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Sync generator (6600 V) leading voltage

Short answer

Full-load current I=1200kVA/(√3×6600V)=105 A/phase, Vph=3810 V. For 0.8 lagging, E=4249 V. For 0.8 leading, solving gives Vt=4560 V/phase, line voltage=7.90 kV.

A 1200 KVA, 6600 V, 3-phase star connected stator of a synchronous generator has an armature resistance of 0.4 Ω/phase and synchronous reactance of 6 Ω/phase. The generator delivers full load current at pf of 0.8 lagging at normal rated voltage. Calculate the terminal voltage for the same excitation and load current at 0.8 pf leading.

Given: 1200 kVA, 6600 V (line), 3-phase, star; R_a = 0.4 Ω/phase, X_s = 6 Ω/phase; full-load current at 0.8 pf lagging at rated voltage. Find the terminal voltage for the same excitation and current at 0.8 pf leading.

Step 1: Rated quantities

I = 1 200 000 / (√3 × 6600) = 104.97 A V_ph = 6600/√3 = 3810.5 V I R_a = 104.97 × 0.4 = 41.99 V; I X_s = 104.97 × 6 = 629.8 V

Step 2: Induced emf (0.8 lagging, cos φ = 0.8, sin φ = 0.6)

E = V + I (R_a + jX_s), where the current lags V by 36.87°:

 E_real = V + IR cosφ + IX sinφ
        = 3810.5 + 41.99(0.8) + 629.8(0.6)
        = 3810.5 + 33.59 + 377.9 = 4222.0 V
 E_imag = IX cosφ - IR sinφ
        = 629.8(0.8) - 41.99(0.6) = 503.9 - 25.2 = 478.7 V
 E = √(4222.0² + 478.7²) = 4249.1 V per phase

Step 3: Terminal voltage at 0.8 leading (same E, same I)

For a leading current:

  • E_real = V_t + IR cos φ − IX sin φ = V_t + 33.59 − 377.9 = V_t − 344.3
  • E_imag = IX cos φ + IR sin φ = 503.9 + 25.2 = 529.1
 4249.1² = (Vt - 344.3)² + 529.1²
 18 054 000 = (Vt - 344.3)² + 279 950
 (Vt - 344.3)² = 17 774 000   ->   Vt - 344.3 = 4216.0
 Vt = 4560.3 V per phase
 VL = √3 × 4560.3 = 7898.7 V
ConditionPhase voltageLine voltage
0.8 lagging3810.5 V6600 V
0.8 leading, same excitation4560 V7899 V

Answer: terminal voltage ≈ 7.90 kV (line), 4560 V per phase.

✓ Answer checked by Claude Opus 5.5 (high effort)

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