Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
Q5 The single-line diagram of a three-phase power system is shown in Figure Q5a, with equipment ratings given in Table Q5a. (a) (b) G₁ G₂ G3 G4 T₁ T₂ T3 TA Draw the zero, positive and negative-sequence reactance diagrams using 1000 MVA, 765 kV base in the zone of Line 1-2, neglect the (A-Y) transformer phase shifts. Calculate the short circuit current as a result of three-phase bolted fault occurred at bus 1. Line 1-2 Line 1-3 Line 2-3 bus 1 0 Υ ΔΕ Line 1-3 Line 1-2 G₂ Line 2-3 bus 2 Da Figure 05a bus 3 Table Q5a T₂ KEY YE my 1000 MVA, 15 kV, X₁-X₂-0.18 p.u., Xo = 0.07, p.u. 1000 MVA, 15 kV, X₁-X2-0.20 p.u., Xo = 0.10, p.u. 500 MVA, 13.8 kV, X₁-X₂-0.15 p.u., Xo = 0.05, p.u. X₂ = 0.05 p.u. 750 MVA, 13.8 kV, X₁-0.3 p.u., X₂-0.4 p.u., Xo = 0.1, p.u. 1000 MVA, 15 kV (A) / 765 kV (Y), XTI = 0.10 p.u. 1000 MVA, 15 kV (A)/ 765 kV (Y), XT2 = 0.10 p.u. 500MVA, 15 kV (Y) / 765 kV (Y), XT3 = 0.12 p.u. 750 MVA, 15 kV (Y)/765 kV (Y), XT4 = 0.11 p.u. Χ = 50 Ω, Χρ = 150 Ω. X = 40 Ω, X = 100 Ω. Χρ = 40 Ω, Χρ = 100 Ω.

Q5 The single-line diagram of a three-phase power system is shown in Figure Q5a, with equipment ratings given in Table Q5a. (a) (b) G₁ G₂ G3 G4 T₁ T₂ T3 TA Draw the zero, positive and negative-sequence reactance diagrams using 1000 MVA, 765 kV base in the zone of Line 1-2, neglect the (A-Y) transformer phase shifts. Calculate the short circuit current as a result of three-phase bolted fault occurred at bus 1. Line 1-2 Line 1-3 Line 2-3 bus 1 0 Υ ΔΕ Line 1-3 Line 1-2 G₂ Line 2-3 bus 2 Da Figure 05a bus 3 Table Q5a T₂ KEY YE my 1000 MVA, 15 kV, X₁-X₂-0.18 p.u., Xo = 0.07, p.u. 1000 MVA, 15 kV, X₁-X2-0.20 p.u., Xo = 0.10, p.u. 500 MVA, 13.8 kV, X₁-X₂-0.15 p.u., Xo = 0.05, p.u. X₂ = 0.05 p.u. 750 MVA, 13.8 kV, X₁-0.3 p.u., X₂-0.4 p.u., Xo = 0.1, p.u. 1000 MVA, 15 kV (A) / 765 kV (Y), XTI = 0.10 p.u. 1000 MVA, 15 kV (A)/ 765 kV (Y), XT2 = 0.10 p.u. 500MVA, 15 kV (Y) / 765 kV (Y), XT3 = 0.12 p.u. 750 MVA, 15 kV (Y)/765 kV (Y), XT4 = 0.11 p.u. Χ = 50 Ω, Χρ = 150 Ω. X = 40 Ω, X = 100 Ω. Χρ = 40 Ω, Χρ = 100 Ω.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question
Q5 The single-line diagram of a three-phase power system is shown in Figure Q5a, with equipment ratings given in Table Q5a. (a) (b) G₁ G₂ G3 G4 T₁ T₂ T3 TA Draw the zero, positive and negative-sequence reactance diagrams using 1000 MVA, 765 kV base in the zone of Line 1-2, neglect the (A-Y) transformer phase shifts. Calculate the short circuit current as a result of three-phase bolted fault occurred at bus 1. Line 1-2 Line 1-3 Line 2-3 bus 1 0 Υ ΔΥ Line 1-3 Line 1-2 G₂ Line 2-3 bus 2 Da Figure 05a bus 3 Table Q5a T₂ KEY YE my 1000 MVA, 15 kV, X₁-X₂-0.18 p.u., Xo = 0.07, p.u. 1000 MVA, 15 kV, X₁-X₂-0.20 p.u., Xo = 0.10, p.u. 500 MVA, 13.8 kV, X₁-X₂-0.15 p.u., Xo = 0.05, p.u. X₂ = 0.05 p.u. 750 MVA, 13.8 kV, X₁-0.3 p.u., X₂-0.4 p.u., Xo = 0.1, p.u. 1000 MVA, 15 kV (A) / 765 kV (Y), XTI = 0.10 p.u. 1000 MVA, 15 kV (A)/ 765 kV (Y), XT2 = 0.10 p.u. 500MVA, 15 kV (Y) / 765 kV (Y), XT3 = 0.12 p.u. 750 MVA, 15 kV (Y)/765 kV (Y), XT4 = 0.11 p.u. Χ = 50 Ω, X = 150 Ω. Χ = 40 Ω, Χρ = 100 Ω. Χρ= 40 Ω, Χρ = 100 Ω.
Transcribed Image Text:Q5 The single-line diagram of a three-phase power system is shown in Figure Q5a, with equipment ratings given in Table Q5a. (a) (b) G₁ G₂ G3 G4 T₁ T₂ T3 TA Draw the zero, positive and negative-sequence reactance diagrams using 1000 MVA, 765 kV base in the zone of Line 1-2, neglect the (A-Y) transformer phase shifts. Calculate the short circuit current as a result of three-phase bolted fault occurred at bus 1. Line 1-2 Line 1-3 Line 2-3 bus 1 0 Υ ΔΥ Line 1-3 Line 1-2 G₂ Line 2-3 bus 2 Da Figure 05a bus 3 Table Q5a T₂ KEY YE my 1000 MVA, 15 kV, X₁-X₂-0.18 p.u., Xo = 0.07, p.u. 1000 MVA, 15 kV, X₁-X₂-0.20 p.u., Xo = 0.10, p.u. 500 MVA, 13.8 kV, X₁-X₂-0.15 p.u., Xo = 0.05, p.u. X₂ = 0.05 p.u. 750 MVA, 13.8 kV, X₁-0.3 p.u., X₂-0.4 p.u., Xo = 0.1, p.u. 1000 MVA, 15 kV (A) / 765 kV (Y), XTI = 0.10 p.u. 1000 MVA, 15 kV (A)/ 765 kV (Y), XT2 = 0.10 p.u. 500MVA, 15 kV (Y) / 765 kV (Y), XT3 = 0.12 p.u. 750 MVA, 15 kV (Y)/765 kV (Y), XT4 = 0.11 p.u. Χ = 50 Ω, X = 150 Ω. Χ = 40 Ω, Χρ = 100 Ω. Χρ= 40 Ω, Χρ = 100 Ω.
Expert Solution
steps

Step by step

Solved in 10 steps with 9 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Three Phase Transformer
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS