b) The bus admittance matrix of a simple power network is given below. Draw the single line diagram marking line parameters.

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
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.50P
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value of x = 0.2, y = 0.8

b)
c)
The bus admittance matrix of a simple power network is given below.
Draw the single line diagram marking line parameters.
Ybus = j
g1
Figure Q4 shows a simple 2 bus bar electric power system, with the
values of X and Y defined below
Last digit of your student number
0, 3, 6
1,4,7
2,5 8,9
g1
-3.78
1.25
2.5
0
V,<0°
1
1.25 2.50
-3.42
1.11
1.11
-4.89
1.0
1.25
Note that the values for X and Y to be used in the calculation depend on the last
digit of your student number.
All per unit quantities are on a base power of 100MVA.
The power flow problem is to be solved using the Newton-Raphson
method.
Take V₁ as the reference voltage of 1.0pu 0°, and choose the initial
voltage at Bus 2 as estimates 82 0 and |V₂| = 1.
-
Obtain the values of the Voltage at Bus 2 at the end of the 1st iteration.
Z =jX
0
1.0
1.25
-2.31]
Value of X
0.1
0.3
0.2
Value of Y
0.6
0.5
0.8
V2₂282
2
P+jQ=Y+j0.3 pu
d2
d2
Transcribed Image Text:b) c) The bus admittance matrix of a simple power network is given below. Draw the single line diagram marking line parameters. Ybus = j g1 Figure Q4 shows a simple 2 bus bar electric power system, with the values of X and Y defined below Last digit of your student number 0, 3, 6 1,4,7 2,5 8,9 g1 -3.78 1.25 2.5 0 V,<0° 1 1.25 2.50 -3.42 1.11 1.11 -4.89 1.0 1.25 Note that the values for X and Y to be used in the calculation depend on the last digit of your student number. All per unit quantities are on a base power of 100MVA. The power flow problem is to be solved using the Newton-Raphson method. Take V₁ as the reference voltage of 1.0pu 0°, and choose the initial voltage at Bus 2 as estimates 82 0 and |V₂| = 1. - Obtain the values of the Voltage at Bus 2 at the end of the 1st iteration. Z =jX 0 1.0 1.25 -2.31] Value of X 0.1 0.3 0.2 Value of Y 0.6 0.5 0.8 V2₂282 2 P+jQ=Y+j0.3 pu d2 d2
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