CL MEDIA (P) LTD. First Edition : 2018 Revised Edition : 2019 © PUBLISHER

Administrative and Production Offices

No part of this book may be reproduced in a retrieval system or transmitted, in any form or by any means, electronics, mechanical, photocopying, recording, scanning and or without the written permission of the publisher.

Published by : CL Media (P) Ltd. A-45, Mohan Cooperative Industrial Area, Near Mohan Estate Metro Station, New Delhi - 110044 Marketed by : G.K. Publications (P) Ltd.

ISBN

: 978-93-89310-94-8

Typeset by : CL Media DTP Unit

A-45, Mohan Cooperative Industrial Area, Near Mohan Estate Metro Station, New Delhi - 110044

For product information : ISBN-93-87444-84-3 Visit www.gkpublications.com or email to [email protected]

Preface GK Publications has been the “publisher of choice” to aspirants preparing for Staff Selection Commissions’ coveted Junior Engineer recruitment examination. Our SSC JE series for this exam has been segmented into study guides and practice papers in Hindi and English modes for Civil, Electrical and Mechanical streams. With last year’s exam going online, we put in effort to help students prepare better with our online test series. Our web portal gave access to full length mock tests which imitated the actual exam and helped students get familiar with online exam environment & improved analytics. Our mechanical stream study guide is vastly popular among candidates taking up Paper II – Conventional of SSC JE. The study guide comprises of numerous practice questions, adhering to the exam pattern, as laid out by SSC. Complete solutions to these questions have been given towards end to help students assess the actual level of their preparation. This book has solved papers of last 11 years, from 2007 to 2017 with exhaustive explanatory solutions to all questions. This valuable practice material help students become acquainted with exam pattern and have better understanding, study and self-practice to face Paper – II with confidence. Timing while practicing with this textbook will also escalate their speed and accuracy. We hope that the book meets expectations of future Junior Mechanical Engineers. Due care has been taken to keep the book error free. However, any and all erroneous statements and feedback to improve the book are always welcome. Wishing a prestigious career to all students, All the Best! Team GKP

Contents 

Exam Pattern



Syllabus

SOLVED PAPERS  2017

1 – 17

 2016

1–8

 2015

1 – 14

 2014

1 – 10

 2013

1 – 13

 2012

1 – 13

 2011

1–6

 2010

1 – 11

 2009

1–9

 2008

1–7

 2007

1–7

 Practice Paper-1

1 – 12

 Practice Paper-2

1 – 13

Exam Pattern Papers

Subject

Paper – I Objective Type (Computer Based Test)

(i) General Intelligence & Reasoning (ii) General Awareness (iii) General Engineering

Maximum Marks

Duration

50 50 100

2 Hours

300

2 Hours

Paper – II Conventional Type (Paper Based Test)

General Engineering

Cut-offs Marks Paper I Mechanical 2017 2016 2015 Category Cut-off Candidates Cut-off Candidates Cut-off Candidates Marks Available Marks Available Marks Available SC 120.00 235 99.00 369 114.75 261 ST 114.50 139 94.50 176 105.50 191 OBC 133.25 413 109.50 1094 125.25 888 OH 113.00 18 87.00 40 100.00 39 HH 83.50 18 54.00 36 80.50 15 UR 136.25 599 115.00 725 131.00 407 Total 1422 2440 1801

Cut-off Marks 102.50 93.75 109.50 93.00 69.00 117.50

2014 Candidates Available 243 123 1068 27 25 649 2135

Paper II Mechanical Category Cut-off Marks SC 268.50 ST 265.50 OBC 299.00 OH 303.50 HH 247.50 UR 164.75 Total

2017 Candidates Available 44 21 110 132 3 9 319

2016 2015 Cut-off Candidates Cut-off Marks Available Marks 210.50 169 50 206.50 81 50 245.75 285 62 205.25 15 40 126.00 16 40 285.75 610 107 1176

2014 Cut-off Candidates Marks Available 149 108 126 55 180 471 124 16 64 10 215 190* 850

Syllabus Theory of Machines and Machine Design Concept of simple machine, Four bar linkage and link motion, Flywheels and fluctuation of energy, Power transmission by belts – V-belts and Flat belts, Clutches – Plate and Conical clutch, Gears – Type of gears, gear profile and gear ratio calculation, Governors – Principles and classification, Riveted joint, Cams, Bearings, Friction in collars and pivots.

