# Synchronous Machines

 Question 1
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).
 A 0.82 B 1.59 C 2.25 D 3.64
GATE EE 2023   Electrical Machines
Question 1 Explanation:
Developed power for motor,
$P_{d e v}=\frac{V_{f}}{Z_{s}} \cos \left(\theta_{s}-\delta\right)-\frac{E_{F}^{2}}{2} \cos \theta_{s}$

At max. power developed,
$\delta=\theta_{\mathrm{S}}$

and this is decide steady state stability of motor.
$\therefore \mathrm{SSSL}, \mathrm{P}_{\text {dev }(max)}=\frac{V E_{F}}{\mathrm{Z}_{\mathrm{S}}}-\frac{\mathrm{E}_{\mathrm{F}}^{2}}{\mathrm{Z}_{\mathrm{S}}} \cos \theta_{\mathrm{s}}$
where, $V=1.0 \mathrm{pu}$

$\mathrm{Z}_{\mathrm{S}}=0.1+\mathrm{j} 0.3=0.316 \angle 71.6^{\circ} \mathrm{pu}$
and $\mathrm{SSSL}=2.5$

$\therefore \quad 2.5=\frac{1 \times E_{F}}{0.316}-\frac{E_{F}^{2}}{0.316} \cos 71.56^{\circ}$
$1.001 \mathrm{E}_{F}^{2}-3.16 \mathrm{E}_{F}+2.5=0$
$\Rightarrow \quad \mathrm{E}_{\mathrm{F}}=1.595 \mathrm{pu}$
 Question 2
A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in volts, is __________. (round off to nearest integer).
 A 258 B 324 C 128 D 245
GATE EE 2022   Electrical Machines
Question 2 Explanation:
Power input$= 10 + 2 = 12 kW$
\begin{aligned} P&=\sqrt{3}VI \cos\phi \\ I_a&=\frac{10 \times 10^3}{\sqrt{3} \times 400 \times 0.8}\\ &=21.65\angle 36.86^{\circ}A \end{aligned}
We have, Emf equation for motor
\begin{aligned} \bar{E_{ph}}&=\bar{V_{ph}}-\bar{I_{a}}\bar{Z_{s}}\\ &=\frac{400}{\sqrt{3}}-21.65\angle 36.83^{\circ}\times 1\angle 90^{\circ}\\ &=244.542\angle -4.062^{\circ}V \end{aligned}

 Question 3
An alternator with internal voltage of $1\angle \delta _{1}\text{p.u}$ and synchronous reactance of $\text{0.4 p.u}$ is connected by a transmission line of reactance $\text{0.1 p.u}$ to a synchronous motor having synchronous reactance $\text{0.35 p.u}$ and internal voltage of $0.85\angle \delta _{2}\text{p.u}$.
If the real power supplied by the alternator is $\text{0.866 p.u}$, then $\left( \delta _{1} -\delta _{2}\right)$ is _________ degrees. (Round off to 2 decimal places.)
(Machines are of non-salient type. Neglect resistances.)
 A 60 B 25.35 C 68.6 D 88
GATE EE 2021   Electrical Machines
Question 3 Explanation: Given real power in (p.u.), $P=0.866$
\begin{aligned} P &=\frac{E V \sin \left(\delta_{1}-\delta_{2}\right)}{X_{\mathrm{eq}}}=\left|\frac{1 \times 0.85}{0.4+0.1+0.35}\right| \sin \left(\delta_{1}-\delta_{2}\right) \\ \left(\delta_{1}-\delta_{2}\right) &=60^{\circ} \end{aligned}
 Question 4
Consider the table given:
$\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}$
The correct combination that relates the constructional feature, machine type and mitigation is
 A P-V-X, Q-U-Z, R-T-Y B P-U-X, Q-S-Y, R-V-Z C P-T-Y, Q-V-Z, R-S-X D P-U-X, Q-V-Y, R-T-Z
GATE EE 2021   Electrical Machines
Question 4 Explanation:
P: Damper bars used in synchronous machine (U) to prevent hunting (X)
Q: Skewed rotor slots used in induction motor (S) t avoid magnetic locking (Y)
R: Compensating winding used in DC machine (V) to neutralize cross magnetizing effects of armature reaction (Z) under main poles (polar zone).
 Question 5
A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and subtransient reactance of 0.2 pu. It is operating at (1+j0) pu terminal voltage with an internal emf of (1+j0.7) pu. Following a three-phase solid short circuit fault at the terminal of the generator, the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is ________ pu.
 A 0.42 B 0.82 C 1.02 D 1.26
GATE EE 2020   Electrical Machines
Question 5 Explanation:
\begin{aligned} &\text{Prefault current,}\\ I_{0}&=\frac{E_{f}-V_{t}}{jX_{d}} \\ &=\frac{1+j0.7-1}{j0.7}=1 \\ &\text{Subtransient induced emf,} \\ {E_{f}}''&=V_{0}+j{X_{d}}''I_{0} \\ &=1+j0.2\times 1=1+j0.2 \\ | {E_{f}''}|&=\sqrt{1^{2}+0.2^{2}}=1.02 \end{aligned}

There are 5 questions to complete.