Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
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Textbook Question
Chapter 9, Problem 9.32P
For the system of Problem 9.13, compute the fault current for the following faults at bus 3: (a) a single line-to-ground fault through a fault impedance
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3.On a double line-to-ground faultImmersive reader
A) The short-circuit currents in the faulted phases are equal.
B) The voltages at the fault point on the faulted phases are zero.
C) The phase not involved in the fault has a zero voltage at the fault point.
D) There are no circulating currents through the neutrals to ground.
E) N. A.
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1. Find the Thévenin equivalent looking into bus 3.
2. Calculate the subtransient fault current for a bolted three-phase fault at bus 3. Pre-fault
voltage is VF 1Z°, and pre-fault current is neglected.
3. Distribute the fault current between the motor and generator.
Please solve Q1(B) ONLY. Q1(A) is for reference.
Chapter 9 Solutions
Power System Analysis and Design (MindTap Course List)
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- The system values are given below. The bus 1 voltage after fault = 1.5 p.u The bus 2 voltage after fault = 1.2 p.u The line admittance between bus 1 and bus 2 (Y12 ) is = 0.8 p.u The post fault current current flow between bus 1 and 2 is ..............arrow_forwardPlease solve Q1(C) ONLY. Q1(A) is for reference.arrow_forwardb) 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…arrow_forward
- 7.In a line-to-ground fault we can expect that A) There are no currents to ground. B) The voltages at the fault point with respect to ground are zero in all three phases. C) The current in the faulted phase flows to ground. D) There are short circuit currents in all three phases. E) N. A.arrow_forwardQ.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' isarrow_forwardQ.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is rarrow_forward
- Q1. Given Zo = 0.3L60°, Z¡ = 0.17L80° and, Z2 = 0.45L120°. Compute the fault current and voltages for a Double Line-to-Ground Fault. Note that the sequence impedances are in per- unit. This means that the solution for current and voltage will be in per-unit. [1] Note: Formulas for the reference For Double line to ground fault the positive, negative and zero sequence circuits are connected in parallel The sequence networks are interconnected, as shown in Fig. 8.9 Zo To Because the sequence currents sum to one node, it follows that Vo I, =-(I, +1,) The current I, is the voltage drop across Z, in series with the parallel combination of Z, and V, I = Z, + Z, +Z, V2 Fig 8.9arrow_forwardThe system values are given below. The bus 1 voltage after fault =6.18 p.u. The bus 2 voltage after fault = 2.54 p.u. The line admittance between bus 1 and bus 2 (Y12 ) is =3.25 p.u Then the post fault current flows between bus 1 and 2 is Post fault current l isarrow_forwardI need the answer as soon as possiblearrow_forward
- 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…arrow_forward8.In a line-by-line failure we can expect Immersive reader A) Currents exist in the ground connections. B) The current values at the ground connections are zero. C) Short-circuit currents exist in all three phases. D) The voltages at the fault point with respect to ground are zero in the faulted phases. E) N. A.arrow_forward1. The subscript d in the generator subtransient reactance refers to: 1. Generator impedance 2. If the available fault current slightly exceeds the breaker published 2. Generator 3. Direct axis 4. 1 and 2 reactance interrupting rating, then it is safe to use the breaker. 1. True | 2. False 3. Maybe 3. The rms symmetrical fault current times an asymmetry factor K, is equal to the ac fault current. | 2. False 4. The most common fault on a 3-phase power system is: 1. True 3. Maybe | 2. DLG 3. L-L 1. SLG 5. All rotating and non-rotating load impedances are usually included in a power system fault study 1. True 2. False 3. Мaybearrow_forward
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