2. The one-line diagram of a three-phase power system is as shown in the figure below. Impedances are marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW-j33.4 Mvar, and at bus 3 is S3 = 77 MW +j14 Mvar. It is required to hold the voltage at bus 3 at 4006 0 kV. Working in per unit, determine the voltage at buses 2 and 1. V3 j0.5 pu V₂ S₂ j0.4 pu S3

2. The one-line diagram of a three-phase power system is as shown in the figure below. Impedances are marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW-j33.4 Mvar, and at bus 3 is S3 = 77 MW +j14 Mvar. It is required to hold the voltage at bus 3 at 4006 0 kV. Working in per unit, determine the voltage at buses 2 and 1. V3 j0.5 pu V₂ S₂ j0.4 pu S3

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.29P: Consider three ideal single-phase transformers (with a voltage gain of ) put together as ...

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Concept explainers

Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If a circuit is interrupted by a failure of a current-carrying wire then an open circuit fault occurs. The open-circuit faults are also called series faults. One or more phases or ground may be involved in a three-phase fault or may occur only between two phases. Current flows into the earth in case of a ground fault or earth fault. For most situations, the prospective short circuit current of a fault that is predictable can be calculated. To limit the loss of service because of a failure, the protective devices can detect fault conditions and operate circuit breakers and also other devices. All phases equally can be affected in a poly-phase system which is known as symmetric electrical fault. The asymmetric fault becomes more complicated to analyze if only some phases are affected. By using methods such as symmetrical components, the power system analysis of these types of faults is often simplified. The main objective of power system protection is to design, detect, and interrupt power system faults.

Question

2. The one-line diagram of a three-phase power system is as shown in the figure below. Impedances are
marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW-j33.4 Mvar,
and at bus 3 is S3 = 77 MW +j14 Mvar. It is required to hold the voltage at bus 3 at 4006 0 kV.
Working in per unit, determine the voltage at buses 2 and 1.
V3
j0.5 pu
V₂
S₂
j0.4 pu
S3

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Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.
Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows: XY15MVA,13.8/138kV, leakage reactance 10 YZ15MVA,138/69kV, leakage reactance 8 With the base in circuit Y chosen as 15MVA,138kV determine the per-unit impedance of the 500 resistive load in circuit Z, referred to circuits Z, Y, and X. Neglecting magnetizing currents, transformer resistances, and line impedances, draw the impedance diagram in per unit.
Equipment ratings for the five-bus power system shown in Figure 7.15 are as follows: Generator G1:Â Â Â Â 50Â MVA,Â 12kV,Â X=0.2 per unit Generator G2: 100Â MVA, 15 kV, X=0.2 per unit Transformer T1: 50 MVA, 10 kV Y/138kVY,X=0.10 per unit Transformer T2: 100 MVA, 15 kV /138kVY,X=0.10 per unit Each 138-kV line: X1=40 A three-phase short circuit occurs at bus 5, where the prefault voltage is 15 kV. Prefault load current is neglected. (a) Draw the positive-sequence reactance diagram in unit on a 100-MVA, 15-kV base in the zone of generator G2. Determine (b) the ThĂ©venin equivalent at the fault, (c) the subtransient fault current in per unit and in kA rms, and (d) contributions to the fault from generator G2 and from transformer T2.
Three zones of a single-phase circuit are identified shown in Figure below. The zones are connected by transformers T1 and T2, whose ratings are also shown. Using base values of 33 kVA and 232 volts in zone 1, Find: 1- Draw the per-unit circuit including the per-unit impedances and the per-unit source voltage. 2- Calculate the load current both in per-unit and in amperes (actual or original value). Vs Zone 1 232.940° Vs G. 38 T₁ 30 KVA 240/480 volts Xeq = 0.10 p.u. Zone 2 Xiine = 4 Ω T₂ 20 KVA 460/115 volts Zload = Xea = 0.10 p.u. Zone 3 u 1+j2.2 Ω 2
Three zones of a single-phase circuit are identified in Figure. The zones are connected bytransformers T1 and T2, whose ratings are also shown.Using base values of 30 kVA and 240 volts in zone 1:a. draw the per-unit circuit and determine the per-unit impedances and the per-unitsource voltage.b. calculate the load current both in per-unit and in amperes. Transformer windingresistances and shunt admittance branches are neglected.
Q2. 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 3
The one-line diagram of a simple power system is shown in Figure below. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. G Stark Item Base MVA Voltage Rating X' x² 20 kV 20 kV 20/220 kV 20/220 kV 100 0.05 0.15 0.15 0.10 0.10 220 kV 0.125 0.125 0.30 0.15 0.25 025 0.7125 0.15 100 100 0.15 0.05 0.10 0.10 0.10 100 0.10 100 100 Lu La 220 kV 0.15 220 kV 0.35 100 A balanced three-phase fault at bus 3 through a fault impedance Zf= jo.I per unit. The magnitude of the fault current in amperes in phase b for this fault is: Select one: A. 345.3 B. 820.1 C. 312500 3888888 产产

Q2. Figure Q2 shows the single-line diagram. The scheduled loads at buses 2 and 3 are as marked on the diagram. Line impedances are marked in per unit on 100 MVA base and the line charging susceptances are neglected. a) Using Gauss-Seidel Method, determine the phasor values of the voltage at load bus 2 and 3 according to second iteration results. b) Find slack bus real and reactive power according to second iteration results. c) Determine line flows and line losses according to second iteration results. d) Construct a power flow according to second iteration results. Slack Bus = 1.04.20° 0.025+j0.045 0.015+j0.035 0.012+j0,03 3 |2 134.8 MW 251.9 MW 42.5 MVAR 108.6 MVAR
Show me how is solution will be if fualt happened before transformers or in one of generator
High voltage engineering subject
400Ω A k=0.9 600Ω B k=0.8 66Ω 600Ω
X=2 Y=8