Please make you hand writing clear as i will study your steps V₁ = 120°1Slack2400 MW320 Mvarj0.0125j0.05300 MW 270 MvarFIGURE 6.20One-line diagram for Problem 6.7.6.7. Figure 6.20 shows the one-line diagram of a simple three-bus power systemwith generation at bus 1. The voltage at bus 1 is V₁ = 1.0/0° per unit.The scheduled loads on buses 2 and 3 are marked on the diagram. Lineimpedances are marked in per unit on a 100-MVA base. For the purposeof hand calculations, line resistances and line charging susceptances are ne-glected.a) Find the bus admittance matrix of the system.b) Find the voltage (magnitude and phase) of bus 3 and the voltage phase of bus 2 using Newton Raphson method: 1) by hand (at least 2iterations)c) Using the values in part (b), find the current in each line, the reactive power at bus 2 and the complex power at bus 1.

Answered: Image /qna-images/answer/3af4c604-29ce-4a77-905e-f13eeed76af1.jpg

Answered: V₁ = 120°1 Slack 2 400 MW 320 Mvar…

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

Chapter7: Symmetrical Faults

Section: Chapter Questions

Problem ACSQ

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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.
6.1. A power system network is shown in Figure 47. The generators at buses 1 and 2 are represented by their equivalent current sources with their reactances in per unit on a 100-MVA base. The lines are represented by n model where series reactances and shunt reactances are also expressed in per unit on a 100 MVA base. The loads at buses 3 and 4 are expressed in MW and Mvar. (a) Assuming a voltage magnitude of 1.0 per unit at buses 3 and 4, convert the loads to per unit impedances. Convert network impedances to admittances and obtain the bus admittance matrix by inspection. j0.25 50.25 -j4 j0.4 j0.1 j0.16 j0.2 -j4+3 4 S3 -j4 S4 FIGURE 47 One-line diagram for Problem 6.1. 100 MW +j25 Mvar 200 MW +j50 Mvar
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
1. FIGURE 52 shows the one-line diagram of a simple three-bus power system with generation at bus I. The voltage at bus l is V1 = 1.0L0° per unit. The scheduled loads on buses 2 and 3 are marked on the diagram. Line impedances are marked in per unit on a 100 MVA base. For the purpose of hand calculations, line resistances and line charging susceptances are neglected a) Using Gauss-Seidel method and initial estimates of Va 0)-1.0+)0 and V o)- ( 1.0 +j0, determine V2 and V3. Perform two iterations (b) If after several iterations the bus voltages converge to V20.90-j0.10 pu 0.95-70.05 pu determine the line flows and line losses and the slack bus real and reactive power. 2 400 MW 320 Mvar Slack 0.0125 0.05 300 MW 270 Mvar FIGURE 52
Figure below shows the single-line diagram of three-bus power system with generation at bus 1 and bus 3. The voltage at bus 1 is V₁ = 1.025/0° per unit. The voltage magnitude at bus 3 is fixed at 1.05 per unit with a real power generation of 250 MW. The scheduled load on bus 2 is marked on the diagram. Line impedances are marked in per unit on a 100 MVA base. Line resistances and line charging susceptances are neglected. By using Gauss-Seidel method and initial estimates of V₂(0) = 1.020° and V3(0) = 1.0520°, determine V₂ and V3. Perform calculation for one iteration. V₁ = 1.025/0° j0.1 Slack Bus j0.2 j0.4 j0.2 3 j0.1 250 MW 150 Mvar |V3|=1.05 P3 = 250 MW
Figure shows the one-line diagram of a simple three-bus power system with generation at buses 1 and 3. The voltage at bus 1 is V1 is 1.025 at an angle of 0◦ per unit. Voltage magnitude at bus 3 is fixed at 1.03 pu with a real power generation of 300 MW. A load consisting of 400 MW and 200 MVAr is taken from bus 2. Line impedances are marked in per unit on a 100 MVA base. (a) Construct Ybus matrix for the system in Figure (b) Using Gauss-Seidel method and initial estimate of V2(0) = 1.0 + j0 and V3(0) = 1.03 + j0 and keeping |V3| = 1.03 pu, determine the phasor values of V2 and V3. Perform two iterations.
6. For a three bus power system assume bus 1 is the swing with a per unit voltage of 1.020 , bus 2 is a PQ bus with a per unit load of 2.0 + j0:5, and bus 3 is a PV bus with 1.0 per unit generation and a 1.0 voltage setpoint. The per unit line impedances are j0.1 between buses 1 and 2, j0.4 between buses 1 and 3, and j0.2 between buses 2 and 3. Using a flat start, use the Newton-Raphson approach to determine the first iteration phasor voltages at buses 2 and 3.
Q6: A five-bus network has generators at buses 1 and 3 rated 270 and 225 MVA, respectively. The generator sub-transient reactances plus the reactances of the transformers connecting them to the buses are each 0.3 pu on the generator rating as base. The turns ratios of the transformers are such that the voltage base in each generator circuit is equal to the voltage rating of the generator. Line impedances in pu on a 100 MVA system base are shown in Fig. below. All resistances are neglected. Using the bus impedance matrix for the network which includes the generator and transformer reactances, find the sub-transient current in a three phase fault at bus 4 and the current coming to the faulted bus over each line. Prefault current is to be neglected and all voltages are assumed to be 1 pu before the fault occurs. (2) 3 j0.126 1.0/0° j0.168 0/0° j0.1111 lll j0.1333 j0.210 ell j0.126 ll j0.252 4 j0.336 hellte
The 6-bus power system network of an electric utility company is shown in the Figure below. The line and transformer data containing the subtransient series resistance and reactance in per unit, and one-half of the total capacitance in per unit susceptance on a 100-MVA base, is tabulated below. The prefault voltage profile of the power system as obtained from four iterations of the newton Raphson power flow method are provided below as well.
Please help
Figure below shows one-line diagram of a simple three bus power system with generation at bus 1. Bus 1 is considered as slack bus. A load consisting of 250 MW and 110 MVAR is taken from bus 2. A load consisting of 128 MW and 35 MVAR is taken from bus 3. Line impedances are marked in per unit on a 100 MVA base. Line susceptances are neglected. G1 0.01 +10.03 V₁ = 1.040° 0.02 +0.04 Select one: O a. None of these O b. 0.9245-j0.025 O c. 0.9245+j0.025 O d. -0.9245-j0.025 e. 0.9638-j0.03 0.0125+j0.025 ·0 P2 Q2 P3 Q3 Start with flat initial estimates of ₂0) = 1 + j0 & V3⁰) = 1 + j0, and keeping |V₂| = 1 pu, find V₂(¹)
In a 7-bus system, the double line fault is occurred at bus -5. During fault the positive sequence potential difference between the bus-4 and bus-1is found to be 0.0175. what is the line impedance which is connected between bus-4 and bus-1 if the current flowing through transmission line is -j 0.148? Select one: O a. -jo.182 O b. +j0.1182 O c. -jo.1182 O d. +j0.182

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