SHORT CIRCUIT STUDIES

Transcription

1 4 SHORT CIRCUIT STUDIES 4.1 INTRODUCTION In a power system short circuits occur once in a while due to lightning, flash over due to polluted insulation, falling of tree branches on the overhead system, animal intrusion and erroneous operations. When the fault current magnitudes are significant, it can cause damage to equipment and explosion if the fault is not cleared for prolonged time. Also, electrical fires and shock hazards to people are possible in a faulted power system. Therefore, it is important to design the power system such that the fault is isolated quickly to minimize the equipment damage and improve personnel safety. Short circuit studies are performed to determine the magnitude of the current flowing throughout the power system at various time intervals after a fault. The magnitude of the current through the power system after a fault varies with time until it reaches a steady state condition. During the fault, the power system is called on to detect, interrupt and isolate these faults, The duty impressed on the equipment is dependent on the magnitude of the current, which is a function of the time of fault initiation. Such calculations are performed for various types of fault such as three-phase, single line to ground fault, double line to ground fault and at different location of the system. The calculated short circuit results are used to select fuses, circuit breakers and protective relays. The symmetrical component model is used in the analysis of the unsymmetrical faults with mutual coupling.

2 The power system components such as utility sources, generators, motors, transformers and cables are represented by impedance value. The short circuit study is performed by representing the electric network in a matrix form. The program places a fault at a required location and performs the short circuit calculations. The short circuit study results are performed according to the applicable industry standards. 4.2 SOURCES OF SHORT CIRCUIT CURRENTS The short circuit current contributions are from the utility sources, generators, synchronous condensers and induction motors. Typical current waveforms during a short circuit are shown in Figure 4.1 for various types of contributing sources. Utility sources - A utility represents the large interconnection of generators, transmission lines and load circuits. The transmission lines, distribution lines and transformers introduce impedance between the generator source and the fault point during a short circuit. Therefore, the source voltage remains unaffected during fault conditions. Generator sources - An in-plant generator contributes to a short circuit and the current decreases exponentially. The generator is driven by a prime mover and an exciter supplies the field; the steady state current will persist unless the circuit is open by a circuit breaker. The generator fault current is determined by three reactance values during various time frames: Xd" - Direct axis sub-transient reactance, during the first cycle Xd' - Direct axis transient reactance, during 1 to 2 seconds Xd - Direct axis reactance, during steady state The total short circuit current (It) of a generator consists of an ac component (lac) and a dc component (Idc). These components are given by equations 4.1 through 4.3. The ac component of the generator fault current is: Iac = X d X d e Td" + xd X d (4.1) The dc component of the generator currents is: r( 1 1 JL Idc = (V2) e'td (4.2)

3 The total generator fault current (It) is: c + Idc (4.3) The reactance values are expressed in per unit in the above equations and the calculated currents will be in per unit. An example is presented at the end of this Chapter to show the generator short circuit currents. Synchronous motor and synchronous condenser - The synchronous machines supply fault current like a synchronous generator. The fault current decays rapidly since the inertia of the motor and load acts as the prime mover with the field excitation maintained. The fault current diminishes as the motor/condenser slows down and the motor excitation decays. Induction motor load - The fault current contribution from an induction motor is due to the generator action produced by the load after the fault. The field flux of the induction motor is produced due to the stator voltage and hence the current contribution decays very rapidly upon fault clearing as the terminal voltage is removed. Total fault current as a function of time - When a short circuit occurs in a system, the circuit impedance decreases appreciably. Therefore, the circuit current increases significantly as shown in Figure 4.2. During an asymmetrical fault, the total current can be treated as the sum of a dc current and a symmetrical ac component. The direct component eventually decays to zero as the stored energy in the system is expended in the form of I R loss. The direct current decay is inversely proportional to the X/R ratio of the system between the source and the fault. Therefore, it is important to analyze the short circuit current during the first cycle, the next several cycles and in the steady state.

