# Synchronous Machines

 Question 1
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 1 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}
 Question 2
A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now _______ A (rounded off to 2 decimal places).
 A 125.29 B 35.45 C 85.12 D 158.36
GATE EE 2020   Electrical Machines
Question 2 Explanation:
At $P_{constant},$
$I_{a1}\cos \phi _{1}=I_{a2}\cos \phi _{2}$
$\cos \phi _{1}=0.9$
$\tan \phi _{1}=0.484=\frac{Q}{P}$
$\Rightarrow \, \, \frac{2Q}{P}=0.9686=\tan \phi _{2}$
$\cos \phi _{2}=0.7182$
$\therefore \, 100\times 0.9=I_{a2}\times 0.7182$
$\Rightarrow \, \, I_{a2}=125.29\: A$
 Question 3
A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW. as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is ________ (rounded off to 2 decimal places)
 A 110 B 150 C 130 D 90
GATE EE 2020   Electrical Machines
Question 3 Explanation:
Assumed full load frequency is 50 Hz
\begin{aligned}\tan \theta &=\frac{50-49.75}{110-100} \\ &=\frac{49.75-49.25}{(x-(110))} \\ \frac{0.25}{10}&=\frac{0.5}{(x-150)} \\ x-110&=\frac{0.5\times 10}{0.25} \\ x&=110+20=130\, MW\end{aligned} Question 4
A three-phase cylindrical rotor synchronous generator has a synchronous reactance $X_s$ and a negligible armature resistance. The magnitude of per phase terminal voltage is $V_A$ and the magnitude of per phase induced emf is $E_A$. Considering the following two statements, P and Q.

P : For any three-phase balanced leading load connected across the terminals of this synchronous generator, $V_A$ is always more than $E_A$.
Q : For any three-phase balanced lagging load connected across the terminals of this synchronous generator, $V_A$ is always less than $E_A$.

