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 7, Problem 7.2MCQ
Even though the fault current is not symmetrical and not strictly periodic, the rms asymmetrical fault current is computed as the rms ac fault current times an "asymmetry factor," which is a function of _______.
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Approximately 75 percentage of the fault are either _____ or permanent LG Faults.
a.
LLG
b.
Transient
c.
Simple
d.
LL
The DC Components components initially average about ______ of the fault current.
a.
0.50 to 0.60 times
b.
infinity
c.
20 to 45 percentage
d.
50 percentage
Please solve Q1(C) ONLY. Q1(A) is for reference.
Chapter 7 Solutions
Power System Analysis and Design (MindTap Course List)
Ch. 7 - Even though the fault current is not symmetrical...Ch. 7 - The amplitude of the sinusoidal symmetrical ac...Ch. 7 - Equipment ratings for the four-bus power system...Ch. 7 - Equipment ratings for the five-bus power system...Ch. 7 - Prob. 7.22PCh. 7 - A three-phase circuit breaker has a 15.5-kV rated...Ch. 7 - Prob. 7.32P
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- 1. 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_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_forward7.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_forward
- In 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_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_forwardcomplet The equivalent reactance of the double circuit line under post fault condition is................than fault condition.arrow_forward
- In a short-circuit test on a 3-pole, 110 kV circuit breaker. Power factor of the fault= 0.4. Recovery voltage= 0.95 times (full line voltage) The breaking current was symmetrical the restriking transient had a natural frequency of 15000 c/s. Determine the average rate of rise of restriking voltage. (The natural is grounded and the fault involves earth). Explain in details.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_forwardHello. Please write here.. Not write in PAPER... To easy Copy itarrow_forward
- What is subtransient and transient fault current? How do they occur in a system?arrow_forwardA fault which gives rise to symmetrical fault currents (equal faults currents with 120-degree displacement) is called a symmetrical fault. Select one: a. False O b. Truearrow_forwardIn a short-circuit test on a 3-pole, 220 kV circuit breaker. Power factor of the fault = 0.8 Recovery voltage, 1.90 times (full line voltage) The breaking current was symmetrical the restriking transient had a natural frequency of 30000 c/s. Determine the average rate of rise of restriking voltage. (The natural is grounded and the fault involves earth).arrow_forward
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