4 UTILITY CONTRIBUTION GENERATOR CONTRIBUTION SYNCHRONOUS MOTOR CONTRIBUTION INDUCTION MOTOR CONTRIBUTION TOTAL SHORT-CIRCUIT CURRENT WAVEFORM Figure 4.1 Decaying Short Circuit Current Waveforms

5 TOTAL DC SHORT-CIRCUIT COMPONENT a CURRENT 4.3 SYSTEM IMPEDANCE DATA Figure 4.2 Total Fault Current Waveform Each component in an electrical system is represented by a suitable impedance value. Then the impedance values are converted to a common base. The system impedance data related issues are presented below. Source data - The utility source is represented by a per unit impedance which is equivalent to the maximum short circuit MVA level available from the utility at the point of common coupling. The minimum source impedance is used in the short circuit current calculations for relay settings. The source impedance data is usually provided by the system/utility. Example The three-phase and one line to ground fault short circuit duties of a 230 kv, 3-phase power system is 671 MVA and 738 MVA respectively. Find the symmetrical component impedance values on a 100 MVA base. Solution - Use an X/R ratio of 20. _~ kv MVA,3ph 671 = 78.8 Ohm X (base) on 100 MVA base = (230 x 230)/ Ohm

6 X, on 100 MVA base = Ohm/529 Ohm P.U. Ri on 100 MVA base = P.U./20 = P.U. X = - =57.5 Ohm 0 MVA, slg MVA,3ph X 0 on 100 MVA base = 57.5 Ohm/529 Ohm = P.U. Ro on 1 00 MVA base = P.U./ P.U. Transmission lines - The transmission lines are represented by positive and zero sequence impedance values. The approach to calculate the transmission line constants and typical values are presented in Chapter 2. In transmission lines it is assumed that the positive and negative sequence impedance values are equal. Cable data - The cable impedance values are presented by the positive and zero sequence impedance values. The calculation procedures are discussed in Chapter 2. Transformer impedance - The transformer impedance values are given in percentage on the self-cooled transformer kva rating and are converted to per unit on the study base. Typical impedance data of two winding transformers up to 2.4 kv are presented in Table 4.1. Table 4.1 Typical Transformer Impedance up to 2.4 kv 3-Phase kva X/R Ratio % Impedance Range The impedance values for the transformers above 2.4 kv are listed in Table 4.2.

7 Table 4.2 Typical Transformer Impedance Above 2.4 kv Ratings Primary kv Primary kvbil % Impedance Range For the calculation of the line to ground fault short circuit currents, the zero sequence connection diagram for certain transformers are required. Synchronous machines - A synchronous machine is represented by the subtransient direct axis reactance. Typical reactance values for various synchronous machines are presented in Table 4.3. Table 4.3 Typical Reactance of Synchronous Machines in Percentage Type of Generator Turbo generator, 2 -pole Turbo generator, 4-pole Waterwheel generator, no damper Waterwheel generator, with damper Synchronous condenser Salient pole motor, high speed Salient pole motor, low speed Xd Xq Xd' Xd" X xo Note: Use average value as typical data and on machine MVA base. Induction motors - The kva rating of induction motors is approximately equal to the hp rating. The sub-transient reactance is given by the locked rotor reactance. Typical locked rotor reactance values of various induction motors are given by: Over 600 V = 0.17 P.U. on machine kva base 600 V and less = 0.25 P.U. on machine kva base The motor resistance can be calculated using the X/R ratio curve given in ANSI Standard 242. Sometimes, the motors in each substation are lumped by type and size and a single per unit impedance is determined based on the lumped kva.

8 4.4 SHORT CIRCUIT CALCULATIONS The following short circuit calculations are generally performed. Three-phase fault - The three-phase conductors are shorted together. The magnitude of the fault current is given by: If = (4.4) L \ Where E is the phase voltage and Zl is the positive sequence impedance. Phase to ground fault - Where one phase conductor is shorted to ground. The fault current magnitude is given by: If = (Z + z E +z ) (L \ +L 2 0 } (4 ' 5) Where Zi, Zi and Zo are the positive, negative and zero sequence impedances respectively. Double line to ground fault - Where two phase conductors are shorted to ground. The positive sequence current magnitude is given by: z /rz +z Z 0 ( 2 (4 ' 6) The three-phase fault currents are the highest and least for phase to ground fault. However, the phase to ground fault current can be the highest under certain circumstances such as: Near solidly grounded synchronous machines. Near the solidly connected wye of a delta/wye transformer of the three phase core/three leg design. A grounded wye/delta tertiary auto-transformer. A grounded wye grounded wye/delta, three winding transformer. Therefore, in systems with the above type of connections, it is necessary to conduct a phase to ground fault calculation. For resistance grounded systems, the phase to ground fault currents will be limited. The following assumptions are made in the short circuit calculations:

