Handbook To Ssc Je Mechanical-1

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Handbook to Junior Engineer

Recruitment Exam Guide o Theory with Exercises o Practice Question Bank

Badboys2

Badboys2

sse

Handbook to Junior Engineer

Recruitment Exam Guide o Theory with Exercises o Practice Question Bank

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1.

Engineering Mechanics and Strength of Materials

2.

Theory of Machines & Machine Desig

A-I - A-40 A-4I - A-82

3.

Thermal Engineering

4.

Fluid Mechanics and Machinery

A-136 - A-173

5.

Production Engineering

A-174 - A-220

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A-83 - A-135

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SECTION A : MECHANICAL

ENGINEERING

I~NfJINI~I~IIINfJ)11~(~II1INI(~S lINI) srl'111~NfJrl'IIf)lf )11Irl'I~III1II..S ENGINEERING MECHANICS It is the branch of Engineering Science which deals with the principles of mechanics along with their applications to the field problems. Engineering Mechanics can be divided into its sub-groups as below Engineering Mechanics

I Statics

Dynamics I

Kinematics

Kinetics

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Statics deals with forces in terms of their distribution and effect on a body at absolute or relative rest. Dynamics deals with the study of bodies in motion. Dynamics is further divided into kinematics and kinetics. Kinematics is concerned with the bodies in motion without taking into account the forces which are responsible for the motion. kinematics deals with the bodies in motion and its causes. Force System: A force system may be coplanar/non-coplanar. In a coplanar force system, all the forces act in the same plane. In a non-coplanar force system, all the forces act in different planar. Classification of force system: (For coplanar forces) Forcel system

J

Coplanar I

L

Collinear

Concurrent

Non-coplanar

l

Parallel

Non-concurrent Non-parallel

Concurrent 1.

Parallel

Non-concurrent Non-parallel

(Complete classification of force system) Coplanar collinear : In this case, all the forces act in the same plane and also have a common line of action. (x-y plane) F3 .' F2 Fl

/

•.•»"

./

-:

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Engineering Mechanics and Strength of Materials

A-2

2.

Coplanar concurrent : In this case all the forces act in the same plane and meet or intersect at a common point. (xy plane)

Law of parallelogram

:

According to law of parallelogram, if two forces are acting at a point and may be showed in magnitude and direction by two adjacent sides of the parallelogram, then the resultant ofthe two forces will be shown by the diagonal of the parallelogram in megnitude and direction. Let 'PI and 'Q' are two forces acting at the point '0' Here 'PI and' a' shows the sides ofthe parallelogram and 'R' is the resultant. p

3.

L?7

Coplanar parallel force: All the forces act in a plane and parallel with each other irrespective of direction.

Q

(x y plane)

Let a = Angle between the two forces 'PI and 'Q' a = Angle between resultant 'R' and one of the force ('Q' in this case) = direction of the resultant then,

---~F2 ---~Fl

Resultant 'R'

4.

Coplanar non-concurrent, non-parallel: In this case, the lines of action of these forces act in the same plane but they are neither parallel nor meet intersect at a common point.

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(xy plane)

= ~p2

+ Q2 + 2PQ cos a

Angle made by resultant 'R'

=

Psina Q + P cos a

or, tan a = ( P sin a Q+pcosa

J :::::>a=tan -1(

P sin a Q+pcosa

J

Lami's theorem: According to Lam is theorem, if three forces are acting at a point and the forces are in equilibrium, then the each ofthe three forces is directly proportional to the sine ofthe angle between the other two forces. Let, P, Q, R = Three forces in equilibrium

y Q a, ~, y = Angles included between three forces P, Q and R then, -P-=_g_=~

sma sin B sin y Moment of a force : It is defined or the product of the magnitude of the force and the perpendicular point from the line of action of the force.

z F} (Non-coplanar concurrent forces)

distance of the

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Engineering Mechanics and Strength of Materials Moment (M) = F x r Couple : Two parallel forces equal in magnitude and opposite in direction and separated by a definite distance are said to form a couple.

A-3

by the resultant of normal reaction with the limiting force of friction with the normal reaction.

21R F

I

d

w=mg From fig : tan Action and Reaction: From Newton's third law, for every action there is a equal and opposite reaction.



= : = Il => = tan -1(FIR)

Angle of repose (<X) : When a body rests on an inclined plane, the angle by which the body is at the verge (just) to start moving in terms as angle of repose. Inclined plane with horizontal

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Surface/ Support R

Some conceptor/terms of friction: Friction: Friction may be defined as the resistive force acting at the surface of contact between two bodies that resist motion of one body relative to another. => Based on the nature of two surfaces in contact, friction in categorised in the following two kinds/types. (a) Static friction: When two contact surfaces are at rest, then the force experienced by one surfuce is termed or static friction. (b) Dynamic friction : When one of the two contact bodies starts moving and the other in at rest, the force experienced by the body in motion is called dynamic friction. R

Cone of friction : It is defined as the right circular cone with vertex at point of contact of two surfaces and axis in the direction of normal reactions.

Cone of friction

F (Frictional for~ ~

Horizontal surface w=mg From the fig : R = w= mg P=F If, P is less than F, the body will not move. But, if P is increased after a stages achieved by limiting force of friction, the body will start moving. Co-efficient of friction (J..t) : It is defined as the ratio of limiting force of friction (F) to the normal reaction (R) between two rigid bodies. u

F

-=>F=IlR R Angle of friction (<1»: It is defined as the angle subtended >

Methods of reducing friction : There are many ways to reduce friction some of them are given as follows: 1. Surfaces ofthe mating parts or contacting surfaces should be smooth 2. Lubrication is also implemented for reducing friction by making surfaces smooth 3. Streamlined shapes should be used because these shapes offers least resistance against air flow or water flow. 4. If the forces are reduced on contacting surfaces, the value of friction is reduced 5. Lesser contact between the mating surfaces also reduces the friction.

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Engineering Mechanics and Strength of Materials

A-4

FREE BODY DIAGRAMS A free body diagram (FBD) is a simplified representation of particle or rigid body that is isolated from it's surroundings, and all applied forces and reactions on the body are put together in a diagram. These diagrams are the simplest abstraction of the external forces and moments acting on a physical object. Creating a free body diagram involves mentally separating the system (the portion of the world you are interested in) from its surroundings (the rest ofthe world) and then drawing a simplified representation of the system. All forces acting on a particle, original body must be considered and equally important. Any force not directly applied on the body must be excluded. Let us consider a system of a beam loaded and supported as shown in Fig.

Methods of analysis: (i) Method joints (ii) Method of sections (iii) Caraphical method Displacement, Speed, Velocity and Acceleration Displacement: Change of position of a body with respect to a certain fixed reference point is termed as displacement. Displacement is a vector quantity. Speed: Rate of change of displacement with respect to its surrounding is called as speed of the body. Since the speed of a bodyis irrespectiveof its direction, therefore it is a scalar quantity. Velocity: The rate of change of position of a bodywith respect to time is called velocity.Velocityis a vector quantity. In other way we can say velocityis the speed of a bodyin a particular direction. Acceleration: The rate of change of velocity of a body with respect to time is called acceleration. A negative acceleration is called retardation. NEWTON'S LAWS OF MOTION

The free body diagram (FBD) of the above system can be drawn Badboys2 as in Fig. ~M

There are three laws of motion known as Newton's laws of motion. 1. Newton's first law of motion: This law statesthat if a bodyis in the state of rest it remains in the state of rest and ifit is in motion it remains in the state of motion until the body is acted upon by some external force. 2. Newton's second law of motion: It statesthattherateofchange of momentum is directlyproportional to the impressed force, and take place in the same direction, in which the force acts. Momentum = mv d(mv) dt

w

--=F

-

IF=m~=mal Beam is subjected to following set of forces after the beam is detached from the supports. (a) Weightofthe beam W acting verticallydownwardsthrough mass centre of the beam. (b) Reaction Rt, normal to the beam at its smooth contact with the corner. (c) Horizontal applied force P and couple M (d) Vertical and horizontal reactions (Ravand Rah)extented at the pin connection at B. => Principle of equilibrium/ Equilibrium conditions : According to the principle of equilibrium, A body, either in co-pl-anar or concurrent or parallel system, will be in equilibrium if the algebric sum of all the external forces is zero and also algebric sum of moments of all the external forces about any point in their plane is zero. So, LF = 0, LM = 0 => Equilibrium equations for non-concurrent forces LF x = 0' LF y = 0 , LM = 0 => Equilibrium equations for concurrent forces LF x = 0, LF y = 0 (only two conditions are required)

where

In = mass

of the body

v = velocity of the body F

= Force

acting on the body = acceleration produced in the body. Newton's third law of motion: This law states that there is always an equal and opposite reaction to every action.

GRAVITATIONAL

LAW

a

3.

Gravitational law is also known as universal law of gravitation. According to this law, Every substance or body has an attractive force with another substance or body and this attractive force is directly proportional to the product oftheir masses and inversely proportional to the square of distance between their centers. This attrative force is directed along the line which joins the centers of bodies. Let MJ and M2be the masses of two bodies and 'R' be the distance between the centers of two bodies. 'F' be the attractive force or force of attraction between those bodies. Now, According to law, F o: MJ X M2

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Engineering Mechanics and Strength of Materials

A-5

Stress developed in one direction ~ uniaxial state of stress Stress developed in two direction ~ biaxial state of stress Stress developed in three direction ~ triaxial state of stress

1 and Foc2 R On combining the above two expressions,

Units of Stress

Foc MIM2 R2

1 Pa = 1 N/m2

SI : Pa, MPa, GPa

F=G MIM2 R2 where, G = universal gravitational constant = 6.67 x 10-11 NM21 kg? ANGULAR DISPLACEMENT The displacement of a body in rotation is called angular displacement.Angular displacement is a vector quantity.Angular displacementO can be measuredin radians, degreesor revolutions. 1 revolution = 21tradians = 360 degrees

MKS : kgf/cm? 1 MPa = 106 N/m2 1GPa = 109 N/m2 NOIE When deformation or strain occurs freely in a direction, • stress developed in that direction is zero. • When deformationis restrictedcompletely,or partially stress is developed. Hence strain is the cause of stress. Type of Stresses

I

ANGULAR VELOCITY The rate of change of angular displacement of a bodywith respect to time is called angular velocity.

de

Normal Stress (acting perpendicular to corresponding plane)

I

0)=-

dt if a bodyis rotating at N r.p.m. then correspondingangular velocity 21tN

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0)=

60 rad/s

I

Axi I

Shear Stress (acting parallel to corresponding plane)

I

I

B di

Direct shear stress

Torsional shear stress

~mg

If the body is rotating 0) rad/s along a circular path of radius r, then its linear velocity (v) is given by v

1kg;:::::0.1 MPa cm

= (l}r

ANGULAR ACCELERATION

Tensile Compressive

STRESS TENSOR

The rate of change of angular velocity is called angular acceleration. It is expressed in radls2

dO)

a=cit

Centripetal and Centrifugal Force Essential force for a circular motion acting radially inwards is called centripetal force which is given by Fe = mv?r where m is the mass of the body w = angular velocity r = radius ofthe circular path As per Newton's third law of motion, the body must exert a force radially outwards of equal.

ox , oy .o z are normal stresses Remaining are shear stress.

STRENGTH OF MATERIALS Load : It is defined as external force or couple to which a component is subjected during its functionality. Stress: It is defined as the intensity of internal resisting force developed at a point against the deformation cuased due to the load acting at the member. , p

·· ..,,·

••• p ----

------

------

,

/

-----_

, , ,

.. ,

•• _•• _ ... _.-

\

L

F d cr=A

Z face

Zxy =Z yx' Zxz =Z zx' Z~ =Z yz Plane Stress problems are those in which the stress acting in one of the mutual perpendicular directions is assumed to be zero .. crz =0 Zxz =0 Zyz =0 [cr]=[crx crxy] cryx cry For a given stress tensor

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Engineering Mechanics and Strength of Materials

A-6

rr =

100 120 10] 20 50 0 [ 10 0 25

Units: MPa

p+-~~~----~-~----------------------------------------------¥------~-!--+p I

I

I

I

:

Zxz = 10 (i.e., shear stress acting onx face along Z direction) Z zy = 0 ox = 100'y o = 50'zo = 25

La

:

I

I

:

L,

~'---------------------+'

:

Elongation of a bar Subjected to axial load P

Oll

'b=~

AE

p

Cv

Elongation of a tapered bar subjected to axial load P

=v: 'bV

s, +cy +cz

Another example of rectangular block is considered

2

Elongation of a prismatic bar under its self weight

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= yL

2E

y = selfweight per unit volume

DIL :

V=/bt 'bV

Cv

: 8

'1.

0/

ob

=--:Y-=T+-';+t

ot

For a sphere,

1

L2 Elongation of a conical bar under its self weight = ~E

Cv

oD

=-

D

D : diameter of sphere

SHEAR STRESS

STRAIN Strain is defined as the ratio of change in dimension to original dimension.

It is defined as the change in initial right angle between two line elements which are parallel to x and y axes respectively. p

Strain

B

B'

I L

Normal Strain

Longitudinal Strain (Slong)

Shear Strain

I

Lateral Strain (Slateral)

Shear strain = Shear angle (
. ..

I

BJ

b

I STRAIN TENSOR Compressive Tensile Strain Strain Consider a rod oflength La subjected to load P

•

Strain tensor is used to define the state of a strain at a point c : normal strain y: shear strain

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Engineering Mechanics and Strength of Materials

•

Shear strain like complementary shear stress are equal in magnitude but opposite in direction.

= [Ex

[EhD

YXY/2] Ey

Y yx/2

A-7

Shear Modulus or Modulus of Rigidity As per Hooke's law, Shear stress ex. shear strain

1't=Gyl

Yxy=Yyx

for a given

Strain Tensor in 3D

Ex [EhD

= Y yxJ2 [

Y2xJ2

1

't, G ex. -.

Y

Y xy/2

Y XZ/2]

Ey

Y yz/2

Y zy/2

Ez

Bulk Modulus (K) K=

o

Normal stress

(J

Relationship Between Elastic Constants

-----------,. .. "

E=2G(1 +J.!) E=3K(1-2J.!)

~_+--____,,_,,"''',...

9KG E= 3K+G --+-~(J

E

1 1 1 1 1 1 1 1

I

1

G=-x-2 1+J.!

........

1 1 1

",...",..,;

E

1

(J

Young's Modulus or Modulus of Elasticity Badboys2 As per Hooke's law upto proportional limit normal

HYDROSTATICSTATEOF STRESS (NO DISTORTION, ONLY VOLUME CHANGES) stress is

directly propotional to longitudinal strain o ex. E)ong o = E = young's modulus E10ng E t => E10ng.J, => 0 I J.. A material having higher E value is chosen EMS = 200 GPa ECI= lOOGPa E

-------------~".",..

-

K=-x-3 1-2J.! Value of any Ee ~ 0 Note: J.!cork =0

= 200 GPa AI 3

K=oo

For

=>

Ev

=

0v

=0

The material neither expands in volume nor contracts in volume. Thus, it is called as incompressible material and for that J.!= 0.5.

Poisson's Ratio ~

used to determine lateral strain theoretically. -lateral strain I J.!= I longitudinal strain

.. (oI)MS < (ol)CI < (0)Ai

Elastic Limit: Maximum value of stress upto which a material can be completely elastic.

I

ProportionalLimit It is the maximum value of stress upto which materials obey

----

Hooke's Law. EL ----------------------

f---------------

B

Lo

1..-

PL

-------------

~

..

Lr

" 'E' is the slope of o vis ~ E diagram

J.!= Load

Engineering Stress (o) = 0" I . ngma x section area True stress =

Instantaneous load Instantaneous

. x section

IfJ.! = 0

oct

=> - = 0 do

---

do

~dr

" ...,

... ,

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A-8 o

F

Engineering Mechanics and Strength of Materials Relationship Between Principal Stress and Principal Strain

~--------------------------------~~ Eng. stress vis Eng. strain curve MS under tension test

A ~ Proportional Limit B ~ Elastic Limit C ~ Upper yield point D ~ Lower yield point F ~ Ultimate point G ~ Fracture point DE ~ Yielding region EF ~ Strain Hardening region FG ~ Necking region Sudden fall of stress occurs from C to D due to slipping of carbon atoms in molecular structure of mild steeL Increase in carbon content increases strength, cast surface hardness and modulus of resilience. Increase in carbon content decreases ductility. For the most metals, its value is between 0.25 to 0.33.

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PROPERTIES OF MATERIALS (a)

(b)

(c)

(d)

(e)

(f)

(g)

Elasticity: It is the property of the material due to which it regains its original shape after the external load is removed after applied. Perfectly elastic bodies are those bodies which returns to their original shape completely. Plasticity: It is the property ofthe material due to which it does not regain its original shape after the removed of external load. Plasticity is the opposite of elasticity of external load. Plasticity is the opposite of elasticity. Ductility: It is the property of the material due to which if can be drawn into thin wires. The length of deformation is very large in a ductile materiaL Brittleness: material is said to be brittle if the length of deformation is very little in tension. A brittle material has lack of ductility. A brittle material tails at a very small deformation. Malleability: It is the property of the material due to which it can be converted into thin sheets in compression. This property is used in forging, rolling etc. Toughness: It is the property ofthe material due to which a maximum amount of energy stored in a material upto fractors. This property is utilized under the action of shock or impact loading. Hardness: It is the property of the material due to which it resists cutting, scractehing, pinetration or inditation.

..

1

E [al - J.l(a2

+ (3)]

El

=

E2

= ~ [a2 - J.l(a3 + ad]

E3

= E [ a 3 - J.l( a 1 + a 2 )]

1

for biaxial state of stress/plane stress problems E3 *- 0 al = E(El) a2 = E E2

a3

= 0 but

+ J.l a2 + J.l a 1

or al

_ E h + J.l2'E2)

-

_

a2 - E

(E2

+ J.l Ed 2·

1- u 1- J.l Work done (stretching wire) : When a wire is stretched, the work is done against internal restoring forces. This work is stored in wire as strain energy. Now, 1 Energy stored in wire, (U) = "2 FI EAI where , F=--- L Energy stored/unit volume of wire

u,

=..!.E (Strain)2 2 or

Uy

= _1_(Stress)2

2E where, E = Young's modulus of elasticity A= Area of cross - section L = length of wire I= increase in length of wire Extension of a tapered bar: Let us consider a circular bar whose diameters are DJ and D2as shownin figure. Let'F' be the tensile load which is applied axially.

F

Let, L = Length of the bar

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Engineering Mechanics and Strength of Materials

A-9

E = Young's modulus of elasticity Extension of tapered bar (8L), 8L =

8L = crILl E

+ cr2L2 + cr3L3 E

1

E

2

3

4FL 1t

=>

Now,

8L

E DID2

Ifbar is of uniform diameter 'D', then,

=

FILl Al EI

+ F2L2 + F3L3 A2E2

A3E3

Ifthe bars are of same material, then EI = E2= E3 = E, then,

~ Elongation of a bar due to self weight: Case (i) For uniform cross - section:

8L

= !.[.s_+.!2_+~] E Al

A2

A3

Composite bars: Let us consider a composite bar which is attached at the top and force F is applied. Now, F= FI + F2 = o.A, + cr2~ As strains in the bars are equal, then

8L

=

wL

2AE where, 8L = Elongation w = Area ofweight of bar A = Area of cross - section E = Young's modulus of elasticity L =Length of bar Case (ii) For coxical bar

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8L

= wL

Types of plans: Various types of beams are given as follows: (a) Cantiliver beam: A cantiliver beam has one of its end is fixed and the other end is free.

2

crE

= Pg L2

6E

L

where, 8L = Elongation w = specific weight of bar material L = Length of bar E = Young is modulus of elasticity P = Mass density of bar material Stresses in bars of variable cross-sections: Let us consider a stepped bar of different lengths and different cross - sections.

F~

~ ~

A" E,

I

A2, E,

IE Let, LI' L2, L3 = Lengths of bars AI' A2, A3 = Area of cross - section of bars EI' E2, E3= Young's moduls of elasticity Here, F = F I = F2= F3 Let, 8L = total change in length

(b)

Simply supported beam: A simply supported beam has both of its ends are supported.

(c)

Overhanging beam: In overhanging beam, the supports are not placed at the end of the beam and also one or both ends are entended over the supports.

I (d)

Fixed beam: In fixed beam, both of its ends are rigidly fixed into the supporting walls.

(e)

Continuous beams: than two supports.

I

I

In continuous beams, there are more

THIN CYLINDERS Pressure vessel is defined as a closed cylindrical or spherical container designed to store gases or liquids at a pressure substantially different from ambient pressure.

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Engineering Mechanics and Strength of Materials

A-tO

Moreover,it can be seen from expressionsof Ehoop and Elong that Elong < Ehoop :. The chances offailure ofthin cylinder is more longitudinally.

Pressure vessel

SHEAR FORCE AND BENDING MOMENT DIAGRAMS Thin d -> 20

Thick d -~ 20 t

d : diameter t: thickness

t

~ ~

SFD and BMD play an important role in design of beams. To design a beam, maximum value of shear force and bending moment are required which are determined from SFD and BMD.

Shear Force Sign Convention X I I I

on the basis of shape of shell Cylindrical Pressure Vessel

Spherical Pressure Vessel

Assumptions for Thin Cylindrical Vessels

G I I I

1

i-ve shear force

X I

C] )-x

D Ci)

y

I I I

I I

concave upwards (+ve) SAGGING BENDING

X

z

,1 __

Ii P

Bending Moment Sign Convention

Stresses are assumed to be uniformly distributed as thickness 't' is small. Radial stresses are neglected.

-----If---~

+ve shear force

i

I I I

pLI

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1

ILp

X

Example of Thin Cylinder: HydraulicCylinder. Example of Thick Cylinder: LPG Cylinder, Steam Pipes. ~

0

Pil

(jlong

+ve bending moment

-

X I

STATEOF STRESS ATA POINT IN THIN CYLINDER pd pd along = 4t' ahoop = 2t Sometimes11 of circumferentialjoint and longitudinaljoint are given. In that case, pd

along

= -4--'

a1 =

21'

a2=

Absolute

~max

=

t 11 eJ

pd

..

ahoop

=

6V

= -y=

2t 11u

HOGGING -ve bending

Beams

I

pd

_ 8d _ pd (2 _ ) d - 4tE ~

Ev

convex upwards -ve HOGGING BENDING

pd

2 = 4t

- 6L _ ~(1 L - 4tE

I

X

pd

al

D Ci)

I I I

4t

Ehoop -

Elong -

CJ

- 2 ) ~

pd 4tE(5-4~)

Statically Determinate Beam

J Cantilever

I

Statically Indeterminate Beam

J

t

J

Simply Supported Beam

Over Hanging Beam

Fixed Beam

~ Propped Cantilever Beam

l Continuous Beam

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Engineering Mechanics and Strength of Materials

A-l1

Beams of Uniform Strength

Bending Stress Normal stresses introduced due to the bending of a shaft / member. Pure Bending: If the magnitude of bending moment remains constant throughout the length of beam, the beam is said to be under pure bending.

A beam is said to be a beam of uniform strength when bending stress developed at each and every cross-section is same.

P-0~~ I

I

MA=:P(CD)

M ~ P(CD)

I I

x

I

I

-ve

ML..--------l

= Z22 = Zxx Ml1 = M22 = Mxx = (crbb = (crb)xx Zl1

Bending Equation

MR : moment of resistance offered by plane of cross-section of beam. (crb) : bending stress at a distance 'y' from Neutral Axis. R : Radius of curvature. E : Young's modulus. INA : area moment of inertia of plane of cross-section about Neutral Axis. From bending eq",

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crb =

11b

(+ve)

As BM = const above beam is under pure bending.

i~

to be used when 'R' is known. M

(crb)max = ±-ZNA . . ~A t => (crb)max "l.. => chances of failure "l.. For a given beam, (crb) ex y This fibre is subjected to compression

(crb)ll ·. (crb) is independent of 'x' . If beam is subjected to transverse shear load, the bending moment varies. ·. (crb) varies. To make beam a beam of uniform strength:(i) depth is varied.

Ix ~L

d = d x

(ii)

depth should be varied parabolically. width 'b' is varied

b = b[~] x

width should be varied linearly. Consider a log, out of which a rectangle is to be cut such that it is strongest in bending. band d ~ arbitrary dimensions of rectangle

NeutralAxis is neither in tension nor in compression

d

This fibre is subjected to tension (crb) =

(crb)max Y y max

.

D ~ diameter of log (given) ·. final dimensions of strongest rectangular cross-section are

A beam offering higher moment of resistance is stronger. I-section beams are strongest as they have high section modulus. Fora giveneross-sectionalareaandmaterialsquarecross-section Shear Stresses in Beams beam is strongerthan circular cross-sectionbeam as Zgquare > Zcircle PAy 't=---

INA' b

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A-12

P A y INA:

shear force on plane of cross-section. area. distance of hatched portion from neutral axis. moment of inertia of entire cross-section about neutral

b

width.

Engineering Mechanics and Strength of Materials ~ I section:

axis.

Consider a Beam of Rectangular Cross-Section Shear stress Distribution

~

T section:

~

Square with DiagonalsVertical:

b 9

By using the above formulae, we get

r max

I----~

= "8 t avg

1'=

.. r a: y2 (parabolic variation) Badboys2 As r f(y2) o:

:. As 's' t t ~ at extreme fibres l' = 0 DEFLECTION OF BEAMS P where, 1'avg = A·

Expression for Maximum Various Cross-Sections

Shear

Stress

Across

Deflection of beams plays an important role in design of beams for rigidity criterion. The expressions of deflections are further used for determination of natural frequencies of shaft under transverse vibrations. For a cantilever beam under any loading condition deflection is maximum at free end. In simply supported beam, deflection is maximum at mid-span (when beam is subjected to symmetric loading only).

Relationship between R, q and Y A = a2

A= bd

3 2

K= 2

K= K=-

For circle,

3

d2y -dx2

4 3

in a circular cross-section

"3

4

3

"2

1'max --=1'NA

9 8

1'avg' 1'NA = =

1.125.

"3

-

1'avg·

Mxx

--

sr.,

R

e=

4

1'max =

---)- For square, circle, rectangle, 1'NA is the maximum shear stress. But in triangular cross-section, it isn't so. In triangular cross-section, 1'max =

e : slope Y : deflection R : radius of curvature

dy dx

JMxx+ C, = EI(:) --> slope equation is obtianed If + C1x + C2 = EI(y) ---)-deflection eq" Mxx

1'avg

Sign Convention ---)- Deflection upwards (+ve)

---)-e ~

+ve

Badboys2

Engineering Mechanics and Strength of Materials Deflection downwards (-ve) Also,

9 (-ve

J?

Wxx

ML EI'

9max = 9B =

'W' at its mid-span.

Alr(---L-/2--~!_B------~OC

d4y __,.4 times integration to = EI dx4----r obtain deflection 'y'

load intensity

Expression for Deflection in Cantilever Beams Case I: Cantilever beam subjected to point load W at free end X

WL2 9B = 9c = 9max = 8EI'

5 WL3 Ymax = Y c = 48 EI

Case VI: Cantilever beam subjected to uniformly distributed load over half its length from fixed end.

W

Aro~o~WN/m

L

7

Yc = Ymax = 384

WL4

ill'

DC

WL3 9max = 9B = 9c = 48EI

Expressions for Deflections in Simply Supported Beam Case I: Simply supported beam subjected to pure bending.

Badboys2 For cantilever, y = Ymaxat x = 0

G4$

AMr-----:

_WL3

..

ML2 Ymax = YB = 2EI

Case V: Cantilever beam of length 'L' subjected to point load

EI d3Y ~ 3 times integration to JC Fxx = dx3 obtain deflection 'y' shear force /

A-13

Ymax=~.

Case II: Cantilever beam subjected to uniformly distributed load

-------'ltF-----C Ll2

1

W N/m

~

:G_) 1

'"

ML 2EI;

9max=9BA= ,

A

Ll2

>1

IE

-------::I~

ML2 Ymax=YC=--· 8ill

Case II: Simply supported beam subjected to concentrated point load 'W' at mid-span. W

I

1

WL4 8EI Case III: Cantilever beam subjected to uniformly varying load

Arl

Ymax=YB=

4$

~

Ll2 L

WN/m

A

L

IB

WL4 Ymax = YB = 30EI Case IV: Cantilever beam subjected to concentrated moment 'M' at free end.

J.I-----"

1

~*~C------------~1B

-L

-DB)

~~.~------------------------~>M

WL

Mmax = 9

max

=9

4; A, B

=-

WL3 Ymax = Yc = 48EI WL2

16EI

Case III: Simply supported beam subjected to uniformly distributed load

Badboys2

Engineering Mechanics and Strength of Materials

A-14

5 WL4 Ymax = Yc = 384 ill

;9max = 9B = 9A =

WI}

---~~::~_~_~~~

24EI

Case IV: Simply supported beam subjected to a concentrated point load acting not at mid-span

~~----~L------~

W

1

Cm~

P

b

Shear Stress Distribution

e: angle of twist shear angle L : distance of cross-section from fixed end <1>:

s/

o

= c

/Y

3EIL

doesn't give max. slope ~

Torsion Equation

2

Wb (a - ab)

c

doesn't give max. deflection

Stiffness of beam =

Load Max. deflection

Higher flexural rigidity is an indicative of higher stiffness of beam but lower deflection and slope. ~

L J 9 : maximum angle of twist.
Now,
R

R9

L A

Maxwell's Reciprocal Theorem

I

(valid for beams under point load and having same L, E and I)

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=

We

,?ILoad at C YeB

~ Deflection at C due to load at B

Deflection at B due to load at C

-B Moreover,
C

Cross-section of a shaft at free end tA =

'tB = 'te = 'tmax

'tA' = 'tB' = 'te' = 'tmax

A

9A' = 9B, = 9C' = 9A = 9B = ge
YeB=

5 WL3 48 ill' WB=W We YeB = WB YBe

T 'tmax = Zp Zp : polar section modulus .

7t

For sohd shafts, Zp = 16 d

3

For hollow shafts, Zp = 1: D3 (1 - K4)

A

D : outer diameter d : inner diameter d K= 5 WL3 Y =---W=W Be 48 EI'

D

~ e

TORSION

Pure Torsion A member of a shaft is subjected to pure torsion when the magnitude oftwisting moment remains constant throughout the length of shaft.

~

A shaft offering higher value of T r' has more strength. Shafts with high value of polar section modulus are preferred. Torsional Rigidity GJ : Torsional rigidity TL 9 = GJ GJ

i => 9 ,j.. => r ,j.. => chances of failure ,j..

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Engineering

Mechanics and Strength of Materials

A-15

~

Torsional Stiffness (q)

2.

T 1 = TA' T2 = TA - T.

3. 4.

TA+Tc=T.

~

T GJ q=-=9 L Torsion of a Tapered Shaft

91 + 92 = <1>0 => 91 = (-92) 3T

4·

=> TA=

= -3.

6.

Shafts in Series

THEORY OF COLUMNS

Column is defined as a vertical structural member which is fixed at both ends and is subjected to an axial compressive load. Strut is defined as a structural member subjected to an axial compressive load. All columns are struts but vice-versa isn't true.

Badboys2 Long Columns (fail due to buckling)

~

CD

T=T1+T2 9

Euler's Formulae Assumptions ~ The self weight of column is neglected. ~ Crushing effect is neglected. ~ Flexural rigidity is uniform. ~ Load applied is truly axial. ~ Length is very large compared to cross-section. . . Pe o: f [E, Imin' end conditions, L2]

=> T = (G1 J1 + G2 J2) L Pe =

..

91 = 92 = 9

CD G1 J1 = G2 J2 Net TM = T (anti-clock) Rxn = T (clock)

1'.

2

E Imin 2

Pe : Euler's buckling load. Imin : min [Ixx and Iyy]. L, : effective length of column. L : actual length of column. L =aL e

1.

1t

Le

Shafts with Both Ends Fixed

Short Columns (fail due to crushing)

As the length of structure, chances of it failing by buckling are more.

Shafts in Parallel T

Medium Columns (fail due to buckling as well as crushing)

f= ~

4 length fixity coefficient

1 a2 (end fixity coefficient)

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Engineering Mechanics and Strength of Materials

A-16

oe : buckling stress Both Ends Both Ends Fixed and Fixed and Hinged Fixed Free Hinged (BF) (F &H) (FF) (BH)

~

1

1

a

1

-

2

J2

1 11=-

1

4

2

0.2

cr

e

n2E =-s2

S t => Pe .,l.. => buckling tendency is increased (S)sc < (SMC)< (SLC) SC : Short Column MC : Medium Column LC : Long Column For steels, if S ~ 30 => short column S > 100 => long column 30 < S ~ 100 => medium column

2 1

-

4

If remaining all other parameters are same, (Pe)BF > (Pe)FH > (Pe)BH > (Pe)FF Which of the following column is stronger?

STRAIN ENERGY METHODS

(1)

Badboys2a4 12'

1

(Pe)l (Pe)2

(2)

nr4 na4 I = -=--=-

I =-

4

2

a4 4n

n2(4)

= 4n 072 = 4n(0.72) 12·

12

_

I

I

1

RR

PE

Pc

P

-0.513

. . (2) is stronger. Rankines formula: It is a combination of Euler and crushing load. It is also known as Rankine Gordon formula.

-=-+-

Strain energy is defined as energy absorbtion capacity of the component during its functionality. Resilience is energy absorbtion capacity of the component within elastic region. Energy absorbtion capacity of a component just before fracture is known as toughness.

SE of bar = work done by load P 1 P2L cr2 crEAL Po =-- =- x AL =--. 2 2AE 2E 2 Strain energy of solid circular shaft subjected to torsion

Strain energy of bar ~

=-

where, PR = Rankine's Load PE = Crippling load by Euler's formula Pc = crushing load h PR were

cry.A = _--=__

l+a(~r where K = radius of gyrotion (minimum) a = Rankini's constant A = Area of cross - section of column Slenderness Ratio ~ Used to compare buckling loads of various columns having same material and same cross-section.

t=-

T Zp '

where T : twisting moment. Zp : polar section modulus for circular

~ =

C~)d3 2

2

SE = _!_ T9 =_!_ T L =2._(AL). 2 2 GJ 4G

x

section.

Badboys2

Engineering Mechanics and Strength of Materials ~

Strain energy of hollow circular d : Inner diameter. D : Outer diameter.

x

A-17 Mxx : moment at section x-x X

section shaft.

I I

A

d K= -

r-

B~~

~;

D

~1(------7)

X

K = 0 for solid K
~x

M

=-M

!

L M2

U~

x

M2L

2Eldx~

2EI

W ~2

SE

= --

4G

(AL) (1 + K2), where

t

T

=

b

Zp

Proof Resilience: It is the maximum strain energy stored up to elastic limit. Modulus of Resilience is proof resilience per unit volume. Modulus of Resilience is the property of material. Proof Resilience is function of volume of component.

(

Wb

Wa

e

e

In this case using the above relation, we get

0"

Wa2b2 U=---

6EIf Theories of failure: Modulus of Toughness

Badboys2 EL PL

Various thories of failure are given as follows:

(a)

Maximum principal stress theory or Rankine's theory According to this theory, material failure will take place when the maximum principal stress exceeds the value of yield stress under a state of complex stress system during of yield stress under a state of complex stress system during a tensile test. J 0"2

~

Modulus of Resilience Two bars A and B are as shown:-

~,.

Now, Let o. (maximum principal stress) and o, ora3 (minimun principal stress) and ay by the yield stress. For design creterion, maximum principal stress must not exceed the working stress (aw) al,2 ~ away

P (A)

For considering yield creterion, a 1 = ± ay or a2 = ± ay This theory is utilized for brittle meterials.

P (B) (b)

2

U

=.!_ P8 = 2P L

A 2 nd2E UB = UI + U2

2p2 (~)

Maximum principle strain theory (St venant's theory) According to this theory, material failure will take place during tensile testing under a three dimensional complex stress starts system when maximum strain value reaches the value of strain due to yielding.

p2 (~)

0"3

nd2 E + 2nd2 E

= 0.5 UA"

STRAIN ENERGY DUE TO BENDING

)2 u=J b (M2EIxx xx dx a

U : strain energy

Badboys2

Engineering Mechanics and Strength of Materials

A-IS Let,

= three principle strains = strain at yielding

6)' 62, 63 6y

61,2,3::;6y

Now, 1

61

= E [ al - Y ( a2 + (3)]

62

= E [ a2 - Y (al + (3)]

63

= E [a3 -y (al +(2)]

1

1

Now,According to failure creterion,

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10"1 - r0"2 - r0"3

=

0" Y

I

(c) Maximum shear stress theroy or Tresca's theory: According to this theory, material failure will take place. It maximum shear stress in complex stress state will be equal to the value of maximum shear stress in simple tension. If al = maximum principal stress a2 = minimum principal stress ay = yield stress then, ay = al -a2 (d) Maixmum strain energy theory : According to this theory, material failure will take place under complex stress state, when total strain on the body or specimen reaches the value of strain energy at elastic limit in simple tension. [(at +a~ +(J~ )-2y(ala2

+a2 a3 +a3aI)] ~ a~

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...,

Engineering Mechanics and Strength of Materials

I···.. 1.

2.

F=G MIM2 R2

(b) F=G

(MIM2)2 R2

MIM2 (MIM2)2 (d) F=G R R Iftwo co-planar forces 'PI and 'Q' are acting at a point and '9' being the angle between them and also resultant 'R' is making an angle a with force Q, then the magnitude of resultant will be equal to : (c)

F=G--

(a)

R=

(b)

Jp2 R = Jp2

(c)

R = ~p2 +Q2 +2PQcos9

Badboys2(d) 3.

EXERCISE

If M 1 and M2 are two masses of two bodies, 'R' is the distance between their centers then which ofthe following expression represents gravitational law 'G' is universal gravitational constant. (a)

+Q2 +2PQsinQ +Q2 -2PQsin9

Jp

R = 2 +Q2 -2PQcos9 In the question number 2, the direction or angle mode by the resultant will be equal to: (a) u = tan-I (b)

(Q:~::se)

a = tan-1 ( Pcos9 ) Q+Psin9

a = tan-1 (

Psin9 ) Q-Pcos9

(d) a = tan-I (

9 P cos ) Q-Psin9

(c)

4.

A-19

A B C --=--=-cos2 a cos2 J3 cos2 y (d) None of these 5. Streamlined shapes offers resistance against air flow or water flow, the magnitude ofthe resistance is : (a) Least (b) Maximum (c) Negative (d) Positive 6. Ifthe forces are reduced on contacting surfaces, the value offriction: (a) increases (b) decreases (c) remains constant (d) None of these 7. The property due to which the material can be drawn into thin wires is knows as : (a) Malleability (b) brittleness (c) Ductility (d) Elasticity 8. The property due to which the material can be converted into thin sheets is known as : (a) Ductility (b) Malleability (c) Hardness (d) Resilience 9. The property of the material due to which the maximum amount of energy stored in a material upto fracture limit is called as: (a) Hardness (b) Resilience (c) Plasticity (d) Toughness 10. The property of the material due to which it resists against indentation is known as: (a) Hardness (b) Toughness (c) Elasticity (d) None of these 11. The work stored in a stretched wire in the form of strain energy per unit volume of wire is given by: (c)

(a)

U; = ..!..E(strain)2 2

(b) Uu =..!..E(strain) 2

1 U, = ..!..E(stress)2 (d) Uu = - E(stress) 2 2 12. IfP = axial load applied,A = cross- sectional are ofuniform circular bar, L = length of the bar, E = Young's modulus of elasticity, then elongation of the bar will be equal to : (c)

Which of the following expressions represents Lami's theorem, ifA, B, C are three are in equilibriumand as shown in figure. A

(a)

PL 2AE

--

(b)

PL AE

-

C --

A B C = -- = -cosu cos J3 cosy

(a)

8L = wL AE

(b) 8L = 2wL AE

A B --=--=-sm u sin J3

(c)

2 8L = wL AE

(d) 8L = wL 2AE

B (a)

(b)

PL2 PL2 (d) -AE 2AE 13. In the above question, if w be the total weight of the bar hanging fixed at one end, then elongation (8L)will be equal to: (c)

--

C sin y

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Engineering Mechanics and Strength of Materials

A-20 14.

Ifin case ofa stepped bars of same material i.e., E = E, = E2 = E3 as shown in figure, then elongation ofthe bar will be

P~ IE

(a)

I A"E

A"E L1

A3,E

~IE

~IE L2

(b) Rate of change of momentum is inversely proportional

~P 22.

~I L3

to impressed force and takes place in the direction to opposite of force acting (c) To every action there is always in equal and opposite reaction (d) None of these If a body is rotating at N rpm, the corresponding angular velocity will be equal to : (a)

P (Ll L2 L31 8L = E Al + A2 + A3)

21tN 120

--rad/s

21tN

(b) -rad/s 360

21tN 21tN --rad/s (d) -rad/s 180 60 If the body is rotating atoi radls along a circular path of radius 'r', then its linear velocity (y) is given by: (a) y= r20) (b) y= rro (c)

(b)

8L = PE(.!1_+~+ Al A2

(c)

P 8L = p(LIAI

L31) A3

23.

+ L2A2 + L3A3) (c)

(d) 15.

16.

18.

8L = PE (LIAI + L2A2 + L3A3)

A beam whose one of its ends is fixed is known as : (a) simply supported beam (b) continuous beam (c) cantilever beam (d) overhanging beam A beam whose both ends are fixed rigidly into the supporting walls is called as : (a) continuous beam (b) fixed beam (c) cantileverbeam (d) None of these A beam whose both ends are supported is known as : (a) simply supported beam (b) fixed beam (c) overhanging beam (d) continuous beam IfPR = Rankin's Load, PE = crippling load by Euler's formula ~nd.P c = crushing load, then Rankin's formula for columns IS given as : (a)

(c) 19.

-

PR

=

1 Pc

---

PE

Pc

(b)

(d)

1 1 -=-+PE Pc 1 1 -=-+PR PE

24.

25.

26.

=

(a)

(')y =

2

(')1

2

+ (')2

(d) o y

= o 21 -

maximum shear angle =

R8

L J

Tr = G8 = "["max J L R

1 PR 1

(c) 27.

Pc

2

o2

Equilibrium equations given for non - concurrent forces are given as : (a) EFx = 0, EFy = 0, EM = (b) EFx = 0, EFy = (c) EFx=O,EFy=O,EM:;tO (d) None of tliese Newton's second law of motion states that: (a) Rate of change of momentum is directly proportional to the impressed force and takes place in the direction of force acting

° °

21.

According to Hook's Law, stress is directly proportional to strain within : (a) Plastic limit (b) yield point (c) elastic limit ofproportionality (d) None of these The value of slenderness ratio (s) for short column ofsteels is in the range of: (a) s~30 (b) s~30 (c) s < 20 (d) None Torsion equation is given as: if8 = maximum angle of twist, J =Polar moment of inertia, T r = Twisting moment,

(d)

R

L

"["max

l= "["max

(a)

wL2 wL2 8max = 8B = --, Ymax =YB =-3EI 5EI

(b)

wL4 wL3 8max = 8B = --, Ymax =YB =-6EI 8EI

(c)

wL6 wL6 8max = 8B = --, Ymax =YB =-6EI 4EI

G8 =!_ L

R

wL4 wL4 8max = 8B = --, Ymax = YB =-2EI 3EI If t = shear stress, G = shear modulus and v = volumn ofthe body then the expression of strain energy stored withing (d)

28.

T, = G8 = "["max L J R

G8

Ifcantilever beam is subjected to uniformly distributed load (UdI), then expressions for deflection are given by:

If (')1' (')2and (')yare maximum principal stress, minimum principal stress and yield stress, then according to maximum shear stress theory, which of the following expression satisfies: (a) (')y=(')1 +(')2 (b) "v" (')1-(')2

(c) 20.

1 1 -=-+PE PR

(d) y= 0)2r

(J)

Badboys2 17.

r

v=-

Badboys2

Engineering Mechanics and Strength of Materials

A-21

the body is given as : '[2 (a) -xV 2G

29.

'[2 r (c) -xV (d) -xV G G If o I' cr2,cr3are three principal stresses, J..l= Poisson's ratio and E = Young's modulus of elasticity, then the expression for strain energy/volume is given by (a) (b)

(c)

30.

31.

[crt + cr~+ cr~+ J..l( crlcr2+ cr2cr3+ cr3crd]

I

="2

IE=21 IE="4

I

Poisson's ratio is described as the ratio of: (a)

Longitudinal strain Lateral strain

(b)

Lateral strain Longitudinal strain

(c)

stress strain

strain (d) stress

33. A cyelindrical elastic body subjected to pure forsion about its axis develops: (a) compressive stress in a direction 45° to the axis (b) shear stress in a direction 45° to the axis (c) tensile stress in a direction 45° to the axis (d) None of these 34. The forces whose line of action lie on the same plane and also must at a point is known as : (a) co-planar non concurrent forces (b) co-planar concurrent forces (c) Non - coplanar concurrent forces (d) Non - coplanar-Non concurrent forces 35.

~('LFH)2 + ('LFy )2 -2('L~)('LFv)

When number of forces are acting on the body and

~

~YF:

39.

tana=-Lftl

(b)

tana=Lftl

LFy

2)H x2)v x LFH x~:)v

(d) tana=2 IfD = diameter of thin cylindrical shell L=length t = thickness ~ = internal pressure, J..l= Poisson's ratio then Hoop strain will be given by: PP(2-!) 2m 3J..l

(c)

PP (l-!':.)

(a)

--

(b)

PiD(1_!:) 3m 3

PP(I-!':.)

(d) 2tE 2 3tE 2 If the both the ends of the column made of mild steel are hinged, then Rankin's constant value will be equal to : 1 7500

(b)

1 6500

--

1 1 (c) -(d) -4500 5500 IfDI = external diameter of short column D2 = internal diameter of short column F = external load applied The highest value of eccentricity will be equal to : (a)

(b)

(c)

(d)

Df-D~ 8D1 Dr-D~

4Dl

40.

Range of Poisson's ratio for steel is given by: (a) 0.21-0.22 (b) 0.23-0.27 (c) 0.37-0.43 (d) 0.57-0.63 41. IfL = original length of specimen, & = Increase in length, then the strain (E) will be equal to : (a)

L FH

and L Fv be the algebric sum of all the horizontal forces and the algebric sum of all the vertical forces, then the resultant will be given by; (b)

LFy

(a)

(a)

38. (b)

(d)

32.

37.

2~ [crt + cr~+ cr~- 2J..l(crlcr2+ cr2cr3+ cr3crd]

IE

(d)

(c) tan a=

_!_[ crt + cr~+ cr~- J..l( crlcr2+ cr2cr3+ cr3crt)] E

(d) _1_[crt +cr~ +crj +2J..l(crlcr2+cr2cr3+cr3crd] 2E Impact strength of a material represents: (a) Hardness (b) Resilience (c) Ductility (d) Toughness If'i' is the actual length of column and IE is the effective length of column, then ifboth ends of a column are fixed, then the effective length (IE) will be equal to: (a)

~('L~)2 + ('LFv )2 + 2('LFH )('LFv)

36. In the above question, the direction or angled of the resultant will be give by:

!

Badboys2

(c) '[

(b) -xV 2G

(c)

42.

L

~L ~L L

(b) Lx Al.

2L (d)

~L

In case of curved beams, the location of the neutral axis does: (a) coincide with the geometric axis (b) lie at the top of the beam (c) lie at the middle of the beam (d) coincidewith normal axis

Badboys2

Engineering Mechanics and Strength of Materials

A-22

43.

44.

45.

46.

47.

48.

In case of curved beams, the bending stresses are distributed in the shape of: (a) Parabola (b) ellipse (c) circle (d) Hyperbola Which ofthe followinghas given maximum principal stress theory (a) Rankins (b) Tresca (c) ST. venant (d) Mohr Which of the following has given maximum shear stress theory: (a) Rankins (b) Tresca (c) Mohr (d) ST. venant Maximum shear stress theory is utilized for which of the followingmaterials. (a) brittlematerial (b) ductilematerial (c) brittle and ductile materials (d) None of these Maximum principal stress theory is used for which of the followingmaterials: (a) ductile and brittle materials (b) ductile materials (c) brittlematerials (d) None of these If llwefficiency of weldedjoint, llR = efficiency of riveted joint, then which of the following relation is true: (a) llw>llR (b) llw
Badboys2 49.

50.

51.

52.

53.

54.

(a)

1 O"bocz

(c)

0"

b

ocz2

(d) O"boc-

1

z2

Neutral plane ofa beam is defined as the plane: (a) whose length changes during deformation (b) whose length does not change during deformation (c) which lies at top most layer (d) None of these In case of a continous beam, which of the following statement is true? (a) It has two supports at ends only (b) It has less than two supports (c) It has more than two supports (d) None of these Stiffness is measured in which of the following: (a) Modulus of elasticity (b) Toughness (c) density (d) ultimate strength Percentage elongation is associated with which of the following terms during tensile test? (a) Malleability (b) creep (c) Hardness (d) ductility

55. Poisson's ratio generally depends on: (a) Material of specimen (b) Area of cross section (c) Magnitude ofload (d) None of these 56. Which of the following has the largest value of Poisson's ratio? (a) Mild steel (b) Rubber (c) ceramics (d) stainless steel 57. A wire is stretched by a load, ifits radius is doubled the young's modulus of elasticity of material of wire will become: (a) trippled (b) doubled (c) No change (d) one fourth 58. The value of Poisson'sratio for aluminium material is equal to: (a) 0.33 (b) 0.43 (c) 0.53 (d) 0.63 59. If o"w= working stress, O"u= ultimate stress then the which ofthe following relation is free? (a) O"w=O"u (b) O"wO"u (d) None of these 60. The point of contraflexure is found to be in which of the following beam? (a) cantilever beam (b) Simplesupportedbeam (c) overhanging beam (d) None of these 61. Alarge plate (uniform)consistingof a rivet hole is subjected to uniformuniaxial tensilestressof 100MPa. The maximum stress in the plate will be equal to :

20emO 10 em

(a) 300MPa (b) 200MPa (c) 100MPa (d) 400 MPa 62. A rod of length I and diameter 'd' is subjected to a tensile load P which of the following do we need to calculate the change in diameter. (a) Young's modulus of elasticity (b) Poisson's ratio (c) Shear modulus (d) Young's modulus of elasticity and shear modulus 63. An elasti~ body is subjected to a tensile stress O"tand a compressive str~ss O"e i~ its perpendicular direction. O"tand 0" eare not equal Inmagnitude,then on the plane ofmaximum shear in the body, there will be : (a) normal stress only (b) shear stress only (c) normal and shear stress both (d) Maximum shear stress only 64. It two principal strains at a point are 1000 x 10-6 and - 600 x 10-6, then the maximum shear strain will be equalto

65.

(a) 1600 x 10--6 (b) 400 x 10-6 (c) 800 x 10-6 (d) 300 x 10-6 In terms of Poisson's ratio (u), the ratio of young's modulas of elasticity(E) to shear modulus (C) of elastic material is : (a) 2 (1- u) (b) 2 (1+ u) (c)

(1- Jl) 2

(d)

(1+ Jl) 2

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Engineering Mechanics and Strength of Materials

A-23

66. If the principal stresses in a plane stress systems are 100 MPa and 40 MPa, then maximum shear stress will be equal to: (a) (c)

30 100

73.

(b) 40 (d) 50

67. A thin cylinder of 100 mm internal diameter and 5mm thickness is subjected to an internal pressure of 10 MPa and a torque of 2000 Nm. The magnitude of principal stresses will be equal to : (a) a1=1098MPa,a2=41MPa (b) 0'1 = 502 MPa, 0'2 = 62 MPa (c) 0'1 =2018,MPa,a2=46MPa (d) 0'1 = 702 MPa, 0'2 = 88 MPa 68. In case of simply supported beam on two end support, the value of bending moment is maximum will be : (a) On the supports (b) at mid - span (c) where there is no shear force (d) where the deflection is maximum 69. A steel cube is subjected to tangential force on its top surface and its bottom is fixed rigidly as shown in figure: then the deformation of the cube will be due to:

Badboys2(a)

Ifthe length of the column is doubled, the value of critical load becomes : (a)

..

(c)

- x ongmal value

(a)

n2EI 4L2

(c) 75.

76.

(b)

1 8

..

- x ongmal value

(b)

n2EI

(d)

n2EI 2L2 n2EI

8L2 16L2 True stress is associated with: (a) Instantaneous cross - sectional area (b) Average cross - sectional area (c) Original cross - sectional area (d) Final cross - sectional area The unit of stress in SI system is given as : (a)

Shear only (b) bending only (c) twisting only (d) Shear and bending both 70. A concentrated load F acts on a simply supported beam of

1 2

- x ongmal value

1 .. 1 .. 1 (d) - x ongmal va ue 4 16 74. For the case of the slender column of length L, flexural rigidity EI built in at its base and free at the top, Euler's critical bucking load will be equal to :

)P

nnlnmllLm

-8F (d) nd2

(c)

N!mm2

(b) N/m2

(c) Kgim2 (d) None of these 77. If aT = True stress, ac = conventional stress, then their relationship is represented by : where E = strain

I

span l at a distance of"3 from the left - end. The bending moment at the point of application ofload is given by : (a)

71.

Fi

(b)

4 FI

(c)

2FI

-

9

-

(a)

nd4 64

(d) 9 3 The second moment of a circular area about the diameter is given by if'd'is the diameter: (b)

nd4 32

nd4 nd4 (c) -(d) -16 8 72. A circular rod of dimeter 'd' and length 3d is subjected to a compressive force F acting at the point as shown in figure. Then the stress value at bottom most support at point A. 3d F

(~

1~) (a)

nd2

(b)

nd2

E

Free Body diagram shows: (a) No forces are acting of the body (b) All the internal forces acting on the body (c) All the internal and external forces acting on the body (d) None of these 79. During tensile testing for cast iron specimen, the stressstrain curve shows: (a) No yield point (b) upper yield point only (c) lower yield point only (d) Both upper and lower yield points 80. In a stress strain curve, the area under stress strain curve upto fracture shows which of the following property: (a) Hardness (b) Ductility (c) Toughness (d) Brittleness 81. IfG = Modulus of rigidity, "C = shear stress, Y= shear strain, aL = Longitudinal stress, E YL = Longitudinal strain, then the expression for 'G' will be given by: (a)

-12F

6F

aT

78.

FI

(c)

ac

-=

Y aL

(c)

(d)

Y "C

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Engineering Mechanics and Strength of Materials

A-24

82.

83.

Three plans on which the principal strains occurs are: (a) Mutually perpendicular to each other (b) Mutually inclined to each other than 90° (c) Inclined at 45° only (d) None of these In case of simply supported beam, the maximum bending moment ofa beam having span 'L' and a concentrated load wat mid-span will be equal to : (a)

wL 3

(b)

wL

91.

(a)

92.

(c)

84.

(a)

(c)

85.

wL3 8 wL2

(b)

(d)

93.

4 94.

16 8 In case of a cantiliver beam, whose span is 'L' and carries a UDL of intensity wlunit length then maximum bending moment will be equal to :

Badboys2(a)

wl2

Wl2 (b) 3

87.

88.

89.

90.

Area under shear force diagram represents the (a) Shear force at a point (b) Bending moment at a point (c) load at a point (d) None of these In shear force and bending moment diagram, if the bending moment is maximum then the shear force at that location will be equal to : (a) Zero (b) maximmn (c) minimum (d) None of these When the uniformly distributed load is applied, then the nature of variation ofthe bending moment diagram will be (a) Hyperbolic (b) linear (c) Parabolic (d) None of these When the concentrated load is applied, then the nature of variation of bending moment will be : (a) Linear (b) Parabolic (c) Hyperbolic (d) Uniform If ah = hoop stress, at = longitudinal stress, then the ratio of ah to at in case of thin cylindrical pressure vessels is equal to (a)

ah = 2 at

(b)

(c)

ah = 8 at

(d) ah = 16 at

~=4 at

PR H

PR PR (d) 4H 8H In a thin spherical pressure vessel, the volumetric strain is given by: (a)

3ae (l-v)

(c)

3ae (l-v)

E

(b)

(c) 95.

(d) 3ae (l+v)

2E

2

(b)

4

1 8

(d) 16

a;

(d) None of these

If T = Torque transmitted, O = angle of twist, J = Polar moment of inertia, then torsional rigidity ofthe shaft will be equal to: (a)

(c) 97.

E

If ac = radial stress, ah = hoop stress, then the radial stress value in a thin spherical vessel will be equal to: (a) Zero (b) 2ah (c)

96.

3ae (1+ v)

Ifwe round a thin cylinder with a wire under the application of tensile stresses, then the hoop stress will be of nature: (a) twisting (b) compressive (c) shear (d) tensile The ratio of maximum shear ("Cmax) to hoop stress (aH) in case of thin cylinderical pressure vessel is equal to (a)

(c)

86.

(b)

2E

wL2

wL2

PR 2H

-

(c)

4

wL wL (d) 2 8 In case of a simply supported beam, whose span is 'L' and carries a UDL at wi unit length, the value of maximum bending moment will be equal to :

IfH = wall thickness, P = pressure, R = mean radius, then maximum shear stress in case of thin cylindrical pressure vessel will be :

T J T

e

(b) T x J

(d) T x

o

Under the action of torsion, the shear stress at the centre of a circular shaft is equal to : (a) maximmn (b) minimum (c) zero (d) None of these 98. If two shafts are connected in paralled position, then (a) angle of twist of both shafts are equal (b) angle of twist of both shafts are unequal (c) torque of both shafts are equal (d) None of these 99. If two shafts are connected in series, then (a) torque of both shafts are equal (b) angle of both shafts are equal (c) shear stress of both shafts are equal (d) torsional stiffness of both shafts are equal 100. Ifs = slendernessratio, then the value of's' for short column should be in the range of : (a) S=32 (b) S< 32 (c) S>32 (d) S~32

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Engineering Mechanics and Strength of Materials 101. IfS = slenderness ratio, then the value of's' for long column should be in range of: (a) S> 120 (b) S< 120 (c) S= 120 (d) S= 60 102. Euler's buckling formula is associated with: (a) Short column (b) long column (c) medium column (d) None of these 103. If'D' is the diameter of a circular column, then radius of gyration (K) will be given by:

104.

105.

106.

107. Badboys2

108.

109.

110.

111.

112.

D

D

(b)2 4 (c) 2D (d) 4D A beam column is described as a column which carries: (a) axialloads only (b) transverse loads (c) axial and transverse loads (d) None of these When two forces are in equilibrium, then which of the following conditions is true. (a) Magnitudes are equal (b) opposite directions (c) collinear in action (d) All of the above In case of co-planar non-concurrent forces, when EH = 0, EV = 0, then the resultant may be: (a) moment (b) couple (c) force (d) None of these When a sphere is placed on a smooth surface, then the reaction will act: (a) inclined to contact plane (b) perpendicular to contact plane (c) horizontal to contact plane (d) All of the above For aquiring equilibrium condition, How many are minimum number of coplaner and non - collinear forces required? (a) 1 (b) 5 (c) 3 (d) 4 If three co-planar and concurrent forces are acting on a rigid body at different points then the body will be in : (a) equilibrium (b) not in equilibrium (c) mayor may not be in equilibrium (d) None of these A body having a weight of 50 N is resting on a rough horizontal floor, then the force of friction acting on the body will be equal to: (a) 50N (b) lOON (c) zero (d) None of these Angle offrictiion is defined as the angle between (a) normal reaction and the resultant of frictional force and normal reaction (b) Normal reaction and frictional force (c) Force on the body and normal reaction (d) None of these The maximum inclination of the plane at which the body just starts to move is termed as : (a) Cone of friction (b) Angle of repose (c) friction angle (d) None of these (a)

A-25 113. The value of frictional force depends on (a) weight of the body (b) area of contact (c) Normal reaction (d) roughness of surface 114. The value of maximum force of friction when the body begins to slide over another body/contacting surface is known to be: (a) limiting friction (b) rolling friction (c) sliding friction (d) None of these 115. IfF = limiting friction, R = normal reaction, then coefficient offriction (u) is given as: (a)

Jl=-

R F

(b) Jl=RxF

F (d) Jl=F2R R 116. If = angle of friction, Jl = coefficient of friction, then which of the following relation is true? (a) Jl = cot (b) Jl = sin (c) Jl = tan (d) Jl = cos 117. If = angle of friction, a = angle of repose, then which of the following relation is true? (c)

Jl=-

(a)

=a

(b)

¢=-

1

a (c) <1>= a2 (d) a= 2<1> 118. If = angle of friction, u= coefficient offriction, then which ofthe following relation is true? (a) =COC1(Jl) (b) =tan-1(Jl) (c) =sin-1(Jl) (d) =cos-1(Jl) 119. When a block of weight w is resting on a rough inclined plane with angle of inclination being 'a', the force offriction will be equal to : (a) wsin (b) w cos (c) wtan (d) w cot 120. If Jls = coefficient of static friction, Jlk= coefficient ofkinefic friction, R = Normal reaction, then frictional force of a moving body with constant velocity will be equal to :

e e

e e

&_

(a)

Jlk

(b)

(c)

Jl~

(d) Jl~

R

R

121. During calculation of shear force, the upward forced to the left the of the section are taken as : (a) Negative (b) Positive (c) Zero (d) None of these 122. In a shear force and bending moment diagrams, Area of load diagrams provides: (a) shear force change (b) bending moment (c) shear force (d) Point of contra flexure 123. There is a cantilever beam whose length is L and it carried a point load F at its true end. Shear force at the center of the beam will be equal to: (a)

(c)

FL

2 F

(b) FL F

(d) 2

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Engineering Mechanics and Strength of Materials

A-26

124. There is a cantilver beam whose length is L and it carries a point load at its free end. Then the bending moment at the centre of the beam will be equal to: 9 (a) --FL (b) - 2 FL 5 FL FL (c) 2 (d) 8 125. A simply supported beam oflength 'I' is carrying a 4dl of intensity w/unit length. Then the bending moment at the centers ofthe beam will be equal to : (a) w z2

(b)

w/2 4

(d)

(c)

wz2 8

w/2 16

126. A simply supported beam oflength 'I' is carrying a Udl of intensity w/unit length. Then the maximum bending moment will be equal to : (a)

wz2

(b)

8

(d)

wz2 2

wz2

135. The unit of moment is: (a) N-m (c)

(a)

Badboys2

(a)

T

-

J

T

r R

=-

M

(b)

-

o

-

J

141.

142.

T

Ymax

=-

143.

144.

0

(c)

J3F

F (d) 4

132. Factor of safety is the ratio of (a) breaking stress to working stress (b) ultimate stress to working stress (c) elastic limit to working stress (d) breaking stress to ultimate stress 133. Effect of a force on the body will depend upon : (a) Direction (b) Magnitude (c) Line of action (d) All of the above 134. The law of parallelogram offorces gives the resultant of: (a) Parallel forces (b) Like parallel forces (c) two coplanar concurrent forces (d) Non coplaner concurrent forces

(d) N/m4

10 3

(b) 5

2 3 (d) 5 10 Tangent of angle offriction is equal to: (a) kinetic friction (b) Limiting friction (c) Frictional force (d) coefficient of friction The coefficient of rolling resistance for a wheel of200 mm diameter which rolls on a horizontal steel roll, is 0.3 mm. The steel wheel carries a load of 600 N. The force required to roll the wheel will be equal to: (a) 90N (b) 180N (c) 45N (d) 270N The ratio oflinear stress to linear strain is given by : (a) Modulus of elasticity (b) Modulus of rigidity (c) Bulk modulus (d) Poisson's ratio the value of Poisson's ratio is always: (a) less than one (b) greater than one (c) equal to one (d) None of these Young's modulus of elasticity for a perfectly rigid bodies (c)

= --

G9 r G9 (d) - = L R L 130. Which of the following is a vector quantity (a) Energy (b) Mass (c) Angle (d) Force 131. Twoforcesof equal magnitude 'F' act an angle of 120 with each other. Then their resultant will be equal to: (a) 2F (b) F (c)

(b) N/m

136. The quantity,which is equal to rate of change ofmomentum is known to be: (a) Force (b) Acceleration (c) Impulse (d) displacement 137. If Dynamic friction = td, static friction = ts' then their relationship will be: (a) tdts (c) td = ts (d) None of these 138. When the applied force is less than the limiting frictional force, the body will : (a) start moving (b) remain at rest (c) slide backward (d) None of these 139. In comparisonto rolling friction,the value of sliding friction will be: (a) less (b) more (c) equal (d) double 140. A bodyof weight 30N rest on a horizontal floor.A gradually increasing horizontal force is applied to the body which just starts moving when the force is 9N. The coefficient of friction between the body and floor will be :

16

127. The greatest value of the Poisson's ratio is equal to: (a) 2 (b) 1 (c) 0.5 (d) 0.25 128. In S.1.system of units, the unit for strain is: (a) Pa (b) KPa (c) GPa (d) None of these 129. Which ofthe followingequation is associated for designing of shaft base on strength is given by:

N/m2

145.

IS:

(a) Zero (b) One (c) infinity (d) None of these 146. Which ofthe following is a diamensionless quantity: (a) stress (b) strain (c) Pressure (d) Modulus of elasticity 147. A cylindricalshellof diameter200 mm and wall thicknessof 5 mm is subjected to internal fluid pressure of lON/mm2. Maximum stress streas developed in the shell will be : (a) 50 Nzmm? (b) 100 Nzmm(c) 200 Nzmm? (d) 400 Nzmm148. The bulk moduless of elasticity: (a) does not increase with pressure (b) increases with pressure (c) independent of pressure (d) None of these

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Engineering Mechanics and Strength of Materials 149. To represent, stress - strain relations for a livearlyelastic homogeneous and isotropic material, minimum number of material constants required are: (a) 2 (b) 3 (c) 1 (d) 4 150. A tension member with a cross- sectionalarea of30 - mmresists a load of 60 KN. The normal stress induced on the plane of maximum shear stress will be : (a) 2 KN/mm2 (b) 4 KN/mm2 (c) 8 KN/mm2 (d) 3 KN/mm2 151. Ifelasticmodulus(E) = 12GPa, shearmodulus(G) = 50 GPa, then the value of Poisson's ration for the material will be eqaul to: (a) 0.1 (b) 0.4 (c) 0.2 (d) OJ 152. Poisson's ratio ofthe material is used in : (a) One dimensional bodies (b) two dimensional bodies (c) three dimensional bodies (d) both band c 153. Hook's Law holds good upto : (a) yield point (b) proportionalitylimit (c) breaking point (d) elastic limit 154. When a cast iron specimen is subjected to tensile test, then the percentage reduction in area will be equal to: (a) 0010 (b) 5% (c) 10% (d) 15% 155. Ifequal and opposite forces are applied to a body tending to elongate, if, then which of the following type of stress is developed? (a) twisting stress (b) compressive stress (c) tensile stress (d) shear stress 156. A 100 kg lamp is supported by a single cable of diameter 4 mm. The stress carried by the cable will be equal to : (a) 40 MPa (b) 78 MPa (c) 48 MPa (d) 88 MPa 157. The modulus of elasticity and rigidity of a material are 200 GPa and 80 GPa respectively.Then the Poisson's ratio will be equal to: (a) 1 (b) 0.55 (c) 0.75 (d) 0.25 158. If a composite bar of copper and aluminium is heated, then the stresses induced in copper and aluminium will be (a) compressive and tensile (b) bending and tensile (c) shear and bending (d) compressive and shear 159. Slowplastic deformation ofmetals under constant loadingl stress as a function of time is known as : (a) Fatigue (b) creep (c) Elastic deformation (d) Plastic deformation 160. The fatigue life ofa part can be improved by: (a) shot peening (b) coating (c) Polishing (d) carburizing 161. Flow stresses are associated with: (a) Breaking point (b) Plastic deformation (c) Fluid motion (d) Fracture stress

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A-27 162. The stress strain curve below represents for:

Strain

163.

164.

165.

166.

