Question 7.10: Calculate the output power of a 12-pole separately excited h...
Calculate the output power of a 12-pole separately excited having 1200 lap-connected conductors each carrying a current if 15 A. The armature is being driven at 300 rpm. The flux per pole is 60 mWb. Resistance if armature circuit is 0.1 Ω.
Current flowing in each parallel path is the same as current flowing through each conductor in the path. Since there are 12 parallel paths in the armature (since A = P = 12), the total armature current is 15 × 12 = 180 A.
Induced EMF, E=\frac{\phi ZNP}{60 A}=\frac{60 \times 10^{-3} \times 1200 \times 300 \times 12}{60 \times 12}
= 360 V
V = E − I_{a} R_{a}
= 360 − 180 × 0.1
= 342 V
Power output = VI = \frac{342 \times 180}{1000} kW = 61.56 kW
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