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Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016
Phone: 011-41013406 Mobile: 8130909220, 9711853908
Email: [email protected], [email protected]
Web: iesmasterpublications.com, iesmaster.org
MECHANICAL ENGINEERINGESE TOPICWISE OBJECTIVE SOLVED
PAPER-I
1995-2019
Typeset at : IES Master Publication, New Delhi - 110016
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016Phone : 011-41013406, Mobile : 8130909220, 9711853908E-mail : [email protected], [email protected] : www.iesmasterpublications.com, iesmaster.org
© 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
Third Edition : 2019
It is an immense pleasure to present topic wise previous years solved paper of Engineering ServicesExam. This booklet has come out after long observation and detailed interaction with the studentspreparing for Engineering Services Exam and includes detailed explanation to all questions. Theapproach has been to provide explanation in such a way that just by going through the solutions,students will be able to understand the basic concepts and will apply these concepts in solving otherquestions that might be asked in future exams.
Engineering Services Exam is a gateway to a immensly satisfying and high exposure job in engineeringsector. The exposure to challenges and opportunities of leading the diverse field of engineering hasbeen the main reason for students opting for this service as compared to others. To facilitate selectioninto these services, availability of arithmetic solution to previous year paper is the need of the day.Towards this end this book becomes indispensable.
Mr. Kanchan Kumar ThakurDirector–IES Master
PREFACE
1. Thermodynamics ..................................................................................... 01 – 126
2. Power Plant Engineering....................................................................... 127 – 272
3. Fluid Mechanics and Machinery ........................................................... 273 – 509
4. Heat and Mass Transfer ......................................................................... 510 – 616
5. Refrigeration and Air Conditioning ....................................................... 617 – 708
6. IC Engine ................................................................................................ 709 – 778
7. Renewable Sources of Energy ............................................................. 779 – 782
CONTENT
SYLLABUSThermodynamic systems and processes; properties of pure substances; Zeroth, First and Second Lawsof Thermodynamics; Entropy; Irreversibility and availability; analysis of thermodynamic cycles relatedto energy conversion : Rankine, Otto, Diesel and Dual Cycles; ideal and real gases; compressibilityfactor; Gas mixtures.
CONTENTS
1. Introduction and First Law .......................................................................... 01 – 36
2. Second Law of Thermodynamics ................................................................ 37 – 58
3. Entropy and Exergy Analysis ..................................................................... 59 – 78
4. Pure Substances ........................................................................................ 79 – 90
5. Ideal and Real Gases ................................................................................. 91 – 99
6. Thermodynamic Relations ....................................................................... 100 – 107
7. Compressible Flow ................................................................................. 108 – 126
UNIT-1 Thermodynamics
IES – 2019
1. A 2 kg of steam occupying 0.3 m3 at 15 bar isexpanded according to the law pv1.3 = constantto a pressure of 1.5 bar. The work done duringthe expansion will be
(a) 602.9 kJ (b) 606.7 kJ
(c) 612.5 kJ (d) 618.3 kJ
2. Certain quantities cannot be located on the graphby a point but are given by a point but are givenby the area under the curve corresponding to theprocess. These quantities in concepts ofthermodynamics are called as
(a) cyclic functions (b) point functions
(c) path functions (d) real functions
IES – 2018
3. 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
4. When the valve of an evacuated bottle is opened,the atmospheric air rushes into it. If theatmospheric pressure is 101.325 kPa and 0.6m3 of air enters into the bottle, then the workdone by the air will be
(a) 80.8 kJ (b) 70.8 kJ
(c) 60.8 kJ (d) 50.8 kJ
5. 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%
6. A steel tank placed in hot environment contains5 kg of air at 4 atm at 30°C. A portion of the airis released till the pressure becomes 2 atm.Later, the temperature of the air in the tank isfound to be 150°C. The quantity of air allowedto escape is
(a) 4.72 kg (b) 4.12 kg
(c) 3.71 kg (d) 3.21 kg
7. 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
8. 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 INTRODUCTION ANDFIRST LAW
IES MASTER Publication
THERMODYNAMICS 3Mechanical Engineering
IES – 2017
9. Statement (I): The specific heat at constantpressure for an ideal gas is always greater thanthe specific heat at constant volume.
Statement (II): Heat added at constant volumeis not utilized for doing any external work.
10. During a constant pressure expansion of a gas,33.3% heat is converted into work while thetemperature rises by 20 K. The specific heat ofthe gas at constant pressure as a proportion ofwork, W is
(a) 8% (b) 10%
(c) 12% (d) 15%
IES – 2016
11. Consider the following processes :
1. Extension of a spring
2. Plastic deformation of a material
3. Magnetization of a material exhibitinghysteresis
Which 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
12. Which of the following statements are correct fora throttling process?
1. It is an adiabatic steady flow process
2. The enthalpy before and after throttling issame
3. In the processes, due to fall in pressure, thefluid velocity at outlet is always more thaninlet velocity
(a) 1 and 2 only (b) 1 and 3 only
(c) 2 and 3 only (d) 1, 2 and 3
13. 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.