Engineering Mechanics and Strength of Materials Equilibrium of Forces, Law of motion, Friction, Concepts of stress and strain, Elastic limit and elastic constants, Bending moments and shear force diagram, Stress in composite bars, Torsion of circular shafts, Bucking of columns – Euler’s and Rankin’s theories, Thin walled pressure vessels.

Thermodynamics Properties of Pure Substances : p-v & P-T diagrams of pure substance like H2O, Introduction of steam table with respect to steam generation process; definition of saturation, wet & superheated status. Definition of dryness fraction of steam, degree of superheat of steam. H-s chart of steam (Mollier’s Chart).

1st Law of Thermodynamics Definition of stored energy & internal energy, 1st Law of Thermodynamics of cyclic process, Non Flow Energy Equation, Flow Energy & Definition of Enthalpy, Conditions for Steady State Steady Flow; Steady State Steady Flow Energy Equation.

2nd Law of Thermodynamics Definition of Sink, Source Reservoir of Heat, Heat Engine, Heat Pump & Refrigerator; Thermal Efficiency of Heat Engines & co-efficient of performance of Refrigerators, Kelvin – Planck & Clausius Statements of 2nd Law of Thermodynamics, Absolute or Thermodynamic Scale of temperature, Clausius Integral, Entropy, Entropy change calculation of ideal gas processes. Carnot Cycle & Carnot Efficiency, PMM-2; definition & its impossibility.

Air Standard Cycles & IC Engines Otto Cycle; Plot on P-V, T-S Planes; Thermal Efficiency, Diesel Cycle; Plot on P-V, T-S Planes; Thermal Efficiency. IC Engine Performance, IC Engine Combustion, IC Engine Cooling & Lubrication.

Power Plant Engineering Simple Rankine Cycle Plot on P-V, T-S, h-s Planes, Rankine Cycle Efficiency with & without Pump work. Boilers; Classification; Specification; Fittings & Accessories : Fire Tube & Water Tube Boilers. Air Compressors & their Cycles; Refrigeration Cycles; Principle of a Refrigeration Plant; Nozzles & Steam Turbines; Fluid Mechanics & Machinery.

Fluid Mechanics Ideal & Real Fluids, Newton’s Law of Viscosity, Newtonian and Non-Newtonian Fluids, Compressible and Incompressible Fluids.

Fluid Statics Pressure at a Point.

Measurement of Fluid Pressure Manometers, U-tube, Inclined Tube.

Fluid Kinematics Stream line, Laminar & Turbulent Flow, External & Internal Flow, Continuity Equation.

Dynamics of Ideal Fluids Bernoulli’s Equation, Total Head; Velocity Head; Pressure Head; Application of Bernoulli’s Equitation.

Measurement of Flow Rate Basic Principles Venturimeter, Pilot Tube, Orifice Meter.

Hydraulic Turbines Classifications, Principles.

Centrifugal Pumps Classifications, Principles, Performance.

SOLVED PAPER 2017

1

Solved Paper 2017 Mechanical Engineering (Paper II) 1. (a) Define the following: (i) Reversible and Irreversible process (ii) External and Internal irreversibility (iii) Intensive and Extensive properties (b) Describe the following: (i) Clausius Statement (ii) Kelvin-Planck Statement (iii) Perpetual motion machine of the second kind (c) Volume of 0.1 m3 of an ideal gas at 300 K and 1 bar is compressed adiabatically to 8 bar. It is then cooled at constant volume and further expanded isothermally so as to reach the condition from where it started. Determine: (i) Pressure at the end of constant volume cooling (ii) Change in internal energy during constant volume process (iii) Net work done and heat transferred during the cycle. Take cp = 14.3 kJ/kg K and cv = 10.2kJ/kg K. (d) A reversible heat engine operates between two reservoirs at temperatures 700°C and 50 ° C. The engine drives a reversible refrigerator which operates between reservoirs at temperatures of 50 ° C and – 25°C. The heat transfer to the engine is 2500 kJ and the net work output of the combined engine refrigerator plant is 400 kJ (i) Calculate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 50°C. (ii) Reconsider (i) given that the efficiency of the heat engine and the C.O.P. of the refrigerator are each 45 percent of their maximum possible values.