9 An unloaded power system is considered. The impedance at the fault location is assumed to be zero. Load characteristics are ignored. Motors are assumed to run at rated speed with rated terminal voltage. Nominal transformer taps will be considered. A symmetrical three-phase power system is considered. The momentary or first cycle, the interrupting and the symmetrical short circuit currents are calculated to meet the circuit breaker selection and relay applications. Momentary or first cycle short circuit current calculations - The momentary or first cycle current is used to evaluate the closing or latching of the medium or high voltage circuit breaker. The circuit impedance of the utility sources, generators, transmission lines and transformers are used for the momentary current calculations. The sub-transient reactance of utility source and the generator are used. The reactance of the synchronous motors and induction motors are modified according to Table 4.4 from IEEE Standard 141. Table 4.4 Multipliers for Short Circuit Current Calculations Type of Rotating Machine LV Studies First Cycle Interrupting Generator, hydro generators with amortisseur windings & condensers Synchronous motors Induction motors Above 1000 hp at 1800 rpm or less Above 250 hp at 3600 rpm All others, 50 hp and above Less than 50 hp 1.0 Xd" 1.0 Xd" 1.0 Xd" 1.0 Xd" 1.0 Xd" Neglect 1.0 Xd" 1.0 Xd" 1.0 Xd" 1.0 Xd" 1.2Xd" Neglect 1.0 Xd" l.sxd" 1.5Xd" l.sxd" 3.0Xd" Neglect The impedance of the equivalent network is established and is converted into per unit at each fault point. The first cycle fault current is calculated using the pre fault voltage and the impedance at the faulted node. The approximate calculated first cycle fault current for comparison with the circuit breaker capability is obtained by using a 1.6 multiplier specified in ANSI C The total three phase symmetrical fault current (Isc tot) is: Isc tot Epu L6 Ibase (4.7) Xpu This current magnitude is compared with the latching current rating of a circuit breaker and the instantaneous rating of the relay settings.

10 Interrupting current calculations - The interrupting fault currents are calculated in accordance with ANSI C for symmetrical current rated circuit breaker applications. The reactance of the rotating machines is chosen and applied as per the multiplying factors listed in Table 4.4. For the calculation of the interrupting currents, both the resistance and reactance of the equivalent circuit is needed. Therefore, the equivalent R and X are calculated at the fault point and the X/R ratio is determined. The interrupting current is calculated using the E/Z method. A multiplying factor is calculated using the X/R ratio and the contact parting time of the circuit breaker. The minimum contact parting times are presented in Table 4.5 obtained from ANSI C Table 4.5 Contact Parting Time for Circuit Breakers Rated Interrupting Time Cycles at 60 Hz Minimum Contact Parting Time, Cycles at 60 Hz The multiplying factors are plotted in ANSI C for two specific conditions. Fault fed predominantly from generators for three-phase faults (8, 5, 3 and 2 cycles). Fault fed predominantly from networks with two or more transformations for three phase and line to ground faults (8, 5, 3 and 2 cycles). The calculated interrupting fault current (lin) is given by: Epu lin = (Multiplying Factor) ibase (4.10) Xpu This interrupting current is to be compared with circuit breaker interrupting capability. It can be seen that the short circuit calculation procedures for the momentary duty and the interrupting duty are different. Short circuit calculations for low voltage circuit breaker applications - The impedance diagram for evaluating the short circuit current at the faulted point developed. The rotating machine impedance values used without multiplying factors are shown in Table 4.4. The resistance values are also calculated using the

11 procedure used in the interrupting current calculations. The fault current is given by E/Z. The low voltage short circuit values are used for the selection of the circuit breakers including the molded case types. Short circuit current calculation for relay applications - For the instantaneous relay setting, the fault current magnitude from the momentary duty is used. For other applications with a time delay, the steady state symmetrical fault currents are calculated and used. The impedance for the transformer, overhead line and the cables are obtained for the steady state conditions. The motor contributions are neglected. For synchronous machines the steady state reactance can be used. In this case it can be seen that the dc component has almost decayed to zero and it is not considered. The required symmetrical fault current is obtained by using the E/Z method. In many power system and industrial set ups, the source impedance is different for peak and off-peak conditions. Therefore, the maximum and minimum fault currents are to be calculated and the relay settings should be verified for both conditions. Impedance diagram - The impedance diagram is derived from the one line diagram by modeling the circuit elements by the respective impedance. The impedance magnitude used for the fault calculation depends on many factors which are discussed in the circuit breaker current rating calculations. This is important for the reactance of the rotating machines (see Table 4.4). Once the fault location and type of short circuit is identified, then the corresponding impedance diagram is developed. This concept is demonstrated with the help of an example. Per unit quantities - For a balanced three-phase system, the relation of three-phase kva, line to line voltage, base current and base impedance are defined as: Base kva Base current = T= (411) V3(BasekV) (BasekVA) 2 Base impedance = (4 12) BaseMVA For changing the P.U. impedance from the given base kv to new base kv: Xnew (Xgiven) 'kv. given ~kv new (4.13) When both the kva and kv are new, then the new P.U. impedance can be calculated using the following equation:

