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GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU Paper Solution Branch- ELECTRICAL ENGINEERING Subject Name- Power System Engineering Paper Code- 7EE5A

GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

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Page 1: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

GLOBAL INSTITUTE OF TECHNOLOGY

JAIPUR

RTU Paper Solution

Branch- ELECTRICAL ENGINEERING

Subject Name- Power System Engineering

Paper Code- 7EE5A

Page 2: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Unit-1

Ans.1 (a). Economic Dispatch Considering Line Losses:

From the law of conservation of power we can write-

Equation

Equation 1.16 depends on the load flow situation. Expression for the total fuel cost is given as:

Subjected to

And

where λ is the lagrangian multiplier.

Page 3: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Equation can be written as-

Where

Li is known as the penalty factor for the ith generator.

(b)

PL=B11P12+2B12P1P2+B22P2

2

The load is at the bus of plant 2 , evidently transmission loss ia not affected by variation of P2

B22=B12=0

When P1=100MW, PL=15MW

15=B11(100)2

Or B11=0.0015MW-1

PL=0.0015 P12

∂PL/∂P1=0.003 P1 ; ∂PL/∂P2=0

L1=1/(1-0.003 P1)

L2=1

The generation at each plant is required to be calculated for λ=60

Page 4: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

(dC1/dP1)L1= λ ; (dC2/dP2)L2= λ

(0.2P1+20)/ (1-0.003 P1)=60 or P1=105.26MW

And 0.15P2+30=60 or P2=200MW

Total load=105.26+200-15=290.26MW

Ans. 1(a)

Page 5: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 6: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Unit-2

Ans . 2 (a)

Page 7: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 8: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 9: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 10: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans . 2 (b)

K.E. = ½*10000*(100π)2

=493.48 MJ

MVA=100/0.85=117.64 MVA

H=K.E./MVA=4.194 MJ/MVA

M=493.48/(180*50)=0.0548 MJ-s/elect. Degree

Page 11: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans . 2 (a)

Swing Equation:

Fig.2.1: Flow of Power in a Synchronous Generator

Te = Ti

Te. ωs = Ti . ωs

and Te. ωs = Ti . ωs = Ps – Pe = 0

Therefore left side of equation is not zero and an accelerating torque comes into play. If Pa is

the corresponding accelerating (decelerating ) power, then

Where D is damping coefficient and θe is the angular position of the rotor. It is more convenient

to measure the angular position of the rotor with respect to the synchronously rotating frame

of reference. Let

δ = θe - ωs.t

Page 12: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Where δ is the power angle of the synchronous machines. Neglecting damping (D = 0) and

substituting equation in equation we get-

Using equation and we get-

Dividing thorough out by G, the MVA rating of the machine

Where

Ans. 2. (b)

Page 13: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 14: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Unit-3

Ans. 3.

Page 15: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 16: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 17: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 18: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 19: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 20: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 21: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 22: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 23: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 24: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Unit-4

Ans. 4. (a) AC Excitation System

The AC excitation system consists of an alternator and thyristor rectifier bridge directly

connected to the main alternator shaft. The main exciter may either be self-excited or

separately excited. The AC excitation system may be broadly classified into two categories

which are explained below in details.

a. Rotating Thyristor Excitation System

The rotor excitation system is shown in the figure below. The rotating portion is being

enclosed by the dashed line. This system consists an AC exciter, stationary field and a

rotating armature. The output of the exciter is rectified by a full wave thyristor bridge rectifier

circuit and is supplied to the main alternator field winding.

The alternator field winding is also supplied through another rectifier circuit. The exciter

voltage can be built up by using it residual flux. The power supply and rectifier control

generate the controlled triggering signal. The alternator voltage signal is averaged and

Page 25: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

compare directly with the operator voltage adjustment in the auto mode of operation. In the

manual mode of operation, the excitation current of the alternator is compared with a separate

manual voltage adjustment.

b. Brushless Excitation System

This system is shown in the figure below. The rotating portion being enclosed by a dashed

line rectangle. The brushless excitation system consists an alternator, rectifier, main exciter

and a permanent magnet generator alternator. The main and the pilot exciter are driven by the

main shaft. The main exciter has a stationary field and a rotating armature directly connected,

through the silicon rectifiers to the field of the main alternators.

The pilot exciter is the shaft driven permanent magnet generator having rotating permanent

magnets attached to the shaft and a three phase stationary armature, which feeds the main

exciter field through silicon rectifiers, in the field of the main alternator. The pilot exciter is a

shaft driven permanent magnetic generator having rotating permanent magnets attached to

Page 26: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

the shaft and a 3-phase stationary armature, which feeds the main’s exciter through 3-phase

full wave phase controlled thyristor bridges.

The system eliminates the use of a commutator, collector and brushes have a short time

constant and a response time of fewer than 0.1 seconds. The short time constant has the

advantage in improved small signal dynamic performance and facilitates the application of

supplementary power system stabilising signals.

