# 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 an

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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

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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

Vt = terminal voltage

Xs = synchronous reactance

Calculation:

Given that

Vph = 4000 / √3 = 2309.4 V and XS = 4 Ω

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

Eph = √(Vcos ϕ + IaRa)2 + (Vsin ϕ + IaXs)2

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

Eph = 2466.5 V

E = √3Eph = 4272 V

$P_{max} = \frac{{E{V_t}}}{{{X_s}}}$

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