2. (a) Give the comparisons between Otto cycle, Diesel cycle and Dual cycle. (b) An air standard Otto cycle is to be designed according to the following specifications. Pressure at the start of the compression process = 101 kPa; Temperature at the start of the compression process = 300 K; Compression ratio = 8; Maximum pressure in the cycle = 8.0 MPa. Find (i) the net work output per unit mass of air (ii) cycle efficiency (iii) MEP (c) Explain the effect of Superheating and Subcooling on vapour compression refrigeration cycle. (d) An air standard Brayton cycle has air entering the compressor at 100 kPa and 27°C. The pressure ratio is 10 and the maximum allowable temperature in the cycle is 1350 K. Determine. (i) temperatures at salient points of the cycle (ii) compressor and turbine work per unit mass of air (iii) net work output and work ratio (iv) thermal efficiency of the cycle (v) specific air consumption in kg/KWh (vi) improvement in the thermal efficiency of the cycle if a regenerator with 100% effectiveness is incorporated in the cycle 3. (a) Define density, specific volume, weight density, specific gravity and Bulk Modulus. (b) A ship weighing 4000 tons and having an area of 465 m2 at water line submerging to a depth of 4.5 m in sea water with a density of 1024 kg/m3 moves to fresh water. Determine the depth of submergence in fresh water. Assume that the sides are vertical at the water line.

2

SOLVED PAPER 2017

(c) What is cavitation? How does it affect the performance of hydraulic machines? (d) The following details refer to a centrifugal pump: Outer diameter : 30 cm, Eye diameter : 15 cm, Blade angle at inlet : 30°, Blade angle at outlet : 25°, Speed 1450 rpm. The flow velocity remains constant. The whirl at inlet is zero. Determine the work done per kg. If the manometric efficiency is 82%, determine the working head. If width at outlet is 2 m, determine the power 0 = 76%. 4. (a) Write short notes on the following : (i )

Stainless steel

(ii) High speed steel (iii) High carbon steel (b) With the help of figure, describe the Shielded Metal Arc Welding process. (c) Explain the different operations performed in grinding machine. (d) Mention the differences between shaper and planer machine tools. 5. (a) Give the classification of kinematic pairs.

(b) An engine, running at 150 r.p.m., drives a line shaft by means of a belt. The engine pulley is 750 mm diameter and the pulley on the line shaft being 450 mm. A 900 mm diameter pulley on the line shaft drives a 150 mm diameter pulley keyed to a dynamo shaft. Calculate the speed of the dynamo shaft, when (i) there is no slip, and (ii) there is a slip of 2% at each drive. (c) Mention the comparison between involute and cycloidal gears. (d) Explain the term height of the governor. Derive an expression for the height in the case of a Watt governor. 6. (a) Three forces of 2P, 3P and 4P act along three sides of an equilateral triangle of side 100 mm taken in order. Find the magnitude and position of the resultant force. (b) A body of weight 300 N is lying on a rough horizontal plane having a coefficient of friction as 0.3. Find the magnitude of the force, which can move the body, while acting at an angle of 25° with the horizontal. (c) Derive the expression for the shear stress in a circular shaft subjected to torsion. (d) Derive the expression for circumferential stress in a thin cylindrical vessel.

SOLVED PAPER 2017

3

EXPLANATIONS 1. (a) (i) A reversible process is one which is performed in such a way that at the end of the process, both the system and the surrounding may be restored to their initial states, without producing any effect in the rest of the universe. Example : Compression of spring The process is said to be an irreversible process if it cannot return the system and the surroundings to their original conditions when the process is reversed. The irreversible process is not at equilibrium throughout the process. Expansion of gas in piston cylinder arrangement. (ii) Externally Reversible Processes  No irreversibilities surroundings.

exist

in

the

 Heat transfer can occur between the system and the surroundings, but only with an infinitesimal temperature difference.

