Question 17.5: A 480-V-rms 200-hp 60-Hz eight-pole delta-connected synchron...
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 Ω. The field current is I_f = 10 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(θ_1) = 0.85, we can determine that
θ_1 = 31.79°
Then, the phase current is
I_{a1} = \frac{P_{dev}} {3V_a \cos(θ_1)}= \frac{200(746)}{ 3(480)0.85} = 121.9 A rms
Thus, the phasor current is
\textbf{I}_{a1} = 121.9∠−31.79° A rms
The induced voltage is
\textbf{E}_{r1} = \textbf{V}_{a1} − jX_s\textbf{I}_{a1} = 480 − j1.4(121.9∠−31.79°)
= 390.1 − j145.0
= 416.2∠−20.39° 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 \textbf{E}_r until \textbf{I}_a is in phase with \textbf{V}_a, as shown in Figure 17.24(b). The
new value of the phase current is
I_{a2} = \frac{P_{dev}} {3V_a \cos(θ_2)} =\frac{ 200(746)} {3(480)} = 103.6 A rms
Then, we have
\textbf{E}_{r2} = \textbf{V}_{a2} − jX_s\textbf{I}_{a2} = 480 − j1.4(103.6)
= 480 − j145.0
= 501.4∠−16.81° V rms
Now the magnitude of \textbf{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 A dc
