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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Chapter 9, Problem 9.57P
To determine
To explain: The reason for the phase angle on one phase is always zeroduring the line to ground fault.
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Q.3
When a line-to-ground fault occurs, the
current in faulted phase 'a' is 100A. The
zero-sequence current in phase 'c' is
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.
Please solve Q1(C) ONLY. Q1(A) is for reference.
Chapter 9 Solutions
Power System Analysis and Design (MindTap Course List)
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- Please solve Q1(B) ONLY. Q1(A) is for reference.arrow_forward1 Xine = j0.30 %3D G M X" = j0.15 X" = j0.20 Xiine = j0.30 Xiine = j0.30 %3D 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.arrow_forward3.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.arrow_forward
- For the following single line diagram of a three-phase system, a three-phase fault occurs at point P and just before the fault the line voltage is 18 kV at P. Find out the transient fault current in pu and original value at point P. Take, the given kVA and kV of the G1 as the new base. Answer follow image.arrow_forwardI need the answer as soon as possiblearrow_forwardIn a short circuit test on 132 kV, 3-phase system, the breaker gives the following results: Power factor of the fault = 0.45. Recovery voltage 0.9 time of full line voltage. The breaking current is symmetrical. The restriking transient has a natural frequency of 15 kHz. Calculate the rate of rise of restriking voltage (RRRV) in the following types of faults: (1) Grounded fault (2) Undergrounded fault.arrow_forward
- Three 6.6 kV altemators of ratings 8 MVA, 2 MVA and 5 MVA have per unit reactances of 0.16, 0.08 and 0.12 respectively and are connected to a common bus bar. A 3-phase fault occurs on a feeder which takes off from the bus. The reactance of the feeder from the bus to the fault point is 0.125 ohm. Find the fault MVA. [87.2 MVA]arrow_forwardIn 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.uarrow_forwardcomplet The equivalent reactance of the double circuit line under post fault condition is................than fault condition.arrow_forward
- 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…arrow_forwardData obtained from a short circuit test on a 132 kV 3-phase circuit breaker with earthed neutral are as follows: -the power factor fault was 0.3. -the recovery voltage was 0.75 of the full line value. -the breaking current was symmetrical. -the restriking transient had a natural frequency of 1600 Hz. Estimate the rate rise of the recovery voltage. Assume that the fault is grounded.arrow_forwardFault current analysisarrow_forward
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