GATE 2023 EE – Question 47
A three-phase synchronous motor with synchronous impedance of $0.1+j0.3$ per unit per phase has a static stability limit of 2.5 per unit. The corresponding excitation voltage in per unit is ___________ (Round off to 2 decimal places).
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Correct answer: 1.57 to 1.59
Explanation
Take the terminal voltage as 1 pu and $Z_s=|Z|\angle\theta$ with $|Z|=0.3162$, $\cos\theta=0.3162$. The developed power of the motor is $P=\dfrac{EV\cos(\theta-\delta)-E^2\cos\theta}{|Z|}$, which is maximum at $\delta=\theta$: $P_{max}=\dfrac{E-0.3162E^2}{0.3162}=2.5$. So $0.3162E^2-E+0.7906=0$, whose discriminant is $1-4(0.3162)(0.7906)\approx0$, giving $E=\dfrac1{2\times0.3162}=1.58$ pu.