7. Prob. 7 below: For the system shown in Figure 3.34, draw an impedance diagram in per unit, by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the generators. 10 kVA 2400 V G Z = j0.2pu T, Transmission Line T2 M Z= (50 + j200)N 20 kVA 2400 V 40 kVA 80 kVA 25 kVA (G2 Z = j0.2pu 2400/9600 V 10/5 kV 4 kV Z= j0.1pu Z = j0.1pu
7. Prob. 7 below: For the system shown in Figure 3.34, draw an impedance diagram in per unit, by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the generators. 10 kVA 2400 V G Z = j0.2pu T, Transmission Line T2 M Z= (50 + j200)N 20 kVA 2400 V 40 kVA 80 kVA 25 kVA (G2 Z = j0.2pu 2400/9600 V 10/5 kV 4 kV Z= j0.1pu Z = j0.1pu
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
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.16MCQ: In developing per-unit circuits of systems such as the one shown in Figure 3.10. when moving across...
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How can I get a Kv base? I don't know why I get the kv base as (5x10^3/10x10^3)x9600
![7. Prob. 7 below:
For the system shown in Figure 3.34, draw an impedance diagram in per unit,
by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the
generators.
10 kVA
G1
Z= j0.2pu
2400 V
T1
Transmission Line
T2
M
Z= (50 + j200)N
20 kVA
40 kVA
80 kVA
25 kVA
2400 V
(G2,
2400/9600 V
10/5 kV
4 kV
Z= j0.2pu
Z= j0.1pu
Z= j0.1pu](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30f3950e-f2e5-46f4-b550-3d529b806bc8%2F660e0160-1b4a-4b98-9741-4192d24219e8%2F87ds5e_processed.png&w=3840&q=75)
Transcribed Image Text:7. Prob. 7 below:
For the system shown in Figure 3.34, draw an impedance diagram in per unit,
by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the
generators.
10 kVA
G1
Z= j0.2pu
2400 V
T1
Transmission Line
T2
M
Z= (50 + j200)N
20 kVA
40 kVA
80 kVA
25 kVA
2400 V
(G2,
2400/9600 V
10/5 kV
4 kV
Z= j0.2pu
Z= j0.1pu
Z= j0.1pu
![The per unit value of the voltage will be
kV actual
kV base
kVpu
kVbase = ( Sx10³
(9600) p.u.
kVbase = 4800V
kVbase = 4.8V
Therefore,
4
kVpu =
s p.u.
4.8 P.u.
kVpu = 0.833p.u.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30f3950e-f2e5-46f4-b550-3d529b806bc8%2F660e0160-1b4a-4b98-9741-4192d24219e8%2Fme71hd1_processed.png&w=3840&q=75)
Transcribed Image Text:The per unit value of the voltage will be
kV actual
kV base
kVpu
kVbase = ( Sx10³
(9600) p.u.
kVbase = 4800V
kVbase = 4.8V
Therefore,
4
kVpu =
s p.u.
4.8 P.u.
kVpu = 0.833p.u.
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