(a) The operation of motor at infinite bus-bars is shown in Fig. 8.45
V_{t}=3300 / \sqrt{3}= 1905 V
I_{a}=\frac{1000 \times 1000}{\sqrt{3} \times 3300 \times 1}= 175 A
\cos \phi=1, \quad \phi=0^{\circ}
Taking the terminal voltage as reference,
\bar{V}_{t}=1905 \angle 0^{\circ} V
\bar{I}_{a}=175 \angle 0^{\circ} A
Then the excitation emf is computed as
E_{f n}=1905 \angle 0^{\circ}-j 175 \angle 0^{\circ} \times 3.24=1905-j 567
which gives E_{f m}=1987 V
Excitation remaining fixed, the maximum power delivered by the motor is
p_{e, \max }=P_{m, \max }( gross )=3 \times \frac{V_{t} E_{f}}{X_{s m}}=3 \times \frac{1905 \times 1987}{3.24 \times 1000}=3505 kW (3-phase)
\omega_{s m}=\frac{120 \times 50 \times 2 \pi}{24 \times 60}=26.18 rad / s
\therefore T_{\max }=\frac{3505 \times 1000}{26.18}=133.9 \times 10^{3} N m
(b) Figure 8.46(a) shows the generator feeding the motor. At rated terminal voltage, unity pf, full-load operation
\bar{V}_{t}=1905 \angle 0^{\circ}
\bar{I}_{a}=175 \angle 0^{\circ}
As calculated before E_{f m}=1905 V
Now \bar{E}_{f g}=1905 \angle 0^{\circ}+j 175 \angle 0^{\circ} \times 4.55
= 1905 + j 796
Or E_{f g}=2065 V
The total series reactance X=X_{s m}+X_{s g}=3.24+4.55=7.79 \Omega
The maximum power output delivered by the motor is
P_{e, \max }=P_{m, \max }(\text { gross })=3 \times \frac{E_{f g} \times E_{f m}}{X}=3 \times \frac{2065 \times 1987}{7.79 \times 1000}=1580 kW (3-phase)
T_{\max }=\frac{1580 \times 1000}{26.18}=60.35 \times 10^{3} Nm
The phasor diagram under condition of maximum power output is drawn in Fig. 8.46(b). For convenience, choosing the motor excitation as reference
At maximum power
Output \delta=90^{\circ}
\bar{E}_{f m}=1987 \angle 0^{\circ} V
\bar{E}_{f g}=2065 \angle 90^{\circ} V
Then \bar{I}_{a}=\frac{\bar{E}_{f g}-\bar{E}_{f m}}{j X}=\frac{2065 \angle 90^{\circ}-1987 \angle 0^{\circ}}{j 7.79}
j \bar{I}_{a} X_{s m}=\frac{-1987+j 2065}{j 7.79} \times j 3.24_{=}=-826.4+j 858.9
Now \bar{V}_{t}=\bar{E}_{f m}+j \bar{I}_{a} X_{s m}=1987-826.4+j 858.9=1160.6+j 858.9
Or V_{t}= 1443.85 or 2500 V (line)
Remark The reduction in P_{e,max} in case (b) compared to case (a) is explained by the fact that V_{t} in this case is only 2500 V (line) compared to 3300 V in case (a).
