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 $0.85\Omega /phase$. When alternator operating at rated condition and unity power factor, the synchronizing torque will be

• A. 2133.5 Nm
• B. 5215 Nm
• C. 6400.5 Nm
• D. 1738.33 Nm

Answer 1

The correct option is C 6400.5 Nm

$I_a=\frac{6\times {10}^6}{\sqrt{3}\times 3.3\times {10}^3}=1049.73A$

$E_f=V_t+j{I}_a{X}_s$

$=\frac{3.3\times {10}^3}{\sqrt{3}}+j\left(1049.73\right)\left(0.85\right)$

$E_f=2103.84\angle {25.1}^{\circ }V$

$\text{Synchronizing power}=\frac{E_{f}V}{X_s}cos\delta$

$=\frac{3\times 2103.84\times 1905.3}{0.85}cos\left({25.1}^{\circ }\right)$

$=3\times 4.267MW/elect.rad$

$=3\times 4.267\times \frac{P}{2}\times \frac{\pi }{180}MW/mech.degree$

$T_P=Torque=3\times 4.267\times \frac{6}{2}\times \frac{\pi }{180}\times \frac{1\times 60}{2\pi \times 1000}$

$=3\times 2133.5Nm$

$=6400.5Nm$