Question

The maximum power delivered by 1500 kW, 3-phase, star connected, 4 kV, 48 pole 50 Hz synchronous motor, with synchronous reactance of 4 Ω per phase and unity power factor:
1. 4271.2 kW
2. 3505 kW
3. 1206.1 kW
4. 2078 kW

Answer 1

Correct Answer - Option 1 : 4271.2 kW

Concept:

For Synchronous motor power is given by

\(P = \frac{{E{V_t}}}{{{X_s}}}\sin \delta \)

For maximum power δ = 90° ⇒ sin 90° = 1

\(P_{max} = \frac{{E{V_t}}}{{{X_s}}}\)

E = generated voltage

V_t = terminal voltage

X_s = synchronous reactance

Calculation:

Given that

V_ph = 4000 / √3 = 2309.4 V and X_S = 4 Ω 

\({I_L} = \frac{{1500}}{{√ 3 × 4000 × 1}} = 216.5\;A\)

E_ph = √(Vcos ϕ + I_aR_a)² + (Vsin ϕ + I_aX_s)²  

Eph = √(2309.4 × 1 + 0)2 + (0 + 216.5 × 4)2  

E_ph = 2466.5 V

E = √3E_ph = 4272 V

\(P_{max} = \frac{{E{V_t}}}{{{X_s}}}\)

\(P_{max} = \frac{{4272 × 4000 }}{{{4}}}=4271.2\;kW\)