Question 11.9: The 45-kVA, 220-V synchronous motor of Example 11.8 is to be......

We will follow the individual steps outlined above.

Step 1. Since the motor is operating at rated speed, the desired terminal voltage will be the rated line-to-neutral voltage of the machine.

V_{\mathrm{a}}=\frac{220}{\sqrt{3}}=127 \mathrm{~V}=1.0  \text {per unit}

Step 2. Setting T_{\text {ref }}=358 \mathrm{~N} \cdot \mathrm{m} and \omega_{\mathrm{m}}=(2 /\text{poles}) \omega_{\mathrm{e}}=40 \pi, the rms armature current can be calculated from Eq. 11.42

I_{\mathrm{a}}=\frac{P_{\mathrm{ref}}}{3 V_{\mathrm{a}}}=\frac{T_{\mathrm{ref}} \omega_{\mathrm{m}}}{3 V_{\mathrm{a}}} \quad \quad \quad (11.42)

I_{\mathrm{a}}=\frac{T_{\mathrm{ref}} \omega_{\mathrm{m}}}{3\left(V_{\mathrm{a}}\right)}=\frac{358 \times(40 \pi)}{3 \times 127}=118 \mathrm{~A}

As calculated in Example 11.8, I_{\text {base }}=118 \mathrm{~A} and thus I_a = 1.0 per unit. This is as expected, since we want the motor to operate at rated torque, speed and terminal voltage, and at unity power factor.

Step 3. Next calculate δ from Eq. 11.43. This calculation requires that we determine the synchronous inductance L_s.

L_{\mathrm{s}}=\frac{X_{\mathrm{s}}}{\left(\omega_{\mathrm{e}}\right)_{\text {rated }}}=\frac{0.899}{120 \pi}=2.38  \mathrm{mH}

Thus

\begin{aligned}\delta & =-\tan ^{-1}\left(\frac{\omega_{\mathrm{e}} L_{\mathrm{s}} I_{\mathrm{a}}}{V_{\mathrm{a}}}\right) \\ \\& =-\tan ^{-1}\left(\frac{120 \pi 2.38 \times 10^{-3} \times 118}{127}\right)=-0.695  \mathrm{rad}=-39.8^{\circ}\end{aligned}

Step 4. We can now calculate the desired values of i_Q  \text{and}  i_D from Eqs. 11.44 and 11.45.

\left(i_{\mathrm{Q}}\right)_{\mathrm{ref}}=\sqrt{2} I_{\mathrm{a}} \cos \delta=128 \mathrm{~A}

and

\left(i_{\mathrm{D}}\right)_{\mathrm{ref}}=\sqrt{2} I_{\mathrm{a}} \sin \delta=-107 \mathrm{~A}

Step 5. \left(i_{\mathrm{F}}\right)_{\text {ref }} is found from Eq. 11.46

\left(i_{\mathrm{F}}\right)_{\text {ref }}=\frac{2}{3}\left(\frac{2}{\text { poles }}\right) \frac{T_{\text {ref }}}{L_{\mathrm{af}}\left(i_{\mathrm{Q}}\right)_{\mathrm{ref}}}=\frac{2}{3}\left(\frac{2}{6}\right) \frac{358}{0.168 \times 128}=3.70 \mathrm{~A}