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
145.80
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
0.50
C
0.80
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.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.