(a) Brittlematerial (b) Hard material (c) Softmaterial (d) ductilematerial Necking pheneomenon in stress-strain curve is observed for: (a) ductilematerial (b) brittlematerial (c) (a) and (b) both (d) None of them When a wire is stretched to double its length, then the longitudinal strain produced in it will be equal to: (a) 0.5 (b) 1 (c) 1.5 (d) 2.0 In a composite bar, the resultant strain produced will be equal to: (a) Sum of the strains produced by individual bars (b) Same as stress produced in each bar (c) Same as strain produced in each bar (d) difference of strains produced by the individual bars The total extension ofthe bar loaded as shown in figure is : lOT~

~

r

~

9T

IE 10mm "IE 10mm "IE lOmm "I

(a)

26xlO ---mm

16x 10 (b) --mm

(c)

6xl0 --mm

32xlO (d) ---mm

AE

AE

AE

AE

167. A composite bar is made up of steel and Aluminium strips each having area of cross = section on cm2 The composite bar is subjected to an axial load of 12000 N. If Esteel= 3 xE AI' then the stress in steel will be equal to: (a) 10N/mm2 (b) 20N/mm2 (c) 30N/mm2 (d) 40 N/mm2 168. If a beam is of a rectangular cross-section, then the distribution of shearing stress across a section is : (a) triangular (b) rectangular (c) Parabolic (d) Hyperbolic 169. The reaction at the prop in a propped cantilever beam subjected to U.D.L will be equal to : (a)

wI 4

(b)

3wl 8

wi wi (d) 8 6 170. Uniformalydistributedload 'WI is acting overper unit length ofa cantilever beam of Sm length. If the shear force at the midpointofbeam is 6 KN. then th value ofw willbe equalto : (a) 2 KN/m (b) 3 KN/m (c) 4KN/m (d) 6KN/m (c)

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A-28

171. The ratio of the compressive critical load for a long column fixed at both ends and a column with one end fixed and the other end being free is : (a) 2:1 (b) 4:1 (c) 8 : 1 (d) 16: 1 172. A simple supported beam PQ oflength 9 m, carries a Udl of 10 KN/m for a distance of 6 m from end P.What will be the reactions forces at P and Q. (a) 40N,20N (b) 20N,20N (c) 30N,20N (d) 80N,40N 173. A simply supported beam of 1 m length is subjected to a Udl of0.4 N/m. The maximumbending moment occuringin the beam will be: (a) O.OSN-m (b) 1N-m (c) 2N-m (d) 4N-m 174. A hollow shaft has external and internal diameters of 10em and Scm respectively. Torsional sectional modulus of the shaft will be: (a) 184cm3 (b) 384cm3 (c) 284cm3 (d) 37Scm3 175. A solid shaft of diameter 20 mm can sustain a maximum shear stress of 400 kg! cm-. The the torque transmitted by the shaft will be equal to : (a) 0.628Kg-cm (b) 62.8Kg-cm (c) 628 Kg-cm (d) 324Kg-cm 176. For designing a connecting rod, which of the following formula is utilized? (a) Rankin's formula (b) Euler's formula (c) both (a) and (b) (d) None of these 177. When a connecting rod is subjected to an axial force, then the buckling of the connecting rod may be with (a) X - axis as neutral axis (b) X - axis or y-axis as neutral axis (c) Z-axis as neutral axis (d) None of these 178. A column which is failed under the application of direct stress is known as : (a) Shortcolumn (b) mediumcolumn (c) long column (d) None of these 179. If L, = buckling load, Lc= crushing load, then which ofthe following relationship is true for long columns? 00 ~>~ ~ ~>~ (c) ~ =Lc (d) None of these 180. In case of compression numbers, they tend to buckle in which ofthe following direction? (a) Maximum cross-section (b) Neutral axis (c) Horizontalaxis (d) Minimum radius of gyration 181. Two books of mass 1 kg each are kept on a table, one over the other. The coefficient of friction on every pair of contacting surfaces is 0.3, the lower book is pulled with a horizontal force F. The minimum value ofF for which slip occurs between the two books is (a) zero (b) 1.06N (c) S.74N (d) 8.83N 182. Ifa system is in equilibrium and the position ofthe system depends upon many independent variables, the principle of virtual work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be

Engineering Mechanics and Strength of Materials (a) -1.0 (b) zero (c) 1.0 (d) infinite 183. A block of mass M is released from point P on a rough inclined plane with inclination angle 9, shown in the figure below. The coefficient of friction is J..!.IfJ..!< tan 9, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is

(a)

(c)

2s g cos 9(tan 9 -1-1) 2s gsin9(tan9-1-1)

2s g cos 9(tan 9 + 1-1)

(b)

2s gsin9(tan9+ 1-1)

(d)

184. A 1 kg block is resting on a surface with coefficient of friction 1-1 = 0.1. A force of 0.8 N is applied to the block as shown in figure. The frictional force is

08±L

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(a) zero (b) 0.8N (c) 0.89N (d) 1.2N 185. A block R of mass 100 kg is placed on a block S of mass ISO kg as shown in the figure. Block R is tied to the wall by massless and inextensible string PQ. If the coefficient of static friction for all surface is 0.4, the minimum force F (in kN) needed to move the block S is

00

Q~

~

Q~

(c) 0.98 (d) 1.37 186. A ball weighing 0.01 kg. hits a hard surfaceverticallywith a speed ofS mls and rebounds with the same speed. The ball remains in contact with the surface for 0.01 second. The average force exerted by the surface on the ball is (a) 0.1N (b) 1N (c) 8N (d) ION 187. A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P, the critical buckling load (Pcr) is given by (a)

E1 Pcr=~ nL

rrEI

(c) Pcr=7

(b) p

c

n2E1 3L2

=--

n2E1 (d) pc =--L2

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Engineering Mechanics and Strength of Materials

A-29

188. For a circular shaft of diameter d subjected to torque T, the

194. A solid circular shaft of diameter d is subjected to a

maximum value of the shear stress is

combined bending moment M and torque T. The material property to be used for designing the shaft using the relation

(a)

.!.i_.J M2 + T2

(b) 16T

tid'

(c)

(d)

189. A 200

x 100 x 50 mm steel block is subjected to a hydrostatic pressure of 15 MPa. The Young's modulus and Poisson's ratio of the material are 200 GPa and 0.3 respectively. The change in the volume ofthe block in mm ' is (a) 85 (b) (c) 100 (d) 110 190. A uniformly loaded propped cantilever beam and its free body diagrams are shown below. The reactions are

195.

so

iillllllllll]l M(illllllilit q

R,

IJIII

5qL

(a)

196.

q

L

1II1II

s, = -8-' R2=

is

nd3

8T nd'

197.

R2

3qL qL2 -8-' M= 8

198.

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3qL 5qL qL2 (b) s, = -8-' R2= -8-' M= 8 5qL 3qL (c) R, = -8-' R2= -8- , M=O 3qL 5qL (d) Rl = -8-' R2= -8- ,M=O

199.

(a) ultimate tensile strength (Su) (b) tensile yield strength (Sy) (c) torsional yield strength (Ssy) (d) endurance strength (Se) Ifthe principal stress in a plane stress problem are crl = 100 MPa, crl = 40 MPa, the magnitude ofthe maximum shear stress (in MPa) will be (a) 60 (b) 50 (c) 30 (d) 20 The state of plane-stress at a point is given by crx= -200 MPa, cry = 100 MPa and txy = 100 MPa. The maximum shear stress in MPa is (a) 111.8 (b) 150.1 (c) 180.3 (d) 223.6 A column has a rectangular cross-section of 10 mm x 20 mm and a length of! m. The slenderness ratio ofthe column is closed to (a) 200 (b) 346 (c) 477 (d) 1000 A thin cylinder of inner radius 500 mm and thickness 10 mm is subjected to an internal pressure of 5 MPa. The average circumferential (hoop) stress in MPa (a) 100 (b) 250 (c) 500 (d) 1000 A solid steel cube constrained on all six faces is heated so that the temperature rises uniformly by LlT. Ifthe thermal coefficient ofthe material is a, Young's modulus is E and the Poisson's ratio is v, the thermal stress developed in the cube due to heating is

a(LlT)E

191. The strain energy stored in the beam with flexural rigidity EI

(a)

(l-2v)

2a(LlT)E (b)

(l-2v)

and loaded as shown in the figure is P

1+-[;

L

t

2L

.;

(c)

3EI 4p2J3

--

(c)

L--+I

~

p2J3 (a)

(b)

a(LlT)E

3a(LlT)E

P

2p2L3

--

3EI 8p2JJ

(d)

3EI 3EI 192. A rod of length L and diameter D is subjected to a tensile load P. Which ofthe following is sufficient to calculate the resulting change in diameter? (a) Young's modulus (b) Shear modulus (c) Poisson's ratio (d) Both Young's modulus and shear modulus 193. The transverse shear stress acting in a beam ofrectangular cross-section, subjected to a transverse shear load, is (a) variable with maximum at the bottom of the beam (b) variable with maximum at the top of the beam (c) uniform (d) variable with maximum of the neutral axis

(1- 2v)

(d)

3(l-2v)

200. For a long slender column of uniform cross-section, the ratio of critical buckling load for the case with both ends clamped to the case with both ends hinged is (a) 1 (b) 2 (c) 4 (d) 8 201. A cantilever beam oflength L is subjected to a moment M at the free end. The moment of the inertia ofthe beam crosssection about the neutral axis is I and the Young's modulus is E. The magnitude ofthe maximum deflection is (a) (c)

ML2 2EI

(b)

ML2 EI

(d)

4ML2 EI

202. The maximum allowable compressive stress corresponding to lateral buckling in a discretely laterally supported symmetrical I-beam, does not depend upon (a) modulus of elasticity (b) radius of gyration about the minor axis (c) span/length of the beam (d) ratio of overall depth to thickness of the flange

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Engineering Mechanics and Strength of Materials

A-30 203. The number of strain readings (using strain gauges) needed on a plane surface to determine th principal strains and their directions is (a) 1 (b) 2 (c) 3 (d) 4 204. The buckling load in a steel column is (a) related to the length (b) directly proportional to the slenderness ratio (c) inversely proportional to the slenderness ratio (d) non-linearly to the slenderness ratio 205. Ifmoment M is applied at the free end of centilever then the moment produced at the fixed end will be (a) M (b) Ml2 (c) 2M (d) zero 206. A thin walled cylindrical pressure vessel having a radius of 0.5 m and wall thickness 25 mm is subjected to an internal pressure of700 kPa. The hoop stress developed is (a) 14 MPa (b) 1.4 MPa (c) 0.14MPa (c) 0.014MPa 207. If J.l = Poisson's ratio G = Modulus of rigidity, K = bulk modulus then (a)

Badboys2(c)

3K-G J.l= 2G+6K

(b)

K-G J.l= 2G+6K

(d)

K-G u= G+3K

3K-2G

u= 2G + 6K

208. For the shear force diagram shown in figure, the loaded beam will be .------.14t

9t

~4m""""I----

4t

209. When a column is fixed at both ends, corresponding Euler's critical load is 2n2EI

(a)

n2EI L2

(b)

(c)

3n2EI L2

(d)

L2 4n2EI L2

210. Consider the beamAB shown in figure below. PatAC ofthe beam is rigid. While part CB has the flexural rigidity EI. Identify the current combination of deflection at end Band bending moment at end A respectively

I:

(

c L

(a)

pC 2PL

(c)

8PC 2PL

B

L ----.j

.j4

(b)

3EI'

3EI'

(d)

pC PL 3EI'

8Pr! PL 3EI'

211. In a stressed body, an elementary cube of material is taken at a point with its faces perpendicular to X and Y reference axes. Tensile stresses equal to 15 kN/cm2 and 9 kN/cm2 are observed on theses respective faces. They are also accompanied by shear equal to 4 kN/cm2. The magnitude of the principal stresses at the point are (a) 12 kN/cm2 tensile and 3 kN/cm2 tensile (b) 17 kN/cm2 tensile and 7 kN/cm2 tensile (c) 9.5 kN/cm2 compressive and 6.5 kN/cm2 compressive (d) 12 kN/cm2tensile and 13 kN/cm2 tensile 212. Under torsion, brittle materials generally fail (a) along a plane perpendicular to its longitudinal axis (b) in the direction of minimum tension (c) along surfaces forming a 45° angle with the longitudinal

axis (d) not in any specific manner 213. A simply supported beam ofspan L and flexural rigidity EI, carries a unit point load at its centre. The strain energy in the beam due to bending is (a)

r!

C

A

f

(a)

48EI

(b)

192EI

(d)

16EI

C (c)

C

96EI

214. The design ofa eccentrically loaded column needs revision when 1.5 tim

(c)

Ai.__

~i~10_t

1.5 tim

(d)

(a)

fe' + f; <1 Ie h

(b)

(c)

fe' + f; > 1 Ie h

(d)

c~f~~uouou~uounuo~ou!3t

f: _h'

<1

Ie h

f: _h'

>1

Ie h

where, calculated average axial compression stress fb = maximum allowable bending compressive stress on the extreme fibre, and

Afo\oaf4t

fc = /m

fe' = calculated

bending stress in the extreme fibre.

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Engineering Mechanics and Strength of Materials

A-31

215. A gun metal sleeve is fixed securely to a steel shaft and the compound shaft is subjected to a torque. If the torque on the sleeveis twice that on the shaft, find the ratio of external diameter of sleeveto diameter of shaft [GivenNs = 2.5 NG] (a) 2.8 (b) 1.6 (c) 0.8 (d) 3.2 216. A column section as indicated in the given figure is loaded with a concentrated load at a point 'P' so as to produce maximum bending stress due to eccentricities about x-x axis and Y- Y axis as 5 t/ m2 and 8t/m2respectively. If the direct stress due to loading is 15t/m2 (compressive) then the intensity of resultant stress at the corner 'B' of the column section is (a) 2 t /m2 (compressive) (b) 12 t/m2 (compressive) (c) 18tlm2 (tensile) (d) 28 t/m2 (compressive) 217. If the principal strusses and maximum shearing stresses are of equal numerical values at a point in a stressed body, the state of stress can be termed as: (a) isotropic (b) uni -axial (c) pure shear (d) gineralized plan state of stress 218. Consider the following statements: 1. 2-D straw applied to a thin plate in its own plane represent the plane straw condition. 2. Under plane straw condition, the strain in direction perpendicular to plane is zero. 3. Normal and shear straw may occur simultaneously on a plane. Which of the above statments is/are correct? (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 3 219. The principal strains at a point in a body, under kiaxial stress state, are 700 x 10-6 and - 40 x 10-6. What will be the maximum shear strain at that point. (a) 110 x 10-6 (b) 300 x 10-6 (c) 550 x 10-6 (d) 150 x 10-6 220. What is the relationship between elastic constarts E, G and K?

223. The expression for the strain energy due to binding of a cantilever beam (length L, modulus of elasticity E and moment of inertia I) is given by :

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KG

(a)

E = 9K+G

(c)

E = K+3G

9KG

(b)

9KG E= K+G

(d)

E = 3K+G

9KG

221. Four vertical columns of same material, height and weight have the same end conditions. Which cross reaction will carry the maximum load ? (a) Solid circular reaction (b) Thin hollow circular section (c) Solid square section (d) I-section 222. A steel speciman 150 mm- in cross section stretches by 0.05 mm over a 50 mm gauge length under an axial load of 30 KN. The strain energy stored in speciman ? (a) 0.75Nm (b) 1.00Nm (c) 1.50Nm (d) 3.00Nm

(a)

p2L3 3EI

(b)

(c)

(d)

224. Consider the following statements: Maximum shear stress induced in a power transmitting shaft is 1. directly proportional to torque being transmitted. 2. Inversely proportional to the cube of itr diameter. 3. directly proportional to itr polar moment of inertia. Which of the statements given above are correct (a) 1, 2 and 3 (b) only 3 (c) 2 and 3 (d) 1 and 3 225. Maximum shear stress in a Mohr's circle (a) in equal to radius of Mohr's circle (b) in greater than radius of Mohr's circle (c) in less than radius of Mohr's circle (d) could be any of the above 226. A shaft is subjected to combined twisting moment T and binding moment M. What is the equivalent binding moment. (a)

]_~M2 +T2

(c)

~IM+~M2

2

l

+T2

227. The ratio of torque carrying capacity solid shaft to that of a hollow shaft is given by : (a) (1 - 0) (b) (1 - k4)-1 (d)

where, k =

o,

Do

Dj = inside diameter of hollow shaft Do = outside diameter of hollow shaft Shaft materials are same 228. The principal stress at a point in 2-D stress system are o 1 and cr2 and corresponding principal strains are Eland E2. If E and J.l denote young's Modulus and Poisson's ratio respectively, then which one of the following is correct. (a)

o = EEl

(b)

E o = --2 I-J.l

(c)

E o = --2 [El- J.lE2] 1- J.l

(d)

o = E[EI

[El+ J.lE2]

- J.lE2]

229. A point in a 2-0 state of strain in subjected to pure shearing strain of magnitude Yxy. radian. Which ohm of the following is the maximum principal strain ? (a)

(c)

Yxy

(b)

Yxy/2

(d)

Yxy/Ji 2yxy

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Engineering Mechanics and Strength of Materials

A-32

230. The SFD for a beam in shown in the fig. The BMD is shown by :

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c

(c)

(d)

SFD C

231. Match list I with list II and select the correct answer using the codes given below the lists : List I List-D A. Clopeyroh'sn theorem 1. Deflection of beam B. Maculayr's method 2. Eccentrically loaded

column C.

Perry's formular

3. 4.

Riveted joints Continuous beam

(b) (d)

ABC 4 1 2 4

Codes: (a) (c)

A 3 4

B

C

2

1

1

2

3 3

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...,

Engineering Mechanics and Strength of Materials

I···~

A-33

HINTS & EXPLANATIONS

5.

(a) Streamlined shapes are those shapes which decreases the amount of friction or resistance against airflow or waterflow. 6. (b) Byreducing the forceson contacting surfaces, friction also decreases as if depends on it. 11. (a) If8L= change in length, L = original length,

33.

(c)

QJ

. 8L Stram=L

41.

(c)

U = '!_E(8L)2

v 2 L where, E = Young's modulus of elasticity. 12. (b) As, we know that,

Load Stress(a) = -------Area of cross - section . () Stress(a) Stram 6 =--E

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.

P

8L

AE

L

Hence,6=T P A

49.

(b)

50.

(a)

LlL

Fatigue is a fracture phenomenon in which material is failed due to cyclic or repeated stresses usually at low values of stress. As we known that, M

= abc (max) x--

I

Ymax

PL AE 13. (d) Let a bar be hanging of weight 'w' EIongation 8L

A cylindrical elastic body subjected to pure torsion about its axis develops tensile stress in direction 45° to the axis. Strain is defined as the ratio of change in length to original length.

=-

where 2 = -_ ,

M

=

I

.

= section modulus

Ymax

ab(max)

Z

x

M

ab(max)

=Z

ab(max)

o: Z

1

Let, w = specific weight of bar F=wAy elongation (8L) = _FL_= ....:....( w_A_y.....:...)_.d_y AE AE

53.

(a)

61.

(a)

64.

(a)

8L= wy8y E

.c=_

Now, Total elongation -

o

--

E

wL2 _- --wL - -2E 2AE

Here, w = WAL. 14. (a) Considering Principal of superposition, 8L=8L] +8L2 +8L3 PLI PL2 =--+--+-EAI EA2

= !(.!i_+~+ E Al

30. (d)

A2

65. (b)

~=2(1+Ji)

PL3 EA3

J A3

Stiffness is defined as the property of the material which resists deflection against load within elastic limit. Given: tensile stress (o.) = 100MPa Maximum stress in the plate = 3 x aT = 3 x 100 =300MPa Given : Principal strains, 6]= 1000 X 10-6 62 =-600 X 10-6 Maximum shear strain (ymax) = 6] - 62 = (1000 x 10-6)-(-600 x 10-6) = 1000 x 1Q-6+ 600 x 1Q-6 = 1600 x lQ-6 As we know that, E=2C (1 + u)

66.

(a)

Given: Principal stresses, ax = 100 MPa, ay =40 MPa

For plane stress system,

L3

Toughness of the material is defined as the maximum amount of energystoredin a material upto fracture under the application of impact loads.

Maximum shear stress

('t"max)

= 100-40 = 60 = 30MPa

2

2

a -a =

x

2

y

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Engineering Mechanics and Strength of Materials

A-34

67.

Using the following relations, For thin cylinders,

(a)

= ~2F2 -2F2 cos300 (0: P = Q = F)

nd3t J=4

R=# R=F

Pdi crN=crC=2t crl' cr2= 1098MPa, 41 MPa 70.

;}~

(b)

A

f

140. (c)

B

tn

~

RA

RB

2F RA =3,RB

F =3

Now, at point If, the bending moment,

N

F~

w=30N N=w=30N when the body just starts moving, P=9N and also, P = F = 9N we know that, F = Il R or IlN

(BM)c = 2F x!....=2Fl

72.

3 3 9 (b) Considering the following relations, o = compressive stress (o) + tensile stress (crt)

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F

F

4F

16F

nd2 4

nd2 16

nd2

nd2

-12F 73.

nd2 (c) For the case of column, P

F Il= N

=0.3 x 600 =180N 147. (b) Given: diameter of shell(D) = 200 mm wall thickness (t) = 5 mm Internal fluid pressure (P) = 10Nrmm' Maximum shearing stress (tmro) PD lOx200 1 4t 4x5 2

L2 Now, If length is doubled, V=2L 2

2

pI = n EI = n EI (2L)2 4L2 L2 -

p- 4L2-"4 pI 74.

= _!_x

P

4 (a) Given: length of column = L

=100N/mm2

150. (a) Given: Area of cross - section (A) = 30 mmLoad (P) = 60 KN Normal stress (crN) = !_ = 60 = 2KN I mm2 A 30 151. (c) Given:E=120GPa G =50GPa Il=?

Considering the followingrelation, E=2G(1 + Il) E Il+ 1 =-

2G

flexural rigidity = EI effective length = 2 x L n2EI n2EI n2EI P--------- L~ - (2L)2 - 4L2 79.

(a)

131. (b)

3

=f..lR

2

-

9

= 30 =10

142. (b) Given: coefficient offriction (u) = 0.3 Load or Reaction (R) = 600 N force required to roll the wheel

= n EI

pI

+ F2 + 2 cos 120

= ~ F2

During tensile testing of cast iron specimen, the stress - strain curve shows no yield point because the length of deformation is very little. Given; P = Q = F, 9 = 120°, R = ~p2 +Q2 +2PQcos9

=~-1=~-1= 120_1 2G 2 x 50 100 u= 1.2-1 =0.2 156. (b) Given: mass(m)= 100kg diameter(d) = 4 mm = 4 x IO-3m Stress (c) =? Il

. n 2 Area of cross-section (A) = 4d =~(4xl0-3)2 4

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Mechanics and Strength of Materials

Engineering

A-35

For long column, (One end fixed and other end is free) = ~ x 16x 10-6 = 4n x 10---{j m' 4 Load = weight (w) = mg = 100 x 9.8= 980N

( p)C L2 -_

2

n EI 4L2

(PC)Ll 4n2EI --=--x--=(pc )L2 L2

w 980 Stress(cr) = - = 6 A 4n x 10=78.03 X 106 = 78 MPa (approx.) 157. (d) Given,E = 200 GPa G=80Gpa

4L2

16

n2EI

1

(PC)L1 : (PC)L2 =16:1 172. (a)

pf'f;lOKN~ J

fl=?

We know that, E=2G(1+fl) 200=2 x 80(1 + u)

:

6~

R

p

9m

R

Q

RP + RQ= 10 x 6 = 60 KN Taking moment about 'P', ~ x 0 + 9 x R = 180 RQ=20KN ~=60-20=40KN

(1+ ) = 200 fl 160 u= 1.25-1 =0.25 166. (a) Given,

173. (a)

lOT" I.____

L

IE

fTITIl

__L...~

10 mm

__

)IE

~__L...

10mm

)IE

~N/m

__,I , 9T

!

__

10mm

)1

IE

Badboys2lOT~lOT

· b di (0.4)(1)2 M axlmum en mg moment = ---

8 =0.05N-m 174. (a) Givenexternal diameter (D) = 10em Internal diameter (D) = 5 em

lO-3=7T~7T 9T~T

.

8L = lOx 10 8L = 7 x 10 8L = 9 x 10 1

)1

1m

AE'

AE'

2

3

-(Di)4Jx2 32

-.o,

AE

3.14[(10)4 -(5)4]x2

8L=8L1+8L2+8L3 100 70 =-+-+AE AE

90 AE

260 26 x 10 =-=--mm AE

n[(Do)4

Sectional modulus (z) =

32 x10 = 183.98cm3 = 184cm3 175. (c) Considering equation,

AE

167. (c) Given:A=3 emP= 12000N Esteel = 3 x EAJ Considering the relations, P=P1+P2 P=P1A1 + P2~ we get, Psteel = 30N/mm2 170. (c) 1.5 m R

T _ Ge _

Q

J L R T=628 kg-em 181. (d)

mg

(SF)R= 6KN=w

x

1.5

w=_i_=4KN 1.5 171. (d) For long column, (fixed at both ends)

(p)C

2

= 4n EI LJ

L2

FBD of Book 1

ByFBD of Book I, LFy=O=> NI =mg So,frictionalforce= flN 1 = fl mg By FBD of Book 2, LFy=0=>N2=N1 +mg=2mg LFx=O

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Engineering Mechanics and Strength of Materials

A-36

182. (b)

=> F~ ~Nl + ~N2 (For slip between two books to occur) F ~ umg + ~ . 2mg ~ 3~ . mg :. Fmin= 3 x 0.3 x 1 x 9.81 = 8.83 N The given statement is the principle of virtual work according to which the partial derivative of total potential energy with respect to each independent variable is zero.

186. (d) FxO.Ol=O.OI {5-(-5)} or

F= ION

187. (d) According to Euler's criterion of buckling load, for pin-ended column oflength L, the critical buckling load is given by

PO'

=

n2 xEI

183. (a) 188. (c)

2

L

T r By-=-=-=> J r

8

T ----

r

L

~d4

d

32

2

16T

Here, all the resolved forces acting on the block, along and perpendicular to inclined plane are shown. :EFN=O => N=Mgcos8 :EFt=O => Mg sin 8- ~N=Ma Mg sin 8 - ~ Mg cos 8 = Ma a = g (sin 8 - ~ cos 8) a = g cos 8 (tan 8 - u) or a = g sin 8 (1 - ~ cot 8) Now, since acceleration is constant so,

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s = ut + "!"at2 2 =>

=>

184. (b)

nd

2s

tN =3a(I_2V) V E

189. (b) By using the relation,

3 x 15 ~V=(200 x 100 x 50) x --5 (1- 2 x 0.3) 2 xl0 = 106 x 22.5 x 10-5 x 0.4 = 90mm3 The given propped beam consists of two parts 1. A cantilever with uniformly distributed load 2. A cantilever with point load (reaction) R2 at end in upward direction. =>

190. (a)

I

q/length

I~

s = 0 + ..!.. g cos 8 (tan 8 - ~ ) t2 2 t=

t=-3 =tmax

k==:::======J

gcos8(tan8-1-l)

F=08;:i~

=> :EFy = 0 Rl + R2 = q x L Also, 81 =~ qL4 For a cantilever with UDL, 81 = 8EI For cantilever with load R2 at end R xL3 8 - ---=-2__

~g

2-

185. (d)

Friction force F, = ~s x N = 0.1 x 9.81 = 0.981 N However, applied force (F = 0.8 N) is less than the static friction (F s), F < F s- so that the friction developed will equal to the applied force F = 0.8 N. Hint: Given FBD, For block'S'

3EI

qxL4 8EI R1 =qL-R2

3 => R2 =-qxL 8

3 5 = qxL-gqL=gqL

Also, moment M = R2 x L - q x L x (~) I

= ~q x L2 _.9.L2 = _ qL2

I I I

~F

I

I

8

2

5 So, finally R1 = gqL,

~R2+~----~----~I------~ I

it (100

+ 150) = 250 kg

3 qL2 R2 = gqL, M = -8-

8

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Engineering Mechanics and Strength of Materials

A-37

F = transverse shear load 'T= transverse shear stress Here, shear stress 'Tis variable and is maximum at the neutral axis.

p

191. (c)

2L--i D~L=t

B

194. (c)

Ra =P PL

PL

For a shaft subjectedto bending moment M and torque T, the equivalent torque is

o~o

Bending moment diagram

Te =~M2+T2

RA+ RB=2P :EMB=O 4PL P> 3L+ pX-RA x4L=0 => RA = --= P 4L .. RB=P BMx=P x X The total strain energy stored is given by

u=f

L(Px)2 xdx (pL)2 x2 JL(Px)2 x dx + + o 2EI 2EI 0 2EI

4p2L3

16T 16~M2 + T2 Induced shear stress is 'T- - ----- nd3 nd3 8 For safe designing 'T::;~

n

Torsional yield strength Factor of safety 195. (c) If maximum and minimum principal stressesare given, the maximum shear stress is given by or

u=-3EI

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192. (d) When the load P is applied in axial or longitudinal direction, increase in length

'T = crl -cr2 = 100-40 = 30MPa max 2 2 196. (c) 0" =

T

- 200 MPa

t"y= 100 MPa

L

1 p

= 100 MPa

'Tmax=

8 = PL = 4PL AE nD2E For, finding the change in diameter (transverse direction), Poisson's ratio v is needed. But modulus of elasticity E is also needed. Now, E= 2G(1 +v) from which vcan be found. Hence, to find 8D, both Young's modulus and shear modulus are needed. 193. (d) The distribution of tranverse shear stress along the vertical height of the beam is given by

t

cry

~t------t t- ------- -------

= ~(-150)2 +(100)2 = 180.27 MPa L 197• (b:\ 81enderness ra ti10 'I = K 'J I\,

bd3 1=12

y

Then K=

L---_

FAy 'T=-I.b 3 3 F where, 'Tmax= "'2'Tmean ="'2b x h

H'

= A)

Moment of'Inertia for rectangular section

hl2

t

( 1= AK2,K l':

{I= fA

J

bd3 12xA

20 x (10)3 = 2.886 12 x 10x 20

3

..

A = 1x 10 = 346.4 2.886 198. (b) P=5MPa,d= 1000mm,t= 10mm 199. (a) We know,

ex = _!_[ crx- v( cry + crz)] E

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Engineering Mechanics and Strength of Materials

A-38 stress is a and for the symmetrical

Let thermal system,

ax=ay=az=a

it will not depend upon the modulus of elasticity. 203. (a)

1

ex = ey = ez = - ( a - )l 2a) E (slenderness ratio )2 e=~(1-2v) E We know e= a~T

Per could be non linearly related to slenderness ratio so better to avoid choice (d).

a (1- 2v) E

~T=

206. (a)

Ea~T

207. (c)

a=....,----...,...

(1- 2v)

As it will be compressive stress. So, 200. (c)

cr

=

2G(1+)l)=3K(1-2)l)

2G+ 2)lG=3 K-6)lK )l(2G+6K)=3 K-2G

We know critical buckling load p

3K-2G

)l = 2G + 6K

2

1[ EI 2

208. (a)

Between A to B, SF = constant :. no load. Between B to C, SF is varying linearly :.UDL Similarly between C & D SF is varying linearly :. UDL.

209. (d)

Euler's critical load, P = 1[2EI

Le

For both ends hinged, L, = L

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We know that E=2G(1 +)l) E=3K(1-2)l)

..

Ea~T (1-2v)

athermal =

a = Pd _ 700x103 x2xO.5 _ x 6h 2t 2 x 25 x 10-3 - 14 10 - 14 MPa.

For both ends clamped, L, =

L

"2

(pcr )ctamped

4

(Per )hinged

1

(Leff

)2

where Leff= effective length ofthe column. 201. (a)

d2y

When both ends are fixed,

We know, EI dx2 = M

Le.ff=0.5L .P ••

210. (a)

dy or EI-=Mx+C1 dx

41[2 El L2

PartAC ofthe beam is rigid. Hence C will act as fixed end.

PI!

C}=O

Again,EIy=

1[2El 1[2El ---or-(O.5L)2 - 0.25L2

thus Os = 3EI

dy At x = 0, dx = 0 as fixed at end ..

cr -

211. (b)

Mx2 -2-+C2

But the bending moment we depend up on rigidity or flexibility of the beam . B.Mat A=2PL. Principal stresses,

Atx = 0, y= 0 (fixed end) So, C2 =0 M

s= 2EI x Ymax =

202. (a)

ML2 2EI

Given

2

a = 15kN/cm2 a x = 9 kN/cm2 tYxy =4kN/cm2

(atx=L)

Since allowable compressive stress depends upon

:. a',2 =12±~(3)2+(4)2 D

y

a} = 17kN/cm2 a2 = 7 kN/cm2

T

I ratio and -; ratio as per I.S. Code 800 : 1984. Therefore,

=12±5kN/cm2

212. (c)

Both of these are tensile in nature. Ductile materials generally fail in shear, therefore, when subjected to torsion, a specimen made of a ductile

Badboys2

Engineering Mechanics and Strength of Materials material breaks along a perpendicular to the longitudinal, when subjected to torsion, a specimen made of a brittle material tends to break along surfaces which are perpendicular to the direction in which tension is maximum i.e., along surface forming at 4?o angle with the longitudinal axis of the specimen. 213. (c) In case of simply supported beam carrying a point load 'P' at the centre, LM2·dx P W = where M = - X 1 2EI' 2 o

A-39

= 700 X 10-6 - (-400 - 10-6) = (700 + 400) X 10-6 = 1100 221. (b)

J

p

U2 2X2 • dx

[2 X

p2 •x3

222. (a)

x

10-6

EIn2 Maximum! Critical load (F) = -2Le So, critical/maximum load a Moment of inertia of section While, we know that M.O'! of thin hollow circular section is maximum. .

"21 P8

Stram energy of bar =

]LI2

:. W; = 2 [ 2x4EI = 2x4EIx3

= ~ x 30 x 103 x 0.05 x 10-3 = 0.75 N-m

Since, P = 1unit

r: w=-96EI

2

223. (b)

Strain energy due to bending (U) =

1

215. (b) Let us use suffix S for steel and suffix g for gun metal.

e

=

T

L IN Since both the steel shaft and gun metal slave are securely fixed,

L (Px)2

J a --dx 2E1

p2 fL x2dx 2E1 a

= -

p2 [x3]L p2 = 2E1 3 a = 2EI

e

L is the same for both.

fLo M2EI dx

L3

x]

p2L3

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---

l=l JsNs

JgNg

224. (d)

6E1 From torsion equation,

Jg Tg Ns -=-·-=2x2.5=5 r, Ts Ng

Tr J

GQ = (')max

~(D4 _d4) 32 =5 ~d4 32

T.

= Tr.R = __ 2_

L

( d)

Tr· -

max

J

D = 1.5651 d

f,g

('T)g

R

(

4)

l1[~ j

T, D J,

r,:-= (T;)s = Tg·d·T. = 2 x 1.5651 x 1/5 =

1

From the above expression, rmax a Tr and rmax a -3 d

0.62604 f

~=1.6 fss 216. (b) Stressat comer B = 15 + 5 - 8 = 12t 1m2 ( compressive) 217. (c) As from the condition given, (')1 = (')2 = r = 'tl = 't2 So, it is the state of pure shear 219. (a) Maximum shear strain, (r) = EI - EI

225. (a)

Maximum

shear

stress

(r max ) =

(')} - (')2

2

«(')}(')2Plane)and «(')}-(')2) = diameter of Mohr's circle.

Badboys2

Engineering Mechanics and Strength of Materials

A-40

227. (b)

.

.

Tr GQ tmax T =L=R

From torrsion equation,

229. (c)

Maximum principal strain = Yxy / 2

tmax should be equal for both shafter.

(;)

solid

= (;)

Hollow THollow

~(D4 32

-D~) 0

I

230. (b)

'fsolid THollow

P~

~ I

Q I

B

228. (b)

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1

Are we know, El

= E[cr1 -1·.ta2] 1

E2

On arranging

cr1 =

= E[cr2 - ~crd

the above equns.

E(El+~E2) 1_~2

,cr2 =

E(E2+~El) 1_~2

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rl'III~f)11Yf)l~ )11I(;IIINI~S lINI) )11I(;IIINI~ 1)I~SIfJN THEORY OF MACHINES

contact between them. The surface of one element slides over the surface of the other. For example: a piston along with cylinder. (b) Higher Pair: In which the links have point or line contact and motions are partly turing and partly sliding. For example: ball bearings, can and follower. Based on the type of mechanical constraint (or mechanical contact) (a) Self Closed Pair: If the links in the pair have direct mechanical contact, even without the application of external force. (b) Force Closed Pair: If the links in the pair are kept in contact by the application of external forces. Based on the type of relative motion between the elements ofthe pair (a) Sliding Pair: A kinematic pair in which each element has sliding contact with respect to the other element. (b) Rolling Pair: In a rolling pair, one element undergoes rolling motion with respect to the other. (c) Turning Pair: In a turning pair, one link undergoes turning motion relative to the other link. (d) Screw Pair: It consists oflinks that have both turning and sliding motion relative to each other. (e) Cylindrical Pair: A kinematic pair in which the links undergo both rotational and translational motion relative to one another. (t) Spherical Pair: In a spherical pair, a spherical link turns inside a fixed link. It has three degrees of freedom.

It is the branch of Engineering

Science, which deals with the study of relative motion between the various parts of machine along with the forces acting on the parts is known as the Theory of Machines (TOM). Kinematic Link: Each resistant body in a machine which moves relative to another resistant body is called kinematic link or element. A resistant body is which donot go under deformation while transmitting the force. Kinematic Pair: Ifthe relative motion between the two elements of a machine in contact with each other is completely or successfully constrained then these elements together is known as kinematic pair.

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CONSTRAINED MOTIONS

Constrained motion (or relative motion) can be broadly classified is to three types. 1. Completely Constrained: Constrained motion in which relative motion between the links of a kinematic pair occurs in a definite direction by itself irrespective of the external forces applied. For example a square bar in a square hole undergoes completely constrained motion. 2. Incompletely Constrained: Constrained motion in which the relative motion between the links depend on the direction of external forces acting on them. These motions between a pair can take place in more than one direction. For example a shaft inside a circular hole. 3. Partially (or Successfully) Constrained Motion: If the relative motion between its links occurs in a definite direction, not by itself, but by some other means, then kinematic pair is said to be partially or successfully constrained. For example a piston reciprocating inside a cylinder in an internal combustion engine.

TYPES OF KINEMA TIC PAIRS Types of kinematic Chains: Usaually, A kinematic chain has a one degree of freedom. The kinematic chains having number of lower fairs are tour are considered to be the most important kinematic chains in which each pair act as a sliding pair or turning pair. Some of them are given as : (a) Four bar chain (b) Single slider crank chain (c) Double slider crank chain The classified of kinematic pairs is listed as below: 1. Based on the nature of contact between the pairing elements. (a) Lower Pair: Links in the pair have surface or area

2.

3.

DEFINITION OF KINEMATIC CHAIN Combination of kinematic pairs joined in such a way that the last link is joined to the first link and the relative motion between them is definite. There are two equations to find out. Whether the chain is kinematic or not. l = 2p-4 where l = number oflinks p = number of pairs also

.

J

3

= -/2

2

where j = number ofjoints To determine the nature of chain we use equation . h 3/ J+-=- -2 where

h

2 2 = No. of higher pairs

Badboys2

Theory of Machines and Machine Design

A-42 If

L.HS > RH.S. then it is a locked chain L.HS. = RH.S. then it is a kinematic chain L.HS. < RH.S. then it is an unconstrained chain

Types: (i) (ii) (iii)

Beam engine Locomotive coupling rod Watts indicator ©©©©mechanism

(b) Single slider crank chain

GRUBLER'S CRITERION In a mechanism total no. of degrees of freedom is given by F = 3(n-l)-2j where n is no. oflinks and j = no. ofjoints (simple hinges) most ofthe mechanism are constrained so F = 1 which produces 1 = 3(n-l)-2j => 2j - 3n + 4 = 0 this is called Grubler's criterion. If there are higher pairs also no. of degrees of freedom is given by F = 3(n-l)-2j-h where h = no. of higher pairs. Also known as Kutz Bach criterion to determine the number of degree of freedom. This statement says that if the higher pairs are present in the mechanism like as slider crank mechanism or a mechanism in which slipping is possible between the wheel and fixed links. Higher pair: When the two element of a pair have a line or point contact when relative motion takes place and the motion between two elements is partly turning and partly sliding. E.g. Cam and follower, bale and bearing, belt and rope drive etc. Number of degree of freedom (movability): The number of independent parameters that define its configuration. The number of input parameters which must be independently controlled in order to bring the mechanism into useful engineering purpose.

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GRASHOF'S CRITERIA Grashof's criteria is applied to pinned four bar linkages and states that the sum ofthe shortest and longest link of a planar four-bar linkage cannot be greater than the sum of remaining two links if there is to be continuous relative motion between the links.

Types : (i) (ii) (iii) (iv) (v)

Pendulum pump Oscillating cylinder engine Rotary internal combustion engine Crank and slotted liver quick return mechanism Whitworth quick return mechanism

(c) Double Slider crank chain Types:

(i) (ii) (iii)

Elliptical trammel Scotch-yoke mechanism Oldham's coupling

Klein's Construction: It is defined as a graphical method to achieve the magnitudes of velocity and acceleration oflinks as well as required points on the links. Klein's construction is drawn on the configuration diagram. And It does not need to be drawn two or three different diagrams. Limitation: It is applicable to slider crank mechanism only.

INSTANTANEOUS CENTRE A point located in the plane (of motion of a body) which has zero velocity. The plane motion of all the particles ofthe body may be considered as pure rotation about the point. Such a point is called the instantaneous centre ofthe body. Ifthere are three rigid bodies in relative planar motion and share three instantaneous centre, all lie on the straight line, called Kennedy's theorem. Instantaneous axis of rotation: The axis passing through the instaneous centre of the body at right angles to the plane of motion is called instantaneous axis of rotation. Axode: The instantaneous centre changes every moment, its locus is called centrods, and the surface generated by the instantaneous axis is called the axode.

Methods to Locate Instantaneous Centre Locating the instantaneous centre of a body depends on the situation given. Following are some examples: (i)

Fig. Linkage shown in Fig. 1 is Grashoftype if

The instantaneous

Va along the (J)

perpendicular to the direction of velocity Va at point A on a rigid body shown in Fig.

s+l
IA = Va (J)

MECHANISM A mechanism is obtained by fixing one ofthe links of a kinematic chain, for example a typewriter. Basically there are two types of a mechanism.

centre I lies at a distance

II I I I

1. Simple mechanism: A mechanism with four links. 2. Compound mechanism: Mechanism with more than four links. Inversion of a Mechanism We can obtain different mechanisms by fixing different links in a kinematic chain, this method is known as inversion of a mechanism.

Inversions of mechanisms: (a) Four bar mechanism

Fig.

Fig.

(ii) The instantaneous centre I is the point of intersection of the lines perpendicular to the direction of velocities at the given points on the body as shown in Fig., we can write as

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Theory of Machines and Machine Design

Va =

v,

=

A-43

co xIA coxIB

where co is the angular velocity with which the body shall appear to rotate about the instantaneous centre I. (iii) If the two links have a pure rolling contact, the instantaneous centre lies on their point of contact. (iv) If the slider moves on a fixed link having straight surface, the instantaneous centre lies at infinity and each point on the slider have the same velocity.

QP

n(n-l) 2

VELOCITY AND ACCELERATION OF MECHANISMS

X

PQ = V~P PQ

at a =~ which is perpendicular to the link PQ PQ

CORIOLIS COMPONENT OF ACCELERATION Ifa particle C moves with a velocityv on a linkAB rotating with angular velocity co, as shown in Fig., then the tangential component ofthe acceleration of the particle C with respect to the coincident point on the linkAB is called coriolis component of acceleration which is given by

To analyse velocity and acceleration ofa mechanism we proceed link by link associated in the mechanism. Let us consider two points P and Q on a rigid link PQ, as shown in Fig. Let point Q of the link moves in clockwise direction relative to point P. In this case the relative velocity of point Q with respect to P would be perpendicular to the line PQ.

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PQ = ( VQP J2 PQ

atQP =axPQ

Kinematic Chain

N=

X

and the tangential component ofthe linear acceleration ofQ with respect to P is given by

Number of Instantaneous Centres in a Constrained If n are the number oflinks in a constrained kinematic chain, then the number of instantaneous centre (N) is given by

= co2

ar

c rv a cc = atcc = 2r.. UI

Fig. where v is the velocity ofthe particle C with respect to coincident pointC.

ACKERMAN STEERING GEAR MECHANISM

Fig. Now ifthe point Q moves with respect to P with an angular velocity co and angular acceleration a, thus velocity has two components, perpendicular to each other. (a) Radial or centripetal component (b) Tangential component These components of velocity can be determined by calculating linear accelerations in radial and tangential directions. Figure shows the link representing both the components of acceleration.

All the four wheels must tum about the same instantaneous centre to fulfill the condition for correct steering. Equation for the correct steering is cot - cot = c/b where c = Distance between the pivots of the front axles b = Wheel base and are angle through which the axis of the outer wheel and inner whel turns respectively. For approximately correct steering, value of c/b should be in between 0.4 and 0.5.

e

e

DAVIS STEERING GEAR MECHANISM According to Davis Steering gear the condition for the correct steering is given by tan a = c/2b where c = Distance between the pivots ofthe front axles b = Wheel base a = Angle of inclination ofthe links to the vertical

FRICTIONAL BEARING t

Clap

Fig. Radial component of the linear acceleration of Q with respect to P is given by

TORQUE

IN PIVOT AND COLLAR

Pivot and Collar bearings are used to take axial thrust of the rotating shaft. While studying the friction in bearing it is assumed that The pressure over the rubbing surfaces is uniformly 1. distributed through out the bearing surface. 2. The wear is uniform throughout the bearing surface. (i) Frictional torque transmitted in a flat bearing is given by

Badboys2

Theory ofMachines and Machine Design

A-44

Slack Side

T = ~ x J.!WR while considering uniform pressure

T2

3

And in case of uniform wear 1 T= -xJ.!WR

2 J.! = Coefficient offriction

where

(ii)

W = Load transmitted to the bearing R = Radius of the shaft Frictional torque transmitted in a Conical Pivot bearing is given by

Driving Pulley

Fig.: Belt drive-open system

2 3

T= - x J.!WR cosec a while considering uniform pressure And in case of uniform wear

I

Driving

I T= - x J.!WR cosec a 2

(iii)

where a = semi angle of the cone Frictional torque trnsmitted in a trapezoidal or truncated conical pivot bearing is given by

Pulley

I

Driver

Pulley

Fig. : Belt drive-cross system When a number of pulleys are used to transmit the power from one shaft to another, then a compound drive is used as shown in Fig.

T = ~ x J.!W [ r ~ - r ~ ] cosec a 3 r1 -r2

Badboys2while considering

uniform pressure. And in case ofuniform wear

1

T = - x J.!W (r1 + r2) cosec a = J.!WR cosec a 2 where r1 and r2 are the external and internal radii of the conical bearing respectively

(iv)

R = r1 + r2 is the mean radius of the bearing. 2 Frictional torque transmitted in a flat collar bearing is given by

- 2 T --XJ.! 3

w[rirf-d-d] ---

Fig.: Belt drive-compound system

Types of Belts There are three types of belts (a) Flat belts: Cross section of a flat belt is shown in Fig. 14. Flat belts are easier to use and are subjected to minimum bending stress. The load carrying capacity of a flat belt depends on its width.

while considering uniform pressure And in case ofuniform wear

T=

"21 x J.!W (rl + r2)

The frictional torque transmitted by a disc or plate clutch is same as that of flat collar bearing and by a cone clutch is same as that of truncated conical pivot bearing.

BELT DRIVE The transmission of power from one rotating shaft to another lying at a considerable distance, is achieved using belts and ropes. Two parallel shafts may be connected by open belt or by cross belt. In the open belt system, the rotation of both the pulleys is in the same direction. If a crossed belt system is used, the rotation of pulleys will be in the opposite direction. Fig. 11and Fig. shows open and crossed system respectively.

Fig. : F at belt The ratio of driving tensions for flat belt drive is given by T

_1 =eJ.l9

T2

=>

2.3 log

UJ

=1'·8

where J.!= coefficient of friction between the belt and the pulley = angle of contact in radians Material used for flat belt is generally leather of various

e

Badboys2

Theory of Machines and Machine Design

(b)

types having ultimate tensile strength between 4.5 to 7 N per cm width. For heavy duty, two or three piles ofleather are cemented and pressed one above the other such belts are called double or triple ply belts. V-belts: Fig. shows the cross section of the V-belts. Vbelts are available in five sections designed A, B, C, D, and E and there are used in order of increasing loads. Section A is used for light loads only and section E is used for heavy duty machines. The angle ofV-belt for all sections is about 40°. In order to increase the power output, several V-belts may be operated side by side. In multiple V-belt drive, all the belts should stretch at the same rate so that the load is equally divided between them. If one of the set of belts break, the entire set should be replaced at the same time. The groove angle in the pulley for running the belt is between 40 to 60°. Due to reduced slipping, V-belts offer a more positive drive. V-belt drives run quietly at high speeds and are capable of absorbing high shock. 0

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Fig. V-belt The ratio of driving tension for the V-beltdrive is given by

.!L =

13) e

e(f,.LCOSeC

T2 c>

2.3 log

where

(c)

GJ

A-45

where

rl and r2 are radii of the two pulleys x = distance between the centres of two pulleys In a crossed belt drive the length of the belt is given by

(r + r )2 I 2 + 2x x When the belt passes from the slack side to the tight side a certain portion of the belt extends and when the belt passes from the tight to slack side the belt contracts. Due to these changes in length, there is relative motion (called creep) between the belt and pulley surfaces. Creep reduces the velocity ofthe belt drive system like slip do. L =

J3 = Semi-angle ofthe groove

Angle of contact in radians V-beltsare usually made of cotton fabric, cards and rubber. Circular belts: The cross section of a circular belt is shown in Fig. The circular belts are also known as round belts. These are employed when low power is to be transmitted, for example in house hold appliances, table top tools and machinery of the clothing. Round belts are made ofleather, canvas and rubber.

(rI + r2) +

Centrifugal Tension The centrifugal tension (Tc)is given by T, = mV2 where m = Mass per unit length of the belt V = Linear velocity ofthe belt The power transmitted can be calculated as below: The total tension on the tight side = T) + Tc The total tension on the slack side = T2+ Tc .. PowerTransmitted = [(TI+Tc)-(T2+Tc)]V =(TI- T2)V Which is equal to the value of powertransmitted given byeffective turning force (TI - T2), that is the centrifugal tension has no effect on the power transmitted. The maximum power transmitted by the belt is given by the maximum total tension in the tight side ofthe belt when it is three times the centrifugal tension. T = 3Tc => T = 3mV2 So velocity for the maximum power transmitted is given by

= 11-9- cosec 13

e=

1t

v

>

)3:

Velocity Ratio The velocity ratio of speeds of driver and driven pulleys is given by 0)2 0))

where

=

N2

=

d + (1- ~J l

t

N) d2 + t 100 dJ, d2 = diameters of driver and driven pulleys 0)1, ffi2 = angular velocities of driver and driven pulleys NI, N2 = rotational speeds of driver and driven pulleys expressed in revoluations per minute (r.p.m.) S = SI + S2+ 0.01SIS2is percentage of total effective slip SI = Percentage slip between driver and the belt S2 = Percentage slip between belt and the follower (driven pulleys)

GEARS AND GEAR DRIVE I

Fig. Circular Belt The ratio of driving tensions in round belts and rope drive is same as V-beltdrive.

A wheel with teeth on its periphery is known as gear. The gears are used to transmit power from one shaft to another when the shafts are at a small distance apart.

Length of Belt

Types of Gears

In an open belt drive system the length of the belt is given by

Commonly used gear are as below: (a) Spur gear: A cylindricalgear whosetooth traces are straight lines parallel to the gear axis. These are used for

L =

1t

(r. + r2) +

(rl - r2)2 + 2x x

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Theory ofMachines and Machine Design

A-46

(b)

(c)

transmitting motion between two shafts whose axis are parallel and coplanar. Helical gear: A cylindrical gear whose tooth traces are straight helices, teeth are inclined at an angle to the gear axis. Double helical gears called herringbone gears. The helical gears are used in automobile gear boxes and in steam and gas turbines for speed reduction. The herringbone gears are used in machinery where large power is transmitted at low speeds. Bevel gear: The bevel gear wheels conform to the frusta of cones having a common vertex, tooth traces are straight line generators of the cone. Bevel gears are used to connect two shafts whose axis are coplanar but intersecting when the shafts are at right angles and the wheels equal in size, the bevel gears are called mitre gears. When the bevel gears have their teeth inclined to the face of the bevel, they are known as helical bevel gears.

(g)

Internal and external gearing: Two gears on parallel shaft may gear either externally or internally as shown in Fig.

_a:0n ..--

Pinion (a)

(b)

Rack

(c)

Fig.

Gear Terminology Terms associated with profile of a gear tooth are illustrated in Fig.

Pitch circle: Essentially an imaginary circle which by pure rolling action gives the same motion as the actual gear. Pressure angle or angle of obliquity: Angle between the common normal to two gear teeth at the point of contact and the common tangent at the pitch point (common point of contact between two pitch circles). It is usually denoted by <1>. The standard pressure

Fig. : Bevel gear Badboys2 (d) Spiral gear: These are identical to helical

(e)

gears with the difference that these gears have a point contact rather than a line contact. These gears are used to connect intersecting and coplanar shafts. Worm gear: The system consists ofa worm basically part of a screw. The warm meshes with the teeth on a gear wheel called worm wheel. It is used for connecting two non-parallel, non-intersecting shafts which are usually at right angles.

1° angles are 14 - and 20°. 2

Addendum: Radial distance of a tooth from the pitch circle to the top of the tooth. Addendum element Working depth

Root or dedendum circle

Fig.

(f)

Fig. : Worm gear Rack and pinion: Rack is a straight line spur gear of infinite diameter. It meshes,both internally and externally, with a circular wheel called pinion. Rack and pinion is used to convert linear motion into rotary motion and vice versa.

Dedendum: Radial distance of a tooth from the pitch circle to the bottom of the tooth. Addendum circle: Circle drawn through the top of the teeth and is concentric with the pitch circle.

Dedendum circle: Circle drawn through the bottom of the teeth. It is also called root circle. Root circle diameter = Pitch circle diameter x cos where is the pressure angle.



Circular pitch: Distance measured on the circumference

of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is denoted by Pc, mathematically P, can be calculated as

P = 1tD

T

c

Fig. : Rack and pinion

where

D T

=

Pitch circle diameter

= Number of teeth on the wheel.

Badboys2

Theory of Machines and Machine Design For two gears to mesh correctly their circular pitch should be same

A-47 from the begining to the end of engagement.

Length of the path of contact: Length ofthe common normal cutoffby the addendum circles of the wheel and pinion.

Arc of contact: The path traced by a point on the pitch circle from

~

= D2

TI

T2

Velocity ratio of two meshing gears is given by VI = 1t DI NI V2 = 1t D2 N2 Linear speed of the two meshing gears is equal So

1t DI

NI

= 1t D2

N2

the beginning to the end of engagement ofa given pair of teeth. It consists of (a) Arc of approach (b) Arc of recess

Arc of approach: Portion of the path of contact from the beginning of the engagement to the pitch point.

Arc of recess: Portion of the path of contact from the pitch point to the end of the engagement ofa pair of teeth. · C ontact R abo

Length of arc of contact

= --=--------

Circular Pitch

Contact ratio is the number pairs of teeth in contact. N2 NI

=

IL

Length of Arc of contact: Length of the arc of contact can be

T2

calculated as

Diametral pitch: It represents the number of teeth on a wheel per unit of its diameter. . . T DIameter pitch P d = D

Badboys2 Module: It represents :::::>

1t

=-

Pc

PcxPd=1t

the ratio of pitch circle diameter (in mm) to the number of teeth. D

1

T

Pd

m=-=-

:::::>

m x Pj==I Recommended series of modules in Indian Standards are 1, 1.25, 1.5, 2,2.5, 3,4, 5, 6, 8, 10, 12, 16, and 20. Modules of second choice are 1.125, 1.375, 1.75,2.25, 2.75,3.5,4.5, 5, 5.5, 7, 9, 11, 14 and 18.

Total depth: Radial distance between the addendum and the dedendum circles of gear. Tooth depth = Addendum

L engt h 0fAr co f contact where



=

Length of path of contact

---=-----==------cos

is pressure angle

Interference: The phenomenon, when the tip of a tooth under cuts the root on its mating gear. It may only be avoided, if the addendum circles ofthe two mating gears cut the common tangent to the base circles between the points of tangency.

Law of gearing: According to the law of gearing, the common normal at the point of contact between a pair of teeth must always pass through the pitch point. Gear Trains Any combination of gear wheels by means of which power and motion is transmitted from one shaft to another is known as gear train. Various types of gear train are 1. Simple gear train: A gear train in which each shaft carries one wheel only. Fig. shows the arrangement of a simple gear train.

+ dedendum

Idle gears

Clearance: Radial distance from the top of the tooth to the bottom of the tooth in a meshing gear. Circle passing through the top of the meshing gear is known as clearance circle. Standard value of clearance is 0.157 m, where m is module. Dedendum = Addendum + 0.157 m = m + 0.157 m = 1.157

m Working depth: Radial distance from the addendum circle to the clearance circle. Working depth = Addendum of first gear + Addendum of second gear

Driven or follower

I I I

i""Il'§""""'~'II"'~III"II"II'§"""lIi Fig. : Simple gear train Velocity ratio

=

Speed of the driving wheel

--=---------==-----

Speed of the driven wheel

Back lash: Difference

between the tooth space and tooth thickness, measured along the pitch circle. In actual practice somebacklash must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion. Path of contact: Path traced by the point of contact of two teeth

no. of teeth on the driven wheel no. of teeth on the driving wheel

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Theory ofMachines and Machine Design

A-48

Train value

= ~ NI

2.

ArmC

= IL T4

Compound gear train: A compound gear train includes two gears mounted on the same shaft as shown in Fig.

Fig. : Epicyclic gear train

Driver

. . NB Velocity RatIo -

Driven or follower

= 1 + -TA TB

Nc

FLYWHEEL

Fig. : Compound gear train Velocity ratio

N N = _I = _1 N6

3.

N2

X

N N _3 x _5 N4 N6

= T2 xT 4 xT 6 T1 x T3 x T5

Reverted gear train: A reverted gear train manifests when

Badboys2 the first driving gear and the last driven gear are on the same axis. Axes are coincidental and coaxial. Fig. shows an arrangement of the reverted gear train. If D1, D2, D3, D4 be the pitch circle diameters of the respective gears and corresponding speeds are N], N2,N3, N4 then D1 + D2

=>

= D3 + D 4 2 2 D1 + D2 = D3 + D4

. . Velocity rano

T2 x T4 = -N1 = --=-_....:.... N4

T1x T3

A wheel used in machines to control the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation. These wheels are known as flywheel. It absorbs energy when crank turning moment is greater than resisting moment and gives energy when turning moment is less than resisting moment. The speed of a flywheel increases during it absorbs energy and decreases when it gives up energy. This way flywheel supplies energy from the power source to the machine at a constant rate throughout the operation. Coefficient of fluctution of energy: Ratio of the maximum fluctuation of energy to the work done per cycle, is called coefficient of fluctuation of energy. Lllimax = Emax- Emin C energy--

LlEmax Wpercyc1e

where

Lllimax = maximum fluctuation of energy Cenergy= coefficient of fluctuation of energy Coefficient of fluctuation of speed: Ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed. .1COmax = COmax - COmin C

s

= .1comax CO mean

Relation between maximum fluctution coefficient of fluctuation of speed. Lllimax = lCOmean (COmax - COmiu) =

1(,)

v-mean

4.

Fig. : Reverted gear train In a clock mechanism a reverted gear train is used to connect hour hand to minute hand in a clock mechanism. Epicyclic gear train: A special type of gear train in which axis of rotation of one or more of the wheels is carried on an arm and this arm is free to rotate about the axis of rotation of one or the other gears in the train. Fig. shows an arrangement of an epicyclic gear train.

Lllimax

=

of energy .1Emaxand

(comax- comin)x co mean COmean

lco~ean c,

... (1)

also energy stored is a flywheel is given by E

=>

1

=

2

"2ICOmean

2E I

... (2)

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Theory of Machines and Machine Design

A-49

from (1) and (2)

Ifhp is the height of porter governor (when length of arms and links are equal). and h., is height of watt's governor then

Lllimax = 2ECs where C, = coefficient of fluctuation of speed. GOVERNORS

~=m+M

The function of a governor is to regulate the mean speed of an engine within mentioned speed limits for varying type of load condition. Terms Used in Governors (a) Height of Governor: Vertical distance from the centre of the ball to a point where arms intersect on the spindle axis. (b) Equilibrium Speed: The speed at which the governor balls, arms etc. are in complete equilibrium and the sleeve does not tend to move upwards or downwards. (c) Sleeve Lift: Vertical distance with the sleeve travels because of change in equilibrium speed. (d) Mean Equilibrium Speed: The speed at the mean position of the balls or sleeve. (e) Maximum and Minimum Equilibrium Speeds: The speeds at the maximum and minimum radius of rotation of the balls, without tending to move either way are known as maximum and minimum equilibrium speeds respectively. IfN 1and N2 are maximum and minimum speeds then

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..

Sensitiveness =

2 (N1 - N2)

___;'----.!_----=c..::....

(N1 + N2)

(f)

Sensitiveness: A governor is said to be sensitive, if its change of speed is from no load to full load may be small a fraction of the mean equilibrium speed as possible and the corresponding sleeve lift may be as large as possible. (g) Stability: If for every speed within the working range there is a configuration of governor balls, then it is said that governor is stable. For a stable governor, the radius of governor balls must increase with increase in the equilibrium speed. (h) Hunting: Fluctuation in the speed engine continuously above and below the mean speed is called hunting. (i) Isochronism: A governor is isochronous provided the equilibrium speed is constant for all radii of rotation of the balls upto the working range. G) Governor Effort: The average force required on the sleeve to make it rise or come down for a given change in speed. (k) Power of Governor: The work done at sleeve for a given percentage change in speed. Mathematically Power = Mean effort x Lift of sleeve Types of Governors Different types of Governors are: (1) Simple governor-Watt type: The simplest type a centrifugal governor is known as watt type or watt governor. Height of the governor is given by 895

h = --

N2

metres

where N = speed of the arm and ball about the spindle axis. (2) Porter governor: It is obtained by modifying a Watt governor with a central load attached to the sleeve. The governor speed increases and decreases as the sleeve moves upwards or downwards respectively.

hw m where m = mass of the ball M = mass of the sleeve (3) Hartnell governor: This is a spring controlled governor. If Is1eeveis the lift of the sleeve and Xcompressionis the compression of the spring then Is1eeve = Xcompression = (r2 - r.)

y_

x r 1 = Minimum radius of rotation r2 = Maximum radius of rotation x = Length of ball arm oflever y = Length of sleeve arm of lever Stiffness of the spring is given by

where

S = S2 - S1 h where S 1 = Spring force at minimum radius of rotation S2 = Spring force at maximum radius of rotation CAMS A rotating machine element which gives reciprocating or oscillating motion to another element called follower is known as cam. These are mainly used for inlet and exhaust values ofl.C. engines, lathes etc. Types of Cams L Radial cam: A cam in which followerreciprocates or oscillates in a direction perpendicular to the axis of the cam. Radial cam is further classified as (a) Reciprocating cam (b) Tangent cam (c)

2.

Circular cam

Cylindrical Cam: A cam in which the follower reciprocates or oscillates in a direction parallel to the cam axis.

CAMS TERMINOLOGY A radial cam with reciprocating roller follower is shown in Fig. L Base Circle: Smallest circle that can be drawn to the cam profile. 2. Trace Point: The reference point on the follower which is used to generate the pitch curve that varies from case to case. For example, in case of knife edge follower, the knife edge represents the trace point and the pitch curve corresponds to the cam profile while in case of roller follower, the centre of the roller represents the trace point 3. Pressure Angle: The angle between the direction of the follower motion and a normal to the pitch curve. Keeping the pressure angle too large will lead to joining of reciprocating follower. 4. Pitch Point: A point on the pitch pitch curve having the maximum pressure angle.

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Theory ofMachines and Machine Design

A-50

and acceleration ofthe follower is given by ar = (1)2(R - r.) cos e where R = Radius of circular flank r1 = Minimum radius of the cam e = Angle turned through by the cam (I) = Angular velocity of the cam

Maximum pressure angle

MACHINE DESIGN

Pitch "_--l.~~~-+-+-e point Pitch point \ \

"-

5. 6. 7.

Cam profile

, "-

" ..........

_----".,.

Fig. Pitch Circle: A circle drawn from the centre of the cam through the pitch points. Pitch Curve: The curve generated by the trace point as the follower moves relative to the cam. Prime Circle: Smallest circle that can be drawn from the centre ofthe cam and tangent to the pitch curve. For a roller follower, the prime circle is larger than the base circle by the radius ofthe roller while in case of knife edge and a flat face follower it is equal. Lift or Stroke: The maximum travel offollower from its lowest position to the topmost position is called life or stroke. Angle of Ascent: It is the angle moved by cam from the time the follower begins to rise till it reaches the highest point. Angle of Descent: Angle during which follower returns to its initial position. Angle of Action: It is the total angle moved by cam from the beginning of ascent to finish of descent. Under Cutting: The situation of a Cam Profile which has an inadequate curvature to provide correct follower movement, is known as under cutting.

Badboys2 8.

9. 10. 11. 12.

A machine which is more economical in the overall cost of production and operation is called a new or better machine. Machine design deals with the creation of new and better machine and also improving the existing machines. Metals selected to design an element of a machine has some mechanical properties associated with the ability of the material to resist mechanical forces and load. The commonly used materials in engineering practice are the ferrous metals which have iron as their main constituent. Various types offerrous metals are shown in Fig.

Grey cast iron (3-3.5% carbon)

Dead mild steel (upto 0.15% carbon)

White cast iron (1.75-2.3% carbon) Mottled cast iron (It is a product between grey and white cast iron in composition, colour and general properties) Malleable cast iron (Cast iron alloy which solidifies in the as-cast condition in a graphite free structure)

Low carbon or mild steel (0.15%-0.45%) carbon Medium carbon Steel (0.45%-0.8% carbon) High carbon Steel (0.8% - 1.5% Carbon)

VELOCITY AND ACCELERATION OF THE FOLLOWER

BASIS OF LIMIT SYSTEM

(a)

In order to control the size of finished part with due allowance for error for interchangeable parts is called limit system. There are generally two basis oflimit system. (a) Hole basis system: In this system the hole is kept as a constant member and different fits are obtained by varying the shaft size. (b) Shaft basis system: In this system the shaft is kept as a constant member and different fits are obtained by varying the hole size.

(b)

Tangent Cam with Reciprocating Roller Follower In the tangent cam flanks of the cam are straight and tangential to the base circle and nose circle. Tangent cams are used for operating the inlet and exhaust values of I.C. engines. Displacement of the follower is given by Yf= (r. + r2)(1- cos e) sec e Velocity of the follower is given by Vf= (I) (r, + r2) sin e sec2 e and acceleration of the follower is given by af = (1)2(r, + r2) (2 - cos' e) sec2 e where r 1 = Minimum value ofthe radius ofthe cam r2 = Radius ofthe roller follower e = Angle turned by the cam, from the beginning of the follower displacement (I) = Angular velocity of the cam Circular Arc Cam with Flat-faced Follower In the circular arc cam the flanks ofthe cam connecting the base circle and nose are of convex circular arcs. Displacement of the flat faced follower is given by Yf= (R - r1) (1- cos e) Velocity of the follower is given by Vf= (I) (R - r.) sin e

Standard

tolerances

The system oflimits and fits comprise of 18 grades of fundamental tolerances according to the Indian standards. These are ITO1, ITO and ITl ... ITl6, these are called standard tolerances. Standard tolerance can be determined in terms of standard tolerance unit (i) microns by using the relation i = 0.45

Vi5 + O.OOlD for grades ITS to IT7.

And for the grades ITO1, ITO and IT 1 as below For ITOl, i (microns) = 0.3 + 0.008D For ITO, i(microns)=0.5+0.012D

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Theory of Machines and Machine Design

A-51

ForlTl, i(microns)=O.8.0.020D where D is the size or diameter in mm.

STATIC LOADING AND DYNAMIC LOADING (a) Static loading: A type ofloading in which the load is applied slowly or increases from nil to a higher value at a slow pace. There are no acceleration produced in the static loading. (b) Dynamic loading: A type ofloading which varies in magnitude as well as direction, very frequently, such type ofloading is called dynamic loading or fluctuating or alternating loads.

SJ. unit ofE is Nzmm-, Hook's law applies to both tension and compression. (iii) Direct shear stress: When a body is subjected to two equal and opposite forces, acting tangentially across the resisting section, as a result of which the body tends to shear off the section, then the stress induced is called shear stress. The strain occured due to the shear stress is called shear strain. Let us consider the two plates held together by means of a rivet as shown in Fig.

STRESS AND STRAIN (a) Stress: Resistive force per unit area to the external force on a body, set up within the body is called stress on that body. (b) Strain: Deformation produced per unit length of a body is called strain.

I I I

~

d (a)

Types of stresses Stresses are classified as (i)

Tensile stress: If a body is subjected to two equal and opposite external pulls, then the stress developed inside the body is called tensile stress

(b)

The direct shear stress induced in the rivet is given as

Pt

where Badboys2

at= A

P t = Axial tensile force in N A = Area of cross-section of the body in at = Tensile stress in N/mm2 the strain produced can be calculated as

r = Ps A

mm?

8/ / where 8/ = change in the length ofthe body or increase in length l = original length of the body e = tensile strain produced (ii) Compressive stress: If the body is subjected to two equal and opposite pushes then the stress developed is called compressive stress. e=

Pc

where

ac= A ac = compressive stress in Nzmm?

Pc = compressi ve force A = area of cross-section of the body in mmCompressive strain is given by e=

/

8/ = decrease in length of the body Hook's law: Hook's law states that when a material is loaded within elastic limit, the stress is directly proportional to strain aoce or a = Ee where a = stress e = strain E = Young's modulus or modulus of elasticity where

e

IS

"['=_sPc 2A If is the deformation produced due to shear stress r then r o: "['=C where C is called modulus of rigidity. (iv) Torsional shear stress: When a body is subjected to two equal and opposite torques or torsional moments acting in parallel planes, the body is said to be in torsion, and the stress produced due to torsion is called torsional shear stress. Let us consider a body of circular cross-section subjected to torque T, which produces a twist of an angle radians as shown in Fig.

e

T

8/ -

a E= -=--

r = direct shear stress in N/mm2 Ps = shear force across the cross-section in N A = cross-sectional area in mm? If the rivet is subjected to a double shear then shear induced

where

Pxl Ax 8/

I

(~{;~~~~~~;~) t~-~

e

T

The torsional shear stress induced at a distance r from the centre is given by "['torsion =

Txr

-1p

where "['torsion = torsional shear stress a distance r in

T = applied torque r = radial distance

Nzmm?

Badboys2

Theory ofMachines and Machine Design

A-52 Ip

moment of inertia of cross-section about centroidal axis. Torsion equation: The shear stress is zero at the centroidal axis of the shaft and maximum at the outer surface. The maximum shear stress at the outer surface of the shaft may be obtained by the equation known as torsion equation given as

d

= polar

1

T

r

Ip

co

C = modulus of rigidity = angle of twist in radians I = length of the cylindrical body (v) Bending stress: When a body is subjected to a transverse load, it produces tensile as well as compressive stresses, as shown in Fig. The bending equation for beams in simple bending is given by

....!,!.-

I

The stress at the surface of contact between the rivet and the plate is given by (Jb

where

e

(J

M

E

Y

I

R

where

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(J = bending stress in N/m2 M = bending moment in Nm y = distance of the extreme fibre from the neutral axis I = rectangular moment of inertia about the neutral axis in m" E = modulus of elasticity R = radius of curvature of the neutral axis

o

where

or

(Jc

= --

P

d-t-n

d t d- t n

= diameter of the rivet = thickness of the plate =

projected area of the rivet

= no. of rivets per pitch length in bearing or

crushing. The bearing stress is taken into account in case of riveted joints, cotter joints, knuckle joints etc. Bearing Pressure: Bearing pressure is localised compressive stress at the area of contact between two components which have relative motion amongst themselves. It is calculated similarly as we did bearing stress. Let us consider a journal supported in a bearing as shown in Fig. Average bearing pressure is given by Pb=

P

-=-

P

A ld where P is load along the radius of the journal I = length of journal in contact d = diameter of the journal I . d = projected area is contact

R

Neutral axis

Stress Concentration From the equation, we have M

(J

y M=

I (JX-

=rr x z

y where z is known as modulus of section. (vi)

Bearing stress or crushing stress: A localised compressive stress at the surface of contact between two members that are relatively at rest is known as bearing stress or crushing stress. Let us consider a riveted joint as shown in Fig.

