Question

The induced emf in a 3.3 kV, 4-pole, 3-phase star-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of \( 2^\circ \) mechanical with respect to the terminal voltage phasor. If the synchronous reactance is \( 2 \, \Omega \), and stator resistance is negligible, then the motor armature current magnitude, in amperes, during loaded condition is closest to: \[ {(round off to two decimal places).} \]

Hint:

In a synchronous machine with negligible resistance, armature current during load can be found using \( I = \frac{|V - E|}{X_s} \), where the angle between \( V \) and \( E \) determines the magnitude of the voltage difference.

Answer 1

Correct Answer: 66.25

Given: Line voltage \( V_L = 3.3 \, {kV} \) Phase voltage \( V = \frac{V_L}{\sqrt{3}} = \frac{3300}{\sqrt{3}} \approx 1905.26 \, {V} \) Induced emf \( E \) is in phase with \( V \) at no-load, and makes an angle \( \delta = 2^\circ \) mechanical with \( V \) when loaded. Electrical angle \( \delta_{{elec}} = 2 \times \frac{{number of poles}}{2} = 4^\circ \) Synchronous reactance \( X_s = 2 \, \Omega \) Stator resistance is negligible Since \( E \) and \( V \) differ by \( \delta_{{elec}} = 4^\circ \), we calculate the current: \[ I = \frac{V - E}{jX_s} \] Using phasor difference magnitude: \[ |I| = \frac{|V - E|}{X_s} \] Assuming \( |V| = |E| \), and angle between them is \( 4^\circ \), use law of cosines: \[ |V - E| = \sqrt{V^2 + E^2 - 2VE\cos(4^\circ)} = \sqrt{2V^2(1 - \cos(4^\circ))} \] \[ \Rightarrow |V - E| = V \sqrt{2(1 - \cos(4^\circ))} = 1905.26 \cdot \sqrt{2(1 - \cos(4^\circ))} \] \[ \cos(4^\circ) \approx 0.99756 \Rightarrow 1 - \cos(4^\circ) \approx 0.00244 \] \[ |V - E| \approx 1905.26 \cdot \sqrt{2 \cdot 0.00244} \approx 1905.26 \cdot \sqrt{0.00488} \approx 1905.26 \cdot 0.06984 \approx 133.5 \, {V} \] Now, \[ |I| = \frac{133.5}{2} \approx 66.75 \, {A} \]