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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

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

Introductory Circuit Analysis (13th Edition)
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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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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