A three-phase, star connected, 20 MVA , 11 kV , 50 Hz alternator produces a short-circuit current equal to full-load current when a field current of 70 A passes through its field winding. The same field current produces an emf of 1820 V (line to line) on open circuit. If the alternator has a

Question

A three-phase, star connected, 20 MVA , 11 kV , 50 Hz alternator produces a short-circuit current equal to full-load current when a field current of 70 A passes through its field winding. The same field current produces an emf of 1820 V (line to line) on open circuit. If the alternator has a resistance between each pair of terminals as measured by D C is 0.16 ohm and the effective resistance is 1.5 times the ohmic resistance, what will be its full-load regulation at (i) 0.707 pf lagging and (ii) 0.8 pf leading.

Accepted Answer

Here, Alternator is three-phase, star connected;

[latex] ext { Rating of alternator }=20 MVA =20 imes 10^{6} VA [/latex]

Terminal voltage (line value), [latex] V_{L}=11000 V[/latex]

Open circuit e m f (line value), [latex] E_{L}=1820 V[/latex]

[latex] ext { Resistance between two terminals }=0.16 ext { ohm } [/latex]

[latex] \qquad ext { Resistance measured/phase }=\frac{0.16}{2}=0.08 ext { ohm }[/latex]

[latex] ext { Effective resistance/phase, } R=1.5 imes 0.08=0.12 ohm [/latex]

[latex] ext { pen circuit } e m f ext { (phase value),}E=\frac{E_{L}}{\sqrt{3}}=\frac{1820}{\sqrt{3}}=1050.8 V [/latex]

[latex] ext { Full load circuit, } I=\frac{20 imes 16^{6}}{\sqrt{3} imes 11000}=1049.7 A[/latex]

[latex] ext { Short circuit current, } I_{s c}=I=1049.7 A[/latex]

[latex] ext { Synchronous impedance/phase, } Z_{s}=\frac{E}{I_{s c}}=\frac{1050.8}{1049.7}=1.001 ohm [/latex]

[latex] ext { Synchronous reactance/phase, } X_{s} =\sqrt{\left(Z_{s} ight)^{2}-(R)^{2}} [/latex]

[latex] =\sqrt{(0.001)^{2}-(0.12)^{2}}=0.994 ohm[/latex]

[latex] ext { Terminal voltage (phase value), } V=\frac{V_{L}}{\sqrt{3}}=\frac{11000}{\sqrt{3}}=6351 V[/latex]

[latex] ext { (i) When } p . f ., \cos \phi=0.707 ext { lagging; } \sin \phi=\sin \cos ^{-1} 0.707=0.707[/latex]

No-load terminal voltage (phase value),

[latex] E_{0} =\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+I X_{s} ight)^{2}} [/latex]

[latex] =\sqrt{(6351 imes 0.707+1049.7 imes 0.12)^{2}+(6351 imes 0.707+1049.7 imes 0.994)^{2}} [/latex]

[latex] =\sqrt{(4616)^{2}+(5533.5)^{2}}=7206 V[/latex]

[latex] \% ext { Reg. }=\frac{E_{0}-V}{V} imes 100=\frac{7206-6351}{6351} imes 100=13.46 \% [/latex]

(ii) When [latex] p . f . \cos \phi=0.8[/latex] leading; [latex] \sin \phi=0.6[/latex]

No-load terminal voltage (phase value),

[latex] E_{0} =\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+L X_{s} ight)^{2}}[/latex]

[latex] =\sqrt{(6351 imes 0.8+1049.7 imes 0 \cdot 12)^{2}+(6351 imes 0.6+1049.7 imes 0.994)^{2}} [/latex]

[latex] =\sqrt{(4616)^{2}+(2767)^{2}}=5381.8 V[/latex]

[latex] \% ext { Reg. } =\frac{E_{0}-V}{V} imes 100=\frac{5381.8-6351}{6351} imes 100=-15.26 \% [/latex]

Question 6.21: A three-phase, star connected, 20 MVA , 11 kV , 50 Hz altern...

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