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 7, Problem 7.31P
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?
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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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- 8.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_forwardI need the answer as soon as possiblearrow_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_forward
- iv. The swing equation can be considered mathematically, as a equation due to the v. The main objects of fault analysis are term.arrow_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_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
- 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_forwardb) 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.arrow_forwardQ2. The single-line diagram of a simple three-bus power system is shown in Figure-2. Each generator is represented by an emf behind the sub-transient reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of Zf = j0.19 per unit. (i) Using Th'evenin's theorem, obtain the impedance to the point of fault and the fault current in (ii) Determine the bus voltages per unit. ) j0.05 j0.075 j0.75 2 j0.30 j0.45 Figure-2: Single line diagram of the power system network for Q2 3arrow_forward
- Please solve Q1(C) ONLY. Q1(A) is for reference.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_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
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