Class: BEE2-A
Time Allowed: 10 mins
Maximum Points: 20
Reg. No:
QUIZ SEVEN - SOLUTION
NOTE: Please draw labelled circuit diagrams where appropriate, show complete calculations, and reasoning
to get maximum marks.
1. A 2200 V, 400kW, 3 phase, Y-connected synchronous generator has a synchronous impedance of 0.3+j3Ω.
Determine:
a. The torque angle [5]
b. Internal generated voltage, EA[15]
if the generator is working at rated conditions and 0.8 pf leading with an efficiency of 94%. [20]
SOLUTION:
Vφ=2200
√3[Yconnection]
cos θ= 0.8 leading →θ= 36.86
η=Po
Pin =400k
0.94 = 425.53kW
P=√3VTILcos θ→IL=425.53k
√32200×0.8= 139.6∠36.86A
XS= 0.3 + j3=3.01∠84.29Ω
EA=Vφ+ (RA+jXS)IA
EA=2200
√3+ (3.01∠84.29)(139.6∠36.86)
EA= 1270.2 + 420.196∠121.5
EA= 1052.84 + j359.610
EA= 1112.56∠18.85V
Torque Angle: 18.85o
1
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