29.MET December 2018 Q.10(b)

Find the synchronous impedance reactance of an alternator in which a given field current produces an armature current of 200 A on short circuit and a generated e.m.f. of 50V on open circuit. The armature resistance is 0.1 ohm. To what induced voltage must the alternator be excited if it is to deliver a load of 100A at a p.f of 0.8 lagging, with a terminal voltage of 200V

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Given

Isc = 200A
Voc = 50V
R = 0.1 ohm
I = 100A
PF = 0.8
V = 200V

To find

1. Synchronous impedance reactance, X
2. Induced voltage, E

Solution

1. Impedance Z = Voc / Isc
                          = 50 / 200
                          =  0.25 ohm

    Impedance2 = Reactance2 + Resistance2

    Z2 = X2 + R2

    X2 = Z2  -  R2

    X = √(0.252 – 0.12)
        = 0.23 ohm

2. E = √((V Cos ø + IR)2 + (V Sin ø + IX)2
       =  √(((200 x 0.8) + (100 x 0.1))2 + ((200 x 0.6) + (100 x 0.23))2
       = 222 V