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1. The third picture is the solution of this problem. But I just don\'t get it w

ID: 2292028 • Letter: 1

Question

1. The third picture is the solution of this problem. But I just don't get it why the field current should be 3.5A.

13.8kV of terminal voltage means (13.8kv/?3) phase voltage(cuz it is Y-connected), so the open circuit voltage should be (13.8kv/?3).

so maybe the current should be 1.5A.

Am I wrong about this???

4-2. A 13.8-kV, 50-MVA, 0.9-power-factor-lagging, 60-Hz, four-pole Y-connected synchronous generator has a synchronous reactance of 2.5 ? and an armature resistance of 0.2 ?. At 60 Hz, its friction and windage losses are 1 MW, and its core losses are 1.5 MW. The field circuit has a dc voltage of 120 V, and the maximum IF is 10 A. The current of the field circuit is adjustable over the range from 0 to 10 A. The OCC of this generator is shown in Figure P4-1. Open-cicuit characteristic 20 18 16 14 12 10 8 10

Explanation / Answer

The Voltage rating given on the name plate is line-line voltage. While finding open circuit charecteristics, We consider line-line voltage on the y-axis and corresponding field current on the x-axis.Remember we consider line-line voltage on y-axis but not phase voltage, as the load side is kept open while finding Open Circuit Charecteristics.Henceforth the given solution is correct. that is Filed current= 3.5A