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MECHANICAL ENGINEERINGESE TOPICWISE OBJECTIVE SOLVED
PAPER-I
UPSC Engineering Services Examination
State Engineering Service Examination & Public Sector Examination.
IES MASTER PUBLICATION Office :Phone : Mobile :E-mail : Web :
F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064, 8130909220, 9711853908 [email protected] www.iesmasterpublications.com
FROM (1995-2018)
Typeset at : IES Master Publication, New Delhi - 110016
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016Phone : 011-26522064, Mobile : 8130909220, 9711853908E-mail : [email protected] Web : www.iesmasterpublications.com
© No part of this booklet may be reproduced or distributed in any form or by any means, electronic,mechanical, photocopying or otherwise or stored in a database or retrieval system without the priorpermission of IES MASTER PUBLICATION, New Delhi. Violaters are liable to be legally prosecuted.
Second Edition : 2018
Mr. Kanchan Kumar ThakurDirector–IES Master
It is an immense pleasure to present topic wise previous years solved paper of EngineeringServices Exam. This booklet has come out after long observation and detailed interaction withthe students preparing for Engineering Services Exam and includes detailed explanation to allquestions. The approach has been to provide explanation in such a way that just by goingthrough the solutions, students will be able to understand the basic concepts and will applythese concepts in solving other questions that might be asked in future exams.
Engineering Services Exam is a gateway to a immensly satisfying and high exposure job inengineering sector. The exposure to challenges and opportunities of leading the diverse fieldof engineering has been the main reason for students opting for this service as compared toothers. To facilitate selection into these services, availability of arithmetic solution to previousyear paper is the need of the day. Towards this end this book becomes indispensable.
PREFACE
CONTENT
1. Thermodynamics 01 – 138
2. Power Plant Engineering 139 – 295
3. Fluid Mechanics and Machinery 296 – 547
4. Heat and Mass Transfer 548 – 659
5. Refrigeration and Air Conditioning 660 – 756
6. IC Engine 757 – 832
7. Renewable Sources of Energy 833–834
Unit No. Pages
S YL L A BU SThermodynamic systems and processes; properties of pure substances; Zeroth, First and Second Laws ofThermodynamics; Entropy, Irreversibility and availability; analysis of thermodynamic cycles related to energyconversion : Rankine, Otto, Diesel and Dual Cycles; ideal and real gases; compressibility factor; Gas mixtures.
Contents1. Introduction and First Law ------------------------------------------------------01–39
2. Second Law of Thermodynamics ---------------------------------------------40–63
3. Entropy and Exergy Analysis ---------------------------------------------------64–85
4. Pure Substances------------------------------------------------------------------86–98
5. Ideal and Real Gases --------------------------------------------------------- 99–107
6. Thermodynamic Relations -------------------------------------------------- 108–116
7. Compressible Flow----------------------------------------------------------- 117–138
1
IES – 2018
1. Which one of the following substances hasconstant specific heat at all pressures andtemperature ?
(a) Mono-atomic gas (b) Di-atomic gas
(c) Tri-atomic gas (d) Poly-atomic gas
2. When the valve of an evacuated bottle isopened, the atmospheric air rushes into it. Ifthe atmospheric pressure is 101.325 kPa and0.6 m3 of air enters into the bottle, then thework done by the air will be
(a) 80.8 kJ (b) 70.8 kJ
(c) 60.8 kJ (d) 50.8 kJ
3. A thermodynamic cycle is composed of fourprocesses. The heat added and the work donein each process are as follows :
Process Heat transfer (J) Work done (J)1 2 0 50(by the gas)2 3 50(from the gas) 03 4 0 20(on the gas)4 1 80(to the gas) 0
The thermal efficiency of the cycle is
(a) 20.3% (b) 37.5%
(c) 40.3% (d) 62.5%
4. A steel tank placed in hot environmentcontains 5 kg of air at 4 atm at 30°C. Aportion of the air is released till the pressurebecomes 2 atm. Later, the temperature of
the air in the tank is found to be 150°C. Thequantity of air allowed to escape is
(a) 4.72 kg (b) 4.12 kg
(c) 3.71 kg (d) 3.21 kg
5. Consider the following statements :
1. Entropy is related to the first law ofthermodynamics.
2. The internal energy of an ideal gas is afunction of temperature and pressure.
3. The zeroth law of thermodynamics is thebasis for measurement of temperature.
Which of the above statements are correct ?