12 Xnew f.,,. kva new (Xgiven) kva ^ given \f-kv. given (4.14) kv new Example The nameplate specifications of a two winding transformer are 40 MVA, 69 kv/13.8 kv, delta/wye-grounded, 7% impedance. This transformer is to be connected in a 13.2 kv distribution system. The system studies are performed on a 100 MVA base. Calculate the transformer impedance on 13.2 kv and 100 MVA base. Solution - The new P.U. impedance can be calculated as: f r\r\\ f i o \^ Xnew = (7.0% J[ J =2.7324% 4.5 COMPUTER-AIDED ANALYSIS There are several programs available to perform the short circuit studies. These programs can be used to perform the following data-related operations. Convert the raw system data to a common base. Prepare one-line diagrams. Determine system impedance for the calculation of momentary, interrupting, symmetrical and relay short circuit currents. The input data to these programs can be entered interactively or presented in ASCII data files or through graphic interface. The output of the short circuit study includes the following. Short circuit input data used in the network analysis. Calculations of three-phase, single line to ground fault, line to line and double line to ground fault currents. For the three-phase, 4 wire system with a neutral conductor, the short circuit currents are required for the line to neutral short circuit. Calculation of appropriate circuit breaker current ratings based on ANSI or IEC standards. Some programs present the short circuit outputs in a one line diagram with the calculated values. Summary of currents at all the buses. The user selects the necessary short circuit results at appropriate buses and

13 compares the results with the circuit breaker ratings. Also, the short circuit currents are compared with the equipment short circuit ratings to ensure safe performance. Example An industrial power plant is shown in Figure 4.3. The 230 kv source has a three-phase circuit current rating of 28,000 A. The step down transformer (Tl) is 100 MVA, 230/24 kv, 0.10 P.U. reactance, delta/wyegrounded with an X/R ratio of 20. There is a 500 MVA standby generator connected to the 24 kv bus. The reactance of the generator is 0.2 P.U with X/R ratio of 20. Transformer T2 is 75 MVA, 24/4.16 kv, delta/wye, 0.11 P.U. reactance with X/R ratio of 30. There are two 2,000 hp, 0.9 power factor, Xd" = 0.2 P.U, Xd 1 = 0.26 P.U. with X/R ratio of 20. The low voltage system contains a transformer T3, 4.16 kv/600 V, 0.08 P.U reactance and X/R There is a low voltage motor at the 600 V bus with a rating of 400 hp, Xd" = 0.3 and X/R = 30. Perform a short circuit study using computer-aided software and determine the short circuit currents at Fl and F2. Also, show the step-by-step calculations. Compare the results. Solution - The reactance of all the passive elements are listed below. A 100 MVA base is used in the calculations. Transformer, Tl, X = (0.10)( 100/100) Transformer, T2, X = (0.11)( 100/75) Transformer, T3, X = (0.08)(100/0.75) = 0.1 P.U. =0.15 P.U. = P.U. Sub-transient reactance of rotating machines 230 kv system short circuit MVA = (>/3 ) (230 kv) (28 ka) = kv source impedance, X = 1.0 (100/11154) = P.U. Generator Gl, X = (0.25) (100/500) = 0.05 P.U. Induction motor Ml, X = (0.20) (100/2) = 10 P.U. Induction motor M2, X = (0.20) (100/2) = 10 P.U. Induction motor M3, X = (0.30) (100/0.4) = 75 P.U. Momentary fault current calculations - For the momentary or first cycle short circuit calculation, induction motors less than 50 hp are omitted. For the 400 hp motor, X = 1.2 Xd" = (1.2) (75 P.U.) = 90 P.U. The reactance of the source, generator and the induction motors are identified in the impedance diagram for the first cycle or momentary short circuit calculations, in Figure 4.4.

