Question : A 100 MW, 60 kV, 3-, 4-pole, 50Hz, Y-connected synchronous ...
P=100 M \omega \quad, V_{L_{1}}=60 k V
R_{a}=0,15 \Omega / {ph} \quad x_{s}=15 \Omega / {p} {h}
\text { Power }=75 \% \text { of } 100=75Mw at 0,8 {lag}
E=v+I2
I=\frac{75 \times 10^{6}}{\sqrt{3} \times 60 \times 10^{3} \times 0,8}=14,073 {kV} / {Ph}
E_{9}=76,54 {~L} 14,073 {kV}
\delta=\sqrt{3} E_{g} \cdot I_a^∗.
\delta= \sqrt{3} \cdot 72,993 L^{4} \cdot 78 \times 902,11 \quad\lfloor 36-86
\delta=75,37+j 92,858 mvA
Q=92,858 MVAR
T=\frac{p}{w} \quad ; w_{s}=\frac{2 \pi}{60} \times \frac{120 F}{1^{\prime}}=\frac{2 \pi}{60} \times \frac{120^{2} \times 50}{4}=50 \pi
\Rightarrow \quad T=\frac{75 \times 10^{6}}{50 \pi}=477,46 \quad k N.m
\text { if the excitation decressed by 2oy }
E g_{\text {new}}=0,8 \times 76,54 kv=61,232 {kv}=35,35 {kv} / {ph}
85,35 \times 10^{3} L 14,073=\frac{60}{\sqrt{3}}+[I_a \cos \phi+jI_a \sin \phi][0,15+j 15]
34.29+j8.595=[\frac{60}{\sqrt{3}}+(I_a cos \phi+j I_a sin \phi)]+j(15 I_a cos \phi)
I_a cos \phi=902.11 \times 0.8=721.688A+0.15 I_a sin \phi
By equating real and imaginary
8.595 \times 10^3=15 \times721.688+0.15 I_a sin \phi
9.23 kv=0.15 I_a sin \phi
I_a sin \phi=-14868.8A
\frac{I_a sin \phi}{I_a cos \phi}=\frac{14868.8}{721.688}⇒tan \phi=20.602
⇒\phi=-87.22
cos \phi=0.0485⇒I_a=\frac{721.688}{0.0485}=14886.3A
\delta=\sqrt{3}I_aI_g=\sqrt{3}\times61.232L14.073\times14886.3 L+87.22
=-309+1548.23 MvA
Q=1548.23 MvA (supplied)
T=\frac{P}{w_s}=477.46 KN.m