(a) 1 and 2 only (b) 1 and 3 only
(c) 2 and 3 only (d) 1, 2 and 3
6. A system absorbs 100 kJ as heat and does60 kJ work along the path 1-2-3. The samesystem does 20 kJ work along the path 1-4-3. The heat absorbed during the path 1-4-3 is
V
P
1 4
32
(a) –140 kJ (b) –80 kJ
(c) 80 kJ (d) 60 kJ
1
Thermodynamics 3
IES – 20177. Statement (I): The specific heat at constant
pressure for an ideal gas is always greaterthan the specific heat at constant volume.Statement (II): Heat added at constant volumeis not utilized for doing any external work.
8. During a constant pressure expansion of agas, 33.3% heat is converted into work whilethe temperature rises by 20 K. The specificheat of the gas at constant pressure as aproportion of work, W is(a) 8% (b) 10%(c) 12% (d) 15%
IES – 20169. Consider the following processes :
1. Extension of a spring2. Plastic deformation of a material3. Magnetization of a material exhibiting
hysteresisWhich of the above process are irreversible?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3
10. Which of the following statements are correctfor a throttling process?1. It is an adiabatic steady flow process2. The enthalpy before and after throttling is
same3. In the processes, due to fall in pressure,
the fluid velocity at outlet is always morethan inlet velocity
(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3
11. Statement (I) : Thermometers using differentthermometric property substance may givedifferent readings except at two fixed points.Statement (II) : Thermodynamic temperaturescale is independent of any particularthermoemtric substance.
12. Statement (I) : First law of thermodynamicsanalyses the problem quantitatively whereassecond law of thermodynamics analyses theproblem qualitatively.Statement (II) : Throttling process isreversible process.
13. Which one of the following statements iscorrect during adiabatic charging of an idealgas into an empty cylinder from a supply main?(a) The specific enthalpy of the gas in the
supply main is equal to the specificenthalpy of the gas in the cylinder
(b) The specific enthalpy of the gas in thesupply main is equal to the specificinternal energy of the gas in the cylinder
(c) The specific internal energy of the gasin the supply main is equal to thespecific enthalpy of the gas in thecylinder
(d) The specific internal energy of the gasin the supply main is equal to thespecific internal energy of the gas inthe cylinder
14. Consider the following systems :1. An electric heater2. A gas turbine3. A reciprocating compressorThe steady flow energy equation can be appliedto which of the above systems?(a) 1 and 2 only (b) 1 and 3 only(c) 1, 2 and 3 (d) 2 and 3 only
15. Consider the following conditions for thereversibility of a cycle :1. The P and T of the working substance
must not differ appreciably, from those ofthe surroundings at any state in theprocess.
2. All the processes, taking place in thecycle, must be extremely slow.
3. The working parts of the engine must befriction-free.
Which of the following conditions are correct?(a) 1, 2 and 3 (b) 1 and 2 only(c) 1 and 3 only (d) 2 and 3 only
16. The property of a thermodynamic system is :(a) A path function(b) A point function(c) A quantity which does not change in
reversible process(d) A quantity which changes when system
undergoes a cycle
Thermodynamics 17
1. (a)
2. (c)
3. (b)
4. (d)
5. (*)
6. (d)
7. (a)
8. (d)
9. (c)
10. (a)
11. (b)
12. (c)
13. (b)
14. (c)
15. (a)
16. (b)
17. (c)
18. (b)
19. (b)
20. (b)
21. (b)
22. (d)
23. (c)
24. (c)
25. (d)
26. (a)
27. (c)
28. (d)
29. (a)
30. (a)
31. (b)
32. (a)
33. (d)
34. (b)
35. (a)
36. (d)
37. (c)
38. (a)
39. (b)
40. (c)
41. (b)
42. (d)
43. (c)
44. (b)
45. (a)
46. (c)
47. (d)
48. (a)
49. (d)
50. (b)
51. (c)
52. (a)
53. (a)
54. (d)
55. (a)
56. (a)
57. (d)
58. (c)
59. (b)
60. (d)
61. (d)
62. (c)
63. (d)
64. (c)
65. (d)
66. (a)
67. (b)
68. (d)
69. (a)
70. (a)
71. (b)
72. (d)
73. (b)
74. (b)
75. (c)
76. (b)
77. (a)
78. (b)
79. (d)
80. (b)
81. (b)
82. (b)
83. (b)
84. (d)
85. (a)
86. (d)
87. (c)
88. (b)
89. (a)
90. (b)
91. (b)
92. (b)
93. (c)
94. (a)
95. (d)
96. (d)
97. (c)
98. (c)
99. (b)
100. (b)
101. (a)
