Draw the fault current direction both in CTs and operating relay coils for internal and external LLL and LLG faults in the Merz Price Differential Protection System, Modified Differential Protection System and Restricted Earth Fault Protection system for Alternator Stator Winding. Its a topic of Switchgear protection in power system

Draw the fault current direction both in CTs and operating relay coils for internal and external LLL and LLG faults in the Merz Price Differential Protection System, Modified Differential Protection System and Restricted Earth Fault Protection system for Alternator Stator Winding. Its a topic of Switchgear protection in power system

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter7: Symmetrical Faults

Section: Chapter Questions

Problem 7.1MCQ

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Draw the fault current direction both in Current transformer and operating relay coils for internal and external double line to line and double line to ground faults. Do the solve for Restricted Earth Fault Protection system for Alternator Stator Winding. Its a topic of Switchgear protection in power system
Draw the fault current direction both in Current transformer and operating relay coils for internal and external double line to line and double line to ground faults in the Merz Price Differential Protection System, Modified Differential Protection System and Restricted Earth Fault Protection system for Alternator Stator Winding. Its a topic of Switchgear protection in power system
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b) A fault occurs at bus 3 of the network shown in Figure Q4. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus…