Question 1 |

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).

258 | |

324 | |

128 | |

245 |

Question 1 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}

\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 2 |

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.)

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.)

60 | |

25.35 | |

68.6 | |

88 |

Question 2 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 3 |

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

\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

P-V-X, Q-U-Z, R-T-Y | |

P-U-X, Q-S-Y, R-V-Z | |

P-T-Y, Q-V-Z, R-S-X | |

P-U-X, Q-V-Y, R-T-Z |

Question 3 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).

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 4 |

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.

0.42 | |

0.82 | |

1.02 | |

1.26 |

Question 4 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 5 |

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).

125.29 | |

35.45 | |

85.12 | |

158.36 |

Question 5 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

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 6 |

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)

110 | |

150 | |

130 | |

90 |

Question 6 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}

\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 7 |

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?

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?

P is false and Q is true. | |

P is true and Q is false. | |

P is false and Q is false. | |

P is true and Q is true. |

Question 7 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 8 |

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 __________.

456.92 | |

145.80 | |

365.24 | |

245.36 |

Question 8 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}

\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 9 |

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

100 A | |

200 A | |

300 A | |

400 A |

Question 9 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}

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 10 |

In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is

0 | |

45 | |

60 | |

90 |

Question 10 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)

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)

There are 10 questions to complete.