Power Systems Test - 3 - PDF Flipbook

Power Systems Test - 3

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GATE
EEE

Power
Systems

Test-03Solutions


POWER SYSTEMS
1. Two lossy capacitors with equal capacitance values and power

factors of 0.01 and 0.02 are in parallel, and the combination is
supplied from a sinusoidal voltage source. The power factor of
the combination is
a) 0.03
b) 0.015
c) 0.01
d) 0.0002
Answer: (b)
Solution:

ϕ1 = cos-1(0.01) = 89.420

ϕ2 = cos-1(0.02) = 88.850

so, ϕ = ϕ1+ϕ2 = 89.1350 ⇒ cos ϕ = 0.015
2

2. The concept of an electricity short, medium and long line is

primarily based on the

a) Nominal voltage of the line

b) Physical length of the line

c) Wavelength of the line

d) Power transmitted over the line

Answer: (c)

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3. For which of the following the excitation control method is
satisfactory?
a) Long lines
b) Low voltage lines
c) High voltage lines
d) Short lines
Answer: (d)
Solution:
The excitation control method is satisfactory only for relatively
short lines. However, it is not suitable for long lines as the
voltage at the alternator terminals will have to be varied too
much in order that the voltage at the far end of the line may be
constant.

4. Power distribution by cable is generally adopted for line length
a) less than 10 km
b) above 10 km
c) less than 50 km
d) above 50 km
Answer: (a)

5. A single load is supplied by a single voltage source. If the
current flowing from the load to the source is 10∠-l500 A and if
the voltage at the load terminals is 100∠600 V, then the
a) Load absorbs real power and delivers reactive power.
b) Load absorbs real power and absorbs reactive power
c) Load delivers real power and delivers reactive power

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d) Load delivers real power and absorbs reactive power.
Answer: (b)
Solution:

= − = −100∠600
−10∠−1500

= −(50+ 86.6)
−8.66− 5

= (8.66 + 5)

So load is RL. Therefore load absorbs real power and also

absorbs reactive power.

6. Which insulation is most widely used for covering wires/cables

used in internal wiring?

a) Paper

b) Wood

c) Glass

d) PVC

Answer: (d)

Solution:

Polyvinyl chloride (PVC) insulation is most widely used for

covering wires/cables used in internal wiring.

3


7. The insulation strength of an EHV transmission line is mainly
governed by
a) Load power factor
b) Switching over-voltage
c) Harmonics
d) Corona
Answer: (b)

8. The magnetic field energy in an inductor changes from
maximum value to minimum value in 5 m-sec when connected
to an a.c. source. The frequency of the source is:
a) 500 Hz
b) 20 Hz
c) 50Hz
d) 200 Hz
Answer: (c)
Solution:

It is given alternating current source.

Inductor stores energy from min → maximum, maximum →

minimum in one half cycle, so total it changes 4 times per cycle.

∴ Time period, = 4 × (5 × 10−3) = 0.02

Hence, = 1 = 1 = 50
0.02

4


9. Which of the following types of wiring preferred for workshop
lighting?
a) Casing Capping wiring
b) Batten wiring
c) Concealed conduct wiring
d) Surface conduit wiring
Answer: (c)
Solution:
Concealed conduit wiring system is preferred for workshop
lighting.

10. Three phase to ground fault takes place at locations F1 and F2
in the system shown in the figure.

If the fault takes place at location F1, then the voltage and the
current at bus A are VF1 and IF1 respectively. If the fault takes
place at location F2, then the voltage and the current at bus A are
VF2 and IF2 respectively. The correct statement about voltages
and currents during at F1 and F2 is
a) VF1 leads IF1 and VF2 leads IF2
b) VF1 leads IF1 and VF2 lags IF2
c) VF1 lags IF1 and VF2 leads IF2
d) VF1 lags IF1 and VF2 lags IF2
Answer: (c)

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Solution:
i) Fault at F1:

For a fault F1: Both generator 1 and generator 2 are supplying
the fault current the voltage at bus A is due to generator 2. The
angle of generator is zero so that the voltage angle at A is
negative. Hence VF1 lags IF1.

Ix fault current will be Ix∠–900
IF1 = –Ix = (1∠1800) Ix∠–900
IF1 = Ix∠90
→ VF1 lags IF1
ii) Fault at F2:

For a fault F2: Both generator 1 and generator 2 are supplying
the fault current the voltage at bus A due to generator 1 the
angle of generator is δ and it is positive so that the voltage angle
at bus A is also positive. Hence VF2 leads IF2.
Now � � � 2� is also ∠–90
⇒ VF2 leads IF2