1. (b) (i) Clausius statement : Clausius statement states "it is impossible for a self acting machine working in a cyclic process without any external force, to transfer heat from a body at a lower temperature to a body at a higher temperature. It considers transformation of heat between two heat reservoirs. This statement tells us that it is impossible for any device, unaided by an external agency, to transfer energy as heat from a cooler body to a hotter body. Consider the case of a refrigerator or a heat pump (Figure)

High-Temperature Thermal Reservoir QH

Refrigerator

W

 There may be irreversibilities within the system. Internally Reversible Processes  No irreversibilities exist within the system.  The system moves slowly and without friction through a series of equilibrium states. It could also be due to turbulence, viscosity, resistance etc.  Irreversibilities may exist in the surroundings, usually due to heat transfer through a finite temperature difference. (iii) Intensive and extensive properties Thermodynamic properties can be divided into intensive and extensive properties. An intensive property is a physical property of a system that does not depend on the system size or the amount of material in the system. An extensive property of a system does depend on the system size or the amount of material in the system. Density, pressure and temperature are intensive properties and volume, internal energy are extensive properties.

QL Low-Temperature Thermal Reservoir When W = 0, (COP)R   and (COP)HP   It is impossible to construct a refrigerator or a heat pump whose COP is infinity. Consider a domestic refrigerator, this device extracts energy as heat from the substance to be cooled and transfers it to the surroundings. The refrigerator is supplied with electric power. Energy transfer as heat from a high temperature body to a low temperature body is a spontaneous process. (ii) Kelvin - Planck statement : Kelvin Planck statement states "It is impossible to construct an engine, which is operating in a cycle produces no other effect except to external heat from a single reservoir and do equivalent amount of work. It considers the transformation of heat into work.

4

SOLVED PAPER 2017

(iii)Perpetual motion machine of second kind (PMM2) : Perpetual motion machines of second kind are those machines that violates the second law of thermodynamics because such machines will absorb continuously heat energy from a single thermal reservoir and will convert the absorbed heat energy completely into work energy. Such machines will have efficiency 100%, or we can say that perpetual motion machines of second kind will have higher efficiency than ideal Carnot cycle and in my opinion, it is not possible. Therefore, we can say that perpetual motion machine of second kind can not exist as such machines violates the second law of thermodynamics.

Heat Source

PMM 2

c v  10.2 kg / kg.v .

Cosider PV Diagram  (i) Pressure at the end of constant volume cooling Cv

 1

P   T Also 2   2  T1  P1 

Consider Process 3 – 1, we have – P3V3 = P1V1

PV 1  0.1  P3  1 1   4.4 Bar V3 0.0227 (ii) Change in internal energy during constant volume Process (V3–V2) mass of gas



V  20.27KJAns

  Ve Sign  decrease in internal energy (iii) Net work done & heat transferred during the cycle:P V  P2 V2 m R  T1  T2  W1 2  1 1   1  1

14.3  1.402 10.2



V  P  W3 1   P3 V3 log e  1   P1 V1 log e  3   P1   V3 

PV = C

P3 V3  P1 V1 

3 1

1 Volume (m3)

V2 / V3

We know that  gas constant R = CP – PV = 14.3 – 10.2 = 4.1 kj/kg k Consider Process 1 – 2 we have 

P1 V1

0.00813  4.1  300  544.5

 20.27 KJ. 1.402  1 W2 – 3 = 0 (Volume remain constant)

2

P3



 0.00813  10.2  300  544.5

Pressure (Bar) 8

1.402 1

 8  1.402    1.815  1

or T2  T1  1.815  300  1.815  544.5 K

P1= 1 Bar

Cp = 14.3 KJ/kg k,



1

 1  1.402  V2  0.1     V2  0.0227 m3  8

Change in internal energy during constant volume process (2 – 3) U3 – U2 = mcv (T3 – T2)