Irregularity in the stress distribution caused by the abrupt change in the shape cross section of a machine component is called stress concentration. It occurs for all kinds of stresses in the presence of filters, notches, holes, keyways, splines, surface roughness or scratches etc. Stress concentration factor: A factor used to associate the maximum stress at the discontinuities of cross-section to the nominal stress is called stress concentration factor. - (Jrnax K t--(Jo

where

K, = stress concentration

factor

= maximum stress at the discontinuity (Jo = nominal stress at the same point

(Jrmx

Stress concentration factor is also known as theoretical or form stress concentration factor.

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Theory of Machines and Machine Design In practical the actual effect of stress concentration is lesser than that calculated by theoretical stress concentration factor, so in actual practice we use fatigue stress concentration factor denoted by Kj which is given by

A-53 Low Highcycle ~-c-y-c~le_'~--~~-----__'~ Infinitelife _______..

en~

Sut

amax.actual

where amax,actual = Actual maximum discontinuity

stress at notch or

11 ~

s;

S~t----t-------'lf---------

Kf
Notch sensitivity: Notch sensitivity is calculated by using the relation

q=

Increase in actul stress K, -lover nominal value -= ---------K, -1 Increase in theoretical stress over nominal value

Fatigue and Indurance Limit A type of failure of a material caused by the repeated stresses below the yield point is called fatigue. Failure is caused due to progressive crack formation which is very fine and is of microscopic size. Fatigue is basically affected by number of cyclic loads, relative magnitude of static and fluctuating loads and the size of component. Endurance limit: It is the maximum value of completely reversed bending stress which a polished standard specimen can withstand without failure, for infinite number of cycles (usually 107 cycles). Following are some empirical relations commonly used in practice. Table

10

d

1

rJ

102 103 104 105 106 1

108

Numberof stresscyclesN

S-N diagram Fatigue Failure Criteria

for Fluctuating

Stress

There are different theories to determine the failure points for steel which can be represented in a graph plotted between the mean stress (am) and variable stress (av) as shown in Fig.

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Material

Empirical Relation

10 1 1

Compressive-+---1

1+---

Steel ae = 0.5 au Cast steel ae = 0.4 au Cast iron ae = 0.35 au Non-ferrous metals and alloys ae = 0.3 au where ae = Endurance limit au = Ultimate tensile strength S-N Diagram: A graph between the faituge strength (s) versus stress cycle (N). With the help of this graph we measure the endurance limit. S-N diagram is shown in Fig. Factor of Safety The ratio of material strength to the working or allowable stress is called factor of safety. Factor of safety is given by Maximum strength of the material FS. = ------=-------Design or working stress of the material Yield point strength FS.ductilematerials= For static loading Working or design stress FS.brittlemateria!s =

Ultimate strength For static loading Design or working stress

Endurance limit FS.fatigueloading = --------Design or working stress

cry Tensite -----. Meanstress (crm)

Inference corresponding to each line are shown in the table. Table Method Name

Mathematical

Gerber Method

_1_ = F.S. au

Relation

(am J2 x F.S. + ~

ae

valid for ductile material Goodman Method

Soderberg Method

Elliptic Method

(F.S.X:: J +S.X ::

J

=1

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Theory ofMachines and Machine Design

A-54

where

cr = mean stress =

cr

m

max

+cr· mm 2

cru = ultimate stress cre = endurable limit for reverse loading stress cr = variable stress = v

c

max

-cr· mm 2

o y = yield point stress FS. = Factor of safety

RIVET JOINTS A rivet is made of a short cylindrical bar with a head integral to it. Reveting is common method ofjoining and fastening because of low cost, simple operation and high production rates. Based on the way in which the plates are connected, rivet joints can be classified into two types of joints listed below. 1. Lap joint: If one plate overlapsthe other and the two plates are riveted together, then this type of joint is called. Lap joint, Fig. shows a cross sectional view of a lap joint.

Badboys2 2.

Lap joint with single riveted But joint: In this joint, plates are kept in a way that their edges touch each other and a cover plate is placed either on one side or both sides of the main plates. Finally the cover plate is riveted with the main plates. There are two types of butt joint. (a) Single strap Butt joint: In this case only one cover plate is used above or below the main plates and then final riveting is done. (b) Double strap Butt joint: In this case instead of one cover plate, two cover plates one on upper side and other on lower side of the main plate emloyed and then final riveting is done. Based on the number of rows of rivets, the butt joints are classified as single or double riveted, triple or quadruple riveted. Crosssectional view of the double riveted joint is shown in Fig.

Double riveted lap joint Depending upon the relative position of the rivets of each row riveting is divided as (1) Chain Riveting: In this riveting, the rivets in the various rows are opposite to each other. Cross sectional view of chain riveting is shown in Fig. (2) Zig-Zig Riveting: In this case the rivets in the adjacent rows are staggired in such a way that every rivet is in the middle of the two rivets of the opposite row. Zig-Zig riveting is shown in Fig.

Doublerivetellap joint (zig-zig riveting) Important Terms used in Riveting (i) Gage line: A line passing through the centres of row of rivets which is parallel to the plate edge. (ii) Pitch: It is the distance from the centre of one rivet to the centre of the next rivet measured parallel to the seam. (iii) Back pitch: The perpendicular distance betweenthe centre lines of the successive rows is known as back pitch. (iv) Diagonal pitch: The distance between the centres of the rivets in adjacent rows of zig-zag riveted joints is called diagonal pitch. (v) Marginal pitch: The distance between the centre of rivet hole to the nearest edge of the plate is called marginal pitch. (vi) Caulking: A process in which, the edges of the plates are given blows to facilitate the forcing down of the edge. Blowing the plate with the help of caulking tool forms a metal to metal contact point. (vii) Fullering: A process in which a more satisfactory joint is made by using a tool which has its thickness near the end equal to the thickness of plate. This gives better joint with clean finish. Failures of Riveted Joint Riveted joints may fail in two ways as below: (i) Failure of Plate (ii) Failure of Rivet (i) Failure of Plate: Plates of the joint can fail in two ways listed below: (a) Tearing of plates at an edge: A joint may fail due to tearing of the plate at an edge during riveting or punching. We can avoid this by keeping the margin, m ~ 1.5 d where d is rivet hole diameter in mm. (b) Tearing of the plate across a row of rivets: The main plate or cover plates may tear off across a row of rivets due to tensile stresses in the main plates. The tearing resistance or pull required to tear off the plate per pitch length is given by Pt = (P - d) t crt whereP t= tearing resistance P = pitch of the rivets d = diameter of the rivet hole t = thickness of the plate crt = tensile stress value permissible for the plate material

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Theory of Machines and Machine Design

A-55

Ifapplied load> P, then tearing of the plate across a row of rivets occurs. (ii) Failure of Rivets: Rivets may fail in two ways listed below. If the plate thickness is less than 8 mm, the diameter of rivet is calculated by equating the shearing resistance to crushing. (a) Shearing of the rivets: If the rivets are unable to resist the tensile stress exerted by the plates, then they are sheared off, this is known as shearing of the rivets. In case oflap joint and single cover butt joint, rivets are in single shear, while in case of double cover butt joint rivets are subjected to double shear forces. The shearing resistance or pull required to shear off the rivet, per pitch length is given by PSsingle= n x ~ x d2 X 1"sfor single shear 4 PSdoubleshear= 2 x PSsingle where n = number of rivets per pitch length 1"= safe permissible shear stress for the rivet material d = diameter of the rivet hole (b) Crushing of the rivets: Ifrivets get crushed off under the tensile stress values then it is known as crushing of the rivets. As a result the joint becomes loose. The crushing resistance or pull required to crush the rivet per pitch length is given by Pc = n . d . t . ac where ac = Permissible crushing stress for the rivet or plate material t = Plate thickness n = Number of rivets per pitch length d = diameter of the rivet hole Efficiency of Riveted Joint Efficiency of riveted joint is the ratio of strength of the joint to the strength ofunriveted solid plate.

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where

Ptransverse= s= I= at =

p

.--fI.~f~s_-r---lr p

Double transverse fillet weld And shear strength of a double parallel fillet weld shown in Fig. is given by PpamUel= 1.414s x I x 1" where 1"= Allowable shear stress

p

p

Double parallel fillet weld Butt Joint: In this joint plates are placed edge to edge order and then welded. Plates are bevelled to V-shape or U-shape if thickness of plate is more than 5 mm. The but joints are designed for tension or compression. Design of Butt joint: We take two cases here, single V-butt joint and double V-butt joint and calculate the tensile strength in each cases. (i) Single V-butt joint: Fig. shows single V-butt joint with thickness of the throat t. (b)

) (

(

) )

Minimum of Pc' Pt and Ps P x t x at P = Pitch of the rivets t = Plate thickness at = Maximum permissible tensile strength of plate

WELDED JOINTS A permanent joint obtained by the fusion of the edges of the two parts to be joined together, with or withiout the application of pressure and a filler material. There are two types of welded joints commonly used listed below: (a) Lap joint or Filler joint: In this joint the plates are overlapped and then welded along the edges. The weld filled is train gular. There are various types of lap joints like single transverse, double transverse and parallel fillet joints. The transverse fillet welded joints are designed for tensile strength whereas the parallel fillet welded joints are designed for shear strength. Design of fillet joint: The tensile strength of a double transverse filled weld shown in Fig. is given by

p

P..-.1 -.J s f.-

11=

where,

1.414 s x I x at Leg or size of weld Length of the weld Allowable tensile stress

)

p

(ii)

r

1']

1

>

)

>

?> :>

p

Single V-Butt joint The tensile strength of the single V-Butt joint is given by P t x I x at where throat thickness or thickness of thinner plate at allowable tensile stress for weldment in N/mm2 replace at by ac in case it is designed for compression I = Length of weld Double V-Butt joint: Tensile strength for doub V-butt joint shown in Fig. is given by P t x I x at where t = tl + t2 P = (tl + t2) x I x at => where I = length of weld = width of plates t 1 = throat thickness at the top

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Theory ofMachines and Machine Design

A-56

t2 = throat thickness at the bottom crt = allowable tensile stress for weldment inN/mm2

pi -

j~ -

t

---Inn -t -:t p

(ii) Rectangular sunk key: A rectangular sunk key is shown in Fig. The width of the key is equal to ~ and 4 thickness is equal to ~ . ,._~ 12

d

w=T

t=_!'

12 1 = 1.5 d

Double V-Butt joint Metric Thread There are various forms of screw threads, metric thread is an Indian Standard (I.S.0) thread having an included angle of 60°, these are two types, coarse threads and fine threads. For a particular value of diameter, coarse threads have large pitch and lead as compared to fine threads. Coarse threads are more in strength and chances of thread shearing and crushing is very less. They are preferred for vibration free applications as they offer less resistance to unscrewing. Fine threads give better adjustment in fitment and are used where high vibrations take place as they offer high resistance to unscrewing. Fine threads are designated as Md x P for example M50 x 5 which indicates an isometric fine thread which has nominal diameter of 50 mm and pitch 5. While in case of coarse threads only Md is mentioned for example M50. Todesignate tolerance grade we use the values of each tolerances like 7 for fine grade, 8 for normal and 9 for coarse grade. For example a bolt thread of 6 mm size of coarse pitch and with allowance on threads and normal tolerance grade is designated as M6-8d.

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KEYS Toprevent the relative motion of the shaft and the machinery part connected to it we use a piece of mild steel called key. Keys are temporary fastenings and are subjectedto considerable crushing and shearing stresses. Different types of keys are listed below. (a) Sunk keys: These keys are designed in such a way that they are half way in the key way of the hub of pulley and half in the key way of the shaft. There are basically five types of sunk keys listed as following: (i) Square sunk key: A square sunkkeyis shownin Fig. If d is the diameter of the shaft width of the square sunk key is equal to d/4 and the thickness is same as width. -f--t

1+-'fJ""+I I

_t.

----

J_~~ _L~2 Shaft cross-section d

w=t=T

length of the key ( = 1.5 d

I

Square sunk key

Rectangular sunk key (iii) Gib-head key: Cross-sectional view of a Gib-head sunk key is shown in Fig.

w=_Q_

4

t=2w=_Q_ 3 6 I =1.5d where d = diameter of shaft

Gib head sunk key (iv) Parallel sunk key: It is a taperless key and may be rectangular or square in cross-section. It is used where the pulley, gear or other mating piece is required to slide along the shaft. (v) Feather key: A special type of parallel key which transmits a turning moment and also permits axial movement. (b) Tangent keys: These keys are fitted in pair at right angles, each key is to withstand torsion in one direction only. Tangent keys are used for heavy duty applications. A crosssectional view of a tangent key is shown in Fig.

Tangent key (c) Saddle keys: These are taper keys fitted in key way and designed such that it is flat on the shaft. (d) Wood ruff keys: This key is made of a piece from a cylindrical disc of segmental cross-section. (e) Round keys: These keys are circular in cross-section and are fitted partly into the shaft and partly into the hub. (f) Splines: When splines are integrated with the shaft which finally fits into the keyways of the hub. These are stronger than a single keyway. Design of Keys A key may fail due to shearing and crushing, it is equally strong in shearing and crushing if following condition satisfies.

Badboys2

Theory of Machines and Machine Design

(iii) Combined loading: When a shaft is subjected to combined

w=~ where

w= t= O"c = r=

A-57

2~ width of the key thickness of the key permissible crushing stress permissible shearing stress

twisting moment and bending moment, then the shaft is designed on the basis of maximum normal stress theory and maximum shear stress theroy and larger size is adopted. According to maximum shear stress theory (Guest's theory) the maximum value of shear stress in the shaft is given by 1 I 2 2 \J (o.,) + 4~ 2

SHAFTS

~max = -

Shafts are used to transmit power from one place to another, these are normally of circular cross-section. Mild steels are hot rolled and then finished to actual size by turning, grinding or cold drawing to manufacture shafts. Alloy steels with composition of nickel, chromium and vanadium is also used to impart high strength. The cold rolled shafts are stronger than hot rolled shafts, but with higher residual stresses.

=

r

max

~

Types of Shafts There can be two types of shafts (a) Transmission shaft such as counter shafts, line shafts, over head shafts, etc. (b) Machine shaft such as crank shaft Design of Shaft Shafts are designed on the basis of (a) Strength: On the basis of strength of the shaft material we design a shaft considering three types of stresses induced in the shafts. (i) Torsionalload (ii) Bending load (iii) Combined torsional and being loads (i) Torsional load: If the shaft is subjected to pure torsional load then torsional shear stress is given by

16Txdo n do - dj

( 4

4) ,

N/m2

=>

n T -- -x~xdo 16

where

do = outer diameter of shaft in m

3

d, = inner diameter of the shaft in m (ii) Bending load: When the shaft is subjected to a bending moment only, then the value of stress induced is given by O"b =

32~ for solid shaft nd where O"b = bending stress and for a hollow shaft 32M

J2

'M2+T2

\j

nd3 M2+T2 = __ ~ 16 max

r, =

nd3 16~max

where T, =equivalent twisting moment = ~ M2 + T2

Now, According to maximum normal stress theory (Rankine's theory the maximum normal stress in the shaft is given by O"b(max) =

-1 O"b + -1 ~ (O"b) 2 + 4~ 2 2 2

,,~_)~ ::3 [~(M M +

_!_ (M + ~M2 + T2 ) = nd3

2

r = 16; N/m2 for solid shaft

~=

= ~ nd3

=>

Badboys2

nd where d = shaft diameter in m T = torsional moment in N-m For a hollow shaft

32M + 4 x (16 T nd3 nd3

_!_ 2

32

Me=

where

+

T2)]

O"b(max)

nd3 --xO"b(max)

32 Me = equivalent bending moment = ~ (M + ~M2 + T2 )

(b) Design of Shafts on the basis of rigidity and stiffness A shaft of small diameter and long length the maximum deflection is expressed as (5max S; 0.75 mm/length in meters also (5max S; 0.06 Lin mm where L = distance between load and bearings in m. These deflections are minimised by using support bearings. If gear is mounted on the shaft then 3 (5max S;

f

where f = gear face with mm If shaft crosses these limits then deflections are minimized by using self aligning bearings. SPUR GEARS When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, its arrangement is called spur gearing, and gear used is spur gear. While designing spur gear it is assumed that gear teeth should have sufficient

Badboys2

Theory of Machines and Machine Design

A-58 strength so that they do not fail under static as well as dynamic loading.

Lewis Equation Lewis equation is used to determine the beam strength of a gear tooth. Each tooth is considered as a cantilever beam which is fixed at the base. The normal force acting on the tip of the gear is resolved into radial and tangential component as shown in Fig. The radial component induces a direct compress stress of small value, so it is ignored. Tangential component FT is duces a bending stress that can break the tooth.

-__ r Tangent to the

=

0.124 _ 0.684 T .

Y for 20° full depth mvolute system

T

0.841 Y for 20° stub system = 0.175 - -T The permissible working stress (O"w)in the Lewi's equation depends upon the material for which, allowable static stress (0"0) may be determined. Allowable static stress is the stress at the elastic limit of the material also known as basic stress. Barth Formula: According to Barth formula, the permissible working stress is given by O"w= 0"0 X C, Cv = velocity factor where

Cv = --- 4.5

- - - base circle

0.912

= 0.154 - --

4.5+v

. at for care full y cut gears operatmg velocities upto 12.5 mls

-- 3

.

.

Maximum value ofbendmg stress

My

= O"w= -

... (1)

I where M is maximum bending moment (i.e. at BC) M= Ftxh

Badboys2

M

from (1)

=

h

O"wI Y

. Now for y for beam of height t

c. lor

. at or dimary cut gears operatmg

velocities upto 12.5 mls

Static Tooth Load ... (2)

M

Ft= -

3+v

Beam strength or static tooth load is given by Fs = O"eb P, Y = O"eb 1t my where O"e= Flexural endurance limit For safety against breakage Fs > FD where FD is the dynamic tooth load which takes place due to inaccurate tooth spacing, irregularities in profiles and tooth deflection under the effect of load.

BEARINGS

t

= -

A machine element which permits a relative motion between the contact surfaces of the members while carrying the load. It supports journal. The bearings are mainly classified as (a) Sliding contact bearings (b) Rolling contact bearings

2

Sliding Contact Bearings

= O"wX bt3 = O"wbt2

F

12 x ~ 6h 2 Now if circular pitch is P, then we can represent t, and h in t

terms of Peas

O"wbK~ Pc

6K2

= O"wbPc K~ 6K2

K2

Let

y=_l_

where

F t = o"wb Pc Y Lewis Equation y = form factor called Lewis form factor b = width of gear face

6K2

Y for 14!.: composite and full depth involute system

2

In these bearings, the sliding takes place along the surfaces of contact between the moving element and the fixed element. These are also known as plain bearings. According to the thickness oflayer ofthe lubricant between the bearing and the journal, sliding contact bearings can be classified as (a) Thick film bearings: Bearings in which the working surfaces are completely separated from each other by the lubricant. These are also called a hydrodynamic lubricated bearings. (b) Thin film bearings: In these bearings although lubricant is present, the working surfaces partially contact each other atleast part of the time. Such type of bearings are also called boundary lubricated bearings. (c) Zero film bearings: Bearings which operate without any lubricant are known as zero film bearings. (d) Hydrostatic bearings: Bearings which can support steady loads without any relative motion between the journal and the bearings because there is externally pressurized lubricant between the members.

Badboys2

Theory of Machines and Machine Design

A-59

Hydrodynamic Journal Bearing Terminology Cross-sectionalview of a hydrodynamicjournal bearing is shown in Fig.

where K is a factor for end leakages for

0.75 <

d (ix) Short and long bearings: Short and long bearings are decided on the basis of the ratio lid. [

If

- < 1 then bearing is said to be short d [ -

Journal I I I I

d

C2=R-r=--=2

2

clearance ratio: Ratio between diametral clearnace to journal diameter .

.

Diametral clearance ratio

C1

= -

d (iv) Eccentricity: It is the radial distance between the centre (0) of the bearing and the displaced centre (0') of the bearing under load. Eccentricity is denoted bye. (v) Eccentricity ratio (Attitude): Ratio of eccentricity to radial clearance is called eccentricity ratio. e E=

-

C2 (vi) Sommerfield number: A dimensionless number used in design of bearings. It's value is given by Sommerfield number

=

1 bearing is called square bearing

[

I

Hydrodynamic journal bearing Diameter of the bearing = D = 2R Diameter of the journal = d = 2r Length of the bearing = l Terminologies associated with a hydrodynamic journal bearing are defined as following. (i) Diametral clearance: Difference between the diameter of bearing and journal is called diametral clearance C1 = D-d (ii) Radial clearance: It is the difference between the radii of bearing and journal D-d C1

Badboys2 (iii) Diametral

!:_ < 2.8, K = 0.002

=

(z;)(~J

- > 1 then bearing is said to be long d (x) Heat generation and rejection in bearing: Due to fluid friction and solid friction heat is generated in the bearing which can be expressed as Qgen= J,.tWV N-m/s where W = load on the bearing V = rubbing velocity in mls Heat rejection is given by Qrejection = Kh A (t, - ta) JIS where Kh = heat dissipation coefficient in W/m2/C A = prejected area of the bearing tb = bearing surface temperature ta = ambient temerature In case of pressure fed bearings ift, is the inlet temperature of oil and to is outlet temperature of the oil then heat rejection is given by Qrejection = pCoil(to- ti) where p = density of oil Coil= specific heat of oil Bearing Characteristic Number The factor ZN is known as bearing characteristic number and P it is a dimensionless number. where Z = Absolute viscosity of the lubricant in kg/m-s N = Speed ofjournal in r.p.m. P = Bearing pressure on the projected bearing area in Nzmm-'

W

P

where N = Journal speed in r.p.m., Z = lubricant viscosity, P = bearing pressure normally we take its value as 14.3 x 106 (vii)Critical pressure in journal bearing: The pressure at which the oil film breaks and metal to metal contact takes place is known as critical pressure. It's value is given by

= -,

.

W = Load on the Journal

[·d The variation of coefficient offriction with respect to the bearing characteristic number is shown in Fig.

t

r-:------,_ Thin film or boundary lubrication 1

(unstable)

3 c

c;J2 ( [+[ d J NI mm

PZN ( d - 4.75 x 106 where

2

N = Journal speed in r.p.m. Z = Absolute viscosity of the lubricant (viii)Coefficient of friction: Coefficient of friction can be expressed as

o ""B

E (5

c

Partial lubrication

"~-+---~--

~

o K ZN P

Badboys2

Theory of Machines and Machine Design

A-60

Variation of coefficientof friction with the bearing characteristic number (

z: J

Rolling Contact Bearings Bearing which operate on the basis of principle of rolling, i.e. the contact between the bearing surfaces is rolling are known as rolling contact bearings. These are also called anti friction bearings as they offer low friction. Mainly there are two types of rolling contact bearings. (i) Ball bearing (ii) Roller bearing Average life (Median life) of a bearing: It is the number of revolutions or number of hours at a constant speed that 50% of a batch of ball bearing will complete or may be exceed and 50% fail before the rated life is achieved. It is denoted by L5o. Life a

1 (Load)? Dynamic load rating: Value of radial load which bearing can suffer for I million revolutions of inner ring with only 10% failure is known as dynamic load rating or basic dynamic capacity or specific dynamic capacity. Rating Life L ~ (~

Badboys2 where

Disc clutch Frictional torque acting on an element dr is given by T, = 2n /-! pr2 dr where p = axial pressure intensity /-! = coefficient of friction For uniform pressure the intensity of pressure is given by P=

P = load C = dynamic basic load rating

J

1/3

IfN is r.p.m. the Life in hours is given by CJ3 106 L= ( x-- hours P 60N P = Cx

dr

J'

P= C (-I L

or

surfaces in contact. Due to friction heat is generated which should be dissipated rapidly. Friction clutches are further classified into (a) Disc or plate clutch (b) Cone clutch (c) Centrifugal clutch (a) Disc clutch: Cross-sectional view of a disc clutch is shown in Fig.

6 ]1/3

W

n(r?-d)

where

rl = external radius of the surface r2 = internal radius of the surface W = axial value of thrust which holds the frictional surfaces together. Total torque transmitted is given by

[rfrf-d- d ]

T = ~ /-!W cosec a 3 (b) Cone clutch: Total torque transmitted in the cone clutch is given by

_!.Q_

[ 60NL

CLUTCHES Clutch is a connection between the driving and driven shafts with the provision to disconnect the driven shaft instantaneously without stopping the driving shaft. Main functions of cluthces are to stop and start the driven member without stopping the driving member, to maintain torque, power and speed, and to eradicate the effects of shocks while transmitting power. Clutches are classified into two types: (1) Positive clutches: These are used where there IS requirement ofpositive drive for examplejaw or claw clutch. (2) Friction clutches: Friction clutch transmits the power by friction without shock. It is used where sudden and complete disconnection of two rotating shafts are necessary, and the shafts are in axial alignment. The power transmission takes place due to two or more concentric rotating frictional

Tcme= ~ /-!W [r? - r~] cosec a 3 r2 - r2 1 2 where

a = semi-angle of frictional surfaces with the

clutch axis. (c) Centrifugal clutch: Total torque transmitted in case of centrifugal clutch is given by T = /-! (C - S) r, x n where C = Spring force acting on shoe = mrwm = mass of shoe r = distance of centre of gravity of shoe from centre w = angular velocity of rotating pulley in rad/s ri = inside radius of pulley rim S = Inward force due to spring-m (W12)r 3 Wl= -w 4 n = number of shoes C - S = mr w- -

_2._ mrw- = }_ mr w16

16

Badboys2

Theory of Machines and Machine Design

I···.. 1.

EXERCISE

A rotating disc of 1 m diameter has two eccentric masses of 0.5 kg each at radii of 50 mm and 60 mm at angular positions of 00 and 1500, respectively. A balancing mass of 0.1 kg is to be used to balance the rotor. What is the radial position of the balancing mass?

2.

(a) 50 mm (b) 120 mm (c) 150mm (d) 280mm The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is (a) 1 (b) 2

3.

Match the items in Column I and Column II.

(c)

3

...,

A-61

(c) (d)

Geneva mechanism is an intermittent motion device Grubler's criterion assumes mobility of a planar mechanism to be one 10. Mobility of a statically indeterminate structure is (a) ::;;-1 (b) zero

(c) 1 11. A double-parallelogram

(d) ?:2

mechanism is shown in the figure. Note that PQ is a single link. The mobility ofthe mechanism is P Q

(d) 4

Column I

ColumnII

Higher kinematic pair 1. crubler's equation Q. Lower kinematic pair 2. Line contact R Quick return mechanism 3. Euler's equation S. Mobility of a linkage 4. Planar 5. Shaper 6. Surface contact (a) P-2, Q-6,R-4, S-3 (b) P-6, Q-2, R-4, S-l (c) P-6, Q-2, R-5, S-3 (d) P-2, Q-6, R-5, S-l Match the items in Column I and Column II. P.

12.

Badboys2 4.

Column I P.

Q.

5.

R S. (a) (c) The

(a) (c) 6.

7.

8.

9.

Addendum Instantaneous centre of velocity Section modulus Prime circle P-4, Q-2, R-3, S-l P-3, Q-2, R-1, S-4 number of inversions for 6

4

ColumnII 1. Cam 2.

Beam

3.

Linkage

Gear (b) P-4, Q-3, R-2, S-l

4.

13.

(d) P-3, Q-4, R-1, S-2 a slider crank mechanism is (b) 5

(d) 3

For a four-bar linkage in toggle-position, the value of mechanical advantage is (a) zero (b) 0.5 (c) 1.0 (d) infinite The speed of an engine varies from 210 rad/s to 190 rad/s. During a cycle, the change in kinetic energy is found to be 400 N-m. The inertia ofthe flywheel in kg-m2 is (a) 0.10 (b) 0.20 (c) 0.30 (d) 0.40 The rotor shaft of a large electric motor supported between short bearings at both deflection of 1.8 mm in the middle of the rotor. Assuming the rotor to be perfectly balanced and supported at knife edges at both the ends, the likely critical speed (in rpm) of the shaft is (a) 350 (b) 705 (c) 2810 (d) 4430 Which of the following statements is incorrect? (a) Gashoffs rule states that for a planar crank-rocker four bar mechanism, the sum of the shortest and longest link lengths cannot be less than the sum of remaining two link lengths (b) Inversions of a mechanism are created by fixing different links one at a time

14.

15.

16.

(a) -1 (b) zero (c) 1 (d) 2 A circular object of radius r rolls without slipping on a horizontal level floor with the centre having velocity V. The velocity at the point of contact between the object and the floor is (a) zero (b) Vin the direction of motion (c) Vopposite to the direction of motion (d) Vverticallyupward from the floor For the given statements: I. Mating spur gear teeth is an example of higher pair. Il. A revolute joint is an example oflower pair. Indicate the correct answer. (a) Both I and IIare false (b) I is true and II is false (c) I is false and II is true (d) Both I and II are true In a mechanism, the fixed instantaneous centres are those centres which (a) Remain in the same place for all configuration of mechanism (b) Large with configuration of mechanism (c) Moves as the mechanism moves, but joints are of permanent nature (d) None of the above Maximum fluctuation of energy is the (a) Ratio of maximum and minimum energies (b) sum of maximum and minimum energies (c) Difference of maximum and minimum energies (d) Difference of maximum and minimum energies from mean value In full depth 114 degree involute system, the smallest number of teeth in a pinion which meshes with rack without interference is (a) 12 (b) 16 (c) 25 (d) 32

Badboys2

Theory of Machines and Machine Design

A-62

17. The two-link system, shown in the figure, is constrained

26.

to move with planer motion. It possesses y

) o

18.

19.

20.

22.

23.

24.

25.

x

(a) 2 degrees of freedom (b) 3 degrees of freedom (c) 4 degrees of freedom (d) 6 degrees of freedom Ifthe ratio of the length of connecting rod to the crank radius increases, then (a) primary unbalanced forces will increase (b) primary unbalanced forces will decrease (c) secondary unbalanced forces will increase (d) secondary unbalanced forces will decrease In a cam mechanism with reciprocating roller follower, the follower has a constant acceleration in the case of (a) cycloidal motion (b) simple harmonic motion (c) parabolic motion (d) 3 - 4 - 5 polynomial motion A flywheel fitted in a steam engine has a mass of 800 kg. Its radius of gyration is 360 mm. The starting torque of engine is 580 N-m. Find the kinetic energy of flywheel after 12 seconds? (a) 233.3 kJ (b) 349.8kJ (c) 487.5 kJ (d) None of these In a slider-crank mechanism, the maximum acceleration of slider is obtained when the crank is (a) at the inner dead centre position (b) at the outer dead centre position (c) exactly midway position between the two dead centres (d) none of these Ifthe rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half the mean radius ofthe former, then energy stored in the later at the same speed will be (a) four times the first one (b) same as the first one (c) one fourth of the first one (d) one and a halftimes the first one What will be the number of pair of teeth in contact ifarc of contact is 31.4 mm and module is equal to 5. (a) 3 pairs (b) 4 pairs (c) 2 pairs (d) 5 pairs The distance between the parallel shaft is 18 mm and they are conntected by an Oldham's couling. The driving shaft revalues at 160 rpm. What will be the maximum speed of sliding the tongue ofthe intermediate piece along its grow? (a) 0.302 m/s (b) 0.604 m/s (c) 0.906m/s (d) None of these Two spur gears have a velocity ratio of 113. The driven gear has 72 teeth of 8 mm module and rotates at 300 rpm. The pitch line velocity will be (a) 3.08m/s (b) 6.12 mls (c) 9.04 mls (d) 12.13 mls

Badboys2 21.

27.

Instantaneous centre of a body rolling with sliding on a stationary curved surface lies (a) at the point of contact (b) on the common normal at the point of contact (c) at the centre of curvature of the stationary surface (d) Both (b) and (c) If Cf is the coefficient of speed fluctuation of a flywheel then the ratio of O)~O)min will be (a)

1-2Cf 1+2Cf

(b)

2-Cf

(c)

l+Cf 1-Cf

(d)

2+Cf 2-Cf

28. A rotor supported at A and B, carries two masses as shown in the given figure. The rotor is

29.

(a) dynamically balanced (b) statically balanced (c) statically and dynamically balanced (d) not balanced A body of mass m and radius of gyration k is to be replaced by two masses m, and m2 located at distances h, and h2 from the CG of the original body. An equivalent dynamic system will result, if (a)

30.

h}+h2=k

(b)

hT+h;=k2

A cord is wrapped around a cylinder of radius 'r' and mass 'm' as shown in the given figure. If the cylinder is releasd from rest, velocity of the cylinder, after it has moved through a distance 'h' will be

31.

(a)

.J2 gh

(c)

l3

(b)

Jib {gh3h \/3

(d) gh There are six gears A, B, C, D, E, F, in a compound train. The number ofteeths in the gears are 20, 60, 30, 80,25 and 75 respectively. The ratio of the angular speeds of the driven (F) to the driver (A) ofthe drive is (a)

(c)

1

1 24 4 15

(b)

8

(d) 12

Badboys2 0

""'"

("I') ("I')

ill

Theory of Machines and Machine Design 32.

33.

34.

35.

In the four-bar mechanism shown in the given figure, links 2 and 4 have equal lengths. The point P on the coupler 3 will generate a/an (a) ellipse 2 3 4 (b) parabola (c) approximately straight line (d) circle p A system of masses rotating in different parallel planes is in dynamic balance if the resultant (a) force is equal to zero (b) couple is equal to zero (c) force and the resultant couple are both equal to zero (d) force is numerically equal to the resultant couple, but neither of them need necessarily be zero. A bicycle remains stable in running through a bend because of (a) Gyroscopic action (b) Corioliss' acceleration (c) Centrifugal action (d) Radius of curved path The maximum fluctuation of energy E[, during a cycle for a flywheel is (a) l( (1)2max - (1)2min) (b) 1/2 1(1)av ((1)2 max - (1)2min)

1

(C)

(a)

40.

41.

42.

43.

2

(d) (where I= Mass moment of inertia of the flywheel (1)av = Average rotational speed K; = coefficient of fluctuation of speed) The road roller shown in the given figure is being moved over an obstacle by a pull 'P'. The value of'P' required will be the minimum when it is

44.

45.

(a)

horizontal

(b) vertical

37.

(c) at45° to the horizontal (d) perpendicular to the line CO Two gear 20 and 40 teeth respectively are in mesh. Pressure

46.

angle is 20°, module is 12 and line of contact on each side of the pitch point is half the maximum length. What will be the height of addendum for the gear wheel (a) 4mm (b) 6mm (c) 38.

8mm

(d)

lOmm

In a slider-bar mechanism, when does the connecting rod have zero angular velocity? (a)

39.

When crank angle

= 0°

(b) When crank angle

= 90°

(c) When crank angle = 45° (d) Never A disc of mass m is attached to a spring of stiffuess k as shown in the figure. The disc rolls without slipping on a horizontal surface. The natural frequency of vibration of the system is

I~

2n: m

(b) _I

2n:

fJ m

(d) I~ I~ 2n: 3m 2n: 2m For a four bar linkage in toggle position, the value of mechanical advantage is (a) 0.0 (b) 0.5 (c) 1.0 (d) 00 What will the normal circular pitch and axial pitch of helical gear if circular pitch is 15 mm and helix angle is 30° (a) 13mmand39mm (b) 26mmand39mm (c) 26mmand 13mm (d) 13mand26mm The speed of an engine varies from 210 rad/s to rad/s. During cycle the change in kinetic energy is found to be 400 Nm. The inertia ofthe flywheel in kgnr' is (a) 0.10 (b) 0.20 (c) 0.30 (d) 0.40 If first and last gear having teeth 30 and 50 respectively of a simple gear train, what will be the train value and speed ratio gear respectively if first gear is driving gear (a) 3/5 and 5/3 (b) 3/5 and 4/5 (c) 5/3 and 3/5 (d) 4/5 and 3/5 The centre of gravity ofthe coupler link in a 4-bar mechanism would experience (a) no acceleration (b) only linear acceleration (c) only angular acceleration (d) both linear and angular accelerations In a four-bar linkage, S denotes the shortest link length, L is the longest link length, P and Q are the lengths of other two links. At least one of the three moving links will rotate by 360° if (c)

lIKes (1) av lKes(1)2av

Badboys2 36.

A-63

47.

W

S+LSP+Q

~

S+L>P+Q

(c)

S+PSL+Q

(d)

S + P > L+ Q

An involute pinion and gear are in mesh. Ifboth have the same size of addendum, then there will be an interference between the (a) tip of the gear teeth and flank of pinion (b) tip of the pinion and flank of gear (c) flanks of both gear and pinion (d) tip of both gear and pinion. ABCD is a four-bar mechanism in which AB = 30 em and CD = 45 em. AB and CD are both perpendicular to fixed link AD, as shown in the figure. Ifvelocity ofB at this condition is V, then velocity of C is

c

o, C)

I

Badboys2

Theory of Machines and Machine Design

A-64

(a) y

(b) Iy

(c) 2_y

(d)

2

48.

49.

3.y

3 The transmission angle is maximum when the crank angle with the fixed link is (a) ff (b) 90° (c) 180° (d) 270° In the given figure, ABCD is a four-bar mechanism. At the instant shown,AB and CD are vertical and BC is horizontaL AB is shorter than CD by 30 cm, AB is rotating at 5 radls and CD is rotating at 2 rad/s. The length of AB is 4

B

c

A

57. The tangential force transmitted (in newton) is (a) 3552 (b) 2611 (c) 1776 (d) 1305 58. Tooth interference in an external involute spur gear pair can be reduced by (a) decreasing centre distance between gear pair (b) decreasing module (c) decreasing pressure angle (d) increasing number of gear teeth 59. Two identical ball bearings P and Q are operating at loads 30 kN and 45 kN respectively. The ratio of the life of bearing P to the life of bearing Q is (a) 8 1116 (b) 27/8 (c) 9/4 (d) 3/2 60. Match the following criteria of material failure, under biaxial stress a 1 and a2 and yield stress ay, with their corresponding graphic representations. List I List II o 2 P.

o

50.

(a) 10cm (b) 20cm (c) 30cm (d) 50cm A link OP is 0.5 m long and rotate about point O. It has a slider at permit B. Centripetal acceleration ofP relative to 0 is 8 m/sec'. The sliding velocityof slider relative to P is 2 mI sec. The magnitude of Coriolis component of acceleration

Maximum normalstress criterion

1.

cry

Badboys2(a) IS

51.

52.

53.

54.

55.

lti m/sec' (b) 8 m/sec/ (c) 32 m/sec' (d) Data insufficient Which one of the following is a criterion in the design of hydrodynamic journal bearings? (a) Sommerfield number (b) Rating life (c) Specific dynamic capacity (d) Rotation factor A cylindrical shaft is subjected to an alternating stress of 100 MPa. Fatigue strength to sustain 1000 cycle is 490 MPa. If the corrected endurance strength is 70 MPa, estimated shaft life will be (a) 1071 cycle (b) 15000 cycle (c) 281914 cycle (d) 928643 cycle 20° full-depth involute profiled 19-tooth pinion and 37tooth gear are in mesh. Ifthe module is 5 mm, the centre distance between the gear pair will be (a) 140 mm (b) 150 mm (c) 280 mm (d) 300 mm The resultant force on the contacting gear tooth in newton is (a) 77.23 (b) 212.20 (c) 225.81 (d) 289.43 A ball bearing operating at a load F has 8000 h oflife. The life of the bearing, in hour, when the load is doubled to 2F IS

(a) 8000 (b) 6000 (c) 4000 (d) 1000 56. Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5, the minimum allowable stress (in MPa) for the gear material is (a) 242.0 (b) 166.5 (c) 121.0 (d) 74.0

Q.

Maximum-distortionenergy criterion

2.

cr ".

cry

Y

- cry cr2

R.

Maximum-shear stress criterion

3.

cr1

- cry - cry

61.

62.

63.

64.

(a) P-2, Q-1, R-3 (b) P-3, Q-2, R-1 (c) P-2, Q-3, R-1 (d) P-3, Q-1, R-2 A solid circular shaft needs to be designed to transmit a torque of 50 N-m. If the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is (a) 8 (b) 16 (c) 24 (d) 32 Stress concentration in cyclic loading is more serious in (a) ductile materials (b) brittle materials (c) equally serious in both cases (d) depends on other factors Feather keys are generally (a) tight in shaft and loose in hub (b) loose in shaft and tight in hub (c) tight in both shaft and hub (d) loose in both shaft and hub For a parallel load on a fillet weld of equal legs, the plane of maximum shear occurs at (a) 22.5° (b) 30° (c) 45° (d) 60°

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Theory of Machines and Machine Design 65. The silver bearings are used almost exclusively in aircraft engines due to their excellent (a) fatigue strength (b) wear resistance (c) corrosive resistance (d) None of these 66. When a shaft rotates in anti-clockwise direction at slow speed in a bearings, then it will (a) have contact at the lowest point of bearing (b) move towards right of the bearing making metal to metal contact (c) move towards left of the bearing making metal to metal contact (d) move towards right of the bearing making no metal to metal contact 67. The most efficient riveted joint possible is one which would be as strong in tension, shear and bearing as the original plates to be joined but this can never be achieved because (a) rivets can not made with same material (b) rivets are weak in compression (c) there should be atleast one hole in the plate reducing its strength (d) clearance is present between the plate and the rivet 68. To resist breaking of the plate in front of the rivet, we make the distance from the centre of the rivet to the edge of the plate at least (a) 1.5 d (b) 2.5 d (c) 2d (d) 3d 69. The uniform pressure theory as compared to the uniform wear theory gives (a) higher frictional torque (b) lower frictional torque (c) either lower or high frictional torque (d) None of these 70. The limiting wear load of spur gear is proportional to (where Ep = Young's modules of pinion material, Eg = Young's modulus of gear material.

A-65

75.

76.

77.

78.

79.

(a)

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(a)

(c)

80.

81.

(c) 71. American standard thread have the angle equal to (a) 55° (b) 60° (c) 29° (d) 58° 72. For overhauling which of the following condition is satisfied? (a) <j):?: a (b) <j):::; a (c) Both (a) and (b) (d) None of the above 73. A radial ball bearing has a basic load rating of 50 kN. Ifthe desired rating life of the bearing is 6000 hours, what equivalent radial load can be bearing carry at 500 rev/min. (a) 18.85 kN (b) 8.85 kN (c) 12.5 kN (d) 14.5 kN 74. The frictional torque transmitted in a flat pivot bearing assuming uniform wear

(a)

IlWR

(b)

3 "4IlWR

(c)

2 -IlWR

(d)

"2IlWR

3

1

where Il = co-efficient of friction W = load over bearing R = radius of bearing The frictional torque for square thread at mean radius while raising load is given by (W = load, ~ = mean radius, <j) = angle of friction, a = helix angle) (a) W~ tan (<j)- a) (b) ~ tan (<j)+ a) (c) WRo tan a (d) WRo tan <j) Which one of the following types of bearings is employed is shafts of gear boxes of automobiles (a) Hydrodynamic journal bearing (b) Multi lobed journal bearing (c) Anti friction bearings (d) Hybrid journal bearings In case of self locking brake the value of actuating force is (a) Positive (b) Negative (c) Zero (d) None of these I.S. specifies which ofthe following total number of grades of tolerances? (a) 18 (b) 16 (c) 20 (d) 22 The theoretical stress concentration factor at the edge of hole is given by

82.

83.

l+(~) 1+3(~)

(b) (d)

1+2m 1+4(~)

Where a = halfwidth (or semi axis) ofellipse perpendicular to the direction of load b = half width (or semi axis) of ellipse in the direction of load In the assembly of pulley, key and shaft (a) pulley is made the weakest (b) key is made the weakest (c) key is made the strongest (d) all the three are designed for equal strength The longitudinal joint in a boiler shell is usually (a) Butt joint (b) Lap joint (c) Butt joint with two cover plates (d) Butt joint with single cover plate To restore stable operating condition in a hydrodynamic journal bearing when it encounters higher magnitude loads (a) Oil viscosity is to be increase (b) Oil viscosity is to be decrease (c) Oil viscosity index is to be increases (d) Oil viscosity index is to be decreases Which of the following graph is correctly represent? 0.12 r-------------,

0.08 Error (e) (in mm) 0.04

o (a) (c)

4 8 12 16 20 24 Module of spur gears (mm)

AandB AandC

(b) Band C (d) A,BandC

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A-66

84. For self-locking which of the following condition is satisfied? (a) ~ a (b) <1>::;; a (c) Both (a) and (b) (d) None of these 85. Which of the following bearing is suitable for fluctuating demands? (a) Needle roller bearing (b) Ball bearing (c) Tapered bearing (d) Cylindrical bearing 86. The S-N curve is a graphical representation of (a) Stress amplitude (SF) versus the number cycle (N) after the fatigue failure on Log-Log graph paper (b) Stress amplitude (SF) versus the number cycle (N) before the fatigue failure on log-log graph paper (c) Number of cycle (N) versus stress amplitude (SF) after the fatigue failure on log-log graph paper (d) Number of cycle (N) versus stress amplitude before the fatigue failure on log-log graph paper 87. Find the diameter of a solid steel shaft to transmit 20 kW at 200 rpm. The ultimate shear stress for the steel may be taken as 360 MPa and factor of safety as 8 (a) 48 mm (b) 68 mm (c) 78 mm (d) 38 mm 88. The efficiency of overhauling screw is (a) ~ 50% (b) ::;;50% (c) equal to 50% (d) none of these 89. Backlash in spur gear is the (a) difference between the dedendum of one gear and the addendum ofthe mating gear (b) difference between the tooth space of the gear and the tooth thickness of the mating gear measured on the pitch circle (c) intentional extension of centre distance between two gears (d) does not exist 90. The ratio of friction radius based upon uniform pressure and uniform wear theory is (Given: Ro = 100 mm and~=25mm) 7 14 21 28 (a) 25 (b) 25 (c) 25 (d) 25 91. A certain minimum number of teeth is to be kept for gear wheel (a) So that gear is of good size (b) For better durability (c) To and interference and under cutting (d) For better strength 92. Which of the following is a positive locking device? (a) Castled nut (b) Lockingbypin (c) Locking by threaded pin (d) Split nut 93. Fatigue strength of a rod subjected to cyclic axial force is less than that of a rotating beam of same dimension subjected to steady lateral force. What is reason behind this? (a) Axial stitfuess is less than bending stitfuess (b) Absence of centrifugal effects in the rod (c) The number of dis-continuities vulnerable to fatigue is more in the rod (d) At a particular time, the rod has only one type of stress whereas the beam has both tensile and compressive stress.

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Theory of Machines and Machine Design 94. In a belt-drive if the pulley diameter is doubled keeping the tension and belt width constant, then it will be necessary to (a) increase the key length (b) increase the key depth (c) increase the key width (d) decrease the key length 95. Deep groove ball bearings are used for (a) heavy thrust load only (b) small angular displacement of shafts (c) radial load at high speed (d) combined thrust and radial loads at high speed 96. Which of the following key is under compression rather than in being shear when under load? (a) Saddle (b) Barth (d) Feather (d) Kennedy 97. Which of the following is maximum capacity bearing? (a) Filling notch bearing (b) Single row bearing (c) Angular contact bearing (d) Self-aligning bearing 98. Reduction of stress concentration is achieved by (a) Additional notches and holes in tension member (b) Drilling additional holes for shafts (c) Undercutting and Notch for member in bending (d) All of above 99. A full journal bearing with a journal of 75 mm diameter and bearing oflength 75 mm is subjected to a load of2500 N at 400 rpm. The lubricant has a viscosity of 16.5 x 10-3 Ns/m2 and radial clearance is 0.03 mm and eccentricity ratio of bearing is 0.27. The value of minimum oil thickness in mm is (a) 0.033 (b) 0.011 (c) 0.044 (d) 0.022 100. A kinematic pair consists of which of the following: (a) two elements that permit relative motion (b) two elements that are connected to each other (c) two elements that do not permit relative motion (d) None of these 101. Which of the following in an inversion of double slider crank chain? (a) Engine indicator (b) Elliptical trammel (c) Quick return mechanism (d) Coupled wheels of locomotive 102. A link that connects double slider crank chain fraces the path of which of the following shape? (a) an elliptical path (b) a circular path (c) a straight path (d) a hyperbolic path 103. Kinematic pair constituted by cam and follower mechanism comes under the category of: (a) Higher pair and open type (b) Lower pair and open type (c) Lower pair and closed type (d) Higher pair and closed type 104. Universal joint is an example of which of the following type of pair : (a) Higher pair (b) Lower pair (c) Rolling pair (d) None of these 105. In case ofa double slider crank chain. How many number of revolute pairs does it have? (a) 2 (b) 4 (c) 6 (d) 3

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Theory of Machines and Machine Design 106. Oldham's coupling is an inversion ofwhich ofthe following type of kinematic chain? (a) Single slider crank chain (b) Double slider crank chain (c) 2 - bar chain (d) 4 - bar chain 107. The number of links in a simple mechanism are: (a) 2 (b) 3 (c) 4 (d) 5 108. Inversion of mechanism is defined as : (a) the process of obtaining by fixing different links in a kinematic chain (b) Turning it upside down (c) Process of obtaining by reversing the input and output motion (d) Changing of higher pair to lower pair 109. For a slider crank mechanism, the velocity and acceleration of the piston at inner dead centre will be : (a) 0 and 0 (b) 0 and - co2r (c) 0 and co2r (d) 0 and> co2r 11o. In a 4 - link kinematic chain, number of pairs (P) and number of links (L) have the following relation: (a) L = 2 P - 1 (b) L = 2P - 4 (c) L = 2 P - 6 (d) L = 2 P 111. Which of the following pair, a ball and socketjoint forms? (a) Spherical pair (b) Rolling pair (c) Turning pair (d) Sliding pair 112. PORS is a four bar mechanism in which PQ = 30 em and RS = 45 em. At any instant, both PQ and RS or perpendicular to timed link PS, if velocity of Q at this situation is Y, then velocity of R will be equal to :

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113.

114.

115.

116.

117.

(a)

2.2y

(c)

i3 y

(b) ly

2

(d) 2y 3 In case of six links mechanism in planar motion, the number of instantaneous centers will be equal to : (a) 30 (b) 10 (c) 15 (d) 6 Ifthe number oflinks in a mechanism is 8, then the number of pairs will be equal to : (a) 6 (b) 12 (c) 3 (d) 18 Coriolis component of acceleration exists whenever a point moves along a path that will have: (a) Rotational motion (b) Linear motion (c) Centrifugal motion (d) None of these A slider on a link rotating with angular velocity 'co'and having linear belocity 'v'. Then the value of coriolis component of acceleration will be equal to: (a) vco (b) v2co vco (c) 2vco (d) 2 In a slider crank mechanism, the length of crank and connecting rod are 0.15 m and 0.75 m respectively. The location of crank is 30° from inner dead center. If the crank rotates at 500 rpm, then the angular velocity of the connecting rod will be equal to : (a) 1.61 rod/s (b) 2.7 rod/s (c) 3.7 rod/s (d) 5.5 rod/s

A-67

118. The maximum and minimum speeds of a flywheel during cycle are Nland N2 rpm respectively. The coefficient if steadiness of the flywheel will be equal to : N} -N2 N} -N2 (a) 2(N} + N2) (b) 2(N} - N2)

119.

120.

121. 122.

123.

124.

125.

N}-N2 Nl +N2 (c) Nl+N2 (d) Nl-N2 The flywheel of a steam engine has mass - moment of inertia 2500 kg - m2.Ifthe angular accelerationis 0.6radls2, the starting torque required will be equal to: (a) 3500 NM (b) 3700 NM (c) 1800 NM (d) 1500 NM Ifthe speed of an engine varies between 390 rpm and 40 rpm in a cycle of operation, the coefficient of fluctuation of speed will be equal to : (a) 0.01 (b) 0.02 (c) 0.05 (d) 0.09 The safe rim velocity of a flywheel is influenced by: (a) Mass of the flywheel (b) energy fluctuation (c) centrifugal stresses (d) speed fluctuation Centrifugal governors are preferred to the inertia type governers because an inertia governor: (a) has more controlling force (b) has less controlling force (c) has high initial and maintenance cost (d) is highly sensitive and more prone to hunting Porter governor is a : (a) Pendulum type governor (b) Dead weight type governor (c) Spring loaded type governor (d) Inertia type governor In case of an isochronous governor, the value of sensitivity is : (a) infinity (b) zero (c) one (d) None of these IfH = height of watt governor co= angular speed for porter governor then, which of the following relation expresses best between the walt and porter governor. (a) H ex:co (b) H ex:co2 1 1 Hex:(d) Hex:2 co co For a walt governor, the angular speed corresponding to the height of 10 cm will be equal to : (take g = 10 m/s2) (a) 10 rad/s (b) 5 rad/s (c) 2 rad/s (d) 1 rad/s Which one of the following governors cannot be isochronous? (a) Hartnell (b) Porter (c) Watt (d) Hartung In a watt governor, the weight of the ball is 50 N and the friction at the sleeve is 10 N, then the coefficient of detention will be equal to : (a) 0 .2 (b) 0.3 (c) 0.4 (d) 0.5 During the dwell period of the cam, the followers: (a) moves in a straight line (b) moves with uniform speed (c) remains at rest (d) does simple harmonic motion (c)

126.

127.

128.

129.

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Theory of Machines and Machine Design

A-68

130. The size of the cam depends upon: (a) Base circle (b) Pitch circle (c) Prime circle (d) None of these 131. The term cifd3y/d93 in a cam-followermotion showswhich of the following parameters? (a) Acceleration of the follower (b) Jerk (c) Displacement (d) Velocity of the follower 132. The deciding factor for designing the size of the cam is : (a) base circle (b) prime circle (c) pitch circle (d) pitch curve 133. Angle moved by the cam during which the follower remains at its highest position is known as: (a) Angle of descent (b) Angle of ascent (c) Angle of action (d) Angle of dwell 134. Cam used for low and moderate speed engines should move with: (a) uniform velocity (b) Harmonic motion (c) uniform acceleration (d) cycloidal motion 135. The contact between cam and follower is to a form a : (a) Higher pair (b) Lower pair (c) Sliding pair (d) Rolling pair 136. For high speed engines, the cam and follower moves with: (a) uniform velocity (b) uniform acceleration (c) cycloidal motion (d) simple harmonic motion 137. The pitch point on the cam exists on: (a) Any point on pitch curve (b) Point on cam pitch curve at which pressure angle is

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rmmmum

138.

139.

140.

141.

142.

143.

144.

(c) Point on cam pitch curve at which pressure angle is maximum (d) Any point on the pitch circle For a simple harmonic motion ofa cam follower, a cosine curve shows : (a) acceleration diagram (b) displacement diagram (c) velocity diagram (d) All of the above Which pair of gears usually has higher frictional losses: (a) Helical gears (b) Spur gears (c) Bevel gears (d) Worm and Worm wheel Axis of a pair of speer gears are 200 mm apart. The gear ratio and number of teeth on pinion are 3 : 1 and 20 respectively. Then the module of the gear will be : (a) 5 mm (b) 10 mm (c) 15 mm (d) 4 mm The outer circle of spur gear is called: (a) pitch circle (b) base circle (c) addendum circle (d) dedeundum circle Which type of gears are used in connecting two co-planar and intersecting shaft? (a) Spur gear (b) Bevel gear (c) Helical gear (d) All of the above The product of module and diameteral pitch is equal to : (a) 1 (b) 4 (c) 3 (d) 0 Axial thrust is minimum in case of: (a) Helical gear (single) (b) Helical gear (double) (c) Bevel gear (d) Spur gear

145. Herring bone gears are usually known to be: (a) spur gears (b) bivel gears (c) single helical gears (d) double helical gears 146. Differential gear is utilized in automobiles for the purpose of: (a) turning (b) reducing speed (c) provide balancing (d) All of the above 147. The efficiency of normal spur gear is usually: (a) upto 75% (b) upto 80% (c) upto 95% (d) above 98% 148. In case of spur gears, the portion of path of contact from the pitch point to the end of the engagement of a pair of teeth is known as :' (a) Arc of contact (b) Arc of approach (c) Arc of recess (d) Arc of departure 149. The ratio of base circle radius and pitch circle radius in an involute gear is equal to: (a) sin <j) (b) cos <j) (c) tan <j) (d) cot <j) 150. In case of an involute toothed gear, involute starts from: (a) base circle (b) Pitch circle (c) addendum circle (d) dedendum circle 151. In case of involute gears, the value of pressure angle genually used will be : (a) 30° (b) 60° (c) 10° (d) 20° 152. The difference between addendum and dedeundum is known as : (a) Backlash (b) Flank (c) Clearance (d) Tooth space 153. Which of the following relation is incorrect for module of gear (m) ? pitch circle diameter number of teeth

circle pitch (b) m=-~-

(a)

m=

(c)

m = diameteral pitch

(d) m=

(a)

-=-

~ N2

(b)

-=-

(d)

I::!L= (T2)2

1t 1t

circular Pitch 154. Which one of the following option is correct to describe the speed ratio of a simple gear train? Where, N, = R.P.M of driving gear N2 = R.P.M of driven gear T 1 = Number of teeth on driving gear T2 = Number of teeth on driven gear ~ T2

~ N2

N2

~ Tl

Tl

155. Creep in a belt occurs due to : (a) uneven contraction and extension of the belt (b) weak material of the pulley (c) weak material of the belt (d) improper crowning 156. In case of belt drivers, the centrifugal tension: (a) reduces the speed of driven wheel (b) reduces the driving power (c) reduces the lengthening of belt under tension (d) reduces friction between the belt and bulley

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Theory of Machines and Machine Design 157. Average tension on the tight side and slack side of a flat belt drive are 700 Nand 400 N respectively. If linear velocity of the belt is 5 mis, the power transmitted will be equal to : (a) 2.5 kw (b) 3.5 kw (c) 1.5kw (d) 4.5kw 158. In a flat belt drive, slip between the driver and the belt is % and that between belt and follower is 3%. Ifthe bulley diameters are same, the velocity ratio of the drive is : (a) 0.96 (b) 0.98 (c) 9.6 (d) 0.99 159. If in case of disc clutch, N 1 = Number of discs on driving shaft N2 = Number of discs on driving shaft then, number of pairs of contact surfaces will be : (a) N 1 - N2 + 1 (b) N 1 - N2 -1 (c) N1+N2+1 (d) N1+N2-1 160. If Jl = actual coefficient of friction in a belt moving in a grooved pulley a: = groove angle. then virtual coefficient of friction will be : (a)

Si~

oc

(b) c~

oc

(c) Jl sin o: (d) Jl cos o: 161. Velocity of belt for maximum power transmission by the belt and pulley arrangement will be equal to :

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(a)

~

(b)

(c)

f:F

(d)

t;:

r:

162. Which of the following clutches is positive type? (a) jaw (b) cone (c) disc (d) centrifugal 163. Which one of the clutch is not a friction clutch? (a) disc clutch (b) cone clutch (c) centrifugal clutch (d) jaw clutch 164. The included angle ofV - belt is generally: (a) 10° - 20° (b) 20° - 30° (c) 30°- 40° (d) 40° - 50° 165. 1fT = Total tension, Tc = centrifugal tension, then a belt can transmit maximum power when the total tension of the drive will be : (a) T = 3 Tc (b) T = 4Tc (c) T = 5Tc (d) T = 7 Tc 166. For a flat open belt drive, the belt speed is 880 m/min and the power transmitted is 22.5 kW. then the difference between the tight side and slack side tensions of the belt drive will be : (a) 3000 N (b) 3040 N (c) 1540 N (d) 1500 N 167. Assertion (A) : A clutch is the best means to connect a driving shaft with a driven shaft for regular power transmission. Reason (R) : A clutch can be frequently engaged and disengaged at operator's will (a) (A) is true, but (R) is false (b) (R) is true, but (A) is false (c) Both (A) and (R) true (d) Both (A) and (R) false

A-69

168. The inner and outer radius of friction surface of a plate clutch are 50 mm and 100 mm respectively. Ifaxial force is 4 KN, then assuming uniform wear theory, the ratio of maximum intensity of pressure to minimum intensity of pressure on cluth plate will be : (a) 2 (b) 4 (c) 8 (d) 10 169. Which one of the clutch is generally used in motor cycles? (a) Single disc wet type (b) Multi disc wet type (c) Single disc dry type (d) Multi disc dry type 170. Which one of the pair is not correctly matched? (a) Clutch - Diaphragm spring (b) Steering gear box - Rock and pision (c) Transmission gear box - Bevel gears (d) Diffemtial - Hypoid gear 171. Elastic modulus of steel is : (a) 70 GPa (b) 210 GPa (c) 270 GPa (d) 310 GPa 172. The diamond riveting is utilized for: (a) structural work (b) boiler work (c) both structural and boiler work (d) None of these 173. The pitch of the rivets for equal number of rivets in more than one row for lap or butt joint should not be less than: (a) d/2 (b) 2 d (c) 1.5 d (d) 3 d 174. Ifthe tearing efficiency of the riveted joint is 35%, then the ratio of diameter of rivet hole to the pitch of rivet is : (a) 0.65 (b) 0.75 (c) 0.85 (d) 0.95 175. A rivet is specified by: (a) shank diameter (b) type ofload (c) length of rivet (d) None of these 176. Which of the following type of material, the rivets are made? (a) brittle (b) ductile (c) high density (d) None of these 177. The shear strength of the rivet is 50 Nrnm", ifthe diameter of the rivet is doubled, then its shearing strength will be equal to: (a) 100 N/mm2 (b) 200 Nzmm(c) 300 Nzmm(d) 400 Nzmm178. The thickness of the boiler plate is 16 mm, then diameter of rivet used in the boiler j oint will be: (a) 28 mm (b) 22 mm (c) 20 mm (d) 24 mm 179. If the tearing efficiency of a riveted joint is 25% then, the ratio of diameter of rivet hole to the pitch of rivets will be equal to: (a) 0.3 (b) 0.6 (c) 0.75 (d) 0.95 180. Lewis equation in case of gears is used to find the: (a) Bending stress (b) Tensile stress (c) compressive stress (d) All of these 181. Centre distance between two involute teeth gears of base radii Rand r and pressure angle <1>, is expressed by : (a) (R + r) sin (b) (R + r) cos (c) (R'+rj tan o (d) (R+r) cot 182. Which type of teeth are normally used and satisfy law of gearing? (a) conjugate teeth (b) cycloidal teeth (c) involute teeth (d) All of these

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Theory of Machines and Machine Design

A-70

183. The maximum efficiency of worm and worm wheel system

193. Which of the following is not the part of roller bearing

will be :

(a) llmax

=

(c)

=

1- sin



1

A-

.

+~n~

(b) llmax

=

(d) llmax

=

1- cos

llmax

1+ sin



1

A-

.

-~n~

194.

1+ cos

1

A-

1+ cos - coso 184. The minimum number of teeth in an involute gear with 1° one module addendum with pressure angle of 142 to avoid under cutting will be equal (a) 10 (b) (c) 30 (d) 185. If '<1>' is the face angle of a bevel following relation is correct? (a) = pitch angle + addendum (b) = pitch angle - addendum (c) = axial pitch (d) = pitch angle 186. If '8' is the root angle of a bevel following relation is correct? (a) 8 = pitch angle + addendum (b) 8 = pitch angle - addendum (c) 8 = pitch angle + dedendum (d) 8 = pitch angle - dedendum 187. In case of spiral gears, maximum

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to : 20 40 gear, then which of the

196. angle angle

gear, then which of the

cos (8 - <1»+ 1 cos(8 + <1»+ 1

(b)

cos (8 + <1»-1 cos(8 + <1»-1

(c)

cos(8+ <1»+1 cos (8 - <1»+ 1

(d)

cos(8+<1»-1 cos (8 - <1»+1

188. The friction torque transmitted in case of flat pivot bearing for uniform ratio of wear will be equal to :

3JlwR

(a) 3 (c) 2

198. Match the following items in List 1 and List 2 List 1 List 2 (A) Worm gears (B) Cross-helical

1. 2.

2

(C) Bevel gears

3.

1

3JlwR

(D) Spur gears

4.

(d)

bearing of radius 'R', then the frictional force is : (a) R (b) 2R 2R

gears

3JlwR

(d)

4R

3 3 190. With a dynamic load capacity of 2.2 KN, a bearing can operate at 600 rpm for 2000 hours. Then its maximum radial load will be equal to : (a) 409.2 N (b) 308.4 N (c) 206.5 N (d) 609.8 N 191. Antifriction bearings are termed as : (a) ball and roller beaing (b) sleeve bearing (c) hydro-dynamic bearing (d) thin lubricated bearing 192. A sliding bearing that can support steady loads without any relative motion between the journal and the bearing is called as : (a) zero - film bearing (b) hydro static lubricated bearing (c) boundary lubricated bearing (d) hydrodynamic lubricated bearing

Parallel shafts Non parallel, intersecting shafts Non - parallel, non intersecting shafts Large speed ratios

Codes:

189. When the intensity of pressure is uniform in a flat pivot

(c)

(b) 4 (d) 5

(b)

4

(c)

197.

angle angle angle angle efficiency is given by :

(a)

(a) Jlw R

195.

(a) shaft (b) inner race (c) outer race (d) cage Which of the following assumption regarding the lubricant film is made in Petroffs equation? (a) It is converging (b) It is diverging (c) It is converging or diverging (d) It is uniform In thrust bearing, the load acts: (a) along the axis of rotation (b) parallel to the axis of rotation (c) perpendicular to the axis of rotation (d) None of these The life of bearing is expressed in : (a) Lacs of revolutions (b) billions of revolutions (c) thousands of revolutions (d) None of these The number of degrees of freedom of a five link plane mechanism with five revolute pairs will be:

D

ABC

(a) (b)

2 4

3 1

1 3

4 2

(c)

4

3

2

1

(d)

3

2

4

199. The arm OA of a epicyclic gear train shown in figure revolves counter clockwise about '0' with an angular velocity of 4 rad/s. Both gears are of same size. The angular velocity of gear c, if the sun gear B is fixed will be equal to :

~c B (a) (c)

4 rad/s 10 rad/s

(b) 8 radls (d) 15 rad/s

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Theory of Machines and Machine Design 200. A flywheel of moment of inertia 9.8 Kg - m2 fluctuates by 30 rpm for a fluctuation in energy of 1936joules. The mean speed of the flywheel is (in rpm) (a) 600 (b) 700 (c) 900 (d) 1200 201. Ifthe ratio of the diameter of rivet hole to the pitch of rivets is 0.25 then the tearing efficiency of the joint will be equal to : (a) 0.25 (b) 0.75 (c) 0.50 (d) 0.65 202. The expected life of a ball bearing subjected to a load of 9800 N and working at 1000 pm is 3000 hours. Then the expected life of the same bearing for a similar load of 4900 N and speed of 2000 rpm will be equal to: (a) 6000 hours (b) 12000 hours (c) 18000 hours (d) 24000 hours 203. If the load on a ball bearing is reduced to half, the life of the ball bearing will be: (a) increases 8 times (b) increases 16 times (c) increases 2 times (d) increases 4 times 204. Spherical roller bearings are normally used: (a) for increased radial load (b) for increased thrust load (c) when there is less radial load (d) to compensate for angular misalignment 205. In thick film hydrodynamic journal bearings, the coefficient of friction : (a) increases with increase in load (b) decreases with increase in load (c) is indepandent of load (d) None of these 206. To restore stable operating condition in a hydrodynamic journal bearing, when if encounters higher magnitudes of loads: (a) oil viscosity decreases (b) oil viscosity increases (c) oil viscosity neither increases nor decreases (d) None of these 207. The dynamic load capacity of 6306 bearing is 22 KN. The maximum radial load it can sustain to operate at 600 rev/ min, for 2000 hours will be equal to : (a) 3.16 KN (b) 4.16 KN (c) 6.21 KN (d) 5.29 KN 208. For full depth of involute spur gears, minimum number of teeth of pinion to avoid interference depends upon : (a) pressure angle (b) speed ratio (c) circular pitch (d) pitch diameter 209. Axial operation claw clutches having self locking tooth profile: (a) can be disengaged at any speed (b) can be disengaged only loaded (c) can be engaged only when unloaded (d) can work only with load 210. According to maximum stress theory of failure, permissible twisting moment in a circular shaft is T. The permissible twisting moment in the same shaft as per the maximum principle stress theory will be equal to :

Badboys2

(a)

T

4

(c) 2T

(b) T T

(d)

6

A-71 211. When the thickness of plates is more than 8 mm, then the

diameter of rivet should be equal to : (a)

d = 6.Jt

(b) d = 4.Jt

(c) d=3.Jt (d) d=2.Jt 212. It we join two plates by using riveting, then the tearing resistance needed for tearing off the plate/pitch length will be given by : (a) (P + d) tFt (b) (P - d) tFt (c) P.d tFt

Pdt (d) Ft

213. A single riveted lap joint has the efficiency of the following range: given by: (a) 45 - 65% (b) 75 - 80% (c) 85 - 90% (d) 30 - 40% 214. IfDj = journal diameter, Fhp = frictional horse power, then the relation between Dj and Fhp associated with journal bearing will be : (a) FhP o: Dj (c)

~p

3 o: Dj

215. If Z = absolute Viscosity of lubricant P = bearing pressure N = journal speed then the bearing characteristic number is given by : (a)

(c)

ZN P ZN2 P

(d) ZNP

216. The main advantage of hydrodynamic bearing over roller bearing is : (a) easy to assemble (b) low cost (c) better load carrying capacity at higher speeds (d) less frictional resistance 217. Increase in values of which of the following results in an increase of coefficient of friction in a hydrodynamic bearing 1. clearance between shaft and bearing 2. shaft speed 3. viscosity of oil Select the correct answer using the codes given below: (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) None of these 218. The Lewis form factor (y) for 20° pressure angle with full depth system is expressed as : (a)

y = 0.154- 0.912 T Y = 0.254 _ 0.952

(b) y= 0.154+ 0.912 T

(d) y = 0.254+ 9.52 T T 219. The stress concentration factor(k) given in Lewis equation has the values for most of the designing purposes is given by: (a) 0.8 (b) 1.8 (c) 1.55 (d) 2.55 (c)

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Theory of Machines and Machine Design

A-72

220. In a kinematic chain, a quaternary point is equivalent to : (a) one binary joint (b) two binary joint (c) three binary joint (d) four binary joint 221. Which mechanism produces intermittent rotary motion from continuous rotary motion 2. (a) Whitworth mechanism (b) Scotch yoke mechanism (c) Elliptical trammel (d) Genera mechanism 222. If T 1 = tension of tight side T2 = tension on slack side Tc = centrifugal tension then, the initial tension developed in the belt resulting into Tc will be given by : TI +2Tc +T2 (a) 2 (c)

TI +2Tc + T2 4

(d)

TI +4Tc + T2 4

223. The spring controlled centrifugal governor is given as : (a) watt governor (b) Pickering governor (c) Porter governor (d) Proell governor 224. If the speed of the engine varies between 430 and 510 rpm in a cycle of operation, the coefficient of fluctuation of speed will be equal to : (a) 0.01 (b) 0.02 (c) 0.04 (d) 0.17 225. In a flywheel, the safe stress is 25.2 MN/m2 and density is 7g/em3. Then the maximum peripheral velocity will be

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(a) 30 (b) 60 (c) 90 (d) 120 226. If Kd = radius of gyration of disc type flywheel K, = radius of gration of rim type flywheel If the diameter is same, then the relation between Kd and K, will be: (a)

K - Kr d - J2

(b) Kd

(c)

A= <j)+e

2 e-<j)

~=2

(b)

() a

1 [R~-R?] -flW 2 R2 -RI

(c)

3flW

2

(d) 2e-<j) 2

229. If e = semi-cone angle, then in a conical pivot bearing under uniform wear, the frctional torque transmitted will be: IflWR IflWR (b) (a) 4 cos O 2 sine IflWR IflWR (c) (d) 4 tan O 4 cote

3 uw [R~ -Rf] (b) 4 cosu R~ -R?

