Sync generator terminal voltage, leading pf

Given: 1200 kVA, 6000 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 (same E) and the same current at 0.8 pf leading.

Equivalent circuit per phase:

   Ia    Ra=0.4   Xs=6j
 --->--/\/\/--(XXX)--+----> Vt (terminal)
   E (induced emf)   |

Step 1: Full-load current and phase voltage

I = S / (√3 V_L) = 1 200 000 / (√3 × 6000) = 115.47 A V_ph = 6000 / √3 = 3464.1 V

Step 2: Induced emf at 0.8 pf lagging

Take V as reference. I = 115.47 ∠−36.87° = 92.38 − j69.28 A. Z_s = 0.4 + j6 Ω.

Using components: I·R = 115.47 × 0.4 = 46.19 V; I·X = 115.47 × 6 = 692.8 V.

For a lagging load, E = V + I Z:

• E_real = V cos0 + IR cos φ + IX sin φ = 3464.1 + 46.19 × 0.8 + 692.8 × 0.6 = 3464.1 + 36.95 + 415.69 = 3916.7 V
• E_imag = IX cos φ − IR sin φ = 692.8 × 0.8 − 46.19 × 0.6 = 554.26 − 27.71 = 526.5 V

E = √(3916.7² + 526.5²) = 3952.0 V per phase (angle 7.66°).

Step 3: Terminal voltage at 0.8 pf leading, same E and same I

Now I = 115.47 ∠+36.87° = 92.38 + j69.28 A. With V_t as reference, E = V_t + I Z, whose components are

• real: V_t + IR cos φ − IX sin φ = V_t + 36.95 − 415.69 = V_t − 378.74
• imaginary: IX cos φ + IR sin φ = 554.26 + 27.71 = 581.97

 E² = (Vt - 378.74)² + 581.97²
 3952.0² = (Vt - 378.74)² + 581.97²
 15 619 000 = (Vt - 378.74)² + 338 690
 (Vt - 378.74)² = 15 280 300  ->  Vt - 378.74 = 3909.0
 Vt = 4287.6 V per phase

Line voltage V_L = √3 × 4287.6 = 7426 V (about 7.43 kV).

The terminal voltage rises (about 23.8 % above rated) because the leading current gives magnetising armature reaction and a voltage rise across the reactance.

Answer: terminal voltage ≈ 7.43 kV (line), 4288 V per phase.