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.

0.42 | |

0.82 | |

1.02 | |

1.26 |

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).

125.29 | |

35.45 | |

85.12 | |

158.36 |

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

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)

110 | |

150 | |

130 | |

90 |

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}

\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?

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 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 __________.

456.92 | |

145.80 | |

365.24 | |

245.36 |

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}

\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

100 A | |

200 A | |

300 A | |

400 A |

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}

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

0 | |

45 | |

60 | |

90 |

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)

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 ________

0.25 | |

0.50 | |

0.80 | |

0.95 |

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 _________.

-20 | |

-25 | |

-15 | |

-10 |

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}

\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

equal to terminal voltage V_t | |

higher than the terminal voltage V_t | |

less than terminal voltage V_t | |

dependent upon supply voltage V_t |

Question 10 Explanation:

Synchonnous motor-leading p.f. over excited.

\therefore \;\;E_f \gt V_t

\therefore \;\;E_f \gt V_t

There are 10 questions to complete.