QUESTION 5a) Obtain the sequence networks reduced to their Thevenin equivalents for the systemshown in Figure 4 in the case of a fault at F. Prefault is VF = 1.05Z 0° per unit.System data in per unit on the same base are given as follows:G1 : XI = 0.2 p.u.G2 : XI = 0.33 p.u.X2 = 0.12 p.u.X2 = 0.22 p.u.Хо - 0.06 р.u.Хо- 0.066 р.u.T1:XI =X2 = Xo = 0.2 p.u.T2:X = X2 = Xo 0.225 p.u.T3 :X1 =X2 = Xo = 0.27 p.u.T4 :X1 = X2 = Xo = 0.16 p.u.L1 :XI = X2 = 0.14 p.u. Xo = 0.3 p.u.L2 :X1 = X2 0.35 p.u. Xo = 0.6 p.u.T1T2L1G1G2L2T4T3Figure 4b) Calculate the per-unit subtransient fault current in phase a, b, c for a bolted threephase to ground short circuit at fault F in Figure 4. Justify your answer.

Answered: Image /qna-images/answer/f5f744ca-e581-4e4d-a262-a715d04cc183.jpg

QUESTION 5 a) Obtain the sequence networks reduced to their Thevenin equivalents for the system shown in Figure 4 in the case of a fault at F. Prefault is VF = 1.05Z 0° per unit. System data in per unit on the same base are given as follows: G1 : Xi = 0.2 p.u. G2 : XI = 0.33 p.u. X — 0.12 р.u. Х 3D0.22 р.u. Xo = 0.06 p.u. Xo = 0.066 p.u. T1:X1 =X2 = Xo = 0.2 p.u. T2 :XI =X2 = Xo = 0.225 p.u. T3: XI =X2 = Xo = 0.27 p.u. T4 :XI = X2 = Xo = 0.16 p.u. Li :XI = X2 = 0.14 p.u. Xo = 0.3 p.u. L2:X1 = X2 = 0.35 p.u. Xo = 0.6 p.u. T1 T2 L1 G1 G2 F L2 T4 T3 Figure 4 b) Calculate the per-unit subtransient fault current in phase a, b, c for a bolted three phase to ground short circuit at fault F in Figure 4. Justify your answer.

QUESTION 5 a) Obtain the sequence networks reduced to their Thevenin equivalents for the system shown in Figure 4 in the case of a fault at F. Prefault is VF = 1.05Z 0° per unit. System data in per unit on the same base are given as follows: G1 : Xi = 0.2 p.u. G2 : XI = 0.33 p.u. X — 0.12 р.u. Х 3D0.22 р.u. Xo = 0.06 p.u. Xo = 0.066 p.u. T1:X1 =X2 = Xo = 0.2 p.u. T2 :XI =X2 = Xo = 0.225 p.u. T3: XI =X2 = Xo = 0.27 p.u. T4 :XI = X2 = Xo = 0.16 p.u. Li :XI = X2 = 0.14 p.u. Xo = 0.3 p.u. L2:X1 = X2 = 0.35 p.u. Xo = 0.6 p.u. T1 T2 L1 G1 G2 F L2 T4 T3 Figure 4 b) Calculate the per-unit subtransient fault current in phase a, b, c for a bolted three phase to ground short circuit at fault F in Figure 4. Justify your answer.

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 9.13P: Consider the oneline diagram of a simple power system shown in Figure 9.20. System data in per-unit...

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5.10 The zero sequence current of a generator for line to ground fault is j2.4 p.u. Then the current through the neutral during the fault is (a) j2.4 p.u. (b) j0.8 p.u. (d) j0.24 p.u. (c) i7.2 p.u.
5.10 The zero sequence current of a generator for line to ground fault is j2.4 p.u. Then the current through the neutral during the fault is (a) j2.4 p.u. (b) j0.8 p.u. (c) j7.2 p.u. (d) j0.24 p.u.
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…
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 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 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 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.
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