Question 16.4: Synchronous-Motor Performance A 480-V-rms 200-hp 60-Hz eight...

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

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

$\cos(\theta_2)=\cos(10.61^\circ)=98.3\% \text{ leading}$