Synchronous motor: pf when input rises to 1500 kW

Given: 6.6 kV, star, $X_s = 20\ \Omega$ per phase (resistance neglected), constant $V$ and constant excitation. Condition 1: input 100 kW at 0.8 leading. Find the power factor at 1500 kW input.

$$V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}$$

Step 1: excitation emf at 100 kW, 0.8 leading

$$I_1 = \frac{100\times10^3}{\sqrt3\times6600\times0.8} = 10.93\ \text{A}, \qquad \vec I_1 = 10.93\angle+36.87^\circ$$

For a motor, $\vec E = \vec V - j\vec I X_s$:

$$\vec E = 3810.5 - j20\,(8.748 + j6.561) = (3810.5 + 131.2) - j174.95 = 3941.7 - j174.95\ \text{V}$$ $$E = 3945.6\ \text{V per phase}\quad (\delta_1 = 2.54^\circ)$$

Step 2: at 1500 kW (E unchanged)

Power per phase $= 500$ kW:

$$P_{ph} = \frac{VE}{X_s}\sin\delta \;\Rightarrow\; \sin\delta = \frac{500\times10^3\times20}{3810.5\times3945.6} = 0.6651, \quad \delta = 41.69^\circ$$

(maximum power $= 3VE/X_s = 2255$ kW, so 1500 kW is below the pull-out limit.)

Current:

$$\vec I = \frac{\vec V - \vec E}{jX_s} = \frac{3810.5 - 3945.6\angle-41.69^\circ}{j20}$$ $$|I| = 138.15\ \text{A}, \qquad \angle I = -18.23^\circ$$ $$\cos\phi = \cos 18.23^\circ = 0.950 \ \text{(lagging)}$$

Check: $\sqrt3\times6600\times138.15\times0.9498 = 1500$ kW.

Answer: power factor at 1500 kW input $\approx 0.95$ lagging (current 138.15 A). The motor moved from leading to lagging because the excitation was fixed while the load increased.