Question 17.5: Power-Factor Control A 480-V-rms 200-hp 60-Hz eight-pole del...

First, we determine the initial value of E_r . Because the initial power factor is cos(θ₁) = 0.85, we can determine that

\theta_1=31.79^\circ

Then, the phase current is

I_{a1}=\frac{P_{\mathrm{dev}}}{3V_{a}\cos{(\theta_1)}}=\frac{200(746)}{3(480)0.85}=121.9\mathrm{~A~rms}

Thus, the phasor current is

\mathrm{I}_{a1}=121.9 \angle -31.79^\circ \mathrm{~A~rms}

The induced voltage is

\mathrm{E}_{r1}=\mathrm{V}_{a1}-jX_s\mathrm{I}_{a1}=480-j1.4(121.9\angle -31.79^\circ)=390.1-j145.0=416.2\angle -20.39^\circ \mathrm{~V~rms}

The phasor diagram for the initial excitation is shown in Figure 17.24(a).
To achieve 100 percent power factor, we need to increase the field current and the magnitude of E_r until I_a is in phase with V_a, as shown in Figure 17.24(b). The new value of the phase current is

I_{a2}=\frac{P_{\mathrm{dev}}}{3V_{a}\cos{(\theta_2)}}=\frac{200(746)}{3(480)}=103.6\mathrm{~A~rms}

Then, we have

\mathrm{E}_{r2}=\mathrm{V}_{a2}-jX_s\mathrm{I}_{a2}=480-j1.4(103.6)=480-j145.0=501.4\angle -16.81^\circ \mathrm{~V~rms}

Now the magnitude of E_r is proportional to the field current, so we can write

I_{f2}=I_{f1}\frac{E_{r2}}{E_{r1}}=10\frac{501.4}{416.2}=12.05\mathrm{~A~dc}