1995-2017 - ies master publication · 2017-12-22 · q–1: a water storage tank 10 m × 10 m × 10...
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1995-2017
Office: F-126, (Lower Basement), Katwaria Sarai, New Delhi-110 016
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MECHANICAL ENGINEERINGESE SUBJECTWISE
CONVENTIONAL SOLVED PAPER-I
First Edition : 2017
Typeset at : IES Master Publication, New Delhi-110016
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016
Phone : 011-26522064, Mobile : 8130909220, 9711853908
E-mail : [email protected], [email protected]
Web : iesmasterpublications.com, iesmaster.org
All rights reserved.
Copyright © 2017, by IES MASTER Publications. No part of this booklet may bereproduced, or distributed in any form or by any means, electronic, mechanical,photocopying, recording, or otherwise or stored in a database or retrieval system withoutthe prior permission of IES MASTER, New Delhi. Violates are liable to be legallyprosecuted.
It is an immense pleasure to present topic wise previous years solved paper of Engineering Services Exam.This booklet has come out after long observation and detailed interaction with the students preparing forEngineering Services Exam and includes detailed explanation to all questions. The approach has been toprovide explanation in such a way that just by going through the solutions, students will be able to understandthe basic concepts and will apply these concepts in solving other questions that might be asked in futureexams.
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 has been themain reason for students opting for this service as compared to others. To facilitate selection into theseservices, availability of arithmetic solution to previous year paper is the need of the day. Towards this endthis book becomes indispensable.
IES Master PublicationNew Delhi
PREFACE
CONTENTS
1. FLUID MECHANICS 01 – 166
2. THERMODYNAMICS 167 – 242
3. HEAT TRANSFER 243 – 336
4. REFRIGERATION AND AIR CONDITIONING 337 – 424
5. POWER PLANT ENGINEERING 425 – 584
6. IC ENGINE 585 – 652
7. RENEWABLE SOURCES OF ENERGY 653 – 658
SYLLABUS
Basic Concepts and Properties of Fluids, Manometry, Fluid Statics, Buoyancy, Equations of Motion, Bernoulli’sequation and applications, Viscous flow of incompressible fluids, Laminar and Turbulent flows, Flowthrough pipes and head losses in pipes. Reciprocating and Rotary pumps, Pelton wheel, Kaplan andFrancis Turbines and velocity traingles.
CONTENTS
1. Fluid Statics and Buoyancy ......................................................................... 01–14
2. Fluid Kinematics ........................................................................................ 15–18
3. Fluid Dynamics and Flow Measurement ...................................................... 19–36
4. Laminar and Turbulent Flow......................................................................... 37–52
5. Boundary Layer Theory, Drag and Lift ......................................................... 53–70
6. Pipe and Open Channel Flow ...................................................................... 71–84
7. Dimensional and Model Analysis ................................................................ 85–113
8. Fluid Jet and Hydraulic Turbines .............................................................. 114–139
9. Pumps and other Hydraulic Machines ...................................................... 140–166
UNIT 1 FLUID MECHANICS
Q–1: A water storage tank 10 m × 10 m × 10 m has a drainage opening on one of the vertical sidesat the bottom which is trapezoidal in shape with a width 2 m at the bottom 4 m at the top and1 m height. A gate of same dimension hinged along the top edge is used to close it. What is theminimum horizontal force required to be applied at the bottom to keep the gate closed if the tankhas full of water in it? Will there by any change in the force required if the tank is only half full?If yes how much? [10 Marks ESE–2014]
Sol–1: The gate is trapezoidal shape of the following dimensions. A1, A2 and A3 represent the area of respectiveportion of gate as shown in figure.
h=1m
b=2m
A1=0.5×1×1 =0.5m2
G
a=4m
A3=0.5×1×1 =0.5m2x1 x3
A2
x2
x
Enlarge view of gate
h=1.0mG
P
B
10m
A
Z–
49 m
HH=10m
The centroid of trapezoidal gate
x =
1 1 2 2 3 3
1 2 3
1 1 10.5 2 0.5x A x A x A 3 2 3A A A 0.5 2 0.5
=
113
3
=
43 3
= 4 m9
This gate is fitted to a wall of size 10m × 10m as shown in figure.
The depth of centroid of gate from top
Z = 499
= 85 m9
The area moment of gate about its centre of gravity,
IG =
2 2 2 23 3 4a 4ab b 4 4 4 2 2h 1 0.2407 m
36 a b 36 4 2
The depth of centre of pressure
H = GIZA Z
=
85 0.2407 9 85 0.0085 9.453 m9 3 85 9
Fluid Statics and BuoyancyCHAPTER1
ESE Subjectwise Conventional Solved Paper-I 3
IES MASTER Publications
Mechanical Engineering
Total pressure force on gate
F = 85gZA 1000 9.81 3 N=278 kN9
Let the applied force at bottom of gate is F0, so
Fo = F (H 9) 125.93 kN
(H 9)
Now the tank is half full,
The centroid, Z =4 404 m9 9
Centre of pressure, H =
40 0.2407 9 40 0.01805 4.4625 m9 3 40 9
Total pressure force, F = gZ A = 401000 9.81 3 N9
= 130.8 kN
Now the force to hold the gate, oF =F (H 4) 60.5 kN
H 4)
Hence the holding force reduces from 125.93 kN to 60.5 kN.
Q–2: A tank with the vertical sides measuring 3m × 3m contains water to a detph of 1.2 m. An oil ofdensity 900 kg/m3 was poured in the tank up to a depth of 0.8 m. The vertical wall can withstandthe trust of 58 kN. Calculate the actual thrust on the wall and centre of pressure. If the oil levelis increased up to 0.9 m, what will be stability of the wall? [10 Makrs ESE–2012]
Sol–2: The size of wall is 3m × 3m. Density of oil, 30 900kg m . Since the oil is lighter than water so it will
float on water as shown in figure.