2

[R~-R?] R2-RI

(d)

2

[R~+R?] R2-RI

3flW

232. If, number of discs on driving shaft = 6 Number of discs on driven shaft = 5 Then the number of pair of contact surface in case of multiple disc clutch will be given by: (a) 7 (b) 8 (c) 9 (d) 10 233. Which one of the following is the correct expression for the torque transmitted by a conical clutch of outer radius RI inner radius r and semi - cone angle a assuming uniform pressure theory? (a)

(c)

flw(R-r) 2cosa

(b)

2flW(R3 _r3) 3sina(R 2 _r2)

2flW(R3 +r3) (d) 3sina(R2+r2)

flw(R+r) 2sina

234. Match List- I and List - II and select the correct answer using the codes given below: List - I List - II A Single plate clutch 1. Scooters B. Multi plate clutch 2. Rolling mills C. Centrifugal clutch 3. Trucks D. Jaw clutch 4. Mopeds Codes:

A <j)-e

p=-

4 uw [R~ -Rf] 3 cosu R~ -R?

_!_ ~w (-R1 +R2) (d) _!_ flW (RI +R2) (c) 2 sm n 2 cos u 231. IfRI = internal radius ofa collar thrust bearing R2 = external radius of a collar thrust bearing w= axial load fl = coefficient of friction then frictional torque by considering uniform wear theory will be given by :

t;

x,

p

(a)

= \/2Kr

(c) Kd = 2~ (d) Kd = 4 227. An external gear with 60 teeth meshes with a pinion of20 teeth, having module being 6 mm, then the centre distance will be equal to (a) 120 (b) 160 (c) 240 (d) 300 228. If ~ = spiral angle, e = angle of shaft, <j)= angle offriction then, which of the following expression is associated with max efficiency for spiral gears? (a)

230. If cone angle = 2d RI = Smaller radius ofpivof R2 = Large radius of pivot then assuming uniform wear theory, then the frictional torque for a truncated conical pivot bearing will be given by:

A

(a) (b) (c) (d)

1 3 3

C D 3 4 2 3 2 4 2 4 4 2

B

235. If, H = followers stroke, w = angular velocity of cam O = cam rotation angle for the maximum follower displacement. then maximum acceleration of cam follower undergoing simple harmonic motion is given by :

Badboys2

Theory of Machines and Machine Design (a) ( C)

~(';y 2H

(b)

(';'f

A-73

3H( n;Y

238. The height of the porter governor is defined as : (a)

m

n;r

(d) ~(

(c)

236. In cam design, the rise motion is given by the SHM s = % ( I - cos

~9)

where h is the total rise,

9

is cam shaft

angle and ~ is the total angle of rise interval. Then the jerk is given by : (a) %(I-COS

~9)

(b) ~-%cos(

~9)

(d) None of these 237. In case of flat belts having negligible centrifugal tension, then the ratio of driving tensions is given by: (a) (c)

_T1

Jl

T2

o

T2

-= Tl

(b)

Tl -=

T2

e

r9

(c)

(c) (c) (d)

(b) (c) (d)

(a) (b)

(a) (d)

(c) (a) (d)

(a) (c) (d)

(a) (d)

(c) (a) (a) (c) (c) (a) (c)

26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

(a) (c) (a) (c)

(a) (a)

(b)

(b)

72

(b)

(c) (b) (a)

73 74 75

(b) (c)

(d)

(c) (c) (a) (a) (a) (c) (c) (d)

(c) (c) (b)

(c) (d) (d)

(a) (a) (d)

N2

N2

K

K

4

(b) 6

K

K

8

(d) 10

KEY

51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

(d)

m

H = m - M x 895

m

e

ANSWER

(b) H=m-Mx995

N2

239. A porter governor can be classified as : (a) inertia type governor (b) pendulum type governor (c) centrifugal governor (d) dead weight governor 240. The cam follower should move with which ofthe following type of motion in case of high speed engines? (a) S.H.M (b) Cycloidal motion (c) Linear motion (d) Uniform motion 241. The shear strength, tensile strength and compressive strength of a rivet joint are 100 N, 120 Nand 150 N respectively. If strength of unriveted plate is 200 N, the efficiency of the rivet joint will be : (a) 60% (b) 70% (c) 50% (d) 40% 242. The usual proportions for the width of the key is equal to: (a)

+r9

Badboys2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

H=m+Mx895

(d) (b)

(a) (d) (b)

(c) (b)

(a) (a) (c) (a) (c) (a) (a) (a) (d)

(b)

76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

(c) (c) (a) (b) (b)

(c) (a) (d)

(a) (a) (b)

(a) (a) (b) (d)

(c) (a) (d)

(c) (d) (b)

(a) (d) (d)

(a)

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125

(b)

(a) (d) (b)

(a) (b)

(c) (a) (d)

(b) (a) (a) (c) (a) (a) (c) (a) (b) (d)

(c) (c) (b) (b) (a) (d)

Badboys2

Theory of Machines and Machine Design

A-74

(a)

126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150

(b)

(a) (c)

(a) (b)

(a) (b) (b)

(a) (c) (c)

(a) (b)

(a) (c) (b)

(a) (b)

(d) (a) (d)

Badboys2

l

(c) (b)

(a)

1

(d)

151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175

l

176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200

(c)

(d) (b)

(a) (b) (c)

(a) (d) (a) (b)

(a) (d) (c)

(a) (c) (b)

(a) (b) (c) (b)

(a) (c)

(a) (a)

1

l

201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225

(b) (b)

(d) (c)

(a) (b) (c)

(a) (b)

(a) (d) (d) (c) (c)

(d) (a) (b)

(a) (d) (a) (d) (c) (c) (b)

(a)

l

(b) (b)

(a) (d) (b) (b)

(d) (a) (c) (b)

(a) (b)

(a) (c)

(a) (c) (c)

(c)

(a) (b) (c)

(a) (d) (c)

(d) (a) (c) (b)

(a) (d) (b) (c)

(a)

(c) (c)

(d) (a) (b)

(d) (b)

1

(c) Since, all the masses lie in the single plane of the disc. So, we have a force polygon.

(a)

(a)

HINTS & EXPLANATIONS 1.

226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242

...,

I

r = mror' = 0.1 r (02N From the above force polygon, R=

~Fl + Fi + 2F}F2 cos 150

0

2

2

(0.025(02) + (0.030 (02) + 2 I I I I I I

x 0.025 x 0.030 (04x (-0.866)

Let (0 be angular velocity of disc . . F1= mlr 1(02.= 0.5 x 0.05 x & = 0.025 (02N F2= m2r2&·= 0.5 x 0.06 x & = 0.030 &N Ifr is the radial position of balancing mass 0.1 kg, so

-----------------~

2.

I' 1 1 1 1

IR

"

, , '

\

5.

,, , \ \ \

= 0.015033 (02. Now 0.1r(02 = 0.015033 (02. => r=0.150m => r= 150mm (c) Accordingto Grubler's criterion,the number of degrees of freedom of a mechanism is given by F = 3(n -1) - 2j - h = 3(8 -1) -2 x 9- 0 = 21-18 = 3 (c) For a 4-bar chain/mechanism like slider-crank mechanism, there are as number of inversions as the number oflinks or bars. These different inversions are obtained by fixing different links one at a item for one inversion.

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Theory of Machines and Machine Design Hence, number of inversions mechanism will be four.

for a slider-crank

A-75

11. (c)

Load to be lifted

6.

(d) Mechanical advantage =

Effort applied Output force

7.

Input force For a four bar linkage in toggle position Effort = 0 .. Mechanical advantage = 00 (a) For flywheel which controls the fluctuations in speed during a cycle at constant output load,

12. (a)

1 (2(02 -(01 2) ~E ="21

~

400 =

±.I(

2102 - 1902 )

14. (a) Type of instantaneous centres: (a) Fixed instantaneous centres (b) Permanent instantaneous centres (c) Neither fixed nor permanent instantaneous centres

~

8.

1= 0.1 kg-m-. (b) The critical or whirling speed of centrally loaded shaft between two bearings

"'e ="'0 =t; "'e = J 0.0018 9.81

Badboys2 9.

10.

=~ = 73.82 rad/ s

2nNc = 73.82 60

N, = 704.96 ::::705 rpm ~ (a) According to Grsashoff's rule for a planar crankrocker four bar mechanism, the sum of lengths of shortest and longest links should be less than the sum of lengths of other two remaining links. So, statement (a) is incorrect and rest are correct. (d) The mobility or degrees of freedom of a plane structure is the number of inputs (i. e., number of independent coordinates required to determine the configuration or position of all the links of the mechanism W.r.t. fixed link. It is determined by Grubler's equation as F = 3(n - 1) - 2j - h where F = degrees of freedom or movability of mechanism n = number oflinks j = number oflower pairs h = numbers of higher pairs Now, a 5-bar chain is the simplest statically indeterminate structure in which link 1 is fixed as shown. Hence to specify the position of all links, two coordinates 91 and 92 are required. So two inputs are required to give a unique output. So, F = 2 or the mobility is 2.

F = 3(n-l) + 2fl -f2. = 3(5 -1)-2 x 5-1 =12-10-1=1 As we know that, velocity at point of contact between object and floor will be (OR. While, radius 'R' will be equal to zero an instantaneous centre is situated at the intersection point of object (radius 'r') and floor.

(a)

Fixed instantaneous centre: They remains in the same place for all configuration of the mechanism.

(b)

Permanent instantaneous centres: They move when the mechanism move, but the joints are of permanent mature.

(c)

Neither centre:-

fixed nor permanent

instantaneous

They vary with the configuration of the Mechanism.

15. (c) The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. The difference between the maximum and

minimum energies is known as maximum fluctuation of energy. ..

AE = Maximum energy - Minimum energy

16. (d) The minimum number ofteeth on a pinion is found on the basis of consideration of avoiding interference. In case of 14Yz° involute system, the minimum number of teeth in a pinion which meshes with rack 2

17. (a)

t . =--=32 nun sin! <j) y

a~-----------------+x

Kutzbach criterion for movabilityof a mechanism, 1 Similarly, for 6-bar or more chains, F > 2 Hence, for a statically indeterminate structures, Mobility ~ 2

Number of degree of freedom = 3 (1- 1) - 2j - h = 3(2 - 1) - 2 x 0 - 1 =3-1=2 Hence, it possesses 2 degree of freedom.

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Theory of Machines and Machine Design

A-76

18. (d) Fp = Primary unbalanced force = mror'cos"

N2 = 300 rpm

F s = Secondary unbalanced force

N2 = T, =.!. N, T2 3

(I)n=~

mrci' =-n-cos28

..:5_ = .!.T

19. (c) For uniform acceleration and retardation the velocity of the follower must change at a constant rate and hence the velocity diagram of the follower consists of sloping straight lines. The velocity diagram represents everywhere the slope ofthe displacement diagram, the latter must be curve whose slope changes at a constant rate. Hence the displacement diagram consists of double parabola. 20.

(a) a = _!__ = mk"

W2

580 = 5.59 rad/s ' 800 x (0.36)2

= WI+ at = O. + 5.59

x

12 = 67.08 rad/s

T2

3

= 24

I

Pitch line velocity = W1r1or W2r2

d2

= 21tN2 -

= 21tx300x--

2

8x72

2 = 542867 mmlmin = 9.04 mls 26. (d) The position ofthe instantaneous centre changes with the motion of the body. Instantaneous centre of a body rolling with sliding on a stationary curved surface lies (i) on the common normal at the point of contact, and also (ii) at the centre of curvature of the stationary surface 27. (d) We know that coefficient of fluctuation of speed (Cs) IS

1 1 KE = -mk2w2 = -x800x(0.36)2 2 2 = 233270N = 233.3 kJ 21.

X

C - (wmax

(67.08)2

S -

(a)

-

wmin)

(ro~;ro~")

or, Cs

+ CSWmin= 2

Wmax

Wmax -

2wmin

wmax = 2+ Cs 2-Cs

wmin

28.

Badboys2 fp = rw2 At IDC

(COS8+ 1 CO:28) 8=0

.. r. ~ rCO>(l+~) At ODC

fp ~ 22.

23.

(c)

-rCO>

8 = 180

0

(l-~)

Energy stored in flywheel is dependent on moment of inertia given by: 1= (w/g)k2 where k = radius of gyration In case of rim type of flywheel, k' = radius offlywheel.

k Since , k' = _2 (c) Arc of contact = 31.4 mm Module (m) = 5 Circular pitch = 1t m = 51t

25.

1

=mk2

mk2

31.4

----s;-

...(iii)

From the equations (i) and (ii) we get mh m+-I_I=m I

h

2

= 2 pairs.

21tN 2 x 1tx 160 = -= 16.75 rad/s (a) 60 60 maximum velocity of sliding = W x d = 16.75 x 0.018 = 0.302 mls (c) T2 = 72 VR= 3

m.h] + m2h; = mk ' (iii) From the equations (ii) and (iii); we get

m. = h,(h, +h2)

=

24.

of gravity. Consider a rigid rotor with the shaft laid on horizontal parallel ways. if it is in static balance, the shaft will not on the ways whatever may be the angular position of the rotor. For this to happen, the centre of gravity of the system of masses must lie at the axis of rotation of the shaft. For the centre of gravity to be at the axis of the shaft, the horizontal and vertical moments of the rotors must be equal to zero 'LWr sin 8 = 0, 'LWr cos 8 = 0 The above equations are also true with the dynamic balance of the inertia forces. Thus if the conditions for the dynamic balance are met, the conditions for static balance are also met. (c) For dynamically equivalent m1 + m2 = m (i) m.h, = m2h2 (ii)

m,h~ +(m,h,)xh2

Arc of contact So, No. of pair of teeth in contact = C· I .h ircu ar pitc

W

29.

(c) Static balance is a balance of forces due to the action

..

mh2 m1 = h + h I

...(iv)

2

From equations (iii) and (iv) mk2

hi(hi + h.) k2 = h1h2

mh2

(hi + h.)

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Theory of Machines and Machine Design 30.

(a)

A-77

Since cylinder falls freely under effect of gravity, it follows basic law of motion and

..

2k

=> 9 + -9

v2 = 2gh and v = ~2gh

31.

2'"

(a) Ratio of angular speeds of F to A TA . Tc . TE

20x30x 25

TB ·To ·TF

60x80x75

J!.

OJ" = 40. (d) In toggle position,

24

32.

(a) Point P being rigidly connected to point 3, will trace

33.

(c) A system of masses rotating in different parallel

same path as point 3, i.e. ellipse.

34.

=0

3m

planes is in dynamic balance ifthe resultant force and the resultant couple are both to zero. This is known as dynamic balancing. (c) A bicycle remains stable in running through a bond because of centrifugal action.

for a four bar linkage, the mechanical advantage will be infinity. 41. (d) = 30° Normal circular pitch = circular pitch x cos = 15 x cos 30° = 13 mm .. Circular pitch AXIal pitch = tan 300

Ijj

42.

(a) We know that

=> 400 = 4001

= '2l( CO max + comin) (comax - comin)

= 1 coay x Cs

X

coay

1 = 0.1 kg-rrr' Tfirst

emax = lco;y Cs

(c) From the figure, it shows that the value of'P' required

43.

(a) Train values =

mT

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40x12

(c) R=-=-2

2

mt r=-=--

44. (d)

20x12

2

...

= 240mm

2

45.

(a)

46.

(a)

= 'VIR 2a _ R 2 cos" 'I' '" - R sin '"'I'

r sin 2

120x;in20

~R; -(240cos20)2

-240sin20

R, = 248 mm

30

3

T = 50 = 5" last

will be minimum when it is at 452 to the horizontal. This can be solved by resolution of forces.

addendum = 248 - 240 = 8 mm ro cos O

47. (b) oicos O If n is large COer = -n Angular velocity is maximum at 9 = 0, 180° Angular velocity is zero at 9 = 90°

39.

tan 30°

llE = ~ l( co~- con => 400 = ~ x [(210)2 -(190)2 ] 1

37.

15

=26mm

35. (d)

36.

15

1 5 Speed ratio = T· 1 3 ram va ue The centre of gravity of the coupler link in a 4 bar mechanism would experience both linear and angular accelerations. According to Grashof s law for a four bar mechanism. The sum of shortest and longest link lengths should not be greater than the sum ofthe remaining two link length. i.e. S + L ~ P + Q An involve pinion and gear in mesh. Ifboth have the same size of addendum, then there will be interference between the tip of the gear teeth and blank ofrenion. This is a phenomenon of interference. We know that, VB= V, CD = 45 em, AB = 30 em

VCD = Vc = W.CD = CD VBA VB W.AB AB Vc =CD VB AB

(c)

CD Vc = VBx-= AB

45

V =-=

30

3 3 Vx-=-V 2

2

. 3 :. Velocity of C = '2 V .

48.

(c) The transmission angle is maximum when crank angle

49.

with fixed link is 180°. The transmission angle is minimum when crank angle with fixed link is 0°. The transmission angle is optimum when crank angle with fixed link is 90°. (b) CD=AB+30cm Rotation of AB, COl= 5 rad/s Rotation of CD, CO2= 2 rad/s So, colAB = co2CD 5 AB = 2 (AB + 30) AB = 20 em

Taking moments about instantaneous centre' A'

I,

e + (kx) r = 0 e

=> (10 + m2) + kx (Or) r

=

0

1 2 -i mr 2J 9 + k (9r2) = 0 => ( '2mr ..

kr'

=> 9+--9=0 3 -mr 2

2

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Theory ofMachines and Machine Design

A-78

50. (a) ac = r0)2 8 = 0.5 x 0)2 0)2= 16 0) = 4 rad/sec Coridis component of acceleration = 20)v =2x4x2=16

51. (a)

52. (c)

The Sommerfield

number defmed as

Z;

(~r is

T = 60x 20000

used in the design of hydrodynamic journal bearings, while rating life, rotation factor and specific dynamic capacity are used for ball and roller contact bearings. It is known that S-N curve becomes asymptotic for 106 cycle, so stress <J at this cycle is known as fatigue or endurance limit of the material.

106.10 N-m

2n(60x30) Tangential

106100

force

= 2122 N

(5 x 20) 12 Resultant (normal) force 2.69 = loglO490

FN =

I I I

---1-------I

2.000 = loglO100

55. (d)

L = (C)3 F

I

1.845 = 10glO70 ---r-------T-Badboys2 I I I I

I I I I

3

X

6

y-2.69 = 2.69-1.845 = -0.28167 x-3 3-6 .. y = 2.69 - 0.28167 (x - 3) y = 3.535 - 0.28167x

56. (b)

Ft x factor =

=>

57. (a) stress of 100

5('9~37)

54. (c)

P

56 2

= 2nNT => 60

59. (b)

= 140mm T

= 166.52 MPa

60 x 15000 2n x 960

1492077 N m . -

Tangential force Ft = D 1 2

84/4 58. (d) Interference is a phenomenon in which the addendum

:~tr~~:~:~::5m[Zp;Zg 1 = 5x-

= 3552.56x1.5 25x0.32x4

= 149207.7 = 3552.56 N

10giON = 5.44964

=

x bny x m

T

..

2 = 3.535 - 0.28167

N = 105.44964 = 281604.53 cycle

53. (a)

<Ja

<Ja

Pitch circle diameter of gear D = m x Z = 4 x 21 = 84 mm

=> x = 5.44964 ..

L oc= _1F3

60P T orque T -- 2nN -

y= 10glO 100 = 2 ..

=>

X-axis 10glON

According to 2-point form, the equation of straight line connecting (6, 10giO 70) and (3, 10giO 490) is

For the shaft subjected to alternating MPa

Ft 2122 2258.1 N cos 20° cos 20° For a ball bearing, the life-load relationship is

=

tip of gear under cuts into the dedendum of base circle of pinion. This tooth interference can be reduced by increasing the number of teeth above a certain minimum number. For example, For 20° full depth involute teeth system, minimum number of teeth to avoid interference is 18. Pp = 30 kN PQ = 45 kN As we know that, L = (~) a , a = 3 for ball bearing

60 P

2n N

Life of bearing P = Lp = (PQ Life of bearing Q LQ Pp

J3

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Theory of Machines and Machine Design

A-79 Let d = diameter of solid shaft . T= Px60 .. 21tN

61. (b)

16T Working shear stress = -3 1td Allowable shear stress Factor of safety = W or kimg sh ear s tr ess

20xl03 x60 =955N-m 21tx 200 From torsion theory, we have torque transmitted by solid shaft (T).

~

955x103 =~x~xd3 16

2x 16T = 140 ~ 1td3

d= 15.4mm= 16mm

1t =16x45xd 99. (d)

3

d=47.6::::48mm

Given data

67. (a) Eccentricityratio

= 0.27 = 1- hO c

hO 0.27= 1- 0.03 ho=0.0219mm ::::0.022mm 101. (b)

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snap head

(a)

Elliptical trammel

(b) (c)

Scotch - Yoke mechanism Oldham's coupling

109 . (d)

70. (d)

Snap head rivet is used for boiler plates. Load stress factor

K = cr~ sin

73. (b)

L= 106 60n

Examples of inversion of double slider crank chain are

Considering the following equations for a slider - crank mechanism when the piston at inner dead centre,

. sin 2a] velocity of piston (v) = rto [ Sma + -2.. 2[ cos 2a] Acceleration of piston (a) = rro cos a + -n-

(C)3

Now, at inner dead centre, a = 0°, then

P

v = rro[ sin 0 + Sin;: 0

Here, n = 500 r.p.m, L= 6000 hours, C = 50 KN

:.(~r

orP=

77. (c)

79. (b)

0) ]

6000 x 60 x 500 = 180

(l

= rro2

[cos 0" + cos:(O)]

C

50 =-=8.85K.N (180)3 5.65

--1

In case of selflocking brake, no external force is required for the braking action. This is not desirable condition in normal application. It is proved by using theory of elasticity that the theoretical stress concentration factor at the edge of hole is given by I +2( ~)

(a)

°

v=O

(l

=

-r.

I)

113. (c) Considering the following relation, Given (N) = 6 Number of instantaneous

87.

= roix O =

Allowable shear stress ~ = 2!L = 360 = 45 N/mm2 fos 8

centres, (N1C) =

where, N = Number oflinks Hence, NIC=

6(6 -1) 2 =3x5=15.

N(N -1) 2

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Theory ofMachines and Machine Design

A-SO

119. (d) Given: Mass moment of Inertia (Is) = 2500 kg-mAngular acceleration (a) = 0.6 rad/sTorque required (r) = Is x a =2500 x 0.6= 1500N-m 120. (c) Given: Speed range of an engine, 111= 390 rpm, 112=410rpm

116. (c)

Coefficient of fluctuation = 112-111 = (112-111) 11m 112+ 111

Let, Ae = coriolis component of acceleration then A, = 2 vw 117. (a) Given: Length of the crank (Lemnk)= 0.15 m Length of connecting rad (Le) = 0.75 m Location of crank (from IDC) (a) = 30° Crank speed (N) = 500 rpm

2 _ 2 (112- 111) 2(410-390) -

ndN

Angular speed of crank (wemnk) =

60

126. (a)

(410+390)

(112+ 111)

= 2x20 = 0.05 800 Heightofthewalt g= 10m/s2

governor (H)= lOCm=O.1

m

3.14 x (2 x Lerank) x 500 H=

60

g 2 (Angular speed)

=> (Angular speed)2 =]_ H

3.14x2xO.15x500 60

Angular speed =

= 7.85 rad/s Considering the following formula,

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{10 = JlOO = 10 rad / s

~OJ

128. (a) Given: weight ofthe ball = 50 N friction at the sluve = ION . . coefficient of detention

r.,

h were,

th

=.!.Q =

0.15 cos30

7.85 x 0.866

~(5)2 _ sin2 30

.J25-0.25

= 7 85 . x

en,

140. (a)

.J24.75

0.2 50 Given: distance between the axis = 200 mm gear ratio = 3 : 1 Number ofteeth on pinion (T p) = 20

Let dg , dp

w = 6.7981 = 6.7981 = 1.37 rad/s e

4.975

Hence Angular velocity of connecting rad (we) is nearest to 1.61 rad/s. So, answer will be 1.61 rad/s 118. (b) Given: maximum speed offlywheel = Nmax= N 1 minimum speed of flywheel = Nmin= N2

= 3:

1

dg = ~ => dg = 3dp dp 1

N, +N2 mean speed of flywheel (Nm) = ___:_-2--=. . then, coefficient of steadmess

dp = 400 = 100 4

=

Nm ----=..:..:'--

...(i)

=> dp + dg = 400

dp + dg 200 2 dp + 3dp = 400

=> 4dp = 400

N1-N2

dp 100 Module ofthe gear (m) = Tp = =5

20

_ (Nl + N2)

-

weight of ball

0.75

h=--=--=5 Lerank

we

frictio at sluve

= -----

2 (Nl - N2)

_ (N1 -N2) - 2(Nl - N2)

143 (a) .

M 0du Ie ()m

=

diameter

d

Number of teeth

T

diiametera I PItC . h (p)d = Number of teeth = -T diameter d

Badboys2

Theory of Machines and Machine Design d T

mXPd =-x-=

T d

A-81

1

157. (c) Given: Average tension (T avg.)l = 700 N and (Tavg.h = 400 N Linearvelocity(v) = 5 m1s Power transmitted (P) = [(T avg.)l - (T avg.h ] x V = (700-400) x 5 = 300 x 5 = 1500 walts = 1.5kw 158. (a) Given: slip between driver and belt (sl) = 1% Slip between belt and follower (s2) = 3% Now, considering the following formula,

. . -~(1-

Velocity ratio - d2

+S2J

50 1 -=-=> 12 4

12 = 200N Imm

2

178. (d) Given: thickness of boiler plate, (t) = 16 mm diameter of rivet (d) = 6.Jt (ift > 15 mm) d = 6M

= 6 x 4 = 24 mm

179. (c) Given: tearing efficiency (11t)= 25% = 0.25 Let, D = diameter of rivet hole P = pitch ofthe rivet,

sl 100

P-D then , TIt =0.25=--=1--P 'I

D

P

166. (c) Given: belt speed (u) = 880m/min. Power transmitter (P) = 22.5 kw Let, T 1 = tension in tight side, T2 = tension in sleek side then, P = (T 1 - T2) x u here, u = 880 m1min. 880

44

60

3

=-m/s=-m/s

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44

3

22.5x1000=(T1-T2)X

(T} - T2) =

22.5 x 1000x 3 44

Let, Pmaximum= maximum intensity of pressure Pminimum= minimum intensity of pressure Pmaximum= Ro = 100 = 2 Pminimum Ri 50

174. (a) Given: tearing efficiency (11t)= 35% = 0.35 Let, D = diameter of rivet hole P = pitch of rivet

(p - D) x t x Ft PxtxFt

0.35 = P - D = 1- D

P

Lf

106

=e·2~OOOJ.3 xl0

6

On solving, we get, w= 609.8 N (approx.)

Degree of freedom (DOF) = 3 (L - 1) - 2J =3(5-1)-2x5 =3x4-10 = 12-10 =2 199. (b) Given, Angular velocity (wA) = 4 radls Angular velocity (wB) = 0 Angular velocity of gear 'c' = wc As both gears are of same size, hence Number of teeth on A (TA) = Number of teeth on B (T B) N ow, using the following relation, wc-wA wB-wA

=_TB=_l

r,

w -4 _c_=_l 0-4

177. (b) Given:Shearstrengthofrivet(11)=50N/mm2

(D)2

x

197. (c) Given: Number oflinks (L) = 5 Number of joints (1) = 5

D = 1-0.35 = 0.65 P

1 IX

(~r

72xl06

P

diamenter of rivet (initial) (D1) = D Final diamenter ofrivet (after doubling) (D2) = 2D As we know that, Shear strength of rivet IX (Diameterj-

=

K = Constant = 3.3

( 1534.1N == 1540N approx)

168. (a) Inner radius (R) = 50 mm Outer radius (RJ = 100 mm axid force(w) =4 KN

11t=

190. (d)

D = 1-0.25 = 0.75 P Given: Dynamic load capacity(C) = 2.2 KN Life of bearing (Lf) = 60 x N x time duration = 60 x 600 x 2000 =72 x 106 Using the following relation,

200. (a)

wc-4=4 wc=4+4= 8radls Given: Moment of inertia (L) = 9.8 kg-m/ fluctuation speed (Nt) = 30 rpm fluctuation energy (E) = 1936 j

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Theory ofMachines and Machine Design

A-82 . 21tNt 2 x 1t X 30 Now, Angular velocity (w) = _= --60 60 =n

rad/s

Now, considering the following relation, E =Lx w x wmean E wmean= Tfxw W mean

1936 =9.8x1t

=629rpm(Approx)

= 629 rpm == 600 rpm (Approx.)

201. (b) Given: diamenter of rivet hole = D Pitch of the rivet = P D -= P

207. (d)

L1 = L2 8 L2 = 8L1 => 8 times. Given: dynamic load capacity = 22 x 103N Speed (N) = 600 rev/min. (time duration) Life (L) = 2000 hours Using the following relation, L

0.25

=

(:r

X

106 rev

we get w = 5.29 KN

. e ffici P-D = 1-- D teanng ICleny ()11t = _-

P

P

= 1-0.25 = 0.75

.

_ 112-111

202. (b) Given: Load (wI) = 9800N

Now, ifload (w2) = 4900 N Speed (N 2) = 2000 rpm then Life (L2) = ?

=

Considering the following, L(w)3 = Constant

J

=>..s. = (W2 3 L2

=.!. 8

1500 1 ~ ="8 => L2 = 1500 x 8 = 12000 hours

225. (b) Given: Safe stress (as) = 25.2 MN/m2 density (p) = 7g/cm3 Safe stress (a J = p x maximum periphered velocity (v2) as = p x v2 =_7_xl06 1000

xu2

2 25.2 xl000 u =---7

u = ..)3600 = 60 m / s 227. (c) Given: Number ofteeth (T) = 60

203. (a) Given, Initially, wI = w

Module (m) = 6 mm Number ofteeth on pinion (T p) = 20

Life = LI

Considering the relation, Lw3=c

160 940 = 0.17

wI

1000 x 60 x 3000 = (4900)3 2000 x 60 x L2 9800

Life=L,

111+ 112

2(510-430) 510+ 430

25.2xl06

After half of load, w 2 =-

112-111

2( 112-111)

111+ 112 2

Speed(N1)= 100rpm (Life) (LI) time duration = 3000 hours

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.

224. (d) Coefficient of fluctuation of speed = ......:..:....___:~ 11m

w 2

Centre distance (D) = m(T + Tp ) = 6 (60 + 20) 2 2 6x80 480 =--=-=240mm 2 2

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rrIII~ll)ll'l~ I~Nf.INI~I~IIINf. THERMODYNAMICS

Surrounding

~system

In the subject of thermodynamics, the inter-relationship among heat, work and system properties are studied. It is also called as the conceptual science of entropy and energy.

(Thermodynamic

Some Thermodynamical Terms in brief (i) Thermodynamic system: A thermodynamical system is an assembly oflarge number of particles which can be described by thermodynamic variables like pressure (P), volume (V), temperature (1). (ii) Surroundings: Everything outside the system which can have a direct effect on the system is called surroundings. The gas cylinder in the kitchen is the thermodynamic system and the relevant part ofthe kitchen is the surroundings. (iii) An adiabatic wall: The wall which prevent the passage of matter and energy. (iv) Diathermic wall: It prevent the passage of matter but allow the passage of energy. An aluminium can is an example of a container whose walls are diathermic. (v) Closed and open system: In a closed system, energy may transfer the boundaries of system but mass does not cross the boundary, while in open system, both mass and energy transfer across the boundary of the system. (vi) An isolated system: In this type of system neither the mass nor the energy can be exchanged with the surroundings. (vii) Equation of state: The relationship between the pressure, volume and temperature of the thermodynamical system is called equation of state. (viii) Properties : A property of a system is any abusable characteristic of the given system various properties of the system depend on the state of the system not on how that state have been reached. (xi) Intensive property of a system or those properties whose values does not depend upon the mass of the system. Eg: Pressure, temperature, viscosity etc., while extensive properties depend upon the mass of the system. Eg: Length, volume etc. (x) Equilibrium: A system is said to be in thermodynamic equilibrium when it does not lead to change its properties (macroscopic) and make balance with its surroundings. There, a system in mechanical, thermal and chemical equilibrium is said to be in thermodynamic equilibrium.

A thermodynamic system is described as a kind of a region available in space and this region is concentrated for the purpose of analysing a problem. The system is considered to be separated from surroundings (external to system) by the boundary of the system. The nature of the boundary may be real or imaginary and it is considered to be flexible i.e., it can change its shape or size. Ifwe combine a system and its surroundings, then it constitutes the universe.

system)

Types of thermodynamic systems: There are three types ofthermodynamic

(a)

systems:

Closed system: A thermodynamic system in which mass is not transferred across system boundary but energy may be transferred in and out ofthe system, is known as closed system. Mass in the piston - cylinder arrangement is the example of a closed system.

(b)

Open system: The open system is defined as a system in which mass as well as energy can be transferred with its surroundings. Open systems are most common. The region where analysis ofthe system is performed is known to be a control volume and the boundary of control volume is known as control surface. Eg: Air compressor

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THERMODYNAMIC SYSTEM

boundary

lnputmass stem boundary Input mass

Exit mass Surroundings Exit energy

(Open system) (c)

Isolated system: In an isolated system, no mass and no energy is transferred across system boundary. ~ystem

boundary

~ Surroundings (No mass transfer No energy transfer)

(Isolated system)

ZEROTH LAW OF THERMODYNAMICS

If objects A and B are separately in thermal equilibrium with a third object C then objects A and B are in thermal equilibrium with each other. Zeroth law of thermodynamics introduces thermodynamic quantity called temperature. Two objects (or systems) are said to be in thermal equilibrium iftheir temperatures are the same. In measuring the temperature of a body, it is important that the thermometer be in the thermal equilibrium with the body whose temperature is to be measured.

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Thermal Engineerging

A-84

FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is based on conservation of energy. According to this law heat Q supplied to a system is equal to the sum ofthe change in internal energyfal.I) and work done by the system (W). Thus we can write Q = ~U+W More about First Law of Thermodynamics 1. Heat supplied to the system taken as positive and heat given by the system taken as negative. 2. It makes no different between heat and work. It does not indicate that why the whole of heat energy cannot be converted into work. 3. Heat and work depend on the initial and final states but on the path also. The change in internal energy depends only on initial and final states of the system. 4. The work done by the system against constant pressure P is W = P~V. So the first law of thermodynamics can be written as Q = /).u + p/).v . 5. Differential form of the first law; dQ = dU+dW or ~ = dU+NV SECOND LAW OF THERMODYNAMICS (i)

Kelvin - Plank Statement: It is impossible to construct an engine that can convert heat completely into work without producing any other effect.According to the statement the efficiency of any heat engine always be less than 100%. Clausius Statement: For a self acting machine, it is impossible to transfer heat from a colder body to a hotter body without the aid of external agency.

Badboys2 (ii)

ENTROPY Entropy is the another thermodynamical variable which many times very useful to understand the system. Entropy is related to the disorder or randomness in the system. Tounderstand this, let us consider two systems as shown in Fig.

Intensive and Extensive properties (a) Intensive properties: Intensive properties are those properties which does not depend on the mass available in the system. Eg : temperature, pressure, etc. (b) Extensive properties: Extensive properties are those properties which depends on the mass available in the system. Eg : Volume,energy,etc. Some terms like specificvolume, specific energy etc. come under the category of specific extensive properties. Thermodynamic equilibrium A systemis said to be in equilibriumwhen there is no driving forces within the system after isolation ofthe system from its surroundings. A system is said to be in thermodynamic equilibrium it satisfies the following three kinds of equilibrium: (a) Mechanical equilibrium (b) Thermal equilibrium (c) Chemicalequilibrium Internal energy or energy of the system The internal energy of the thermodynamic system is regarded as the combination of all kinds/forms of energy of the system. These all forms of energy include kinetic energy, potential energy vibrational energy, rotational energy etc. If dU is the internal energy of the system, then, dU=MC~T where, M = mass in kg, C = specific heat - capacity, ~T = change in temperature. Energy can also be considered as a property of a thermodynamic system. Consider a system that undergoes a change from state 'A' to state 'B' and the systemundergoes a cyclic process. A

f

~B

--~V

System 1 System 2 If SI and S2 are the entropies of the system I and 2 respectively at any temperature, then SI < S2. (i) Entropy is not a conserved quantity. (ii) Entropy can be created but cannot be destroyed. (iii) Entropy of the universe always increases. If a system at temperature T is supplied a small amount of heat ~Q, then change in entropy of the system can be defined as M=

~Q

T

for constant T

For a system with variable T, we have

(Total work done)eyeIe = (Total heat) cycle WI+W2=QI +Q2 QI-WI =W2-Q2 I~Q=dU+ 8wl Specific heat of constant volume (C v) .

It IS defined as the rate of change of internal energy with

respect to temperature keeping the volume as constant. C =(dU) V dT p Specific heat of constant pressure (C ) It is defined as the rate of change of errthalpy with respect to temperature keeping the pressure as constant.

sf dQ ~S

= Sf -

s, =

fT

s,

The second law of thermodynamics may be stated in terms of entropy as: It is impossible to have a process in which the entropy of an isolated system is decreased.

C =(dH) p dT v THERMODYNAMICAL PROCESSES Any process may have own equation of state, but each thermodynamical process must obeyPV = nRT.

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Thermal Engineerging 1.

A-85

Isobaric Process: Ifa thermodynamic system undergoes physical change at constant pressure, then the process is called isobaric.

p ......------to--~

p ......... :-~

(-~Vr

00

p

P

M

B

(v) Work done: W = P~V= 0 (vi) First law ofthermodynamics in ischoric process Q ~U+W=~U+O or Q ~U

........ -...........,

»c.sr

3.

'W=-P~V

W = +Pav

'---~----~-

....v

V Expansion

Compression

(i)

Isobaric process obeys Charle's law, VOC T

(ii)

dP Slope of P ~ V curve, dV = O.

Isothermal Process: A thermodynamical process in which pressure and volume of the system change at constant temperature, is called isothermal process. p

(iii) Specific heat at constant pressure C;=

5R

7R

p

2 for monoatomic and Cs= Tfordiatomic

(iv) Bulk modulus of elasticity: As P is constant, M and

=0

M

(-A:f

B

Badboys2 (v) Work done:

(i)

An isothermal process obeys Boyle's law PV = Constant. (ii) The wall of the container must be perfectlyconducting so that free exchange of heat between the system and surroundings can take place. (iii) The process must be very slow, so as to provide sufficient time for the exchange of heat. (iv) Slope of P - V curve: For isothermal process PV = Constant After differentiating w.r. t. volume, we get

W = P~V=nR~T (vi) First law ofthermodynamics in isobaric process Q= ~U+W

= ~U+P~V

= ~U+nR~T

= nCv ~T + nR~T = n( Cv + R)~T = nCp~T (vii) Examples:Boilingof water and freezingofwater at constant pressure etc. 2. Isochoric or Isometric Process: A thermodynamical process in which volume ofthe system remain constant, is called isochoric process.

P+V dP dV

dP -P V or tan O = dV (v) Specific heat at constant temperature: As~T=O, or

T

j

W=O

An isochoric process obeysGay - Lussac' sLaw, P oc T

dP (ii) Slope of P - V curve, dV

=

4.

(vi) First law ofthermodynamics in isothermal process. As ~T 0, :. ~U=O Q ~U+W=O+W or Q W Adiabatic Process: An adiabatic process is one in which pressure, volume and temperature ofthe systemchange but heat will not exchange between system and surroundings. p

00

(iii) Specific heat at constant volume Cv

=

-P

V

C =

o'-----------·v (i)

0

3R 5R 2 formonoatomicand Cv = 2 fordiatomic

(iv) Bulk modulus of elasticity : As Vis constant, ~V= 0

p

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Thermal Engineerging

A-86

(i)

Adiabatic process must be sudden, so that heat does not get time to exchange between system and surroundings. (ii) The walls of the container must be perfectlyinsulated. (iii) Adiabatic relation between P and V PVY =k (iv) Adiabatic relation between Vand T & P and T For one mole of gas PV=RT,

= k,

(Rnv

k

y

Isothermal expansion: If Qj is the heat absorbed from the source and WI is the work done, then, QI

2.

we get

W2 =

3.

Q2 ~

nR(lj - Tf ) y-I

nR(Ii - T2) y-I

W3~nRT2fn(~)

~ nRT2fn(~:J (As flU

=

=

0)

~ -nRT2fn(~J

k RY = another constant

or

AU ~ 0)

J

P

4.

Q

nflT

Adiabatic compression: If W4 is the work done during the adiabatic compression, then W4

0

= nflT = 0

(vi) First law ofthermodynamics in adiabatic process Q L1U+ W As Q 0, :. Sll=> W or UI-Uj -W .. UI Uj-W P-V Diagram Representing Four Different Processes

nR(T2 -Ii) y-I

nR(lj - Tf) y-1

-nR

(Ii- T2) y-I

Net work done in the whole cycle W =Wj+W2+W3+W4

(V

-- n R'T'II -LJ~n - 2J + nR(Ii -T2) Vi y-I

(V

II - 3J -nR__:__:_----='-'(Ii -T2) n RT 2~n V4 y-I

In the adiabatic expansion B ---+ C

...p

v:

1', y-1

1 2

500K 400K

300K

1. 3.

(As

RT

V

Specific heat:C

nR1Jfn(~ )

Isothermal compression: If Q2 is the heat reject to the sink and W3 is the work done during the process, then

k

Badboys2(v)

WI ~

Adiabatic expansion: If W2 is the work done during the adiabatic expansion, then

k R = new constant

or

~

nR1Jfn(~

V

or P=

Substituting in PVf

Also

RT

1.

T Vy-1 2 3

or

...(ii)

Similarly in the adiabatic compression D ---+ A

L------------___.v Isobaric Process 2. Isothermal Process Adiabatic Process 4. Isochoric Process

T Vy-1 2 4

1', v,y-l 1 1

...(iii)

CARNOT CYCLE

or

Camot cycle has four operations. Thermodynamic coordinates after each operation are shown in Fig. Initially at A coordinates are r; Vj,Tj.

From equations (ii) and (iii) , we have

p

...

or

V2

5_

V3 VIV3

V2V4

V2

Also

VI

V4

...(iv)

V3 V4

Efficiency of carnot engine Workdonebyengine (W) 11 Heat absorbed by engine from source ({1)

o-------------+v

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Thermal Engineerging

A-87

nR[Tjfn( ~ )-T2fn(~)] nRTjfn(~: ) As

r, - T2

----

11

_ 1 --T2

11

Enthalpy: Enthalpy is regarded as the total energy of a thermodynamic system. It is defined as the sum of internal energy and product of pressure and volume. It is an intensive property of the system. It is describe as: H=U+PV or h= u+ pv where, H = Total enthalpy U = Total internal energy P = Pressure, V = Volume(Total) h = specific enthalpy u = specific internal energy v= specific volume Enthalpy is also considered as a function of temperature for the case of perfect gases. Hence, it can be written as : L1H=MCpL1T where, L1H= H2- HI = enthalpy difference L1T = T2- T I = temperature difference C, = specific heat at constant pressure Dryness fraction (X) : Dryness fraction is defined as the ratio of the mass of vapour or dry saturated steam to the total mass of wet steam or mass of mixture (water in saturated form in liquid region as well as vapour region).

Badboys2

My Dryness fraction (X) = My + ML . where, My= mass of steam or vapour ML= mass of Liquid (saturated) Drynessfractionis utilized to calculatethe quantity ofliquid or vapour phase within the mixture. The value of dryness fraction varies between 0 and 1. It also determines the quality ofthe steam.

can never be realised because dissipative forces cannot be completelyeliminated. Irreversible Process Any process which cannot be retraced in the reverse direction exactly is called an irreversible process. Most of the processes occurring in the nature are irreversible processes. Examples: (i) Diffusion of gases. (ii) Dissolution of salt in water. (iii) Rusting of iron. (iv) Sudden expansion or contraction ofa gas.

AVAILABILITY AND REVERSIBILITY Available Energy The sources of energy can be divided into two groups (l) High grade energy (2) Low grade energy The conversion of high grade energy to shaft work is exempt from the limitations of the second law, while conversion of low grade energy is subject to them. High Grade Energy (1) Mechanicalwork (2) Electricalenergy (3) Water power

Low Grade Energy (1) Heat or thermalenergy (2) Heat derived fromnuclear

fissionor :fusion (3) Heat derived from combustion

of fossilfuels

(4) Windpower (5) Kineticenergyof a jet (6) Tidalpower

Available Energy Referred to a Cycle. The maximum work outputobtainablefrom a certain heat input in a cyclic heat engine is called the Available Energy (A.E.), or the available part ofthe energy supplied. The minimum energy that has to be rejected to the sink by the second law is called the Unavailable Energy (U.E.), or the unavailablepart of the energy supplied. AE.+U.E. or AE. = Ql - U.E. For the given r. and T2,

11rev. =

T

1- T~

REVERSIBLE AND IRREVERSIBLE PROCESSES Reversible Process Any processwhich can be made to proceedin the reverse direction by variation in its conditions such that any change occurring in anypart ofthe directprocessis exactlyreversedinthe corresponding part of reverse process is called a reversible process. Examples: (i) An infinitesimally slow compression and expansion of an ideal gas at constant temperature. (ii) The process of gradual compression and extension of an elastic spring is approximately reversible. (iii) A workingsubstancetaken alongthe completeCarnot's cycle. (iv) The process of electrolysis is reversible if the resistance offered by the electrolyte is negligibly small. A complete reversible process is an idealised concept as it

Available and unavailable energy in a cycle. For a given T}, 11rev. will increase with the decrease of Tj. The lowestpracticable temperature of heat rejection is the temperature of the surroundings, To. u.E. = Ql- Wmax

Badboys2

Thermal Engineerging

A-88

hrrnx

Wmax

..

TO 1-Tl

and

(1-~~)QI

AE. Qxy- To(Sy- Sx) or u.E. Qxy- Wmax or u.E. = To(S, - Sx) The unavailable energy is thus the product of the lowest temperature of heat rejection, and the change of entropy of the system during the process of supplying heat. Wmax

Availability of a Given System It is the maximum useful work (total work minuspdV work) that is obtainable in a process in which the system comes to equilibrium with its surroundings. It depends on the state of both the system and surroundings. Let U, S, and V be the initial values ofthe internal energy,entropy, and volume of a systemand Uo,So,and Votheir [mal values when the system has come to equilibrium with its environment. The system exchanges, heat only with the environment, and the process may be either reversible or irreversible, the useful work obtained in the process W ::;;(U-ToS+poV)-(Uo- ToSo+PoVo) Let <1>= U - ToS+ PoV where is the availability function and is a compositeproperty of both the system and its environment, with U, S, and V being properties of the system at some equilibrium state, and Toand Po the temperature and pressure of the environment. (In the Gibbs function, G = U - TS + P V,T, and p refer to the system). The decrease in the availability function in a process in which the system comes to equilibrium with its environment is -o=(U - ToS+ PoV)-(Uo- ToSo+PoVo) .. W::;;-o Thus the useful work is equal to or less than the decrease in the availability function.

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Irreversibility of the Process

total volume of the gas. Therefore volume of the gas is equal to volume of the vessel. (4) The molecules of gases are in a state of random motion, i.e., they are constantlymoving with all possible velocitieslying between zero and infinity in all possible directions. (5) Normally no force acts between the molecules. Hence they move in straight line with constant speeds. (6) The molecules collide with one another and also with the walls ofthe container and change there direction and speed due to collision. These collisions are perfectly elastic i.e., there is no loss of kinetic energy in these collisions. (7) The molecules do not exert any force of attraction or repulsion on each other except during collision. So, the molecules do not posses any potential energy.Their energy is wholly kinetic. (8) The collisions are instantaneous i.e., the time spent by a molecule in a collision is very small as comparedto the time elapsed between two consecutive collisions. (9) Though the moleculesare constantlymoving from oneplace to another,the averagenumber ofmoleculesper unit volume of the gas remains constant. (10) The molecules inside the vessel keep on moving continuously in all possible directions, the distribution of molecules in the whole vessel remains uniform. (11) The mass of a molecule is negligibly small and the speed is very large, there is no effect of gravity on the motion of the molecules. Ifthis effect were there, the density of the gas would have been greater at the bottom of the vessel.

Equation of State or Ideal Gas Equation The equation which relates the pressure (P), volume (V) and temperature (T) of the given state of an ideal gas is known as ideal gas equation or equation of state. i.e., PV = nRT where R = universal gas constant Numerical value ofR = 8.31joule mol-1 kelvirr ' n = no. of moles of gas

Behaviour of Real Gases

The actual work done bya system is always lessthan the idealized reversible work, and the difference between the two is called the irreversibility ofthe process. 1= Wmax- W This is also sometimesreferredto as 'degradation' or 'dissipation' . For a non-flow process between the equilibrium states, when the system exchangs heat only with the environment .. I~O I = To [(1\S)system + (~S)SUlT.J Similarly, for steady flow process, I = To(1\Ssystem + ~SSUlT) The same expression for irreversibility applies to both flow and non-tlowprocesses. The quantity To(~Ssystem + 1\SSillT) represents an increase in unavailable energy (or energy).

The gases actually found in nature are called real gases. 1. Real gases do not obey gas laws 2. These gases do not obey the ideal gas equation PV=nRT 3. A real gas behaves as ideal gas most closely at low pressure and high temperature. 4. Equation of state for real gases is given by Vander waal's equation

BEHAVIOUR OF IDEAL AND REAL GASES

According to first law ofthermodynamics, heat given to a system (~Q) is equal to the sum of increase in its internal energy (~U) and the work done (~W) by the system against the surroundings. i.e., 1\Q= ~U + 1\W Heat (~Q) and work done (~W) are the path functionsbut internal energy (~U) is the point function.

Behaviour of Ideal Gases The behaviour of ideal gases is based on the following assumptions of kinetic theory of gases : (1) All the molecules of a gas are identical. The molecules of different gases are different. (2) The molecules are rigid and perfectlyelastic spheres of very smalldiameter. (3) Gas molecules occupy very small space. The actual volume occupied by the molecule is very small compared to the

(p+ :2)tV

-nb) =

nRT

Here a and b are Constant called Vander waal's constant.

ANALYSIS OF THERMODYNAMIC CYCLES RELATED TO ENERGY CONVERSION

Work Let us consider a gas or liquid contained in a cylinder equipped with a movable piston, as shown in Fig. Supposethat the cylinder has a cross-sectional area A and the pressure exerted by the gas at the piston is P.

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Thermal Engineerging

A-89

(iii) If the closed loop is traced in the anticlockwise direction, the expansion curve lies below the compression curve (Wx <Wy),the area of the loop is negative.

I-dx F=P~ The force exerted by gas on the piston F = PA If the piston moves out a small distance dx, the work done dW = Fdx = PAdx = PdV where dV = Adx, is the change in volume of the gas. The total work done by the gas when its volume changes from ~ to Vf VI W =

-------;-;,D-----t> V

L___~A _

_.v

W= area A RCA

Fig.(i)

Fig. (ii)

W= - area A RCT>F.FA

Fig.(iii)

PROPERTIES OF PURE SUBSTANCES 1.

It is a single substance and has a uniform composition. It

V;

2. 3.

has constant chemical composition through its mass. It has a same colour, taste and texture. It has a fixed melting point and boiling point.

P(Vf -VJ =Pi1V

Cyclic Process and Non-cyclic Process If a system having gone through a change, returns to its initial state then process is called a cyclic process. If system does not return to its initial state, the process is called non-cyclic process.

Badboys2

'-----!-f-;-'

JiV = area A RC[)F.A

{J

f PdV

If the pressure remain constant while volume changes, then the work done W =

p

!'

p

p

Types of Pure Substances Two different types of pure substances are : (i) Element: An element is a substance which cannot be split up into two or more simpler substances by usual chemical methods of applying heat, lighting or electric energy, e.g., hydrogen, oxygen, sodium, chlorine etc. (ii) Compound: A compound is a substance made up of two or more elements chemically combined in a fixed ratio by weight e.g. H20 (water), NaCI (sodium chloride) etc. P-T DIAGRAM OF A PURE SUBSTANCE

L----~--

..v

(a) cyclic process

L-------

...

v

(b) Non-cyclic process

Work done in Cyclic Process Suppose gas expands from initial state A to final state B via the pathAXB. p

o'----,:r:c-) -------'-c'--.v The work done in this expansion Wx = + areaAXBCDA N ow gas returns to its initial state B via path B YA. Work done during this compression Wy -area BYADCB The net work done W Wx+ Wy areaAXBCDA -areaBYADCB +areaAXBYA Thus for a cyclic process (i) Work done in complete cycle is equal to the area ofthe loop representing the cycle. (ii) Ifthe closed loop is traced in the clockwise direction, the expansion curve lies above the compression curve. (Wx >Wy), the area ofloop is positive.

If the heating of ice at - 10°C to stream at 250°C at the constant pressure of 1atm is considered 1-2 is solid (ice) heating, 2-3 is melting of ice at O°C,3-4 is the liquid heating, 4-5 is the vaporization ofwater at 100°C, and 5-6 is the heating in the vapour state. The process may be reversed from state 6 to state 1 upon cooling. The curve passing through the 2, 3 points is called the fusion curve and the curve passing through the 4, 5 points (which indicated the vaporization or condensation at different temperature and pressure) is called the vaporization curve. The vapour pressure of a solid is measured at different temperatures, and these are plotted as a sublimation curve. These three curves meet as the tripple point as shown in the figure. The slopes of sublimation curve and vaporization curves for all substance are positive and slope of the fusion curve for more substance is positive but for water, it is negative. The triple point of water is at4.58 mm ofHg and 273.16 K whereas that of CO2 is at 3885 mm ofHg and 216.55 K.So when solid CO2 ( dry ice) is exposed to 1 atm pressure, it gets transformed into vapour, absorbing the latent heat of sublimation from surroundings. Phase equilibrium diagram on P-T coordinates.

1

F.. U's3:m I',2.,3__.,a-.n-e-

-T

Badboys2

Thermal Engineerging

A-90 T-s diagram for a pure substance Consider heating of the system of 1 kg of ice at-5°C to steam at 250°C. The pressure being maintained constant at 1 atm. Entropy increases of the system in different regimes of heating. 2500C

f

(i)

6_

l00'C ---------------

Entropy increase of ice as it is heated from -5°C to O°C at 1 atm. (Cpice = 2.093 kJ/kg-K) ilSl = S2- Sl dQ

=j-= T

T2=273 me dT

j

_p

T1=268

T

273

= mcp

en 268

Fig. (a) p-v- T surface for water which expands a freezing

273

= 1 x 2.093

en 268

Badboys2 (ii) Entropy

= 0.0398 kJ/-K increase of ice as it melts into water at O°C (latent heat offusion of ice = 334.96 kJ kg) ilS2 = S3- S2

mL

334.96

... (wherem = lkg) Entropy increase of water as it is heated from O°C to 100°C

(c Pwater

= 4.187 kJ/kg-K)

T3

ilS3 = S4- S3 = m cp

= 1 x 4.187t'n (iv)

en T2

(:~) p ~T

This equation forms the basis of the h-s diagram of a pure substance, also called the Mollier diagram. The slope of the constant pressure curve on the enthalpy-entropy diagram is equal to the absolute temperature. When this slope is constant, the temperature remains constant. Iftemperature increases, slope of the isobar increases. The constant pressure curve for different pressure can be drawn on the h-s diagram as shown in the figure. States 2,3,4 and 5 are saturation curves.

373 273 = 1.305 kJ/-K

Entropy increase of water as it is vaporized at 100°C, absorbing the latent heat of vaporization (2257 kJ/kg) ilS4 = S5- S4

mL

=T= (v)

Fig. (b) p-v- T surface of a substance which contracts on freezing h-s diagram or Mollier diagram for a pure substance. From the first and second laws of thermodynamics, the following property relations are obtained: Tds= dh-vdp or

= T = -----ri3 = 1.232 kJ/-K (iii)

Fig. (a) shows a substance like water that expand up freezing. Fig. (b) shows substances other than water which contract upon freezing. Any point on the p-v- T surface represents an equilibrium state of the substance. The triple point line when projected to the p- T plane becomes a point.

2257 273 =6.05kJ/kg-K

... (wherem=

1 kg)

Entropy increase of vapour as it is heated from 100°C to 250°C at 1atm.

--7

523

= 1 x 2.093

en 373

=0.706kJ/-K

p-v- T surface for the pure substance. The relation between pressure, specific volume and temperature can be understood with the help ofP-v- T diagram.

(s) Entropy

Figure shows the phase equilibrium diagram of a pure substance on the h-s co-ordinates indicating the saturated solid line, saturated liquid lines and saturated vapour line, the various phases and the transition (liquid + vapour or solid + liquid or solid + vapour) zone.

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Thermal Engineerging

A-91

STEAM TABLES In steam table, properties of water are arranged as a function of pressure and temperature. Saturates steam: Temperature table Specific volume, m3/kg

Internal Energy KJIKG

Temp.

Pressure

Sat

Sat

Sat

°C

kPa,MPa

Liquid

Vapour

Liquid

T

P

Vr

Vg

0.01

0.6113

0.001000 206.132

5

Enthalpy KJIKg

Entropy KJIKgK

Sat.

Sat

Sat

Evap.

Vapour

Liquid

Evap. Vapour Liquid

Or

Org

ug

hr

0.00

2375.3

2375.3

0.00

2501.3 2501.3 0.0000 9.1562

9.1562

0.8721

0.001000 147.118 20.97

2361.3

2382.2

20.98

2489.6 2510.5 0.0761 8.9496

9.0257

10

1.2276

0.001000 106.377 41.99

2347.2

2389.2

41.99

2477.7 2519.7 0.1510 8.7498

8.9007

15

1.7051

0.001001 77.925

62.98

2333.1

2396.0

62.98

2465.9 2528.9 0.2245 8.5569

8.7813

20

2.3385

0.001002 57.790

83.94

23319

2402.9

83.94

2454.1 2538.1 0.2966 8.3706

8.6671

25

3.1691

0.001003 43.359

104.86 2304.9

2409.8

104.87 2442.3 2547.2 0.3673 8.1905

8.5579

30

4.2461

0.001004 32.893

125.77 2290.8

2416.6

125.77 2430.5 2556.2 0.4369 8.0164

8.4533

35

5.6280

0.001006 25.216

146.65 2276.7

2423.4

146.66 2418.6 2565.3 0.5052 7.8478

8.3530

40

7.3837

0.001008 19.523

167.53 2262.6

2430.1

167.54 2406.7 2574.3 0.5724 7.6845

8.2569

45

9.5934

0.001010 15.258

188.41 2248.4

2436.8

188.42 2394.8 2583.2 0.6386 7.5261

8.1647

50

12.350

0.001012 12.032 209.30 2234.2

2443.5

209.31

2382.7 2592.1 0.7037 7.3725

8.0762

55

15.758

0.001015

9.568

230.19 2219.9

2450.1

230.20

2370.7 2600.9 0.7679 7.2234

7.9912

60

19.941

0.001017

7.671

251.09 2205.5

2456.6

251.11

2358.5 2609.6 0.8311 7.0784

7.9095

65

25.033

0.001020

6.197

272.00 2191.1

2463.1

272.03

2346.2 2618.2 0.8934 6.9375

7.8309

70

31.188

0.001023

5.042

292.93 2176.6

2469.5

292.96

2333.8 2626.8 0.9548 6.8004

7.7552

75

38.578

0.001026

4.131

313.87 2162.0

2475.9

313.91

2321.4 2635.3

1.0154 6.6670

7.6824

80

47.390

0.001029

3.407

334.84 2147.4

2482.2

334.88

2308.8 2643.7

1.0752 6.5369

7.6121

85

57.834

0.001032

2.828

355.82 2132.6

2488.4

355.88

2296.0 2651.9

1.1342 6.4102

7.5444

90

70.139

0.001036

2.361

376.82 2117.7

2494.5

376.90

2283.2 2660.1

1.1924 6.2866

7.4790

95

84.554

0.001040

1.982

397.86 2102.7

2500.6

397.94

2270.2 2668.1

1.2500 6.1659

7.4158

100

0.10135

0.001044 1.6729 418.91 2087.6

2506.5

419.02

2257.0 2676.0

1.3068 6.0480

7.3548

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Sat

Sf

Sat Evap.

Vapour

Srg

Sg

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Thermal Engineerging

A-92

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Badboys2

Thermal Engineerging

A-93

P=10 kPa. (45.81)

P = 100 kPa. (99.62)

P=50 kPa.(81.33)

T

v

u

h

s

v

u

h

s

v

u

h

s

Sat

14.674

2437.9

2584.6

8.1501

3.240

2483.8

2645.9

7.5939

1.6940

2505.1

2675.5

7.3593

50

14.869

2443.9

2592.6

8.1749

100

17.196

2515.5

2687.5

8.4479

3.418

2511.6

2682.5

7.6947

1.6958

2505.6

2676.2

7.3614

150

19.513

2587.9

2783.0

8.6881

3.889

2585.6

2780.1

7.9400

1.9364

2582.7

2776.4

7.6133

200

21.825

2661.3

2879.5

8.9037

4.356

2659.8

2877.6

8.1579

2.1723

2658.0

2875.3

7.8342

250

24.136

2736.0

2977.3

9.1002

4.821

2735.0

2976.0

8.3555

2.4(0)

2733.7

2974.3

8.0332

300

26.445

2812.1

3076.5

9.2812

5.284

2811.3

3075.5

8.5372

2.6388

2810.4

3074.3

8.2157

400

31.053

2968.9

3279.5

9.ffJ76

6.2W

2968.4

3278.9

8.8641

3.1026

2967.8

3278.1

8.5434

500

35.679

3132.3

3489.0

9.8977

7.134

3131.9

3488.6

9.1545

3.5655

3131.5

3488.1

8.8341

600

40.295

3302.5

3705.4

10.1608

8.058

3302.2

3705.1

9.4117

4.0278

3301.9

3704.7

9.(1)75

700

44.911

3479.6

3928.7

10.4028

8.981

3479.5

3928.5

9.6599

4.4899

3479.2

3928.2

9.3398

800

49.526

3663.8

4159.1

10.6281

9.~

3663.7

4158.9

9.8852

4.9517

3663.5

4158.7

9.5652

54.141

3855.0

4396.4

10.8395

10.828

3854.9

4396.3

10.0967 5.4135

3854.8

4396.1

9.7767

HID

58.757

4053.0

4640.6

11.0392

11.751

4052.9

4640.5

10.2964 5.8753

4052.8

4640.3

9.9764

1100

63.372

4257.5

4891.2

11.2287

12.674

4257.4

4891.1

10.4858 6.3370

4257.3

4890.9

10.1658

1200

67.987

4467.9

5147.8

11.4090

13.597

4467.8

5147.7

1O.6(i)2 6.7986

4467.7

5147.6

10.3462

1300

72.ffJ3

4683.7

4400.7

11.5810

14.521

4683.6

54ffi.6

10.8382 7.2ffJ3

4683.5

54ffi.5

10.5182

Badboys2 <XX>

-

P = 200 kPa (120.23)

-

-

-

-

-

-

-

P = 400 kPa (143.65)

P =300 kPa (133.55)

Sat.

0.88573

2529.5

27~.6

7.1271

0.60582

2543.6

2725.3

6.9918

0.46246

2553.6

2738.5

6.8958

150

0.95964

2576.9

2768.8

7.2795

0.63388

2570.8

2761.0

7.0778

0.47084

2564.5

2752.8

6.9299

200

1.08034 2654.4

2870.5

7.5056

0.71629

2650.7

2865.5

7.3115

0.53422

2646.8

2860.5

7.17~

250

1.19880 2731.2

2971.0

7.7005

0.7%36

2728.7

2%7.6

7.5165

0.59512

2726.1

2964.2

7.3788

300

1.31616 2800.6

3071.8

7.8926

0.87529

28~.7

3059.3

7.7022

0.65484

2804.8

3066.7

7.5(i)1

400

1.54930 29fl)'7

3276.5

8.2217

1.03151 2965.5

3275.0

8.0329

0.77262

2964.4

3273.4

7.8984

500

1.78139 3130.7

3487.0

8.5132

1.18669 3130.0

3486.0

8.3250

0.88934

3129.2

3284.9

8.1912

600

2.01297

3301.4

3704.0

8.7769

1.34136 3300.8

3703.2

8.5892

1.00555 3300.2

3702.4

8.4557

700

2.24426

3478.8

3927.7

9.0194

1.49573 3478.4

3927.1

8.8319

1.12147 3477.9

3926.5

8.6987

800

2.47539

3663.2

4158.3

9.2450

1.64994 3(i)2.9

4157.8

9.0575

1.23722 3662.5

4157.4

8.9244

Badboys2

Thermal Engineerging

A-94

The efficiency of the Rankine cycle is given by

RANKINE CYCLE This is a reversible cycle. When all the following four processes are ideal, the cycle is an ideal cycle called Rankine cycle. Flow Diagram of Rankine Cycle

(hI -h4)-(h2

(h, - h4) where,

High pressure, high temperature steam

-h3)

Q 1 = heat transferred to the working fluid

Q2 = heat rejected from the working fluid W T = work transferred from the working fluid

Process 1

W p = work transferred into the working fluid r

~,

:

RANKINE CYCLE WITH REHEATER

I I

Boiler

The flow diagram for the ideal Rankine cycle with reheat is shown in fig.

I I I I I I

I I

I I I

I

Air _J T and Combusfuel tion products Process High pressure water

Simple steam power plant Process 1 : Reversible constant pressure heating process of water to form steam in steam boiler. Process 2: Reversible adiabatic expansion of steam by turbine. Process 3: Reversible constant process of heat rejection as the steam condenses till it becomes saturated liquid. This is by condenser. Process 4: Reversible adiabatic compression of the liquid ending at the initial pressure by the pump. Rankine cycle Plot on p-v, t-s and h-s Planes

Badboys2

Pump Wp

1"

In this cycle, the expansion of steam from the initial state 1to the condenser pressure is carried out in two or more steps depending upon the number of reheats used. In this case efficiency,

QI = h, -h6s + h3 - h2s; Q2 = h4s - hs WT = h, - h2s + h3 - h4s; Wp = h6s - hs

(a)

In practise, the use of reheat only gives a small increase in cycle efficiency, but it increases the net work output by making possible the use of higher pressures, keeping the quality of steam at turbine exhaust within a permissible limit. By increasing the number of reheats, still higher steam pressures could be used, but the mechanical stresses increase at a higher proportion than the increase in pressure, becuase ofthe prevailing high temperature.

... s (c)

Badboys2

Thermal Engineerging

A-95

RANKINE CYCLEWITH REGENERATOR AND REHEAT The effect of reheat alone on the thermal efficiency of the cycle is very small. Regeneration or the heating up of feedwater by steam extracted from the turbine enhances the efficiency of the cycle. Flow diagram of Rankine Cycle with regeneration and reheats shown in Fig.

Process 1: Air is compressed reversibly and adiabatically. Process 2: Addition of heat reversibly at constant pressure. Process 3: In the turbine, air expands reversibly and adiabatically. Process 4: From the air heat is rejected reversibley at constant pressure. Efficiency of Brayton Cycle:

11= 1- Q2

= 1-

T4 - Tl T3 - T2

Q} As

Now

...(i)

Q 1= heat supplied = mcp (T 3 - T2) Q2 = heat rejected = mcp (T4 - T 1)

T2

T

(Y-l)/Y

P2

3.

T; = ( Pt J

= T4

(Since PI = P3' and P4 = PI)

T4 -1 = T3 -1 T} T2 or

T4 - T1 T1 ( PI T3 - T2 = T2 =

\ (Y-l)/y

lpJ

(

I

If r k = compression ratio = [from Eq. (i)]

= V

v2

\ y-l

l~)

I

/v2 the efficiency

becomes

Y-l

( J

11= 1- :~

Badboys2

or

11Brayton =

1-

1

...(ii)

y-l rk

If r p = pressure ratio = P2/P 1 the efficiency may be expressed in the following form also Here, WT = (hI -h2) + (1-m1)(h2 - h3) + (1-m1)(h4 -h5) + (I-ml -m2)(h5 - h6) + (1-ml -m2 -m3)(h6 - h7) kJ/kg. Wp= (I-m} -m2 -m3)(h9-hg) + (I-ml -m2)(hll -hlO) + (1- m.) (h13 - h12) + I(h15 - h14) kJ/kg Q1 = (hI - h15) + (1- m1)(h4 - h3) kJ/kg and Q2 = (I-ml -m2 -m3) (h7 - hg) kJ/kg The energy balance of heaters 1, 2 and 3 give m 1h2 + (1 - m 1) h 13= 1 x h 14 m2h5+(I-ml-m2)hl1 =(I-ml)h12 m3h6 + (I-m} -m2-m3) h9= (I-m1-m2)h10 from which m I> m2 and m3 can be evaluated.

BRAYTON CYCLE It is the air standard cycle for the gas turbine power plant.The flow diagram of Brayton cycle is shown in Fig.

(

11= or

,(y-l)/y

l-l:~)

11 Brayton -

1 (rp

i

1 y-l )/y

The efficiency of the Brayton cycle, therefore, depends upon either the compression ratio or the pressure ratio.

