Q1. Given Z, = 0.3L60º, Z¡ = 0.17L80° and, Z2 = 0.45L120°. Compute the fault current and voltages for a Double Line-to-Ground Fault. Note that the sequence impedances are in per- unit. This means that the solution for current and voltage will be in per-unit. [1] Note: Formulas for the reference For Double line to ground fault the positive, negative and zero sequence circuits are connected in parallel The sequence networks are interconnected, as shown in Fig. 8.9 Zo Because the sequence currents sum to one node, it follows that Vo I, =-(I, +1;) The current I, is the voltage drop across Z, in series with the parallel combination of Zo and z, 1Z0°( V, I, = Z, + ZZ, Z, +Z, V: Fig 8.9

Q1. Given Z, = 0.3L60º, Z¡ = 0.17L80° and, Z2 = 0.45L120°. Compute the fault current and voltages for a Double Line-to-Ground Fault. Note that the sequence impedances are in per- unit. This means that the solution for current and voltage will be in per-unit. [1] Note: Formulas for the reference For Double line to ground fault the positive, negative and zero sequence circuits are connected in parallel The sequence networks are interconnected, as shown in Fig. 8.9 Zo Because the sequence currents sum to one node, it follows that Vo I, =-(I, +1;) The current I, is the voltage drop across Z, in series with the parallel combination of Zo and z, 1Z0°( V, I, = Z, + ZZ, Z, +Z, V: Fig 8.9

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 9.19P

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Equipment ratings for the four-bus power system shown in Figure 7.14 are as follows: Generator G1: 500 MVA, 13.8 kV, X=0.20 per unit Generator G2: 750 MVA, 18 kV, X=0.18 per unit Generator G3: 1000 MVA, 20 kV, X=0.17 per unit Transformer T1: 500 MVA, 13.8/500YkV,X=0.12 per unit Transformer T2: 750 MVA, 18/500YkV,X=0.10 per unit Transformer T3: 1000Â MVA, 20/500YkV,X=0.10 per unit A three-phase short circuit occurs at bus 1, where the prefault voltage is 525 kV. Prefault load current is neglected. Draw the positive-sequence reactance diagram in per unit on a 1000-MVA, 20-kV base in the zone of generator G3. Determine (a) the ThĂ©venin reactance in per unit at the fault, (b) the subtransient fault current in per unit and in kA rms, and (c) contributions to the fault current from generator G1 and from line 1-2.
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…
Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is r
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.
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. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…
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…
3.On a double line-to-ground faultImmersive reader A) The short-circuit currents in the faulted phases are equal. B) The voltages at the fault point on the faulted phases are zero. C) The phase not involved in the fault has a zero voltage at the fault point. D) There are no circulating currents through the neutrals to ground. E) N. A.
Q3: - A three-phase fault occurs at bus-1 of the network shown in figure below. The pre-fault voltage is 1.05 PU and the pre-fault load current is neglected. (45%) i) determine the bus impedance matrix. ii) calculate the sub-transient fault current and the contribution to the fault current from the transmission line. T1 T2 T.L j0.1 PU 10.1 PU j0.105 PU G jXdg j0.15 PU M jXdm $0.2 PU
A 30kV, 50MVA generator having a solidly grounded neutral and Z1 = 0.25pu, Z2=0.15pu and Z0=0.05pu. a. Calculate the fault current and voltages for a single line to ground fault. b. Sketch the phasor diagram for the fault voltages.
1 Xine = j0.30 %3D G M X" = j0.15 X" = j0.20 Xiine = j0.30 Xiine = j0.30 %3D 1. Find the Thévenin equivalent looking into bus 3. 2. Calculate the subtransient fault current for a bolted three-phase fault at bus 3. Pre-fault voltage is VF 1Z°, and pre-fault current is neglected. 3. Distribute the fault current between the motor and generator.
complet The equivalent reactance of the double circuit line under post fault condition is................than fault condition.
Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is