3. Static Excitation System

In this system, the supply is taken from the alternator itself through a 3-phase star/delta

connected step-down transformer. The primary of the transformer is connected to the

alternator bus and their secondary supplies power to the rectifier and also feed power to the

grid control circuit and other electrical equipment.

Page 27: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

This system has a very small response time and provides excellent dynamic performance.

This system reduced the operating cost by eliminating the exciter windage loss and winding

maintenance.

Ans. 4. (b)

Page 28: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 29: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans. 4. (a)

DC Excitation System

Page 30: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

The DC excitation system has two exciters – the main exciter and a pilot exciter. The exciter

output is adjusted by an automatic voltage regulator (AVR) for controlling the output terminal

voltage of the alternator. The current transformer input to the AVR ensures limiting of the

alternator current during a fault.

When the field breaker is open, the field discharge resistor is connected across the field winding

so as to dissipate the stored energy in the field winding which is highly inductive.

The main and the pilot exciters can be driven either by the main shaft or separately driven by

the motor. Direct driven exciters are usually preferred as these preserve the unit system of

operation, and the excitation is not excited by external disturbances.

The voltage rating of the main exciter is about 400 V, and its capacity is about 0.5% of the

capacity of the alternator. Troubles in the exciters of turbo alternator are quite frequent because

of their high speed and as such separate motor driven exciters are provided as standby exciter.

Advantage-

It is more reliable.

It is compact in size; it can brought anywhere due to its low weight.

Disadvantage-

High cost

Page 31: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans. 4, (b)

Page 32: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 33: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Unit-5

Ans. (a)

Page 34: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 35: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 36: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 37: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 38: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans. 5 (b)

Page 39: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 40: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans. 5 (a). Phase Shifting Transformer

Page 41: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Page 42: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Ans. 5. (b).

(i) Shunt Capacitor

Shunt Capacitors have several uses in the electric power systems. They are utilized as sources

of reactive power by connecting them in line-to-neutral. Electric utilities have also

connected capacitors in series with long lines in order to reduce its impedance. This is

particularly common in the transmission level, where the lines have length in several hundreds

of kilometers. However, this post will generally discuss shunt capacitors.

Shunt capacitors are usually called “power factor correction capacitors,” although they also

serve other functions and provide multiple benefits, which will be discussed in the succeeding

paragraphs. Also, they are used at all voltage levels from end-user utilization to extra high

voltages.

Page 43: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Shunt capacitors, either at the customer location for power factor correction or on the

distribution system for voltage control, dramatically alter the system impedance variation with

frequency. Capacitors do not create harmonics, but severe harmonic distortion can sometimes

be attributed to their presence.

A shunt capacitor at the end of a feeder results in a gradual change in voltage along the feeder.

Ideally, the percent voltage rise at the capacitor would be zero at no load and rise to maximum

at full load. However, with shunt capacitors, percent voltage rise is essentially independent of

load. Thus, automatic switching is often employed in order to deliver the desired regulation at

high loads, but prevent excessive voltage at low loads. Moreover, capacitor switching may

result in transient over voltages inside customer facilities.

Applications

Utilities use shunt capacitors at distribution and utilization voltages to provide reactive power

near the inductive loads that require it. This reduces the total current flowing on the distribution

feeder, which improves the voltage profile along the feeder, frees additional feeder capacity,

and reduces losses. In fact, substation transformers experience lower loadings when utilities

install sufficient capacitors on the distribution system. The reduced loadings not only improve

contingency switching options on the distribution system, but also extend equipment life and

defer expensive additions to the system.

(ii) Series Compensation

Definition: Series compensation is the method of improving the system voltage by

connecting a capacitor in series with the transmission line. In other words, in series

compensation, reactive power is inserted in series with the transmission line for improving

the impedance of the system. It improves the power transfer capability of the line. It is mostly

used in extra and ultra high voltage line.

Page 44: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Advantages of Series Compensation

Series compensation has several advantages like it increases transmission capacity, improve

system stability, control voltage regulation and ensure proper load division among parallel

feeders. These advantages are discussed below.

Increase in Power Transfer Capability – The power transfer over a line is given by

where P1 – power transferred per phase (W)

Vs – sending-end phase voltage (V)

Vr – receiving-end phase voltage

XL – series inductive reactance of the line

δ – phase angle between Vs and Vr

If a capacitor having capacitance reactance Xc is connected in series with the line, the

reactance of the line is reduced from XL to ( XL– Xc). The power transfer is given by

where, The factor k

is known as a degree of compensation or compensation factor. Thus, per unit compensation is

given by the equation percentage compensation is given by the equation

Page 45: GLOBAL INSTITUTE OF TECHNOLOGY JAIPUR RTU …...Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan) Solution 7th Sem University Examination

Global Institute of Technology, Jaipur ITS-1, IT Park, EPIP, Sitapura Jaipur 302022 (Rajasthan)

Solution 7th Sem University Examination 2019

Subject- Power System Engineering, Code- 7EE5A, Semester- 7th /Year- 4th

Where XL = total series inductive reactance of the line per phase

XC = capacitive reactance of the capacitor banck per phase

In practice , k lies between 0.4 and 0.7. For k = 0.5,