Question 17.4: Synchronous-Motor Performance A 480-V-rms 200-hp 60-Hz eight......

a. The speed of the machine is given by Equation 17.14:

n_s = \frac{120 f}{P}             (17.14)

n_s = \frac{120 f}{P} = \frac{(120(60)}{8} = 900 rpm

\omega _s = n_s \frac{2 \pi}{60} = 30 \pi = 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}}{\omega _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 (\theta _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 I_{a1} is positive. Thus, we have
I_{a1} = 28.78 \angle 25.84° A rms

Then from Equation 17.42, we have

V_a = E_r + jX_sI_a               (17.42)
E_{r1} = V_{a1} − jX_sI_a = 480 − j1.4(28.78\angle 25.84°

= 498.9 \angle -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

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°)

Because E_r is constant in magnitude, we get

E_{r2} = 498.9 \angle -8.360° V rms

(We know that E_{r2} lags V_a = 480∠0° because the machine is acting as a motor.) Next, we can find the new current:

I_{a2} = \frac{V_a- E_{r2}}{jX_s} = 52.70 \angle 10.61° A rms

Finally, the new power factor is
\cos (θ_2) = \cos (10.61°) = 98.3\% leading