Please solve in detail Considering the circuit below, the terminal impedance and the fault current for a bolted fault at F are:jo-05 Pu0000J0-055 Pu0000jo.15 Pujo-1PUFV=1.00 PUV=IPU(VOLTS)Select one:a.Zthj 0.355 pu; IF = -j2.816 pub. Zth j0.087 pu; IF =C.== -j11.5 puZth j 0.355 pu; IF = -j5.633 pud. Zth j0.087 pu; IF = -j22.98 pu=

Step 1: Step 2: Step 3: Step 4:

Answered: Considering the circuit below, the…

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 7.21MCQ

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Discuss causes of insulation failures in high voltage engineering. High voltage engineering
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
Discuss the concept of fault analysis in power systems. What are the types of faults and their effects?
The reactance data for the power system shown in below figure in per unit ona common base is as the following: X²_x² j0.1 G₁ to j0.25 Ti Select one: O a. No right answer O b.-12.19 pu O c.-14.986 pu O d.-) 3.806 pu 2 j0.3 + AL j0.05 j0.25 Assume that the pre fault voltage at bus 2 = 1 pu. For bolted line (b) to line (c) fault at bus 2, the fault current is = j0.5 3 j0.25 T₂ 4 38 + A j0.25 j0.1 G₂ j0.05
Q2\ Two generators G1 & G2 are rated 15 MVA, 11 kV and 10 MVA, 11 kV respectively. The generators are connected to a transformer as shown in the following figure. Find the sub- transient current in each generator when a three-phase fault occurs on the high voltage side of the transformer. A/Y 15 MVA, 11 kV, X"gl=j0.10 pu G2 15 MVA, 11/66 kV, Xr-j0.06 pu 10 MVA, 11 kV, X"g2-j0.10 pu
The direct electrical connection of the windings allows transient over voltages to pass through the auto transfonner more easily, and that is an important disadvantage of the autotransformer. (a) True (b) False
stribution feeder has 600/5 C100 CT. A three phase fault of l=10200A occurs at F as shown in Calculate the voltage drop developed across CT if the phase relay burden resistance Z₁ = 0.1 , the lead resistance R₁ = 0.5f, and the CT resistance Rg=0.40 Will this fault current lead to CT saturation? If not at what fault level, the CT will saturate? GOM 600:5 O/C R Fig. 2 10200A
Q3/ For the system shown in figure. The generator is supplying 1.0 pu MW power under prefault condition. When three-phase fault occur at point F. What should the value of fault reactance must be to make the critical clearing angle equal to 58.9. Sc j0.1 j0.1 10.21 BE j0.38 38 j0.21 IEgl-12 pu 10.19 F 10.19 IVI-1.00 pu j0.1 j0.1 Q4/ Figure below shows the one-line diagram
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 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…
Explain the concept of fault analysis in power systems. How is fault current calculated, and what protection mechanisms are in place?

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