Problem 6a- Explain the reasons for applying system grounding in electric power systems.Consider the industrial plant distribution bus which is supplied by a utility source with per-unitsequence reactances as shown in Figure (4-a). Assume that all reactances are given on a 5,000-kVAbase, and that the plant's bus voltage is 4,160-V.b- Determine the fault current for a single line to ground fault on phase A of the bus.Assume now that an ungrounded 5,000 kVA generator is added at the distribution bus as shown inFigure (4-b).c- Determine the fault current for a single line to ground fault on phase A of the bus.Assume now that the utility transformer is grounded through a 8 2 grounding resistor as shown inFigure (4-c.)d- Determine the value of the grounding resistor in per unit.e- Determine the fault current for a double line to ground fault on phases B and C of the bus. Utility5,000KVAUtilityGenX=0.15X= 0.06X- 0.15X = 0.15X = 0.06X = 0.15X- 0.1Xo = 0.04X= 0.1Plant Distribution BusPlant Distribution Bus(a)(b)5,000KVAUtilityulyGenX= 0.06x,-0.15X = 0.06X = 0.15Xo = 0.04X- 0.1%3DPlant Diatribution Bus(c)Figure (4) Single-line diagrams for Problem 6

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Answered: Explain the reasons for applying system…

Engineering

Electrical Engineering

Explain the reasons for applying system grounding in electric power systems.

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 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 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 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
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. 0.2 p.u. 0.04 p.u. 0.25 p.u. 0.2 p.u. 0.2 p.. 0.2 p.u. 0.06 p.u. 0.25 p.u. Figure (3) Single-line diagram ele ver ele 888 ele 0.06 p.u. 0.25 p.u.
answer both i will rate.... a. In sequence networks of power system, why does voltage sources appear only in the positive sequence? b. Describe the main differences in fault protection objectives when dealing with a temporary fault vs a permanent fault.
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…
Describe the concept of fault current analysis in power systems and its role in protecting electrical equipment.
Describe the process of fault current analysis in power systems and its significance in protecting electrical equipment from damage.
Q2 Figure Q2 shows a single line diagram of a power system and the associated data of this system are given in Table Q2. The pre-fault load current and A-Y transformer phase shift are neglected. (a) (b) If a Single Line-to-Ground (S-L-G) fault occurs at Bus 5 and the pre-fault voltage is 1.0 pu, calculate the subtransient fault current in Ampere. (c) (d) (e) Using base of 100 MVA and 11 kV at generator G₁, construct the positive sequence, negative sequence and zero sequence networks with their corresponding component values indicated. G₁ Recalculate (b) if the neutral on HV side of T3 is solidly grounded. Repeat part (b) with Line-to-line (L-L) fault. What will happen to L-L fault current in (d) if the neutral on the HV side of T3 is solidly grounded? Bus 1 T₁ Bus 4 Line 1 Line 2 Figure Q2 Bus 5 T2 T3 Bus 2 Bus 3 G₂ G3
Discuss the concept of fault analysis in power systems. What are the types of faults and their effects?
Draw the fault current direction both in Current transformer and operating relay coils for internal and external double line to line and double line to ground faults. Do the solve for Restricted Earth Fault Protection system for Alternator Stator Winding. Its a topic of Switchgear protection in power system