ELECTRICAL ENGINEERING ESE TOPICWISE CONVENTIONAL SOLVED PAPER-II
1987-2019
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IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : info.publications@iesmaster.org Web : iesmasterpublications.com
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First Edition
: 2018
Second Edition : 2019
Typeset at : IES Master Publication, New Delhi-110016
PREFACE
Engineering Services Exam (ESE) is one of most coveted exams written by engineering students aspiring for reputed posts in the various departments of the Government of India. ESE is conducted by the Union Public Services Commission (UPSC), and therefore the standards to clear this exam too are very high. To clear the ESE, a candidate needs to clear three stages - ESE Prelims, ESE Mains and Personality Test. It is not mere hard work that helps a student succeed in an examination like ESE that witnesses lakhs of aspirants competing neck to neck to move one step closer to their dream job. It is hard work along with smart work that allows an ESE aspirant to fulfil his dream. After detailed interaction with students preparing for ESE, IES Master has come up with this book which is a one-stop solution for engineering students aspiring to crack this most prestigious engineering exam. The book includes previous years' solved conventional questions segregated subject-wise along with detailed explanation. This book will also help ESE aspirants get an idea about the pattern and weightage of questions asked in ESE. IES Master feels immense pride in bringing out this book with utmost care to build upon the exam preparedness of a student up to the UPSC standards. The credit for flawless preparation of this book goes to the entire team of IES Master Publication. Teachers, students, and professional engineers are welcome to share their suggestions to make this book more valuable.
IES Master Publication New Delhi
CONTENTS
1.
ELECTRICAL MACHINES AND POWER TRANSFORMERS ................. 01 – 188
2.
POWER SYSTEM ................................................................................ 189 – 388
3.
ANALOG ELECTRONICS .................................................................... 389 – 518
4.
DIGITAL ELECTRONICS ..................................................................... 519 – 576
5.
MICROPROCESSOR ........................................................................... 577 – 619
6.
COMMUNICATION SYSTEM ................................................................ 620 – 681
7.
POWER ELECTRONICS ...................................................................... 682 – 787
8.
CONTROL SYSTEM ............................................................................ 788 – 966
9.
SYSTEM AND SIGNAL PROCESSING ............................................... 967 – 978
UNIT 1
ELECTRICAL MACHINES AND POWER TRANSFORMERS SYLLABUS
Magnetic Circuits-Analysis and Design of Power transformers. Construction and testing. Equivalent circuits. Losses and efficiency. Regulation. Auto-transformer, 3-phase transformer. Parallel operation. Basic concepts in rotating machines. EMF, torque, basic machine types. Construction and operation, leakage losses and efficiency. D.C. Machines. Construction, Excitation methods. Circuit models. Armature reaction and commutation. Characteristics and performance analysis. Generators and motors. Starting and speed control. Testing, Losses and efficiency. Synchronous Machines. Construction. Circuit model. Operating characteristics and performance analysis. Synchronous reactance. Efficiency. Voltage regulation. Salient-pole machine, Parallel operation. Hunting. Short circuit transients. Induction Machines. Construction. Principle of operation. Rotating fields. Characteristics and performance analysis. Determination of circuit model. Circle diagram. Starting and speed control. Fractional KW motors. Single-phase synchronous and induction motors.
CONTENTS
1.
Transformers ................................................................................................ 02-41
2.
Basic Concepts of Rotating Electrical Machines .......................................... 42-45
3.
DC Machines ............................................................................................... 46-88
4.
Induction Machines and Single Phase Motors ........................................... 89-149
5.
Synchronous Machines ............................................................................ 150-188
1
Transformers
Q–1:
The star-star connection of transformers is most economical for large, low-voltage transformers (True/False). [8 Marks ESE–1987]
Sol.
False: Star-star connection of transformer is used for
V
small high voltage transformers. This is due to the fact that both of the windings are strenged by only
3
V 1 3 K
1 57.74% = ×100 of line to line voltage. The turns 3
per phase are also minimum. Q–2:
Explain purpose and action of a delta-connected tertiary winding in Y-Y connected transformers. [8 Marks ESE–1989]
Sol.