Which of the following options is correct?
 A P is false and Q is true. B P is true and Q is false. C P is false and Q is false. D P is true and Q is true.
GATE EE 2020   Electrical Machines
Question 4 Explanation: For lagging p.f. load : For all lagging power factor loads: $E_A \gt V_A$
For unity p.f. load: Still we can see : $E_A \gt V_A$
For 'slighlty' load, phasor diagram will be quite similar to that of unity p.f. load, thus $E_A$ will be greater than $V_A$. Thus P is false. Question 5
A 220 V (line), three-phase, Y-connected, synchronous motor has a synchronous impedance of (0.25+j2.5)$\Omega$/phase. The motor draws the rated current of 10 A at 0.8 pf leading. The rms value of line-to-line internal voltage in volts (round off to two decimal places) is __________.
 A 456.92 B 145.8 C 365.24 D 245.36
GATE EE 2019   Electrical Machines
Question 5 Explanation:
For synchronous motor,
\begin{aligned} E_g &=V_t-IZ \\ V_t &=\frac{220}{\sqrt{3}} \; \text{V (phase)} \\ Z &=(0.25+j2.5) \Omega \\ I &= 10\angle \cos ^{-1} 0.8A\\ E_g&= \frac{220}{\sqrt{3}}-(0.25+j2.5) \times 10\angle \cos ^{-1} 0.8\\ E_g&= 141.658 \angle -8.728^{\circ} \; \text{V (phase)}\\ E_g &= 245.36 \; \text{V(line)} \end{aligned}
 Question 6
A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at a power factor of 0.5 leading is
 A 100 A B 200 A C 300 A D 400 A
GATE EE 2019   Electrical Machines
Question 6 Explanation:
$P=VI \cos \phi$
For the same voltage and load,
$I \cos \phi =constant$
$I_1 \cos \phi _1=I_2 \cos \phi _2$
at unity power factor,
\begin{aligned} I_1&=200A \\ 200 \times 1 &=I_2 \times 0.5 \\ I_2&=400A \end{aligned}
 Question 7
In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is
 A 0 B 45 C 60 D 90
GATE EE 2018   Electrical Machines
Question 7 Explanation:
Salient pole synchronous motor power and torque relations per phase:
$P=\frac{E_bV}{X_t}\sin \delta +\frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \;Watts$
$T=\frac{60}{2 \pi N_S}\left [ \frac{E_bV}{X_d} \sin \delta + \frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \right ]\; N-m$
The second term is reluctance power or reluctance torque, which is directly proportional to $\sin 2\delta$.
Therefore, reluctance torque will be maximum.
When, $\delta =45^{\circ}$
$\because \;\;\sin2 (45^{\circ})\Rightarrow \sin 90^{\circ}=1$ (Maximum)
 Question 8
A 3-phase 50 Hz generator supplies power of 3MW at 17.32 kV to a balanced 3-phase inductive load through an overhead line. The per phase line resistance and reactance are 0.25$\Omega$ and 3.925$\Omega$ respectively. If the voltage at the generator terminal is 17.87 kV, the power factor of the load is ________
 A 0.25 B 0.5 C 0.8 D 0.95
GATE EE 2017-SET-2   Electrical Machines
Question 8 Explanation: \begin{aligned} V_S-V_R&=I(R \cos \phi +X_L \sin \phi) \\I &=\frac{P_R}{V_R \cos \phi } \\ V_S-V_R&= \frac{P_R}{V_R \cos \phi } (R \cos \phi +X_L \sin \phi )\\ \tan \phi &= \left ( \frac{V_SV_R-V_R^2-RP_R}{X_LP_R} \right )\\ \phi &=\tan ^{-1} \left [\frac{V_SV_R-V_R^2-RP_R}{X_LP_R} \right ]\\ &\;\Rightarrow \;\text{substitute all values} \\ \text{power factor} &= \cos \phi =0.8 \; \text{lag} \end{aligned}
 Question 9
A 25 kVA, 400 V, $\Delta$-connected, 3-phase, cylindrical rotor synchronous generator requires a field current of 5A to maintain the rated armature current under short-circuit condition. For the same field current, the open-circuit voltage is 360 V. Neglecting the armature resistance and magnetic saturation, its voltage regulation (in % with respect to terminal voltage), when the generator delivers the rated load at 0.8 pf leading, at rated terminal voltage is _________.
 A -20 B -25 C -15 D -10
GATE EE 2017-SET-2   Electrical Machines
Question 9 Explanation:
25 kVA, 400 V, $\Delta -$synchronous generator
\begin{aligned} \therefore \;\; I_{rated} &=\frac{25000}{\sqrt{3}\times 400}=36.085A \\ I_{phase}&=\frac{36.085}{\sqrt{3}}=20.83A \\ X_3 &=\left.\begin{matrix} \frac{V_{OC}}{I_{a_{rated}}} \end{matrix}\right|_{I_{f_{same}}} \\ \therefore \;\;X_s &= \frac{360}{20.83}=17.28\Omega /phase\\ E &=\sqrt{(V \cos \phi +I_aR_a)^2+(V \sin \phi -I_aX_s)^2} \\ &=\sqrt{(400 \times 0.8)^2 +(400 \times 0.6 - 20.83(17.28))^2} \\ &=\sqrt{(102400)+(240-360)^2} \\ &= 341.76V\\ \therefore \;\; \% V.R. &= \frac{E-V}{V} \times 100\\ &=\frac{341.76-400}{400} \times 100=-14.56\% \end{aligned}
 Question 10
If a synchronous motor is running at a leading power factor, its excitation induced voltage ($E_f$) is
 A equal to terminal voltage $V_t$ B higher than the terminal voltage $V_t$ C less than terminal voltage $V_t$ D dependent upon supply voltage $V_t$
GATE EE 2017-SET-2   Electrical Machines
Question 10 Explanation:
Synchonnous motor-leading p.f. over excited.
$\therefore \;\;E_f \gt V_t$
There are 10 questions to complete.