2p.y = 2a = 76 0.4+3.42 나 0.2 +0.85 Open circuit tests carried out on a 3.5 MVA, 4.16 kV, 60 Hz, three phase star connected synchronous generator gave the following results: EPM: Vpnt Ip.y (2p.u) ► If (A): 10 ▸ Ę₁ (KV): 15 3.5 4.16 5.7 20 25 30 6.25 7.95 =1+120° 6.2 +10.8 1+1 (0.87 176. 1 +0.87 £76.7 ➤ In the short circuit test a field current of 20A was needed to circulate full load current. The armature winding 1+0.2+10 resistance is 1.20 /phase, the field winding resistance is 500 and the rotational loss is 250kW. Determine the power angle, the voltage regulation and the efficiency when the generator delivers rated load at terminal voltage of 4.4 kV and at 0.8 power factor leading. ▸ Draw the power flow diagram for the generator and include all values of power flows. 1.2 + Jo E =14743

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 2.13MCQ
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2p.y = 2a =
76
0.4+3.42
나
0.2 +0.85
Open circuit tests carried out on a 3.5 MVA, 4.16 kV, 60
Hz, three phase star connected synchronous generator
gave the following results:
EPM: Vpnt Ip.y (2p.u)
► If (A):
10
▸ Ę₁ (KV):
15
3.5 4.16 5.7
20
25
30
6.25
7.95
=1+120° 6.2 +10.8
1+1 (0.87 176.
1 +0.87 £76.7
➤ In the short circuit test a field current of 20A was needed
to circulate full load current. The armature winding 1+0.2+10
resistance is 1.20 /phase, the field winding resistance is
500 and the rotational loss is 250kW. Determine the
power angle, the voltage regulation and the efficiency
when the generator delivers rated load at terminal voltage
of 4.4 kV and at 0.8 power factor leading.
▸ Draw the power flow diagram for the generator and
include all values of power flows.
1.2 + Jo
E =14743
Transcribed Image Text:2p.y = 2a = 76 0.4+3.42 나 0.2 +0.85 Open circuit tests carried out on a 3.5 MVA, 4.16 kV, 60 Hz, three phase star connected synchronous generator gave the following results: EPM: Vpnt Ip.y (2p.u) ► If (A): 10 ▸ Ę₁ (KV): 15 3.5 4.16 5.7 20 25 30 6.25 7.95 =1+120° 6.2 +10.8 1+1 (0.87 176. 1 +0.87 £76.7 ➤ In the short circuit test a field current of 20A was needed to circulate full load current. The armature winding 1+0.2+10 resistance is 1.20 /phase, the field winding resistance is 500 and the rotational loss is 250kW. Determine the power angle, the voltage regulation and the efficiency when the generator delivers rated load at terminal voltage of 4.4 kV and at 0.8 power factor leading. ▸ Draw the power flow diagram for the generator and include all values of power flows. 1.2 + Jo E =14743
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