Question 17.4: 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 50 hp and a power factor of 90 percent leading. The synchronous reactance is X_s = 1.4 Ω. a. Find the speed and developed torque. b. Determine the values of \textbf{I}_a, \textbf{E}_r , and the torque angle. c. Suppose that the excitation remains constant and the load torque increases until the developed power is 100 hp. Determine the new values of \textbf{I}_a, \textbf{E}_r , the torque angle, and the power factor.
a. The speed of the machine is given by Equation 17.14:
n_s = \frac{120f}{ P} (17.14)
n_s = \frac{120f}{ P} = \frac{120(60)}{ 8} = 900 rpm
ω_s = n_s\frac{ 2π}{ 60} = 30π = 94.25 rad/s
For the first operating condition, the developed power is
P_{dev1} = 50 × 746 = 37.3 kW
and the developed torque is
T_{dev1} = \frac{P_{dev1}}{ ω_s} =\frac{ 37,300 }{94.25} = 396 Nm
b. The voltage rating refers to the rms line-to-line voltage. Because the windings are delta connected, we have V_a = V_{line} = 480 V rms. Solving Equation 17.44 for I_a and substituting values, we have
P_{dev} = P_{in} = 3V_aI_a \cos(θ) (17.44)
I_{a1} = \frac{P_{dev1}} {3V_a \cos(θ_1)} = \frac{37,300 }{3(480)(0.9)} = 28.78 A rms
Next, the power factor is \cos(θ_1) = 0.9, which yields
θ_1 = 25.84°
Because the power factor was given as leading, we know that the phase of \textbf{I}_{a1} is
positive. Thus, we have
\textbf{I}_{a1} = 28.78∠25.84° A rms
Then from Equation 17.42, we have
\textbf{V}_{a} = \textbf{E}_{r} + jX_s\textbf{I}_a (17.42)
\textbf{E}_{r1} = \textbf{V}_{a1} − jX_s\textbf{I}_a = 480 − j1.4(28.78∠25.84°)
= 497.6 − j36.3
= 498.9∠−4.168° V rms
Consequently, the torque angle is δ_1 = 4.168°.
c. When the load torque is increased while holding excitation constant (i.e., the values of I_f , B_r, and E_r are constant), the torque angle must increase. In Figure 17.21(b),we see that the developed power is proportional to sin(δ). Hence,
we can write
\frac{\sin(δ_2)}{ \sin(δ_1)} = \frac{P_2} {P_1}
Solving for \sin(δ_2) and substituting values, we find that
\sin(δ_2) = \frac{P_2} {P_1} \sin(δ_1) = \frac{100 hp}{ 50 hp} sin(4.168°)
δ_2 = 8.360°
Because E_r is constant in magnitude, we get
\textbf{E}_{r2} = 498.9∠−8.360° V rms
(We know that \textbf{E}_{r2} lags \textbf{V}_{a} = 480∠0° because the machine is acting as a motor.)
Next, we can find the new current:
\textbf{I}_{a2} = \frac{\textbf{V}_a − \textbf{E}_{r2}} {jX_s} = 52.70∠10.61° A rms
Finally, the new power factor is
\cos(θ_2) = cos(10.61°) = 98.3% leading
