Consider the radial power system shown below. Assume that the generator is ideal, with a terminal voltage of 1.0 pu. The system data aregiven in the accompanying table. If the line-to-line fault currents flowing in circuit breaker B2 for faults at buses 2 and 3 are respectively OAand 9.53A, then the corresponding faulted phase voltages in pu are equal to:235From To Positive sequenceimpedanceB₁B₂B3B4120.01+j0.05230.003+j0.04340.008+j0.0450.01+j0.05Select one:O a. 0V, 0.22VO b. 0V, 0.44VC.OV, 0.33VOd. None of these

Answered: Consider the radial power system shown…

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

Chapter9: Unsymmetrical Faults

Section: Chapter Questions

Problem 9.4MCQ

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What are the hazards of phase failure and phase reversal?
Consider the oneline diagram of a simple power system shown in Figure 9.20. System data in per-unit on a 100-MVA base are given as follows: The neutral of each generator is grounded through a current-limiting reactor of 0.08333 per unit on a 100-MVA base. All transformer neutrals are solidly grounded. The generators are operating no-load at their rated voltages and rated frequency with their ENIFs in phase. Determine the fault current for a balanced three-phase fault at bus 3 through a fault impedance ZF=0.1 per unit on a 100-MVA base. Neglect -Y phase shifts.
b) A fault occurs at bus 4 of the network shown in Figure Q3. 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, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) = j0.1X p.u. V₁ = 120° p.u. V₂ = 120° p.u. (i) (ii) 0 jX(1) = j0.2 p.u. 1 jx(2) j0.2 p.u. 2 jX1(0) = j0.25 p.u. jXT(1) jXT(2) 종 3 j0.1X p.u. JX3(1) j0.1Y p.u. j0.1X p.u. JX3(2) j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0 = x = 1, jX2(1) j0.2Y p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V3 = 120° p.u. Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: y = 7 = = jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from…
Q2. The single-line diagram of a simple three-bus power system is shown in Figure-2. Each generator is represented by an emf behind the sub-transient reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of Zf = j0.19 per unit. (i) Using Th'evenin's theorem, obtain the impedance to the point of fault and the fault current in (ii) Determine the bus voltages per unit. ) j0.05 j0.075 j0.75 2 j0.30 j0.45 Figure-2: Single line diagram of the power system network for Q2 3
b) A fault occurs at bus 2 of the network shown in Figure Q3. 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, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. JX20 /0.1X p.u. jXa2) 0.1X p.u. JX20 j0.2Y p.u. V,= 120° p.u. V, 120° p.u. V, 120° p.u. jX4-70.2X p.u. jX2 j0.2X p.u. jX o 0.2Y p.u. jXncay J0.25 p.u. jXna J0.25 p.u. 3 jXno0.3 p.u. jXTu) /0.2Y p.u. jXra j0.2Y p.u. - j0.2Y p.u. Xp-10.1X p.u. jXa j0.1X p.u. jXp0)- j0.05 p.u. 0 Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXac1) = j0.22 p.u., jXac2) = j0.22 p.u., and jXaco) = j0.23 p. u. X-2 Y=8 (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 2.
Description In the particular case of figure below derive both the critical clearing angle and the critical clearing time. P, = Pmaz sin d Pm A1 do der Smar A generator having H = 6.R MJ/MVA is delivering power of 1.0 per unit to an infinite bus through a purely reactive network when the occurrence of a fault reduces the generator output power to zero. The maximum power that could be delivered is 2.5S per unit. When the fault is cleared, the original network conditions again exist. Determine the critical clearing angle and critical clearing time. (Roll=PQRS)
b) A fault occurs at bus 3 of the network shown in Figure Q3. 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, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. jx2(1) = j0.1X p.u. jx2(2) = j0.1X p.u. jX2(0)=j0.2Y p.u. = V₁ 120° p.u. V₂ = 120° p.u. V₁ = 120° p.u. jx)=j0.1X p.u. jX2)= j0.1X p.u. jxo)=j0.1Y p.u. jx(1) = j0.25 p.u. 2 X (2) = 0.25 p.u. 3 jx1(0)=j0.3 p.u. jXT(I)=j0.1Y p.u. jX3(1)=j0.1X p.u. = jXT(2) j0.1Y p.u. jx13(2)=j0.1X p.u. jXT(0) = j0.1Y p.u. jX3(0) = j0.05 p.u. 0 0- Figure Q3. Circuit for problem 3b). (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…
Three 15MVA, 30kV synchronous generators A, B, and C are connected via three reactors to a common bus bar, as shown in Figure below. The neutrals of generators A and B are solidly grounded, and the neutral of generator C is grounded through a reactor of 2. Q. The generator data and the reactance of the reactors are tabulated below. A line-to-ground fault occurs on phase a of the common bus bar. Neglect prefault currents and assume generators are operating at their rated voltage. Determine the fault current in phase a. GB Gc Item GA GB Gc 0.25 pu 0.155 pu 0.056 pu 0.20 pu 0.155 pu 0.056 pu 0.20 pu 0.155 pu 0.060 pu 6.0 ? REACTOR Reactor 6.0 6.0
The one line diagram of a simple three bus power system is shown in figure Each generator is represented by an emf behind the subtraction reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. Athree phase fault occurs at bus 3 through a fault impedance of Zf=j0.19 per unit. a-using thevenin’s theorem obtain the impedance to the point of fault and the fault current in per unit. b-determine the bus voltage and line currents during fault
b) The one-line diagram for a simple power system is given in Figure 1(b). į. Draw the impedance diagram of the system. i. Construct the bus admittance matrix, Xeus and bus impedance matrix, Zous of the system. I. Evaluate the fault current, bus voltages and line current during the flow when a balanced three-phase fault occurs on bus 1. bus 1 bus 2 j65 N G G 25 MVA 25 MVA 13.8 kV 15% 13.2/69 kV 11% 25 MVA 69/13.2 kV 11% 15 MVA 13 kV 15% Figure 1(b) m
A 15,000 KVA, 6.9 KV generator, star connected has positive, negative and zero sequence reactances of 0.25, 0.25 and 0.08 p.u. respectively. A reactor of 0.06 p.u. reactance on the generator rating is placed in the line from neutral to ground: A double line to ground fault occurs at the terminals of the generator when it is operating at its rated voltage and unloaded. Find the unitial r.m.s. line and ground wire current for a solidly grounded fault, and also the voltage to ground of the unfaulted line. (UPSC-1975) Ans: 1-1,-4970 A; I, 5480 A; V. 69.7 KV
13. A generator rated 30 MVA, 11 kV has a direct-axis subtransient reactance of 0.3 p.u. the negative and zero sequence reactances are 0.45 and 0.15 p.u. respectively. The neutral is solidly grounded. Determine the subtransient current in the generator when a line to ground fault occurs at the terminals of the generator with the generator running on no load at rated voltage. Ignore resistance. (

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