Question 11.8: Consider again the 45-kVA, 220-V, six-pole synchronous motor......

a. We must first calculate L_{\text {af }}. From Example 11.7, we see that the motor produces rated open-circuit voltage (220-V rms, line-to-line) at a field current of 2.84 A. From Eq. 11.34

E_{\mathrm{af}}=\frac{\omega_{\mathrm{e}} L_{\mathrm{af}}i_{\mathrm{F}}}{\sqrt{2}}\quad \quad \quad (11.34)

L_{\mathrm{af}}=\frac{\sqrt{2} E_{\mathrm{af}}}{\omega_{\mathrm{e}} i_{\mathrm{F}}}

where E_{\text {af }} is the rms, line-to-neutral generated voltage. Thus

L_{\mathrm{af}}=\frac{\sqrt{2}(220 / \sqrt{3})}{120 \pi \times 2.84}=0.168  \mathrm{H}

Rated torque for this six-pole motor is equal to

\begin{aligned}T_{\text {rated }} & =\frac{P_{\text {rated }}}{\left(\omega_{\mathrm{m}}\right)_{\text {rated }}}=\frac{P_{\text {rated }}}{\left(\omega_{\mathrm{e}}\right)_{\text {rated }}(2 / \text { poles })} \\ \\& =\frac{45 \times 10^{3}}{120 \pi(2 / 6)}=358 \mathrm{~N} \cdot \mathrm{m}\end{aligned}

Thus, setting T_{\text {ref }}=T_{\text {rated }}=358 \mathrm{~N} \cdot \mathrm{m} and \left(i_{\mathrm{F}}\right)_{\text {ref }}=2.84 \mathrm{~A}, we can find i_Q from Eq. 11.40 as

(i_{\mathrm{Q}})_{\mathrm{ref}}=\frac{2}{3}\left(\frac{2}{\text { poles }}\right) \frac{T_{\text {ref }}}{L_{\mathrm{af}}\left(i_{\mathrm{F}}\right)_{\mathrm{ref}}} \quad \quad \quad (11.40)

i_{\mathrm{Q}}=\frac{2}{3}\left(\frac{2}{\text { poles }}\right) \frac{T_{\text {ref }}}{L_{\mathrm{af}}\left(i_{\mathrm{F}}\right)_{\mathrm{ref}}}=\frac{2}{3}\left(\frac{2}{6}\right) \frac{358}{0.168 \times 2.84}=167 \mathrm{~A}

Using the fact that \theta_{\mathrm{me}}=\omega_{\mathrm{e}} t and setting i_D = 0, the transformation from dq0 variables to abc variables (Eq. C.2 of Appendix C) gives

\left[\begin{array}{l}S_{\mathrm{a}} \\S_{\mathrm{b}} \\S_{\mathrm{c}}\end{array}\right]=\left[\begin{array}{ccc}\cos \left(\theta_{\mathrm{me}}\right) & -\sin \left(\theta_{\mathrm{me}}\right) & 1 \\\cos \left(\theta_{\mathrm{me}}-120^{\circ}\right) & -\sin \left(\theta_{\mathrm{me}}-120^{\circ}\right) & 1 \\\cos \left(\theta_{\mathrm{me}}+120^{\circ}\right) & -\sin \left(\theta_{\mathrm{me}}+120^{\circ}\right) & 1\end{array}\right]\left[\begin{array}{c}S_{\mathrm{d}} \\S_{\mathrm{q}} \\S_{0}\end{array}\right] \quad \quad (C.2)

i_{\mathrm{a}}(t)=i_{\mathrm{D}} \cos \left(\omega_{\mathrm{e}} t\right)-i_{\mathrm{Q}} \sin \left(\omega_{\mathrm{e}} t\right)=-167 \sin \left(\omega_{\mathrm{e}} t\right)=-\sqrt{2}(118) \sin \left(\omega_{\mathrm{e}} t\right)  \mathrm{A}

where \omega_{\mathrm{e}}=120 \pi \approx 377  \mathrm{rad} / \mathrm{sec}. Similarly

i_{\mathrm{b}}(t)=-\sqrt{2}(118) \sin \left(\omega_{\mathrm{e}} t-120^{\circ}\right)  \mathrm{A}

and

i_{\mathrm{c}}(t)=-\sqrt{2}(118) \sin \left(\omega_{\mathrm{e}} t+120^{\circ}\right)  \mathrm{A}

The rms armature current is 118 A and the machine base current is equal to

I_{\text {base }}=\frac{P_{\text {base }}}{\sqrt{3} V_{\text {base }}}=\frac{45 \times 10^{3}}{\sqrt{3}  220}=118 \mathrm{~A}

Thus the per-unit machine terminal current is equal to I_{\mathrm{a}}=118 / 118=1.0 per unit.

b. The motor terminal voltage can most easily be found from the rms phasor relationship

\hat{V}_{\mathrm{a}}=j X_{\mathrm{s}} \hat{I}_{\mathrm{a}}+\hat{E}_{\mathrm{af}}

In part (a) we found that i_{\mathrm{a}}=-\sqrt{2}(118) \sin \left(\omega_{\mathrm{e}} t\right)  \mathrm{A} and thus

\hat{I}_{\mathrm{a}}=j 118 \mathrm{~A}

We can find E_{\text {af }} from Eq. 11.34 as

E_{\mathrm{af}}=\frac{\omega_{\mathrm{e}} L_{\mathrm{af}} i_{\mathrm{F}}}{\sqrt{2}}=\frac{120 \pi \times 0.168 \times 2.84}{\sqrt{2}}=127 \mathrm{~V}  \text { line-to-neutral}

and, thus, since \left(\hat{E}_{\mathrm{af}}\right)_{\mathrm{rms}} lies along the quadrature axis, as does \hat{I}_a,

\hat{E}_{\mathrm{af}}=j 127 \mathrm{~V}

The base impedance of the machine is equal to

Z_{\text {base }}=\frac{V_{\text {base }}^{2}}{P_{\text {base }}}=\frac{220^{2}}{45 \times 10^{3}}=1.076  \Omega

and the synchronous reactance of 0.836 pu is equal to X_{\mathrm{s}}=0.836 \times 1.076=0.899  \Omega. Thus the rms line-to-neutral terminal voltage

\begin{aligned}\hat{V}_{\mathrm{a}} & =j X_{\mathrm{s}} \hat{I}_{\mathrm{a}}+\hat{E}_{\mathrm{af}}=j 0.899(j 118)+j 127 \\& =-106+j 127=165 \angle 129.9^{\circ}  \mathrm{V} \text { line-to-neutral }\end{aligned}

or V_{\mathrm{a}}=287 \mathrm{~V} rms line-to-line = 1.30 per unit.

Note that the terminal voltage for this operating condition is considerably in excess of the rated voltage of this machine, and hence such operation would not be acceptable. As we shall now discuss, a control algorithm which takes advantage of the full capability to vary i_F, i_D, \text{and}  i_Q can achieve rated torque while not exceeding rated terminal voltage.