A series of 3 load centers are supplied from a single generator. A simple current grading protection scheme is used to protect the network from over-current faults. Which of the following combinations of per-unit breaker settings are representative of a practical system? O A. Scheme A B C ~ * * 3pu 4pu 5pu OB. Scheme A B C * * 3pu 3pu 3pu * * 5pu 4pu 3pu A B C EEL 3pu 1pu ~ OC. Scheme ~ O D. Scheme

A series of 3 load centers are supplied from a single generator. A simple current grading protection scheme is used to protect the network from over-current faults. Which of the following combinations of per-unit breaker settings are representative of a practical system? O A. Scheme A B C ~ * * 3pu 4pu 5pu OB. Scheme A B C * * 3pu 3pu 3pu * * 5pu 4pu 3pu A B C EEL 3pu 1pu ~ OC. Scheme ~ O D. Scheme

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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Concept explainers

Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If a circuit is interrupted by a failure of a current-carrying wire then an open circuit fault occurs. The open-circuit faults are also called series faults. One or more phases or ground may be involved in a three-phase fault or may occur only between two phases. Current flows into the earth in case of a ground fault or earth fault. For most situations, the prospective short circuit current of a fault that is predictable can be calculated. To limit the loss of service because of a failure, the protective devices can detect fault conditions and operate circuit breakers and also other devices. All phases equally can be affected in a poly-phase system which is known as symmetric electrical fault. The asymmetric fault becomes more complicated to analyze if only some phases are affected. By using methods such as symmetrical components, the power system analysis of these types of faults is often simplified. The main objective of power system protection is to design, detect, and interrupt power system faults.

Question

A series of 3 load centers are supplied from a single generator. A simple current grading protection scheme is used to protect the network from over-current faults. Which of the following combinations of per-unit breaker settings are representative of a practical system?
O A. Scheme
A
в
C
*
X
*
Зри
4pu
5pu
OB. Scheme
A
В
с
*
*
Зри
Зри
Зри
A
в
с
ГЕЕ
5pu 4pu
Зри
A
в
с
*
*
*
Зри
2pu
N
OC. Scheme
O D. Scheme
г
*
1pu

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Consider the oneline diagram of a simple power system shown in Figure 9.20. System data in per-unit on a 100-MVA base are given as follows: The neutral of each generator is grounded through a current-limiting reactor of 0.08333 per unit on a 100-MVA base. All transformer neutrals are solidly grounded. The generators are operating no-load at their rated voltages and rated frequency with their ENIFs in phase. Determine the fault current for a balanced three-phase fault at bus 3 through a fault impedance ZF=0.1 per unit on a 100-MVA base. Neglect -Y phase shifts.
Equipment ratings for the five-bus power system shown in Figure 7.15 are as follows: Generator G1:Â Â Â Â 50Â MVA,Â 12kV,Â X=0.2 per unit Generator G2: 100Â MVA, 15 kV, X=0.2 per unit Transformer T1: 50 MVA, 10 kV Y/138kVY,X=0.10 per unit Transformer T2: 100 MVA, 15 kV /138kVY,X=0.10 per unit Each 138-kV line: X1=40 A three-phase short circuit occurs at bus 5, where the prefault voltage is 15 kV. Prefault load current is neglected. (a) Draw the positive-sequence reactance diagram in unit on a 100-MVA, 15-kV base in the zone of generator G2. Determine (b) the ThĂ©venin equivalent at the fault, (c) the subtransient fault current in per unit and in kA rms, and (d) contributions to the fault from generator G2 and from transformer T2.
33
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 line to ground fault occurs on the terminals of an alternator, sequence voltages and sequence currents during this fault are Vao = -0.3 pu Val = -0.4 pu Va2 = -0.7 pu Iao = 1a1 = 1a2 = -j2 pu If zero sequence impedance is j0.05pu then what will be neutral to ground impedance placed approximately,
Equal area criteria is a method used to calculate: Select one: O a. None of the answers b. Critical clearing angle and time for transient faults O c. Critical clearing voltage angle for transient faults d. Transient stability current limit O e. Steady state stability power limits
Q-5-) 60 Hz generator supplies 0.50 Pmax power to a busbar with infinite power over a conveying line. When a fault occurs, the reactance value between the generator and the infinite bus becomes 400% times the value before the fault. When the fault is isolated, the maximum power that can be transferred generator to the endless bus is 75% of the original (initial) maximum value. Accordingly, using t area criterion method, calculate the critical clearance angle of the system. 00
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.
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)
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…
(9) An unloaded generator with a pre-fault voltage 1 pu has the following sequence impedance: Zo = 0.15 pu, Z₁ = Z₂ = 0.25 pu The neutral is grounded with a reactance of 0.05 pu. The fault current in pu for a single-line to ground fault is (a) 3.75 pu (b) 4.28 pu (d) 7 25 pu