14 SOURCE, 23O kv Gl 24 kv 4.16 kv M2 Ml M3 Figure 4.3 One-Line Diagram for Example P.U Figure 4.4 Impedance Diagram for the Momentary Duty Calculation The symmetrical first cycle or momentary short circuit current at the faulted point Fl is calculated as: Ibase at the 4.16 kv bus = 100 MVA/( >/3 )(4.16 kv) ka 1 Isym = (13.879kA) =78.3 ka The peak value of the momentary short circuit current is (1.6 x 78.3 ka) ka.

15 Calculation of the interrupting short circuit current - The impedance diagram is shown in Figure 4.5. The reactance, X/R ratios and the resistances for the circuit breaker interrupting current calculations are listed below. Transformer Tl, X/R = 30, R = (0.10/30) Transformer T2, X/R = 30, R = (0.15/30) Transformer T3, X/R = 30, R = (10.67/30) 230 kv system, X/R = 20, R - ( /20) Generator Gl, X/R = 20, R = (0.05/20) Motor Ml, X = (1.5 Xd") = (1.5 x 10) Motor Ml, X/R = 20, R = (15/20) Motor M2, X/R = 20, R = (15/20) Motor M3, X = (3 Xd") = 3 x 75 Motor M3, X/R = 30, R = (225/30) E/X = (13.879kA) = 77.32kA J = P.U. = P.U P.U. = P.U. = P.U. = 15 P.U. = 0.75 P.U. = 0.75 P.U. = 225 P.U. = 7.5 P.U P.U 1795 P.U Figure 4.5 Impedance Diagram for the Interrupting Duty Calculation

16 P.U Figure 4.6 Resistance Diagram for the Interrupting Duty Calculation The resistance circuit for the interrupting duty calculation is presented in Figure 4.6. Using the reduced R, the X/R ratio is calculated at the faulted point Fl. The X/R ratio at the faulted point Fl = / = The corresponding NCAD ratio is determined from C The NCAD ratio for X/R ratio of is The interrupting current = (77.32 ka) (0.96) = 74.2 ka Short circuit calculation for low voltage circuit breaker applications - The impedance diagram for evaluating the short circuit current at the faulted point F2 is shown in Figure 4.7. The impedance at F2 = ( j ) P.U. = P.U. The base current at 600 V is ka. The short circuit current at F2is ka P.IM P.U P.U P.U 3.56 P.U 2.5 P.U Figure 4.7 Impedance Diagram for the Low Voltage Short Circuit Current Study

17 Computer-aided analysis - The data is entered interactively to the program and the input data listings and the output summary are presented below. The software used in the calculations was the AFAULT program from the SKM System Analysis. The present version of the program is called Power tools for Windows with the particular option [9]. The program performs the calculations by placing various faults at each node for different fault duties. The equipment voltage rating, reactance and the reactance to the 100 MVA base are displayed. The input data is classified into various sections as discussed below. Source data as per List From Bus, to Bus, Voltage, Base MVA, Xd", X/R, X on 100 MVA Base. List 4.1 Source Data From Bus To Bus kv Base MVA Xd" P.U X/R Zon 100 MVA Base P.U Source Generator 1 Motor M 1 Motor M2 Motor M3 Bus 1 Bus 5 Bus 9 Bus 10 Bus ( J ) ( j 0.05) (0.5 +j 10) (0.5 +j 10) (2.5+J75) Feeder data as per List 4.2. The assumed impedance values are low and are not used in the step-by-step calculations. From Bus, to Bus, Voltage, Length in Feet, Zl or Z2 in P.U, ZO in P.U. List 4.2 Feeder Data From Bus Bus 1 Bus 3 Bus 4 Bus 4 Bus? Bus 8 Bus 9 To Bus Bus 2 Bus 4 Bus 5 Bus 6 Bus 8 Bus 9 Bus 10 Line Length kv Feet ZlorZ2, P.U. ZO, P.U (0.0 +j 0.01) (0.0 +j 0.03) (0.0 +j 0.01) (0.0 +j 0.03) (0.0 + j ) (0.0 + j ) (0.0 +j ) (0.0 + j ) (0.0 +j ) (0.0 + j ) (0.0 + j ) (0.0 +j ) (0.0 +j ) (0.0 +j 0.026) Transformer data as per List Primary Bus, Connection, Voltage, Secondary Bus, Connection, Voltage, MVA Base, Z1/Z2 on 100 MVA Base, ZO on 100 MVA Base.

