can you explain the calculations please Example 1For the system shown in figure, the lineimpedances are as indicated in per uniton 100MVA base.Obtain the bus admittance matrix of thepower network.Solution-Y 12-Y 13Y21+Y23-23bus[Y, Y1112 13==İY₂1 22 23|=|21131 32 33112 0.02+j0.0410.01+j0.0310.0125+ j0.025Y 13 =Y 23Y12 +Y13-Y21- 31= 10-j20 = y₂1=10-j30=Y31=16-j32=y32GV₁=1.05 20⁰ puY31+ Y32bus20.0125+j0.025 pu45.2MVAR0.02.jo.04 pu0.01+ j0.03 puT3138.6MW45.2MVAR20-j50 -10+j20-10+j30]-10+j20 26-j52 -16+j32L-10+j30 -16+j32 26-j62=0.01+j0.03 pu30.02 + j0.04 puV₁=1.0520 pu138.6MW20.0125 jo.025 pu256.6MW110.2MVAR256.6MW110.2MVAR

Answered: Image /qna-images/answer/2779d9a3-e3a4-4c3d-8712-7aefe38764ce.jpg

For the system shown in figure, the line impedances are as indicated in per unit on 100MVA base. Obtain the bus admittance matrix of the power network. G V₁=1.05/0° pu 0.02+j0.04 pu 0.01+j0.03 pu 3 2 0.0125+j0.025 pu 45.2 MVAR 138.6 MW 256.6 MW 110.2 MVAR

For the system shown in figure, the line impedances are as indicated in per unit on 100MVA base. Obtain the bus admittance matrix of the power network. G V₁=1.05/0° pu 0.02+j0.04 pu 0.01+j0.03 pu 3 2 0.0125+j0.025 pu 45.2 MVAR 138.6 MW 256.6 MW 110.2 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 7.21P

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Question

can you explain the calculations please

Example 1
For the system shown in figure, the line
impedances are as indicated in per unit
on 100MVA base.
Obtain the bus admittance matrix of the
power network.
Solution
-Y 12
-Y 13
Y
21+Y23-23
bus
[Y, Y
11
12 13
=
=İY₂1 22 23|=|
21
1
31 32 33
1
12 0.02+j0.04
1
0.01+j0.03
1
0.0125+ j0.025
Y 13 =
Y 23
Y12 +Y13
-Y21
- 31
= 10-j20 = y₂1
=10-j30=Y31
=16-j32=y32
G
V₁=1.05 20⁰ pu
Y31+ Y32
bus
2
0.0125+j0.025 pu
45.2
MVAR
0.02.jo.04 pu
0.01+ j0.03 pu
T
3
138.6
MW
45.2
MVAR
20-j50 -10+j20-10+j30]
-10+j20 26-j52 -16+j32
L-10+j30 -16+j32 26-j62
=
0.01+j0.03 pu
3
0.02 + j0.04 pu
V₁=1.0520 pu
138.6
MW
2
0.0125 jo.025 pu
256.6
MW
110.2
MVAR
256.6
MW
110.2
MVAR

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