The O C C and S C C are shown in Fig. 6.63. The open circuit phase voltage in volt are
\frac{800}{\sqrt{3}}, \frac{1143}{\sqrt{3}}, \frac{1500}{\sqrt{3}}, \frac{1760}{\sqrt{3}}, \frac{2000}{\sqrt{3}}, \frac{2173}{\sqrt{3}}, \frac{2350}{\sqrt{3}}, \frac{2476}{\sqrt{3}}, \frac{2600}{\sqrt{3}}
or =462,660,866,1016,1155,\\1255,1357,1430,1501
Full-load phase voltage
V=\frac{2000}{\sqrt{3}}=1155 V
Full load line current
\quad I_{L} =\frac{k V A \times 1000}{\sqrt{3} \times V_{L}} \\ =\frac{1000 \times 1000}{\sqrt{3} \times 2000}=288.7 A
Full-load phase current,
\quad I=I_{L}=288.7
(a) At power factor 0.8 lagging (neglecting leakage reactance since not given)
E =\bar{V}+\bar{I} \bar{R}=1155+\left(288.7 \angle-\cos ^{-1} 0.8\right) \times 0.2
=1155+(57.74 \times 0.8-j 57.74 \times 0.6)
=1155+46.2-j 34.64
=1201.2-j 34.64=1201.7 \angle-1.65^{\circ} V
Here
\delta=-1.65^{\circ} V
From the O C C, the field current required to produce the voltage of 1201.7 V is 32 A. Therefore o a=I_{f_{1}}=32 A. This current leads the voltage vector O E by 90^{\circ} or leads the terminal voltage vector O V by \left(90-\delta=90-1.65^{\circ}=88.35^{\circ}\right) 88.35^{\circ}.
\bar{I}_{f_{1}}=I_{f_{1}} \angle 90^{\circ}-1.65^{\circ}=32 \angle 88.35^{\circ}=(0.92+i 31.98) A
From the S C C, the field current required to produce full-load current of 288.7 A is 29 A. Therefore o b=I_{f_{2}}=29 A . For \cos \phi=0.8, \phi=36.87^{\circ}
From the phasor diagram shown in Fig. 6.64.
\begin{aligned}\bar{I}_{f_{2}} &=I_{f_{2}} \angle 180^{\circ}-\phi=29 \angle 180^{\circ}-36.87^{\circ} \\&=28^{\circ} \angle 143.13^{\circ} A =-23.2+j 17.4 \\\bar{I}_{f} &=\bar{I}_{f_{2}}+\bar{I}_{f_{1}}=-23.2+j 17.4+0.92+j 31.98 \\&=-22.28+j 49.38=54.18 \angle 114.3^{\circ} A\end{aligned}
From the O C C, the open circuit phase voltage corresponding to the field current of 54.18 A is 1555 V
\therefore Percentage voltage regulation =\frac{E_{o}-V}{V} \times 100=\frac{1555-1155}{1155} \times 100= 3 4 . 6 3 \%
(b) At power factor 0.8 leading
\bar{E} =\bar{V}+\bar{I} R
=1155+\left(288.7 \angle+\cos ^{-1} 0.8\right) \times 0.2
=1155+46.2+j 34.64
=1201.2+j 34.64=1201.7 \angle+1.65^{\circ} V
From the phasor diagram shown in Fig. 6.65.
\begin{aligned}I_{f_{1}} &=I_{f_{1}} \angle 90^{\circ}+\delta=32 \angle 90^{\circ}+1.65=32 \angle 91.65^{\circ} A \\&=-0.92+j 31.98 A \\I_{f_{2}} &=I_{f_{2}} \angle 180^{\circ}+\phi \\&=29 \angle 180^{\circ}+36.87^{\circ}=29 \angle 216.87^{\circ} A =-23.2-j 17.4 A \\\bar{I}_{f} &=\bar{I}_{f_{2}}+\bar{I}_{f_{1}}=-0.92+j 31.98-23.2-j 17.4 \\&=-24.12+j 14.58 A =28.18 \angle 31.15^{\circ} A\end{aligned}
\begin{array}{l}\text { From the O.C.C., the open-circuit phase voltage corresponding to a field current of } 28.18 A \text { is }\\1120 V \text { . }\\\text { Percentage voltage regulation }=\frac{E_{o}-V}{V} \times 100=\frac{1120-1155}{1155} \times 100=-3.03 \% \end{array}