W=Q

1. (c) V1 = 0.1 m3T1 = 300k

Cp

1/ 

1  105  0.1 P1 V1 m   0.00813 kg RT1  4.1  1000  300

Q

(P3)   

P   V2  V1  1   P2 



P2 V2





 4  4  1  105  0.1log e  = 14816 N – M(J)  1  = 14.82 kJ Net work done = W1 – 2 + W2 – 3 + W3 – 1 = (– 20.27) + 0 + 14.02 = 5.45kJ Ans for a cycling Process;

     w

 Heat transferred during the complete cycle = – 5.45 kJ Ans

SOLVED PAPER 2017

1. ( d) Combined engine Refrigerator Plant  Engine Source Temp.

(T1 )  700C  700  373  973K

Heat Transfer to the Refrigerator

 Q 4  4198.62 kJ Ans W2 = Q3 – Q4  1270  Q3  4198.62

Engine Sink Temp.

 Q3  5468.62 kJ

(T2 )  50C  50  273  323K Heat Transfer to engine (Q1) = 2500 kJ Refrigerator Source temp.

Therefore Q2 + Q4 = Net heat supplied to the reservoir at 50 C

830  4198.62  5028.62 kJ

(T4 )  25C  25  273  248K

(b) For irreversible engine with

Refrigerator Sink temp.

   0.45  (rev. )

(T3 )  50C  50  273  323K Net work transfer (W1–W2) = 400 kJ

Q4 & (Q2 + Q3)

T  T2 W1 Q1  Q2   (a) rev  1 T2 Q1 Q1

   0.45(rev. )

COP  0.45COP  rev.

rev  0.6680 Sour ce Q1

  COP  0.45  3.3067

Source Q4 Q4

Q1 W1

& COP  0.45   COPrev .

   0.45  0.6680    0.3006

973  323 rev  973

H.E.

5

W2

Referi.

  COP   1.488 

W1 W1  0.3006   W1  751.5 KJ Q1 2500

W1  Q1  Q2  751.5  2500  Q 2  Q2  1748.5 kJ  W1  W1  400 kJ (given)

Q2

Q3 Sink

Sink

rev 

W1 W1  0.668   W1  1670 kJ Q1 2500

Q1  2500 kJ given &

W1  Q1  Q2  1670  2500  Q 2

 Q2  830 kJ again W1–W2 = 400 (given) 1670 – W2 = 400  W2  1270 kJ

COPrev.  Q

Q4 Q T4  4  W2 T3  T4 3  Q4

248

COPrev .  (323  248)  again  COP   rev

COPrev .  3.3067

Q4 Q4  3.306  Q3  Q 4 W2 1270

 Q4  3.306  1270

751.5  W2  400  W2  351.5 kJ Q

COP  W4  2

1.488 

Q4  Q 4  523.03 kJ Ans 351.5

W2  Q3  Q4  351.5  Q3  523.03  Q3  838.53KJ Therefore Q2 + Q3 = 1748.5 + 838.53

Total Heat  2584.03 kJ Ans 2. (a) Comparison of Otto, Diesel and Dual Cycles: Reference of Same Compression Ratio and Heat Addition: The Otto cycle 1-2-3-4-1, the Diesel cycle 1-23'-4'-1 and the Dual cycle 1-2-2"-3"-4"-1 are shown in p-V and T- diagram in Fig. (a) and (b) respectively for the same compression ratio and heat input.

SOLVED PAPER 2017

6

Otto cycle : 1-2-3-4-1 Diesel cycle: 1-2-3'-4'-1 Dual cycle: 1-2-3"-4"-5"-1

3

Reference of Same Compression Ratio and Heat Rejection:

3

4"

3"

3'

2

4' 5" 2 4 T

P

3"

3

4" 3'

3'

P 2

4" 4' 4

4' 45"

1

1

1

3"

2"

V 1-2-3-4 Otto cycle, 1-2-3'-4' Diesel cycle and 1-2-2"-3"-4" Dual cycle.

