:  This type of fault occurs rarely in practice.  This type of fault is the most severe type of all faults and it imposes more heavy duty on the circuit breakers. ACTIVE LEARNING ASSIGNMENT 34

 Transients on a transmission line: ACTIVE LEARNING ASSIGNMENT 4

5  Transients on a transmission line {cont.}:

ACTIVE LEARNING ASSIGNMENT 6  Transients on a transmission line {cont.}:

 Circuit Breaker Selection:  A typical circuit breaker operating time is given in Figure.  Once the fault occurs, the protective devices get activated.  A certain amount of time elapses before the protective relays determine that there is overcurrent in the circuit and initiate trip command. This time is called the detection time.  The contacts of the circuit breakers are held together by spring mechanism and, with the trip command, the spring mechanism releases the contacts. ACTIVE LEARNING ASSIGNMENT 7

 When two current carrying contacts part, a voltage instantly appears at the contacts and a large voltage gradient appears in the medium between the two contacts.  This voltage gradient ionizes the medium thereby maintaining the flow of current. This current generates extreme heat and light that is called electric arc.  Different mechanisms are used for elongating the arc such that it can be cooled and extinguished. Therefore the circuit breaker has to withstand fault current from the instant of initiation of the fault to the time the arc is extinguished.  Two factors are of utmost importance for the selection of circuit breakers. These are: 1. The maximum instantaneous current that a breaker must withstand and 2. The total current when the breaker contacts part. ACTIVE LEARNING ASSIGNMENT 8  Circuit Breaker Selection {cont.}:

 However the instantaneous current following a fault will also contain the dc component.  In a high power circuit breaker selection, the sub transient current is multiplied by a factor of 1.6 to determine the rms value of the current the circuit breaker must withstand.  This current is called the momentary current. The interrupting current of a circuit breaker is lower than the momentary current and will depend upon the speed of the circuit breaker.  The interrupting current may be asymmetrical since some dc component may still continue to decay.  Breakers are usually classified by their nominal voltage, continuous current rating, rated maximum voltage, K -factor which is the voltage range factor, rated short circuit current at maximum voltage and operating time.  The K -factor is the ratio of rated maximum voltage to the lower limit of the range of the operating voltage.  The maximum symmetrical interrupting current of a circuit breaker is given by K times the rated short circuit current. ACTIVE LEARNING ASSIGNMENT 9  Circuit Breaker Selection {cont.}:

 Short circuit of a Synchronous machine (On no load)  Fig. shows a typical response of the armature current when a three-phase symmetrical short circuit occurs at the terminals of an unloaded synchronous generator.  It is assumed that there is no dc offset in the armature current. The magnitude of the current decreases exponentially from a high initial value. The instantaneous expression for the fault current is given by: ACTIVE LEARNING ASSIGNMENT10

 where V t is the magnitude of the terminal voltage, α is its phase angle and  Xd” is the direct axis subtransient reactance  Xd’ is the direct axis transient reactance  Xd is the direct axis synchronous reactance  Td” is the direct axis subtransient time constant  Td’ is the direct axis transient time constant  In the expression we have neglected the effect of the armature resistance hence α = π/2. Let us assume that the fault occurs at time t = 0.From eqn we get he rms value of the current as :  which is called the subtransient fault current. The duration of the subtransient current is dictated by the time constant T d . As the time progresses and T d  < t < T d, the first exponential term of eqn. will start decaying and will eventually vanish. However since t is still nearly equal to zero, we have the following rms value of the current ACTIVE LEARNING ASSIGNMENT11

 This is called the transient fault current. Now as the time progress further and the second exponential term also decays, we get the following rms value of the current for the sinusoidal steady state  In addition to the ac, the fault currents will also contain the dc offset. Note that a symmetrical fault occurs when three different phases are in three different locations in the ac cycle. Therefore the dc offsets in the three phases are different. The maximum value of the dc offset is given by  where T A is the armature time constant. ACTIVE LEARNING ASSIGNMENT12

 Short circuit of a loaded synchronous machine:  The synchronous machine is loaded when short circuit occurs.  Fig. shows a synchronous machine operating under steady state condition supplying a load current Io at terminal voltage Vo.  Eg is the induced emf under loaded condition. Xd is direct axis synchronous reactance of the machine.  When a short circuit occurs at the terminals of machine a short circuit current starts flowing through it. As it is time dependent it changes from sub-transient to transient magnitude.  The induced emfs during sub-transient and transient period are given as: Eg”=Vo+jIoXd”Eg’=Vo+jIoXd’  For Synchronous motor induced emf is given by: Em”=Vo-jIoXd”Em’=Vo-jIoXd’ ACTIVE LEARNING ASSIGNMENT 13

14  Short circuit of a loaded synchronous machine {cont.}:

 Algorithm for short circuit studies: Algorithm adopted for this type of analysis consists of following steps:  STEP 1: Obtain pre-fault voltages at all buses and currents in all lines through a load flow study.  STEP 2: Find Bus impedance matrix by inverting the bus admittance matrix.  STEP 3: Choose MVA base, KV base & calculate I base.  STEP 4: Specify the faulty bus and obtain current at the faulty bus and bus voltages during fault at all buses.  STEP 5: Find current flows in each line of the system.  STEP 6: Calculate SCMVA rating of circuit breaker(choose acc. to the fault current magnitude ) for each line & at each bus. ACTIVE LEARNING ASSIGNMENT 15

 References: 1. D.P. Kothari, I.J. Nagrath “Modern Power System Analysis”, McGraw Hill Education (INDIA) Pvt. Ltd., Fourth Edition, Eighth Reprint, 2013, ISBN : C.L. WADHWA “ELECTRICAL POWER SYSTEMS”, NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS, SIXTH EDITION, Reprint, 2014, ISBN : J.B. Gupta “A Course in POWER SYSTEMS”, S.K. KATARIA & SONS PUBLISHERS, Eleventh Edition, Reprint, 2015, ISBN : V.K. METHA, ROHIT METHA “PRINCIPLES OF POWER SYSTEMS”, S.CHAND & Co. PVT. LTD., First Multicolour Edition, Reprint, 2014, ISBN : system/ui/Course_home-6.htm system/ui/Course_home-6.htm system/ui/Course_home-6.htm /M pdf 3/M pdf 3/M pdf ACTIVE LEARNING ASSIGNMENT 16

17 ANY QUESTIONS? ANY QUESTIONS?