BRAYTON CYCLE WITH REGENERATOR REHEATER

AND

In the regenerator, the temperautre of air leaving the compressor is raised by heat transfer from the turbine exhaust. The maximum temperature to which the cold air at 2 could be heated is the temperature of the hot air leaving the turbine at 5. This is possible only in an infinite heat exchanger. In the real case, the temperature at 3 is less than that at 5. We+---

Compressor

...(iii)

Turbine

Heat exchanger

(a)

Badboys2

Thermal Engineerging

A-96

4

_____.

S

(b)

.. s (b)

Effectiveness of the regenerator: E

t3 = ---

t2

actual temperature = ------=----------

Ts - T1

rise of air

maximum possible rise of temperature

Efficiency of Brayton cycle with regenerator.

Badboys2

Q} 11= 1- Q2

= 1-

T6 - T} T} [(T2 lTd-I] T4 - T3 = 1- T4 1- (Ts I T4)

T} T2 [1-(T} IT2)] 1-(Ts IT4)

f

·r;

= 1- T4

T} y-1/y 11--1 --y T4 p

~v

For a fixed ratio of (T /T 4) the cycle efficiency drops with increasing pressure ratio.

(c)

Effect of Reheat on Brayton Cycle: In the cycle 4-5-6-4', rp is lower than in the basic cycle 1-2-3-4', so its efficiency is lower. Therefore, the efficiency of the cycle decreases with the use of reheat. But T 6 is greater than T'4. Therefore, ifregeneration is employed, there is more energy that can be recovered from the turbine exhaust gases. So when regeneration is employed in conjunction with reheat, there may be a net gain in cycle efficiency.

AIR STANDARD CYCLES

1. 2.

3. (a)

Air standard cycles are utilized for the purpose of analysing the working ofInternal combustion Engine. It is considered to be an idealized cycle and the following assumptions are made for analysis: The working fluid is taken as air for the cycle and assumed to be a perfect gas. The engine is assumed to working in a closed cycle. There is no change of mass of the working medium. All the processes that conclude the cycle is of reversible nature.

Badboys2

Thermal Engineerging 4. 5. 6. 7.

A-97

Heat is considered to be provided from a constant High temperature source and not be chemical combustion. Heat loses are negligible. The value of specific heats of workings substance remain constant throughout the cycle. Kinetic energy and Potential energy remain constant throughout the process.

AIR-STANDARD OTTO CYCLE The air-Standard Otto Cycle of spark-ignition (SI) engine. It is named after N.A.Otto a German engineer who first built a fourstroke engine in1876. In most (S-I) engines, the piston executes four complete strokes within the cylinder and crankshaft completes two revolutions for each, thermodyamic cycle. These engines are called four-stroke internal combustion engine.

The schematic diagram of each stroke is shown in fig.

1. Induction

IVO

2. Compression Both value closed

Jl~

3. Expansion

4. Exahust

EVO

~It

rl~

Spark

0

0 0

o TDC or inner

dead centre Badboys2

(IDC)

BDC or outer _____. dead centre

(ODC)

(a)

(b)

Indicator diagram:

(c)

(d)

on the piston which moves to the right and the pressure and temperature of the gases decrease. Process 5-6, Blow-down : The exhaust valve opens, and the pressure drops to the initial pressure. Process 6-1, Exhaust: With the exhaust valve open, the piston moves inwards to expel the combustion products from the cylinder at constant pressure.

The efficiency of air-standard otto cycle 2,6

11 = 1- Q2

=1

Q1

mcy (T4 - T}) mCy(T3 - T2)

1 DC

ODC ---.~ V Process 1-2, Intake: The inlet valve is open, the piston moves to the right, admitting fuel-air mixture into the cylinder at constant pressure. Process 2-3, Compression: Both the valves are closed, the piston compresses the combustible mixture to the minimum volume. Process 3-4, Combustion: The mxiture is then ignited by means of a spark, combustion takes place, and there is an increase in temperature and pressure. Process 4-5, Expansion: The products of combustion do work

Y-1

Process 1-2,

Process 3-4,

T2 T1

= ~

( v2 J

=1

(T4 - T}) (T3 - T2) ...(i)

Badboys2

Thermal Engineerging

A-98

The indicator diagram of diesel cycle or 3

4

r Y-l .

V2

( J

Fromeq. (1) 11= 1- ~

11otto = 1-

or

--_'.~

1 y-l

...(ii)

rk

whe-e rk is called the compression ratio and given by

rk

=

Volume at the beginning of compression Volume at the end of compression

VI = ~ V2 v2

The efficiency of the air standard Otto cycle is thus a function of the compression ratio only. The higher the compression ratio, the higher the efficiency. It is independent of the temperature levels at which the cycle operates. The net work output for an otto cycle

Badboys2

V

Process 1-2, Intake: The air valve is open. The piston moves out admitting air into the cylinder at constant pressure. Process 2-3, Compression : The air is then compressed by the piston to the minimum volume with all the valves closed. Process 3-4, Fuel injection and combustion: The fuel valve is open, fuel is sprayed into the hot air, and combustion takes place at constant pressure. Process 4-5, Expansion: The combustion products expand, doing work on the piston which moves out the maximum volume. Process 5-6, Blow-down: The exhaust valve opens, and the pressure drops to the initial pressure. Pressure 6-1, Exhaust: With the exhaust valve open, the piston moves towards the cylinder cover driving away the combustion products from the cylinder at constant pressure. The above processes constitute an engine cycle, which is completed in four strokes of the piston or two revolutions of the crank shaft. The efficiency of diesel engine:

Now,

11= 1- 21._ = 1- mcv (T4 - T1) = 1 (T4 - TI) QI mCp(T3 - T2) Y (T3 - T2) Here, El_ = P4 = rp (say) P2 PI Wnet = PlVl (P3V3 _ P4V4 _P2V2 +IJ y-l PlVl PlVl PlVl

W

= PIVI (r -1) (£Y-l_l) net y- 1 P k

Ql = Q2 _ 3 = mcp (T3 - T2) = heat supplied Q2 = Q4-1 = me, (T4 - T 1)= heat rejected Efficiency in terms of compression ratio, expansion ratio and cutoffratio. Compression ratio,

vI v2

V4 v4 -_-e - V3 - v3

Expansion ratio,

[

Cut-off ratio,

[

V3 v3 -_-c - V2 - v2

It is see that

rk=re·rc

Process 3-4

T4 =(~JY-I T3 v4

DIESEL CYCLES It is a compression-ignition (CI) engine proposed by Rudolph Diesel in 1890s. It is very similar to SI engine differing mainly in the method of initiating combustion. In diesel cycle or combustion engine during the compression stroke only air is compressed however in SI engine air-fuel mixture is compressed.

VI rk =-=V2

rr

l

y-l

rc T4 =T3--l

rYk

Process 2-3

T2 = P2v2 =~=_ T3 P3v3 v3

rc

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=1- Q2 =1- mCp(Tc-TB)+mcp(TD-TC) Q1 mCv(TE- TA) 11=1

Process 1-2 T1=T2

T3

1

1

.--=-.-"y-1 r "y-1 "k e J.k

Substituting the values of T l' T2 and T4 in the expression of efficiency. T

y-1

l__ T3 _1_ 3 . ry-1 re ry-1 k

11= 1

1 11Diese1 = 1--. Y

As r c > 1, =

-IJ Y1 (ry~ -1

k

1

rJ-l

y-1 . --1 rk re-

is also greater than unity. Therefore, the

efficiencyof the Diesel cycle is less than of the Otto cycle for the Badboys2 same compression ratio. Dual cycles Dual cycle is also called as limited pressure cycle. In this cycle, the addition of heat is done partly at constant pressure values and partly at constant volume values.

1 C~D ~ ())

Isentropic process

B

~ tZl

E

£

1\

Volume(V)

~

Process A - B : Reversible adiabatic compression Process B - C : Constant volume heat addition Process C - D : Constant pressure heat addition Process D - E : Reversible adiabatic expansion Process E - A : Constant volume heat rejection constant volume T

I

B ~

S

Heat supplied, Q, = meP (T, - TB + meP (TD - TC) Heat rejected, Q2 = mcv (TE - TA) We know that, efficiency (11)

Q1-Q2 Q1

(TE-TA) (Tc -TB)+y(TD -Tc)

where, y=specific heat ratio. The compressionratio (C.P), Expansion ratio (E.R) and cut - offratio (C. R) can be given as follows: C.R= VA ,E.R= VE ,C.R= VD VB VD Vc Comparison of Otto, diesel and dual cycles The three cycles are compared on the basis of their compression ratio, heat input, work output, etc. (a) For same heat input, the temperature attrained is maximum forOtto cycleand minimum for dieselcycle. Also for same heat input, efficiency of Otto cycle is maximum while that ofDieselcycleis minimum. (b) For same maximum pressure and same heat input, the efficiency of Diesel cycle is more than Dual cycle. (c) For same pressure and temperature, the efficiency of Diesel cycle is more than dual cycle. Steam boilers Steam boiler is basically a closed vessel into which water is heated until the water is converted into steam at required pressure by combustion of fuel. In this, fuel is generally burnt in a turnace and hot gases are produced. These hot gases come in contact with water vessel and the heat of hot gases is transfered to water and steam is produced. This steam is fed through pipes to the turbine of thermal power plant. Applications (i) Steam boilers are utilized as generators for the production of electricity in the energy sector. (ii) Steam boilers are used in agriculture and soil steaming. (iii) Steam boilers are also used for heating the building in cold weather. Classification of steam boilers Steam boilers are classified based on the following basis: (i) Steam pressure (ii) Firing method (iii) Tube contents (iv) Circulation of water (v) Heat source (vi) Stationary or Portable (vii) Position (viii)Passage of gas (ix) Draught nature (i) Steam pressure: Steam boilers are classified according to pressure as : follows: (a) Low pressure boiler: It is described as a boiler which developes pressure of the steam whose value of below 80 bar. Examplesare Cochran, Lancashire,Locomotive boilers etc. (b) High pressure boiler : It is defined as the boiler in which steam is developedat more than 80 bar pressure. Examples are Babcock and Wilcox, Benson,Lamount etc.

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(ii) Firing method: Steam boilers are classified according to firing method as follows: (a) Internally fired boiler: In this type, the boiler shell contains the turnace and the turnance is totally covered by water cooled surfaces. Examples are Lancashire boiler, Locomotive, Scotch etc. (b) Externally fired boiler : In this type, the region of turnace is constructed outside the boiler shell. Example: Babcock and wilcox boiler.

(viii)Passage of gas:

(iii) Tube contents :

Difference between fire tube boiler and water tube boiler:

Steam boilers are classified as according to the contents of the tube as follows: (a) Fire tube boiler : In this type, the hot gases are available inside the tubes and the water surrounds the tube. The combustion of hot gases takes place and the products of combustion pass through the fire tubes (surrounded by water). The excess heat of hot gases is transferred to water and transformed into steam. The exhaust gases are discharge through chimney. Examples are Cohran, Lancashire, locomotive boilers etc. (b) Water tube boilers: In this type, water flows inside the tubes while the hot gases passes outside the tubes. The tubes are generally surrounded by the products of combustion of hot gases : Examples are Babcock and wilcox, stirling boilers etc.

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Steam boilers are classified according to gas passage as : (a) Single - Pass (b) Multi - Pass

(ix) Draught nature: Steam boilers are classified according to draught nature as follows: (a) Natural draught boilers: In this type, Natural gas or air circulation develops draught. (b) Forced draught boilers: In this type, mechanical systems like tans develops draught. Fire Tube Boiler Hot gas es are inside the tubes while water out side the tubes Firing is done internally

Water tube boiler Water is available inside the tube while hot gases outside the tubes. Firing is done externally.

Operatinal pressure 16 bar

Operational pres sure is high as upto 100 bar

is upto

Less amount of steam is generated It requires large floor area for g rven power It is not generally used in large power plants Boiler shell diameter is large for given power Parts are not accessible easily for the purpose of maintenance Efficiency is less

High amount of steam is generated It requires les s floor area for the same power It is used in large power plants Boiler shell diameter is small for the same amount of given power Parts are easily accessible for the purpose of maintenance Efficiency is high

Initial cost is less

Initial co s t is high

It has a large ratio ofwater content to s team capacity

It has a comparatively small ratio ofwater content to s team capacity Quick in evaporation

(iv) Circulation of water: Steam boilers are classified as according to water circulation as follows: (a) Natural circulation: In this type, water circulation within the boiler takes place by natural convection current produced by the application of heat. Example are Lanca-shire, Locomotive, Babcock and wilcox boiler etc. (b) Forced circulation: In this type, circulation is done by means of forces pumps along with natural circulation for the purpose of increasing the circulation. Examples are Lamont, Velox, Benson boilers etc.

(v) Heat source: It may be of the following types; (a) Combustion offuel in solid, liquid or gaseous form (b) Electrical energy (c) Nuclear energy (d) Hot waste gases (by - products of chemical processes) (vi) Stationary or Portable (a) Stationary boilers: Stationary plants use stationary type of boilers. (b) Portable boilers : These are those boilers which are easily de-assembled and transported from one place to other place.

(vii) Position: According to position, follows (a) Horizontal (c) Inclined

steam boilers may be classified (b) Vertical

Slow in evaporation

Boiler Mountings and accessories: Boiler mountings and accessories are fittings and devices which are utilized for safe and efficient operations. (i) Boiler Mountings: These are components which are mounted on the boiler body for the purpose of safety and control of the steam generation. Different Boiler mountings are given as follows: (a) Water level indicator: It provides and indication of the water level in the boiler constantly. It is also termed as water gauge. (b) Pressure gauge: It is utilized for the purpose of measurement of pressure inside the vessel. It is mounted on the top front of boiler shell. (c) Safety values: It is utilized for the purpose of release of excess amount of steam in a condition when steam pressure must have at least two safety values various types of safety values are given below: ---+ Dead weight safety value ---+ Liver safety value

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x.toi

Thermal Engineerging ~ Spring loaded safety value ~ High steam and Low water safety value. (d) Fusible plug: It provides a protection of the boiler against any kind of damage because of overheating for law water level. It is fixed over the combustion chamber. (e) Blow - off cock: It discharge a water partion during operation of boiler to blowout mud or scale at regular intervals. It is also used for emptying boiler for the purpose of cleaning, inspection and repair. (f) Feed check value: It provides a proper control of water supply to the boiler and to provide the prevention of escaping of water from the boiler if the pump pressure is less. (g) Stop value : It provides a regulation of the flow of steam from one steam pipe to other steam pipe or from boiler to steam tubes. (h) Boiler Accessories: Various types of boiler accessories are given as follows: (a) Feed pumps: It is used for delivering feed water to the boiler. Rotary pump and Reciprocating pumps are generally used as feed pumps. (b) Injector: It is utilized for feeding water into the boiler. It is employed in vertical and locomotive boilers. It can also be used in place of feed pumps where the space availability is less. It is very low in cost, simple and its thermal efficiency is very high. (c) Economiser: It is described as a device which uses the waste heat of hot flue gases for the purpose of heating the feed water which is further supplied to the boiler.

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% saving in fuel consumption

Cp(TB-TA) H_H

dynamic action of steam turbine plays an important role during the working of steam turbine. The steam is impinged with a high pressure via nozzle and because of this impingement, a specific amount of heat energy is transformed into kinetic energy. Now, the high velocity steam particles strike on the blade of turbine resulting in change in direction of motion and develops a momentum. Steam flows with high velocity like jets and moving parts of turbine transforms these high velocity sets into mechanical work which rotates the shaft of generator and thus rotation of shaft gives rise to the generation of electricity.

Classification of turbine: Turbines are classified based on the principle of operation given as follows; (a) Simple Impulse turbine: In Simple impulse turbine, rotar is connected to the shaft which provides useful power. It is a rotating element of turbine which consists of moving blades. In the middle portion of the turbine, the nozzle and blades are available. Nozzle works as a passage for steam flow where high pressure energy of steam is converted into kinetic enegry. Simple impulse turbine is also known as De Lavel turbine. In this type, only one set of blades is fixed to the wheel through which conversion of kinetic energy of steam into mechanical work takes place. It has a very large ratio of expansion i.e., the velocity of steam is very high (1000 m/s).

x 100

w

where, C, = specific heat at constant pressure T B = Temperature of heated feed water T A = initial feed water temperature H= Enthalpy H w = Sensible heat of water (d) Air Pre-heater : It is used for the purpose of increasing or raising the air temperature before entering if into the furnace. The position of air - preheater is always after economiser. Tube type, plate tube and storage type of air preheaters are used. (e) Superheater: It is a device which is used for the purpose of increasing the steam temperature which is higher at its saturation temperature. It can be utilized in fire tube and water tube boilers. Due to superheater, consumption of steam is reduced and efficiency of the steam plant is increased. (f) Steam separator: It is device which separates the increased water molecules from the steam which is passing to the turbined. (g) Steam trap: It is device which is utilized for the purpose of draining the condensed steam from steam pipes, steam separators etc. So that no steam could be escaped.

Steam turbines: In steam turbines, conversion of high pressure and high temperature steam into mechanical energy takes place. The

(b) Pressure Compounded impulse turbine : In case of simple impulse turbine, high velocity steam flows through moving blades produces a high amount of rotational speed which should be avoided for practical purposes. In this type, a number of simple impulse turbines in series are mounted on a single common shaft. Each simple impulse turbine is said to be the stage of the turbine. Each stage consists of its own nozzles and blades. The steam coming out from the boiler passes through first nozzle where its pressure is decreased and velocity is increased. This high velocity steam is directed towards the moving blades of first stage result absorption of all of its velocity. Hence the steam pressure does not is being apsorbed. Thus the total

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pressure absorbed is equally (nearly) divided in stages which reduces the steam velocity entering into moving blade.

Pressure compounding (c) Simple velocity compounded impulse turbine: This type of turbines consistsof a setof nozzles and moving blades rows which are connected the shaft whereas the fixed blade rows are connected to casing. Ithas in general, moving and fixed blades. The steam at a very high pressure coming out from boiler is expanded in the nozzle where pressure energy is transformed into the kinetic energy.The steam at high velocity is impinged on the first stage of moving blades and steam flows through blades loosing some amount of velocity due to the fact that some part of momentum imported by blade. High kinetic energy is absorbed and steam velocity remains constant while passing through fixed blades. The process is repeated fill all of steam energy is being absorbed.

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(e) Reaction turbine: In the reaction turbines, steam from the boiler first flows through guiding mechanism and then flows through the moving blades. The kinetic energy is minimised. The pressure energy does not change before striking. When the steam passes through moving blades, then is a difference of pressure between inlet and outlet tips due to w_hichpressure is decreased while passing through movmg blades. The fixed blades are considered to change the steam direction and at the same time allowing to get expanded to higher velocity values. The steam pressure decreases when if passes over the moving blades. Fixed

P -I+-dH(,+ll+-dH ....................

Entropy (~) ~

3

2-+1

Vr2>Vr

j

Steam flow through turbine blades: (some important formulas) )1

(d)

Velocitycompounding Pressure velocity compounded turbine: Itis the combinationarrangement of pressure compounding and velocity compounding. The decrease in total pressure of steam is split in stages and velocity of each stage is compounded. A High amount of pressure is decreases in this type of turbine and so lesser number of stages are required which gives rise to design of a smaller turbine for similar value of pressure drop. The efficiency is very low and it is very rarey used nowadays.

VW2

Velocity triangle of steam turbine

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11. Maximum efficiency of reaction turbine;

Let us consider the following: u = velocity (Linear) of blade

11max

v WI= velocity of whirl at inlet

= flow velocity at inlet

v f2

= flow velocity of output

Vrl

= relative velocity at inlet

vr2

= relative velocity at output

VI = Absolutive velocity of steam at inlet v 2 = Absolute velocity of steam at output u = nozzle angle J3 = Angle with which discharged steam makes with tangeal of wheel = Inlet angle of moving blade = exit angle of moving blade co = steam weight which flowes through blade Qv = volume of steam flowing through blades D = diameter of blade drum

e

1.

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w

Work done on blades/sec. W

Horse power

=

g

2cos2 a 2

l s-cos U Comparison between Impulse and reaction turbine:

VW2= velocity of whirl at outlet v fl

=

= -(VWl g

- VW2)X u

u(VW1-VW2) 75

Impulse turbine Only kinetic energy is utilized for the purpose of rotation of turbine

Reaction turbine Kinetic energy and Pressure energy both are utilized for the purpose of rotation of turbine

Water flows over the nozzle followed by striking the blades Pres sure is decreas ed in nozzle and not in moving blades

Guide mechanism directs the water for the purpose of flowing over turbine Pressure is decreased in fixed blades (nozzles) and also in moving blades

The type of the blades is profile Low power is

The type of the blade is aerofoil High power is produced

Low efficiency For same power generations ifrequires less space

High efficiency For same power generations, it requires more space

Internal combustion Engine: (IC Engine) : An I.e. Engine is defined as a heat engine in which the combustion of fuel is occured in the presence of air and this results in releasing the energy in the cylinder of an engine.

3.

Comparison between I.C. Engine and Steam angine :

v2

4.

Energy supplied to blades/kg of steam

5.

Blade efficiency =

6.

Stage efficiency

= 2~

Steam Engine In steam engine, combustion takes place outside the engine. Steam engines are operated at temperature values of around 600° C

2u(vW1 -VW2) 2 vI

work done on blade

(11) = Total energy supplied per stage

(11) =

It requires cooling due to high operational temperatures The enhaust of steam The exhaust of an l.C. Engine engine is utilized as teed is exited to the atmosphere. water Instantaneous use for Instantaneous use of I.e. steam engine is not Engine is possible pos sible

11 = blade efficiency x nozzle efficiency Axial force on wheel

8.

If blade speed

=

=u=

w (v fl - Vf2 ) g vI CoSU

2

.fhen efficiency

impulse turbine 9. If blade speed = u = vcosa, then efficiency reaction turbine is maximum. 10. Maximum efficiency of impulse turbine, cos2 a ( vr1 cos '\

11max

=

-l 2

vr2 cosn

+ 1)

Operation one. Engines done at temperature values of about 2400°C

It does not require cooling

(VWl-VW2)u g x ~h x J

7.

LC. Engine In r.c Engine, Combustion takes place inside the engine.

of

of

Weigh t to power ratio is Weight to power ratio is low high Steam engine has a very I.e. Engine has higher value low value ofefficiency of efficiency i.e.30-36% i.e. 15-20%

Classification ofI.C. Engine: An I.e. Engine can be classified as follows: (i)

According to the type of fuel used

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(a) Diesel engine (b) Petrol engine (c) Cras engine (ii) According to method of igniting the fuel (a) Spark Ignition Engine (b) Compression Ignition Engine (iii) According to number of strokes/cycle (a) Two stroke cycle engine (b) Four stroke cycle engine (iv) According to cycle of operation (a) Diesel cycle engine (b) Oho cycle engine (c) Dual cycle engine (v) According to member of cylinders (a) Single cylinder engine (b) Multi cylinder engine (vi) According to cooling system (a) Air cooled engine (b) water cooled engine (c) oil cooled engine (vii) According to cylinder position (a) Horizontal engine (b) vertical engine (c) V -engine (d) Radial engine (viii) According to speed of the engine (a) Low speed engine (b) Medium speed engine (c) High speed engine Four stroke cycle engine: Most of the I.C. Engines work on the basis of four stroke cycle engine. Following is the sequence of operations are given in four stroke cycle engine: (a) Suction Stroke: To start with, the piston is near to Top dead center (TD.S) and the inlet value is open and exhaust value is closed.As the piston moves from T.D.C to B.D.C, the charge is to rush in and fill the space created by piston. The charge consists of air-fuel mixture. The admission of charge inside cylinder continuous until the inlet value is closed. (b) Compression stroke: In this stroke, both values are closed and the piston moves from B.D.C to TD.C. The charge is compressed upto compression ratio of 5 : 1 to 9 : 1 and the pressure and temperature at the end of compression are 6 - 12bar and 250 - 300° C respectively. (c) Power or Expansion stroke: When the piston reaches TD.C. position, the charge is ignited by causing an electric spark by spark plug. During combustion process, chemical energy of fuel is released and there is a rise in temperature and pressure of gases. The temperature of gases increases to 1800 - 2000°C and pressure reaches 30-40 bar. Now the combustion products expand and push the piston down the cylinder. The reciprocating piston motion is converted into rotary motion of crankshaft by a connecting rod and crank. During expansion,pressure decreasesdue to increase in volume of gases and heat absorption by cylinder walls. (d) Exhaust stroke: In this stroke, the exhaust value is opened at the end ofworking strokes when the piston is at

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B.D.C. But actually exhaust value begines to open when about 85% of working stroke is completed. A pressure (45 bar) forcesabout 60% ofburnt gases into exhaustmaxifold at high speed. The remaining burnt gases are cleared off the swept volume when the piston moves from B.D.C to T.D.C.The exhaust value opens and inertia of flywheeland other moving parts push the piston back to TD.C, forcing the exhaust gases out through the open exhaust value. At the end of exhaust stroke, the piston is at TDS and one operating cycle has been completed. Engine components: There are various components of an I.C. Engine which are grven as: (a) Cylinder: In this, the burning offuel takes place and power produced. (b) Cylinder head: One end of the cylinder is covered by it and if also consists of the values. (c) Piston: The other end of the working space of cylinder is coveredby the piston. Powerproduced due to combustion products is transmitted to the crank shaft. (d) Connecting rod: It changes and transmits the translatory motion of the piston to rotating crank bin during working stroke. (e) Crank shaft: Crank shaft is used to transmit the work from the piston to driven shaft. (f) Crank webs (g) Main bearings (h) Crank pin and bearing (i) Fuel Nozzle: It is used for delivering fuel into combustion chamber through an injection system having pump. G) Piston rings: It provides a gas high seal between the piston and the liners. (k) Intake value: It permists the fresh air to enter. (1) Exhaust value : The combustion products are exhausted through this value after working. (m) Cam - shaft (n) Cam (0) Rocker arms (P) Value - springs (q) Crank case : It holds the cylinder piston and crank shaft. (r) Flywheel (s) Bed plate (t) Cooling water jackets Carburettor It is a device which is used for the purpose of atomising and vaporising the fuel and if also mixes fuel and air in different preportions for the charging purposes. Air Fuel mixture: There is a range of air-fuel mixtures through which combustion takes place. The limits of these ranges of airfuel mixtures are given below: Upperlimit => 20 : 1 Lower limit => 7: 1to 10: 1 Two- stroke cycle Engine: In two stroke cycle engine, the following four operations are given as follows: (a) air-induction (b) air compression and fuel injection

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(c) expansion (d) release and exhaust Engine Performance: Engine perfomance given an indication the efficiencywith which the conversion of chemical energy of fuel into usefuel mechanical work takes place. Some certain parameters are used for the purpose of evaluating the engine performance which are given below: (i) Indicated Power : It is defined as the total power produced due to the fuel combustion in the combustion chamber. PmxLxAxNxKxh Indicated Power (lP) = 60

kw

where, Pm= mean effective pressure (KPa) L = length of stroke (m) A = Piston area (m/) N = Number of revolution of crank shaft (rpm) 1 K = 2" for 4 - stroke engine and 1 for 2 - stroke engine

(vii) Volumetricefficiency: It is defined as the ratio of mass of charge actually inducted to the mass of charge given by swept volume at ambient temperature and pressure. (ix) Relative efficiency: It is defined as the ratio of Indicated thermal efficiency to the air standard cycle efficiency. l1lTH

l1Rel. =-11cycle

(x) Fuel- Air ratio: . Mass of consumed fuel Fue I - Ai r ratio = -------Mass of air taken inside during the same period of time (xi) Relative fuel air ratio: · fu I . . (lL.) R e1ative e - aIr ratio .._K

n = number of cylinders

Actual fuel- air ratio = -------Stoichiometric fuel

air ratio

Pm·LANKn Indicated Horse Power (IHP) = 4500 (ii) Brake Power: When an engine produced power at the output shaft, then this power is termed as brake power.

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Brake Power (BP) =

2nNT

kw 60 x 1000 Where, T=Torque(N -m) N= Speed(rpm) (iii) Frictional Power : It is the difference of Indicated power and brake power. Frictional power (FP) = IP - BP (iv) Indicated thermal efficiency: Indicated Power Input Fuel Energy

11

=------

IlH

LP mf xCL where, m, = mass of fuel supplied (kg/s) CL = Lower calorific value of the fuel (v) Brake thermal efficiency: 11

Brake Power Input fuel energy

=-----BTH

B.P mfxC (vi) Mechanical efficiency: 11mech.

Brake power Indicated Power

B.P IP B.P BP+FP

Specific fuel consumption(SFC) = Fuel consumed(gm/hr) Power produced (xii) Mean-piston speed

s=21N where, I= stroke length N = Cranck shaft (rpm) (xiii) Specific output: Specific output = ~(kw AxL

/ m3)

n 2 where, A=-d 4 d = bore diameter (m) L = length of stroke (m) (xiv)Calorific value offuel : It is defined as the heat quantity developed due to its combustion at constant pressure and under normal conditions. It is the amount of thermal energy developed by complete combustion of a fuel. Difference between four - stroke and two storke cycle: Four stroke cycle Two stroke cycle Cycle is completive in four strokes Heavy flywheel is required Engine is heavy for same power output Less cooling and lubrication is needed High initial cost High thermal efficiency

Cycle is combusted in two strokes Light flywheel is required Engine is light for same power output High cooling and lubrication is needed Low initial cos t Lower thermal efficiency Volumetric efficiency is Volumetric efficiency is more less

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Difference between SI and CI Engine: S.I Engine C.I. Engine Ifperforms

on OHO cycle

Compression ratio ranges from 5 to 10 Carburetor supplies fuel Maintenance cost is low but running cost is high Spark plug is used Thermal efficiency is low

Ifperforms on diesel cycle Compression ratio ranges from 13 to 27. Fuel injector supplies fuel Maintenance cost is high but running cost is low No Spark plug is used Thermal efficiency is high

Engine Lubrication and cooling: (a) Engine Lubrication: Lubrication is defined as a method in which oil is provided between two moving surfaces having relative motion between them. Lubrication is employed to reduce friction betweenmoving parts. There is an oil film made which acts like a cushion for moving parts and absorbs heat from the parts. The basic characteristic oflubricant is viscosity, oiliness, chemical stability,Adhesiveness, film strength, flash point first point etc. Kinds oflubricants (i) Oils ---+ mineral oils ---+ fatty oils ---+ synthetic ---+ multigrade oils (ii) Greases ---+ Lubricating grease Eg.Aluminium, Calcium, Sodiumgreases. Lubrication system The following lubrication system are as : (a) Wet Sump Lubrication system: It consists of a sump which contains an oil supply. This sump is connected to the bottom of case of engine. ---+ Splash system ---+ Full pressure system ---+ Semi pressure system

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(b) Dry sump lubrication system In this system, oil supply is carried in a separats tank. Scavenger pumps are used to return the oil to the tank. It is generally used in radial engines or high capacity engines. (c) Engine cooling: Engine cooling is very necessary to maintain the temperature ofthe engine low other it affectsthe behaviour of fuel combustion and also shortens the life of engine. Cooling systems: There are generally two types of cooling systems used as : (i) Air - cooling (ii) Water or liquid cooling (i) Air cooling: In air cooling system, air flows through the outsides of cylinder barrel and out surface area is increased by the use oftimes. The heat dissipated amount to air depends upon amount of air flowing through cooling tins, tin surface area and thermal conductivity of material of the fins used. Applications: Small engines, Industrial and agricultural engines. (ii) Water or liquid cooling: In this system,water or liquid is made to circulate around the cylinders and thus absorbing heat from the walls of the cyclinder and cylinder head. Coolant absorbs heat while passing through the engine and lubricates the water pump. => Methods for circulating water • Thermo syphon cooling • Forced cooling • Pressurised water cooling • Evaporative cooling • Thermostat cooling => Various components of water cooling system • Water jacket • Water pump • Thermostat • Fan • Radiators • Radiator cap Applications: Industrial cooling towers, Marine vessel thyristors ofHVDC value etc.

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...,

Thermal Engineerging

I···.. 1.

2.

A-t07

EXERCISE

Which of the following is not a property of thermodynamic system? (a) Pressure (b) Energy (c) Heat (d) Volumes Match List - I and List - II and Give answer the following codes: List- I List- II A. Heat 1. Point function B. Energy 2. Path function C. Entropy 3. Second Law of thermodynamics 4. F=C-P+2 D. Cribbs function Codes: ABC (a) 234 (b) 4 2 3 (c) 2 1 3 (d) 1 4 2

10. The fixed points for celcius temperature scale are: (a) Ice point as 0° C (b) Steam point as 100°C (c) Both ice and steam points as O°C and 100°C respectively (d) Triple point of water as 0.Q1° C 11. For the calculation of real temperature in thermodynamics, the values of absolute zero temperature is known to be: (a) 273°C (b) -273°C (c) o-c (d) 373°C 12. Which of the following gives the correct relation between centigrade and fahrenheit scales? (c = degree centigrade, F = degree Fahrenheit)

D

(a)

4 3

13.

3.

5.

6.

7. 8.

9.

5 (b) C=-(F+32) 9

5 9 (d) C=-(F-32) F=-(F+32) 9 5 The gas constant 'R'is equal to the: (a) sum of two specific heats (b) difference of two specific heats (c) product of two specific heats (d) ratio of two specific heats According to which law, all perfect gases change in volume by II 273 rd oftheir original volume at 0° C for every 1° C change in temperature while pressure is kept constant. (a) Joule's Law (b) Boyle's like (c) Charles's law (d) Gay - Lussca Law Properties of substances like pressure, temperature and density in thermodynamic co-ordinates are: (a) Path function (b) Point function (c) Cyclic function (d) Real function For which of the following substances, the internal energy and enthalpy are the functions of temperature only? (a) Saturated steam (b) Water (c) Perfect gas (d) None of these Ifa graph is plotted for absolute temperature as a function of entropy, then the area under the curve would give: (a) amount of heat supplied (b) amount of work transter (c) amount of heat rejected (d) amount of mass transfer Heat is being supplied to air in a cylinder fiiled with frictionless piston held by a constant weight. The process will be: (a) Isochoric (b) Isothermal (c) Isobaric (d) Adiabatic During an adiabatic process, the pressure P of a fixed mass of a ideal gas changes by ~P and its volume v changes by (c)

Badboys2Stirling and Ericsson cycles are:

4.

5 C = -(F -32) 9

(a) irreversible cycles (b) quasi - static cycles (c) semi - reversible cycles (d) reversible cycles In thermodynamic cycle, heat is rejected at: (a) constant volume (b) constant pressure (c) constant enthalpy (d) constant temperature A system is taken from state "X" to state "Y" along with two different paths 'A' and 'B'. The heat absorbed and work done by the system along these paths are QA'QBand WA' WBrespectively. Which of the following is the correct relation? (a) QA+WA=QB+WB (b) QA-WA=QB-WB (c) QA +QB=WA +WB (d) QA =QB Which one of the following is an extensive property of a thermodynamic system? (a) Pressure (b) Density (c) Volume (d) Temperature Which of the following cycle consists of three processes? (a) Ericsson cycle (b) Stirling cycle (c) Alkinson cycle (d) None of these The first type of perpetual motion machine is one, which: (a) does not work without internal energy (b) works without any external energy (c) can completely convert heat into work (d) cannot completely convert heat into work Zeroth law ofthermodynamics deals with: (a) concept of temperature (b) enthalpy (c) entropy (d) external and internal energy both

14.

15.

16.

17.

18.

19.

~ V. THe value of

~V

V is given by :

Badboys2

Thermal Engineerging

A-lOS (a)

v ~P

(b)

1 ~P v P

~P P

(d)

v.-

1 P

(c) v20.

21.

23.

24.

25.

T3/2

(d)

32.

v

Which of the following gases has the heighest value of characteristic gas constant (R)? (a) Nitrogen (b) Oxygen. . (c) Carbon-di-oxide (d) Sulpher-di-oxide Molecular kinetic energy of the gas is proportional to: (a) T (b) P

(c) 22.

~v

33.

y-1

(a)

27.

28.

29.

30.

31.

The internal energy of the perfect as depends on (a) temperature, entropy and specific heats (b) temperature only (c) termperature, pressure and specific heats (d) temperature, enthalpy and specific heats Heat and work are: (a) Point function (b) System properties (c) Path function (d) None of these Which one of the following physical quantity is constant in the Gay - Lussac's Law? (a) Pressure (b) Volume (c) Temperature (d) Weight An ideal gas as compared to a real gas at very high pressure occupies: (a) More volume (b) Samevolume (c) Less volume (d) None of these . Which Law states that the internal energy of the gas IS a function of temperature? (a) Law ofthermodynamics (b) Joule's Law (c) Boyle's Law (d) Charle'sLaw The application of gas laws are limited to : (a) gases and liquid (b) steam and liquid (c) gases alone (d) gases and vapours Mean strquare molecular speed is : (a) directly proportional to density (b) inversely proportional to density . (c) directly proportional to the square root of denslt~ (d) inversely proportional to the square root of density Temperature of a gas is produced due to : (a) its heating value (b) kinetic theory of molecules (c) repulsion of molecules (d) attraction of molecules According to kinetic theory of gases, the absolute zero temperature is attained when: (a) volume of gas is zero (b) pressure of gas is zero (c) kinetic energy of molecules of gas is zero (d) None of these . . The work in a closed system undergoing an isentropic process is given by : (b)

(:~F

y-1

(b)

(~~F

TI/2

y-1

(c) 34.

35.

Badboys2 26.

Which of the following in an irreversible process? (a) Isothermal process (b) Isentrobic process (c) Isoparic process (d) Isenthalpic process In a reversible adiabatic process, the ratio (T 1 : T2) will be equal to:

y -1 ) -mR(T1-T2 2

36.

37.

38.

39.

40.

41.

42.

(~~)2y

y

(d)

(~~) y-1

An engine which takes 105 MJ at a temperature of 400 K, rejects 42 MJ at a temperature of 200 K and delivers 15 KWh mechanical work. Is this engine possible ...? (a) Possible (b) Not possible . (c) data insufficient (d) Cannot be predicted A mixture of gases expands from 0.03 m' to 0.06 m' at constant pressure of IMP a and absorbs 84 KJ heat during the process. The change in internal energy of the mixture will be equal to: (a) 54KJ (b) 64KJ (c) 30KJ (d) 75KJ Actual expansion process in a throttling device is : (a) Reversible adiabatic expansion (b) Isenthalpic expansion (c) Isothermal expansion (d) Irreversible First Law of thermodynamics depicts the relation between (a) heat and work (b) heat, work and system properties (c) various properties of the system (d) various thermodynamics processes The first Law of thermodynamics was given by : (a) Obert (b) Keenan (c) Joule (d) Newton For non - flow closed system, the value of net energy transferred as heat and work equals changes in : (a) Enthalpy (b) Entropy (c) Internal energy (d) None of these During throlling process, which ofthe following not good : (a) Enthalpy does not change (b) Enthalpy changes (c) Entropy does not change (d) Internal energy does not change A closed gaseous system undergoes a reversible process during which 20 kcal are rejected, the volume changing from 4 m' to 2 m' and the pressure remains constant at 4.2 kg/em'. Then the change in internal energy will be equal to: (a) + 10 kcal (b) -10 kcal (c) 0 (d) +5kcal If Q = Heat content of a gas, u = Internal energy P = Pressure, V = Volume, T = Temperature then which of the following statement is applicable to perfect gas and is also true for an irreversible process?

Badboys2

Thermal Engineerging

43.

A-t09

(a) dQ=dU+qdV (b) Tds=dU+PdV (c) dQ=Tds (d) dQ=Tds+dU Which of the following equation given the correct expression for the work done by compressing a gas isothermell y? Ify = Heat capacity ratio and all other parameters have their usual meaning.

(a)

[

)YY-I]-

_Y_P1V1 (P2 Y -1 PI

1

constant volume

T (b)

1

P2 (b) mRTI/nPI T

mR(l- ~~)

(d)

(c) MCp (T2- Tl)

A Frictionless heat engine can be 100% efficient only it the exhaust temperature is : (a) equal to its input temperature (b) less than its input temperature (c) O°C (d) OK 45. Kelvin Plank's Law deals with : (a) conservation of energy (b) conservation of heat (c) conservation of mass (d) conversion of heat into work 46. The second Law of thermodynamics defines: (a) Heat (b) Entropy (c) Enthalpy (d) Internal energy 47. A machine produces 100 KJ of heat to spend 100 KJ heat. This machine will be known as : (a) PMM-l (b) PMM-II (c) PMM-III (d) PMM-N 48. Third Law ofthermodynamics is : (a) an extension of second law (b) an extension of zeroth law (c) an extension of first law (d) an independent law of nature 49. In which cycle, all the four processes are not reversible? (a) Vapour compression cycle (b) Joule cycle (c) Carnot cycle (d) None of these 50. For thermodynamic cycle to be irreversible, it is necessary that,

(c)

1

constant volume

44.

Badboys2

(a)

dQ = 0 T

(b)

dQ

T



"-

2

""

(d)

1

\v

1['

4

./

"'

3 S

1

constant volume

--~"'S 52. According to clausius statement: (a) Heat flows from hot substance to cold substance (b) Heat flows from hot substance to cold substance unaided (c) Heat flows from cold substance to hot substance with aid of external work. (d) (b) and (c) both above 53. Which one of the following is the correct sequence for the air standard efficiencies of different gas power cycles at a definite compression ratio? (a) 11oHo>11diesel > 11dual (b) 11oHo>11dual > 11diesel (c) 11diesel > 11oHo > 11dual (d) 11dual> 11oHo > 11diesel 54. The air standard efficiency of otto cycle is given by : (a)

1 11= 1- (r)y-I

1 (b) 11= 1- (r)y+1

(c)

1 11= 1+ (r)y-I

(d) 11=1---

1 y-I (r)Y

55. A carnot engine working between 36°C and 47°C temperature, produces 150 KJ of work. Then the heat added during the process will be equal to: (a) 1500KJ (b) 3000KJ (c) 4360KJ (d) 6000KJ 56. Which of the following is the expression for Joule Thompson coefficient? (a)

(!\

(b) (:)

(c)

(:)p

(d) (:\

T (a)

2

T

<0

dQ > 0 (d) dQ~ 0 T In thermodynamic cycles, the otto cycle is represented by which of the following T - S diagram? (c)

51.



--~S

t

Badboys2

x.uo 57.

58.

59.

60.

61.

62.

Thermal Engineerging The entropy of the universe is: (a) increasing (b) decreasing (c) constant (d) unpredictable The unit of entropy is given as : (a) kg/J'K (b) J/kg.m (c) J/kgOK (d) J/sec In a statistical thermodynamics, entropy is defined as : (a) Reversible heat transfer (b) Measure of reversibility of a system (c) A universal property (d) Degree of randomness The change in entrophy is zero during: (a) Reversible adiabatic process (b) Hyperbolic process (c) Constant pressure process (d) Polytropic process Steam coming out of the whistle of a pressure cooker is : (a) dry saturated vapour (b) wet vapour (c) super heated vapour (d) ideal gas At critical point, any substance: (a) will exist in all the three phases simultaneously (b) will change directly from solid to vapour (c) will lose phase distinction between liquid and vapour (d) will behave as an ideal gas The ratio of two specific heats of air is equal to : (a) 1.41 (b) 2.41 (c) 4.14 (d) 0.41 Dryness fraction of steam is defined as :

Badboys2 63. 64.

(a)

(b) (c)

(d)

Mass of dry steam Mass of water vapourin suspension Mass of water vapour in suspension Mass of dry steam

69.

70.

71.

72.

73.

Mass of water vapour in suspension Mass of water vapour in suspension + Mass of dry steam

The critical point for water is : (a) 374°C (b) 373°C (c) 273°C (d) 323°C The amount of heat requiredto raise the temperaturevolume is known as: (a) specific heat at constant pressure (b) specific heat at constant volume (c) kilo -joule (d) None of these Critical pressure is the pressure of steam at (a) exitofsteamnozzle (b) either at inlet or at outlet of steam nozzle (c) inlet of steam nozzle (d) throat of steam nozzle As the pressure increases, the saturation temperature of vapour: (a) increases (b) decreases (c) increases first then decreases (d) decreases first then increases In steam tables, the entropy is shown as zero for, (a) saturated vapour at atmospheric pressure (b) saturated liquid at atmospheric pressure (c) saturated vapour at 0° C (d) saturated liquid at 0° C Expression for the specific entropy of wet steam is: (a) Sg+Xsf

(b) hf+XT

~

~

L

L

74.

75.

Mass of dry steam Mass of dry steam + Mass of water vapour

Triple point is a point of a pure substance at which: (a) liquid and vapour exist together (b) solid and liquid exist together (c) Solid and vapour exist together (d) solid, liquid and vapour phase exist together 66. Sensible heat is the needed to (a) vaporise water into steam (b) change the temperature of a liquid or vapour (c) convert water into steam and super heat it (d) measure dew point temperature 67. If H = enthalpy of dry air, H, = enthalpy of water vapour w = ~pecifichumidity, then the enthalpy of moist air will be equal to: (b) H +wH (a) Ha+ n, (c) Ha-wHy (d) H:w-H: 65.

68.

76.

~+X~

~+XT

The latent heat ofvapourization of a fluid at look K is 2560 KJ/kg. Then change of entropy associated with the evaporation will be equal to: (a) 6.86 KJ/kgOK (b) 9.86 KJ/kgOK (c) 25.6 KJ/kgOK (d) -25.6 x 103 KJ/kgOK When wet steam undergoes adiabatic expansion, then (a) its dryness fraction may increase or decrease (b) its dryness fraction increases (c) its dryness fraction decreases (d) its dryness fraction does not change Critical pressure for steam is equal to :

W nObM

~ nl~

(c) 163bar (d) 184bar 77. Throttling calorimeter is used for the measurement of: (a) very low dryness fraction upto 0.7 (b) very high dryness fraction upto 0.98 (c) dryness fraction of only low pressure (d) dryness fraction of only high pressure 78. A wet vapour can be completely associated/specified by which of the following? (a) Pressure only (b) Temperature only (c) Specific volume only (d) Pressure and dryness fraction 79. It a given temperature, the enthalpy of superheated steam is always: (a) less than enthalpy of saturated steam (b) greater than enthalpy of saturated steam

Badboys2

Thermal Engineerging

80.

81.

82.

83.

84.

85.

A-11I

(c) equal enthalpy of saturated steam (d) None of these The enthalpy of vapour at lower pressure depends (a) Temperature (b) Volume (c) Temperature and volume (d) Neither temperature nor volume For high boiler efficiency,the feed water is heated by: (a) regenerator (b) convective heater (c) super heater (d) economiser Locomotive type of boiler is: (a) horizontal multi - tubular fire tube boiler (b) horizontal multi - tubular water tube boiler (c) vertical tubular fire tube boiler (d) water wall enclosed turnace type Lancashire boiler is a : (a) water tube boiler (b) fire tube boiler (c) locomotive boiler (d) high pressure boiler Water - tube boilers are those in which : (a) water passes through the tubes (b) flue gases pass through tubes and water around it (c) work is done during adiabatic expansion (d) there is change in enthalpy The capacity of the boiler is defined as : (a) The volume offeed water inside the shell (b) The volume of steam space inside the shell (c) The maximum pressure at which steam can be generated (d) Amount of water converted into steam from 100 C to 110 C in one hour. The type of safety value recommended for high pressure boiler is(a) Dead weight safety value (b) Liver safety value (c) Spring loaded safety value (d) None of these Boiler rating is generally defined in terms of: (a) maximum temperature of steam (b) heat transfer rate KJ/hr (c) heating surface (d) heating output kg/hr Boiler mountings are necessary for: (a) operation and safety of a boiler (b) increasing the efficiency of boiler (c) Both (a) and (b) (d) None of these Bab cock and wilcox boiler is (a) water tube type (b) fire tube type (c) Both (a) and (b) (d) None of these Range of high pressure boilers are: (a) blow 80 bar (b) above 80 bar (c) 40 - 80 bar (d) 60 - 80 bar A boiler mounting used to put - off fire in the fuel when water level in the boiler falls below a safe limits. (a) Blowoffcock (b) Stop value

92.

93.

94.

95.

96.

97.

Badboys2

98.

0

0

86.

87.

88.

89.

90.

91.

99.

100.

101.

102.

103.

104.

(c) Feed check value (d) Fusible plug Steam in boiler drum is always (a) wet (b) dry (c) superheated (d) Both (a) and (b) Air pre heater: (a) increases evaporation capacity of boiler (b) increases the efficiency of boiler (c) enables low grade fuel to be burnt (d) All of the above In a Lancashire boiler, the economiser is located: (a) before air pre-heater (b) After air preheater (c) Between feed pump and boiler (d) Not used Locomotive boiler produces steam at : (a) Medium rate (b) Low rate (c) Veryhigh rate (d) Verylow rate For the same diameter and thickness of a tube, fired tube boiler as compared to water tube boiler has: (a) More heating surface (b) Less heating surface (c) Same heating surface (d) No heating surface Ifcirculation of water in a boiler is made bypump, then it is known as: (a) Forced circulation boiler (b) Natural circulation boiler (c) Internally fired boiler (d) Externallyfiredboiler The device used to empty the boiler, when required and to discharge the Need, scale of sediments which are accumulated at the bottom of the boiler is known as : (a) Safety value (b) Stop value (c) Fusible value (d) Blow- offcock Maximum heat is lost in boiler due to: (a) Unburnt carbon (b) Flue gases (c) Incomplete combustion (d) Moisture in fuel What salts of calcium and magnisium cause tempeorary hardness of boiler feed water? (a) Sulphides (b) Carbides (c) Nitrates (d) Bi - carbonates Bluding in turbine means: (a) leakage of steam (b) extracting steam for preheating feed water (c) removal of condensed steam (d) steam doing no useful work In parson's steam turbine, steam expands in: (a) partly in nozzle and blade (b) blades only (c) nozzles only (d) None of these Pressure compounding can be done in the following type turbines: (a) Impulse turbines (b) Reaction turbines (c) Both impulse and reaction turbines (d) None of these Blade efficiency of steam turbine is equal to: (a) V (Vw,- Vw2) / 2g (b) 2V (Vw,- Vw2) / V/

Badboys2

x.nz 105.

106.

107.

108.

109.

Thermal Engineerging (c) V(Vw,+Vwz)/g (d) 2Vrvw, + Vw2)/V,2 Steam turbines are governed by the following methods: (a) Throttle governing (b) Nozzle control governing (c) By - Pass governing (d) All of these For Parson's reaction turbine, degree of reaction is: (a) 50% (b) 100% (c) 75% (d) 25% In flow through a nozzle, the mach number is more than unity : (a) in converging section (b) at the throat (c) in diverging section (d) can be in any section of the nozzle depending upon the nozzle profile and geometry When steam flows over moving blades of an impulse turbine: (a) Pressure drops and velocity increases (b) Pressure remains constant and velocity decreases (c) Both pressure and velocity remain constant (d) Both pressure and velocity decrease Curtis turbine is an example of: (a) velocity compounded impulse steam turbine (b) pressure compounded impulse steam turbine (c) pressure - velocity compounded impulse steam turbine (d) reaction steam turbine Stage efficiency of steam turbine is equal to :

Badboys2 110.

(a)

Blade efficiency Nozzle efficiency

(b)

Nozzle efficiency Blade efficiency

(c) Nozzle efficiency x blade efficiency (d) None of these 111. The reason of compounding of steam turbine is : (a) To reduce rotor speed (b) To reduce exit losses (c) To improve efficiency (d) All of the above 112. For a single stage impulse turbine, having nozzle angle 'a', then maximum blade efficiency under ideal condition is given by:

(a)

cosa

--

2

(b)

sin a

2

(c) tan a (d) cot a 113. Shock effect in a nozzle is felt in : (a) divergent portion (b) straight portion (c) convergent portion (d) throat 114. In a steam turbine, the critical pressure ratio for in dry saturated steam is given by : (a) 0.545 (b) 0.577 (c) 0.585 (d) 0.595 115. Steam nozzle converts: (a) Heat energy to potential energy (b) Heat energy to kinetic energy (c) Kinetic energy to heat energy (d) Potential energy to heat energy 116. For maximum discharge, ratio of the pressure at the exit and at inlet of the nozzle (PiP,) is equal to:

n

(a)

(c)

~

P2

=C!Jn-1 =

PI

(_2_) n-i-I

(_2_)

(b) P2 = PI n-l

n

n-I

n

n+1

117. Multi-stage turbines are: (a) Reaction type (b) Pressure compounded (c) Velocity compounded (d) All of these 118. If11BT = Brake thermal efficiency 111T= Indicated thermal efficiency 11m = Mechanical efficiency then, which ofthe following relation is correct?

(c)

11BT

=~

11m IT

119. De- Laval turbine is: (a) Pressure compounded impulse turbine (b) Simple single wheel impulse turbine (c) Velocity compounded impulse turbine (d) Simple single wheel reaction turbine 120. The reason for inter cooling in multistage compressors is : (a) To minimize the work of compression (b) To cool air delivery (c) To cool the air during compression (d) None of these 121. Work done by prime mover to run the compressor is minimum if the compression is : (a) adiabatic (b) isothermal (c) isentropic (d) polytropic 122. Reciprodcating air compressor is best suited for: (a) Lar ge quan ti ty of air at high pressure (b) Small quantity of air at low pressure (c) Small quantity of air at high pressure (d) Large quantity of air at low pressure 123. In a Brayton cycle, the air enters the compressor at 3000 K and maximum temperature of cycle is 12000 K. Then the thermal efficiency of cycle for maximum power output will be:

W

~%

~

~%

(c)

50% (d) 90% 124. An axial flow compressor will be having symmetrical blades for the degrees of reaction: (a) 25% (b) 50% (c) 75% (d) 100010 125. A reciprocating compressor having 0.20 m bore and stroke runs at 600 rpm. If the actual volume delivered by compressor is 4m3 I min. Its volumetric efficiency is about (a) 70% (b) 75% (c) 80% (d) 85% 126. The centrifugal type of rotary compressor is used in (a) Boilers (b) Gas turbines (c) Cooling plant (d) None of these 127. If the compressor ratio is increased, then the volumetric efficiency of a compressor :

Badboys2

Thermal Engineerging

A-l13

(a) decreases (b) increases (c) constant (d) None of these 128. Volumetric efficiencyof air compressor is defined as :

(a)

Stroke volume Clearance Volume

(b) Airactually delivered Amount of piston

139.

140.

displacement

129.

130.

131.

(c) Compression ratio (d) Index of compressor performance Centrifugal compressor works on the principal of: (a) conversion of Pressure energy into kinetic energy (b) conversion of kinetic energy into pressure (c) centripetal action (d) developing pressure directly Air is compressed by a double stage compressor (with complete intercooling), from 1 bar pressure 127°C temperature to 36 bar pressure. Then the inter-stage pressure for the minimum work of the compressor: (a) 6 bar (b) 12bar (c) 18bar (d) 24 bar An engine operators between temperature limits 900 K and T2 and another between T2 and 400 K. For both of the engines to be equally efficiency,the value ofT 2should be : (a) 600K (b) 700K (c) 625K (d) 750K A heat engine develops 60 kw of work having an efficiency of 60%, then the amount of heat rejected will be equal to: (a) 400kw (b) 40kw (c) 20kw (d) 200kw In carnot cycle, addition and rejectionofheattakesplaceat : (a) constant pressure (b) constant temperature (c) constant volume (d) constant speed A heat engine receives 1120 KJ of heat and rejects 840 KJ of heat while operating between two temperature limits of 560 K and 280 K. If indicates that the engine operates on the following cycle: (a) Reversible cycle (b) Irreversible cycle (c) Impossible cycle (d) Unpredicable cycle Area under T - S diagram represents: (a) Heat transfer for reversible process (b) Heat transfer for Inreversible process (c) Heat transfer for all processes (d) Heat transfer for adiabatic processes A heat pump working an reversed camot cycle has COP of 5. Ifit work as refrigerator taking 1 kw of work input the refrigerating effectwill be : (a) 1kw (b) 2kw (c) 3kw (d) 4kw Tworeversiblerefrigerator's are arranged in series and their COP are 4 and 5 respectively. Then the COP of composite refrigeration system will be : (a) 1.5 (b) 2. (c) 4 (d) 3.7 An ideal refrigerator is operating between a condenser temperature of -37°C and an evaporator temperature of -3 ° C. Ifthe machine is functioning as a heat pump, its COP will be equal to:

Badboys2 132. 133.

134.

135.

136.

137.

138.

141.

142.

143.

144.

145.

146.

147.

(a) 7.75 (b) 8.75 (c) 9.75 (d) 10.75 In a refrigeration system, the refrigerant gain heat: (a) Compressor (b) Condenser (c) Expansion value (d) Evaporator The Bell- columan refrigeration cycle uses: (a) Hydrogen as a working fluid (b) Air as a working fluid (c) carbon di - oxide as a working fluid (d) Any inert gas as a working fluid A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 Kl/kg, Enthalpy after throttling = 125 Kl/kg, Enthalpy before compression= 375 Kl/kg. Then the COP of refrigeration is: (a) 5 (b) 10 (c) 6 (d) 9 A camot refrigeration cycle has a COP of 4. The ratio of higher temperature to lower temperature will be equal to : (a) 2.5 (b) 2 (c) 2.8 (d) 1.25 The refrigeration system works on: (a) First Law ofthermodynamics (b) Second Law ofthermodynamics (c) Zeroth Law of therodynomics (d) None of these One ton of refrigeration is equal to: (a) 210KJ/min (b) 21 KJ/min (c) 420 KJ/min (d) 20 KJ/min If a heat pump cycle operates between the condensar temperatureof + 27°C and evaporatortemperatureof-23°C, then the COP of camot will be equal to : (a) 6 (b) 12 (c) 5 (d) 1.2 Which of the following has minimum freezing point? (a) Freon-12 (b) Freon-22 (c) Ammonia (d) Carbon di-oxides In actual refrigeration systems, the compressor handles vapour only. This process commonly reflered to as : (a) Gas compression (b) Phase compression (c) Dry compression (d) wet compression

148. The expression 0.622(

J

Py is associated with: Pt -Py (a) Relative humidity (b) Specifichumidity (c) Degree of saturation (d) Partial pressure [where, Py = Vapour pressure, P, = total pressure = Pa + PyJ 149. In the absorption refrigeration cycle, the compressor of vapour compression refrigeration cycle is replace by : (a) Liquid pump (b) Generator (c) absorber and generator (d) Absorber, Liquid pump and generator 150. When water Lithium bromide is used in a vapour absorption refrigeration system, then: (a) they together act as refrigerant (b) water is the refrigerant (c) Lithium bromide is refrigerant (d) None of these

Badboys2

Thermal Engineerging

A-114

151. The wet bulb depression is zero when relative humidity equals: (a) Zero (b) 50% (c) 75% (d) 100% 152. In winter air conditioning, the inside design conditions are given by the following: (a) 21°CDBT,600IoRH (b) 21°CDBT,50%RH (c) 25°CDBT,50%RH (d) 25°CDBT,600IoRH 153. Piston rings are generally made offollowing material: (a) cast iron (b) mild steel (c) aluminium (d) carbon steel 154. Which of the following is not an internal combustion engine? (a) 3 - stroke pertrol engine (b) 4 - stroke petrol engine (c) Diesel engine (d) Steam engine 155. Compression ratio of an IC engine is given as : ifV c = cylinder volume, Vs = swept volume (a)

1+ Vs Vc

(b)

1- Vs Vc

Badboys2(c)

1+ Vc Vs

(d) 1- Vc

Vs

156. Two stroke engines have: (a) valves (b) Ports (c) both (a) & (b) (d) None of these 157. Detonation is said to take place in the engine when: (a) sudden acceleration is imparted to the engine (b) temperature rise too high (c) high pressure waves are set up (d) combustion of fuel takes place without spark provided to it 158. Number of working strokes/minute for a two stroke cycle engine as compared to speed ofthe engine in rpm, is : (a) same (b) half (c) double (d) four times 159. For the same compressionratio and heat input, the efficiency of an oHo cycle engine as compared to diesel engine is : (a) less (b) more (c) equal (d) None of these 160. The ratio of the stroke to the crank radius in an LC. engine IS given as : (a) 1: 1 (b) 3: 1 (c) 2: 1 (d) 1: 2 161. The correct sequence of stroke in a four stroke engine is : (a) suction, compression, exhaust, expansion (b) expansion, compression, suction, exhaust (c) suction, compression, expansion, exhaust (d) suction, expansion, compression, exhaust 162. Stoichiometric air-fuel ratio by volume for compression of methane in air is : (a) 15: 1 (b) 9.53: 1 (c) 17.2:1 (d) 10.5:1 163. The term air injection, associatedwith fuel injection system ofan IC engine means injection of:

(a) Air only (b) Liquid fuel only (c) Both (a) and (b) (d) Solid fuel and air 164. If the compression ratio is increased in SI engine, the knocking tendency will : (a) increase (b) decrease (c) Not be affected (d) Cannot be predicated 165. The function of carburetor is : (a) Atomise and vaporise the fuel and to mix it with air in proper ratio (b) Refining the fuel (c) Increase the pressure of the fuel vapour (d) Inject petrol in cylinder 166. In an engine working on otto cycle, air fuel - mixture is compressed from 400 C.C to 100C. C. ifr = 1.5, then the thermal efficiencyof cyclewill be : (a) 50% (b) 65% (c)

60%

(d) 90%

167. Mixture formation in a carburetor is based on the principle of: (a) Pascal's Law (b) Buoyancyprinciple (c) Ventureprinciple (d) Pitot tube principle 168. Octane number of a gasoline is a measure of its: (a) Knocking tendency (b) Ignition delay (c) Smoke point (d) Ignition temperature 169. In four cylinder petrol engine, the standard firing order is: 1-3-3-4 (b) 1-4-3-3 ~ 1-3-3-4 ~) 1-3-4-3 170. Which of the followingis not a part ofMPFI petrol engine? (a) Fuel injector (b) Carburetor (c) MAP sensor (d) Electronic control unit 171. Fuel injector is used in : (a) steam engines (b) gas engines (c) spark ignition engines (d) compression ignition engines 172. Compression ratio of diesel engine may have a range of: (a) 8 to 10 (b) 16to 30 (c) 10to 15 (d) 40 to 50 173. For diesel engine, the method of governing is employed is: (a) Quality governing (b) Quantity governing (c) Hit and miss governing (d) None of these 174. In diesel engine, the governor controls: (a) Fuel pressure (b) Fuel volume (c) Fuel flow rate (d) Fuel temperature 175. The firing order of six cylinder diesel engine is (a) 1-3-5-3-4-6 (b) 1-5-3-6-2-4 (c) 1-4-3-6-3-5 (d) 1- 6 - 3 - 5 - 3 - 4 176. The ignition of fuel in a diesel engine is caused by: (a) spark plug (b) compressed fuel (c) heat resulting from the compressed air that is supplied for the combustion (d) air fuelmixing

W

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Thermal Engineerging

A-115

177. The curve shown in the given figure is characteristic diesel engines: Then 'Y' axis shows: Y

(c) 70em3 (d) 84em3 187. Unit of brake specific fuel consumption is: (a) kg-hr-kw (b) kglkw-hr (c) kw - hr / kg (d) kg - hr / kw 188. Indicated mean effective pressure is : Area of indicator diagram x spring scale (a) Length of base of indicator diagram (b)

--+------~x o Power

(c)

178.

179.

(a) Efficiency (b) Specific fuel consumption (c) Air fuel ratio (d) Total fuel consumption Value overlapping happens: (a) completelybeforeT.D.C (b) completelyafter T.D.C (c) partially before T.D.C and partially after T.D.C (d) completelybefore B.D.C Morse test is used for multicylinder spark ignition engine to determine: (a) Thermal efficiency (b) Mechanical efficiency (c) Volumetricefficiency (d) Relative efficiency Actual power generated in engine cylinder is known as : (a) brake horse power (b) Indicated horse power (c) One boiler horse power (d) fractional horse power Flash point for diesel fuel should be: (a) Minimum49° C (b) Maximum49°C (c) Minimum99°C (d) Maximum99°C The ratio of brakepowerto indicatedpowerof an I.C. engine is called: (a) Thermal efficiency (b) Mechanical efficiency (c) Volumetricefficiency (d) Relative efficiency The primary winding of ignition coil consist of: (a) few turns of thin wire (b) many turns of thin wire (c) few turns of thick wire (d) many turns of thick wire How many power stroke/second will take place in a four stroke petrol engine rotating at 3000 r.p.m? (a) 25 (b) 50

180. Badboys2 181.

182.

183.

184.

(c)

100

190.

191.

192.

193.

194.

195.

196.

(d) 200

185. If P = mean effective pressure, L = length of stroke N = ~peed of the engine (rps), A = Bore area Then, the indicated power of four stroke engine will be equal to: (a) Pm.LAN

189.

(b)

197.

Pm·LA N

Pm·LAN Pm·LAN (d) 2 4 186. Ifthe bore diameter stroke length, and compression ratio of a single cylinder engine are 7 em, 8 em and 8 em respectively. Then the clearance volume will be equal to : (a) 44em3 (b) 50em3 (c)

198.

199.

Area of indicator diagram Length of base of indicator diagram x spring scale Length of base of indicator diagram x spring scale Area of indicator diagram

Area of indicator diagram (d) Length of base of indicator diagram Most important properly of an IC engine lubricant is: (a) density (b) viscosity (c) thermal conductivity (d) none of these The process of scavenging is associated with: (a) two stroke engine (b) four stroke engine (c) gas turbine (d) compressor The function oflubrication in engine are: (a) Lubrication and cooling (b) cleaning, sealing and noise reduction (c) efficiency increase (d) Both (a) and (b) Ifthe lubricant for an automobile to be used under subzero temperatures is to be selected, then which of the following properties will get priority consideration? (a) calorific value (b) pour point (c) specific gravity (d) carbon content Control of maximum oil pressure in the lubrication system is attected by : (a) oil filter (b) Pressure switch (c) Pressure relief value (d) Pump motor What technique is adopted forthe lubrication of the cylinder of a scooter engine? (a) splash lubrication (b) Forced feed lubrication (c) Gravity feed lubrication (d) Petrol lubrication Using lubricants on engine parts reduces: (a) motion (b) force (c) Acceleration (d) Friction The water pump generally employed for cooling of engine of a vehicle is: (a) Gear type (b) vane type (c) centrifugal type (d) Riciprocating type The advantage of using pressure cap on the radiator (a) Evaporation of coolant is increased by its used (b) It prevents vaccum formation in the system (c) By using this, atmospheric pressure is always maintained in the system (d) Boilining point temperature ofthe coolant is decreased by its use Radiator tubes are generally made up of: (a) Brass (b) Steel (c) Cast iron (d) None of these Which of the following lubrication system is used in a car engine generally?

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Thermal Engineerging

A-116

200.

201.

202.

203.

(a) Petrol (b) Splash (c) Pressure (d) Dry sump For a good quality of lubricant, consider the following statements: 1. change in viscosity should be minimum with the change in temperature 2. pecific heat should be low 3. Flash point should be high The true statements from the above are: W 1&2 ~ 2&3 (c) 1 & 3 (d) 1,2 & 3 In pressure lubrication system of an engine, the maximum oil pressure is controlled by : (a) oil pump (b) oil pressure gauge (c) oil pressure reliefvalue (d) oil filter A body of weight 100 N falls freely a vertical distance of 50m. The atmospheric drag force is O.5N. For the body,the work interaction is : (a) -25J (b) +25J (c) + 500J (d) - 500 J Match List - I and List - II and answer according to the codes given below: List- I List-II (A) cetane number 1. Ideal gas (B) Approach and range 2. vander walls gas

Badboys2(C)

ABC (a) 4 5 (b) 3 4 (c) 1 2

(a)

1

4

(b) 2

(c) 4 (d) 3 208. Which among the following relation is valid only for

reversible process undergone by a pure substance? (a) oQ=Pdv+dU (b) oQ=du+sw (c) Tds = dU + ow (d) Tds = Pdv + dU 209. Consider a refrigerator and a heat pump working on the reversed camot cyclebetween the same temperature limits. Which of the following is correct? (a) COP of refrigerator = COP of heat pump (b) COP of refrigerator = COP of heat pump + 1 (c) COP of refrigerator = COP of heat pump - 1 (d) COP of refrigerator = Inverse of COP of heat pump 210. A solar energy based heat engine which receives 80 KJ of heat at 100°Cand rejects70 KJ ofheat to the ambientat 30°C is to be designed. Then the thermal efficiencyof heat engine will be: (a) 12.5% (b) 24.5% (c) 40% (d) 70%

4

5

3.

S.1.Engine

4.

C.IEngine

5. 6.

Cooling towers Heat exchangers

D 2 1 1 2 4 6

always equal to : (a) zero (c)

213.

3

214.

when applied to: (a) A closed system undergoing a reversible adiabatic process (b) An open systemundergoing an adiabatic process with negligiblechangesin kineticenergyand potentialenergy (c) A closed system undergoing a reversible constant volume process (d) A closed system undergoing reversible constant pressure process 205. A steel ball of mass 1 kg of specific heat 0.4 Kj is at a temperature of 60° C. Itis dropped into 1kg water at 20° C. The final steady stats temperature of water is (a) 23.5°C (b) 30°C (c) 40°C (d) 42.5°C 206. For reversible adiabatic compression in a steady flow process, the work transfer/mass is

f pdv

(b)

f vdP

Cp Cv

p-

T(:;)

v ] is

(b) R (d) Rf

212. During a phase change ofa pure substance:

204. The first Law ofthermodynamics takes the form w = -Ll D

(a)

f

211. For an ideal gas, the expression [ T(:;)

(OT) :;t 0 oP h (D) dh = CpdT, even when pressure, varies

(d)

f

(c) Tds (d) sdT 207. A condenser ofa refrigeration system rejects heat at a rate of 120 kw, while the compressor consumes of power of 30 kw. The coefficientofperformance of the systemwill be:

215.

216.

