Question 7.3: A 10-MVA 13.8-kV 60-Hz two-pole V-connected three-phase alte......
A 10-MVA 13.8-kV 60-Hz two-pole Y-connected three-phase alter-nator has an armature-winding resistance of 0.07 Ω per phase and a leakage reactance of 1.9 Ω per phase. The armature-reaction EMF for the machine is related to the armature current by
E_{a r}=-j19.91I_{a}Assume that the generated EMF is related to the field current by
E_{f}=60l_{f}(a) Compute the field current required to establish the rated voltage across the terminals of a load when the rated armature current is delivered at 0.8 PF lagging.
(b) Compute the field current needed to provide the rated terminal voltage to a load that draws 100% of the rated current at 0.85 PF lagging.
The terminal voltage (line to neutral) is given by
V_{t}={\frac{13,800}{\sqrt{3}}}=7967.43\mathrm{{\scriptsize~V}}(a) The rated armature current is obtained as
I_{a}=\frac{10\times10^{6}}{\sqrt{3}(13,800)}=418.37{\mathrm{~A}}From the equivalent circuit of Fig. 7.12,
E_{r}=V_{t}+I_{a}(R_{a}+j X_{1}) \\ \equiv7967.43+(418.37\angle{\mathrm{-cos}}^{-1}0.8)(0.074{\mathrm{+j1.9}})Thus
E_{f}=8490.35\angle4.18~ deg{V}From the specifications,
E_{a r}=-j(19.91)(418.37\angle-\cos^{-1}0.8~\mathrm{deg})Thus
{E}_{a r}=-8329.75\angle53.13~\mathrm{deg}\,\mathrm{V}Now, the field excitation is
E_{f}=E_{r}-E_{a r}Thus we obtain
E_{f}=8490.35\angle4.18~\mathrm{deg}+8329.75\angle53.13~\mathrm{deg} \\ =15,308.61\angle28.40~{\mathrm{deg}}~\mathrm{V}From the given relation, we obtain the required field current:
|I_{f}|={\frac{15,308.61}{60}}=255.14{\mathrm{~A}}(b) In this case we have
I_{a}=(418.37)(1.0)\angle-\mathrm{cos}^{-1}0.85\ \mathrm{deg} \\ =418.37\angle-31.79\;\mathrm{deg}\;\mathrm{A}As a result,
E_{r}=7967.43+(418.37\angle-31.79\;\mathrm{deg})(0.07+j1.9) \\ =8436.94\angle4.49~\mathrm{deg}~\mathrm{V}We now calculate
E_{a r}=-j(19.91)(418.37\angle-31.79\;\mathrm{deg}) \\ =-8329.74\angle58.21\ \mathrm{deg}\ \mathrm{V}As a result,
E_{f}=E r-E_{a r} \\ =8436.94\angle4.48~\mathrm{deg}+8329.74\angle58.21~\mathrm{deg} \\ =14,957.72\angle31.16\,\,\mathrm{deg}\,\mathrm{V}The field current is thus obtained as
\left|I_{f}\right|={\frac{E_{f}}{60}}=249.30~\mathrm{A}