Power System
Exam Duration: 45 Mins Total Questions : 30
In a nuclear reactor, chain reaction is controlled by introducing
- (a)
iron rods
- (b)
cadmium rods
- (c)
graphite rods
- (d)
brass rods
For variable heads of near about but less than 30 m. Which type of turbines is used in hydropower stations?
- (a)
Pelton
- (b)
Kaplan
- (c)
Francis
- (d)
None of these
Match List I (Classification of head) with List II (Types of turbine) and select the correct answer using the codes given below the lists.
List I | List II |
P. Low head, 2-15 m | 1. Propeller or Kaplan |
Q. Medium head, 15-70 m | 2. Kaplan or Francis |
R. High head, 70-500 m | 3. Pelton |
S. Very high head > 500 m | 4. Francis or Pelton |
- (a)
P Q R S 1 3 4 2 - (b)
P Q R S 4 2 1 3 - (c)
P Q R S 1 2 4 3 - (d)
P Q R S 4 3 1 2
A single-phase 2 wire system supplies a load of 1000 kW. The voltage between the conductors is 11000 V. Return wire is earthed. Now, a third wire is added in the system and supply made is 3 phase. For same power factor the transmitted power will be
- (a)
1500 kW
- (b)
3000 kW
- (c)
2450 kW
- (d)
3100 kW
The bus admittance matrix of a power system is given as\(\begin{matrix}1 \\ 2 \\ 3 \end{matrix}\overset { \begin{matrix} \ \ \ \ 1\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ & \ \ \ \ \ \ \ \ 3 \end{matrix} }{ \left[ \begin{matrix} -j\ 50 & +j\ 10 & \ +j\ 5 \\ +j\ 10 & \ -j\ 30 & \ \ +j\ 10 \\ +j\ 5 & +j\ 10 & -j\ 25 \end{matrix} \right] } \).The impedance of line between bus 2 and 3 will be equal to
- (a)
+ j 0.1
- (b)
- j 0.1
- (c)
+ j 0.2
- (d)
- j 0.2
Series compensation on EHV lines is resorted to
- (a)
iprove the stability
- (b)
reduce the fault level
- (c)
improve the voltage profile
- (d)
as a substitute for synchronous phase modifier
For a synchronous phase modifier the load angle is
- (a)
\({0 }^{ \circ }\)
- (b)
\({25 }^{ \circ }\)
- (c)
\({30 }^{ \circ }\)
- (d)
None of these
At an industrial sub-station with a 4MW load, a capacitor of 2 MVAR is installed to maintain the load power factor at 0.97 lagging. If the capacitor goes out of service, the load power factor becomes
- (a)
0.85 lag
- (b)
1.00 lag
- (c)
0.80 lag
- (d)
0.90 lag
In a thermal power plant, the feed water coming to the economizer is heated using
- (a)
high pressure steam
- (b)
low pressure steam
- (c)
direct heat in the furnace
- (d)
flue gases
A load centre is at an equidistant from the two thermal generating stations C, and Cz as shown in the figure. The fuel cost characteristics of the generating stations are given by
\({ F }_{ 1 }=a+b{ P }_{ 1 }+c{ P }_{ 1 }^{ 2 }\ Rs/ h \ \\{ F }_{ 2 }=a+b{ P }_{ 2 }+2c{ P }_{ 2 }^{ 2 }\ Rs/ h\)
where, P1 and P2 are the generation in MW of C1 and C2 respectively. For most economic generation to meet 300 MW of load P1 and P2, respectively are
- (a)
150, 150
- (b)
100, 200
- (c)
200, 100
- (d)
175, 125
A single-phase overhead line consists of two conductors of radius 1 cm each. Also, spacing between conductor is 1.25 m and frequency is 50 Hz. Then, reactance per km will be
- (a)
\(0.64 \ \Omega /km\)
- (b)
\(0.64 \ k\Omega /m\)
- (c)
\(0.7 \ \Omega /m\)
- (d)
Data is insufficient
Ring main distribution system is preferred to a radial system, because
- (a)
it is lessexpensive
- (b)
voltage drop in the feeder is less
- (c)
power factor is higher
- (d)
supply is more reliable
In a 4 bus system the circuit breakers and the various zones of protection are shown in figure. If the circuit breakers E, F and C trip, the location of the fault is on
- (a)
1
- (b)
3
- (c)
2
- (d)
4
The surge impedance of a 400 km long overhead transmission line is 400\(\Omega\). For a 200 km length of the same line, the surge impedance will be
- (a)
200\(\Omega\)
- (b)
800\(\Omega\)
- (c)
400\(\Omega\)
- (d)
100\(\Omega\)
High voltage DC (HVDC) transmission is mainly used for
- (a)
bulk power transmission over very long distances
- (b)
interconnecting two systems with the same nominal frequency
- (c)
eliminating reactive power requirement in the operation
- (d)
minimizing harmonics at the converter stations
The insulation resistance of cable of length 10 km in MQ. For a length of 100 km of the same cable, the insulation resistance will be
- (a)
1M\(\Omega \)
- (b)
10 M\(\Omega \)
- (c)
0.1 M\(\Omega \)
