Question 16.4: Synchronous-Motor Performance A 480-V-rms 200-hp 60-Hz eight...
Synchronous-Motor Performance
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 \ \Omega.
a. Find the speed and developed torque.
b. Determine the values of \pmb{\text{I}}_a,\pmb{\text{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 \pmb{\text{I}}_a,\pmb{\text{E}}_r, the torque angle, and the power factor.
a. The speed of the machine is given by Equation 16.14:
n_s=\frac{120f}{P}=\frac{120(60)}{8}=900 \text{ rpm} \\ \omega_s=n_s\frac{2\pi}{60}=30\pi=94.25 \text{ rad/s}
For the first operating condition, the developed power is
P_{\text{dev1}}=50 \times 746=37.3 \text{ kW}
and the developed torque is
T_{\text{dev1}}=\frac{P_{\text{dev1}}}{\omega_s} =\frac{37,300}{94.25}=396 \text{ Nm}
b. The voltage rating refers to the rms line-to-line voltage. Because the windings are delta connected, we have V_a = V_{\text{line}} = 480 \text{ V rms}. Solving Equation 16.44 for I_a and substituting values, we have
P_{\text{dev}}=P_{\text{in}}=3V_aI_a\cos(\theta) \quad \quad \quad \quad \quad (16.44) \\ I_{a1}=\frac{P_{\text{dev1}}}{3V_a\cos(\theta_1)} =\frac{37,300}{3(480)(0.9)}=28.78 \text{ A rms}
Next, the power factor is \cos(\theta_1)=0.9, which yields
\theta_1=25.84^\circ
Because the power factor was given as leading, we know that the phase of \pmb{\text{I}}_{a1} is positive. Thus, we have
\pmb{\text{I}}_{a1}=28.78 \underline{/25.84^\circ} \text{ A rms}
Then from Equation 16.42, we have
\pmb{\text{V}}_a=\pmb{\text{E}}_r+jX_s\pmb{\text{I}}_a \quad \quad \quad \quad \quad (16.42) \\ \pmb{\text{E}}_{r1}=\pmb{\text{V}}_{a1}-jX_s\pmb{\text{I}}_a=480-j1.4(28.78 \underline{/25.84^\circ}) \\ \quad \quad \quad =497.6-j36.3 \\ \quad =498.9 \underline{/-4.168^\circ} \text{ V rms}Consequently, the torque angle is \delta_1 = 4.168^\circ.
c. When the load torque is increased while holding excitation constant (i.e., the values of I_f, B_r, \text{ and } E_r are constant), the torque angle must increase. In Figure 16.21(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{(\delta_2)} and substituting values, we find that
\begin{matrix} \sin(\delta_2)&=&\frac{P_2}{P_1}\sin(\delta_1) =\frac{100 \text{ hp}}{50 \text{ hp}}\sin(4.168^\circ) \\ \delta_2&=&8.360^\circ \end{matrix}
Because E_r is constant in magnitude, we get
\pmb{\text{E}}_{r2}=498.9 \underline{/-8.360^\circ} \text{ V rms}
(We know that \pmb{\text{E}}_{r2} \text{ lags } \pmb{\text{V}}_a=480 \underline{/0^\circ} because the machine is acting as a motor.)
Next, we can find the new current:
\pmb{\text{I}}_{a2}=\frac{\pmb{\text{V}}_a-\pmb{\text{E}}_{r2}}{jX_s} =52.70 \underline{/10.61^\circ} \text{ A rms}
Finally, the new power factor is
\cos(\theta_2)=\cos(10.61^\circ)=98.3\% \text{ leading}
