Test: Electrical Machines- 7 - Electrical Engineering (EE) MCQ

P = P_max sin δ
1 = 2 sin δ
δ = 30°

For alternator

F = 50 Hz
For I.M

N_r = (1 - S) Ns = (1 - .03) × 750
∴ N_r = 727.5 RPM

∴ A 10-pole synchronous motor must be coupled to a 12-pole synchronous generator to accomplish this frequency conversion.

For generator

When E_f cosδ < V, i.e under excited generator, then reactive power (Q_out) is negative i.e supplying leading VARs and therefore operating at leading power factor, thus behaving like a capacitor.

In star connection, the third harmonic component of the three phases cancel out at the line terminals because they are co phased.
∴ the line emf will be consisting of the fundamental and fifth harmonic only
Fundamental component E₁ = 300 V
5^th harmonic component, E₅ = 0.1 × 300 = 30 V
Emf per phase, 

= 301.5 V
rms line voltage, 

Generated voltage of salient pole synchronous generator

Direct component of current
I_d = I_a sin 30.96 = 0.5145 PU
∴ |E_F| = 1.17 + .5145 × 0.3 = 1.324 PU

Given
Slots/pole/phase m = 8
Conductor/slot = 9
Coil/pole = 24
Slot = 24 × 4 = 96

Total no. of conductor z = 96 × 8 = 768
Conductor/phase = 256
Number of turn = 128

= 0.9556

E_ph = 4.44 × 0.9556 × 0.965 × 0.28 × 50 × 128 = 7337.09 V

Coil span is 16 – 1 = 15 slots
Number of empty slots = 18 – 15 = 3 slots

Pitch factor for 3^rd harmonic, 
= 0.707
Pitch factor for 5^th harmonic 
= 0.259
 

= 132.79 × 0.9136 + 75.307 × 4 sin (24 + 45°)
= 121.316 + 281.19
= 402.506
Percentage voltage regulation