- (d)
0.01 M\(\Omega \)
A 240 V single-phase AC source is connected to a load with an impedance of 10 L60° Q. A capacitor is connected in parallel with the load. If the capacitor supplies 1250 VAR, the real power supplied by the source is
- (a)
3600 W
- (b)
2880 W
- (c)
2400 W
- (d)
1200 W
Buses for load flow studies are classified as
1. The load bus
2. The generator bus
3. The slack bus
The correct combination of the pair of quantities specified having their usual meaning for different buses is
- (a)
Load bus Generator Slack bus P, |V| P, Q P, \(\delta \) - (b)
Load bus Generator Slack bus P, \(\delta \) Q, |V| Q, \(\delta \) - (c)
Load bus Generator Slack bus |V|, Q P, \(\delta \) P, Q - (d)
Load bus Generator Slack bus P, Q P, |V| |V|, \(\delta \)
Two sections of a transmission line are connected through 800 \(\mu\)H inductance. Each line has a surge impedance of 350\(\Omega \). A 500 kV, 2 us rectangular surge travels along the line towards the inductance. Maximum value of transmitted wave will be
- (a)
207 kV
- (b)
500 kV
- (c)
414 kV
- (d)
621 kV
A 400 V, 40 kW, 3-phase, 50 Hz induction motor runs at power factor of 0.72 lag with an efficiency (11)85%. Find the capacitance per phase of mess connected capacitor bank necessary to rise the power factor of the supply to 0.98 lag.
- (a)
240 \(\mu F\)
- (b)
237.4 \(\mu F\)
- (c)
235.48 \(\mu F\)
- (d)
243.49 \(\mu F\)
A star connected 440 V, 50 Hz alternator has per phase synchronous reactance of 10 \(\Omega \). It supplies a balanced capacitive load current of 20 A, as shown in per phase equivalent circuit of the given figure. It is desirable to have zero voltage regulation. The load power factor should be
- (a)
0.92
- (b)
0.47
- (c)
0.39
- (d)
0.82
A Buchholz is used
- (a)
protection of a transformer against all internal fauIts
- (b)
protection of a transformer against external fauIts
- (c)
protection of a transformer against both internal and external faults
- (d)
protection of induction motors
HVDC transmission is preferred to EHV-AC because
- (a)
HVDC terminal equipment are inexpensive
- (b)
VAR compensation is not required in HVDC systems
- (c)
system stability can be improved
- (d)
Harmonic problem is avoided
The use of high speed circuit breakers
- (a)
reduce the short circuit current
- (b)
improve system stabiIity
- (c)
decrease system stabiIity
- (d)
increase the short circuit current
In the protection of transformers, harmonic restraint is used to guard against
- (a)
magnetizing inrush current
- (b)
unbalanced operation
- (c)
lightning
- (d)
switching over voltages
Consider the transformer connections in a part of a power system shown in the figure. The nature of transformer connections and phase shifts are indicated for all but one transformer.
Which of the following connections and the corresponding phase shift 8, should be used for the transformer between A and B
- (a)
Star-star (\(\theta \) = 0°)
- (b)
Star-delta (\(\theta \) = 30°)
- (c)
Delta-star (\(\theta \) = 30°)
- (d)
Star-zigzag (\(\theta \) = 30°)
A 75 MVA, 10 kV synchronous generator has Xd = 0.4 pu. The Xd value (in pu) as a base of 100 MVA, 11 kV is
- (a)
0.578
- (b)
0.279
- (c)
0.412
- (d)
0.44
In the circuit shown in figure, what value of C will cause a unity power factor at the AC source?
- (a)
68.1\(\mu\)F
- (b)
165\(\mu\)F
- (c)
0.681\(\mu\)F
- (d)
6.81\(\mu\)F
At a 220 kV substation of a power system, it is given that the three-phase fault level is 4000 MVA and single line to ground fault level is 5000 MVA. Neglecting the resistance and the shunt suspectance of the system.The positive sequence driving point reactance at the bus is
- (a)
\(2.5\Omega \)
- (b)
\(4.033\Omega \)
- (c)
\(5.5\Omega \)
- (d)
\(12.1\Omega \)
A three-phase alternator generating unbalanced voltage is connected to an unbalanced load through a three-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltage of the alternator are Ec = 10\({ \angle 0 }^{ \circ }\) V, Eb = 10\({ \angle -90 }^{ \circ }\) V, E = 10\({ \angle 120 }^{ \circ }\) V. The positive4 sequence component of the load current is
- (a)
\({ 1.310\angle -107 }^{ \circ }A\)
- (b)
\({ 0.332\angle -120 }^{ \circ }A\)
- (c)
\({ 0.3996\angle -120 }^{ \circ }A\)
- (d)
\({ 3.510\angle -81 }^{ \circ }A\)