Question

A 3-phase, 20 mva, 10 kv alternator has internal reactance of 5% and negligible resistance. The external reactance per phase to be connected in series with the alternator so that steady current on short circuit does not exceed 6 times the full load current is

Answer 1

The external reactance is 0.375 Ω

Explanation:

Full load current I = 2 x 10^6 / √ 3 x 10 x 10^3  = 1154.7 A

Voltage per phase = V = 10 x 10^3 / √ 3 = 10,000 / √ 3 volts

Total percentage reactance required = Full load current / short circuit current x 100

Total percentage reactance required =  1/8 x 100 = 12.5 %

External percentage reactance required = 12.5 - 5 = 7.5 %

Let X Ω be the per phase external reactance required.

Now percentage reactance = IX / V x 100

7.5 = 1154.7 X / 10,000 / √ 3 x 100

X = 7.5 x 1000 / √ 3 x 100 x 1154.7 = 0.375 Ω

Thus the external reactance is 0.375 Ω

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