Question 16.5: Power-Factor Control A 480-V-rms 200-hp 60-Hz eight-pole del...
Power-Factor Control
A 480-V-rms 200-hp 60-Hz eight-pole delta-connected synchronous motor operates with a developed power (including losses) of 200 hp and a power factor of 85 percent lagging. The synchronous reactance is X_s = 1.4 \ \Omega. The field current is I_f = 10 \text{ A}.
What must the new field current be to produce 100 percent power factor? Assume that magnetic saturation does not occur, so that B_r is proportional to I_f.
First, we determine the initial value of E_r. Because the initial power factor is \cos(\theta_1)=0.85, we can determine that
\theta_1=31.79^\circ
Then, the phase current is
I_{a1}=\frac{P_{\text{dev}}}{3V_a\cos(\theta_1)} =\frac{200(746)}{3(480)0.85}=121.9 \text{ A rms}
Thus, the phasor current is
\pmb{\text{I}}_{a1}=121.9 \underline{/-31.79^\circ}\text{ A rms}
The induced voltage is
\begin{matrix} \pmb{\text{E}}_{r1}&=& \pmb{\text{V}}_{a1}-jX_s\pmb{\text{I}}_{a1}&=&480-j1.4(121.9 \underline{/-31.79^\circ}) \\ &=&390.1-j145.0 \\ &=&416.2\underline{/-20.39^\circ}\text{ V rms}\end{matrix}
The phasor diagram for the initial excitation is shown in Figure 16.24(a).
To achieve 100 percent power factor, we need to increase the field current and the magnitude of \pmb{\text{E}}_r \text{ until } \pmb{\text{I}}_a is in phase with \pmb{\text{V}}_a, as shown in Figure 16.24(b). The new value of the phase current is
I_{a2}=\frac{P_{\text{dev}}}{3V_a\cos(\theta_2)}=\frac{200(746)}{3(480)} =103.6 \text{ A rms}
Then, we have
\begin{matrix} \pmb{\text{E}}_{r2}&=& \pmb{\text{V}}_{a2}-jX_s\pmb{\text{I}}_{a2}&=&480-j1.4(103.6) \\ &=&480-j145.0 \\ &=&501.4\underline{/-16.81^\circ}\text{ V rms}\end{matrix}
Now the magnitude of \pmb{\text{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 \text{ A dc}