Purpose and action of a delta connected tertiary winding in YY connected transformers: Only YY connected transformers are rarely used owing to their drawbacks in supplying unbalanced load effectively between their line and neutral. In this type of transformer the phase voltage gets distorted by the presence of third harmonics, as an easy path for developed third harmonics emf to cause flow of 3rd harmanic current, as well as zero sequence current. This zero sequence current flow may cause the phase voltage to be unbalanced. Hence by using a delta connected tertiary winding, the zero sequence currents are provided a low reactance path and voltage balance on unbalanced loads. This delta connected winding also provides a path for third harmonic current to flow.
Q–3:
The maximum efficiency of a 100 kVA single phase transformer is 98 percent and occurs at 80 percent full load at 0.8 p.f. If the leakage impedance of the transformer is 5 percent, find the voltage regulation at rated load at 0.75 p.f. lagging. [12 Marks ESE–1991]
Sol.
Maximum efficiency, of a 100KVA single phase transformer is 98% occure at 80% of full load at 0.8 pf. Leakage impedance of transformer = 0.05 efficiency =
output output losses
1 losses = 1 output
output at maximum efficiency = 100×1000×0.8×0.8 = 64000W
Electrical Engineering
ESE Topicwise Conventional Solved Paper-II
3
1 1 64000 = 1308 watt. Transformer loss = 0.98
At maximum ,
ohmic loss = core loss, Pc =
1308 = 654 W 2
100 ohmic losses at full load = 654 80
repu =
2
= 1022 W
1022 = 0.01022 = x 100000
Zepu = 0.05 xepu =
Zepu2 – repu2 = 0.0489 = x
voltage regulation = r cos 2 x sin 2 = 0.01022×0.75 + 0.0489×0.66 = 3.99% Q–4:
With necessary circuit diagram brief sumpner’s test. Explain how the test results are useful in determining equivalent circuit and efficiency of the transformer. [16 Marks ESE–1992, 1998]
Sol.
Sumpner’s test on transformer
wattmeter
In the figure shown above two identical A + + transformers are used for conducting this test. The primary windings of the both the transformers are connected in parallel and Main supply V source – + + – supplied through main power supply. The secondary windings of both the transformers – – are connected in series and supplied through an auxillary source. The primary of the transformers connected in parallel are applied wattmeter with rated terminal voltage and secondaries of transformers are allowed to pass through the rated current in the secondary windings. The wattmeter connected in the primary will read Auxillary the No load (or) core losses of the transformer. supply The wattmeter connected in secondary windings of the transformer will read copper losses of the transformer. Thus this is equivalent to conducting of the open circuit test and short circuit test simultaneously. During this time the exact temperature rise of the transformers can be estimated that is the reason this is also called as heat run test. This test is used to find out the primary and secondary impedances, No load Resistance and reactance. Thus we can calculate the efficiency of the transformer by this test. FL
where
KVA cos KVA cos i cu
i = iron losses of transformer cu = copper losses of the transformer FL = efficiency at full load.
Disadvantage: The accuracy of this test depends on the two identical transformers.
IES MASTER Publication
4
Electrical Engineering
Electrical Machines and Power Transformers
Q–5:
Explain the working of a transformer at no load and loaded conditions. [20 Marks ESE–1993]
Sol.
Transformer at no load and loaded conditions: (i) No load working: •
When the transformer is under no load condition the phasor of the system is as shown in figure.
•
The magnetic flux being common to both primary and secondary
V1
is drawn first. Induced emf E1 and E2 lag by 90º and are shown
Ie
Ic
as in figure.
•
V1' is treated as the voltage drop in the primary, in the direction of primary current.
•
No load primary current Ie is the exciting current of the transformer and can be resolved into two components.
Im
Ex1 = –j Iex1 E1,E2 = V2
i.e. Magnetising component to provide magnetic flux, and coreloss component or power component. •
–j Iex1 V1Ie I V1 = –E1
Total voltage drop in primary at no load is Ie(r1+jX1) = IeZ1, Where Z1 is the primary leakage impedance.
Therefore V1 = V1' Ier1 jX1 Ie l1
l1
•
At no load V1' and V1 are very nearly equal.
(ii)
Full load working : l1
I2
I
V2
V1 l2
•
The secondary circuit of the transformer is considered first and then the primary circuit, for developing phasor under load.
•
When switch is closed I2 current flows in the secondary winding which lags secondary voltage V2 by angle 2 .
•
The secondary mmf I2N2 gives rise to a leakage flux l 2 which links only the secondary and not the primary.
•
Flux l 2 is in phase with I2, it. induces Ex2 in the secondary winding.