Generation - Scenario 2

SITUATION

A compound DC generator rated 20 kiloWatts, with resistances in its armature winding at 0.05 ohm, series field winding at 0.025 ohm, and shunt field winding at 100 ohms, operates at full load with 250 Volts at its terminals.

With the machine configured as short shunt, what is the internal electromotive force produced by the generator?

ANALYSIS

1.) Schematic diagram, short shunt compound:

2.) One-line diagram:

o|---EMF---R,ar---|---R,se---V,L---|>
                  |
                 R,sh
                  |
                  v

3.) Power drawn by load: P,L = 20kW (full load)

4.) Voltage at the generator terminals: V,t = 250V

5.) Resistances

5.1.) Armature winding resistance: R,ar = 0.05 ohm

5.2.) Series field winding resistance: R,se = 0.025 ohm

5.3.) Shunt field winding resistance: R,sh = 100 ohms

6.) Short shunt configuration

6.1.) Load current is same as series field current: i,L = i,se

6.2.) Load voltage is same as generator terminal voltage: V,L = V,t = 250V

6.3.) Series field current, load branch:

i,L = P,L / V,L
i,se = P,L / V,t
i,se = 20kW / 250V
i,se = 80A

6.4.) Shunt field voltage, load-side loop:

-V,sh + i,se * R,se + V,L = 0
V,sh = 80 * 0.025 + 250
V,sh = 252V

6.5.) Shunt field current:

i,sh = V,sh / R,sh
i,sh = 252V / 100 ohms
i,sh = 2.52A

6.6.) Armature current:

i,ar = i,sh + i,se
i,ar = 2.52 + 80
i,ar = 82.52A

6.7.) Internal electromotive force:

-EMF + i,ar * R,ar + V,sh = 0
EMF = 82.52 * 0.05 + 252
EMF = 256.126V

CONCLUSION

This short shunt compound DC generator produces 256.126 Volts of internal EMF to operate at 250V terminal voltage with a full load of 20kW.

For comparison, the next scenario examines the generator's internal EMF at no-load condition.