102. (d)
103. (c)
104. (d)
105. (a)
106. (b)
107. (b)
108. (b)
109. (a)
110. (b)
111. (b)
112. (d)
113. (c)
114. (a)
115. (b)
116. (b)
117. (c)
118. (b)
119. (b)
120. (c)
121. (d)
122. (c)
123. (d)
124. (a)
125. (a)
1 8 ESE TopicWise Objective Solved Paper-I 1995-2018
Sol–1 (a)Sol–2 (c)
Work done by air
P=101.325 kPadv = 0.6 m3
= Pdv
= 101.325 × 0.6
= 60.8 kJ
Sol–3 (b)Thermal efficiency,
= 2
1
Q 501 1 37.5%Q 80
Sol–4 (d)
5 kg4 atm30ºC
Initial state
2 atm150ºC
final state
em
From mass conservation,
i em m = cv 2 1d m m mdt
(1) No inlet, so im 0
em = 1 2m m ...(i)
(2) m1 = 5kg (given)
(P1V = mRT1)
m2 =2 2 1 1
2 2 1
P V P m RTRT RT P
=2 5 303 1.79 kg
423 4
From equation (i) em = 5 – 1.79 = 3.21 kg
Sol–5 (*)
Sol–6 (d)For process 1 – 2 –3
Q123 =U3 – U1 + W123
U3 – U1 = Q123 – W123 = 100 – 60 = 40 kJ
For process 1 – 4 – 3
Q143 = U3 – U1 + W143
= 40 + 20 = 60 kJSol–7: (a)
The specific heat of ideal gas of constantpressure is the sum of specific heat atconstant volume and a constant term. Thisconstant term comes due to work donecomponent. This work done componentis missing when heat is added at constantvolume.
Sol–8: (d)The work done during expansion,
W = 0.333 QBecause heat added,
Q = pC T
Because work done,
W = p0.333C T
pC
W =
1 10.333 T 0.333 20
pC 100 15%W 0.333 20
Sol–9: (c)A proces is reversible if both the systemand surroundings are restored to initialcondition. Thus, plastic deformation of a
Thermodynamics 19
material and magnetization of a materialexhibiting hysteresis are irreversibleprocesses. The spring after extension willget back to its original position after removalof load, hence it can be considered asreversible.
Sol–10: (a)Throttling process involves the passage ofa higher pressure fluid through a narrowconstriction. This process is adiabatic, andthere is no work interaction. Hence,Q = 0 & W = 0
PE = 0(Inlet and outlet are at the same level)
KE = 0(KE does not change significantly)
Applying the SFEE, h1 = h2
Thus, enthalpy remain constant
Further, the velocity of flow is kept low andany difference between the kinetic energyupstream and downstream is negligible. Theeffect of the decrease in pressure is anincrease in volume.
Sol–11: (b)
Thermodynamic or absolute temperaturescale is independent of any workingsubstance. The fact that the efficiency of areversible heat engine cycle depends onlyon the temperature of the two reservoirsmakes it possible to establish such a scale.If the temperature of a given system ismesured with thermometers using differentthermoemetric properties, there isconsiderable difference among the readings.
Sol–12: (c)
Throttling process is a irreversible processas the entropy of the fluid increases duringthe process. The first law of thermodynamicsonly gives a quantitative estimate of theheat and work interaction between thesystem and surroundings, however, it doesnot state about quality of energy. It is thesecond law of thermodynamics which dealswith the low grade and high grade energyand concepts of availability.
Sol–13: (b)
For charging the tank,
mphp =m2u2 – m1u1
If the tank is initially empty, then m1 = 0
mphp =m2u2
Since mp = m2, hence, hp = u2
Sol–14: (c)
The reciprocating compressor can beconsidered as steady flow system providedthe control volume includes the receiverwhich reduces the fluctuation of flow.
For electric heater, at steady state
W = Q
I
W C.V.
I
Q
Sol–15: (a)A reversible process must be quasi-staticand frictionless.Further, a heat engine cycle in which thereis a temperature difference (i) between thesource and the working fluid during heatsupply, and (ii) between the working fluidand the sink during heat rejection, exhibitsexternal thermal irreversibility. Thus, P andT of the working substance must not differ,appreciably from those of the surroundingsat any state in the process.
Sol–16: (b)
The propety of a thermodynamic system isa point function, since for a given state,there is a definite value for each property.The change in a thermodynamic propertyof a system in a change of state isindependent of the path the system followsduring the change of state and dependsonly on the initial and final states of thesystem.
