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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 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.
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
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)
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
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)
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
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
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
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
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
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)
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
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.
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
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
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
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
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
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
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
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
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
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
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
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.
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)
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
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
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
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)
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
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)
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
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
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
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
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)
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
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
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
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.
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.
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
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,