Example 5: The following information is given for a 300-kW, 600-V, long-shunt compound generator : Shunt field resistance = 75 Q, armature resistance including brush resistance = 0.03 0, commutating field winding resistance 0.011 2, series field resistance = 0.012 N, divertor resistance 0.036 Q. When the machine is delivering full |load, calculate the voltage and power generated by the armature. Solution: Power output = 300,000 W Output current = 300,000/600 = 500 A Ish = 600/75 = 8 A 8 A 500 A 1. 0.012 2 I, = 500 + 8 = 508 A Since the series field resistance and divertor 600 V 0.011 2 resistance are in parallel their combined resistance 0.012x 0.036 is = 0.012+0.036 Total armature circuit resistance = 0.03 + 0.011 + 0.009 = 0.05 2 Voltage drop 508x0.05 = 25.4 V Voltage generated by armature = 600 + 25.4 625.4 V 0.009 2 0.03 2 Power generated = 625.4x508 = 317,700 = 317.7 kw using matlab E00 U SL

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Example 5: The following information is given for a 300-kW, 600-V, long-shunt
compound generator : Shunt field resistance 75 Q, armature resistance including brush
resistance = 0.03 0, commutating field winding resistance 0.011 2, series field
resistance = 0.012 N, divertor resistance 0.036 Q. When the machine is delivering full
|load, calculate the voltage and power generated by the armature.
Solution:
Power output = 300,000 W
Output current = 300,000/600 = 500 A
Ish = 600/75 = 8 A
8 A
500 A
1.
0.012 2
I, = 500 + 8 = 508 A
Since the series field resistance and divertor
600 V
0.011 2
resistance are in parallel their combined resistance
0.012x 0.036
0.009 2
is =
0.012+0.036
Total armature circuit resistance
= 0.03 + 0.011 + 0.009 = 0.05 2
Voltage drop 508x0.05 = 25.4 V
Voltage generated by armature = 600 + 25.4 625.4 V
0.03 2
Power generated = 625.4x508 = 317,700 = 317.7 kw using matlab
U SL
Transcribed Image Text:Example 5: The following information is given for a 300-kW, 600-V, long-shunt compound generator : Shunt field resistance 75 Q, armature resistance including brush resistance = 0.03 0, commutating field winding resistance 0.011 2, series field resistance = 0.012 N, divertor resistance 0.036 Q. When the machine is delivering full |load, calculate the voltage and power generated by the armature. Solution: Power output = 300,000 W Output current = 300,000/600 = 500 A Ish = 600/75 = 8 A 8 A 500 A 1. 0.012 2 I, = 500 + 8 = 508 A Since the series field resistance and divertor 600 V 0.011 2 resistance are in parallel their combined resistance 0.012x 0.036 0.009 2 is = 0.012+0.036 Total armature circuit resistance = 0.03 + 0.011 + 0.009 = 0.05 2 Voltage drop 508x0.05 = 25.4 V Voltage generated by armature = 600 + 25.4 625.4 V 0.03 2 Power generated = 625.4x508 = 317,700 = 317.7 kw using matlab U SL
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