ELECTRICAL WIRING:RESIDENTAL-6 PLANS
18th Edition
ISBN: 9781305098329
Author: MULLIN
Publisher: CENGAGE L
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Chapter 28, Problem 15R
The utility company has provided a letter to the contractor stating that the available fault current at the line side of the main service-entrance equipment in a residence is 17,000 amperes RMS symmetrical, line to line. In the space following each statement, write in “Meets Code” or “Violation” of 110.9 and 110.10 of the Code.
- a. Main breaker has a 10,000-ampere interrupting rating; branch breakers have a 10,000-ampere interrupting rating. _______
- b. Main current-limiting fuse has a 200,000-ampere interrupting rating; branch breakers have a 10,000-ampere interrupting rating. The panel is marked “Series-Rated.” _______
- c. Main breaker has a 22,000-ampere interrupting rating; branch breakers have a 10,000-ampere interrupting rating. The panel is marked “Series-Connected.” _______
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Hello. Please write here.. Not write in PAPER... To easy Copy it
Approximately 75 percentage of the fault are either _____ or permanent LG Faults.
a.
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b.
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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.
Chapter 28 Solutions
ELECTRICAL WIRING:RESIDENTAL-6 PLANS
Ch. 28 - What is the minimum size service for a one-family...Ch. 28 - Prob. 2RCh. 28 - Prob. 3RCh. 28 - Prob. 4RCh. 28 - Prob. 5RCh. 28 - Prob. 6RCh. 28 - Prob. 7RCh. 28 - List the standard sizes of fuses and circuit...Ch. 28 - Prob. 9RCh. 28 - Using the method shown in this unit, what is the...
Ch. 28 - State four possible combinations of service...Ch. 28 - Which Code section states that all overcurrent...Ch. 28 - All electrical components have some sort of...Ch. 28 - Arc-fault damage is closely related to the value...Ch. 28 - The utility company has provided a letter to the...Ch. 28 - While working on the main panelboard that is...Ch. 28 - Repeat problem 16 for a main service panel that...Ch. 28 - NEC ______________________________ generally...Ch. 28 - Prob. 19RCh. 28 - Prob. 20RCh. 28 - Overcurrent devices must be accessible. (True)...
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- A three-phase circuit breaker has a 15.5-kV rated maximum voltage, 9.0-kA rated short-circuit current, and a 2.50-rated voltage range factor. (a) Determine the symmetrical interrupting capability at 10-kV and 5-kV operating voltages. (b) Can this breaker be safely installed at a three-phase bus where the symmetrical fault current is 10 kA, the operating voltage is 13.8 k V, and the (X/R) ratio is 12?arrow_forwardPlease solve Q1(C) ONLY. Q1(A) is for reference.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
- Please read the question carefully and provide the solution as asked in the questionarrow_forwarda) Classify the four categories of 3 phase power system faults we experience in our electrical power industry. b) Explain the emergence of such power systems fault as required in (question 1a)? c) Point out each type of fault specified under (a). d) Produce the (1D) design of the three-phase star connected faults stated under(b)arrow_forwardA 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,arrow_forward
- complet The equivalent reactance of the double circuit line under post fault condition is................than fault condition.arrow_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_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
- ques -4arrow_forwardIn which of the following faults symmetrical component method is not required a. Line to line fault O b. Double line to ground fault lon O c. Single line to ground fault O d. Three phase balanced fault Clear my choice pagearrow_forwardIn a power system with negligible resistance, the fault current at a point is 800 p. u. The series reactance to be included at the fault point to limit the short circuit to 5.00 p. u. is (a) 3.00 p.u (b) 0.200 p.u () 0.125p.u8 (d)0.075 p.uarrow_forward
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