6


11. The incremental cost curves in Rs/MWhr for two generators
supplying a common load of 700 MW are shown in the figures.
The maximum and minimum generation limits are also
indicated. The optimum generation schedule is:

a) Generator A: 400 MW,
Generator B: 300 MW

b) Generator A: 350 MW,
Generator B: 350 MW

c) Generator A: 450 MW,
Generator B: 250 MW

d) Generator A: 425 MW,
Generator B: 275 MW

Answer: (c)
Solution:
Incremental fuel cost of generator ‘A’ for maximum power
generation = 600 Rs/ MWhr
Incremental fuel cost of generator ‘B’ for maximum power
generation = 650 Rs/ MWhr

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As the incremental fuel cost for maximum generation of
generator ‘A’ is less than t the incremental fuel cost for
minimum generator ‘B’ is hence we can operate the generator
‘A’ at its maximum output of 450 MW and the remaining will
be generated by generator ‘B’.
12. As per recommendation of ISI, the maximum number of points
of lights, fans and socket outlets that can be connected in one
sub-circuit is
a) 8
b) 10
c) 15
d) 20
Answer: (b)
Solution:
As per recommendation of ISI, for a sub-circuit the maximum
number of points recommended is 10.
13. What is the maximum number of point of light, fan and socket-
outlets that can be connected in one sub-circuit?
a) Ten
b) Four
c) Six
d) Twelve
Answer: (a)

8


14. The color of the light given out by a sodium vapour discharge
lamp is
a) pink
b) bluish green
c) yellow
d) blue
Answer: (c)
Solution:
The filaments of the lamp sputter fast moving electrons of
sodium atoms, excite to higher energy levels and the electrons
thus excited relax by emitting the characteristic mono chromatic
bright yellow light.

15. Total instantaneous power supplied by a 3-phase ac supply to a
balanced R-L load is
a) Zero
b) Constant
c) Pulsating with zero average
d) Pulsating with non-zero average
Answer: (b)
Solution:
Let the phase voltage be Va = Vm sinωt
Vb = Vm sin(ωt – 1200)
Vc = Vm sin(ωt – 2400)

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Let the phase of load be ‘ϕ’, As it is a RL load, ‘ϕ’ is a lagging

angle.

Phase current ia = im sin(ωt – ϕ)

ib = im sin(ωt – 120 – ϕ)

ic = im sin(ωt – 240 – ϕ)

Instantaneous total 3 phase power = (Vm sinωt) (im sin(ωt – ϕ)) +
(Vm sin(ωt – 1200)) (im sin(ωt – 120 – ϕ)) + (Vm sin(ωt – 2400))

(im sin(ωt – 240 – ϕ))

= VmIm [cos(ωt − ωt + ϕ) − cos(2ωt − ϕ) + cos(ωt – 120 – ωt +
2

120 + ϕ ) − cos(ωt − 120 + ωt − 120 − ϕ) + cos(ωt – 240 – ωt +

240 + ϕ ) − cos(ωt – 240 + ωt − 240 − ϕ )]

= VmIm {3 cos ϕ [cos(2 − ϕ) + cos(2ωt − 240 − ϕ) +
2

cos(2ωt − 480 − ϕ)]}

= 3 VmImcosϕ − [cos(2 − ϕ) + cos(2 − 120 − ϕ) +
2

cos(2 − 240 − ϕ)]

= 3 Vm Im cos ϕ − 0 [∵ all are 1200 displacement vectors]
√2 √2

= 3VIcosϕ

16. The active and reactive power of an inductive circuit are 60 W

and 80 VAR respectively. The power factor of the circuit is:

a) 0.8 lag

b) 0.5 lag

c) 0.6 lag

d) 0.75 lag

Answer: (c)

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Solution:
According to power triangle,

tan = Opposite side = 80 = 1.333
Adjacent side 60

= tan−1(1.333) = 53.120

cos θ = cos(53.120 ) = 0.6 lag

17. Electric power for illumination in locomotive is provided by

a) main steam engine

b) small turbo-generator

c) battery

d) solar cells

Answer: (b)

Solution:

The electric power is provided by the small turbo generators.

18. The figure shows a two-generator system supplying a load of

PD = 40 MW, connected at bus 2.

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The fuel cost of generators G1 and G2 are: C1 (PG1) = 10,000
Rs/MWhr and C2 (PG2) = 12,500 Rs/MWhr and the loss in the
line is Ploss(pu) = 0.5 P2G1(pu), where the loss coefficient is
specified in pu on a 100 MVA base. The most economic power

generation schedule in MW is

a) PG1 = 20, PG2 = 22
b) PG1 = 22, PG2 = 20
c) PG1 = 20, PG2 = 20
d) PG1 = 0, PG2 = 40

Answer: (a)

Solution:

For economic load dispatch,

1 1 = 2 2
1 2

=1 1 1 2
�1− 2 � 2
1− 1 1

From the given data,

= 2(0.5) 1 = 1 = 0
1 2

1 × 10000 = 1 × 12500
(1− 1) (1−0)

1 − PG1 = 100 ⇒ PG1 = 0.2pu
125

⇒ PG1 = 0.2 ∗ 100 = 20MW

But PL = 0.5P12

= 0.5 (0.2)2

= 0.02 pu

PL = 0.02 × 100 = 2MW

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PG1 + PG2 = PD + PL
20 + PG2 = 40 + 2
⇒ PG2 = 22MW

19. In suburban services as compared with urban service

a) the coasting period is smaller but free running period is

longer

b) the coasting period is smaller.

c) the coasting period is longer.

d) the coasting period and free running periods are same.