1.2m
3m0.8mP1 B
Oil
F
A
P2
P3
E D C
The pressure force due to oil,
P1 = o1 1gh.A 900 9.81 0.8 0.8 3=8475.84N2 2
The increase in pressure force on lower zone i.e. waterdue to pouring of oil,
P2 = o gh.A 900 × 9.81× 0.8 × 1.2 × 3 = 25427.52 N
The pressure force due to water,
P3 = 1 1gh.A 1000 9.81 1.2 1.2 3 = 21189.6 N2 2
Total pressure force on wallP = P1 + P2 + P3 = 8475.84 + 25427.52 + 21189.6 = 55092.46 N = 55.093 kNSince the total force is less than critical thrust of 58 kN so the wall in safe.
The centre of pressure from bottom,
x =3 2 1
FD BC ABP P P BC3 2 3
P
=
0.821189.6 0.4 25427.52 0.6 8475.84x 1.23
55092.96
=36163.6
55092.86 = 0.6564 m
Fluid Mechanics 4
From top of wall = 3 – 0.6564 = 2.3436 m
Now the depth of oil is increased to 0.9m from 0.8m
Ptotal = 1 2 3P P P
=900 9.81 0.9 0.9 3 900 9.81 0.9 1.2 3 21189.6
2
= 10727.235 + 28605.96 + 21189.6 = 60522.795 N = 60.523 kN
This total force is more than critical thrust (58 kN) and the wall will fail.
Q–3: A hydraulic lift of the type commonly used for greasing automobiles consists of a 280 mmdiameter ram that slides in a 280.18 mm cylinder. The annular space between the ram and cylinderis filled with oil having a kinematic viscosity of 0.00042 m2/s and specific gravity of 0.86. If the rateof travel of the ram is 0.22 m/s, find the frictional resistance when 2m of the ram is engaged inthe cylinder. [5 Marks ESE–2011]
Sol–3: The schematic of ram in cylinder, V=0.22m/sec
D =280.18mm0
D=280mm
Ram
Oil
h
Kinematic viscosity of oil, V = 0.00042 m2/sec.
Specific gravity, = 0.86
The shear stress,
= 2 2
3du V 0 V 0.00042 860 0.22 N m 882.93 N/mdy h h 0.09 10
The force on piston/ram i.e. friction resistance,
F = .A = 0 DL = 882.93 × × 0.28×2 = 1553.3 N
Q–4: A solid, half-cylinder-shaped log of 0.48 m radius and 2.5 m long, floats in water with the flat faceup.(i) If the immersion depth of the lowest point is 0.3, what is the uniform specific weight of the
log?(ii) The log tilts about its axis (zero and net applied force), by less than 22°. Is it in stable
equilibrium? Justify your answer with a sketch and logic.(iii) If the log tilts by 18° (left side down; zero net applied force), what is the magnitude and sense
of any moment that results? [15 Marks ESE-2011]
Sol–4: The half cylinder in water,
B
P
h = 0.3m
N
R = 0.48m
GB
0Ml =
2.5m
A
Let the specific density of wood is , then in floatingcondition, Buoyancy force,
FB = W
Area ABP × Length 0g = 2R L g2
0Area ABP× =
2R2
...(i)
The area ABP is required to be calculated separately as
Area ABP = Area OAPBO – Area OAB
=
22 R 2 Area of triangle OAN
2
= 2 1R 2 AN ON2
ESE Subjectwise Conventional Solved Paper-I 5
IES MASTER Publications
Mechanical Engineering
where, cos =ONAO
= R h
R
= h1R
= 0.310.48
= 68º
Area ABP = 2 168 0.48 2 OA sin ON180 2
= 1.187 × 0.482 – 0.48 × sin68 × (0.48–0.3)
= 0.27344 – 0.080107 = 0.1933m2
From equation (i),
0.1933 × 1000 =
20.482
=
32
193.3 2 534.12 kg/m0.48
Density of wood, = 534.1 kg/m3
Specific density = 0.5341
Distance of centre of Gravity of log cross-ections
OG = 4R3
= 4 0.48
3
= 0.204m
The location of centre of Buoyancy (B) is centre ofgravity of log cross-section inside water (B).
A
P
B
G2
0.07m
x0.06m
0.18mR
O
M
G1
N
So, location of centre of gravity of cross-section OABP (OG2) as,
x =
2 sin 2 0.48 sin68R 0.25 m3 3 1.187
NG2 = OG2 – ON = x – ON = 0.25 – 0.18 = 0.07mNow to get location of centre of Buoyancy ‘B’ inside cross section. Take the moment of areas OAB andABG about G2.
BG2 =
1 21 2 Area OAB NG NGArea of OAB×G GArea ABP Area ABP
= 0.18 0.48sin68 0.06 0.07
0.1933
= 0.054m
Distance of centre of Buoyancy from plane surface of log,OB = ON + NG2 + G2B
= 0.18 + 0.07 + 0.054 = 0.304 m
Distance between centre of gravity ‘G’ and centre of Buoyancy, BGB = OB – OG = 0.304 – 0.204 = 0.1 m
Moment of area of surface of log in the plane of water.
I = 31 L AB12
= 31 2.5 2 Rsin6812
= 0.147 m4
Volume of displaced liquid,V = Area ABP × length
= 0.1933 × 2.5 = 0.48325 m3
Metacentric hight GM = I 0.147BM GB GB 0.1V 0.48325
= 0.3042 – 0.1 = 0.2042 m
