A 1000 kVA, 6.6 kV, 50 Hz, Y-connected synchronous generator has a no-load voltage of 11.4 kV at a certain field current. The generator gives rated terminal voltage at full load 0.75 lagging power factor at the same field current. Calculate:
(a) The synchronous reactance (armature resistance being negligible);
(b) the voltage regulation;
(c) the torque angle;
(d) the electrical power developed; and (e) the voltage and kVA rating, if the generator is reconnected in delta.
\phi=-\cos ^{-1} 0.75=-41.4^{\circ}
\bar{V}_{t}=6.6 / \sqrt{3} \angle 0^{\circ}=3.81 \angle 0^{\circ} kV
E_{f}=V_{o c}=11.4 / \sqrt{3}=6.58 kV
\bar{E}_{f}=6.58 \angle \delta kV
Refer to circuit diagram of Fig. 8.111 from which we can write
6.58 \angle \delta^{\circ}=3.81 \angle 0^{\circ}+j X_{ s } 0.0875 \angle-41.4^{\circ} (i)
Equating real and imaginary parts of this equation yields
6.58 \cos \delta=3.81+0.058 X_{s} (ii)
6.58 \sin \delta=0.0656 X_{s} (iii)
Eliminating X_{s} between Eqs (ii) and (iii), we get
6.58 \cos \delta-5.82 \sin \delta=3.81 (iv)
Or \cos \left(\delta+41.4^{\circ}\right)=0.434 \Rightarrow \delta=22.8^{\circ}
(a) From Eq. (iii) using the value of \delta as obtained above
X_{s}=\frac{6.58 \sin 22.8^{\circ}}{0.0656}=38.9 \Omega(b) Voltage regulation =\frac{11.4-6.6}{6.6} \times 100=72.7 \%
(c) Torque (power) angle \delta=22.8^{\circ}
(d) Electrical power developed =3 E_{f} I_{a} \cos (\delta-\phi)
=3 \times 6.58 \times 0.0875 \cos \left(22.4^{\circ}+41.4^{\circ}\right)=763 kWAlso directly
electrical power developed =3 V_{a} I_{a} \cos \phi=3 \times 3.81 \times 0.0875 \times 0.75=750 kW
This checks the result as there are no losses in armature resistance \left(R_{a} \approx 0\right)
(e) Machine reconnected in delta
Voltage rating =6.6 / \sqrt{3}=3.81 kV
Current rating =0.0875 \sqrt{3} kA
kVA \text { rating }=\sqrt{3} \times 3.81 \times 0.0875 \sqrt{3} \times 1000=1000 kVA (same as in star connection)
