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

\omega_s=n_s\frac{2\pi}{60}=30\pi =94.25\mathrm{~rad/s}

For the first operating condition, the developed power is

P_{\mathrm{dev1}}=50\times 746=37.3\mathrm{~kW}

and the developed torque is

T_{\mathrm{dev1}}=\frac{P_{\mathrm{dev1}}}{\omega_s}=\frac{37,300}{94.25}=396\mathrm{~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_{\mathrm{dev}}=P_{\mathrm{in}}=3V_aI_a\cos{(\theta)}       (17.44)

I_{a1}=\frac{P_{\mathrm{P_{dev1}}}}{3V_a\cos(\theta_1)}=\frac{37,300}{3(480)(0.9)}=28.78\mathrm{~A~rms}

Next, the power factor is cos(θ₁) = 0.9, which yields

\theta_1=25.84^\circ

Because the power factor was given as leading, we know that the phase of I_a1 is positive. Thus, we have

\mathrm{I}_{a1}=28.78\angle 25.84^\circ \mathrm{~A~rms}

Then from Equation 17.42, we have

\mathrm{V}_a=\mathrm{E}_r+jX_s\mathrm{I}_a          (17.42)

\mathrm{E}_{r1}=\mathrm{V}_{a1}-jX_s\mathrm{I}_a=480-j1.4(28.78\angle 25.84^\circ)=497.6-j36.3=498.9\angle -4.168^\circ \mathrm{~V~rms}

Consequently, the torque angle is δ₁ = 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{(\delta_2)}}{\sin{(\delta_1)}}=\frac{P_2}{P_1}

Solving for sin(δ₂) and substituting values, we find that

sin{(\delta_2)}=\frac{P_2}{P_1}\sin{(\delta_1)}=\frac{100\mathrm{~hp}}{50\mathrm{~hp}}\sin{(4.168^\circ)}

\delta_2=8.360^\circ

Because E_r is constant in magnitude, we get

E_{r2}=498.9\angle -8.360^\circ \mathrm{~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:

\mathrm{I}_{a2}=\frac{\mathrm{V}_a-\mathrm{E}_{r2}}{jX_s}=52.70\angle 10.61^\circ \mathrm{~A~rms}

Finally, the new power factor is

\cos{(\theta_2)}=\cos{(10.61^\circ)}=98.3% \mathrm{~leading}