(a) dG=O (b) dP= 0 (c) dH = 0 (d) dU= 0 At the triple point of pure substance,the number of degrees of freedom is : (a) 2 (b) 1 (c) 0 (d) 4 A vessel of volume 1m' contains a mixture ofliquid water and steam in equilibrium at 1.0 bar. Given that 90% ofthe volumeis occupiedbythe steam, then the fractionofmixture will be equalto: Ifat 1bar,Vf= 0.001 mvkg, V = 1.7mvkg (a) 5.27 x 10-3 (b) 7.27 x 10-4 g (c) 8.29 X 10-3 (d) 9.23 x 10-3 The following data is provided for a single stage impulse steam turbine : Nozzle angle = 20°, Blade velocity = 200 m/s Relative steam velocity at entry = 350 m/s, Blade inlet angle = 30°, Blade exit angle = 25°, Itblades friction is neglected, the work done/kg steams is : (a) 124KJ (b) 164KJ (c) 174KJ (d) 184KJ if V N and a are the nozzle exit velocity and angle in an impulse turbine, then the optimum blade efficiencyis given by: (a) VN sin a

(b)

VN sina 2

(d)

VN cosa 2

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Thermal Engineerging

A-117

217. For a single stageimpulseturbine with rotor diameter of2 m and speed of 3000 rpm when the nozzle angle is 20°, the optimum velocity of steam in mls is : (a) 334 (b) 668 (c) 356 (d) 711 218. The figure below shows a thermodynamic cycleundergone by a certain system. Then the mean effectivepressure in NI

m';

:

(a)

220.

---~

Pin I I I I

223.

224.

0.01 0.03 (a) 1.5kPa (b) 3 kPa (c) 4.5kPa (d) 6kPa In orderto burn 1kg ofCH4 completely,the minimumnumber of kg of oxygenneeded is (take atomic weights ofH, C and o as 1, 12 and 16 respectively) (a) 3 (b) 4 (c) 5 (d) 6 An IC engine has a bore and stroke of 2 units each. The area to calculate heat loss can be taken as : (a) 4n (b) 5n (c) 6n (d) n Atmospheric air from 40° C and 60% relative humidity can be bought to 20° C and 60% relative humidity by: (a) cooling and dehumidification process (b) cooling and humidification process (c) Adiabatic saturation process (d) sensible cooling process The use of Refrigerant R - 22 for temperature below-30°C is not recommended due to its: (a) good miscibility with lubricating oil (b) Poor miscibility with lubricating oil (c) low evaporating temperature (d) None of these IfAir-Fuel ratio ofthemixture in petrol engine is more than 15: 1,then, (a) NO is reduced (b) CO2 is reduced (c) HCxisreduced (d) CO is reduced An aircraft is flying at an altitude where the air density is half the value at ground level with reference to the ground level, the air fuel ratio at this altitude will be : (a)

..fi

(b)

8i

(c) 2 (d) 4 225. The silencer of an internal combustion engine (a) reduces noise (b) decrease break specific fuel consumption (c) Increase break specific fuel consumption (d) has no effect on its efficiency 226. Nitrogen at an initial state of 10 bar, 1 m3 and 300 K is expanded isothermally to a final volume of2 m3. The p-V-T relation is

(p+ :2 Jv =

=2

P Pout Pin

I~

(C)

P Pout Pin

V

Vc

V

-n I~ (d)

P Pout Pin

I I I

V

Vc

I

Vc

V

228. The following four figures have been drawn to represent a fictitious thermodynamic cycle, on the p-Vand T-S planes.

T

P

• • • • Fig. 1

P

V

Fig. 2

S

Fig. 4

S

T

Fig. 3

V

According to the first law of thermodynamics, equal areas are enclosed by (a) Figures 1 and 2 (b) Figures 1 and 3 (c) Figures 1 and 4 (b) Figures 2 and 3 229. Match items from groups I, II, III, IV and V. r

Group I

RT, wherea>O. EHeat

The final pressure

(b)

V(mj

Badboys2 221.

222.

p Pout

I I I I

219.

(a) will be slightly less than 5 bar (b) will be slightlymore than 5 bar (c) will be exactly 5 bar (d) Cannot be ascertained in the absence of the value of a 227. A p- V diagram has been obtained from a test on a reciprocating compressor.Which ofthe followingrepresents that diagram?

Group II

Group In Group IV

Group V

When Differential Function Phenomenon added to the system, is G Positive I Exact KPath M Transient

F Work H Negative J Inexact

L Point

N Boundary

I

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Thermal Engineerging

A-11S F-G-J-K-M (b) E-G-I-K-M E-G-I-K-N F-H-I-K-N (b) F-H-J-L-N (b) E-G-J-K-N E-H-I-L-M F-H-J-K-M 230. A 100 W electric bulb was switched on in a 2.5 m x 3 m x 3 m size thermally insulated room having a temperature of20°C. The room temperature at the end of24 h will be (a) 321°C (b) 341°C (c) 450°C (d) 470°C 231. The above cycle is represented on T-S plane by (a)

234. A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used V = specific volume and p = pressure of the gas The specific work required to be supplied to the compressor for this gas compression process is 2

p

(a) 3

p V = constant

W

2 1m

(a)

T

3

VI (b)

236.

V

T

3\12 ~2 2£J 2// 3

Badboys2

S

(c)

T

(d)

237.

S

T

3

238.

232. In a steady-state steady-fltwprocess taking place in a deJce with a single inlet and a single outlet, the work done per unit mass flow rate is given by W =

_fOlutlet In et

Vdp ,where V is the

specific volume and p is the pressure. The expression for W given above is (a) valid only if the process is both reversible and adiabatic (b) valid only if the process is both reversible and isothermal (c) valid for any reversible process (d)

. . must b e W = incorrect; It

fOutlet Inlet

f Vdp 1

(c) VI(P2 -PI) (d) -P2 (VI - V2) 235. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m'. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ) during this process will be

(2007, 2m) 100 kPa

2

(b)

1

y

400 kPa

f pdV

239.

240.

p dV

233. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is true at the end of above process? (a) The internal energy of the gas decreases from its initial value but the enthalpy remains constant (b) The internal energy ofthe gas increases from its initial value but the enthalpy remains constant (c) Both internal and enthalpy ofthe gas remains constant (d) Both internal and enthalpy of the gas increase

241.

&TI

~

~OO

(c) 554.67 (d) 8320.00 Ifa closed system is undergoing an irreversible process, the entropy of the system (a) must increase (b) always remains constant (c) must decrease (d) can increase, decrease or remain constant Consider the following two processes: I. A heat source at 1200 K loses 2500 kJ ofheat to sink at 800 K. Il. A heat source at 800 K loses 2000 kJ of heat to sink at 500 K. Which ofthe following statements is true? (a) Process I is more irreversible than Process II (b) Process IIis more irreversible than Process I (c) Irreversibility associated in both the process is equal (d) Both the processes are reversible A mono-atomic ideal gas (y= 1.67; molecularweight=40) is compressed adiabatically from 0.1 MPa, 300 K to 0.2 MPa. The universal gas constant is 8.314 kJ mor' K-I . The work of compression of the gas (in kJ/kg is (a) 29.7 (b) 19.9 (c) 13.3 (d) zero One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is (a) equal to entropy change of the reservoir (b) equal to entropy change of water (c) equal to zero (d) always positive A turbo-charged four-stroke direct injection diesel engine has a displacement volume of 0.0259 m3 (25.9 L). The ending has an output of950 kW at 2200 rpm. The mean effective pressure in MPa is closest to (a) 2 (b) 1 (c) 0.2 (d) 0.1 The values of enthalpy of steam at the inlet and outlet of a steam turbine in a Rankine cycle are 2800 kJ/kg and 1800 kJ/ kg respectively. Neglecting pump work, the specific steam consumption in kglkW-h is (a) 3.60 (b) 0.36 (c) 0.06 (d) 0.01

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Thermal Engineerging

A-119

242. The crank radius of a single-cylinder IC engine is 60 mm and the diameter ofthe cylinder is 80 mm. The swept volume of the cylinder in cm ' is (a) 48 (b) 96 (c) 302 (d) (m 243. The contents ofa well-insulated tank are heated by a resistor of23Q in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy (Al,') during the process in kWare (a) Q=O, W=-2.3,ilU=+2.3 (b) Q=+ 2.3, W= 0, ilU=+2.3 (c) Q=-2.3, W=0,ilU=-2.3 (d) Q=O, W=+2.3,ilU=-2.3 244. An ideal gas of mass m and temperature T I undergoes a reversible isothermal process from an initial pressure PI to final pressure P2. The heat loss during the process is Q. The entropy changes ilS of the gas is (a)

(c)

(d) zero

245.A cylinder contains 5 m3 of an ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 5 bar. The work in kJ required for this process is (a) 804.7 (b) 9532 (c) 981.7 (d) 1012.2 246.Specific enthalpy and velocity of steam at inlet and exit of a steam turbine, running under steady state, are as given below. Specific enthalpy Velocity (kJ/kg) (m/s) Inlet steam condition 3250 180 Exit steam condition 2360 5 The rate of heat loss from the turbine per kg of steam flow rate is 5 kW. Neglecting changes in potential energy of steam, the power developed in kW by the steam turbine per kg of steam flow rate, is (a) 901.2 (b) 9112 (c) 17072.5 (d) 17082.5 247. The maximum theortical work obtainable, when a system interacts to equilibrium with a reference environment, is called (a) Entropy (b) Enthalpy (c) Energy (d) Rothalpy 248. An isolated system is one, which (a) permits the passage of energy and matter across the boundaries (b) permits the passage of energy only (c) does not permit the passage of energy and matter aeross it (d) permits the passage of matter only 249. The measurement of thermodynamic property known as temperature, is based on (a) Zeroth law ofthermodynamics (b) First law ofthermodynamics (c) Second law ofthermodynamics (d) None of the above

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250. Which thermometer is independent of the substance or material used in constructions? (a) Mercury thermometer (b) Alcohal thermometer (c) Ideal gas thermometer (d) Resistance thermometer 251. A perpetual motion machine of the first kind i.e. a machine which produces power without consuming any energy is (a) possible according to first law ofthermo-dynamics (b) impossible according to first law ofthermo-dynamics (c) impossible according to second law of thermo-dynamics (d) possible according to second law of thermo-dynamics. 252. In Rankine cycle, regeneration results in higher efficiency because (a) pressure inside the boiler increases (b) heat is added before steam enters the low pressure turbine (c) average temperature of heat addition in the boiler Increase (d) total work delivered by the turbine increases 253. A process, in which the working substance neither receives nor gives out heat to its surroundings during its expansion or contraction, is called (a) isothermal process (b) isentropic process (c) polytropic process (d) adiabatic process 254. If 8Q is the heat transferred to the system and 8w is the work done by the system, then which of the following is an exact differential (a) fQ (b) 8W (c) 8Q+8W (d) 8Q-8W 255. The ratio of specific heats of a gas at constant pressure and at constant volume (a) varies with temperature (b) varies with pressure (c) is always constant (d) none of the above 256. The piston of an oil engine, of area 0.0045 m3 moves downward 75 mm, drawing in 0.00028 m3 offresh air from the atmosphere. The pressure in the cylinder is uniform during the process at 80 kPa, while the atmospheric pressure is 101.325 kPa. Find the displacement work done by the air finally in the cylinder. (a) 13J (b) 18J (c) 21 J (d) 27 J 257. Saturated liquid at a higher pressure PI having hfl = 1000 kJ/kg is throttled to a lower pressure P2. The enthalpy of saturated liquid and saturated vapour are 800 kJ/kg and 2800 kJ/kgrespectively. Find the dryness fraction of vapour after throttling. (a) 0.1 (b) 0.2 (c) 0.8 (d) 0.9 258. For which of the following situations, zeroth law of thermodynamics will not be valid? (a) 50 cc of water of at 25°C are mixed with 150 cc of water at 25° C (b) 500 cc of milk at 15°C are mixed with 100 cc of water at 15°C (c) 5 kg of wet steam at 100°C is mixed with 50 kg of dry and saturated steam at 100°C. (d) 10 cc of water at 20°C are mixed with 10 cc of sulphuric acid at 20°C.

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Thermal Engineerging

A-120 259. The compression ratio of a gas power plant cycle corresponding to maximum work output for the given temperature limits ofT minand Tmaxwill be

(a)

(Truax Tmm

t

y Y+1)

y

(b)

(Tmin ) 2(Y-l)

(Truax Tmln

y

y

y-l

(d)

(Tmm Tmax

260. The Carnot cycle consists of two reversible adiabatic processes and (a) two reversible isothermal processes (b) two reversible constant pressure processes (c) two reversible constant volume processes (d) one reversible constant pressure processes 261. Equal volume of all gases, at the same temperature and pressure, contain equal number of molecules. This is according to (a) Charle's law (b) Avagadro's law (c) Joule's law (d) Gay Lussac law 262. In the polytropic process equation, p v» = constant, if n = 1, the process is called (a) constant pressure process (b) constant volume process (c) constant temperature process (d) none of these 263. For a reversed Carnot cycle, which figure represents the variation of T L for different values of COP for a constant value ofT H = 300 K (say)?

Badboys2

~ 0

\OOK

TH = 300 K

~ 0

U

U

(a)

(b) TL,oK

( (d)

a )

y2) (Y - b) = RT

lP-

~ 0

U

U

(c)

(d) TL'°K

•

TH=300K

~ 0

\J TL'°K

264. In steam power plant the heat supplied to boiler is 3608 kJf kg. The enthalpies at the entry and exit of turbine are 2732 kJ/kg and 335 kJ/kg respectively. Ifthe efficiency of power plant is 64% then the efficiency of turbine will be (a) 0.93 (b) 0.94 (c) 0.95 (d) 0.96 265. Vander Waal's equation of state ofa gas is (a) pY=nRT

(p+ ;2 }V+b)=RT

266. In which case the work done is negative? (a) A rigid steel vessel containing steam at a temperature of 110°C is left standing in the atmosphere which is at a temperature of32°C (b) One kg of air flows adiabatically from the atmosphere into a previously evacuated bottle. (c) A rigid vessel containing ammonia gas is connected through a valve to an evacuated rigid vessel. The vessels, the valve and the connecting pipe are well insulated. The valve is opened and after a time, conditions through the two vessels become uniform. (d) A mixture of ice and water is contained in an insulated vertical cylinder closed at the top by a non-conducting piston, the upper surface is exposed to the atmosphere. The piston is held stationary while the mixture is stirred by means of a paddle-wheel protruding through the cylinder wall as a result some of the ice melts. 267. At STP, 8.4 litre of oxygen and 14 litre of hydrogen mix with each other completely in an insulated chamber. Calculate the entropy change for the process assuming both the gases behave like an ideal gas (a) 2.48kJ (b) 5.49kJ (c) 7.85kJ (d) zero 268. For an ideal gas the expression

[T(;)P-T(;)vN!;)p -T(;l.s

always

equal to

TL,oK

TH = 300 K

(b)

a )

y2) (Y - b) = RT

lP+

lTmaJ

y-l

(c)

( (c)

(a)

zero

Cp Cv

(b)

(c) R (d) RT 269. The pressure p of an ideal gas and its mean kinetic energy E per unit volume are related by the relation (a)

1 3

p=-E

(b)

p=-

3E

2

2 E p=-E (d) p=3 3 270. An ideal gas expands isothermally from volume vIto v 2 and then compressed to original volume vI adiabatically initial pressure is PI and final pressure is P 3. The total work done by gas is w, then (a) P3>P1,w>0 (b) P3 PI' w < 0 (d) P 3 = PI' w = 0 271. If during a process, the temperature and pressure of system are related by (c)

y-I

T2=(P2JY TI

then the system consists of

PI

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Thermal Engineerging

A-121

(a) any gas undergoing an adiabatic process (b) an ideal gas undergoing a polytropic process (c) any pure substance undergoing an adiabatic process (d) an ideal gas undergoing a reversible adiabatic process 272. In a gas turbine, hot combustion products with the specific heats Cp= 0.98 kJ/kgK, andCv=0.7638 K enter the turbine at20 bar, 1500K exits at 1bar. The isoentropic efficiencyof the turbine is 0.94. The work developed by the turbine per kg of gas flow is (a) 686.64kJ/kg (b) 794.66kJ/kg (c) 10009.72kJ/kg (d) 1312.00kJ/kg 273. The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressure p I and P2 are as shown in the figure. It can be concluded that

(c) R-T-3,P-S-l,P-T-4,Q-S-5 (d) P-T-4,R-S-3,P-S-l,P-S-5 277. Given below is an extract from steam tables. Enthalpy (kJ/kg) Specific volume (m3/kg) (bar) Saturated Saturated Saturated Saturated liquid vapour liquid vapour

Temperature (0C)

Psat

45

0.09593

0.001010

15.26

188.45

2394.8

342.24

150

0.001658

0.001658

1610.5

2610.5

Specificenthalpy of water in kJ/kg at 150bar and 45°C is (a) 203.60 (b) 200.53 (c) 196.38 (d) 188.45 278. A thin layer of water in field is formed after a farmer has a watered it. The ambient air conditions are: temperature 20°C and relative humidity 5%. An extract of steam tables is given below. Temperature (0C) Saturation pressure (kPa)

T----+

(a) PI represents monoatomic gas and P2 represents diatomic gas (b) the adiabatic index for PI is higher than that for P2 (c) the pressure PI is greater than the pressure P2 (d) none of the above 274. One kilomole of an ideal gas is throttled from an initial pressure of 0.5 MPa to kO.l MPa. The initial temperature is 300 K. The entropy change of the universe is (a) 13.38kJ/K (b) 4014.3kJ/K (c) 0.4621kJ/K (d) -0.0446kJ/K 275. The inversion temperature Ti ofa gas is related to the Van der Waal' s constants as

Badboys2

(a)

T-~ 1-

27R.b

(b)

T _ 27R.b 18a

T _ 2R.b 2a I(d) TI = R.b 8a 276. Group I shows different heat addition processes in power cycles. Likewise, Group II shows different heat removal processes. Group III lists power cycles. Match items from Group I, II and III. (c)

Group I Group II Group ill I P. Pressure S. Pressure 1. Rankine cycle constant constant Q. Volume T. Volume 2. Otto cycle constant constant R. Temperature U. Temperature 3. Carnot cycle constant constant 4. Diesel cycle 5. Brayton cycle (a) (b)

P-S-5,R-U-3,P-S-l, Q-T-2 P-S-l, R-U-3,P-S-4,Q-T-2

- 15 -10 0.1

0.26

-5

0.01

5

10

0.4

0.61

0.87

1.23

15

20

1.71 2.34

Neglecting the heat transfer between the water and the ground, the water temperature in the field after phase equilibrium is reached equals (a) lO.3°C (b) -lO.3°C (c) -14.5°C (d) 14.5°C 279. Which combination ofthe following statements is correct? The incorporation ofreheater in a stream power plant P. always increases the thermal efficiency of the plant. Q. always increases the dryness fraction of steam at condenser inlet. R always increases the mean temperature of heat addition. S. always increases the specific work output 280. Which one of the following is not a necessary assumption for the air-standard Otto cycle? (a) All processes are both internally as well as externally reversible (b) Intake and exhaust processesare constant volume heat rejection processes (c) The combustion process is a constant volume heat addition process (d) The working fluid is an ideal gas with constant specific heats. 281. In an air-standard Otto cycle, the compression ratio is 10. The condition at the beginning of the compression process is 100kPa and 27°C. Heat added at constant volume is 1500 kJ/kg while 700 kJ/kg of heat is rejected during the other constant volume process in the cycle. Specific gas constant for air = 0.287 kJ/kg-K. The mean effectivepressure (in kPa) of the cycle is (a) 103 (b) 310 (c) 515 (d) 1032 282. The thermal efficiency of an air-standard Brayton cycle in terms of pressure ratio rpand y (= c/c) is given by 1__ 1_ (a)

(c)

[y-I

p

1 1--

[I/y P

1-_!_ (b)

rY

(d) 1

(y-l)/y rp

p

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Thermal Engineerging

A-122

283. Which one of the following pairs of equations describes an irreversible heat engine? (a)

~8Q > 0 and ~8Q T
(b)

~8Q <0 and ~8Q T
(c)

~8Q>0

and ~8Q T>O

(d)

~8Q<0

and ~8Q T>O

284. For a gas turbine power plant, identify the correct pair of statements. P. Smaller in size compared to steam power plant for same power output Q. Starts quickly compared to steam power plant R Works on the principle of Rankine cycle S. Good compatibility with solid fuel (a) P, Q (b) R, S (c) Q,R (d) P,S 285. A diesel engine is usually more efficient than a spark ignition engine because (a) diesel being a heavier hydrocarbon, releases more heat per kg than gasoline (b) the air standard efficiency of diesel cycle is higher than the otto cycle, at a fixed compression ratio (c) the compression ratio of a diesel engine is higher than that of an SI engine (d) self ignition temperature of diesel is higher than that of gasoline 286. Rankine cycle efficiency for a power plant is 29%. The camot cycle efficiency will be (a) less (b) more (c) equal (d) none of these 287. Diesel cycle consists of (a) two adiabatic and two constant volume process (b) two adiabatic and two constant pressure process (c) two adiabatic, one constant pressure and one constant volume processes (d) two isothermal, one constant pressure and one constant volume processes 288. A Camot refrigeration system requires 1.5 kW per ton of refrigeration to maintain a region at - 30°C. The COP of system will be (a) 1.69 (b) 2.33 (c) 2.79 (d) 3.44 289. Brayton cycle can not be used in reciprocating engines for same adiabatic compression ratio and work output because (a) it requires large air-fuel ratio (b) it is less efficient (c) large volume oflowpressure air cannot be efficiently handled (d) all of these 290. The relative humidity is defined as the (a) mass of water vapour present in 1 m3 ofdry air (b) mass of water vapour present in 1 kg of dry air (c) ratio of actual mass of water vapour in a unit mass of dry air to the mass of water vapour in the same mass of

Badboys2

dry air when it is saturated at the same temperature and pressure (d) ratio of actual mass of water vapour in a given volume of moist air to the mass of water vapour in the same volume of saturated air at the same temperature and pressure. 291. The correct representation of a simple Rankine cycle naT s diagram is

(a)

(b)

i

1

T

T

(c)

(d) 1

4

S

--+

S ____.

292. With reference to air standard Otto and Diesel cycles, which ofthe following statements are true? (a) For a given compression ratio and the same state of air before compression. Diesel cycle is less efficient than an Otto cycle. (b) For a given compression ratio and the same state of air before compression. Diesel cycle is more efficient than an Otto cycle. (c) The efficiency of a Diesel cycle decreases with an increase in the cut-offratio. (d) The efficiency of a Diesel cycle increases with an increase in the cut-offratio. 293. A refrigerating machine working on reversed Carnot cycle takes out 2 kW per minute of heat from the system while between temperature limits 0000 K and 200 K. COP and Power consumed by the cycle will be respectively: (a) 1 and 1 kW (b) 1 and2kW (c) 2andlkW (d) 2and2kW 294. The bypass factor, in case of sensible cooling of air, is given by (a)

tdl - td3 td2 -dd3

(b)

td2 - td3 tdl -dd3

td3 - tdl (d) (c) td2 - dd3 where tdl = Dry bulb temperature of air entering the cooling coil, td2 = Dry bulb temperature of air leaving the cooling coil, td3 = Dry bulb temperature of cooling coil 295. An engine working on air standard otto cycle has a cylinder diameter 10 em and stroke length of15 em. IfV cis 196.3 cm3 and heat supplied is 1800 kJ/kg, the work output will be (a) 1080.78kJ/kg (b) 1282.68kJ/kg (c) 973.44kJ/kg (d) 1172.56kJ/kg 296. Efficiency of a diesel cycle with approach to otto cycle, when (a) diesel engine will operate at high speed (b) cut off period of diesel cycle is reduced to zero

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Thermal Engineerging

297.

298.

299.

300.

A-123

(c) diesel fuel is balance with petrol (d) none of these A engine is working on air standard diesel cycle.The engine has bore 250 mm, stroke 375 mm and clearance volume is 1500 cm'. If the cut off value is 5% of stroke volume the efficiency of engine will be (a) 53.25% (b) 60.5% (c) 64.89% (d) 67.75% Number of processes in a Rankine cycles are (a) 3 (b) 4 (c) 5 (d) 6 The comfort condition in air conditioning are at (a) OODBTandO%RH. (b) 20°CDBTand60%RH (c) 30°CDBTand 80%RH. (d) 40°CDBTand 90%RH. The dual combustion cycle consists of two adiabatic processes and (a) two constant volume and one constant pressure processes (b) one constant volume and two constant pressure processes (c) one constant volume and one constant pressure processes (d) two constant volume and two constant pressure processes The air standard diesel cycle is less efficient than the Otto cycle for the (a) same compression ratio and heat addition (b) same pressure and heat addition (c) same rpm and cylinder dimensions (d) same pressure and compression ratio An otto cycletakes in air at 300 k. The ratio of maximum to minimum temperature is 6 for maximum work output the optimum pressure ratio will be (a) 7.48 (b) 8.37 (c) 8.93 (d) 9.39 The mean effective pressure of an Otto cycle can be expressedas where (~P = Pressurerise during heat addition)

Badboys2 301.

302.

303.

(~).llth (a)

(c)

(Y-I)(r-l) (~).llth (y-l)(r-l)

(b)

(Y-l)(r-l) (~).llth

(d)

(Y-l)r

304. The diesel engine and otto engine has same compression ratio. The cut offratio of diesel engine is S. The air standard efficiency of these cycles will be equal when (a) Sf - r( s - 1) = 0 (b) Sf - r( s - 1) + 1= 0 (c) Sf - r(s - 1)- 1= 0 (d) Sf - (s - 1) - r = 0 305. Brayton cycle consists of sets of processes (a) isentropics and constant volume (b) isentropics and constant pressure (c) isothermal and constant pressure (d) isothermal and constant volume 306. For a given set of operating pressure limits of a Rankine cycle the highest efficiency occurs for (a) Saturated cycle (b) Superheated cycle (c) Reheat cycle (d) Regenerative cycle

307. For the same maximum pressure and temperature (a) Otto cycle is more efficient than diesel cycle. (b) Diesel cycle is more efficient than Otto cycle. (c) Dual cycleis more efficientthan Otto and diesel cycles (d) Dual cycleis less efficientthan Otto and Diesel cycles. 308. Ifcompression ratio of an engine working on otto cycle is increasedfrom5to 6, its air standardefficiencywillincreaseby (a) 1% (b) 20% (c) 16.67% (d) 8% 309. An air standard diesel cycle at fixed compression ratio and fixedr (a) thermal efficiency increases with increase in heat addition and cut offratio (b) thermal efficiency decreases with increase in heat addition and cut off ratio (c) thermal efficiencyremains the same wethe increase in heat addition and cut offratio (d) none of these 310. In an air standard otto cycle, the pressure in the cylinder at 30% and 70% ofthe compression stroke are 1.3bar and 2.6 bar respectively. Assuming that compression follows the law PV 1.3 = constant,what will be the air standard efficiency of cycle (a) 36% (b) 42% (c) 46% (d) 48% 311. The stroke and bore ofa four stroke spark ingition engine are 250 mm and 200 mm respectively.The clearancevolume is 0.001 m3. If the specificheat ratio y = 1.4,the air-standard cycle efficiency ofthe engine is (a) 46.40% (b) 56.10% (c) 58.20% (d) 62.80% 312. An engine working on otto cycle having compression ratio of 5. The maximum and minimum pressure during the cycle are 40 bar and 1 bar respectively. The mean effective pressure of cycle will be (a) 7 bar

(b) 7.89bar

(c) 9.04bar (d) 11.79bar 313. A heat pump works on a reversed cannot cycle. The temp in the condenser coil is 27°C and that in the evaporator coil is -23°C. For a work input of lkW, how much is the heat pumped? (a) 1 kW (b) 5 kW (c) 6kW (d) None of these 314. What is sol-air temperature? (a) It is equal to the sum of outdoor air temperature and absorbed total radiation divided by outer surface convective heat transfer coefficient (b) It is equal to absorbed total radiation divided by convective heat transfer coefficient at outer surface. (c) It is equal to the total incident radiation divided by convective heat transfer coefficient at outer surface. (d) It is equal to the sum of indoor air temperature and absorbed total radiation divided by convective heat transfer coefficient at outer surface. 315. In a Brayton Cycle, what is the value of optimum pressure ratio for maximum net work done b/w temperature. T I and T3,where T3is the maximum temperature and Tl is the minimum temperature?

Badboys2 0

('I') ('I') ('I')

Thermal Engineerging

A-124

(a)

(c)

fp=(~t

y-l

Y

(b)

fp=(~ t

y Y-1)

(d)

fp=(~r fp=(~

2(y-l)

Process in figure

Name of the process

P.

0-1

i

Q. R

0-2 0-3

ill.

S.

0-4

IV.

T.

0-5

v.

ii.

J-y-

316. Match list I (processer with) list II (Type) for Bell coleman or Joule or Reverse Brayton cycle for gas cycle refrigeration and select the correct answer using the codes given below the lists. List! List II A. Compression 1. Isobaric B. Heat rejection 2. Isothermal C. Expansion 3. Isentropic D. Heat absorption 4. Isenthalpic Codes: ABC D (a) 3 1 4 2 (b) 3 I 3 1 (c) 3 2 3 2 (d) 3 1 2 2 317. Match list I with list II and select the correct answer using the codes given below the lists. List I List II A. Pelton turbine 1. Specificspeedfrom 300 to 1000+ axialflowwith fixed runners vanes B. Prancis turbine 2. Specific speed from 10 to 50 + tangential flow C. Propeller turbine 3. Specific speed from 60 to 300 + mixed flow D. Kaplan turbine 4. Specificspeedfrom 300 to 1000+ axialflowwith adjustablerunner vanes Codes: ABC D (a) 2 1 3 4 (b) 4 1 3 2 (c) 2 3 1 4 (d) 4 3 1 1 318. Centrifugal pump have which ofthe following advantages? 1. low initial cost 2. compact, occupying less floor space 3. easy handling of highly viscous fluid (a) 1,2 and 3 (b) 1and 2 (c) 1and 3 (d) 2and3 319. Various psychrometric processes are shown in the figure below.

Badboys2

w (kg/kg)

(a)

Chemical dehumidification Sensible heating Cooling and dehumidification Humidificationwith steam injection Humidificationwithwater injection

P-i,Q-ii,R-iii,S-iv,T-v (b) P-ii,Q-i,R-iii,S-v,T-iv (c) P-ii,Q-i,R-iii,S-iv,T-v (d) P-iii,Q-iv,R-v,S-i,T-ii 320. For a typical sample of ambient air (at 35°C, 75% relative humidity and standard atmospheric pressure), the amount of moisture in kg per kg of dry air will be approximately? (a) 0.002 (b) 0.027 (c) 0.25 (d) 0.75 321. The statements concern psychrometric chart. 1. Constantrelativehumiditylines are uphill straightlines to the right. 2. Constant wet bulb temperature lines are downhill straight lines to the right. 3. Constant specific volume lines are downhill straight line to the right. 4. Constant enthalpy lines are coincident with constant wet bulb temperature lines Which of the following statements are correct? (a) 2and3 (b) 1and 2 (c) 1and 3 (d) 2 and 4 322. In a Pelton wheel, the bucket peripheral speed is 10m/s,the waterjet velocityis 25 mls and volumetric flowrate of thejet is 0.1 m3/s. Ifthe jet defletion angle is 120° and the flow is ideal, the power developed is (a) 7.5kW (b) 15.0kW (c) 22.5kW (d) 37.5kW 323. Dew point temperature is the temperature at which condensation begins when the air cooled at constant (a) volume (b) entropy (c) pressure (d) enthalpy 324. The stroke and bore of a four stroke spark ignition engine are 250 mm and 200 mm respectively.The clearance volume is 0.001 m'. Ifthe specificheat ratio y = 1.4, the air-standard cycle efficiency of the engine is (a) 46.40% (b) 56.10% (c) 58.20% (d) 62.80% 325. If a mass of moist air in an airtight vessel is heated to a higher temperature, then (a) specific humidity of air increases (b) specific humidity of air decreases (c) relative humidity of air increases (d) relative humidity of air decreases

o, C)

I

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Thermal Engineerging

A-125

I···..

ANSWER KEY

1 (c) 26 (b) 51 (c) 76 (b) 101 (b) 126 (b) 151 (d) 176 (c)

•

••

,

1

201

(c)

226 (b) 251

(b)

276

(a)

301 (a)

2 (c) 27 (c) 52 (d) 77 (b) 102 (a) 127 (a) 152 (a) 177 (d) 202

(a)

227 (d) 252

(c)

277

(d)

302 (d)

3

(d) 28 (d) 53 (b) 78 (d) 103 (a) 128 (b) 153 (a) 178 (c)

203

(a)

228 (a) 253

(d) 278

(c)

303 (b)

4

(a) 29 (b) 54 (a) 79 (b) 104 (b) 129 (b) 154

(d) 179 (b)

204

(b)

229 (d) 254

(d) 279

(b)

304 (a)

5 (b) 30 (c) 55 (c) 80 (a) 105 (d) 130 (a) 155 (a) 180 (b)

205

(a)

230 (b) 255

(c)

280

(b)

305 (b)

6 (c) 31 (a) 56 (a) 81 (d) 106 (a) 131 (a) 156 (b) 181 (a)

206

(b)

231 (c) 256

(d)

281

(d)

306 (d)

7 (c) 32 (d) 57 (a) 82 (a) 107 (c) 132 (b) 157 (c) 182 (b)

207

(c)

232 (c) 257

(a)

282

(d)

307 (b)

8 (b) 33 (a) 58 (c) 83 (b) 108 (b) 133 (b) 158 (a) 183 (c)

208

(a)

233 (c) 258

(d) 283

(a)

308 (d)

9 (a) 34 (b) 59 (d) 84 (a) 109 (a) 134 (b) 159 (b) 184 (a)

209

(c)

234 (b) 259

(a)

284

(a)

309 (b)

10 (c) 35 (a) 60 (a) 85 (c) 110 (c) 135 (a) 160 (b) 185 (d) 210

(a)

235 (a) 260

(a)

285

(c)

310 (c)

11 (b) 36 (b) 61 (b) 86 (c) 111 (a) 136 (d) 161 (c) 186 (a)

211

(b)

236 (a) 261

(b)

286

(b)

311 (c)

12 (a) 37 (b) 62 (c) 87

(d) 112 (a) 137 (b) 162 (b) 187 (b)

212

(a)

237 (b) 262

(c)

287

(c)

312 (c)

13 (b) 38 (c) 63 (a) 88 (c) 113 (a) 138 (a) 163 (c) 188 (a)

213

(c)

238 (a) 263

(b)

288

(b)

313 (c)

39 (c) 64 (c) 89 (a) 114 (b) 139 (d) 164 (a) 189 (b)

214

(a)

239 (d) 264

(d)

289

(c)

314 (a)

15 (b) 40 (a) 65 (d) 90 (b) 115 (b) 140 (b) 165 (a) 190 (a)

215

(a)

240 (a) 265

(c)

290

(d)

315 (b)

(a)

316 (b)

14 (c) Badboys2

16 (c) 41 (c) 66 (b) 91 (d) 116 (a) 141 (a) 166 (a) 191 (d) 216

(d) 241 (a) 266 (b,d) 291

17 (a) 42 (b) 67 (b) 92 (a) 117 (d) 142 (d) 167 (c) 192 (b)

217

(b)

242 (d) 267

(b)

292

(a, c)

317 (c)

18 (c) 43 (b) 68 (a) 93 (d) 118 (a) 143 (b) 168 (d) 193 (c)

218

(a)

243 (a) 268

(c)

293

(c)

318 (d)

19 (b) 44 (d) 69 (b) 94 (a) 119 (c) 144 (a) 169 (d) 194 (d) 219

(b)

244 (b) 269

(c)

294

(b)

319 (b)

20 (a) 45 (d) 70 (d) 95 (c) 120 (a) 145 (a) 170 (b) 195 (d) 220

(b)

245 (a) 270

(c)

295

(c)

320 (b)

21 (a) 46 (b) 71 (a) 96 (b) 121 (b) 146 (b) 171 (d) 196 (c)

221

(a)

246 (a) 271

(d) 296

(b)

321 (a)

22 (b) 47 (b) 72 (d) 97 (a) 122 (c) 147 (c) 172 (b) 197 (b)

222

(b)

247 (c) 272

(a)

297

(b)

322 (b)

23 (c) 48 (d) 73 (d) 98 (d) 123 (a) 148 (b) 173 (a) 198 (a)

223

(a)

248 (c) 273

(c)

298

(b)

323 (c)

24 (b) 49 (a) 74 (c) 99 (b) 124 (b) 149 (d) 174 (c) 199 (c)

224

(b)

249 (a) 274

(c)

299

(b)

324 (c)

25 (a) 50 (b) 75 (a) 100 (d) 125 (d) 150 (b) 175 (b) 200 (c)

225

(a)

250 (c) 275

(d) 300

HINTS & EXPLANATIONS 5.

(b) Considring the followingdiagram, ~l'

1

C S Q) !ZJ !ZJ Q)

1-0

P-.

..., (a)

12. (a) As we know that, F-32 C = -100 180

-

C= 100(F_32)=~(F_32) 180 9 ~2

Volume (V) ~ hence, (QA -wA) = (QB -wB)

5 C =-(F -32) 9 13. (b) Let, Cp = specific heat at constant pressure

325 (d)

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Thermal Engineerging

A-126

19.

C, = specific heat at-constant volume then, Cp - C, = R (gas constant) where, R = 287 Jzkg" K (b) As we knonw that, during an adiabatic process, PvY = constant PIVIY=P2 V/=C Here, PI =P,P2=M>, V1=Y, V2=tN

T2=47°C=47+ 273 =320K work produced (wp) = 150 KJ T2- T1 (320- 309) Now, efficiency (11)=T= 320

11

11= = 0.0344 320 We also know that,

Hence tN = _!_ ( dP) 'V y P 28.

thenvoc-~-=1

VI

.JP

30.

Here,

Q1

105

=

42 200

21 = 80

150 QA

Q

74.

3

Q1

Q2 T2

55.

PI

"2 KT

T; = 400

Now,

41.

V2

Here, T =0 Hence, K.E = 0 (b) Given: Q1 + 105 MJ, Q2 = 42 MJ T1=400K, T2=200K

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35.

0.0344=

~2 -

(c) Given: T = 0 Now considering kinetic energy equation of gases, (kinetic energy) K.E =

34.

. () work Produced ( w p ) efficiency 11 = ---------,-----'--,-:::--:Heat added (QA)

(d) Let v = mean square molecular velocity p = density,

(c)

=~=4360KJ A 0.0344 Given Latent heat ofvapourization Temperature (T) = 100 K change of entropy (ds) =

dQv

T

= 2560 = 25.6 KJ IK Ok 100 g

123. (a) Given: In a Brayton cycle, Initial temperature (T) = 300 K Final or maximum temperature (Tf) = 1200 K

=~ 100

* Q2

T; ~

So, Engine is not possible. (a) Given: Initial volume (v) = 0.03 m' Final volume (v2) = 0.06 m' Pressure (P) = 1 MPa = 1 x 103 kPa Heat absorbed (ow) = -84 KJ Now, using first Law of thermodynamics, dQ=ow+du du = dQ - Ow= -Pdv - dw =-103(0.06-0.03)-(-84) =-103 x 0.03 + 84 =-30+ 84= 54KJ (c) Given: Initial volume (v) = 4m3 Final volume (vf) = 2 m' Pressure (P) = 4.2 kg/emwork done (w)p = Pdv =4.2 x 104 x (4-2) =4.2x2xl04 = 8.4 X 104 joule As the system undergoes a reversible process, then, Heat (Q)H= 20 x 4.2 X 104 = 8.4 x 104 joules change in internal energy = w D - QH =8.4x 104-8.4 x 104=0 (c) Given, TI = 36° C = 36 + 273 =309K

(dQ) = 250

Tf-Ti)

thermal efficiency (11thermal ) = ( ~

x 100

= 12000 - 300 x 100

1200 900 =--xl00= 75% 1200 125. (d) Given: Bore diameter (D) = 0.20 m stroke (L) = 0.25 m speed (N) = 600 rpm Actual volume delivered = 4m3/min = 0.0667 mvs swept volume (v) = ~ D2L = ~ x (0.2)2 x 0.25 =0.00785m3 Now using the following relation, Volumetric efficiency (11vot)=

Volume of air induced swept volume

we get, 11vol = 85%

131. (a) Given: First conditions: TI = 900 K, T2=T2 Second conditior, T I = T2' T2= 400 K Efficiency for first, condition, (First Engine) 111= T2 - T1 = T2 - 900 = 1- 900 T2 T2 T2 Efficiency for second condition, (second Engine) _ T2 -T1 _ 400-T2 112T2 400

-1-

T2 400

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Thermal Engineerging

A-127

136. (d) Given: coefficient of performance (COP)p = 5

Now, 11,= 112

work input (W,/p)= 1 kw

1- 900 =1- T2 T2 400

(COP)p=~=Qa=5

T2 = '-'360000 = 600K

132. (b) Given: work developed / produced (W) = 60 kw Efficiency (11)= 60% or 0.6 considering the following formula, . () Effirciency 11 =

11=-

1

WI/P

T22=360000

work Produced (w ) ( ) Amount of Heat Generated QG

Q =5kw Now, (COP)p-(COP)R =w= 1 5 -(COP)R = 1 (COP)R = 5-1 =4 Now,

(COP)

R

= 4 = Qar = Qar W 1 liP

Q =4x 1=4kw COP of refrigerator' l' = (COP), = 4 COP of refrigerator '2' = (COP)2 = 5 Now, COP of composite refrigeration system (COP)c;

w

137. (b)

QG 0.6 = 60 QG

(COp)

60 QG =-=100kw 0.6 Now, Amount of heat rejected (QR) = Qa - w = 100-60=40kw. 134. (b) Given: Heat received (Q,) = 1120kT Heat rejected (Q2) = 840kJ Temperature Limits, T, = 560 K, T2= 280 k Now, considering clausing inequality,

840 280

2

(T)= 37°C =37+273=31OK Evaporator temperature (T E) = - 3°C =-3 +273 =270K

(COp) Refrigerator=

270

TE Tc-TE

310- 270

270 = 6.75 40

(COp) Refrigerator= (COp) Refrigerator+ 1

Ql *- Q2 Hence, -T T 1

4x5 1+4+5

(COP)c = 2

3

-=-=-

=

138. (a) Given: condensortemperature

Ql = 1120 = 14 = ~ Tl 560 7

T2

(COP)l x (coph 1+ (COP)l + (coph

20 (COP)c =-= 10

Badboys2

Q2

= c

2

Hence, engine operates on Irreversible cycle 135. (a) Considering the following Temperature Entropy(s) diagram:

(T) -

=6.75+1 =7.75 141. (a) Given: Enthalpy after compression (~) = 425 KJ/kg Enthalpy after throttling (H4) = 125 KJ/kg Enthalpy before compression (H) = 375 KJ/kg Here,H3=H4 Considering the following block diagram,

Throttle

Now, we know that, L\Q =

f T.ds

f T.ds

2-

375 -125

B

QA-B =

Hl-H4 ()COP Refrigerator= H H

= T.(SB - SA)

A

dQ = T.ds Hence, Area under T - S diagram shows heat transfer for reversible process.

1

250

425-375 50 (COP).Refrigerator =5 142. (d) Given: coefficient of performance (COP) = 4 Let T, = Higher temperature T2= Lower temperature

Badboys2

Thermal Engineerging

A-12S

COP =

_1_ COP

____.:!l__ T1-T2

= Tl -

T2 =..!L-l T2 T2

7Vc = 307.72

307.72

=> Vc = -7-

V =43.96cm3=44cm3

202. (a) Weight of body = 100N,dragforce(Fo)=0.5N

..!L= ~= 1.25 T2 4

145. (a) Given: TJ = 27° C TJ =27+273=300k T2=-23°C T 2 =-23+273=250k

(

_ ____:!i_ _ 300 = 300 = 6 ) COP cornot - Tl - T2 - 300 - 250 50

166. (a) Given: Total volume = 400 cc

=O.4(T

compressed volume = 100 cc r = 1.5 400 Comression ratio (C.R) = 100 = 4

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I thermal efficiency of cycle ( 11thermal ) = 1- (

)r-l

C.R =1

1 (4)1.5-1

1 11thermal = 1- (4)0.5

distance covered (s) = 50 m work done (w) = Fox 50 =0.5 x 50 =25J Now, work interaction is work is performed by the system on the surrounding i.e. - 25 1. 205. (a) Given: mass of steel ball (m) = 1 kg Specific heat (C) = 0.4 KJ/kg temperature (~T) = 60° C For system energy equilibrium/conservation, dU=O mC ~T=1 x 0.4 x (T-60) -60) For water, temperature = 20° C mass (m) = 1kg specific heat (cw) = 4.18 KJ/kg m C ~T= I x4.18+(T-20) now. dU:'mC ~T+m C ~T=O , 0.4 (T _s60)+4.18 (T-:_20) =0 O.4T-24 +4.18T -83.6=0 4.58T -107.6 = 0 4.58T= 107.6 T = 107.6 = 23.49°C ::;:: 23.50C 4.58

1 = 1-(4)112 207. (c) = 1__ 1_= I-.!.=

.J4

1-0.5

2

11thermal = 0.5 or 50%

184. (a) Given: speed (N) = 3000 rpm

186. (a)

3000 3000 Power stroke / s = __ = _= 25 2 x 60 120 Given: Bore diameter (D) = 7 em Stroke length (L) = 8 em compression ratio (C.R) = 8 Swept volume (V) =

"4 D 2L 7t

() Heat rejected COP system = Power Consumed

210. (a)

= 120 = 4 30 Given: Heat Temperature Heat rejected Temperature

received (Q) = 80 KJ (TJ) = 100°C (Q2) = 70 KT (T2)= 30° C

Thermal efficiency ( 11th. ) = = 80-70

Ql -Q2 100 Ql x

xlOO

80 =~x(7)2x8 4 =307.72cm3 considering the following relation, C.R

Vs -v, = ____::__-=Vc

Where, Vc = clearanace volume

= !Qxl00 80 =12.5% 214. (a) Given volume of vessel (v) = 1 m' steam volume (Vs) = 0.9 m' water volume (V w) = 0.1 m' At 1 bar, Vf= 0.001 mvkg V g = 1.7m3/kg

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Thermal Engineerging

A-129 =0.03KN-m

Vs 0.9 steam mass ( ms ) = - = - = 0.5294 kg Vg 1.7

. () work done Mean effectIve pressure MEP = ---volume

( ) v:

0.1 water mass mw = = -= 100 kg Vf 0.001

0.03 0.02

0.5294 Now, Dryness fraction =

ms

+

=0.005266 = 5.266 X 10-3 = 5.27 X 10-3 215. (a) Given: a=20°, ~ =200 mis, VS1 = 350mls ~l = 30°, ~e = 25° As, VS1 = VS2 = 350 mls use the following diagram and formula,

IE

VW1

>I<

3

0.5294+100

mw

)1

VW2

2 = 1.5 KN/m2 220. (b)

= 1.5 kPa Given: Bore (d) = 2 unit, stroke (L) = 2 unit

Total area for heat loss = ~d2 + ndl. = ~(2)2 + n (2)(2)

4

4

=n+4n =5n 226. (b)

T = constant (

Thus, ~~ Ub(VwJ + VW2) work done I kg = ---'------"1000 We get, work done/kg = 124.79 KJ = 124KJ 217. (b) Given: Rotor diameter (d) = 2 m speed (N) = 3000 rpm nozzle angle (a) = 20°

'1

a

(

a

'1

lp, + vfjV =lp2 + vljV t

2

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ndN

Blade velocity

(Ub) = 60 =

1

=> P2 =10x2+

n x 2 x 3000 227. (d)

60

=314m/s 2ub

Now, optimum velocity of steam ( vopt. ) =-cos

v

218. (a)

o;

229. (d)

2x314 2x314 -----opt. - cos 20° - 0.9396 230. (b)

= 668.4 mls :::::: 668 mls Given

KN/m

2

5

A

1'---"","

131. (c) .____.___-........__--+

0.01

3

V (m )

0.03

In the graph above, In dABC, AB = 3 KN/m2, BC = 0.03 - 0.0 1 =0.02m3 Now, work done = Area of dABC

1

=-xABxBC 2 = _!_ x 3 x 0.02

2

232. (c)

a a a 1x2 - 22 = 5+4

Asa>O, p> 5 In reciprocating compressor, at initial point ofsuction and final point of compression a little higher value of pressure is required to open the inlet and outlet value respectively. Heat is positive when added to system, is in exact differential path function and boundary phenomenon. Work is negative, inexact differential, path function and transient phenomenon. Heat generated by bulb = 100 x 24 x 60 x 60 J = 8.64 x 106 J . . Heat dissipated = (L x v) x [Cy (T - 20)] .. 100 x 24 x 60 x 60=(1.20 x 3) x 2.5 x 3 x Cy(T -20) 0.32 x 106 = Cy (T - 20) = 1000 x 1.004 (T -20) => T=338.72°C First of all, process (1-3) is adiabatic, means a vertical line in T-8 diagram. As given figure is clockwise for (1-2-3) so from Figures 1 and 2, clockwise (1-2-3) will be selected. Under steady-state flow conditions, d W=-dH+dQ ...(i) Also, in reversible process, T. d 8 = dH - Vdp ... (ii) =>- Vd P = - dH + T. d8 From Eqs. (i) and (ii), we get dW=-Vdp Integrating both sides, we get W=-jVdp

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Thermal Engineerging

A-130

233. (c)

234. (b)

Enthalpy of balloon is given by H=U+Pr Initially balloon kept in insulated and evacuatedroom. => No heat transfer from outside. ~e = 0 Also, gas does not have to do any work against any external pressure. => ~ W=O From 1stlaw, => ~Q=~U+~W ~U=O Further, between initial and final states, total energy or enthalpy remains same for the gas. The change in pressure and volume is suchthat their product remains constant. Hence, h also remains constant. dH=dU +d(pVO) dH = dU + pdV + Vdp

f dH = f dq + f Vdp

T2=396.30K W

8.314 x(300- 396.30) 1.67-1 W

~kf;+~p.E+ W =-fVdp

kmol = 1194J/mol Imol=Mg=40 g For 40 g, W=1194

239. (d)

240. (a)

2

~Ssystem+ ~Ssurrounding~ 0 AL= Vs(Sweftvolume)= 0.0259m3, Pem =? N=2200rpm 1 For 4-stroke diesel engine, K = "2

1

235. (a)

1194 For 1kg W= --xl000 , 40 =29.7 kJ/kg In every case, entropy of universe is always positive. (L\S)universe~ 0

W=-fVdp

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= 1194__!!_

m

~=Q+fVdp Q-~=-fVdp

R(TI-T2) y-l

m

P=950kW=950

Work is done on the system.

Pem xVs xNx950xl03 =

= 8.31kJ 237. (b)

Due to internal friction produced in irreversible process, entropy of the system increases. From Clausius inequality, Cyclic integral of di < 0 for irreversible process

io'w

P = Pem xALxNxK 60

V2 Wisothermal= PIVIenVI x x 6 0.030 -0.8 0.015 10 en 0.015

236. (a)

x

60

1 2

950 xl 03 x 60 x 2 Pem = 0.0259 x 2200 =2 x 106 Pa=2MPa 241. (a)

H1

Ql + Q2 + Q3 + ....+ Qn < 0 Tl T2 T3 Tn For process I, 2500 _ 2500 = -1.042 1200 800 For process II,

238. (a)

2000 _ 2000 = -1.5 800 500 Process II is more irreversible than process I. In adiabatic process,

HI =2800kJ/kg H2= 1800kJ/kg Work done = HI - H2= (2800 -1800) kJ/kg = 1000kJ/kg Then, specific steam consumptionm

W = PIVI - P2V2 y-l

3600 = 1000 =3.60kg/kW-h

y-l

_]_= T2

(RL]r P2

=> 300 = T2

0.67 (~)1.67 = (0.5)0.4012 2

242.(d)

d=80mm Strokelength L = 2 x Crank radian =2 x60= 120mm Then, swept volume Vs= A x L = ~d2 xL 4

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Thermal Engineerging

A-131 258. (d)

= ~ x (80)2 x 120 244. (b)

two systems which are equal in temperature to a third system, they are equal in temperature to each other '.

=603cm3 We have dQ = dU + pdU mRTdV or T dS = mCvdT + --V-dS = 8+mR dV V (as isothermal process, dT = 8) or

263. (b)

fIdS=mRfdVI V

or

Accordingly when 50 cc of water at 25°C are mixed with 150cc of water at 25°C, the resulting temperature ofthe mixture will be 25°C, Same analogy applies to situations in (b) and (c). However,this argument is not valid when water and sulphuric acid, initially at the sametemperature,are mixed.Heretemperaturewill rise due to chemical reaction - the change is often violent. For reversed Carnot cycle, COP =

TL h-TH

For a fixed value of TH' as TL increases,COP also increases but not linearly. In fact COP decreases with increasing differencebetween operating temperatures.

~S=mR£n V2 VI ( PI' ~S=mR£nlp2)

According to zeroth law of thermodynamics, "when

(asPIVI =P2V2)

264. (d)

11=

WTurbine QSuppIied

(2732-335) 0.64 = 11T --'----3-6-08---'245. (a) 11T=0.96 266. (b,d) Signs of work for the four cases are given below (a) 0 (b) '-' ve (c) 0 (d) '-' ve

246. (a)

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and 8=-5 KW m= 1kg/s 247. (c) In thermodynamics, energy or available energy of a system in the maximum useful work possible during a process that brings a system into equilibrium with surroundings (heat reservoir).

267. (b)

x2 =

n2 = 0.625 nl +n2

~S = -

256. (d)

257. (a)

R (0.375 £n 0.375 + 0.625 £n 0.625)

= 0.66 R = 5.49 Jik

L....-------------+S

254. (d)

0 .375

14 n2 =-=0.625 22.4

3

The object of the regenerative feed heating cycle is to supply the working fluid to the boiler at same state between 2 and 2' (rather than at state 2) there by increasing the average temperature of heat addition to the cycle. du = 8Q- 8W Since du is the property and it is exact differential so 8Q- 8W is the exact differential. Here we have to find out the work done an the air in the cylinder. work = change in volume due to piston displacement x pressure inside the piston = 0.0045 x 0.075 x 80 x 103 =27 joule. In throttling process enthalpy remains constant. h, =h2 1000= 800+x (2800- 800) x=O.1

R (nl£nxl + n2 £n X2)

8.4 nl =--= 22.4

252. (c)

T

~S = -

269. (c)

1 221 2 pV =-mnc = -.-mnc 332 or

3p = _!_ mnc2 = E 2

or

270. (c)

2

V

2E

p=-

3

3

1

p

l~

i,, ,, , ,, ,,, ,, ,,

:2 ,,, ,, ,, ,, ,,, ,,

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Thermal Engineerging

A-132 279. (b)

P3 > PI

272. (a)

Wgas < 0 As we know that slope of isothermal process in PV diagram is less than slope of adiabatic process in PV diagram. Thats why P3 > PI and from the process it is clear that work done is negative. Cp = 0.98, = 0.7638 PI =20bar, T3 = 1500k P2 = lbar, 11=0.94

Incorporation ofreheater in a steam power plant always increases dryness fraction of steam at condenser inlet and always increases specific work output.

280. (b) End of combustion

c,

p

~ Exhaust valve open

I

11= Cp =~=1.28305 0.7638

c,

Patm

11-1 0.28305 T4 = (P4 , ----;- => = (20) 1.28305 T3 lp3) 1500 1

l

Intake

L...-----1f----------+----.v

T4 = 29047434k. 11= T3 - T4' T3 - T4

Badboys2 274. (c)

=> 0.94 =

1500- T4 1500-2904.7434

~suniverse = ~Ssyst + DSsurrounding

. P2 =R IogJ-=8.314 PI

0

(Throtteled)

I

~----;------------------;--~v

0.1

og0.5

TDC BDC Air-standard auto cycle with four reversible processes 1-2; isentropic compression 2-3; V = constant heat addition 3-4; isentropic expansion 4-1; V = constant heat rejection From the first figure, it can be seen that intake and exhaust are not constant volume processes.

~suniverse= 13.38 kJ/k Gases become cool during Joule Thomson's expansion only if they are below a certain temperature called inversion temperature T I: The inversion temperature is the characteristic of each gas. It is related to the Van der Waals' constants 'a' and 'b' by the relation

p

T _ 2a

281. (d)

1 - R.b 276. (a)

3

T'4= 2820.45 Work w= Cp (T3-T'4) w= 0.98(1500 - 2820.45) w= 1294.049 kJ/kg. ~ssurrounding = ~Su = ~Ssys

275. (d)

BDC

TDC p

Pressure constant heat addition and pressure constant heat removal are Brayton cycles. Constant

temperature

heat addition

and constant

temperature heat removal are Carnot cycles Pressure constant heat addition and pressure constant heat removal are Rankine cycles.

v

Volume constant heat addition and volume constant heat removal are Otto cycles. 277. (d)

F or a given saturation pressure, iftemperature is lower than the saturation temperature liquid or compressed

then it is subcooled

liquid. For 150 bar pressure

saturation temperature is 342.24. But as temperature is lower than that, thus it is compressed liquid at 45°C, specific enthalpy would be 188.45 kJ/kg.

Given, PI = 100 kPa Tl =27+273=300K Heat supplied (process 2-3) Qs = 1500 kJ/kg Heat rejected (process 4-1 ) QR = 700 kJ/kg Gas constant for air, R = 0.287 kJ/kg-K

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Thermal Engineerging

A-133 V3- v2 = 0.05 (VI - v2)

.. VI Compression ratio, r = 10 = V2

(:~ -I)=O.OS[ :~ -I]

Now, mean effective pressure is given by Work done Pmean = ----Swept volume V4 Vt Now -=-=10 'V3 V2 => VI = 10V2 Also swept volume VS=VI-V2 => VS=0.9VI Initiallyfor air

~diesel = 1- (r

...(i)

" _ nRT _lxO.287x300 vt - ~ 100

285. (c)

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288. (b)

The air standard diesel cycle is less efficient than the Otto cycle, given the same compression ratio and heat addition. However, it is more efficient than the Otto cycle with the same peak pressure and heat addition.

302. (d)

3

Qs - QR

W 800 Pmean = Vs = 0.7749 = 1032.39kPa

For same compression ratio and the same heat supplied, otto cycle is most efficient and diesel cycle is least efficient. In practice, however, the compression ratio of the Diesel engine ranges between 14 and 25 whereas that of the otto engine between 6 and 12. Because of its higher efficiency than the otto engine.

TI =300k T3 =6 TI T3 = 1800k we know that for maximum work output T2T4=TIT3

COP = Refrigeration effect = ~ = 2.33 Work done 1.5 200

293. (c)

300-200

2

T2 = ~TtT3

T L = Lower temperature T H = Higher temperature

RE

2

Power= --=COP 295. (c)

1t

Vs =-d

4

2

2 1t

L = -(10)

4

JTIT3 T4 = JTtT3 T3 =

=lkW 2

x15=1177.5cm

T2

= .)1800 x 300 = 734.84 k t

3 ~=(T2Jr-t v2 TI

r = 1+ ~ = 1+ 1177.5 = 1 +5.99 ~7 Vc 196.3

W

1l=-=1--Qs

1

r= (~ )r-I =9.39.

1

(r)y-l 303. (b)

~=1-

1800

(7iA-1

t

W

Pm=Vs ll+n

W = 973.44 kJ/kg 297. (b)

Vs = ~(25)2 (37.5) = 18398.43 em3 4 vc = v2 = 1500 em3 r = 1+~ = 1+ 18398.43 = 13.26 Vc 1500

= 60.5

301. (a) =0.861 m3/kg

.. Vs=0.9x 0.861 =0.7749m3/kg Work done in cycle W = Heat supplied - Heat rejected = = 1500 - 700 = 800 kJ/kg

=>

)~-Il~(ct_~; l

1 1(l.61)l.4_1] lldiesel = 1- (13.26)OAll.4 x 0.01

PIVl =nRTI

..

rc -1 = 0.05 [12.26] rc = 1.61

h

W

W

= - = ----Qs mcv(T3-T2)

_ W +n - mR~T --

y-1

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Thermal Engineerging

A-134 more efficient.

w = 11+nmR.1T = 11+nM>vc y-1

:.

(Y-1) 308. (d)

11DieseI>11DuaI> 110tto

11= 1__

1_=

M

.1n

-xl00= n

Vs =(r-l)vc Pm

=

11+n(.1p)vc

-(I-r)-xl00 r

= (r-1)-x

-...;,...:=........:...__~-=----

M

6-5 = (1.4-1)x-5-x100

11+n(M» = (y-l)(r-l)

310. (c)

(r)~-l=1- (r)~-I[r(:=~)]

I I I

I I --4--------1 >I P 11 V' I 2

sr-1 = r( s - 1) sr-I - r( s - 1) = 0 Efficiency of ideal regenerative cycle is exactly equal to that of the corresponding Camot cycle. Hence it is

maximum,

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307. (b)

=0.08 x 100=8%

11otto= 11diesel

1-

306. (d)

100

r

(Y-1)(r-1)vc

304. (a)

l-(r)y-1

(r)Y-I

70% V2 =1 v2' = 1 +0.7 (r-l) =0.7r +0.3 v2' = 1 +0.3 (r-l) =0.3r+0.7

Following figures shows cycles with same maximum pressure and same maximum temperature. In this case, otto cycle has to be limited to lower compression ratio to fulfil the condition that point 3 is to be a common state for both cycles. T -S diagram shown that both cycles will reject the same amount of heat.

3

4

11=1---

=1 V

=1.7

1.3

1.7

0.3r+0.7 r=4.68

P

V---+

.'. =(2.6)1.3

0.7r+0.3

1

(r)Y-I 1

--1-:-4-:---:-1 = 0.46 = 46%

(4.68) .

311. (c)

vc=O.OOI m3 Ys

7t 2 3 = -x 0.200 x 0.250m

4

= 1-[ S Thermal efficiency= 1

30%

lIx ~=(_!i_J v2 P2

2

v,1 11

0.001+~xO.2002 Qrejected = 1 Constant Qsup plied Qsupplied

Thus the cycle with greater heat addition Qsuppliedis

312. (c)

]1.4-1 =58.2%

0.001

Given PI = 1 bar P3 =40bar r=5

xO.250

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Thermal Engineerging

A-135

320. (b)

i.e., 0-3 Humidification and steam injection - temperature increases and w increases to i.e., 0-5 Humidification and water injection - temperature decreases but w increases i.e., 04 On a psychrometric chart 75% RH

P2 =Pl( 11=1---

:J 1

(r r'

Qs

0.025

=1.(5)14 =9.51 bar

= 1---

1

(5)°.4

=0.4746

R(T3-T2) r-1

= Cv (T3 - TI) = _____..:.._--=----~

321. (a)

v2 (P3 - P2) (40 - 9.51) x vc = = 76.255vc r-1 1.4-1 1+~=5

322. (b)

Vc

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Vs

= 4vc

W 11=-

30°

On a psychrometric chart Constant relative humidity lines are uphill curve not straight, to the right. Constant WBT lines straight downhill to the right. Constant specificvolumedownhill straight to the right. Constant enthalpy lines are not coincident to WBT. P=2u(v-u)(1 +cos<j» x flowrate = 2 x 10(25-10) (1 + cos 120°) x 0.1 = 20 x 15 x 0.5 x 0.1 = 15kW

323. (c)

Qs

T 0.4746=

W 76.225vc

Tdew

W = 36.17vc Pm = W = 36.17vc = 9.04 bar Vs 4vc 313 . ()c

324. (c)

=~ 319. (b)

s

· ..c. I ( ) QI QI H mt: usmg rormu a cop = W = Q2-Ql

T2-Tl Chemical dehumidification- temperature increases,w decreases, i.e., 0-2 Sensible heating - straight horizontal line towards right, i.e.0-1

Air is cooled at constant pressure to make unsaturated air to saturated one, Given, StrokeL = 250 mm Diameter, d=200mm Clearance volume Vc=0.00Im3 Now, swept volume 1t

Vs =-d

4

11-1-(

2

1t

L=-(200)

2

4

Vc

VC+VS

x250cc

JY-1

Substituting values, w (kg/kg)

1.4-1

11-1-

0.001 [ 0.001+~(0.2)2 x (0.25) ]

= 0.582or 58.2%

Wc)

Cooling and dehumidification-temperature decreases and w also decreases

325. (d)

Relative humidity of the air decreases.

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I~I.(. JI)) )11~(~IlllNI(~S llN)) )lll(~IIINI~11Y Fluid: Fluid is a substance which has the property tendency to flow under the action of shear and tangential forces. Liquids and gases both are fluids. Ideal and Real fluids: • In ideal fluids, there is no viscosity and no surface tension and are incompressible. • In real fluids, viscosity, surface tension together exist and are compressible along with density. Classification of fluids : Fluids can be classified on the basis of the following: Based on density and viscosity (i) Ideal fluid: An ideal fluid is describedas a fluid which is in compressible and also has zero viscosity and constant density. (ii) Real fluids: A real fluid is described as a fluid which is compressible and viscous by nature. The density of real fluid are variable and while in motion, an amount of resistance is always offered by these fluids. (iii) Newtonian fluids: Newtonian fluidss are defined as fluids those obey Newton's law of viscosity. The density of these fluids may be constant or variable. The viscosity is calculated according to Newton's law of viscosity as : du t=f..ldy where, r = shear stress f..l= viscosity offluid du/dy = velocity gradient Examples are, water, ethyl alcohol, benzene etc. (iv) Non-Newtonion fluids: Non-newtonian fluids are defined as fluids those do not obey Newton's laws of viscosity. The density of these fluids may be constant or variable and the viscosity of these fluids does not remain constant. Examples are Gels, Solutions of polymers, pastes etc. (v) Compressible fluids: A compressible fluid is defined as the fluid which reduces its volume when an external pressure is applied. All the fluids available in nature are compressible. (vi) In-compressible fluids: Incompressible fluids are defined as the fluids whose density does not change when the value of pressure changes. There is no effect of pressure on the density of fluid. In these fluids, density remains constant and viscosity remains non-zero. (vii) Inviscid fluid: Inviscid fluid is the fluid which has zero viscosity and density may be constant or variable.

p =

2.

1.

Density (p) : It is defined as mass per unit volume of substance.

V

Specific Weight (co) : It is defined as weight per unit volume of substance. co=

3.

mg =pg V

Relative density Specific gravity (Sg) : It is defined as ratio of density of fluid to the density of standard fluid. It may also be defined as the ratio of specific weight of the fluid to the standard weight of fluid. S _ weight of fluid g - weight of standard fluid Density of Fluid Density of standard Fluid Ex: Oil ofSg of 0.8 => Poil = 800 kg/m' Specific volume (v) : It is expressed as the volume per unit mass of fluid.

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FLUID PROPERTIES

m

Sg =

V

v=-=-

4.

I

m P Compressibility(13)

Hydrostatic law: It states that rate of increase of pressure in a vertical direction is equal to weight density of fluid at that point.

.

P

Mathematically, pressure head (h) = pg dV

13

=

_____:y_ = _!_ dp dp P dp

Liquids are highly incompressible :. dp = 0 dp Gases are highly compressible as P ex p.

5.

Bulk Modulus of Elasticity (K) It is defined as reciprocal of compressibility.

VISCOSITY It is the property of fluid by virtue of which one layer resists the motion of another adjacent layer. i.e. its resistence to shearing stresser.

Newton's Law of Viscosity The viscous shear stress between two layers at a distance 'y' du from the surface can be written as: r ex dy

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Fluid Mechanics and Machinery

A-137

du r = Ildy 'Il' is co-efficient of dynamic viscosity / viscosity. Il is a property of fluid called dynamic viscosity and is a function of temperature only. Fluids which obeyNewton's law of viscosity are known as Newtonian fluids. as

•

•

• •

•

If Il is high => velocitygradient du is less => highly viscous dy fluid. du If Il is low => velocity gradient dy is high => easy to flow fluid.

Adhesive forces are attractive forces between the molecular ofa liquid/fluid and the molecular of a solid boundary surface in contact. • Property of a liquid. • The basic cause of surface tension is the presence of cohesive forces. • It is a property by virtue of which liquids want to mnimize their surface area upto maximum extent.

Icr=~IN/m Wetting and Non-Wetting Liquids •

It is the mutual property of liquid-surface.

•

If adhesion »» cohesion, Liquid wets the surface. If cohesion > > > > adhesion, No wetting For wetting, angle of contact (8) should be acute and for non-wetting angle of contact (8) should be obtuse. For pure water 8 = 0°. For Mercury-glass, 8 = 130° to 140°.

Kinematic Viscosity (v) It is expressed as the ratio of dynamic viscosity (u) and density of fluid (p).

v=1:

p Units SI ~ m2/s CgS ~ Stokes/cmvs 1 stokes = 1Q-4 m2/s Effect of temperature and pressure on viscosity: • Viscosity of liquids decrease but that of gases increase with increase in temperature. • In ordinary situations, effect of pressure on viscosity is not so significant but in case of some oils, viscosity increase with increase in pressure.

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RHEOLOGY It is the branch of science in which we study about different types of fluids Bingham plastic

•

• •

CAPILLARITY When a tube of very fine diameter is immersed in a liquid, there will be rise or fall of liquid level in the tube depending upon whether the liquid is wetting with the tube or non-wetting. The rise or fall ofliquid level in the tube is a phenomenon known as capillarity. h : rise of liquid level in tube o : surface tension r : radius of capillary tube p : density of liquid 8 : angle of contact 2 c cos 8 h=--pgr

Rheopactic

'"'" ~~--....0:::

For an annular capillaryhaving external radius r2and inner radius r.,


,.<:1 r:.rJ.

Newtonian

r

h ~

Ideal fluid

(~~) Examples • Newtonian: Water, air • Dilatent : Butter, starch solution • Psuedo plastic : Paints • Bingham plastic: Gel, cream • Thixotropic: Printer's ink and enamel SURFACE TENSION (0') Cohesive and Adhesive forces: Cohesive forces are intermolecular attraction of forever between molecular of same liquid/fluid.

=

2 cr cos 8 pg(r2 - r1)

Pascal's law: It states that pressure intensity at any point in a liquid of rest, is same in all directions. If P ,P and P are the pressure in x, y & z - direction acting on ~ flhid element, at rest, then P, = py = pz· PRESSURE MEASUREMENT DEVICES L

II.

BAROMETER It is a device made by Torricelli and is used to measure local atmospheric pressure. PIEZOMETER • It is a device used for measurements of moderate pressure (gauge) of liquids only. • Piezometer cannot measure the pressure of gas.

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Fluid Mechanics and Machinery

A-138

m

MANOMEIER • used for measurement of high pressure. • It makes the use of a manometric fluid.

-+

specific depth of the fluid. Fluid pressure does not depend on the shape or area of the container. Pressure is a scalar quantity as it has only megnitude but no direction.

