(a) X_{s}(\text { adjusted })=\left.\frac{V_{ rated } / \sqrt{3}}{I_{S C}}\right|_{\text {At } I_{f} \text { corresponding to } V_{O C}=V_{\text {rated }}}
Rated armature current
\sqrt{3} V_{\text {rated }} I_{a}(\text { rated })=10 MVA
I_{a}(\text { rated })=\frac{10 \times 10^{3}}{\sqrt{3} \times 13.8}=418.4 A
At I_{f}=842 A , \quad I_{S C}=\frac{418.4}{226} \times 842=1558.8 A
The X_{s} \text { (adjusted) }=\frac{13.8 \times 10^{3} / \sqrt{3}}{1558.8}=5 \Omega
Base values ( MVA )_{ B }=10,( kV )_{ B }=13.8
X_{s}( pu )=5 \times \frac{10}{(13.8)^{2}}=0.2625
R_{a}=0.75 \Omega
\bar{Z}_{s}=0.75+j 5=5.06 \angle 81.5^{\circ} \Omega
Z_{s}=5.06 \Omega, \quad \theta=81.5^{\circ}, \quad \alpha=8.5^{\circ}
(b) (i) pf = 0.9 lagging ; \phi=\cos ^{-1} 0.9=25.84^{\circ}
P_{e}=8.75 MW
\tan \phi=\frac{Q_{e}}{P_{e}}
Q_{e}=P_{e} \tan \phi=8.75 \tan 25.84^{\circ}
= 4.24 MVA, positive as it a lagging.
To find the field current, we need excitation emf. On per phase basis
V_{t}=13.8 / \sqrt{3}=7968 V
\frac{P_{e}}{3}=V_{t} I_{a} \cos \phi
\frac{8.75 \times 10^{6}}{3}=7968 I_{a} \times 0.9
Or I_{a}=406.7 A , \quad \bar{I}_{a}=406.7 \angle-25.8^{\circ}
=2058 \angle 55.7^{\circ} V
Generator equation \bar{E}_{f}=\bar{V}_{t}+\bar{I}_{a} \bar{Z}_{s}
Or \bar{E}_{f}=7968 \angle 0^{\circ}+2058 \angle 55.7^{\circ}
=9128+j 1700=9285 \angle 10.5^{\circ} V
The phasor diagram is drawn in Fig. 8.70
E_{f}=9285 or 16.04 kV (line)
From the modified air-gap line
I_{f}=\frac{842}{13.8} \times 16.04=978.7 A
Ohmic loss
3 I_{a}^{2} R_{a}=3 \times(406.7)^{2} \times 0.75
= 0.372 MW
P_{m}(\text { in })=8.75+0.372=9.122 MW
(ii) From Eq. (8.83)
P_{m}(\text { in })=\frac{E_{f}^{2}}{Z_{s}^{2}} R_{a}+\frac{V_{t} E_{f}}{Z_{s}} \sin (\delta-\alpha) (i)
Field current adjusted to
I_{f}=842 A
E_{f}=V_{O C}=V_{t} \text { (rated) }=7968 V =7.968 kV
P_{m} \text { (in) }=9.122 / 3=3.04 MW per phase
Substituting in Eq. (i)
3.04=\frac{(7968)^{2}}{(5.06)^{2}} \times 0.75+\frac{(7.968)^{2}}{(5.06)} \sin (\delta-\alpha)
118=\frac{(7968)^{2}}{(5.06)^{2}} \sin (\delta-\alpha)
\sin (\delta-\alpha)=0.094
\delta-\alpha=5.4^{\circ}
\delta=5.4^{\circ}+8.5^{\circ}=13.9^{\circ}
Reactive power output (Eq. 8.82b)
Q=-\frac{V_{t}^{2}}{Z_{s}^{2}} X_{s}+\frac{V_{t} E_{f}}{Z_{s}} \cos (\delta+\alpha)
Or Q=-\frac{(7.968)^{2}}{(5.06)^{2}} \times 5+\frac{(7.968)^{2}}{(5.06)} \cos \left(13.9^{\circ}+8.5^{\circ}\right)
Or Q=-0.82 MVAR / phase
Or Q=2.46 \text { MVAR, leading }
