Consider the system shown in the single-line diagram of Figure (3). All reactances are shown in per unit to the same base. Assume that the voltage at both sources is 1 p.u. a- Find the fault current due to a bolted-three-phase short circuit at bus 3 b- Find the fault current supplied by each generator and the voltage at each of the buses 1 and 2 under fault conditions 0.06 p.u. 0.04 p.u. 0.2 p.u. 0.25 p.u. 0.2 p.u. 10 0.2 P.M. 0.2 p.. ele ele G₂ 0.06 p.u. 0.25 p.u. ele ele ele 0.06 p.u. 0.25 p.u.

Consider the system shown in the single-line diagram of Figure (3). All reactances are shown in per unit to the same base. Assume that the voltage at both sources is 1 p.u. a- Find the fault current due to a bolted-three-phase short circuit at bus 3 b- Find the fault current supplied by each generator and the voltage at each of the buses 1 and 2 under fault conditions 0.06 p.u. 0.04 p.u. 0.2 p.u. 0.25 p.u. 0.2 p.u. 10 0.2 P.M. 0.2 p.. ele ele G₂ 0.06 p.u. 0.25 p.u. ele ele ele 0.06 p.u. 0.25 p.u.

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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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 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…
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.
Problem 5 Consider the system shown in the single-line diagram of Figure (3). All reactances are shown in per unit to the same base. Assume that the voltage at both sources is 1 p.u. a- Find the fault current due to a bolted- three-phase short circuit at bus 3. b- Find the fault current supplied by each generator and the voltage at each of the buses I and 2 under fault conditions. 0.04 p.u. 0.2 p.u. 0.06 p.u. 0.2 p.u. 0.25 p.u. G, 0.2 p.u. 0.2 p.u. 0.06 p.u. 0.06 р.и. 3 0.25 p.u. 0.25 p.u. G, Figure (3) Single-line diagram for Problem 5 ele
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…
In the figure below, all impedances are in per unit on a common MVA base. The generators are operating on no load at their rated voltage. A three-phase fault occurs at bus 1 through a fault impedance of Zf = jo.o8 per unit. Find Vi during the fault in per unit. X, = 0.1 Ooto X 0.1 X =0.2 X-0.1 O a. vi(n=0.4 p.u O b. vif)=0.5 p.u OC V1(0=0.6 pu VI=0.7 p.u
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 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
The one-line diagram of a simple power system is shown in Figure below. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. G Stark Item Base MVA Voltage Rating X' x² 20 kV 20 kV 20/220 kV 20/220 kV 100 0.05 0.15 0.15 0.10 0.10 220 kV 0.125 0.125 0.30 0.15 0.25 025 0.7125 0.15 100 100 0.15 0.05 0.10 0.10 0.10 100 0.10 100 100 Lu La 220 kV 0.15 220 kV 0.35 100 A balanced three-phase fault at bus 3 through a fault impedance Zf= jo.I per unit. The magnitude of the fault current in amperes in phase b for this fault is: Select one: A. 345.3 B. 820.1 C. 312500 3888888 产产
This question is about electrical power material and I want to solve it in 20 minutes
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)
Figure 2 shows the circuit of a simple power system. The ratings of the generators, transformers and the motors are as shown in the figure. The reactance of lines 1 and 2 are 40 ohms and 50 ohms respectively. Assume that system is unloaded, and the generators are generating their rated voltage. Select the generator G₁ ratings as base values. Choose the fault location in bus 1 And give the below answers For a symmetrical three phase fault on the bus (a).Determine the Thevenin's equivalent circuit which can be used to calculate the fault current. (b) Calculate the fault current contribution from the generator connected to the bus on which you have selected for fault calculation