2. Why synchronous impedance method does not provide the exact value of voltage regulation of an alternator?

A three-phase, 4-pole, 1,500 rpm alternating having a double layer-windings with 18 turns per coil. The slots per pole per phase are 10 and the coil span is 22 slots. The spread phase is 60 degrees and the flux per pole is 0.0625 Weber. If the armature windings are wye connected. a) Determine the generated voltage per phase and the line to line voltages. b) If the alternator is delivering a three - phase load of 1,200KW at 0.8 lagging power factor what is the line current of the alternator

Draw a complete circuit diagram of 1 phase synchronous alternator including the field excitation circuit. Then, draw the phasor diagram of the generator a general leading power factor. Include the phase resistance in the diagram.

A 9.9 kV, Y-connected, 3 phase alternator has a synchronous reactance of 8 ohms per phase and negligible resistance. At a certain field current the armature is known to induce 8.2 kV per phase. Calculate the power that can be developed by the alternator at this excitation.

5)This MCQ QUESTION FROM BASIC POWER ELECTRICAL ENGINEERING course.

The standard practice of synchronous generator construction now a days is to have Cylindrical field Rotating armature Either of the above Rotating field In an alternator, the armature reaction influences the magnitude of No load loss Wave form of generated voltage. Speed of the machine Flux Terminal voltage per phase

PLEASE SOLVE THIS WITH COMPLETE & HANDWRITTEN SOLUTION, THANK YOU  !

A single phase, 2000 V alternator has armature resistance and reactance of 0.8 ohms and 4.94 ohms respectively. The voltage regulation of the alternator at 100 A load at 0.8 leading power factor is ??

A 2MVA, 2.3kV three – phase Y – connected alternator operates at rated kVA at a power factor of 80%. The dc armature winding resistance between terminals is 0.08 ohm. The field takes 70 A at 125 V from an exciter equipment. Friction and windage loss is 20kW, iron losses 36kW and stray load losses are 2.0 kW. Calculate the efficiency of the alternator. Assume the effective armature winding resistance is 120% of the DC resistance. a. 94.0 % b. 92.1 % c. 93.2 % d. 89.5 %

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