Changing the line voltages into phase voltages, we have,
\frac{1500}{\sqrt{3}}, \frac{2200}{\sqrt{3}}, \frac{2700}{\sqrt{3}}, \frac{3200}{\sqrt{3}}, \frac{3550}{\sqrt{3}}, \frac{3800}{\sqrt{3}}, \frac{4000}{\sqrt{3}}, \frac{4100}{\sqrt{3},}, \frac{4150}{\sqrt{3}}
866,1270,1560,1848,2050,2190,2310,2370,2400
Plot the O C C curve by taking field A T_{S} along the x -axis and terminal phase voltages along the y -axis as shown in Fig. 6.61.
Full load current, \quad I_{f l}=\frac{k V A \times 1000}{\sqrt{3} V_{L}}=\frac{1500 \times 1000}{\sqrt{3} \times 3300}=263 A
Resistance between terminal =0.5 \Omega.
In star connected windings, the resistance per phase =\frac{0.5}{2}=0.25 \Omega.
Effective resistance per phase, R=1.4 \times 0.25=0.35 \Omega
Leakage reactance per phase,
X_{L}=1.5 \Omega
The phasor diagram is shown in Fig. 6.62. for finding the voltage E. The terminal voltage V is taken as reference phasor.
Rated phase voltage, V=\frac{3.3 \times 1000}{\sqrt{3}}=1905 V
\begin{aligned}E &=\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+I X_{s}\right)^{2}} \\&=\sqrt{\left(1905 \times 0.8+263 \times(0.35)^{2}+(1905 \times 0.6+263 \times 1.5)^{2}\right.}\end{aligned}
=\sqrt{(1616)^{2}+(1537.5)^{2}}=2230 V
Corresponding to 2230 volt, the field ampere turns from O C C curve is 7150 ATs.
Armature reaction A T s per pole per phase are =1.4 \times A T_{S} per pole per phase
=1.4 \times \frac{I \times \text { Turns per phase }}{\text { No. of poles }}=1.4 \times \frac{263 \times 55}{8}=2525 ATs
The field A T s as obtained from the O C C curve corresponding to normal open circuit voltage are called no-load ATs. These ATs (oa) are drawn at right angle to the voltage vector O E. The armature reaction ATs per pole per phase (2525) is drawn parallel opposition to load current 1 . i.e., a b is drawn parallel opposition to I. The resultant of o a, and a b is given by o b. The angle between o a and a b is (90+\phi)
\begin{aligned}\therefore \quad o b &=\sqrt{o a^{2}+a b^{2}-2 o a \times a b \times \cos \left(90+36.87^{\prime \prime}\right)} \\&=\sqrt{(7150)^{2}+(2525)^{2}+2 \times 7150 \times 2525 \times \cos \left(53.13^{\prime \prime}\right)} \\&=\sqrt{(7150)^{2}+(2525)^{2}+2 \times 7150 \times 2525 \times 0.6}=8900 ATs\end{aligned}
Corresponding to 8900 ATs, the open circuit voltage E_{o} is 2380 from the O C C curve. Hence when the load is switched off the open circuit voltage is 2380 volt. Now the speed has increased from 760 to 770 rpm but we need to have open circuit voltage corresponding to 760 rpm, therefore E_{0} corresponding to 760 rpm,
E_{o} =\frac{2380 \times 760}{770}=2350 V
\% \text { age Regulation } =\frac{E_{o}-V}{V} \times 100=\frac{2350-1905}{1905} \times 100= 2 3 . 4 \%