S V From the T-S diagram, it can be seen that Area 5-2-3-6 = Area 5-2-3'-6' = Area 5-2-2"- 3"-6" as this area represents the heat input which is the same for all cycles. All the cycles start from the same initial state point 1 and the air is compressed from state 1 to 2 as the compression ratio is same.

3

3"

3'

2" T

1

It is seen from the T-s diagram for the same heat input, the heat rejection in Otto cycle (area 5-1-4-6) is minimum and heat rejection in Diesel cycle (5-1-4'-6') is maximum. Consequently, Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and Dual cycle having the efficiency between the two

4'

2 4

4"

5

5" 5' S For the same compression ratio and same heat input 6

The Diesel cycle is more efficient that Dual cycle, which in term is more efficient that Otto cycle.

2. (b) Air Standard otto cycle Otto (Petrol) cycle

QH 3

Expa nsion

P QH

Wc

4

2

WT

T

Qc Comp.

QC

1

V

S

QH = Heat Supplied at Constant Volume Qc = Heat Rejected at Constant Volume Wc = Work done in compression stroke (-) WT = Work done in Power stroke (+)

V1  8 (given) V2

•

Compression Ratio 

•

T1  27 C  27  273  300 K

•

P1= 101 kPa

•

T2  V1   T1  V2 

 1

(Isentropic compression)

V   T2  T1   1   V2 

 1

 T2  300   8 

  T2  689.21K

0.4

 1.4 

SOLVED PAPER 2017

Pressure Ratio,

P2  V1   P1  V2 

1.4

 P4  435.27 kPa

 P2  101   8 

1.4

T4  V3   T3  V4 

P2  1856.29 kPa

•

Process 2-3 addition x-

1

 1  T4  3009.05     8

0.4

 T4  1309.74 K

Constant Volume heat

WT  1220 KJ / kg

 Q  Q4  U  Cv   T3  T2 

Wc  c v   T1  T2   718   300  689.2

P3 T3   Isochoric Pressure  P2 T2

 Wc  279.44kJ / kg So,  W.D. net  WT  WC  1220  279.44

P3 = max. pressure = 8 MPa (given) = 800 kpa

  W  net  940.55 kJ / kg. Ans

8 0 0 0  6 9 8 .2 1 1 8 5 6 .2 9  T3  3009.5 K 

7

T2 

(b) Cycle efficiency   

 1309.74  300   100  1   3009  698.2  

(i) The net work output/mass of air = ?

WT  W D  Comp.rej  Wnet  WT  WC We  W D

th  56.4% Ans

WT  Cv   T3  T4 

v1 

 WT  710  3009.05  T4 

Rair  T1 287  300  P1 101  103

 v1  0.852 m 3 / kg

WT  718  3009.05  1309.74 

v1 0.852   0.106 m 3 / kg 8 8 (c) MEP (Mean effective Pressure)

WT  1220 kJ/kg

v2 

For 4-cycle 

V  P4  V3      P4  P3   3  P3  V4   V4   1  P4  8000     8

Wnet  T  T1  1 4 Q in  T3  T2 



MEP 

1.4

Wnet 940.55  Vmax  Vmin 0.852  0.106

MEP  1260.79 kPa Ans

2. (c) Sub cooling : Refrigeration is the process of extraction of heat, or the transmission of heat by mechanical methods, from one location to another. Subcooling in refrigeration implies cooling the refrigerant in liquid state, at uniform pressure, to a temperature that is less than the saturation temperature, which corresponds to condenser pressure or some pressure P. Effects of Sub cooling : Refrigeration can be improved when a liquid refrigerant is subcooled by a circulation of cold water in the heat exchanger. As a general rule, a 1% increase in refrigeration can be achieved for every 2 degrees of liquid subcooling obtained. Due to this characteristic, designs of condensers are changed to achieve obtain liquid subcooling. Production of flash gas is reduced during the process of expansion.

4

Valve 1

C Condenser

Pressure 3

E

Condensation

4

3

P

Compressor

Evaporator

Subcooling

L

L+V

2 1 Evaporation

V

Compression

2 Superheating

h Enthalpy