Answer: (c)
20. The connected load of a consumer is 2 kW and his maximum

demand is 1.5 kW. The demand factor of the consumer is

a) 0.75

b) 0.375

c) 1.33

d) 1

Answer: (a)

Solution:

Demand factor = Max demand = 1.5 = 0.75
Connected load 2

21. The rated voltage of a 3-phase power system is given as:

a) peak line to line voltage

b) rms phase voltage

c) peak phase voltage

d) rms line to line voltage

Answer: (d)

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Solution:
The rated voltage of 3-phase power system is rms line to line
voltage.
22. Electric power for illumination in locomotive is provided by
a) main steam engine
b) small turbo-generator
c) battery
d) solar cells
Answer: (b)
Solution:
The electric power is provided by the small turbo generators.
23. A 3-phase transmission line is shown in figure:

Voltage drop across the transmission line is given by the
following equation:


�∆ � = � � � �

Shunt capacitance of the line can be neglect. If the line has
positive sequence impedance of 15 Ω and zero sequence in
impedance of 48Ω, then the values of Zs and Zm will be

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a) Zs = 31.5 Ω; Zm = 16.5 Ω
b) Zs = 26 Ω; Zm = 11 Ω
c) Zs = 16.5 Ω; Zm = 31.5 Ω
d) Zs = 11 Ω; Zm = 26 Ω
Answer: (b)
Solution:
If Zs & Zm be the self & mutual impedances of a line
respectively positive sequence impedance = Zs – Zm = 15 Ω
Zero sequence impedance = Zs + 2 Zm = 48 Ω

∴ Zm = 11 Ω, Zs = 26 Ω
24. The distribution losses that the utility suffers while transferring

power from generating station to the consumer is accounted
under:
a) Maintenance cost
b) Fixed charges
c) Running charges
d) Cost of fuel
Answer: (c)
Solution:
The distribution losses that the utility suffers while transferring
power from generating station to the consumer is accounted
under running charges.

15


25. During a disturbance on synchronous machine, the rotor
swings from A to B before finally settling down to a steady state
at point C on the power angle curve. The speed of the machine
during oscillation is synchronous at point(s)
a) A and B
b) A and C
c) B and C
d) only at C
Answer: (a)
Solution:
When the synchronous machine swings from A to B due to
disturbance and settles at a point ‘C’, then during this process,
rotor angle ‘δ’ increases until synchronous speed is achieved
and the mechanical input and electrical output are balanced. The
rotor will be having synchronous speed at point ‘A’ and also at
point ‘B’ before it finally settles.

26. The capacity factor of a plant is given by
a) maximum load/average load
b) average load/maximum load
c) average load/plant capacity
d) maximum load/plant capacity

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Answer: (c)

Solution:

Capacity factor of a plant = Average load
Plant capacity

27. The purpose of choke in a fluorescent tube is to:

a) increase voltage momentarily

b) decrease current

c) increase current

d) decrease voltage momentarily

Answer: (a)

Solution:

There are two filaments, one at each end of the tube light. Light

is emitted only when the gas inside the tube is ionized. For that

very high voltage is needed but once the gas is ionized, the

voltage can be reduced to great extent for the normal operation

of tube light. The purpose of the choke is to provide a very high

voltage initially between the filaments (across the two ends of

the tube light). Again once tire gas in the tube is ionized the

choke provides a low voltage.

28. If the power factor is high, then the consumer maximum KVA

demand:

a) increases

b) becomes zero

c) remains constant

d) decreases

Answer: (d)

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Solution:
If power factor is high then power factor angle is low, i.e., cosϕ
high ⇒ ϕ is low ⇒ θ is low

So, S = � 2 + 2 is low

Here, cosϕ1 < cosϕ2 (∵ϕ1 > ϕ2)
So, S1 > S2 (For constant P)
29. The insulation level of a 400 kV, EHV overhead transmission
line is decided on the basis of
a) Lightning over voltage
b) Switching over voltage
c) Corona inception voltage
d) Radio and TV interference
Answer: (b)
Solution:
In any transmission line lightning over voltages is more severe
compared to switching over voltages. The lighting voltages are
external voltages. Hence the insulation is provided for switching
over voltage.

18


30. The power factor of a spot welding machine is expected to be
around
a) 0.3 to 0.5 leading
b) unity
c) 0.8 lagging
d) 0.3 to 0.5 lagging
Answer: (d)
Solution:
Spot welding machine operates on high inductive load i e. it
operates on very low lagging power factor i.e. 0.3 to 0.5 lagging.

As ϕlag is high, cosϕ (lag) is low

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