Answer to Question #212795 in Electrical Engineering for joe

Question

 A 2,500 kVA, three – phase, 60 Hz, 6.6 kV wye connected alternator has a field resistance of 0.45 Ω and an armature resistance of 0.05 Ω per phase. The field current at full load 0.85 pf, is 200 A. The stray power losses amount to 82 kW. Calculate the efficiency of the alternator at full load, 0.85 pf lagging.

Answer 1

We have to calculate efficiency at full load

For this we have to calculate losses as the alternator is working at full load

operating power = (2500*0.85) KW

= 2125KW

Losses in field resistance = I²R_f

= (200)² * 0.45

= 18KW

For armature losses

Full load losses = I_fl = 2500 *10³/$3*6.6*10$ ³

= 218.69A

So for armature losses = 3I²_flR_a (since armature is 3 phase)

= 3*(218.69)²*0.4

= 57.39KW

Hence efficiency = operating power/operating power + total losses

= 2125/(2125 + 18+57.39)

= 0.93104*100

= 93.104%