V:56) 1|0E1j0.2j0.2SB2H = 5 sec.j0.2j0.2E2j0.2G2H = 6 sec.For the system shown, assume E1 = 2,17 16.95° pu, and E2 =2.58 12.82° pu. Assume the electrical power (Pe) from G2 after clearingshort-circuit is written saPez - Pmax sin ög2-Determine value of Pmax.For the system shown, assume E1 = 2.3 (69° pu, and E2-2.43 /76° pu. Assume the electrical power (Pe) during short-circuit fromgenerator G1 is written asPej = Pmax sin õg1.Determine the value of Pmax

calculate reactance, X=j0.2+j0.2+j0.2||j0.4 =j0.4+j0.2×0.40.6 =j0.4+j0.080.6 =j0.533 Ω…

For the system shown, assume E1 = 2.17 16.95° pu, and E2 =2.58 12.82° pu. Assume the electrical power (Pe) from G2 after clearing short-circuit is written sa Pez Pmax sin ôg2. Determine value of Pmax.

For the system shown, assume E1 = 2.17 16.95° pu, and E2 =2.58 12.82° pu. Assume the electrical power (Pe) from G2 after clearing short-circuit is written sa Pez Pmax sin ôg2. Determine value of Pmax.

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

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.10P: For the circuit element of Problem 2.3, calculate (a) the instantaneous power absorbed, (b) the real...

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Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.

Question

V:56)

1|0
E1
j0.2
j0.2
S
B2
H = 5 sec.
j0.2
j0.2
E2
j0.2
G2
H = 6 sec.
For the system shown, assume E1 = 2,17 16.95° pu, and E2 =2.58 12.82° pu. Assume the electrical power (Pe) from G2 after clearing
short-circuit is written sa
Pez - Pmax sin ög2-
Determine value of Pmax.
For the system shown, assume E1 = 2.3 (69° pu, and E2-2.43 /76° pu. Assume the electrical power (Pe) during short-circuit from
generator G1 is written as
Pej = Pmax sin õg1.
Determine the value of Pmax

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