Atmospheric Pressure p

SIMPLE U TUBE MANOMETER

y

-I-h- -- --

Atmospheric pressure is defined as the normal pressure exerted by the atmospheric air on the surfaces which are in contact with air.

Gauge Pressure

- --- --

Gauge pressure is defined as the difference between the absolute pressure and the pressure exerted by the atmosphere i.e. atmospheric pressure.

PI + pgy - Pn gh = 0

Absolute Pressure Absolute pressure is defined as the sum of fluid pressure and atmospheric pressure. It is an actual pressure at a given specific point.

Vacuum Pressure +-- DIFFERENTIAL U TUBE MANOMETER

Vacuum pressure is defined as the pressure which is below the atmospheric pressure. Absolute pressure

t

w

i

Badboys2 INCLINED TUBE MANOMETER In this type, the masuring leg is inclined at angle of 10°. The inclination is provided for the purpose of improving the accuracy and sensitivity of the results. This type of manometer is utilized for the purpose of measurement of very small pressure difference.

1

, Gauge pressure I'

r

Atmospheric pressure

, Gauge pressure

Pressure

I'

absolute pressure

~

II

Zero Absolute pressure ~

Relations Among Different Kinds of Pressures Let, Patm= Atmospheric pressure Pabs= Absolut pressure Pgauge= Gauge pressure Pvac= Vacuum pressure

Ipabs= PatIn+ PgaugeI Ipvac= Patm- PabsI Hydrostatic Force on a Plane Surface (Inciined tube manometor) FLUID STATICS

-+

-+

-+

In fluid statics, the behaviour or characteristics ofthe fluid is studied when the fluid is at rest Pressure is described as the normal force applied by a fluid/unit area. The unit of pressure N/m2 which is also termed as Pascal. In case of a fluid, Pressure acts in all the directions. In static liquid. The value of pressure increases with the increasing depth. At any point in a fluid, pressure is directly proportional to the fluid density and depth in the fluid. Pressure (P)

-+

pAHg

= ---;::- = pHg

hence from the above expression, pap and p aH The pressure of fluid is equal in all directions at any

y""

'"

cp : Centre of pressure cg : Centre of gravity F : hydrostatic force acting on the plane surface inclined to free surface. . SIn

h

hcp

h

e= = = =-- = y ycp y F

=

pghA

hcp

=

- leg sirr' 8 h + ---=-,=--hA

leg: moment of inertia of the plane surface about c.g

Badboys2

Fluid Mechanics and Machinery For a horizontal surface, O = 0°

11ep = 11

=>

A-139 HYDROSTATIC FORCES ON CURVED SURFACES Consider a curved surface as shown in the figure.

v

For a vertical surface, O = 90° -

=>

-

leg

h ep =h+- hA

Vertical Surfaces (9 = 90°) (i)

d/2

h leg

bd3 12

=

A=

F : Hydrostatic force acting on the curved portion FH: Horizontal component of F Fe : Vertical component of F FH = pghA

bd

Fy

= pgV

F=~~

(ii)

+F~

V = Volume till the free surface STABILITY OF SUBMERGED •

Badboys2

d

h

3

bd

A=

2

bd2

--

leg

36

(iii)

• • •

Stability of Floating Body Metacentric point (M): When a body is given a small angular displacement which is floating in a liquid in a state of equilibrium. It starts oscillating about some point (M), known as metacentric point. • IfM lies above G, the body is in stable equilibrium. • If M and G coincide, the body is in neutral equilibrium. • IfM lies below G, the body is in unstable equilibrium.

d

leg

~d4

64

A = ~d2

4

(iv)

:l. ':"

~

GM =-

2 =

Centre of Buoyancy: B Centre of Gravity: G IfB lies above G, the body is in stable equilibrium. IfB and G coincide, the body is in neutral equilibrium. IfB lies below G, the body is in unstable equilibrium.

Metacentric Height (GM)

d

h

BODY

l+-~11a

1 -BG V

I:

Moment of inertia of the face of the body intersected by free surface V : Volume of the fluid displaced. BG : Distance between centre of buoyancy and centre of gravity. GM : Metacentric height For Stable equilibrium GM > 0 For neutral equilibrium GM = 0 For unstable equilibrium GM < 0

Buoyancy

I

a h

leg

2 =

A =

a4 12 a2

When the bodies are immersed partially or fully in a fluid, the resultant hydrostatic force acts on the body in the vertical upward direction. This force is known as upthrust or buoyant force. FB : buoyant force FB = pgV V = volume of the fluid displaced by body

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Fluid Mechanics and Machinery

A-140

Centre of Buoyancy It is the point at which upthrust or buoyant force is acting on the body and is exactly same as the centre of gravity of displaced fluid.

Floatation For floatation of body, the density of the body must be equal to or less than density ofliquid i.e. ps s p density of density of solid liquid NOTE: For a completely submerged body, the centre of buoyancy doesn't change. However, for a floating body the centre of buoyancy changes when the orientation of body changes.

FLUIDKINEMATICS • •

There are two approaches to kinematics of a fluid flow i.e. Lagragian approach and Eularian approach. In classical fluid mechanics, Eularian approach is considered.

Different Types of Flow 1. Steadyflow If the properties in the flow are not changing with respect to time, such a flow is known a steady flow.

Badboys2 2. Uniformflow

•

Continuity equation: If states if no fluid in added/removed from the pipe in any length then mass passing across different reactions will be equal. Mathematically, for reaction (1 - 1) and (2 - 2), PlA1V1= P2A2V2 2

2

for incompressible fluid, A1V1= A2V2•

Continuity equation in cartesian - co-ordinates ~ (pu) + ~ (p v) + ~ (pw) + ap

ax

ay

Incompressibleflow If the density of the fluid doesn't change with respect to

pressure, the flow is known as incompressible flow.

4.

•

•

5.

Rotationaland Irrotationalflow If the fluid particles are rotating about their centre of mass, the flow is known as rotational flow. If the fluid particles aren't rotating about their centre of mass, the flow is known as irrotational flow. Laminar and turbulent flow: In Laminar flow, individual particles move in a zig-zag way. For Reynold's number (R ). If Re < 2000, flow in laminar If Re > 4000, flow in turbulent If 2000 < Re < 4000, flow may be laminar/turbulent Rate of flow / Discharge (Q): Q = Area x Average velocity Q=AxV

Internal and External flows : ---j. In case of an internal flow, it is surrounded or bounded by solid boundaries. Due to these solid boundaries the development of boundary layer is restricted. E.g: Flow through pipe ---j. In case of external flow, the fluid flows over the bodies which are immersed in an un-bounded fluid and hence the boundary layer develops freely in single direction. Eg : flows over air foild, turbine blades etc.

at

=

0

Acceleration of A Fluid Particle ~

A

A

A

V = u i-i v j-r w k ~

~

ev a=at

~

av ev =u-+v-+wax ay

~

ev az

J.. Convective acceleration

If the properties (velocityat any given time) is not changing with respect to space, such a flow is known as uniform flow.

3.

az

au au a =u-+v-+w-+-

ax ay av av a =u-+v-+w-+Y ax ay aw aw a =u-+v-+w-+x

z

ax

ay

au

~

+

av at

J.. L-...J temporal or local acceleration

au

az

at av av az at aw aw az at

a=~a2x+a2y+a2z For uniform flow, convective acceleration = 0 For steady flow

H _1

local/temporal acceleration = 0 For steady and uniform flow, total acceleration = 0 Consider a tank as shown in figure For the figure, convective acceleration = 0 temporal acceleration = 0 (if H is constant) temporal acceleration =1= 0 (if H is verying)

_

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Fluid Mechanics and Machinery

A-141

Stream Line It is an imaginary line drawn in such a waythat the tangent drawn

at any point on this line gives the direction of velocity vector of the fluid particle at that point. Equation of streamline in differential form

dy Id -~I

= Y= dx dx v Slope of equipotential line

where

Stream Function (w) •

It is defined only for 2D flows and is a function of space and time.

a\jJ ay a\jJ

PATH LINE

U= --

It is the actual path traced by a fluid particle.

V=-

ax

STREAK LINE It is the locus of all fluid particles at a moment which have passed

•

through a given point.

Rotational components in flow

V = ui = v} wz

There is no boundation on equation.

VORTICITY

Badboys2 It is defined as double of angular velocity. (Circulation per unit of enclosed area) Vorticity = 20)

as it satisfies continuiting

Equistream Line It is a line obtained byjoining points having same stream function values.

1 (av au 1 =-l---) 2 ax ay

where Wz is the net rotation of fluid particle about its own centre of mass. If'w, = 0 => flow is irrotational If'w, *- 0 => flow is rotational

\jJ

dy =~ dx u Slope of equistream line ( dY) dx

x ( dY) tjl=constant

dx

\jI=constant

=_~

x~

V

U

= (- 1)

:. Equistream and Equipotential lines are orthogonal to each other.

Cauchy-Riemann EqD In irrotational flows,

m

CIRCULATION

u= _ ax =_ a.v ~ ay B

IB

ax

= a.v1... (I) ay

It is defined as the line integral of velocity vector along a closed

loop.

r=~v.dr F = (Vorticity)Area

Velocity Potential Function (cj» • •

Velocity potential function q, is a function of space and time. It is defined in such a way q, that u v

aq, ax aq, =-ay aq, =--

w =--

az

•

where u, v and w are the components of velocity vector in x, yand z direction. q, only exists in irrotational flow. For this, q, must satisfy laplace equation i.e.

1'12 q, = 01 Equipotential Line It is a line joining the points having same potential function values.

v=-:=~

~ 1:=-~1

... (2)

Equations (1) and (2) are known as Cauchy-Riemann equations. FLUID DYNAMICS The following types of energies are involved in fluid dynamics. (a) Kinetic energy: Kinetic energy is defined as the energy which is because of motion of the body. (b) Potential energy : Potential energy is defined as the energy due to elevation of the body above the specified I arbitrary datum. (c) Pressure Energy : Pressure energy is defined as the energy due to pressure above datum (d) Internal energy: Internal energy is defined as the energy related with the inter-molecular altratiction of forces or internal state of matter. It can be stored as nuclear energy, thermal energy, chemical energy etc. Some expressions regarding above energies kinetic energy (k.E) = ..!..mv2 2 where, m = mass of the body, v = velocity of the body (b) Potential Energy (P.E) = mgH where, m = mass of the body H = elevation of the body from datum g = 9.8 m/s? (c) Pressure energy: Pressure energy (PEnergy) = VH (a)

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Fluid Mechanics and Machinery

A-142

DifferentKindsofHeads (a) Head: It is described as the amount of energy per unit weight. (b)

Kinetic head: It is defined as the kinetic energy per unit weight. kimetre . h ead =

kinetic energy weight of the body

-------=::..::...._-

P V2 The Bernoulli's Eq=in such a casecan be written as-1 + -21 + Zt pg g P2

vi

=-+-+Z2 +hf pg 2g where hf : head losses encountered as the fluid flows from point 1 to 2.

Applications of Bernoulli's Equation mg 2

(c)

kinetic Head = ~ 2g Potential head : It is defined as the potential energy per unit weight. . Ih d Potential energy Potentia ea = ------=..::...weight of the body mgH mg Potential head = H (d) Pressure Head: It is defined as the fluid pressure per unit specific weight. fluid pressure P =specificweight pg (e) Total head: It is defined as the sum of kinetic head, potential head and pressure head. Total head = kinetic head + potential head + Pressure head Pressure head =

Badboys2

v2 HT=-+H+2g

(f)

v=~2gH Flow sensors

Flow Measurement Devices Venturimeter

P

pg

It is a highly accurate device used for measurement of discharge.

Euler's Equation of Motion The • • •

Following are the applications of Bernoulli's equation given below: (a) Sizing of pumps: In case of pumps, kinetic energy is converted into pressure energy according to Bernoulli's eqaution. (b) Ejectors: In ejectors, pressure energy of the fluid is converted into velocity energy for the purpose of entraining suction fluid. The mixed fluid is recompressed by converting velocity enegry into pressure energy. This process is based on Bernoulli's equation. (c) Pitot tube: It is utilized for the purpose of measuring the fluid flow velocity. The principle of pitot tube is based on the Bernoulli's equation. (d) Carburetor: Carburetor also works on the basis of Bernocelli's equation. When the velocity of air is increased, it lowers the static pressure and increases the value of dynamic pressure. (e) Siphon: A siphon is a device used for the purpose of removing a liquid from its container. The velocity expression is given as following:

CD

Euler's equation considers the following assumptions Flow is irrotational Flow is laminar Flow is invicid.

I~

+ V dv + g dz

= 0 I~

Throat (minimum cross-sectional area)

Euler's Eqn for steady flow

Integrating the above equation. We obtain Bernoulli's equation

P

CD

Converging section 2gh QTH = A2Al A2 _ A2

V2

- + - + Z= constant (Head form) pg 2g 1 P + - pV2 + pgz = constant 2 For Bernoulli's equation, there are two more assumptions i.e. • flow is steady • flow is incompressible Under the five assumptions stated above, the summation of all energies (Pressure, Kinetic and Potential) per unit volume remains constant at each and every point in a flow. BERNOULLI'S EQUATION FOR REAL FLUID In real fluids, viscous shear stresses are present due to which energy is not conserved.

1

Diverging section

2

Q = Cd QTH Coefficientof discharge (it's value vasics between 0.96 0.98)

Cd=~ h : piezometric head difference between 1 and 2 hL : head loss

Orifice meter : The principle of an orifice meter is same as that of venturimeter. In this type, the cross-section of the flowing stream is reduced while passing through the orifice, the value of velocity head is increased at the expense of pressure head. Bernoulli's equation

Badboys2

Fluid Mechanics and Machinery provides a basis for the purpose of correlation maintained between increase in velocity head with the decrease in pressure head. An orifice meter is also termed as pipe orifice or orifice plate, In can be easily installed in the pipeline. A thin circular plate along with a hole in it is placed in the orifice meter. The diameter 1 of an orifice meter is generally kept"2 times the pipe diameter orifice meter is most commonly used for the purpose of measuring the flow offluid in pipes having fluids of single phase. ~2

I'

Vena confracta

:rl~~e

A-143

Working principle: A pitot tube consists of a tube which points directly into the flow of fluid. The liquid flows up the tube and after attaining equilibrium, the liquid is reached at a height above the free surface of the water stream. Now, neglecting friction, Po- P = Hpg where, Po = stagnation pressure P = static pressure Velocity (v) = ~2gH Flow Through Pipe Bends • The main aim of this chapter is to determine the forces. • The pipe bend is horizontal. Hence, there wouldbe no effect of weight. Consider a pipe bend as shown,

Direction of flow -+-1---+

Differential manometer

Fluid (system)

Orifice Meter Discharge is given as :

Fluid

Cc·~d2J2gH

Badboys2

Q=

~

or Q-CdAo~2gH where,

P1 A1

Cd = (coefficient of discharge)

l-C~(~r A = Area of cross-section of orifice meter D = diameter of pipe at section (1) d = diameter of pipe at section (2) Cc = contraction coefficient of water jet ~ Cd (coefficient of discharge) depends upon the Reynold's number (Rc) Pitot tube: A pitot tube is a device which is used for the purpose of measuring the velocity of fluid flow. It has a wide applicability such as for calculating the speed of air of an aircraft, speed of water of boat and also for measuring the velocities of liquid, air or gas in various industrius applications. A pitot tube is utilized for measuring the local velocity at a point in the flow stream.

Pitot tube

Fx, Fy are the horizontal and vertical forces acting on the fluid element. By momentum equation, Fxand Fycan be found PIAl - P2A2cos e + Fx=mV2 cos e - mVI Fy- P2A2sin

e = mV2sin e

F = \jIF2x + F2 Y Vortex flows •

When a certain mass of fluid is rotating with respect to some different axis, such a flow is known as Vortex flow. • There are 2 types of vortex flow (i) Free vortex (ii) Forced vortex FREE VORTEX • No external torque is required. Hence angular momentum remains conserved. 1 V: velocity Vocr r: radius FORCED VORTEX • External torque is required to maintain its angular velocity at a constant value. w= constant V o: r NOTE: • Free vortex flows are irrotational flows and thus, Bernoulli's equation can be applied.

Badboys2

Fluid Mechanics and Machinery

A-144

•

Forced vortex flows are rotational flows and hence, Bernoulli's equation cannot be applied.

(ii) Exit Losses

Fundamental Equation of Vortex Flows dp = pw2 r dr - pg dZ General equation and can be applied between any two points For free surface, dp = 0 => pw? r dr = pg dZ Integrating the above equation we get,

Iz = • •

V12

h.=

2g

(iii) Losses Due to Sudden contraction

w:t I

A pipe is a closed contour which carries fluid under pressure. When fluid flows through pipe, it encounters losses. These losses can be broadly categorized into (i) Major losses (ii) Minor losses

CD

Major Losses •

h.=

These losses are due to friction. The losses are evaluated by Darcy-Weishback Equation.

vi [_1 _1]2 2g Cc

A2

CC=A

fLV2 ...(1)

2gd

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f: friction factors L : length of pipe

V : velocity in pipe of fluid d : diameter of pipe f= 4 f friction coefficient The above equation (1) is valid for both laminar and turbulent flow.

3

If'C, is not given, hj= 0.5 Vll2g Head loss occurs after Venacontracta as boundry layer separation occurs.

(iv) Entrance Losses

NOTE: Head loss is independent of pipe orientation. It depends only on details of the flow through the duct. For fully developed laminar flow, f= 64/Re where Re : Reynold's No.

O.5V2 h--f2g BEND LOSSES

pVD Re=-J..l

KV2 h--f- 2g

V = velocity D = diameter m = dynamic Viscosity

K = Constant which depends upon angle of bend and its radius of curvature.

Minor Losses • •

Bernoulli's Equation, momentum Eq" are used to determine these losses. The magnitude of minor losses is very loss.

FLOW THROUGH BRANCHED PIPES

Pipes in Series •

In series, discharge (Q) remains same but head is divided.

(i) Losses Due to Sudden Enlargement ,~

~:

N_ CD I

®

......_

-'

121 ~

13, d31 I : I

Badboys2

Fluid Mechanics and Machinery Q=Q] =Q2=Q3 h-= (hd]

=>

2g

+ L2Vj + L3V}]

d1

Shear between fluid layers e = Il du/dy(x-dir.)

Entrance Length

+ (hf)2 + (hfh

h=~[LIV?

A-145 •

d2

d3

Dupit's Equation A pipe ofuniform diameter is said to be equivalent to compound pipe if it carries same discharge and encounters same losses.

The distance in downstream from the entrance to the location at which fully developed flow begins is called L entrance length for laminar flow in pipes. ~

= 0.06 R,

L, = entrance length D = diameter of pipe Steady Laminar Flow in Circular Pipes

~ ! --! -~---PIPES IN PARALLEL •

~P+.?Pdx "[:

In parallel arrangement, discharge gets divided.

Q=Q1 +Q2

(hf)1 = (hfh SYPHON

• Siphon is a long bend pipe used in carrying water from a Badboys2 reservoir at higher level to another reservoir at lower level.

•

shear stress R: radius of pipe M : dynamic viscosity of fluid

1" :

9+~---t-~• •

The height point of siphon is called summit. No section of the pipe will be more than 7.6 m above the hydraulic gradient line. When absolute pressure of water becomes less than 2.7 m gases come out from water and get collected at the summit thereby providing an obstruction to flow.

POWER TRANSMISSION THROUGH PIPE \J

ap

ax : pressure

gradient

u : velocity at a distance 'r ' from cente

t=(-:H u = - _1_

(ap)

(R 2 _ r2) 4M from above expression of 'u', we can conclude that velocity is varying parabolically.

ax

1 ( = - 4M

Umax

ap) ax

2

'1

l-2-) U max

Umax

.fi ,= U = U

P1 - P2 P theoretical = pQgH P actual = pQ (H - hf) where hf are the head losses in pipe. pQ (H - hf) 11= pQgH

=

i.e. average velocity equals the local

32MUL D2

LAMINAR FLOW BETWEEN

TWO PARALLEL PLATES

Case I : One plate is moving with a velocity of 'U' while the other is stationary.

--r-bj

= ~ I·

d~t-'--::I;;?::-

Laminar Flow in Pipes At low velocity of real fluids, viscosity is dominant. The flow of fluid takes place in form oflaminar. This laminated flow is known as laminar. No slip at boundary Flow is rotational No mixing of fluid layers

-u

(ap)

Uy 1 u=--b 2M

ax

(by-y

2

)

Case II : When both plates are at rest

Features of Laminar Flow • • •

( 1t R

velocity. Pressure drop (P, - P2) in a given finite length 'L'

H

•

2

R ,Q =

u=-2 at r = R/

for maximum efficiency 1he

Ox

head losses remain same but

u

=-

2~ (:)

(by - y2) (Poiseuille

flow)

Badboys2

Fluid Mechanics and Machinery

A-146

= __ 1

u

_

U

(ap)b

2

8J.l ax

max

•

large velocity gradients existing in it. Velocity within the boundary layer increases from zero to main stream velocity asumptically.

2

=3 Umax

TURBULENT FLOWS •

•

•

Boundary layer

In turbul ent flow, there is continuous mixing of fluid particles and hence velocity fluctuates continuously. u' and v' are fluctuating components of velocity

2 ( du ,2 1: =

1: = turbulent

shear stress

I: mixing length, 1= 0.4 y, y is distance from pipe wall Mixing length is the length in transverse direction where in fluid particles after colliding loose excess momentum and reach the momentum as of local environment.

•

Umax

- U

= 5.75 10glO (R/y)

---=..:..:,:=..:....___

V*

V. : Shear velocity V. u

(

y,

=t

Badboys2 • V* = 5.75 10glO ly') •

y' = 0' I 107 (for smooth pipes) y' = Kl30 (for rough pipes) Reynold's condition for rough & smooth pipes

• •

ReR < 4

y=o

au

y=O

-=0

a2u

ay2

ax

•

o IX X 1/2 'x' is the distance from leading edge of the plate. As x increase, boundary layer thickness increases. The transition from laminar to turbulent flow is decided by Reynold's No. R, ::;5 x 105 => flow is laminar R, > 6 x 105 => flow is turbulent

•

> O.

Displacement thickness (0) 0*

0'

=

1 o

(ll- ~ ') dy UOC)

Momentum thickness (8)

=> transition

8=

Thickness of Laminar Sublayer (0')

1

~(ll-

~)'

o UOC)

UOC)

dy

Energy thickness (OE)

11.6v

8

=--

fo

V*

II-

u (

OE = -U

Hydrodynamically Rough and Smooth Boundaries

87 < 0.25

ap

For separation of boundary layer,

In turbulent flow in pipes, average velocity equals local velocity at y = 0.223 R.

K

=0

ay

o : boundary layer thickness UOC) : free stream velocity Nominal thickness is the thickness of boundary layer for which J.l= 0.99 UOC) In case ofa converging flow (aPlax = - ve), the boundary layer growth is retarded.

=> smooth pipe

From Nikuradsee's

u = 0.99Uo

=> rough pipe

4 < ReR < 100 •

u=O

•

v ReR > 100

Conditions

y= 0 y=o at

ldy)

pu'v' = P I

Boundry

OC)

experiment,

Shape factor (H)

=> smooth boundary

=

u2 -2

,

j dy

Uoc>

0* e

Von Karman's Momentum Integral Equation Assumptions K

87 > 6

=> rough boundary

K 6 .. 0.25 < - < => transition

•

Flow is 2D, incompressible

•

-=0

dP

~

d8

dx

pU~

dx

0'

BOUNDARY LAYER THEORY • •

The concept of boundary layer was first introduced by L. Prandtl. Boundary layer is a layer in the vicinity of the surface with

and steady

where 8 : momentum thickness 1:0 : plate shear stress p : density UOC) : free stream velocity

Drag force (FD)

Badboys2

Fluid Mechanics and Machinery It is the force exerted by the fluid in a direction parallel to relative motion. A zero angle of incidence, of the plate the drag force is due to shear force. C _-fx _

A-147 (b) Medium head and small quantity of flowing water (c) Low head and larger quantity of flowing water Based on the specific speed of the turbine (a) Low specific speed turbine (specific speed < 60) (b) Medium specific speed turbine (specific speed : 60 to 400) (c) High specific speed turbine (specific speed: above 400)

~

to

1 2 -pU 2

00

CD = average drag coefficient Cfx = local drag coefficient For air flow over a flat plate, velocity (U) and boundary layer thickness (8) can be expressed as

~ =Hi)-Hir

Basic Definitions of Hydraulic Turbines ~ Impluse turbine : In this type, only kinetic energy is available at the inlet of turbine. Eg : Pelton wheel turbine

~

Reaction turbine : In this type, kinetic energy and pressure energy both available at the inlet of turbine. Eg : kaplan turbine, Francis turbine Radial flow turbine : In this type, the flowing of water is in the radial direction through the runner. Inward radial flowturbine: In this type, the flowing of water is from outward to inward radially. Outward radial flowturbine : In this type, the flowing of water is from inward to outward radially. Axial flow turbines : In this type, the flowing of water is through the runner along the direction parallel to the rotational axis of the runner. Mixed flow turbine : In this type, the flowing of water is through the runner in radial direction but leaves in the direction parallel to the rotational axis of the runner. Yangential flow turbine : In this type, the flowing of water is along the tangent of the runner.

~ ~

4.64 x

8 = ~Rex

~

TURBOMACHINERY

~

The conversion of energy carried by water into electrical energy is carried out by the turbo-generator. In this a rotating turbine driven by the water and connected by a common shaft to the rotor of a generator. Any turbine consists of a set of curved blades designed to deflect the water in such a way that it gives up as much as possible of its energy. The blades and their support structure make up the turbine runner, and the water is directed on to this either by channels and guide vanes or through a jet, depending on the type of turbine. The efficiency of any turbomachine

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~ ~

ComparisonbetweenImpulseTurbineand ReactionTurbine

P[Power output)

11 = ---~--~=--..:....._-1000 x Q x g x H(Power input) where, Q = flow rate of the falling water the number of cubic metres per second g = acceleration due to gravity H = effective head Mass of a cubic metre of fresh water = 1000 kg .. mass falling per second = 1000 x Q

Classification of Hydraulic Turbines The hydraulic turbines are classified based on the following basis: ~ Based on the type of energy at inlet (a) Impulse turbines (b) Reaction turbines ~ Based on the direction of flowing water (a) Tangential flow turbines (b) Axial flow turbines (c) Radial flow turbines • Inward radial flow turbines • Outward radial flow turbined (d) Mined flow turbines ~ Based on the Head of water and water quantity available (a) High head and small quantity of flowing water

Reaction turbine (i) A part of energy offluid is converted into kinetic energy before entering the fluid into turbine.

(ii) There are no losses in (ii) There are losses in flow flow regulations regulations (iii) The whole unit is placed above the tailrace

(iii) The whole unit is submerged in water below tailrace (iv) Blades are in acting (iv) Blades are in acting mode only when they are mode at all the time in front ofno:zzle

Hydraulic Turbines In hydraulic turbines, the conversion of hydraulic energy into mechanical energy takes place. This mechanical energy is utilized for running an electrical generator which is directly connected with the shaft of the hydraulic turbine. Thus, finally, the conversion of mechanical into electrical energy takes place.

Impulse turbine (i) In this, the conversion of potential energy into kinetic energy takes place by nozzle before entering to turbine

Pelton Wheel Turbine The pelton wheel is an impulse turbine. The Pelton wheel turbine with water flow from moving cup (b) and actual motion of water and cup (c) are shown in fig. below. (8)

(b):) ~)

(e);::

v=Q

water

Fig. : Pelton wheel turbine: (a) vertical section ; (b) water flow as seen from moving cup; (c) actual motion of water and cup The volume rate of flow Q corresponding to head H

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Fluid Mechanics and Machinery

A-148

Vw = velocity of whirl at outlet

Q= A~(2gH)

2

u = peripheral velocity

where A = area ofthe jet g = acceleration due to gravity. The input power to the turbine P

Vr1 = relative velocity at inlet Vr2 = relative velocity at outlet

= 1000 x Q x g x H = 1000 x A~2gH x g x H

VI = absolute velocity at inlet V 2 = absolute velocity at outlet

(.: Q = A~2gH ) Specific Speed Ns

=

1] x ~

2

P

r;;

H x"H

Where,

11 = rate of rotation (in rpm) H = effective head (in m) P = available power (in kW) The range of specific speed for pelton wheel is 10-80 Main parts of a pelton turbine : The following are the main parts are given of a pelton turbine: (a) Nozzle and spear: Spear controls the amount of water that strikes the buckets. (b) Runner: It consists of circular shaped disk. On the periphery of this circular disk, number of buckets are fixed evenly, buckets have the shape of hemi - spherical cup and divided by the splitter which divides the water jet into two parts, Runner is made up of cast iron or stainless steel etc. (c) Casing: Itacts as cover and prevents the water splashing. It is made up of cast iron and steel etc. (d) Breaking jet: It strikes the back of vane and utilized for stopping runner in a very short duration oftime. Velocity triangle for Pelton wheel :

Vf2 = velocity of flow at outlet Efficiencies : (a) Hydraulic efficiency : It is defined as the ratio of work done / second by jet of water to the input energy / second.

l1H

*

u

Vu

2

2

p

)1

V?

Maximum hydraulic efficiency (l1max.) (b)

(c)

(d)

=

1+ coso 2

Mechanical efficiency : It is described as the ratio of power available at the shaft and the power produced by the wheel. l1mech.

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IE

2U(VWI ± VW2 ) = -....:....._-:----....:....

Pshaft

= Q

H

P W W The value of mechanical efficiency varies between 0.97 to 0.99. Volumetric efficiency : It is defined as the ratio of volume of water actually strikes the buckets and total volume of water supplied by the jet to the turbine. Overall efficiency : It is defined as the product of hydraulic efficiency (l1H)' mechanical efficiency (l1mech) and volumemetric efficiency (11 vol). 110 = l1H x l1mech x 11vol.

VU1

Some important formulas : Gross head = HG = difference between head race and tail race Net head = Hnet = ~ - hf- h

Francis Turbine Francis turbines are by far the most common type in presentday medium or large-scale plants. They are used in installations where the head is as low as two metres or as high as 300. These are radial-flow turbines. Francis turbine is completely submerged, it can run equally well with its axis horizontal or vertical. Francis turbine is shown in figure below

fLv2 where, hf

=--

2gdp

here, f = frictional factor L = Length of penstock v = mean velocity in penstock d, = diameter of penstock h = height of nozzle above the tail race ~

Work done/second

= (VWI ± VW2 ).u pavl

~

work done / weight

= (VWl ± VW2 ).g

u

where, Vw = velocity of whirl at inlet 1

Fig. : Francis turbine: (a) cut-away diagram; (b) flow across guide vanes and runner Francis turbines are most efficient when the blades are moving nearly as fast as the water, so high heads imply high speeds of

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Fluid Mechanics and Machinery

A-149

rotation. T he range

~ 0f

W.D/s = pQd[VW1Ul ± VW2U2]

l

~) speerifiIC speeds, ( Ns = n x ~~)

for •

Francis turbine is 70-500.

Main parts of a Francis turbine : (a) Penstock: It is a tube of large diameter through which (b)

(c)

(d) (e)

Work done per second: (W.D/s)

water from dams reaches to the inlet of the turbine. Spiral casing : It is a closed passage. The diameter of the spiral casing is decreases along the flowing direction. The area of spiral casing is maximum at inlet and minimum (nearly zero) at outlet. Guide vanes : It is an aerofoil like shape vane which is fixed between two rings and a part of pressure energy is converted into kinetic energy by guide vanes. Runner: It is connected to the shaft of the turbine. Draft tube: It is defined as tube which expands gradually and it discharges water passing through the runner to the tail race.

For radial discharge,

VW2 = 0 , then

W.D/s = pQdVwI ul

~

Hydraulic efficiency : It is defined as the ratio of workdone per second on the runner and the energy at inlet/ second. 11H =

~

~

VWIUl±VW2·U2 gH

Mechanical efficiency : It is defined as the power to the power developed by the runner. as 11mech. Volumetric efficiency (11vol): It is defined quantity of fluid working on the runner to the of fluid supplied.

ratio of shaft It is denoted as the actual total quantity

Actual fluid quantity

Velocity triangle of Francis turbine :

11vol . = T ota I flU1·d quantity . ~

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Overall efficiency (110): It is defined as the ratio of shaft power to the input power. It may also be defined as the product of hydraulic efficiency, mechanical efficiency and volumetric efficiency. 110

=

shaft power Input power

or 110 = 11H x 11mech. x 11vol.

KAPLAN TURBINES Kaplan turbine is a axial flow or propeller type turbine which has adjustable blades. It is an inward flow reaction turbine, i.e., the working fluid changes pressure as it moves through the turbine and gives up its energy. Axial-flow turbine and runner of kaplan-turbine shown in figure below.

Some important formulas : ~

Net head (H) :

•

At the exit of penstock,

• •

At the exit of draft tube

.

P , H = -- P -

v2

2g

Zl --

v2

2g

p

v2

P

2g

At the exit of draft tube when the position of draft tube is at the tail race. H=--

•

p

+-

At the exit of pen stock when the position of draft tube is at the tail race. H=-+z.+-

•

P

H =-+zl

Fig. : A Propeller or axial-flow turbine Main parts of a kaplan turbine: (a) Scroll casing: It is the casing in which guiding the water (b)

v2

2g

If the velocity at the exit of draft tube is negligible, then Net head, at the exit of penstock.

H=(~+Z<J

(c) (d) (e)

and controlling of passage of water takes palce. Guide vanes: The water is directed at a suitable angle by the guide vanes. The guide vanes also works for the purpose of regulation the water quantity which is to be supplied to the runner. Stay ring: The stay ring guides the water from scroll casing to the guide vanes. Runner blades: Runner blades are connected to the hub and there is an axial flow ofwater through the runner. Draft tube : It is utilized for the purpose of connecting

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Fluid Mechanics and Machinery

A-150 runner exit to the tail race. The draft tube is a kind of a pipe which has a gradually increasing area used for discharging water from turbine exit to tail race. There are generally four types of draft tubes gi ven as : • • • •

Conical draft tube Single elbow tube Moody spreading tube Elbow draft tube with circular inlet and rectangular outlet.

The range of specific speed of Kaplan turbine is 350-1000. The figure given below shows the ranges of head, flow rate and power of different types of turbine.

Velocity Triangle of Kaplan Turbine

1000

.¬

100

1 .c

I MW

10

Francis and similar turbines

20 kW

flowrate 1m) $-1

Fig. : Ranges of application of different types of turbine. Note the overlap at the boundaries

VELOCITY DIAGRAMS

Blade

Shift Disc

--*)~--

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e,

= vane tip angle a, ~ = angle which absolute velocity makes with tangential directions

Degree of reaction (R): It is defined as the ratio of change in static energy in the rotor to the total energy transfer.

Fixed Blade

(Vr} - Vr{l (uf-ufl R

l

2g

)

+l

2g

)

= -----=--------=-H

Run away speed: It is defmed as the speed when the turbine has maximum value of discharge while running. Ranges ofrun away speed for different turbines are given below: Pelton turbine = l.8 to l.9 Kaplan turbine = 2.5 to 3 Francis turbine = 2 to 2.2 N

Cavitation and cavitation factor: Cavitation is defined as a process which occurs in a flowing liquid. In this process, the cavities are formed and grown. After that these cavities collapse in a high pressure zone at the time when there is fall of to vapour pressure or below vapour pressure. Cavitation affects the output and efficiency. Due to cavitation, output and efficiency both decreases. In reaction turbines, the locations where the cavitation occurs, are given as follows: (a) Cavity may occur at the exit of the convex side ofrunner. (b) Cavity may also occur at the inlet of draft tube.

TI I I I I I I

I I I I I U2

-l---~~====~====~~I ---:+I

Vwl

---+!

~---------~VW----------~

V w2

Here, a = guide blade angle or runner vane angle at inlet

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Fluid Mechanics and Machinery e = wheel vane angle = vane angle at outlet Vf1

. tan e =

tan a = --

Vwl '

Vf1 Vwl -u

A-151

V

; tan = ____f1_ u

Unit Quantities The unit quantities provide the speed, discharge and power for a particular turbine by keeping the head of I m (assumed) considering the same efficiency unit quantities provide a suitable information regarding the prediction of performance of turbines. (a) Unit speed (Nu) : The turbine speed working under unit head is known as unit speed N

Nu=

(b)

JH

Unit discharge (Q,) : The turbine discharge working under unit head is known as unit discharge.

Qu= (c)

(b)

unit speed (NJ Unit speed vs unit power Constant speed curves or operating characteristics curves: In this type, various tests are performed at a constant speed by varying the head and adjusting the discharge. The following curves are drawn. -+ Power vs discharge -+ efficiency vs discharge -+ efficiency vs unit power -+ maximum efficiency vs % Full load

Q JH

Unit Power (Pu) : The turbine power produced while working under a unit head is known as unit power P Pu = H3/2

Performance of Turbines

Badboys2 The performance of turbines should be studied for the purpose of providing information regarding the performance ofturbine. For the purpose of studying the turbine performance, characteristic curves are used and drawn on the basis of actual tests. There are following there kinds of characteristic curves are used: (a) Constant head curves : These are also known as main characteristics curves. In this type, head is kept constant, and the speed of turbine is varied by varying the flow rates after the adjustment percentage of gate opening. The curves drawn under constant head are given as : -+ Unit discharge vs unit speed -+ Unit power vs unit speed -+ Overall efficiency vs unit speed ~;:::i 100% CI

% full load

(Operating characteristic curves) (c)

Constant efficiency or Muschel curves : In these curves, the data obtained from constant head and constant speed curves are drawn for the purpose of finding the constant efficiency zone. H = constant

Full opening

25%

'-'

(l)

eo ~ ,J::I

75%

:.a

50%

.~

25%

o fZl

.......

- ~.•.

, 'Best performance curves Nu

Muschelcurves Unit speed (Nu) (Unit speed vs unit discharge)

unit speed (Nu) Unit speed vs efficiency

Centrifugal Pumps In centrifugal pumps, the conversion of mechanical energy into Hydraulic or pressure energy by the application of centrifugal force. The flow of water is in radial outward direction. The principle on which it works is forced vortex flow. Common applications are sewage,petroleum and petrochemical pumping. -+ Working principle: Centrifugal pumps work based on the principle of forced vortex flow in which rotation of a certain mass by the external torque rise in pressure head takes place. The energy is converted due to two main parts

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Fluid Mechanics and Machinery

A-152

ofthe pump. i.e. impeller and casing. The driver energy is converted into kinetic energy by impeller and the kinetic energy is converted into the pressure energy by the diffuser. Main Parts of Centrifugal Pump A centrifugal pump consists of two main parts: (a) Rotating component : -+ Impeller -+ Shaft (b) Stationary component : -+ casmg -+ casing cover -+ bearings (a) Rotating components: -+ Impeller: It is the main rotating component which works for the purpose of providing centrifugal acceleration to the fluid. -+ Shaft: It works for the purpose of transmitting torques which are encountered during starting and during operation. It also works as a supporting member for the impeller and other rotating components. (b) Stationary components : -+ Casing: In this, the conversion of kinetic energy into pressure energy takes place. Generally three kinds of casings are used given as : • Volute casing: It is utilized for higher heads • Vortex casing : In this, reduction of eddy currents takes place. • Circular casing : It is utilized for lower head. -+ Diffusers are employed in multistage pumps. Priming In this process, suction pipe, casing and delivery pipe is filled upto delivery value with water. It is also utilized for removing air from the above mentioned parts. -+ Positive priming : In this type, the speed of processing is increased -+ Negative priming: In this type, the speed of processing is decreased. Classification of Centrifugal Pumps Centrifugal pumps may be classified based on the following: (a) Based on working head: -+ Low head centrifugal pumps : These are generally single stage centrifugal pumps. The working ofthese pumps is generally below the 15 m head. -+ Medium head centrifugal pumps : The working of these pumps is generally at the heads lying between 15 mand45 m. -+ High - head centrifugal pumps : In high head centrifugal pumps, the value of head enceeds 45 m. These are generally multistage pumps having guids vanes. (b) Based on specific speed : Specific speed of a centrifugal pump is defined as the speed of identical pump which provides unit discharge with unit head.

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NJQ

specific speed (N s) = ~

H

Type of pump

specific speed

Radial flow 10- 25 Mixed flow 70 - 135 Axial- flow, propeller 100 - 425 (c) Based on types of casing : -+ Volute chamber pump: It has a spiral shaped casing. In this type, the sectional area increases from tongue to delivery pipe in a uniform way. -+ Vortex chamber pump : In this type, there is an uniform increasing area which is given between the outer periphery of impeller and the volute casing. -+ Diffuserpump: In this type, the guids vanes are placed at the impeller vanes-outlet. In this, due to enlarge area of cross - section of guids vanes, the velocity of water increases and the pressure decreases. It provide improved value of efficiency. (d) Based on direction of flow of water : -+ Radial flow : In this type, the flow of water in the impeller is totally in radial direction. -+ Mixed flow: In this type, the change of direction of flow of water from radial flow to the combination of a radial flow and axial flow takes place due to which flow area is enhanced. -+ Axial flow: In this type, high discharge and lowheads is used. Eg : Irrigation. (e) Number of entrances to the impleller : -+ Single suction pump : In this type, suction pipe is arranged only one side of impeller. -+ Double suction pump : In this type, suction pip is arranged on the both of the sides ofthe impeller. Due to which, discharge is increased. (f) Based on disposition of the shaft : -+ Horizontal shafts: In this type, horizontal shafts are used in centrifugal pumps. -+ Vertical shafts : In this type, vertical shafts are utilized if there is a lake of space available. (g) Based on number of stages : -+ Single stage : In this type, only one impuller is connected to the shaft. -+ Multi stage: In this type, a number of impellers are mounted on the same shaft and enclosed in the same casing. Velocity Triangles for Centrifugal Pumps

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wVw2u2g 11mech. =

(c) (Forward facing Vanes)

(Radial Vanes)

(Back-ward facing vanes)

Ps

Overall efficiency (110) : It is defined as the ratio of power output to power input to the pump or shaft. wHm

110 =-p-

(d)

Volumetric efficiency (l1vol) : It is defined as the ratio of actual discharge to the sum of actual discharge and rate of leakage. QA

(e)

where, QA = Actual discharge, QL = rate of leakage Hydraulic efficiency : It is defined as the ratio of manometric head to the theoretical head. 11N=-

Work Done by Impeller on the Water WD

[VW2 .u2 - VWlul ]

Characteristic

= .:::...._--=------=---=

where, W.D = work done VWl = velocity of whirl at inlet

Badboys2VW2 = velocity of whirl at outlet ul = tangential velocity of impeller at inlet u2 = tangential velocity of impeller at outlet Ifwater comes radially, a = 0°, and VW1 = 0

v:

= __

Curves

These curves are utilized for the purpose of predicting the performance of centrifugal pump working under different head, rate of flow and speed. The main characteristic curves are as : ~ Main characteristic curve ~ Operating characteristic curve ~ Muschel or constant efficiency curve ~ Main characteristic curve :

g

then work done

Hm HT

u2

2_

g

Heads in Centrifugal Pumps (a) Suction head: It is defined as the vertical height of centre line of the centrifugal pump, which is above the water surface to the pump. (b) Delivery head : It is defined as the distance between centre line ofthe centrifugal pump and the surface of water in the tank to which water is to be delivered. (c) Static head: It is defined as the sum of suction head and delivery head. (d) Manometric head: It is defined as the head against which work is done by the centrifugal pump. Efficiencies (a) Manometric efficiency: It is defined as the ratio of manometric head with the head provided by impleller, 11mano. =

11mano.

(b)

Manometric head (Hm ) (VW2U2)/ g

Fig. (3)

(Main Characteristic Curves) In fig (1), for a given speed, when discharge increases, Hm decreases and for a given discharge, greater is speed, large is~. In fig. (2) for a given speed, as discharge increases, Ps increase. In fig. (3) higher is the speed, higher is the maximum efficien cy. ~ Operating characteristic curves : In these curves, it the speed is kept constant, the variation of manometric head, power and efficiency with respect to discharge provides operating characteristic curves.

•

• •

(VW2U2)/ g

Mechanical efficiency (l1mech) It is defined as the ratio ofpower delivered by the impeller with the power input to shaft.

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Fluid Mechanics and Machinery

A-154

~

efficiency

Muschel or constant efficiency curve

a '._/

lip power

--g

(])

...s:::

~

E§

~ (Operating

I---.+--l-_

Discharge (Qd) characteristic

curves)

Rate of flow (Lifi/s)-----7

...,

(Muschel curves)

EXERCISE

IIIIII~ 1.

The velocity components in the x and y direction of a twodimensional potential flow are u and v respectively, then

au. aX

IS

6. For the continuityequation given V.; = 0 to be valid, when

equal to

Ov (a)

Ox

Badboys2(c)

Ov

(b)

Ov

7.

Ox

Ov

ay

(d)

ay

2. The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u = uo(1 - 4r2/D2), where r is the radial distance from the centre. If the viscosity of the fluid is u, the pressure drop across a length L of the pipe is (a)

/-!uOL D2

(c)

8/-!uoL D2

8.

3. A two-dimensional flow field has velocities along x and y

4.

directions given by u = x2t and v = - 2xyt respectively, where t is time. The equation of streamlines is (a) x2 y = constant (b) xy2 = constant (c) xy= constant (d) not possible to determine In a two-dimensional velocity field with velocities u and v along the x and y directions respectively, the convective acceleration along the x-direction is given by

au

au

ax

ay

Ov u-+v-

au

(a)

u-+v-

(c)

ax

ay

au

Ov

Ox

ay

au

au

Ox

ay

(b) u-+v(d) v-+u-

5. For a Newtonian fluid (a) (b) (c) (d)

shear stress is proportional to shear strain rate of sheer stress is proportional to shear strain shear stress is proportional to rate of shear strain rate of shear stress is proportional to rate of shear strain

9.

10.

11.

; is the velocity vector, which one of the following is a necessary condition? (a) Steady flow (b) Irrotational flow (d) Incompressible flow (c) Invescid flow Match the following: P. Compressive flow U. Reynolds number Q. Free surface flow V. Nusselt number R. Boundary layer flow W. Weber number S. Pipe flow X. Froude number y. Match number T. Heat convection z. Skin friction coefficient (a) P-U; Q-X; R-V; S-Z; T-W (b) P-W; Q-X; R-Z; S-U; T-V (c) P-Y; Q-W; R-Z; S-U; T-X (d) P-Y; Q-W; R-Z; S-U; T-V A hydraulic turbine develops 1000 kW power for a head of 40 m. If the head is reduced to 20 m, the power developed (in kW) is (a) 177 (b) 354 (c) 500 (d) 707 A phenomenon is modelled using n dimensional variables with k primary dimensions. The number ofnon-dimensional variable is (a) k (b) n (c) n-k (d) n + k The maximum velocity of a one-dimensional incompressible fully developed viscous flow, between two fixed parallel plates is 6 mls. The mean velocity (in mls) of the flow is (a) 2 (b) 3 (c) 4 (d) 5 A pump handling a liquid raises its pressure from 1 bar to 30 bar. Take density of the liquid as 990 kg/m '. The isentropic specific work done by the pump in kllkg is (a) O. 10 (b) 0.30 (c) 2.50 (d) 2.93

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Fluid Mechanics and Machinery 12. A streamline and an equipotential line in a flow field (a) are parallel to each other (b) are perpendicular to each other (c) intersect at an acute angle (d) are identical 13. For steady, fully developed flow inside a straight pipe of diameter D, neglecting gravity effects, the pressure drop dP over a length L and the wall shear stress 'tw are related by (a)

't

(c)

't

w

dPD =-4L

(b)

'tw

~PD2 =--24L

w

dPD =-2L

(d)

't

4dPL =-D

w

14. Biot number signifies the ratio of (a) convective resistance in the fluid to conductive resistance in the solid (b) conductive resistance in the solid convective resistance in the fluid (c) inertia force to viscous force in the fluid (d) buoyancy force to viscous force in the fluid 15. A flow field which has only convective acceleration is (a) a steady uniform flow (b) an unsteady uniform flow (c) a steady non-uniform flow (d) an unsteady non-uniform flow 16. Match Group A with Group B:

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Group A

Group B

P:

Biotnumber

1

Q:

Grashof number

2

R:

Prand tl number

3

s:

Reynolds number

4

Ratio of buoyancy to viscous force Ratio of inertia force to viscous force Ratio of momentum to thermal diffusivities Ratio of internal thermal resistance to boundary layer thermal

(a) P-4, Q-l, R-3, 8-2 (b) P-4, Q-3, R-l, 8-2 (c) P-3, Q-2, R-l, 8-4 (d) P-2, Q-l, R-3, 8-4 17. Consider the following statements regarding streamline(s): (i) It is a continuous line such that the tangent at any point on it shows the velocity vector at that point (ii) There is no flow across streamlines dx h di fferentra. I equation . f 0 a (iii) - = -dy = -dz. IS tel u v w streamline, where u, v and ware velocities in directions x, y and z, respectively (iv) In an unsteady flow, the path of a particle is a streamline. Which one ofthe following combinations of the statements is true? (a) (i), (ii), (iv) (b) (ii), (iii), (iv) (c) (i), (iii), (iv) (d) (i), (ii), (iii) 18. Consider a velocity field V = K ( yi + xk), where K is a constant. The vorticity, 0z, is (a) -K (b) K (c) -K/2 (d) K12

A-155 19. Length of mercury column at a place at an altitude will vary with respect to that at ground in a (a) linear relation (b) hyperbolic relation (c) parabolic relation (d) manner first slowly and then steeply 20. A type of flow in which the fluid particles while moving in the direction of flowrotate about their mass centre, is called (a) steady flow (b) uniform flow (c) laminar flow (d) rotational flow 21. A-2d flow having velocity V = (x + 2y + 2) i + (4 - y)j will be (a) compressible and irrotational (b) compressible and not irrotational (c) incompressible and irrotational (d) incompressible and not irrotational 22. Buoyant force is (a) resultant of upthrust and gravity forces acting on the body (b) resultant force on the body due to the fluid surrounding it (c) resultant of static weight of body and dynamic thrust of fluid (d) equal to the volume ofliquid displaced by the body 23. If cohesion between molecules of a fluid is greater than adhesion between fluid and glass, then the free level of fluid in a dipped glass tube will be (a) higher than the surface of liquid (b) same as the surface of liquid (c) lower than the surface of liquid (d) unpredictable 24. In a pipe pitot tube arrangement the static stagnation head is 20 m and static head is 5m. If the diameter of pipe is 400 mm. Find the velocity of flow of water in pipe (a) 17.15 mls (b) 22.22 mls (c) 38.76 mls (d) 42.85 mls 25. Depth of oil having specific gravity 0.6 to produce a pressure of 3.6 kg/ern! will be (a) 40 em (b) 36 em (c) 50 em (d) 60 em 26. The capillary rise in a narrow two-dimensional slit ofwidth 'w'is (a) half of that in a capillary tube of diameter 'w' (b) two-third of that in a capillary tube of diameter 'w' (c) one-third of that in a capillary tube of diameter 'w' (d) one-fourth of that in a capillary tube of diameter 'w' 27. For a turbulent flow in pipe the value of y at which the point velocity is equal to the mean velocity of flow, is (y is measured form pipe axis). (a) 0.772 R (b) 0.550 R (c) 0.223 R (d) 0.314 R 28. Pressure in Pascals at a depth of 1m below the free surface of a body of water will be equal to (a) 1 Pa (b) 98.1 Pa (c) 981 Pa (d) 9810 Pa 29. A circular disc of diameter' d' is slowly rotated in a liquid of large viscosity J..lat a small distance h from a fixed surface. The minimum torque required to maintain an angular velocity 0) will be (a)

J..l1td40) 32h

(c)

J..l1td30) 32h

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Fluid Mechanics and Machinery

A-156

30.

31.

32.

Viscosity of a fluid with specific gravity 1.3 is measured to be 0.0034 Ns/m2. Its kinematic viscosity, in m2/s, is (a) 2.6 x 1O-{i (b) 4.4 x 10-{i (c) 5.8 x 1O-{i (d) 7.2 x 10-{i The shear stress at a point in a glycerine mass in motion if the velocity gradient is 0.25 metre per sec/per meter, will be (a) 0.0236kg/m2 (b) 0.02036kg/m2 (c) 0.0024kg/m2 (d) none of these The general equation of continuity for three-dimensional flow of a incompressible fluid for steady flow is

40.

41.

When pressure p, flow rate Q, diameter D, and density d, a dimensionless group is represented by (a)

pQ2 dD4

(b)

__ P__ dQ2D4

(c)

pD4d Q2

(d)

pD4 dQ2

Maximum wall shear stress for laminar flow in tube of diameter D with fluid properties J..land p will be 32000J..l2

(a)

(c)

33.

34.

ou+av +ow =0 ox Oy oz

au ov Ow -+-+-=1

ax

Oy

oz

(b)

(d)

Ou=av =Ow =0 ox Oy oz

(a)

au av Ow -+-+-=u.v.w

(c)

ax

oz

Oy

where u, v and ware components of velocity in x, y and z directions respectively. The pressure in meters of oil (specific gravity 0.85) equivalent to 42.5 m of water is (a) 42.5m (b) 50m (c) 52.5m (d) 85m The cause ofturbulence in fluid flow may be (a) high Reynold number (b) abrupt discontinuity in velocity distribution (c) critical Reynold number (d) existence of velocity gradient without abrupt discontinuity .. h . 02<j> 02<j>. kn For an irrotational flow t e equation -2 + -2 IS own

42.

43.

Badboys2 35.

36.

37.

ax

(a)

~P

44.

45.

46.

O'J..lY

(c) 38.

39.

Pressure force on the 15 em dia head light of an automobile travelling at 0.25 m/s is (a) IO.4N (b) 6.8N (c) 4.8N (d) 3.2N A fire engine supplies water to a hose pipe L m long and D mm in diameter at a pressure P kPa. The discharge end of the hose pipe has a nozzle of diameter d fixed to it. Determine the diameter d of nozzle so that the momentum ofthe issuing it may be maximum

_ (D3)1I4

(a) d- -

8L

(b)

_ (D5)1I4

3)1/5 (c) d=

( ~L

( ~L

(d)

8000J..l2 pD3

Air flows over a flat plate 1 m long at a velocity of 6 m/s. The shear stress at the middle of plate will be [Take S = 1.226 kg/nr', v = 0.15 x 10-4 m3/s (0.15 stokes) for air] (a) 84.84 x 10-3 N (b) 92.69 x 10-3 N (c) 67.68 x 10-3 N (d) 103.45 x 10-3 N The friction head lost due to flow of a viscous fluids through a circular pipe of length L and diameter d with a velocity v, and pipe friction factor 'f is 4fL

(a)

-.d

(c)

2g

v2

(b)

2g

4f L y2 nd2• 2g

47.

Value of coefficient of compressibility for water at ordinary pressure and temperature is (a) 1000 kg/cur' (b) 2100 kg/cnr' (c) 2700 kg/cnr' (d) 21,000 kg/cnr' Crude oil of kinematic viscosity 2.25 stokes flows through a 20 em diameter pipe, the rate of flow being 1.5 litres/ sec. The flow will be (a) laminar (b) turbulent (c) uncertain (d) None of these Pseudo plastic is a liquid for which (a) dynamic viscosity decreases as the rate of shear increases (b) Newton's law of viscosity holds good (c) dynamic viscosity increases as the rate of shear increases (d) dynamic viscosity increases with the time for which shearing forces are applied. Match List I with List II and select the correct answer using the codes given below the lists.

A B. C.

L~t I (Loss)

L~t II (Parameter

Leakage Loss Friction Loss Entrance Loss

1. 2. 3.

responsible)

Zero at design point Proportional to head Proportional to half of relative velocity square.

Codes

5 )115 (d) d=

pD3

l6000J..l2

d- -

8L

(b)

y2

Oy

as (a) Bernoulli's equation (b) Cauchy Riemann's equation (c) Euler's equation (d) Laplace equation. A control volume refers to (a) a closed system (b) a specified mass (c) an isolated system (d) a fixed region in space The pressure coefficient may take the form

6400J..l2

pD3

(a) (c)

A

B

C

1 1

2

3

3

2

(b) (d)

A 2 2

B

C

3 1

3

1

Badboys2 0

""'"

("I') ("I')

Fluid Mechanics and Machinery 48.

49.

50.

51.

The velocity potential in a flow field is = 2xy. The corresponding value of stream function is (a) (y2 _ x2) + constant (b) (x2 - y2) + constant 122 (c) -(x - y ) + constant (d) 2 (x - y) + constant 2 Consider the following The components of velocity u and v along X and Y directions in a two dimensional flow problem of an incompressible fluid are (i) u = x2 cos Y ; v = - 2x sin y (ii) u = x + 2; v = 1 - 4 (iii) u = xyt ; v = x3 _ y2t12 (iv) In u = xty; v = xy - ylx Which of that will satisfy the continuity equation ? (a) 1,2 and 3 (b) 1,2 and 4 (c) 2, 3 and 4 (d) 1,2, 3 and 4 Consider the following statements regarding bernaulli's theorom for fluid flow 1. Conservation of energy 2. Steady flow 3. Viscous flow 4. In compressible flow Which of the above statements is/are correct? (a) 1,2 and 4 (b) 1 only (c) 2,3 and 4 (d) 1,2, 3 and 4 An ideal fluid is defined as the fluid which: (a) is compressible (b) is in compressible (c) is in compressible and non-viscous (d) has negligible surface tension Which of the following is the unit ofkinematic viscosity? (a) N- s/m? (b) m2/s (c) kg/s m2 (d) m/kg-s Poise is the unit of: (a) dynamic viscosity (b) kinematic viscosity (c) mass density (d) velocity gradient Surface tension is a phenomenon because of: (a) viscous forces only (b) Adhesion between liquid and solid molecules (c) Difference in magnitude between the forces due to adhesin and cohesion (d) cohesion only In case of a static fluid: (a) only normal stresses can exist (b) linear deformation is small (c) fluid pressure is zero (d) resistance to shear stress is small Gauge pressure is equal to: (a) absolute pressure + atmospheric pressure (b) absolute pressure - atmospheric pressure (c) atmospheric pressure - absolute pressure (d) atmospheric pressure - vaccum It pressure intensity is 1.006 MN/m2 and specific gravity of sea water is 1.025, then the depth of a point below water surface in sea will be equal to : (a) 10 m (b) 1000 m (c) 100 m (d) 1m An oil of specific gravity 0.7 and pressure 0.14 kgf I cm-. Then the weight of the oil will be equal to : (a) 70 em of oil (b) 2 m of oil (c) 20 em of oil (d) 80 cm of oil

Badboys2 52.

53. 54.

55.

56.

57.

58.

A-157

59.

Compressibility (B) is equal to : (If k is bulk modulus) (a)

~ = _!_ k

(b) ~=k

~=_1 (d) ~ = k2 k2 60. Kinematic viscosity is equal to : (a) Dynamic viscosity x density Dynamic vis cosity (b) Density Density (c) Dynamic vis cosity (c)

(d)

Dynamic viscosity x Density 61. Void ratio does not depend upon: (a) Liquid limit (b) Volume (c) Bulk volume (d) Porosity 62. The height of the water column corresponding to a pressure equivalent to 60 m of mercury column will be equal to: (a) 8160 em (b) 816 em (c) 81.6 em (d) 7996.0 em 63. The difference of pressure between inside and outside of a liquid drop is (a) ~P=Txr

65.

66.

67.

68.

69.

M>=!

r

(d) M>= 2T 2r r Falling drops of water become spherical in shape due to the property of : (a) adhesion (b) cohesion (c) surface tension (d) viscosity The pressure at a depth of 5 km below the surface of sea water considering specific gravity of water to be 1.3, will be equal to : (a) 63765 Pa (b) 637654 Pa (d) 1.48 x 108 Pa (c) 1.27 x108Pa Capilary action is due to : (a) viscosity of liquid (b) cohesion of liquid particles (c) surface tension (d) None of these A piece of wood having weight 5 kg floats in water with 60% of its volume under the liquid. Then the specific gravity of the wood will be equal to: (a) 0.83 (b) 0.60 (c) 0.71 (d) 0.72 Hydrostatic low states that the rate of increase of pressure in vertical direction is equal to : (a) fluid density (b) fluid specific weight (c) fluid weight (d) fluid specific gravity A rectangular tank of square cross-section (2m x 2m) and height 4 m is completely filled up with a liquid. Then the ratio oftotal hydrostatic force on any vertical wall to its bottom is equal to: (a) 2.0 (b) 4.0 (d) 1.0 (c) 6.0 (c)

64.

(b)

M>=_!_

o, C)

I

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Fluid Mechanics and Machinery

A-15S 70.

71.

72.

73.

74.

75.

(c) [ML T-1] (d) [ML2T] A piezometer tube is used for the purpose of measurement of: (a) Low pressure (b) High pressure (c) Moderate pressure (d) Vacuum pressure Atmospheric pressure is 1.03 kg/em- and vapour pressure is 0.03 kg / cm-, then the air pressure will be equal to : (a) 1.03 kg/ern(b) 1.06 kg/ern(c) 0.53 kg/ern(d) 1 kg/cm2 In case of Rotameter, which one of the following statement is correct? (a) Float has a density lower than the density of flowing fluid (b) Float has a density equal the density of flowing fluid (c) Float has a density greater than the densityofflowing fluid (d) None of these If gauge pressure = 21 bar, atmospheric pressure = 1.013 bar then the value of absolute pressure will be equal to : (a) 21 bar (b) 22.013 bar (c) 20.012 bar (d) 21.018 bar Water at 20° C is flowing through a 20 em diameter pipe. The kinematic viscosity of water is 0.0101 stoke. It the Reynold's number is 2320, then the velocity with which the water will be flowing through the pipe will be: (a) 1.117 cm/s (b) 2.228 cm/s (c) 4.677 cm/s (d) 5.67 cm/s A certain liquid has 5 tonnes mass and having a volume of 10m3. Then the mass density of the liquid will be : (a) 500 kg/m' (b) 1000 kg/m' (c) 50 kg/m' (d) 5000 kg/m' The volumetric change ofthe fluid caused by a resistance is known as: (a) volumetric strain (b) volumetric index (c) compressibility (d) stress A glass tube of3 mm diameter is immersed in water which is at 20° C. The surface tension for water is 0.0736 N/m. The contact angle for water is 0°. Then the value of capillary rise or depression will be equal to : (a) 20 mm (b) 10 mm (c) 30 mm (d) 36 mm The desirable properties for any practical fluids are: (a) should be viscous (b) should posses surface tension (c) should be compressible (d) All of the above A liquid compressed in a cylinder has initially a volume of 20 m3 at a pressure of 100 Pa. It the new volume is 40 m3 at a pressure of 50 Pa, then the bulk modulus of elasticity will be equal to : (a) 20 Pa (b) 40 Pa (c) 50 Pa (d) 70 Pa Match List - I and List - II and select the correct answer using the codes given below: List - I List - II A capilarity 1. cavitation B. vapour pressure 2. Density of water C. viscosity 3. shear forces D. specific gravity 4. surface tension

Badboys2 76.

77.

78.

79.

80.

81.

Codes:

Viscosity has the following dimensions: (a) [ML T2] (b) [ML-1 T-1]

82.

83.

84.

85.

A B C D (a) 4 1 3 2 (b) 2 3 4 (c) 4 3 2 (d) 2 4 1 3 The reading ofthe pressure gauge fitted on a vessel is 25 bar. The atmospheric pressure is 1.03 bar and the value of g is 9.81 m/s-', The absolute pressure in the vessel will be equal to : (a) 23.97 bar (b) 24 bar (c) 26.03 bar (d) 27.04 bar A metal piece having density exactly equal to the density of fluid is placed in the liquid. The metal piece will : (a) will be wholly immersed (b) sink to the bottom (c) float on the surface (d) will be partially immersed Resultant force on a floating body will act: (a) vertically upwards through meta centre (b) vertically down wards through metal centre (c) vertically upwards through centre of buoyancy (d) vertically downwards through centre of buoyancy For a floating body,match List - I with List - II and select the correct answer from the codes given below: List - I List - II A Meta - centre is above 1. stable equilibrium the centre of gravity B. Meta - centre is below 2. unstable equilibrium the centre of gravity C. Meta centre and centre 3. Neutral equilibrium of gravity coincides Codes:

ABC 2 3

(a)

(b)

2

3

3 2 (d) 2 1 3 86. Find the buoyant forceacting on the aluminium cubewhich is suspended and immersed in a jar pilled with water when it is given that the side of the cube is 5.0 em. (a) 2.45 N (b) 1.25 N (c) 3.25 N (d) 6.25 N 87. The centre of pressure of a plane submerged surface: (a) should coincide with centroid of surface (b) should coincide with centroid of pressure prism (c) may be above or below centroid (d) cannot be above mentioned 88. The vertical distance of centre of pressure below the e.g of the inclined plane area (submerged in liquid) is : (c)

(a) (c)

leg

sin2 e

A'x' leg

tan

2

e

(b) (d)

leg

cos2 o

A'x' leg

cot

2

o

A'x' A'x' where, Q = inclination of plane area. x = distance ofc.g of plane area from free liquid surface.

Badboys2

Fluid Mechanics and Machinery 89.

90.

91.

92.

93.

The velocity distribution in turbulent flow is a function of the distance y measured from the boundary surface and friction velocity V*, follows as : (a) Parobolic Law (b) logarithimic Law (c) hyperbolic Law (d) linear law The Boundary layer on a flat plate is called laminar boundary layer if the value of Reynold's number (Re) is : (a) Re < 5 x 105 (b) Re < 2000 (c) Re<4000 (d) None of these The continuity equation in fluid mechanics is a mathermatical statement employing the principle of: (a) conservation of energy (b) conservation of mass (c) conservation of momentum (d) None of these The velocity at which the flow changes from laminar to turbulent for a given fluid at a given temperature is termed as: (a) maximum velocity (b) minimum velocity (c) average velocity (d) critical velocity In laminar, incompressible flow in a circular pipe the ratio between average velocity to maximum velocity will be equal to:

Badboys2(a)

2

(b) 4

(c)

3 4

I (d) ~

94. An oil having kinematic viscosity of 0.25 stokes flows through a pipe of 10em diameter. The flow will be critical at a velocity of about: (a) 1.5 mls (b) 0.5 m/s (c) 2.5 m/s (d) I mls 95. In a turbulent flow through a pipe, the shear stress is : (a) Maximum at centre and decreased linearly towards the wall (b) Maximum at centre and decreased logarithimically towards the wall (c) Maximum midway between the centre - line and the wall (d) Maximum at the wall and decreases linearly to zero at the centre 96. An oil with specific gravity 0.85 and viscosity 3.8 poise flows in a 5 em diameter horizontal pipe at 2 m/s. The Reynold's number will be approximately equal to : (a) 224 (b) 2240 (c) 22.4 (d) 22400 97. At what distance (r) from the centre of pipe of radius (R) does the avergae velocity occur at laminar flow: (a) r = 0.304 (b) r = 0.707 (c) r = 0.808 (d) r = 0.609 98. With the same cross-sectional area and placed in the turbulent flow, the largest drag will be experienced by : (a) A sphere (b) A streamlined body (c) A circular disc held normal to the flow of direction (d) A circular disc held parallel to the flow of direction

A-159 The shear stress in a turbulent pipe flow: (a) varies parabolically with radius (b) is constant over the pipe radius (c) is zero at centre and increases linearly to the wall (d) None of these 100. The concept of stream function which is based on the principle of continuity is applicable to : (a) Three - dimensional flow (b) Two - dimensional flow (c) One - dimensional flow (d) None of these 101. The most essential feature of the turbulent flow is : (a) Large discharge (b) Small discharge (c) High velocity (d) Velocity and pressure at a point shows irregular fluctuations at high frequency 102. The flow is said to be subsonic ifthe value of mach number

99.

IS :

103.

104.

105.

106.

107.

(a) Mach No. = I (b) Mach No. < 1 (c) Mach No. > 1 (d) Mach No. = 2 Ratio of inertia force to surface tension is termed as : (a) Mach number (b) Froude number (c) Reynold's Number (d) Weber's number If the free steam fluid velocity (v) is 20 mls and the pipe diameter (d) is 1m, if the dynamic density (p) is 0.150 kg/m! and the fluid viscosity is 0.0000122, then the Reynold's number (R) will be equal to : (a) 245902 (b) 235902 (c) 434904 (d) 324906 Laminar flow developed at an average velocity of 5 m/s occurs in a pipe of 10 em radius. Then the velocity at 5 em radius will be equal to : (a) 10 m/s (b) 7.5 mls (c) 5 mls (d) 2.5 mls Stanton diagram is a graph of: (a) Friction factor versus Reynolds number (b) Friction factor versus log of Reynolds number (c) Log of friction factor versus Reynolds number (d) Log offriction factor versus log of Reynolds number The experimental determination of coefficient of velocity is given as : (a) (c)

c,

=J?;;.

(b)

Cv=~

C

= J4XH2

(d)

c, =~~~

v

y

108. Application of Bernoulli's equation requires that: (a) the duct is two - dimensional (b) the flow is laminar (c) the duct is frictionless (d) the fluid is nonviscous and incompressible P v2 109. It -+-2 +z= constant is a Bernoulli's equation, with pg g v2 their usual meanings then the term 2g represents: (a) kinetic energy (b) Pressure energy (c) kinetic energylunit weight (d) None of these

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Fluid Mechanics and Machinery

A-160

11O. Match the List - I and List - II and selectthe correct answer using the codes below: List - I List - II A Froude number 1. pi pu2 B. Mach Number 2. U I (gd) C. Weber's Number D. Euler's Number

3. 4.

U/Jid pLU2/cr

119. Energy loss in flow through nozzle as compared to venturi meter is : (a) Same (b) More (c) Less (d) Unpredictable 120. Using Blasius equation, the friction factor for turbulent flow through pipes varies as :

P 5. 6.

111.

