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 short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus 3.

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 short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus 3.

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 7.3MCQ: The amplitude of the sinusoidal symmetrical ac component of the three-phase short-circuit current of...

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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. 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…
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)…
The p.u. impedance value of an alternator corresponding to base values of 13.2 kV and 30 MVA is 0.2 p.u. Then the p.u. impedance value of an alternator for the new base values of 13.8 kV and 50 MVA is......
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 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…
A short transmission line connects a step-up transformer on the source side with a series impedance of j0.5 ohms (referred to the primary) to a step-down transformer on the load side with a series impedance of 50 ohms (referred to the primary). The turn ratio of the step-up transformer is 1:10 and a no-load primary voltage of 23 (line-to-line) kV. The step-down transformer has a turn ratio of 35:2 and it has a Y-connected, balanced load of 0.490 +j 0.0816 ohms connected to its secondary side. Assuming an ideal transmission line, compute the load voltage (line-to-neutral).
A short transmission line connects a step-up transformer on the source side with a series impedance of j0.5 ohms (referred to the primary) to a step-down transformer on the load side with a series impedance of 50 ohms (referred to the primary). The turn ratio of the step-up transformer is 1:10 and a no-load primary voltage of 23 (line-to-line) kV. The step-down transformer has a turn ratio of 35:2 and it has a Y-connected, balanced load of 0.490 +j 0.0816 ohms connected to its secondary side. A capacitor bank of -j0.171 ohms is added parallel to the load. Assuming an ideal transmission line, the load voltage (line-to-neutral) would be: O A. 7360 V O B. 6.2 kV O C. 12750 V O D. 4806 V O E. None of the other choices are correct
Three zones of a single-phase circuit are identified in the figure. The zones are connected by transformers T₁ and T2, whose ratings are also shown. Using base values of 100 kVA and 240 volts in zone 1, draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. Zone 1 Zone 2 Vs = 220/0° volts 3---38 T, 30 KVA 240/480 volts M 0.10 p.u. Xoa Xune = 2 fl T T₂ 20 kVA 460/115 volts Xeg = 0.10 p.u. Zone 3 ww Zload = 0.9 - 10.20
A short transmission line connects a step-up transformer on the source side with a series impedance of j0.5 ohms (referred to the primary) to a step-down transformer on the load side with a series impedance of 50 ohms (referred to the primary). The turn ratio of the step-up transformer is 1:14 and a no-load primary voltage of 17 (line-to-line) kV. The step-down transformer has a turn ratio of 18:1 and it has a Y-connected, balanced load of 0.463 +j 0.0772 ohms connected to its secondary side. Assuming an ideal transmission line, the real power absorbed by the load would be: A. 42 MW B. None of the other choices are correct C. 54 MW D. 37.4 MW E. 28 MW
A short transmission line connects a step-up transformer on the source side with a series impedance of j0.5 ohms (referred to the primary) to a step-down transformer on the load side with a series imnedance of 50 ohms (referred to the primary). The turn ratio of the step-up transformer is 1:14 and a no-load primary voltage of 17 (line-to-line) kV. The step-down transformer has a turn ratio of 18:1 and it has a Y-connected, balanced load of 0.463 +j 0.0772 ohms connected to its secondary side. Assuming an ideal transmission line, the load voltage (line-to-neutral) would be: y50 O A. 14.2 kV O B. 6.52 kV O C. 17.2 kV O D. 5.07 kV O E. None of the other choices are correct
estion 9 The fault current level of a circuit breaker used in a transmission line with phase voltage 2.2 kV and breaking capacity 50 MVA is yet ......kA. vered -ked out of Flag question estion 10 The voltage between lines in a transmission line is the maximum value of the sine waveform. yet vered Select one: -ked out of O True Flag question O False