18 List 4.3 Transformer Data From Bus To Bus Primary Conn Primary Sec kv Conn Sec kv Zl orz2, P.U. ZO, P.U. Bus 2 Bus 6 Bus 8 Bus 3 Bus? Bus 11 Del Del Del Wye Wye Wye (0.03 +j 0.10) ( j ) (0.36 +j ) ( j 0.10) ( JO. 1467) (0.36 +j ) The program calculates the short circuit components at each bus location for the various fault duties along with the summary for each type of calculations. The summary of the typical output listing is presented below. Fault current for the low voltage circuit breaker applications as per List 4.4. The list includes the Bus Number, Bus Name, Voltage, Fault Current for Three- Phase, X/R Ratio, Fault Current for SLG, X/R Ratio. List 4.4 Fault Current for the Low Voltage Circuit Breaker Applications Bus# Bus Name Voltage Volts 3 Phase Fit, ka 3 Phase X/R SLG ka SLG X/R BUS 1 BUS 2 BUS 3 BUS 4 BUSS BUS 6 BUS 7 BUSS BUS 9 BUS 10 BUS Fault current for momentary or first cycle as per List 4.5. The list includes the Bus Number, Bus Name, Voltage, Fault Current for 3-Phase, X/R Ratio, Fault Current for SLG, X/R Ratio. The fault current magnitude is 1.6 times the short circuit at the fault location. Fault current for interrupting as per List 4.6. The output contains the Bus Number, Bus Name, Voltage, decrement ratio, Fault Current for three-phase, X/R Ratio, Fault Current for SLG, X/R Ratio.

19 Bus# List 4.5 Fault Current for the Momentary Duty Bus Name BUS1 BUS 2 BUS 3 BUS 4 BUSS BUS 6 BUS 7 BUSS BUS 9 BUS 10 Voltage Volts Phase E/Z* Phase X/R SLG E/Z* SLG X/R List 4.6 Fault Current for the Interrupting Duty Bus# Bus Name Voltage Volts Ratio 3 Phase E/ZKA 3 Phase X/R SLG E/ZKA SLG X/R BUS 1 BUS 2 BUSS BUS 4 BUSS BUS 6 BUS 7 BUS 8 BUS 9 BUS The results from the step-by-step calculations and the computer-aided analysis are compared at the faulted points Fl (node 8) and F2 (node 11). Description Calculated Computer-Aided Momentary at Fl (node 8) ka ka Interrupting at F1 (node 8) 74.2 ka ka Low voltage short circuit at F2 (node 11) ka ka In the step-by-step calculations, the cable lengths are ignored. Therefore, the calculated short circuit currents are higher than the values from the computer-aided analysis.

20 4.6 LIMITING THE SHORT CIRCUIT CURRENTS Using series reactors, high impedance transformers and high resistance grounding can control the short circuit current in the power system. The series reactor can be used in the generator circuits, bus bars, feeders and in the shunt capacitance circuits. There are advantages and limitations to these approaches. With the application of shunt capacitor banks for power factor correction, there is always the inrush current issue during energization. Also, the outrush current from the capacitor banks is a concern when a line circuit breaker closes in to a nearby fault. In order to limit both the inrush and outrush currents series reactors are used. Three schemes of series reactors for shunt capacitor application are discussed. Scheme 1: Series reactor with each capacitor bank - Such a scheme is shown in Figure 4.8. In order to satisfy the criteria (Iph. f) to less than 2.0E+7, there will be two reactors with two capacitor banks. Scheme 2: Series capacitors for inrush and outrush requirements - The required scheme is shown in Figure 4.8. The reactor size for each capacitor bank will be small to limit the inrush current. A third reactor will be used to limit the outrush current. Scheme 3: Reactor to limit outrush current and breaker to limit the inrush current - The inrush current can be controlled by using circuit breaker with controlled switching or by using closing resistor/inductor. The outrush current can be controlled by using a series reactor. Such a scheme is shown in Figure 4.8. Example In order to demonstrate the circuit breaker selection and the application of series reactor for the current limiting a shunt capacitor bank, a case is presented. The circuit breaker is chosen to meet this application and the short circuit current magnitudes are calculated if the required current specifications are met. Then a reactor is chosen in series with the circuit breaker and the procedure is repeated. The system is a 230 kv, 60 Hz, three-phase with a short circuit rating of 40 ka. The circuit breaker has to be selected for capacitor switching application. The capacitor is available in two banks each of which is 60 MVAR.