14. Statement (I) : First law of thermodynamicsanalyses the problem quantitatively whereassecond law of thermodynamics analyses theproblem qualitatively.
Statement (II) : Throttling process is reversibleprocess.
15. Which one of the following statements is correctduring adiabatic charging of an ideal gas into anempty cylinder from a supply main?
(a) The specific enthalpy of the gas in thesupply main is equal to the specif icenthalpy of the gas in the cylinder
(b) The specific enthalpy of the gas in thesupply main is equal to the specific internalenergy of the gas in the cylinder
(c) The specific internal energy of the gas inthe supply main is equal to the specificenthalpy of the gas in the cylinder
(d) The specific internal energy of the gas inthe supply main is equal to the specificinternal energy of the gas in the cylinder
16. Consider the following systems :
1. An electric heater
2. A gas turbine
3. A reciprocating compressor
The 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
17. Consider the fol lowing conditions for thereversibility of a cycle :
1. The P and T of the working substance mustnot differ appreciably, from those of thesurroundings at any state in the process.
2. All the processes, taking place in the cycle,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
18. The property of a thermodynamic system is :
(a) A path function
(b) A point function
(c) A quantity which does not change inreversible process
(d) A quantity which changes when systemundergoes a cycle
IES MASTER Publication
THERMODYNAMICS 17Mechanical Engineering
Sol–1: (d) 1.31 1P V = 1.3
2 2P V
1
2
PP =
1.32
1
VV
V2 =
11.3
11
2
P VP
V2 = 1
1.3 3 315bar 0.3 m 1.7634 m1.5 bar
Pv1.3 = constant = C (say)
P = 1.3C
V
work done W =2 2
1 1
V V
1.3V V
CPdv dVV
=22
11
VV 0.31.3
VV
VCV dV C0.3
= 0.3 0.32 1
C V V0.3
= 1.3 0.3 1.3 0.32 2 2 1 1 1
1 P V V P V V0.3
= 1 1 2 22 2 1 1
P V P V1 P V P V0.3 0.3
= [15×105 × 0.3 – 1.5 × 105 × 1.7634]/0.3= 618.3 kJ
Sol–2: (c)Path functions like heat transfer and worktransfer cant be represented by points in thegraph, but can be given by the area under thecurve.Ex: heat transfer will be area under curve inT-s diagram work transfer will be area undercurve in P-V diagram.
Sol–3: (a)Sol–4: (c)
Work done by air
= Pdv
= 101.325 × 0.6
P=101.325 kPadv = 0.6 m3
= 60.8 kJ
Sol–5: (b)Thermal efficiency,
= 2
1
Q 501 1 37.5%Q 80
Sol–6: (d)
2 atm150ºC
final state
5 kg4 atm30ºC
Initial 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–7: (*)Sol–8: (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–9: (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 component ismissing when heat is added at constantvolume.
EXPLANATIONS
18 ESE Topicwise Objective Solved Paper-I 1995-2019 Mechanical Engineering
Sol–10: (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–11: (c)A proces is reversible if both the system andsurroundings are restored to initial condition.Thus, plastic deformation of a material andmagnetization of a material exhibitinghysteresis are irreversible processes. Thespring after extension will get back to itsoriginal position after removal of load, hence itcan be considered as reversible.
Sol–12: (a)Throttling process involves the passage of ahigher 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–13: (b)
Thermodynamic or absolute temperature scaleis independent of any working substance. Thefact that the efficiency of a reversible heatengine cycle depends only on the temperatureof the two reservoirs makes it possible toestablish such a scale. If the temperature ofa given system is mesured with thermometersusing different thermoemetric properties, thereis considerable difference among the readings.
Sol–14: (c)
Throttling process is a irreversible process asthe entropy of the fluid increases during theprocess. The first law of thermodynamics onlygives a quantitative estimate of the heat andwork interaction between the system andsurroundings, however, it does not state aboutquality of energy. It is the second law ofthermodynamics which deals with the lowgrade and high grade energy and concepts ofavailability.
Sol–15: (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–16: (c)
The reciprocating compressor can beconsidered as steady flow system providedthe control volume includes the receiver whichreduces the fluctuation of flow.
For electric heater, at steady state
W = Q
I
W C.V.
I
Q
Sol–17: (a)A reversible process must be quasi-static andfrictionless.Further, a heat engine cycle in which there isa temperature difference (i) between the sourceand the working fluid during heat supply, and(ii) between the working fluid and the sinkduring heat rejection, exhibits external thermalirreversibility. Thus, P and T of the workingsubstance must not differ, appreciably fromthose of the surroundings at any state in theprocess.
Sol–18: (b)
The propety of a thermodynamic system is apoint function, since for a given state, there isa definite value for each property. The change
