Determine the voltage regulation by zero power factor method of a 500 kVA , 6600 V, three-phase, star-connected alternator having a resistance of 0.075 ohm per phase, when delivering a current of 500 A at power factor (i) 0.8 lagging (ii) 0.707 leading and (iii) unity. The alternator has the following open circuit and full-load zero power factor curves:
\begin{array}{lccccccc}\text { Field current in A: } & 24 & 32 & 50 & 75 & 100 & 125 & 150 \\\text { Open circuit terminal voltage in } V: & 1400 & - & 4500 & 6400 & 7500 & 8100 & 8400 \\\text { Saturated curve, zero pf in } V: & 0 & 0 & 1900 & 4200 & 5750 & 6750 & 7100\end{array}Question 6.30: Determine the voltage regulation by zero power factor method...
The O C C and Z P F C are plotted as shown in Fig. 6.70 . At rated terminal voltage of \frac{6600}{\sqrt{3}} V, draw a horizontal line at B. Take B C=O A=32 A. This field current O A is the field current required to circulate full-load current on short circuit. Draw a line C D parallel to O G (the initial slope of O C C ) to meet O C C at D. From point D draw a perpendicular D F on the line B C. Here B C D is the Potier’s triangle. From Potier’s triangle, Field current required to overcome armature reaction on
\operatorname{load}=F B=26 A\text { and } \quad F D=\frac{800}{\sqrt{3}}=462 V
\begin{array}{l}\text { Where } F D \text { represents voltage drop in leakage reactance at full-load current of } 500 A \text { (given) }\\\begin{array}{ll}\text { Now, } & I X_{L}=462 \\\therefore & X_{L}=\frac{462}{500}=0.924 \Omega\end{array}\end{array}
Draw the phasor diagram, as shown in Fig. 6.71, where,
O V=\text { terminal phase voltage, } V=\frac{6600}{\sqrt{3}}=3810 V
(i) When p f . \cos \phi=0.8 lagging; \sin \phi=\sin \cos ^{-1} 0.8=0.6
O E=E=\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+1 X_{L}\right)^{2}}
=\sqrt{(3810 \times 0.8+500 \times 0.075)^{2}+(3510 \times 0.6+500 \times 0.924)^{2}}
=\sqrt{(3085)^{2}+(2748)^{2}}=4131 V
\text { From } O C C \text { , the field current corresponding to } 4131 V \text { (i.e., } \frac{7156}{\sqrt{3}} V \text { ) }
o a=I_{f_{1}}=92 A \text { (it leads vector } O E \text { by } 90^{\circ} \text { ) }
Field current corresponding to armature reaction
\left.a b=I_{f_{2}}=B F=23 A \quad \text { (it is parallel to load current } O I\right)
Total field current I_{f}=O b=\sqrt{o a^{2}+a b^{2}-2 o a \times a b \times \cos \left(90^{\circ}+36.87^{\circ}\right)}
=\sqrt{(92)^{2}+(23)^{2}-2 \times 92 \times 23 \cos 126.87^{\circ}}
=\sqrt{8464+529+2539}=107.4 A
Corresponding to this field current of 107.4 A, the terminal voltage from O C C is 7700 V (line value)
E_{0} =\frac{7200}{\sqrt{3}}=4446 V \text { (phase value) }
\% \operatorname{Reg} =\frac{E_{0}-V}{V} \times 100=\frac{4446-3810}{3810} \times 100=16.69 \%
\text { (ii) When } p.f., \cos \phi_{1}=0.707 \text { leading; sin } \phi_{1}=\sin \cos ^{-1} 0.707=0.707 ;\phi_{1}=45^{\circ} \text { leading. }
\text { The phasor diagram is shown in Fig. } 6.72.
\begin{array}{c}O E^{\prime}=E^{\prime}=\sqrt{\left(V \cos \phi_{1}+I R\right)^{2}+\left(V \sin \phi_{2}-L X_{L}\right)^{2}} \\=\sqrt{(3810 \times 0.707+500 \times 0.075)^{2}+(3810 \times 0.707-500 \times 0.924)^{2}} \\=\sqrt{(2731)^{2}+(2232)^{2}}=3527 V\end{array}
\text { From } O C C \text { , the field current corresponding to } 3527 V \left(\text { i.e. } \frac{6109}{\sqrt{3}} V \right)
o a^{\prime}=I_{f_{1}}^{\prime}=72 A
Field current corresponding to armature reaction
a^{\prime} b^{\prime}=I_{f_{2}}^{\prime}=B E=23 A \text { (parallel to load current } O I^{\prime} \text { ) }
Total field current,
\begin{aligned}I_{f}^{\prime} &=o b^{\prime}=\sqrt{\left(o a^{\prime}\right)^{2}+\left(a^{\prime} b^{\prime}\right)^{2}-2 o a^{\prime} \times a^{\prime} b^{\prime} \times \cos \left(90^{\circ}-45^{\circ}\right)} \\&=\sqrt{(92)^{2}+(23)^{2}-2 \times 92 \times 23 \times \cos 45^{\circ}} \\&=\sqrt{(8464+529-1496}=86.6 A\end{aligned}
Corresponding to this field current of 86.6 A, the terminal voltage from O C C is \frac{5000}{\sqrt{3}}=2887 V
\% R_{e g}=\frac{B_{0}^{\prime}-V}{V} \times 100=\frac{2887-3810}{3810} \times 100
=24.23 \%
(iii) When p f, \cos \phi_{2}=1 ; \sin \phi_{2}=\sin \cos ^{-1} 1=0 ; \phi_{2}=0^{\circ}
Draw the phasor diagram as shown in Fig. 6.73.
O E^{\prime \prime} =E^{\prime \prime}=\sqrt{(V+I R)^{2}+\left(I X_{L}\right)^{2}}
=\sqrt{(3810+500 \times 0.075)^{2}+(500 \times 0.924)^{2}}
=\sqrt{(3847)^{2}+(462)^{2}}=3875 V
\text { From } O C C \text { , the field current corresponding to } 3875 V \left(\text { i.e. } \frac{6712}{\sqrt{3}} V \right)
o a^{\prime \prime}=I_{f_{1}}^{\prime \prime}=82 A
Field current corresponding to armature reaction
a^{\prime \prime} b^{\prime \prime}=I_{f_{2}}^{\prime \prime}=B F=23 A
(parallel to load current O I^{\prime \prime} i.e., in phase with \left.O V\right)
Total field current,
I_{f}^{\prime \prime}=o b^{\prime \prime}=\sqrt{\left(o a^{\prime \prime}\right)^{2}+\left(a^{\prime \prime} b^{\prime \prime}\right)^{2}-2 o a^{\prime \prime} \times a^{\prime \prime} b^{\prime \prime} \times \cos (90 \pm 0)}
=\sqrt{(80)^{2}+(23)^{2}-0}=85.2 A
Corresponding to this current of 85.2 A, the terminal voltage from O C C is \frac{6750}{\sqrt{3}}=3897 V
\% \operatorname{Reg}=\frac{E_{0}^{\prime \prime}-V}{V} \times 100=\frac{3897-3810}{3810} \times 100=2.26 \%
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