21 Scheme 1 L1, L2 - Series Reactors C1, C2 - Capacitance L3 L1 -C1 L2 Scheme 2 L1, L2, L3- Series Reactors C1, C2 - Capacitance L3 L2 Scheme 3 L2, L3 - Series Reactors C1, C2 - Capacitance C2 Figure 4.8 Series Reactor Schemes for Current Limiting Solution - The circuit breaker is intended to switch 120 MVAR shunt capacitor banks and should meet the performance criteria described in ANSI C37.06 [7]. The desired performance specifications of the circuit breaker to meet the capacitor switching application (definite purpose) of the 230 kv systems are: Nominal voltage rating Maximum voltage rating Rated current Three-phase short circuit rating = 230 kv = 242 kv = 2000 A = 40kA

22 Rated capacitive current (breaking duty) TRY 188 kv peak - peak voltage base Inrush current Transient frequency = 400 A = 443 kv (2.4 PU) = 20 ka peak = 4,250 Hz Without current limiting reactor - When a system fault occurs near the capacitor bank location, the electrical energy stored in the capacitor bank discharges through the low fault impedance with considerable magnitude and at high frequency. Such a system is shown in Figure 4.9. The size of the 230 kv capacitor bank is 60 MVAR, three-phase. System Fault 'pk Capacitor Bank Figure 4.9 Fault Outside the Circuit Breaker Without Series Reactor The expected outrush current magnitude and frequency, for a single 60 MVAR, 230 kv capacitor bank is given by: /-< _ MVAR 60 2 >r(60) xkv 2 2 ;r(60) x = 3 MFD V 230 kvx / r Pk TI _ V/V3, A pk --_--^= kA.T _ 2 ^L f xc 2 >/36.1//Hx3//F = 15.3kHz An inductance of 10^ H for the bank and JU H/ft with a 100 feet cable length is used (IEEE C ).

23 The calculated outrush current magnitude of 54.1 ka is much higher than the allowable IEEE Standard C37.06 (Table 3A) value of 20 ka. Also, the frequency of the outrush current is higher than the allowed value of 4,250 Hz. The outrush current from the capacitor bank needs to be controlled using current limiting series reactors. The minimum series reactor needed to limit the (I p k x f) product to less than 2.0 x 10 is given by: V pk 'min 7 2n x2 xlo With current limiting reactor - The equivalent circuit with a series reactor in the shunt capacitor circuit is shown in Figure For the proposed 230 kv, 60 MVAR bank the minimum reactor needed is: T _ min ~ 230 kvx 2 ;r(2x!0 7 ) = 1.5 mh For the high outrush current to occur, a breaker must close into a fault very close to the 230 kv substation. A series inductor of 3 mh is selected for the 230 kv circuit and the corresponding Ipk and the frequency of oscillation are given by: = 6kA f = 2 A f xc 2 ^3000 yuhx3 = 1.68 khz This peak current and the frequency of oscillation are below the ANSI C37.06 values. Therefore, the circuit breaker is acceptable for the energization of the 230 kv, 60 MVAR shunt capacitor bank.

24 Reactor Circuit Breaker C1 = 60 MVAR Figure Capacitor Circuit with Series Reactor From this example, it is clear that series reactors can be useful in shunt capacitor circuits to limit the fault current magnitudes and in the protection of circuit breakers. Example Consider a 800 MVA, 13.8 kv generator with the following parameters: Xd" Xd? Xd = P.U P.U. =1.820 P.U. TdO" = TdO 1 = TdO =0.330 Calculate the total fault currents at 1.5 cycles. Solution - (IT) Xd" = (0.210)1 -^- = P.U. v j X'd = (0.330) = P.U. 800 f 100^1 Xd =(1.820) = P.U. \800j The ac components of the generator short circuit currents are: Id" = v^ xday = P.U.

