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.
Question 6.21: A three-phase, star connected, 20 MVA , 11 kV , 50 Hz altern...
Here, Alternator is three-phase, star connected;
\text { Rating of alternator }=20 MVA =20 \times 10^{6} VA
Terminal voltage (line value), V_{L}=11000 V
Open circuit e m f (line value), E_{L}=1820 V
\text { Resistance between two terminals }=0.16 \text { ohm }
\qquad \text { Resistance measured/phase }=\frac{0.16}{2}=0.08 \text { ohm }
\text { Effective resistance/phase, } R=1.5 \times 0.08=0.12 ohm
\text { pen circuit } e m f \text { (phase value),}E=\frac{E_{L}}{\sqrt{3}}=\frac{1820}{\sqrt{3}}=1050.8 V
\text { Full load circuit, } I=\frac{20 \times 16^{6}}{\sqrt{3} \times 11000}=1049.7 A
\text { Short circuit current, } I_{s c}=I=1049.7 A
\text { Synchronous impedance/phase, } Z_{s}=\frac{E}{I_{s c}}=\frac{1050.8}{1049.7}=1.001 ohm
\text { Synchronous reactance/phase, } X_{s} =\sqrt{\left(Z_{s}\right)^{2}-(R)^{2}}
=\sqrt{(0.001)^{2}-(0.12)^{2}}=0.994 ohm
\text { Terminal voltage (phase value), } V=\frac{V_{L}}{\sqrt{3}}=\frac{11000}{\sqrt{3}}=6351 V
\text { (i) When } p . f ., \cos \phi=0.707 \text { lagging; } \sin \phi=\sin \cos ^{-1} 0.707=0.707
No-load terminal voltage (phase value),
E_{0} =\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+I X_{s}\right)^{2}}
=\sqrt{(6351 \times 0.707+1049.7 \times 0.12)^{2}+(6351 \times 0.707+1049.7 \times 0.994)^{2}}
=\sqrt{(4616)^{2}+(5533.5)^{2}}=7206 V
\% \text { Reg. }=\frac{E_{0}-V}{V} \times 100=\frac{7206-6351}{6351} \times 100=13.46 \%
(ii) When p . f . \cos \phi=0.8 leading; \sin \phi=0.6
No-load terminal voltage (phase value),
E_{0} =\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+L X_{s}\right)^{2}}
=\sqrt{(6351 \times 0.8+1049.7 \times 0 \cdot 12)^{2}+(6351 \times 0.6+1049.7 \times 0.994)^{2}}
=\sqrt{(4616)^{2}+(2767)^{2}}=5381.8 V
\% \text { Reg. } =\frac{E_{0}-V}{V} \times 100=\frac{5381.8-6351}{6351} \times 100=-15.26 \%
Related Answered Questions
Here, Rated power = 50 kVA = 50 \times 10...
From the given data, draw O C C between phase volt...
\begin{array}{l}\text { Here, Rating }=10 k...
\text { Here, } \quad \text { Rated power }...
\text { Converting three-phase terminal lin...
Reducing all line voltages to phase voltage, for a...
Changing the line voltages into phase voltages, we...
The O C C and S C C are shown in Fig. 6.63. The op...
The O C C and Z P F C are plotted as shown in Fig....