Question 7.4: A 10-MVA three-phase Y-connected 13,800-V two-pole 60-Hz tur......
A 10-MVA three-phase Y-connected 13,800-V two-pole 60-Hz tur-bogenerator has an armature resistance of 0.06 Ω per phase and a synchronous reactance of 1.8 Ω per phase. Find the excitation voltage at no load, half load, and full load for power factors of (a) 0.8 leading, (b) unity, and (c) 0.8 lagging with rated terminal voltage.
We have
Z_{s} = 0.06 + j1.8=1.801\angle88.09~\mathrm{deg}The full-load current is obtained as
I_{a\!f}=\frac{10\times10^{6}}{\sqrt{3}(13,800)}=418.37~\mathrm{A}The terminal voltage per phase is
V_{t}={\frac{13,800}{\sqrt{3}}}=7967.434\,\mathrm{V}We thus have
E_{f}=7967.434+\sigma(418.37)(1.801)\angle(88.09\ \mathrm{deg}+\phi)where σ is the fraction of the full load considered and \phi is the phase angle.
(a) .For a power factor of 0.8 leading,
\phi=36.87\,\,\mathrm{deg}At no load, \sigma=0,
E_{f}=7967.434\,\mathrm{V}At half load, \sigma=0.5,
E_{f}=7967.434+0.5(418.37)(1.801)\angle124.961\ \mathrm{deg} \\ =7757.702\angle2.281\ \mathrm{deg}\ \mathrm{V}At full load, \sigma=1, and we calculate
E_{f}=7560.935\angle4.685\ \mathrm{deg}\ \mathrm{V}Note that |E_{f}| is decreased as loading is increased.
(b) For a unity power-factor load, \phi=0: at no load \sigma=0, E_{f}=7967.434\;\mathrm{V}. At half load, \sigma=0.5,
E_{f}=7967.434+0.5(418.37)(1.801)\angle88.09\ \mathrm{deg} \\ =7988.863\angle2.701\ {\mathrm{deg V}}At full load, we get
E_{f}=8027.935\angle5.383\mathrm{~deg V}Note that |E_{f}| is increased as loading is increased.
(c) For a power factor of 0.8 lagging, \phi=-36.87\,\mathrm{deg.} At half load, we have
E_{f}=7967.434+0.5(418.37)(1.801)\angle51.221\ \mathrm{deg} \\ =8208.65\angle2.05\;\mathrm{deg}\;\mathrm{V}At full load, we have
E_{f}=8459.772\angle3.981\mathrm{~deg V}Again, we note the increase in |E_{f}| as loading is increased.