25 Id' - = P.U. Id = V y = P.U. The time constants of the ac currents are given by: " ^ (TdO") = s = cycles T'd = Xd. X, (TdO') = s = cycles The ac component of the generator fault current is: t t Iac = (Id"-Id')e"Td" +(I'd-Id) e ~Td' +W At 1.5 cycles, the ac current components are: lac Ibase lac = ( x ) + ( x 0.982) = on 100 MVA base 100MVA = 4.18kA V3(13.8kV) = ( P.U.) (4.18 ka) = ka The dc component of the generator currents is: Idc = (A/2) (38.10) e ~Td Where Td in cycles is (0.33 x 60) 19.8 cycles. Idc at a time oft =1.5 cycles is: Idc = P.U. on a 100 MVA base = ( P.U. x 4.18 ka) = ka

26 The total generator fault current (It) is: It = Vlac 2 + Idc 2 = V = ka Therefore, in the generator circuit there are both ac and dc current components present as shown. PROBLEMS 1. The three-phase short circuit rating of a 230 kv, 60 Hz system is 670 MVA. The single line to ground fault rating is 600 MVA. Calculate the source impedance values on a 100 MVA base. State the assumptions made, if any. 2. The three phase short circuit rating of a 345 kv source is 20,000 MVA and the single line to ground short circuit MVA is 15,200 MVA. Calculate the sequence impedance of the source in P.U. on a 100 MAV base. 3. A 75 kva, 14 kv/4.16 kv, delta/wye-grounded, 6% impedance transformer is to be used in a 13.8 kv distribution system. The system studies are performed on a 100 MVA base. Calculate the transformer impedance for the new bases. 4. Calculate the total fault currents due to a 3-phase fault at the open terminals of a generator at 1.6 cycles. The name plate specifications of the generator are 500 MVA, 13.8 kv with the following parameters: Xd" =0.200 P.U. TdO" =0.030 Xd' =0.350 P.U. TdO' =8.000 Xd =1.800 P.U. TdO = What are the different types of faults in a power system? What is the role of a neutral conductor in the power system? 6. Why is it necessary to select a circuit breaker based on the short circuit current ratings? 7. Consider an industrial power system with a 115 kv source from a substation (Bus 1) to the next substation (Bus 2) connected through an overhead line. Bus 2 is connected to Bus 3 through an underground cable of 2,000 feet length. A step down transformer at Bus 3 supplies an electrical motor at 13.8 kv. The necessary system data are as follow:

27 Source impedance on 100 MVA base is: Z, =Z 2 = ( j ) P.U. Zo = ( j ) P.U. Overhead line impedance between Bus 1 and Bus 2 on 100 MVA base: Z = ( j ) P.U. Cable impedance between Bus 2 and Bus 3: Zi = ( j ) Ohm/1000 feet ZO = ( j ) Ohm/1000 feet Transformer impedance at Bus 3: MVA = 120 Voltage ratio =115 kv/4.16 kv Connection = Wye/Delta Reactance = 12% Induction motor drive: Rating =100MW Voltage = 13.8kV Power factor =0.8 Draw the one-line diagram of the system and state the assumptions. Calculate the short circuit currents at each node and select the circuit breaker ratings at appropriate locations. Also calculate the voltage drop at various locations. If the voltage profile is not acceptable, suggest suitable remedial actions. Compare the calculated results with a computer program output. 8. Refer to Example 4 regarding the energization of a 60 MVAR capacitor bank. Using the same circuit breaker, an additional 60 MVAR shunt capacitor bank is to be connected for power factor correction purposes. This capacitor bank is to be energized through a circuit switcher as shown in Figure The value of the series reactor is 20 mh. Calculate the Ipk and frequency of oscillation without series reactor and with series reactor. Discuss the circuit breaker suitability for the given application.

28 Circuit Switcher with Series Reactor Reactor Circuit Breaker C1 = 60 MVAR C2 = 60 MVAR Figure 4.11 Back-to-Back Capacitor Switching with a Circuit Switcher REFERENCES 1. ANSI/IEEE Standard: 141, IEEE Recommended Practice for Electrical Distribution for Industrial Plants, 1996 (Red Book). 2. ANSI/IEEE Standard: 399, IEEE Recommended Practice for Power System Analysis, 1990 (Brown Book). 3. ANSI Standard: C37.10, American National Standard Requirements for Transformers 230,000 Volts and Below. 4. ANSI Standard 242, IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems, ANSI/IEEE Standard C37.010, IEEE Application Guide for AC High Voltage Circuit Breakers Based on a Symmetrical Current Basis, ANSI/IEEE Standard C37.13, IEEE Standard for Low Voltage AC Power Circuit Breakers Used in Enclosures, ANSI Standard C37.06, AC High Voltage Circuit Breakers Rated on a Symmetrical Current Basis, ANSI Standard C37.012, Application Guide for Capacitance Current Switching for AC High Voltage Circuit Breaker Rated on a Symmetrical Current Basis, Power Tools for Windows, SKM System Analysis, Inc., Manhattan Beach, California.