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Question

# A 6 MVA, 6 pole, 50 Hz, 3 phase alternator is connected to an infinite bus of 3.3 kV and runs at 1000 rpm. The synchronous reactance of the machine is 0.85Ω/phase. When alternator operating at rated condition and unity power factor, the synchronizing torque will be

A
2133.5 Nm
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B
5215 Nm
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C
6400.5 Nm
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D
1738.33 Nm
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Solution

## The correct option is C 6400.5 NmIa=6×106√3×3.3×103=1049.73A Ef=Vt+jIaXs =3.3×103√3+j(1049.73)(0.85) Ef=2103.84∠25.1∘V Synchronizing power =EfVXscosδ =3×2103.84×1905.30.85cos(25.1∘) =3×4.267MW/elect.rad =3×4.267×P2×π180MW/mech.degree TP=Torque=3×4.267×62×π180×1×602π×1000 =3×2133.5 Nm =6400.5 Nm

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