Q:28) toLOAD1S,- 75 MVAcosp=0.95 backXH = j 144,4 ohm125 MVALOAD2Sy= 25 MVAcosp=0.9 back150 MVA200 MVA380/36 kVX12=j0.1 puUG= 20 kV20/400 kVXu=j0.2 puXa=j0.1 pua-)Uhas380 kV ,S base=100 FOR MVA, CALCULATE PU EMPEDANCE VALUES AND DRAW PU EMPEDANCEbaseDIAGRAMb-) WHEN V1=1 20° PU, with RĘCYÇLING method find real V4 value.C-) Calculate the electromotive force of the generator3.

Answered: Image /qna-images/answer/14d7dd28-3d1a-4603-aa06-c38fc75bb2f8.jpg

2 LOAD1 Sy= 75 MVA cosp=0.95 back XH = j 144,4 ohm 125 MVA LOAD2 Sy= 25 MVA cosp=0.9 back 150 MVA 200 MVA 380/36 kV X12=j0.1 pu UG= 20 kV 20/400 kV Xu=j0.2 pu Xa=j0.1 pu -) Upas 380 kV ,S base=100 FOR MVA, CALCCULATE PU EMPEDANCE VALUES AND DRAW PU EMPEDANCE 3.

2 LOAD1 Sy= 75 MVA cosp=0.95 back XH = j 144,4 ohm 125 MVA LOAD2 Sy= 25 MVA cosp=0.9 back 150 MVA 200 MVA 380/36 kV X12=j0.1 pu UG= 20 kV 20/400 kV Xu=j0.2 pu Xa=j0.1 pu -) Upas 380 kV ,S base=100 FOR MVA, CALCCULATE PU EMPEDANCE VALUES AND DRAW PU EMPEDANCE 3.

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...

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Concept explainers

Three-Phase Transformers

Three-segment transformers are a type of transformer used to transform voltages of electrical systems into three ranges. Two type transformers are shell-type transformer and core type transformer. In brief, it could be described because of the exquisite kinds of configurations.

Transformer

Ever since electricity has been created, people have started using it in its entirety. We see many types of Transformers in the neighborhoods. Some are smaller in size and some are very large. They are used according to their requirements. Many of us have seen the electrical transformer but they do not know what work they are engaged in.

Question

Q:28)

to
LOAD1
S,- 75 MVA
cosp=0.95 back
XH = j 144,4 ohm
125 MVA
LOAD2
Sy= 25 MVA
cosp=0.9 back
150 MVA
200 MVA
380/36 kV
X12=j0.1 pu
UG= 20 kV
20/400 kV
Xu=j0.2 pu
Xa=j0.1 pu
a-)Uhas380 kV ,S base=100 FOR MVA, CALCULATE PU EMPEDANCE VALUES AND DRAW PU EMPEDANCE
base
DIAGRAM
b-) WHEN V1=1 20° PU, with RĘCYÇLING method find real V4 value.
C-) Calculate the electromotive force of the generator
3.

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