Generation - Scenario 4

Consider this generator again from the previous scenario, but this time, with half of its rated capacity attached. This post examines what will happen to its internal EMF at half-load condition.

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 HALF 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 = 10kW (HALF 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 = 10kW / 250V
i,se = 40A

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

-V,sh + i,se * R,se + V,L = 0
V,sh = 40 * 0.025 + 250
V,sh = 251V

6.5.) Shunt field current:

i,sh = V,sh / R,sh
i,sh = 251V / 100 ohms
i,sh = 2.51A

6.6.) Armature current:

i,ar = i,sh + i,se
i,ar = 2.51 + 40
i,ar = 42.51A

6.7.) Internal electromotive force:

-EMF + i,ar * R,ar + V,sh = 0
EMF = 42.51 * 0.05 + 251
EMF = 253.1255V

CONCLUSION

This short shunt compound DC generator produces 253.1255 Volts of internal EMF to operate at 250V terminal voltage with a HALF LOAD of 10kW.