112.

p2U2 U/C

Codes: A B C D (a) 3 6 4 1 (b) 2 4 3 (c) 1 3 4 6 (d) 2 4 5 6 A pivot tube is used for measuring: (a) pressure of flow (b) velocity of flow (c) flow rate (d) total energy Venturimeter is used to measure flow of fluids in pipes when pipe is: (a) horizontal (b) vertical, flow downwards (c) vertical, flow upwards (d) In any position Which of the following device is used to measure flow on the application of Bernoulli's theorem: (a) venturimeter (b) orifice plate (c) Pivot tude (d) All of these For a nozzle to convert sup sonic flow into a super sonic flow, must be : (a) convergent type (b) divergent type (c) convergent-divergent type (d) of uniform cross-sectional area Flow through a supersonic nozzle is an example of: (a) Isolated system (b) open system (c) closed system (d) insulated system In the Navier, stoke equation, the forces considered are: (a) Gravity, pressure and viscous (b) Pressure, viscous and turbulence (c) Gravity, pressure, and turbulence (d) Pressure, gravity, turbulence and viscous The line which provides the sum of pressure head, datum head and kinetic head of a flowing fluid in a pipe with respect to some reference line is known as : (a) Hydraulic gradient line (b) Total friction line (c) Total energy line (d) None of these Which of the following flow measurement device is independent of density? (a) Electro - magnetic flow meter (b) Orifice plate (c) Turbine meter (d) Venturi - meter

(a)

Re

(b) ReO.5

(c)

ReO.33

(d) ReO.25

121. Which of the following parameter is measured using orifices? (a) Velocity (b) Pressure (c) Flow rate (d) Both pressure and velocity 122. If Cy = coefficient of velocity, Cc = coefficient of contraction C, = coefficient of resistance, then coefficient of discharge (Cd) is equal to (a) c, x Cc (b) c, x c, (c) Cy+ Cc (d) Cy-Cr 123. For a viscous flow, the coefficient of friction is given by:

Badboys2 113.

114.

115.

116.

117.

118.

(c) f=~

124.

125.

126.

127.

128.

129.

(d) f=~ Re Re Loss of head due to friction to maintain 0.05 m3/s of discharge ofliquid (specific gravity 0.7) through a steel pipe 0.2 m diameter and 1000 m long, taking coefficient of friction 0.0025 will be equal to : (a) 6.44 m (b) 9.45 m (c) 10.8 m (d) 12.3 m If velocity of flow through a pipe is doubled, then the head loss due to friction becomes : (a) two time (b) four times (c) eight times (d) half For maximum power, transmission through a pipeline, the frictional head loss equals: (a) HI 3 (b) HI 4 (c) HI 6 (d) HI 9 The maximum velocity of one dimensional incompressible flow between two fixed parallel plates is 6 mis, then the mean velocity of the flow is : (a) 2 m/s (b) 4 m/s (c) 6 m/s (d) 9 m/s The surge tanks are used in a pipeline to: (a) reduce frictional loss in pipe (b) ensure uniform flow in pipe (c) relieve the pressure due to water hammer (d) reduce cavitation Two pipe systems are said to be equivalent, when in two systems: (a) head loss and discharge are same (b) friction factor and length are same (c) length and diameter are same (d) length and discharge are same

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Fluid Mechanics and Machinery 130. Ifa draft tube is used with a Francis turbine (installed above tail race level), the pressure at the runner outlet: (a) is equal to atmospheric pressure (b) is above atmospheric pressure (c) is below atmospheric pressure (d) depends upon turbine speed 131. Francis turbine is a : (a) tangential flow reaction turbine (b) axial flow reaction turbine (c) radial flow reaction turbine (d) mixed flow reaction turbine 132. Francis turbine is best suited for: (a) all types of heads ( b) medium head (34 - 180 m) (c) low head ( upto 30 m) (d) High head (above 180 m) 133. The value of speed ratio in kaplan turbine is generally kept as: (a) 1.5 (b) 2.5 (c) 1 (d) 0.5 134. Mean diameter ofrunner ofa pelton wheel turbine is 200 mm and least diameter of jet is 1 em, then the jet ratio and number of buckets will be : (a) 20, 25 (b) 200, 115 (c) 20,40 (d) 25,65 135. The specific speed of a turbine is represented as :

Badboys2

(a)

NHS/4 JP

(b)

NJP

(c)

NJP PHS/4

NJP (d) gHs/4

HS/4

136. Ifthe jet ratio in a Pelton turbine wheel is 18, the number of buckets will be equal to : (a) 24 (b) 30 (c) 34 (d) 40 137. An impulse turbine is used for: (a) Low head of water (b) High head of water (c) Medium head of water (d) High discharge 138. A kaplan turbine produces 3000 kw power under a head of 5 m and a discharge of 75 m3/s. Then the overall efficiency will be equal to : (a) 82% (b) 90% (c) 94% (d) 96% 139. The function of hydraulic turbine is to convert water energy into : (a) Heat energy (b) Electrical energy (c) Atomic energy (d) Mechanical energy 140. If 'a' is the angle of blade fip at outlet, then maximum hydraulic efficiency of an impulse turbine is : l-cosa 1+ cosa (a) (b) 2

(c)

l+cos2 a 2

A-161 142. Ajet of water ofO.002m2 area moving with a velocityof15 mls strikes on a series of blades moving with a velocity of 6 m/s. The force exerted on the blades will be : (a) 0.18N (b) 270N (c) 27 N (d) 180 N 143. In kaplan turbine, the number of blades is generally equal to : (a) 2 to 4 (b) 4 to 8 (c) 8 to 16 (d) 16 to 24 144. Run away speed of a hydraulic turbine is : (a) Full load speed (b) The speed at which turbine runner will be damaged (c) The speed it the turbine runner is allowed to revolve freely without load and with the wicket gates wide open (d) The speed corresponding to maximum overload permissible 145. Impulse turbine is generally fitted: (a) little above the tail race (b) at the level of tail race (c) slightly below the tail race (d) about 2 to 5 m below the tail race 146. In a reaction turbine: (a) flow can be regulated without loss (b) water may be allowed to enter a part or whole of the wheel circumference (c) the outlet must be above the tail race (d) there is only partially conversion of available head to velocity head before entry to runner 147. The condition for maximum efficiency of reaction turbine is given by: (b) Vb

149.

150.

151.

2

(d)

l+sina 2

141. Which of the following turbine does not requir a draft tube? (b) kaplan turbine (a) Pelton turbine (c) Francis turbine (d) Propeller turbine

V2 (d) Vb _ --- 1 cosa cos2 a In a reaction turbine, a stage is represented by: (a) each row of blades (b) number of discharge of steam (c) number of casings (d) None of these A pelton wheel operates with a speed of 600 rpm, speed ratio of 0.44 and a net head of300 m. Then the diameter of the wheel will be : (a) 0.82 m (b) 1.08 m (c) 1.51 m (d) 2.14 m Compared to cylindrical draft tube, a tapered draft tube: (a) prevents hammer blow and surges (b) responds betler to load fluctuations (c) converts moke kinetic heads into pressure head (d) prevents cavitation even under reduced discharge The degree of reaction ofa kaplan turbine is: (a) equal to 1 (b) equal to 180 (c) Greater than zero but less than 1 12 (d) Greater than 112but less than 1 The discharge through a turbine is (a) directly proportional to HI/2 (b) inversely proportional to H1I2 (c) directly proportional to H3/2 (d) inversely proportional to H3/2 (c)

148.

152.

= V? cos a

Vb

VI

=--

2

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Fluid Mechanics and Machinery

A-162

153. To maximize the work output at turbine, the specific volume of working fluid should be: (a) As small as possible (b) As large as possible (c) constant throughout the cycle (d) None of these 154. Cavitation depends upon: (a) vapour pressure which is the function oftemperature (b) absolute pressure or barometric pressure (c) suction pressure which is height of runner outlet above tail race level (d) All of the above 155. Open type impeller centrifugal pump is used for the purpose of: (a) mixture of water, sand, pebbles and clay (b) sewage treatment (c) water treatment (d) None of these 156. The discharge through a single acting reciprocating pump is given by: (b) (c)

Qd

=

ALN 120

Qd

_ 2ALN -

165. Air vessel is used in reciprocating pumps to obtain:

166.

167.

168.

169.

60

(a)

158.

159.

160.

161.

162.

163.

164.

(a) at the top (b) at the bottom (c) at the centre (d) None of these In order to avoid cavitation in centrifugal pumps: (a) The suction pressure should be high (b) The delivery pressure should be high (c) The suction pressure should be low (d) The delivery pressure should be low A centrifugal pump lifts 0.013 m3/s water from a depth of 32 m. If the pump motor consumes 6 kw and density of water is 1000 kg/m ', then the overall efficiency of the pump will be : (a) 88% (b) 75% (c) 69% (d) 79% The vanes of a centrifugal pump move due to: (a) Pressure energy of water (b) kinetic energy of water (c) Both (a) and (b) (d) Power supplied by Prime mover The maximum value ofthevane exit angle for a centrifugal pump impeller is equal to : (a) 10° to 15° (b) 15° to 20° (c) 20° to 25° (d) 25° to 30° The vanes of a centrifugal pumps are usually: (a) curved forward (b) curved backward (c) radial (d) None of these A pump is defined as a device which converts: (a) hydraulic energy into mechanical energy (b) mechanical energy into hydraulic enegry (c) kinetic energy into mechanical energy (d) none of these N. P. N. S. H stands for: (a) Net Positive Supply Head (b) Net Power Supply Head (c) Net Positive Suction Height (d) Net Positive Suction Head

Q = 2ALN

(b)

Q = ~N

(d) Q= 2ALN

60

(d) Qd = ALN (c)

157. In a centrifugal pump, the liquid enters the pump:

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(a) Rise in delivery head (b) Reduction of suction head (c) Increase in supply of water (d) continuous supply of water at uniform rate Which of the following provides the correct relationship between power (P) required to run a centrifugal pump and diameter (d) of its impeller? (a) P ex: d5 (b) P ex: d4 ( C ) P ex: d2 (d) P ex: d The specific speed of a pump is defined as the speed of unit of such a size that it : (a) requires unit power to develop unit head (b) delivers unit discharge at unit power (c) delivers unit discharge at unit head (d) produces unit power with unit head available For small discharge at high pressure, which of the following preferred? (a) Mixed flow (b) Reciprocating (c) Axial flow (d) Centrifugal The discharge through a double acting reciprocating pump is where N is rpm :

Q = ALN

170. Centrifugal pump is started with its delivery value is :

171.

172.

173.

174.

175.

(a) kept 50% open (b) irrespective of any position (c) kept fully closed (d) kept fully open Pick up the wrong statement about centrifugal pump: (a) Head is proportional to (diameter f (b) Discharge is proportional to diameter (c) Head is proportional to (speed)? (d) Power is proportional to (speed)? The multistage centrifugal pumps are used to obtain: (a) High head (b) High discharge at high head (c) High efficiency with high discharge (d) Pumping of viscous fluids A centrifugal hydraulic pump is used to force water to an open tank through a pipe having a diameter of 2 decimeters. Given tha the pump is 4 km away from the tank, the average speed of water in the pipe is 2 m/s. After neglecting the other minor losses, evaluate the absolute discharge pressure at the pump exit if it is to maintain a constant head of5 m in the tank. (Assume Darcy's friction factor of 0.01 for the pump). (a) 5.503 bar (b) 55.03 bar (c) 44.911 bar (d) 0.449 bar In a centrifugal pump when delivery value is fully closed, the pressure of fluid inside the pump will (a) becomes zero (b) reduce (c) increase (d) remain unaltered In a double acting reciprocating pump, there are: (a) One suction value and one delivery value (b) Two suction values and two delivery values (c) One suction values and two delivery values (d) Two suction values and two delivery values

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Fluid Mechanics and Machinery 176. A centrifugal pump will start delivering water only when the pressure rise in the impeller is equal to or greater than the: (a) manometric head (b) kinetic head (c) static head (d) velocity head 177. A centrifugal pump has the following specifications. Speed = 1000 rpm Flow = 1200 m3/s Head = 20 m Power= 5 HP If the speed is increased to 1500 rpm, then the flow will become: (a) 1800 m3/s (b) 2700 m3/s (c) 1200 m3/s (d) 3600 m3/s 178. Casting of a centrifugal pump is designed to minimise: (a) Friction Loss (b) Cavitation (c) Static (d) Loss of kinetic energy 179. A mercury manometer is used to calculate the static pressure at a point in a water pipe as shown in fig. The level difference of mercury in two limbs is 10 mm. Then the gauge pressure at that point will be equal to :

A-163

184. In a flow field, the stream lines and equipotential lines : (a) are parallel (b) cut at any angle (c) area orthogonal everywhere in the field (d) cut orthogonal except at the stagnation point 185. For a fluid element in a two - dimensional flow field (xy) if it will undergo: (a) Translation and deformation (b) Translation and rotation (c) Translation only (d) Deformation only 186. Existence of velocity potential implies that (a) Fluid is in continuum (b) Fluid is irrotational (c) Fluid is ideal (d) Fluid is compressible 187. The velocity components in x and y directions are given by u = A

Manometer

(a) 1236 Pa (b) 1333 Pa (c) Zero (d) 98 Pa 180. The force F required to support the liquid of density d and the vessel on top is :

y3- x2y, V =

xi -~4 v". Then, the value of 'A'

for the possible flow field involving an incompressible fluid is : (a)

Badboys2

x

H)

(b)

m

(c) 4 (d) 3 188. In a case of a turbulent flow of a fluid through a circular tube (as compared to the case oflaminar flow at the same flow rate), the maximum velocity is shear stress at the wall is , and the pressure drop across a given length is . The correct words for the blanks are: (a) Higher, higher, higher (b) Higher, lower, lower (c) Lower, higher, higher (d) Lower, higher, lower 189. The parameters which determines the friction factor for the turbulent flow in a rough pipe are : (a) Reynolds number and relative roughness (b) Mach number and relative roughness (c) Froude number and relative roughness (d) Froude number and Mach number 190. The discharge in m3/s for laminar flow through a pipe of diameter 0.04 m having a centre line velocity of 1.5 mls is: 3n 3n (a) 50 (b) 10000

--:-: r 1

H

~_~ _~

----------------------------_. ------------

Vessel Liquid

-_..-

Friction less piston of area =A (a) gd (hA- H) (b) gdHA (c) gdHA2 (d) gd (H- h)A 181. Kinematic viscosity of air at 20° C is given to be 1.6 x 10-5 m2/s. Its kinematic viscosity at 70° C will be : (a) 2.2 x 10-5 m2/s (b) 4.2 x 10-5 m2/s (c) 3.4 x 10-5 m2/s (d) 6.6 x 10- 5 m2/s 182. The velocity potential function for a source varies with the distance 'r' as : 1 (a) (b) r r

(c)

r2

(d) 1m

183. Stream lines, path lines and streak lines are virtually identical for : (a) Uniform flow (b) Flow of ideal fluids (c) Stead flow (d) Non-uniform flow

3n 3n (c) 2500 (d) 5000 191. The predominant forces acting on an element of fluid in the boundary layer over a flat plate in a uniform parallel stream are (a) viscous and pressure forces (b) viscous and body forces (c) viscous and inertia forces (d) inertia and pressure forces 192. Prandtl's mixing length in turbulent flow signifies: (a) the average distance perpendicular to the mean flow covered by the mixing particles (b) the ratio of mean free path to characteristic length of the flow field (c) the wavelength corresponding to the lowest frequency present in the flow field (d) the megnitude of turbulent kinetic energy

Badboys2

Fluid Mechanics and Machinery

A-164

193. The necessary and sufficient condition which brings about separation of boundary layer is : (3)

:

>0,

!~

<0

(b) :

<0,

!>O

dP < 0 dU > 0 (d) dP > 0 dU < 0 (c) dy "dx dy "dx 194. In a fully developed laminar flow in the circular pipe, the head loss due to friction is directly proportional to : (a) (mean velocity)? (b) (mean velocity)! (c) (mean velocityr' (d) (mean velocity)" 195. The discharge velocity at the pipe exit as shown in figure is:

(a)

2

(b) 1

(c) 4 (d) 3 200. If I = moment of inertia of the plane of the floating body at water surface v = volume of body submerged in water BG = distance between the centre of gravity G and centre of buoyancy B then, The metacentric height GM is expressed as : (a)

GM=_!_-BG

(c)

GM= BG--

(b) GM=_!_+BG

V

V

I

V

= 0,V2'V = 0 (c) V2<1> *- 0,V2'V = 0 (a)

196.

(b)

~2gH

fiijl

(c) ~2g(H + h) (d) ~2g(H - h) Which of the following forces act on a fluid at rest? 1. gravity force 2. hydrostatic force 3. surface tension 4. viscous force Select the correct answer using the codes given below : (a) 1,2,3and4 (b) 1,2and3 (c) 1 and 2 (d) 3 and 4 The normal stress is same in all directions at a point in a fluid only when : (a) the fluid has zero viscosity and is at rest (b) one fluid layer has no motion relative to an adjacent layer (c) the fluid is frictionless (d) the fluid is frictionless and incompressible The depth of fluid is measured in vertical z-direction, x and yare other two directions and are mutually perpendicular the static pressure variation in the fluid is given by: dP dP (a) - = g (b) - = P dz dz dP dP (c) -=pg (d) -=-pg dz dz (The symbols have their usual meanings) A stepped cylindrical container is filled with a liquid as shown in figure. The container with its axis vertical is first placed with its large diameter downwards and then upwards. The ratio in the forces at the bottom in the two cases will be :

Badboys2 197.

198.

199.

IE

d

;;.1

1 2d

>1

V2<1>

(b) V2<1> *- 0,V2'V *- 0

(d) V2<1> = 0,V2'V *- 0 202. The parameters for ideal fluid flow around a rotating circular cylinder can be obtained by superposition of some elementary flows. Which one of the following set would result into such a flow? (a) source, vortex and uniform flow (b) doublet, vortex and uniform flow (c) sink, vortex and uniform flow (d) vortex and uniform flow 203. How could magnus effect be simulated as a combination? (a) uniform flow, irrotational vortax and doublet (b) uniform flow and doublet (c) uniform flow and vortex (d) uniform flow and line source 204. The velocity distribution in a turbulent boundary layer is given by: uNo = (y/8)ln, then the displacement thickness is equal to : 8 (a) 8 (b) 7 (c)

8 8

(d) 28

205. In the flow past bluft bodies: (a) pressure drag is smaller than friction drag (b) pressure drag occupies the major part of total drag (c) friction drag occupies the major part of total drag (d) pressure drag is less than that of stream lined body 206. Match List - I and List - II and select the correct answer using the code below: List - I List - II A Dynamic viscosity 1. ML2T-3 B. Moment of momentum 2. ML-l T-2 C. Power 3. ML-lil D. Volume modulus 4. ML2T-2 of elosticity 5. ML2T-l Codes: ABC D (a) 3 5 2 (b) 4 2 3 (c)

IE

V

201. For irrotational and incompressible flow, and, if = velocity potential, 'V = stream function, then which of the following is correct?

H

(a)

I

(d) GM=-xBG

(d)

1 2

2 4

3 5

4 1

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Fluid Mechanics and Machinery 207. Consider the following energies associated with Pelton turbine 1. Mechanical 2. Kinetic 3. Potential (a) 1 - 2 - 3 (b) 2 - 3 - 1 (c) 3-2-1 (d) 1-3-2 208. Consider the following types of water turbine 1. Bulb 2. Francis 2. Kaplan The correct sequence of order in which the operating head decreases while developing the same power is : (a) 3 - 1 - 2 (b) 1 - 3 - 2 (c) 2-3-1 (d) 3-2-1 209. Critical speed of turbine is defined as : (a) the speed at which natural frequency of vibrations becomes equal to the number of revolutions while the time is same (b) the speed at which the turbine stops functioning (c) the speed at which natural frequency becomes larger than the number of revolutions of same time (d) None of these 210. The dimension of the specific speed of the turbine is : (a) F1I2L-3/41312 (b) F-112L3/4T312 (c) FI/2 L 415 T 3/4 (d) F2/3L -3/4 T312 211. Consider the following turbines: 1. Kaplan 2. Pelton 3. Francis The correct sequence in increasing order of the specific speed of these turbines is : (a) 1, 2, 3 (b) 2, 3, 1 (c) 3,2, 1 (d) 1,3,2 212. The power, speed and discharge of a hydraulic turbine are respectively proportional to : (a) HlI2,HlI2,H3/2 (b) HlI2,H312, H1I2 (c) HlI2,HlI2,H1I2 (d) H3/2, HII2,H1I2 213. Which one of the following turbines is used in underwater power stations? (a) Pelton turbine (b) Deriaz turbine (c) Tubular turbine (d) Turbo - impulse turbine 214. Consider the following components of a centrifugal pump: 1. impeller 2. Suction pipe 3. foot value and strains 4. delivery pipe The correct sequence of these components through which the fluid flows is : (a) 1 - 2 - 3 - 4 (b) 3 - 2 - 1 - 4 (c) 1 - 2 - 4 - 3 (d) 2 - 4 - 1 - 3 215. The volute casing ofa centrifugal pump: 1. eliminates head loss due to change in velocity after exit from impeller 2. directs the flow towards the delivery pipe 3. converts a part of velocity head to pressure head 4. gives a constant velocity of flow Select the correct answer using the codes given below: (a) 2and3 (b) 1 and 2 (c) 1, 2 and 4 (d) 1, 2 and 3

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A-165 216. If NSI= specific speed of pump handling large discharges at low heads NS2= specific speed of pump handling low discharges at high heads, then: (a)

NS1 > NS2

(b) NS1 = NS2

(c) NS1 < NS2 (d) NS1 ~ NS2 217. The power is absorbed by a hydraulic pump is directly proportional to : (a) N (b) N2 (c) N3 (d) N4 Where (N = rotational speed of pump) 218. The minimum net positive suction head (NPSH) required for a hydraulic pump is to: (a) prevent cavitation (b) increase discharge (c) increase efficiency (d) increase suction head 219. Which of the following pump is not a positive displacement pump? (a) reciprocating pump (b) centrifugal pump (c) vane pump (d) lobe pump 220. If Tin= input torque, Tip = Out put torque, then In case of a hydraulic coupling, (a) T!p (b) Tjp (c) Tjp (d) None of these 221. Jet pumps are generally used in process industry for their: (a) larger capacity (b) high efficiency (c) capacity to transport gases, liquids and mixtures of boths (d) None of these 222. The specific speed ofa hydraulic pump is the geometrically similar pump working against a unit head and: (a) Delivering unit quantity of water (b) Consuming unit power (c) Having unit velocity of flow (d) Having unit radial velocity 223. In centrifugal pump,cavitation is reduced by : (a) increasing the flow velocity (b) Reducing discharge (c) Throttling the discharge (d) Reducing suction head 224. Water is pumped through a pipe line to a height of 10m at the rate of 0.1 m3/s. If frictional and other losses amount to 5m, the pumping power required in kw will be : (a) 14.7 (b) 8.7 (c) 9.7 (d) 10.8 225. The most probable value of speed ratio of kaplan turbine is : (a) 0.45 (b) 0.15 (c) 2.0 (d) 1.18 226. A series of normal flat vanes are mounted on the periphery of a wheel, the vane speed being v. For maximum efficiency,the speed of liquid jet striking the vane should be: (a) 2v (b) 3v (c) 4v (d) 5v

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Fluid Mechanics and Machinery

A-166

227. The unit discharge, Quand unit speed,Nucurve for a turbine is shown is shown in figure, then the curve 'A' shows:

A ~Nu

228.

229.

230.

231.

(a) kaplan turbine (b) Pelton turbine (c) Francis turbine (d) Propeller turbine A pelton turbines work under a head of 405 m and runs at 400 rpm. The diameter of its runner will be: (if'k., = 0.45) (a) 1.92 m (b) 2.91 m (c) 3.60 m (d) 4.30 m Constant efficiency curves of turbines are drawn between (on both of the axes) : (a) power and speed (b) efficiency and speed (c) efficiency and power (d) efficiency and head Specific energy is minimum at a depth of flow is known (a) critical depth (b) normal depth (c) sub-critical depth (d) alternate depth In case of performance of turbines, the main characteritics are associated with: (a) constant head and variable speed (b) variable head and constant speed (c) constant head only (d) variable head only Mushel characteristitics are studied in the performance of turbine : Mushel characteristics are related to : (a) constant speed (b) constant head (c) constant efficiency (d) None of these

Badboys2 232.

3

6 8

13

16 18

233. If h, = atmospheric pressure head h, = vapour pressure head h, = suction head h = working head of turbine then cavitation factor is defined as : (a)

ha + hv + hs h

(b)

ha - hv - hs h

(c) ha + h,+ h, (d) ha - h,- h, 234. Which of the following type of turbine requires a heavy duty governor? (a) Francis turbine (b) kaplan turbine (c) (a) and (b) both (d) None of these 235. Hydraulic radius is equal to : Area (a)

(wetted perimeter)

2

wetted perimeter area (c) square root of area (b)

(d)

Area wetted perimeter

-

236. It u = average velocity, L = length of plate, J.l= coefficient of viscosity for the viscous low between two parallel plates, the pressure drop per unit length is equal to : -

(a)

(c)

12 J.lu d2 12 J.l2ii d2

(b)

12J.l2ii d -

(d)

6J.lu d2

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Fluid Mechanics and Machinery

A-167 161

(c)

181

(a)

201

162

(d)

202

163

(b) (b)

182

143

(a) (b) (b)

183

(c)

203

(a) (b) (a)

144

(c)

164

(d)

184

(c)

204

145

(a)

165

(d)

185

(d)

166

(a)

186

(a) (b)

205

146 147

167

(c)

187

168 169

(b) (a)

188

149

(a) (a) (b)

189

(c)

150

(c)

170

(c)

(d)

151

(d)

171

152

172

153

(a) (b)

173

154

(d)

174

(c)

194

155

175

157

(c)

177

(b) (a) (a)

195

156

(a) (a)

138

(b) (a) (a) (b) (a) (b) (a)

(b) (a) (a)

158

(a)

178

139

(d)

159

(c)

179

140

(a)

160

(d)

180

121

(c)

141

122

142

127

(a) (b) (a) (b) (a) (b)

128

(c)

148

129

(a)

130 131 132

123 124 125 126

133 134 135 136 137

176

221

(c)

222

(a)

223

(d)

(c)

224

(a) (c)

206

(b) (a)

225 226

(a)

(d)

207

(c)

227

(c)

(c)

208

(c)

228

209

191

(c)

211

(a) (a) (b)

229

190

(a) (b)

231

(a) (b) (a) (a)

192

212

(d)

232

(c)

213

(c)

233

214

234

(b) (b)

235

(d)

216

(b) (a) (a)

236

(a)

197

(a) (a) (b) (b) (b) (a)

217

(c)

(d)

198

(d)

218

(a) (b)

199

(c)

219

200

(a)

220

(a) (b) (a)

193

196

210

215

230

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HINTS & EXPLANATIONS

IIIIII~ 1.

2.

(d)

For two-dimensional flow, continuity equation has to be satisfied. au Ov 00_ Ov ax +v ay = 0 => & - - ay

From equation (ii) putting value of

Here,

uav = 2

c is a numerical constant taking it zero.

From Eqs. (i) and (ii),

3.

We know, :

= v = - 2xyt

4. ...(i)

'" =-x2 yt For the equations of streamlines, '" = constant :. -x2yt = constant For a particular instance, x2y= constant (a) In a two-dimensional velocityfluid with velocitiesu, v along x and y directions. Acceleration along x direction ax

a", and - = -u = -

ax

2

x t

...(ii)

'" = x2yt + f(y)

au ax

=u-+v-

au ay

'------v----'

...(iii)

Differentiating Eq. (iii) w.r.t. y (iv) we get

8\Jf ay =-x2t+f'(Y)

= aconvective+ 3temporalorlocal

convective acceleration

Integrating Eq. (i), we get

...(iv)

in equation (iv)

=> '" = -x2 yt + C

... (ii)

16J...luoL ~p=---=-D2 (a) Given u x2t and v = - 2xyt

alfl ay

,

we get, -x2t = - x2t + f'(y) Since, f(y) = c

(d) Pressure drop across a straight pipe of length L is given by 32J...lvavL ~p = D2 ...(i) 1

Uo

•••

Since,

au az = 0

+

au az

w-+v~

au at

temporal acceleration

: for 2-dimensional flow.

au

au

:. Convective acceleration = u ax + v ay

.

I

Badboys2

Fluid Mechanics and Machinery

A-168

5.

(c) For Newtonian fluid, dy Shear stress a: dy

12. (b)

A streamline and equipotential line in a flow field are perpendicular to each other (because, when (slope) 1 and (slope), are multiplied and we get, (slope), x (slope), = -I 13. (a) Assumption:

dy . . where, dy = velocity gradient

(i)

y

rl

1------+1

-y

I------------W

V

at

changing in the direction of the flow.

dy Now tOCdy dx oc-dt dy [dx / dy]

I+----L------+I Pressure is constant along the vertical axis. Pressure along horizontal axis does change. ~P. P2.p] <0 Apply N2M (2nd) over the length I => P]m2 - (P, -I~PI) m2 - 2m lie

= rate of change of shear strain. dt (d) Continuity equation

v· v=O

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7.

8.

~P Zr L 6L Neither P nor I depend as on r. ---

Ow

Ox Oy az Multiplying by density on both sides, we get,

p(auOx + Oy Ov + OwJ= az

r

So, - is independ at then (t = r where (is on stat)

0

E

This is the equation for compressible flow. (b) Compressive flow Weber number Free surface flow Froude number Skin friction Boundary layer flow coefficient Pipe flow Reynolds number Heat convection Nusselt number (b) The relation between head and power is given by Pu = Power/Head 1.5 Thus,

Powe~ 1

(HeadI)·5

Powe~

At center r= 0, t =c D

~ =~x twxr => TW = ~PD L r D 4L 21. (d) u=x+2y+2 v=4-y

au ax

Ov = I-I =0 Oy As it satisfy the continuity equation incompressible flow so this is incompressible -+-

Wz=

24.

H:-:]

=

~[O-21

x

and since rotational component is not zero so flow is not irrotational. (a) Stagnation head = 20 m Static head = 5 m Dynamichead=20-5= 15m

Specific work done W = Vdp = (30-I)xI05 990xl000

Now, -2g = 15 u=17.15m/s

6/3 = 4 m/s 27. m

-dp p

2.93 kl/kg

for

=-1

u2

2

Thus, vavg= (2/3) x vrnax.= 2 11. (d) p] = I bar, P2 = 30 bar p = 9900 kg/nr'.

W = dp m p

0 =0

Zr wr

=~

vavg

x

t=--

... (i)

1 (Head2) .5

Given, power] = 1000 kW, Head] = 40 m Head, = 20 m Putting values in the Eq. (i) and solving, we get Power, = 353.6 = 354 kW 10. (c) In case of two fixed parallel plates, when the flow is fully developed, the ratio vrnax/ vavgis given by vrnax

x

~ ~

dt (dx/dy)

au Ov => -+-+-=0

D~

Pl~

oc"":""'_-----'-

6.

a

at D = 0

(ii) Fully developed the ~ D - 0 ; properties are not

+ dv ~ fluid flow

1

where,

Flow is steady y (i.e.)

v-u (c) ~

(y)k

= 5.75 10glO

+ 3.75

for v= u

(f)

5.75 + 10glO .1_ R

=

+ 3.75 = 0

0.223 => y = 0.223 R

Badboys2

Fluid Mechanics and Machinery 29.

(a)

A-169

Consider an element of disc at a distance r and having width dr.

dr

,r,\"

\~:\ \\" 1\('",,:, \,I,\,:, \\:~ Linear velocity at this radius = reo du Shear stress = Mdy torque = shear stress x area x r =

't

2 7tr dr x r

du

= M- 2m2dr

~p

dy assuming that gap h is small so that velocity distribution may be assumed linear dy

rro h

dT = M- 2m2dr

Badboys2

~p L

h

J2 7tMCOr3dr = M7tdco

d/2

o

4

h

'to =

dy

= 6.30

x

10-4 m2/s

= 7.91

4

x

105

rx;

= 5 ..{~

- 1.328 CD----

_

~RcL

2

45.

= 5

l x 0.15 X 10-4 6

10-3 m = 7.91 mm

x

x 105 (for middle

1.328 -2 - 1 ~4x 105 .

FD = 2 x 1 x 1 x

IfH is the head causing the flow, then V2 6LV2 H= _2 +--' 2g 2gD

x

point of plate)

10-3

6V2

Cf--

2

1226(6)2 = 2 x 1 x 1 x 2.1 x 10-3 x --2 = 92.69 x 10-3 N (a) Kinematic viscosity, U = 2.25 dia of pipe, d = 20 em Rate of flow = 1.5 liters/sec Now to find the flow we must know the reynolds number

Re=So

dM -=0 d6

6xl

6xO.5 Rex= 0.15 X 10-4 = 2

=~d2V

'4

pD3

6= 5~lx0.15xl0-4

7td2V; g 4 from continuity equation

4

64000M2

pDoD2

Hence the boundary layer is laminar over the entire length of the plate.

W

= ~D2V

flow, maximum

32M X 2000M2

VL

w 39. (b) Momentum of issuing jet is M = -QV2 g

Q

In laminar

42. (b) R, = --;- = 0.15 X 10-4

du du Shear stress = fl = pv dy dy = 129.3 x 6.30 x 10-4 x 0.25 (Asp = 129.3) = 0.02036 kg/rrr'

M=

be maximum so V

pD3

· gra dilent = -du = 0.25 m/ sec meter 31. (b) Ve Iocrty U

L should

16000M2

32h

Kinematic Viscosity,

be maximum

pVD attained when -= 2000 M

27tMCO r3dr

= --

'to to

should be maximum. velocity will be

rco h

du

T=

For

,

Now

U=

,

vd v 15000

(~)x(20)2

Re =

47.75 em/sec.

ud 20x47.75 - = v 2.25

= 424.4

Badboys2

Fluid Mechanics and Machinery

A-170

48.

R, = 424.4, means Reynoldsnumber ofthis flow is less then 2000(424.4< 2000) Hence the flow is "Laminar" (a) Given here, = 2xy, considering the following relation, -a a\jl =>4= -=-ax ay

62.

then,

65.

a\jl (a a\jl '1 _ a => - I --) -ax -" ay' ax ax a= ~(2xy) ax ax Similarly, a= 2x = ay

_ a\jl ax

On integrating, a\jl = 0+ ay

-

67.

= 2y = _ a\jl ay

C'

f a\jl = -2x => ax

2 \jI

= - 2x + C (y) z

Badboys2 49. (a)

2y2 _2x2 2y2 (y) = -2 +cl> lthen, 'I' = --+-+Cl

2

2

= y2 _ x2 + c1 or y2 - x2 + constant If flow in 2D, continuity equation becomes,

\jI

ay av ax ay So, for (i), u = x2 cos y, V = - 2x sin y

-+- =0

ay av ax + ay = 2xcosy-2xcosy

=0

au av for (ii), ax + ay = 1 - 1 = 0 au av 2yt for (iii) ax + ay = yt-T = yt - yt = 0 au av 1 1 for (iv) -+= -+x--= x ax ay x x 57.

H _ _E__13.6x103 x60x10-2 x9.81 - pg 1000x 9.81

=8.16m = 816 em (a) Given: Depth (h) = 5 km = 5 x 1000 = 5000 m Specific gravity = 1.3 P = pgh = 1.3 x 9.8 x 5000 P=63700Pa Valueof'P' IS closed to 63765 Pa' (b) Weight of wood piece (m) = 5 kg Weight of wood piece = weight of liquid displaced S = 60% of volume (v) x 1000 =~xl000 100 S = 600 v 513 v=-=-m 600 120 density of wooden piece (p] = Mass(m) Volume(v)

(y)

f c'(y) = f 2y C

(b) Let H = height of water column,

(c) Given: Pressure intensity (P) = 1.006 MN/m2 Specific gravity = 1.025, density (p) = 1.025 x 103 Let 'H' be the depth of point below water surface in sea we know that, P=pgH 6

H = _E_ = 1.006x10 = 100.04m ~ 100 m pg 1025x9.8 58. (b) Given: specific gravity of oil = 0.7 Pressure (p) = 0.14 kgf I emDensity of oil (Poil) = 0.7 x 1000 = 700 kg/m' H=_E_= 0.14x9.81 pg 700x9.81 0.14 x 9.81 x 10000 700 x9.81 1400 = 700 = 2 m of oil.

5 =-=600kg/m 1 120

3

Specific gravity of wood =

density of wood piece d . f ensity 0 water

=~=0.6 1000 69. (d) Hydrostatic force on vertical walls, PI = pgHI (i) P2 = pgll, (ii) Here, heights of vertical walls, HI =H2=4m then

_!l_ = pgHl =!!L 'P2

pgH2

H2

_!1_=i=l P2 4 70. (b) From Newton's Law of viscosity, (J.!) Shear stress (r) = J.!x velocity gradient du 't = J.!xdy 't

N

J.!= du =-2 _

m

dy

S

J.!= J:£_s = [MLr2][T] m2 [L2] J.!= [ML- 1 T- 1] 72. (d) Given: Atmospheric Pressure (Patm)= 1.03kg! em? Vapourpressure (Pv) = 0.03 kg I cm2 Air pressure (Pa) = ? P = Patm-P v = 1.03-0.03 = 1kg/ern74. (b) (fauge pressure (Pg) = 21 bar Atmospheric pressure (PatIn)= 1.013bar Absolute pressure (Pab)= 't Pab= Pg + Patm = 21 + 1.013 =22.013 bar

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Fluid Mechanics and Machinery 75. (a) Given: diameter of pipe (D) = 20 em kinematic viscosity (v) = 0.0101 stoke Reynold'snumber (Re)= 2320 Let v = velocity of flowing water,

A-171 1

= +2

94.

Re = vD = vx20 v 0.0101

76.

2320 = v x 20 =? v = 2320 x 0.0101 0.0101 20 v = 1.1716 em/s ~ 1.117 cm/s (a) Given: Mass ofliquid = 5 tonnes = 5 x 103 kg volume= 10m3

Re= vD Il P

. .. mass of liquid mass density of liquid (p) = ---~volume

= 5 x 103 = 5 x 102 78.

10 = 500 kg/m' (b) Given: diameter of glass tube (d) = 3 mm surface tension (crT)= 0.0736 N/m contact angle for water (a) = 0° 4crT cosa Then capilary rise (H) = wd 4 x 0.0736 x cos 0°

4 x 0.0736 x 1

kg-f xd 9.81 x 10-6 x 3 = 10.4 mm (approx.) (c) Initial volume(vI) = 20 m3 Initial presssure (P1) = 100Pa Final volume (v ) = 40 m3 Final pressure 2)= 50 Pa Change in volume (dv) = V2- VI = 40 - 20 = 20 m3 Change in pressure (dP) = PI - P2= 100- 50 = 50 Pa change in pressure Bulk modulus of elasticity (k) = V oume lumetrifIC stram .

80. Badboys2

(p

82.

86.

93.

= dP =~=50Pa dv 20 vI 20 (c) Given: Gauge Pressure (Pg) = 25 bar Atmospheric pressure (Patm) = 1.03 g=9.81 m/s2 Absolute pressure (Pabs) = Pg + P t =25+ 1.03 am = 26.03 bar (b) Given: Side of the cube ~a)= 5 em = 5 x 10-2m Volume of cube (v) = (a) =(5 x 10-2)3 = 125x l0-6m3 Buoyant force acting on the cube =v.pg = 125x 10---6 x1000x 10 (Assuming g = 10m/s2) = 125x 10-2 = 1.25 N (a) Average velocity (Vavg) =

Maximum velocity (Vmax) Vavg. Vmax.

ap (Rf

ax =

-R~) 81l

ap(Rf -R~) ax 41l

(b) Given: kinematic viscosity (v) = 0.25 stokes diameter of pipe (D) = 10em for a critical velocity, Reynold's number should be between 2000 and 4000 . PvD Reynold's number ( Re) = -Il

= vD v 2

96.

2000= vxl0x10 =?V= 2000xO.25 =0.5m/s 0.25 IOx102 (a) Given: specific gravity = 0.85 Viscosity (u) = 3.8 poise N

= 0.38-2 s m

Diameter (D) = 5 cm flow velocity (v) = 2 m/s density (p) = 0.85 x 1000 = 850 kg/m'' pvD Reynold's number (Re) =-Il 850 x 2 x 5 x 10-2 0.38 = 223.7 = 224 104. (a) Given: fluid velocity (v) = 20 m/s pipe diameter (d) = 1 m dynamic density (p) = 0.150 kg/m' fluid viscosityru) = 0.0000122 pvd 0.15x20x1 Reynold's number ( Re ) = = ---Il 0.0000122 =2458901.6 ~ 245902 105. (b) Given: Average velocity (Vavg) = 5m/s 1 pipe radius (R) = 10 em = 10m = O.1m 1

another radius (r) = 5 cm = 20 m = 0.05 m According to velocity distribution, Umax. =2V avg. =2x5=10m/s

Vavg =Umax [1- ~:] =1+

(~~O~n

= 10[1- 0~~~5]= 10[0.75] Vavg. = 7.5 m/s 124. (a) Given: discharge (Qd) = 0.05 m3/s, f= 0.0025 Specific gravity = 0.7 diameter of pipe (d) = 0.2 m length of pipe (L) = 1000m Considering the following formula for head loss, H _ 4fLv2 L2gd for discharge (Qd) = Area x velocity

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Fluid Mechanics and Machinery

A-I72

149. (b) Given: wheel speed (N) = 600 rpm 005=Axv= .

=~d2 4

Speed ratio (k) = 0.44 net head (H) = 300 m

xv

0.05 = ~(0.2)2 xv 4 4 x 0.05

v=

Speed ratio (k) =

H ow,

dl ea

oss

u _ 1tDN

1.59m/s

1tx (0.2)2 N

b

,,2gH where,

(H) 4xO.0025xl000x(I.59)2 L = 2xl0x0.2

-

k=

60

1tDN 60~2gH

= 6.32 m (which is near to 6.44 m)

125. (b)

0.44 =

4fL V2 Head loss (HL) = --2gd HL ocv2 HLI vr HLI --=-:::::>--=--=HL2 v~ HL2 HL2 = 4

X

(v)2

1

(2v)2

4

159. (c)

HLI = 4 times

127. (b) Given: MaximuI? velocity (vmax) = 6 m/s We know that the Ratio Maximum velocity ( v max.) _ Mean velocity (v mean)

Badboys2 134. (a)

vmax.

- 2

1000 x 9.8 x 0.013 x 32

2

6 3 2x6 --=-:::::>vmean =--=4m/s vmean 2 m Given: mean diameter of runner (DQ1)= 200 mm Least diameter of jet (dL) = 1 em = 10 mm

. . ( ) Dm 200 jet ratio m =-=-=20 dL 10

1

Number of buckets (Nb) = 15+2m

1000x 6 = 0.679 = 67.91 == 69% (approx.) over all efficiency (110) is nears to the value 69%. 173. (a) Diameter of pipe (d) = 2 decimeter, Length (L) = 5

km Average speed of water (v) = 2m/s con stan t head (H) = 5 m Darcy's friction factor (t) = 0.01 Let Pabs= absolute discharge pressure at pump exit. Pabs 4tLv HL (head Loss) = pg = 2gd

20

= 15+2

= 15 + 10 = 25 Number of buckets = 15 + 0.5 m = 15 + 0.5 x 18

= 15 + 9 = 24 138. (a) Given: Power developed (P) = 3000 kw

m3/s

Overall efficiency of turbine

('10) = (

Ql Nl -=-:::::>--=-Q2 N2 p ) pgQdH 1000

P xl000 pgQdH

2

p _ 4fLv2p 4 x 0.01 x 5000 x (2)2 x 1000 abs 2d 2 x 0.2 Pbs = 5.503 bar 177. (a) (ftven: Initial parameters: Q1 = 1200 m3/s, N1 = 1000 rpm Final parameters: Q2 = ? , N2 = 1500 rpm for centrifugal pump, Q a: N

136. (a) Given: jet ratio (m) = 18

Head (H) = 5 m discharge (Qd) = 75

0.44 x 60.J2 x 9.8 x 300 D=-------3.14 x 600 = 1.07 m or 1.08 (approx.) Water lifted I s (Qd) = 0.013 m3/s depth (h) = 32 m Power consumption (P ) = 6 kw water density (p) = 1O~Okg/m-' pgQd h overall efficiency (110) = 1000 x Pc

i

=i

vmean

3.14 x D x 600

60·J2 x 9.8 x 300

3000 x 1000 1000x9.8x75x5

= 0.815 ~ 0.82 ~ 82% 142. (b) Given: Area of jet of water (A) = 0.002 m2 Striking velocity (v) = 15 mls blade velocity (vb) = 6 m/s Force exerted on the blades = p A v (v- Vb) = 1000 x 0.002 x 15 (15 - 6) = 270 N

1200

1000

Q2

1500

_ 1200 x 1500 Q2 1000

1800

m3/s

179. (a) Difference of mercury level (H) = 10 mm = 10 x 1O-3m

180. (b)

density of water (Pwater)= 1000 kg/m ' density of mercury (Phg)= 13600 kg/m ' gauge pressure (pg) = (PHg- Pwater)gH =(13600-1000) x 9.8 x 10 x 10-3 = 1234.8 Pa~ 1236 Pa(approx.) As we know that, Force = Pressure x Area

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Fluid Mechanics

and Machinery

A-173

=dxgxHxA =dgHA 181. (a) Given: kinematic viscosity of air = 1.6 x 10-5 m2/s considering the following relation, Il a: (T)1I2 ... (i) p o:

T1

...(ii)

(T)3/2 we get, kinematic viscosity at 70° C = 2.2 x 10-5 m2/s VOC

187. (d) Given u = ')..xy3 - x2 y, v= xy2

FA = Pressure x Area = w x ~(2d)2 x 2h = 2whnd2 4 Case II : When container place with its large diameter upwards: then force (FB) will be : FB = Pressure x Area 2

=wx~(d)2 4

-"43Y 4

For the case of incompressible

flow,

FA

x2h=

wnd h 2

2whnd2 = 4

FB

wnd2h

2 204. (c) Given: Velocity distribution is given as :

au+av=o

ax ay au ax = ')..y3 -2xy, au ay = 2yx

~=(x..JI/7 = 3y

3

UO

8

displacement thickness (8*) is given as :

')..y3- 2xj + 2yx - 3y3 = 0 =0

')..y3 - 3y

y5 (')..- 3) = 0 ')..-3=0 ')..=3 190. (b) Given: diameter of pipe (d) = 0.04 m line velocity (vrnax) = 1.5 mls Ratio of average velocity with maximum velocity is 2. Then, Vavg: Vmax = 2: 1

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1 [7 817]8 8 1 [7S"x8

8

[Y]o- 81/7 S"y

vrnax 1.5 vavg. =-2-=T=0.75m/s

=8-

7

=8-S"8

8

4

=0.0004nm2

height (H) = 10m flow rate (Q) = 11m3Is

discharge (Qd) = A x vavg.

Losses due to friction and others (HLf) = 5m

=O.0004nx 0.75 =0 0003nm3/s .

3n

= --m 10000

Pumping power (p)= PgQ(H+HLf 1000

3

Is

195. (b) Considering Bernoulli's equation, between section (A) and section (B), 2 PYA Pv~ PA +--+pgzA = PB +--+pgzB 2 2 Here, PA=PB=P, YA=O'ZA =HI' ZB=(H-h)

y2 Hence, P + 0 + pgH = P +E..___!!_+ pg(H - h)

2

y2 pgH = E..___!!_+ pg(H - h)

2

y2 pgH = E..___!!_ + pgH - pgh

2

y2 pgh = E..___!!_

2

2gh =v~

199. (c)

8/7]

1/7

224. (a) Given:

Area (A) = ~d2 = ~(0.04)2

4

0

YB =figh Case I : When container placed with its large diameter down ward: then force (FA)will be :

)kw

= 1000 x 9.8 x O.l (10 + 5) kw 1000 P = 14.7 kw 226. (a) For maximum efficiency, jet velocity = 2 x wheel speed =2v 228. (a) Given: Head (H) = 405 m speed (N) = 400 rpm ~ = 0.45 Speed of wheel

(Il)

ndN rtd x 400 = -6-0 = -6-0-

20 u=-nd

3

....(i)

and also u = kn ~2gH = 0.45 •../2 x 9.81 x 405 u = 40.1 mls ....(ii) equation (i) and (ii), 20 - x n x d = 40.l

3

d = 40.1 x 3 = 1.92 m 20 x 3.14

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1)llf)I)'Jf~rl'If)~ 1~~f.I~I~I~11I~f. STEEL Iron contains carbon in two forms: (free form) and (combined form). But in steel, carbon is present in chemically combined form which is limited up to 1.5%. Beyond this percentage of carbon, categorized into cast iron. Or we may say steel is a mixture of iron and chemically combined carbon from 0.15%-1.5%. Some other elements are also present in steel like sulphur, silicon, phosphorous and manganese etc. Classification of Steel: These steel are known as plain carbon steel. According to percentage of carbon, it may be classified as under: Dead Mild Steel - below 0.15% carbon. Mild Steel (low carbon steel) - carbon from 0.15%-0.3%. Medium Carbon Steel- 0.3%-0.8% carbon. High Carbon Steel- 0.8%-1.5% carbon.

Badboys2 (a) (b) (c) (d)

Classification of Steel according Manufacturing Process: (a) Killed Steel: It is a well de-oxidised steel and the steel has been completely deoxidised by the addition of an agent such as silicon or aluminium, before casting, so that there is practically no evolution of gas during solidification. Killed steels are characterised by a high degree of chemical homogeneity and freedom from porosity. The main disadvantage of killed steels is that it suffers from deep pipe shrinkage defects. This steel is denoted by 'K'. (b) Semi-Killed Steel: It is a secondary level de-oxidised steel than the killed steel and does not show the degree of properties like killed steel. Most of steel carrying carbon 0.15% to 0.25% comes in this category. Generally it is free from blow holes and having a sound outer surface. It is most widely used in structural work. (c) Rimmed Steel: Generally dead mild steel or we may say steels having 0.5% carbon are rimmed and partially deoxidised. Due to rimmed, it consists a good surface finish. It is mostly used in rolling, deep drawing and spinning, etc. It is denoted by 'R'. Capped steels: Capped steel starts as rimmed steel but part way through the solidification the ingot is capped. This can be done by literally covering the ingot mold or by adding a deoxidizing agent. The top of the ingot then forms into a solid layer of steel, but the rim of the rest of the ingot is thinner than in rimmed steel. Also there is less segregation of impurities. The yield of rimmed and capped

steel is slightly better than that of semi-killed steel. Effect of Alloying Elements on Steel Alloying Effect on Steel Element Chromium It promotes hardness, toughness and corrosion resistance. Silicon Improves elasticity,magnetic permeability and decrease hysteresis losses. Nickel Improves corrosion resistance, toughness, ductility, deep hardness and tensile strength. Cobalt Improves toughness, hardness, tensile strength and thermal resistance.s Manganese Minimise the bad effect of sulphur and increase strength and toughness also. Tungsten Increases toughness, hardness, shock resistance, wear resistivity and red hot hardness, etc. Molybdenum Improves thermal resistance, wear resistance, red hot hardness and hardness etc. Vanadium Promotes elastic limit, shock resistance, ductility and tensile strength etc. Titanium Promotes hardness. Niobium Decrease hardness and promotes fine gram growth, impact strength and ductility etc. Aluminium It acts as a de-oxidizer and promotes fine growths Copper It Increase corrosion resistance and strength etc. Boron It improves hardenability. Steel Alloys: Along with the carbon, all steels may be alloyed by mixing some other elements in various proportions to improve following most common properties of steel. Some of them are given below: (a) To improve hardness, toughness, wear resistance, corrosion resistance, ductility and red hot hardness, etc. (b) Sometimes alloying is done to improve grain structure. Classification: Steel alloys may be classified into many types on the basis of different properties. Some of them are given below: (a) Internal Structure: On the basis of internal structure steel alloys.

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Production Engineering (b) According to Application: Structural steel, Tool steel and Special Alloys steel. (c) Principle Alloying Element: Nickel steel, Manganese steel, Tungsten steel and Chromium steel etc.

Special Steel Alloys Stainless Steel: It is alloy of steel containing chromium as principal alloying element along with other elements like Nickel and Manganese, etc. Generally, chromium present in stainless steel is about 12%. The chromium present in stainless steel reacts with oxygen present in atmosphere and makes a strong layer of chromium oxide which is a highly corrosive resistant in nature. On the basis of structure, stainless steel may be classified into following: (a) Austenitic Stainless Steel: Austenitic steel (not temperable): Cr= 16.5 - 26%, Ni = 7 - 25%, Mo if used 1.5 - 4.5%, C = max 0.07%. It contains about 10%-12% chromium, 7%10% Nickel, 2% Manganese and 1%-2% Silicon and some other elements in minor quantity like Molybdenum and Titanium etc. Its hardness and strength may be improved by cold working only, not by any heat treatment etc. It is highly corrosion resistant and non-magnetic in nature. These alloys are highly resistant to many acids including hot and cold nitric acid and at temperature above 1200°F, are stronger and scale-less than any other plain chromium alloys. It consists good ductility and weldability, etc.

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(b) Martensitic Stainless Steel: Martensitic steel (temperable) : Cr = 12 - 18%, Mo ifused = 1.3 - 2%, C = max 0.25%. It contains 10%-14% chromium along with 0.08%-1.5% carbon and some other elements. The carbon dissolves in austenite which when quenched, provides martensitic structure. It consist comparatively less corrosion resistivity and good strain resistivity. It responds good for heat treatment. (c) Ferritic Stainless Steel: Ferritic steel (partial temperable) Cr = 12 -30%, Mo if used = 1.3 - 2.5%, C = max 0.08%. It contains about 12%-18% and 25%-30% chromium without any other major alloying elements. Sometimes (1%-15%) manganese and upto 1% silicon is added. Generally, low carbon steel is employed to make ferritic stainless steel. It consists poorer ductility and formability along with good weldability having good corrosion resistivity. It is mostly used in utensils, cutlery, surgical instruments and furnace parts, etc.

High Speed Steel (H.S.S.): It is a well known tool steel and possesses the best combination of all properties. Ferritic austenitic steel ( partial temeperable) : Cr = 17 - 27%, Ni = 4 6%, Mo ifused = 1.3 - 2%, C = max 0.10%. Which are essential for a good cutting tool for working at higher speed. These are hardness, toughness, wear resistance, hot hardness. High speed and cutting feed may result in production of high temperature at job and tool steel. So, it requires to be retain its properties like hardness and toughness etc. at generated high temperature. This property of retaining hardness and toughness etc. at elevated temperature is known as red hot hardness. High speed steels belongs to the Fe-C-X multicomponent alloy system where X represents chromium, tungsten, molybdenum, vanadium and cobalt. Generally, the X component is present in

A-175 excess of 7%, along with more than 0.60% carbon. High speed steel may use with almost 2-3 times higher cutting speed than high carbon cutting tool. High speed steel may retain its hardness upto 600°C approximately. According to the alloying elements, high speed steel may be divided into following: (a) Plain High Speed Steel: It contains 18% tungsten, 4% chromium, 1% vanadium, 0.7% carbon with rest percentage of iron(Fe). It consists good red hot hardness, wear and shock resistivity. It is commonly used for making cutting tools for lathe machines, planner machines, shaper machines and drilling machines, etc. Such HSS tool could machine (tunn) mild steel jobs at speed only up to 20 - 30 mlmin. (b) Cobalt High Speed Steel: It contains about 20% tungsten, 12% cobalt, 4% chromium, 2% vanadium, 0.8% carbon and rest iron. It improves red hardness and retention of hardness of the matrix. (c) Vanadium High Speed Steel: It is simply plain high speed steel containing higher percentage of vanadium which provides better abrassive resistance than plain high speed steel. It forms special carbides of supreme hardness, increase high temperature wear resistance, retention of hardness and high temperature strength of the matrix. (d) Molybdenum High Speed Steel: It contains 6% molybdenum, 6% tungsten, 4% chromium, 2% vanadium and rest iron. It shows better cutting properties. It improves red hardness, retention of hardness and high temperature strength of the matrix, form special carbides of great hardness. Tungsten High Speed Steel: It contains 0.73% carbon, 18% tungsten, 4% chromium, 1% vanadium andrestiron(Fe).1t improves red hardness, retention of hardness and high temperature strength of the matrix, form special carbides of great hardness.

HEATlREAlMENTPROCESS PRACnCE

USED IN ENGINEERING

Heat treatment is an operation or combination of operations involving heating at a specific rate, soaking at a temperature for a period of time and cooling at some specified rate. The aim is to obtain a desired microstructure to achieve certain predetermined properties (physical, mechanical, magnetic or electrical).

The important principle of heat treatment are as follows: (a) Phase transformation during heating. (b) Effect of cooling rate on structural changes during cooling. (c) Effect of carbon content and alloying elements.

Objectives of heat treatment (heat treatment processes): (a) To increase strength, harness and wear resistance (bulk hardening, surface hardening). (b) To increase ductility, toughness and softness (tempering, recrystallization, annealing). (c) To obtain fine grain size (recrystallization annealing, full annealing, normalizing). (d) To remove internal stresses induced by differential deformation by cold working, non -uniform cooling from high temperature during casting and welding (stress relief annealing).

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A-176

(e) Toimprove surface properties (surface hardening, corrosion resistance-stabilizing treatment and high temperature resistance-precipitation hardening, surface treatment). (t) To improve cutting properties of tool steels (hardening and tempering). (g) To improve magnetic and electrical properties (hardening, phase transformation). (h) Toimprovemachinability(fullannealing and normalizing). The process of heat treatment may be classified into following types:(a) Annealing: Annealing is basicallyknown as metal softening process in which the metal heated upto its critical temperature or 30°-50°C above its critical temperature and then allows to cool at a specific rate like in full annealing process metal allowed to get cool in furnace. Normally at the rate of 10°-30°C per hour decrement of temperature of furnace. Annealed is done for one of the following purpose:(a) To reduce hardness. (b) To relive internal stresses. (c) To improve machinability. (d) To facilitate further cold working by restoring ductility. (e) To produce the necessary microstructure having desired mechanical, magnetic and other properties. Types of annealing process: (a) Full annealing: It is defined as the steel to austenite phase and then cooling slowly through the transformation range when applied to steel. Full annealing is called as annealing. (b) Box annealing: Annealing a metal or alloy in a sealed container under condition that minimize oxidation. The material is packed with cast iron chips, burnt charcoal. It is also called as black annealing or pot annealing. (c) Bright annealing: Annealing in a protective medium is to prevent surface discoloration is called bright annealing. The protective medium is obtained by the use of an inert gas, such gas, argon or nitrogen or by using reducing gas atmosphere. (b) Normalizing: Normalizing: Steel is normalized by heating 50 to 60°C (90 to 110°F) into the austenite-phase field at temperatures somewhat higher than those used by annealing, followed by cooling at a medium rate. For carbon steels and low-alloy steels, normalizing means air cooling. Many steels are normalized to establish a uniform microstructure and grain size. The faster cooling rate during normalizing results in a much finer microstructure, which is harder and stronger than the coarser microstructure produced by full annealing. Steel is normalized to refine grain size, make its structure more uniform, make it more responsive to hardening, and to improve machinability. When steel is heated to a high temperature, the carbon can readily diffuse, resulting in a reasonably uniform composition from one area to the next. The steel is then more homogeneous and will respond to

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Production Engineering the heat treatment more uniformly. The properties of normalized steels depend on their chemical composition and the cooling rate, with the cooling rate being a function of the size of the part. Although there can be a considerable variation in the hardness and strengths of normalized steels, the structure usually contains a fine microstructure. This process is almost similar to annealing except in this process metal is heated 40°-50°C above its critical temperature and holding time is very shorter than annealing like (15 minutes) and then cooled down at room temperature in still air. This process improves impact strength of metal and removed internal stress of metal. It also increase mechanical properties of metal like softness, mechanibility and refine grain structure of metal. (c) Hardening: In this process, metal is heated 30°-50°C more than its critical temperature and hold at that temperature up to a specific time, then cooled rapidly by quenching in water, oil or salt bath. This process increase hardness of metal. (d) Spheroidizing: Spheroidizing: - To produce steel in its softest possible condition with minimum hardness and maximum ductility, it can be spheroidized by heating just above or just below the A 1 eutectoid temperature and then holding at that temperature for an extended period of time. Spheroidizing can also be conducted by cyclic processing, in which the temperature of the steel is cycled above and below the A 1 line. This process breaks down lamellar structure into small pieces that form small spheroids through diffusion in a continuous matrix. Surface tension causes the carbide particles to develop a spherical shape. Because a fine initial carbide size acceleratessperoidization, the steel is often normalized prior to spheroidizing. This is the process of producing a structure in which the cementite is in a spheroidal distribution. If the steel is heated slowly to a temperature just below the critical range and held for a prolonged period of time, then structure will be obtained the globular structure obtained given improved machinability of steel. (e) Tempering: This process may be defined as opposite of hardening. In this process, hardened steel is re-heated below critical temperature and allows to cool as slow rate which increase the softness and decrease the hardness and brittleness. This process increases toughness and ductility of steel. This process enables transformation of some martensite into ferrite and cementite.Tempering is used to reach specific values of mechanical properties, to relieve quenching stresses, and to ensure dimensional stability. It usually follows quenching from above the upper critical temperature; however, tempering is also used to relieve the stresses and reduce the hardness developed during welding and to relieve stresses induced by forming and machining. The exact amount ofmartensite transformed into ferrite and cementite will depend upon the temperature to which the metal is re-heated. When the hardened steel is reheated to a temperature between 100°-200°C, then

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Production Engineering carbon is precipitated out from martensite to form a carbide called epsilon. This leads to the restoration of BCC structure in the matrix. Further, heating between 200°C350°C enables the structure to transformation to ferrite and cementite. Classification of Tempering: The tempering process may be classified as given below:(i) Low temperature tempering: In this tempering process, steel is re-heated after hardening between temperature range 150°C-200°C. This process mostly used for tempering carbon tool steel, low alloy steel, surface hardened parts and measuring tools, etc. This process increase toughness and ductility and reduce internal stress. (ii) Medium temperature tempering: In this process, hardened steel is re-heated between temperature range 300°-450°C and retained at this temperature for a specific time and then allowed to cooled down at room temperature. In this process, martensite and austenite transformed into secondary troosite, which causes increase toughness and reduction of hardness. This process also improves ductility but reduce its strength. This process is used for the steel which supported to used with impact load like hammer, coils and cheals, etc. (iii) High temperature tempering: In this process, hardened steel is re-heated between temperature range 500°-650°C, then holding for a certain time and after which allows to cooled down at room temperature. This process removes internal stress completely and provide a micro structure which have good strength and toughness. This process is mostly used for crank shafts, gears and connecting rods etc.

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(f) Case Hardening: This process is also known as Carburising. In this process, the hardness is increased only at outer surface. In this process, steel is heated upto red hot and then immersed into high carbon reason which causes a production of surface having high carbon contents. This results increase hardness surface. The carbon is infused into surface of steel by diffusion from carbon monoxide gas at elevated temperature ranges between 870°-950°C due to having carbon contents at surface of steel. The hardness increased up to limited depth of steel and below this outer surface, a soft and tough core maintained automatically. Case hardening is any of several processes applicable to steel that change the chemical composition ofthe surface layer by absorption of carbon, nitrogen, or a mixture of the two and, by diffusion, create a concentration gradient on the surface. The processes commonly used are carburizing and quench hardening, cyaniding, nitriding, and carbonitriding. (g) Nitriding: It is also a case hardening but carbon is replaced by Nitrogen. It provides equal advantage for both carbon-alloy steel and other alloys steel. It can be used for

A-177

complete hardening and selected portion hardening. Even heat treated parts can be skin hardened through this process.Alloyshaving Chromium,Aluminium, Molybdenum and Vanadium responds best by this process. This process can be achieved by using gas nitriding and salt bath nitriding. But in both process, steel is heated below critical temperature. But this process is comparatively slower than other hardening process. This process is mostly used for hardening drills remains and milling, cutting tool in which the hardness at this shank is normally required less than this cutting edge. (h) Cyaniding: This is also a case hardening process which

is used for low and medium carbon steels. In this process, carbon and nitrogen absorbed at surface which cause the hardness at the surface only. In this process, steel is heated in between range of 800°-950°C temperature in a molten salt bath. The types and proportion of cyanide salt in preparing the molten salt bath depends upon the amount of carbon contents needed at the metal surface. Mostly a mixture of sodium cyanide, sodium chloride and sodium carbonate is used in equal ratio. The hardness induced in the case of metal is due to the formation of compounds of nitrogen and carbon absorbed at surface. In this process, a low temperature is used normally 120°1500C. (i) Induction Hardening: Induction hardening is a form of

heat treatment in which a metal part is heated by induction heating and then quenched. The quenched metal undergoes a martensitic transformation, increasing the hardness and brittleness of the part. Induction hardening is used to selectively harden areas of a part or assembly without affecting the properties of the part as a whole. Induction heating is a non-contact heating process which utilises the principle of electromagnetic induction to produce heat inside the surface layer of a work-piece. By placing a conductive material into a strong alternating magnetic field electrical current can be made to flow in the material thereby creating heat due to the I2R losses in the material. In magnetic materials, further heat is generated below the curie- point due to hysteresis losses. The current generated flows predominantly in the surface layer, the depth of this layer being dictated by the frequency ofthe alternating field, the surface power density, the permeability of the material, the heat time and the diameter of the bar or material thickness. By quenching this heated layer in water, oil or a polymer based quench the surface layer is altered to form a martensitic structure which is harder than the base metal. Itis fastest method of hardening in which metal contain medium or high cast are hardened by this process. In this process, metal are placed under a high frequency (2000 cycles/sec) alternating current. When this high frequency current is passed through the metal, the carbon

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contents present in metal itself increases its hardness after passing the current. The metal itself is quenched into liquid bath. This process takes 1-5 seconds only. This can be done on a specific area and to whole components normally automobile parts like crank shaft, gears and tappet pins are hardened by this process. The depth of hardness and degree of hardness depends upon the voltage and frequency of current. In hardening of large component, slower frequency current is used. (j) Flame Hardening: In this process, a high intensity oxy-acetylene flame is used to heat the steel. After heating steel above critical temperature steel is quenched to air or water bath. Jet can be used but this process is limited with medium and high carbon steel. This process can be made manual or fully computerised and automatic. Flame hardening consists of austenitizing the surface of steel by heating with an oxyacetylene or oxyhydrogen torch and quenching immediately. A hard surface layer of martensite forms over a softer interior core. (k) Laser Beam Hardening: It is a surface hardening process and almost similar to flame and induction hardening. In this process, medium and high carbon steel is coated with absorbtive media like Zinc or Manganese Phosphate and then a Laser Beam is passed through that which causes production of heat inside the metal and after passing the laser beam, metal is quenched into water or oil bath. It is a faster method and can be easily done on complete or localised area of metal. This process can be manual or fully automatic. (I) Heat Treatment of Non-Ferrous Metal: The mainly used heat treatment process for Non-ferrous metal is strain hardening, dispersion hardening but most popular method is age hardening / precipitation hardening. Age Hardening: In this process, non-ferrous alloys are heated into a single phase solid solution. On account of their decreasing solid solubilitywith lowing the temperature their structure is transformed into two distinct phase. After which these metal allows to cooled down at rapid rate which caused structure is a super-saturated solid solution. When this alloy metal is heating at a predetermined temperature again the solute atoms precipitate of supersaturated solid solution. This process results in increasing hardness. This one of the reason due to which this process is known as precipitation hardening also.

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Defects in Heat Treatment Process: Due to some reasons, some defects may arise during heat treatment process. (a) Oxidation: If during heat treatment process, atmosphere is oxidising then oxidation may occurs, which can be prevented by controlling heating atmosphere or using carburesing agents. (b) Cracks: Sometime, cracks in metal during quenching, this may happens due to having unproper designs of object, too much hardness or delay in tempering after quenching. This defectmay be controlled by improving

(c)

(d)

(e)

(f)

part design, maintain proper hardness and doing tempering on proper timing. Soft Spots: This defect may arise due to localised de-carburisation, heterogeneous initial structure or formation of bubbles during quenching process. This defect may be controlled by quenching properly in suitable solution and ensuring the heterogeneous structure of metal. Coarse Grain Structure Formation: This defect may arise due to heating at elevated temperature to a long period then specified. This can be controlled by heating at specified temperature up to proper time. Shape Distortion: This defect caused by non-uniform heating. This can be controlled by heating gradually up to specified temperature. Holes Formation: This defect is caused due to bubble formation during quenching which can be controlled by carefully quenching and using specified quenching media / solution.

METAL CASTING Metal casting is a process in which molten metal is poured (in liquid state) into a mould. There molten metal acquiresthe desired shape and size. Which is made previously in the mould after some time when metal gets solidified it is removed from mould. Casting is the oldestmethod of shaping metal and non-metals. In earlier time most popular casting method was "Sand Casting" in which desired shape article is pressed in to sand and when the article removed from sand it leaves an impression or cavity in the sand. Which is exactly according to the shape of article after removal of article molten metal is poured in this cavity formed in the sand. The article used to make cavity in sand is known as pattern and the cavitymade in sand is known as mould. Advantage of Casting 1. Casting is a cheap, fast and economicalmethod ofproducing any shape of metal and Non-metals. 2. Large and heavy structures can be made easily by casting method. 3. For identicalmass productioncasting is verysuitablemethod. 4. Dueto productionofminimum scrap,wastageofraw material isminimised. 5. Complex shape can be made easily by casting method with low production cost and in less time investment. 6. Casting is suitable for metal, non-metal and alloys. 7. Insertion of any objectof same material or dissimilar metal is easier in casting method. 8. Some mechanical properties achieved in casting process are distinct from any other manufacturing method. Someimportant terms (A) Mould. (B) Pattern (C) Core. (A) Mould: It may be defined as a shape made up of sand, Die Steel, Ceramic, and rubber etc. in which desired cavity is produced with the help of suitable pattern. According to the material used, in making cavity, the material can represent the mould's name like, ifsand is prime material then it will be known as sand mould, and rubber mould if rubber is prime material in masking mould. Sand mould may further be classified in following types :

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Production Engineering (i) Green sand mould :- The mold contains well prepared mixture of sand, water (moisture) and binder (clay), as name resemble green is not actually green colour but normally natural sand used in wet condition having suitable percentage of moisture and clay. (ii) Skin dried mould:- It is more expensive mould having additional binding material with Green sand, which enables it less collapsibility, but higher finishing and produce better dimensional accuracy. This additional bonding material used in this mould is dried by using torch etc. (iii) Dry- Sand Mould :- It is mould silica sand which is mixed with organic binder and baked in suitable ovan. Where its moisture content is reduced due to which it provides lower collaspibility. These moulds are used for better dimensional accuracy because its formation is more time consuming. Where as additional heat and bonding material, involvement causes reduction in production quantity and increase in production cost. (iv) No-Bake Mould:- The sand is mixed with liquid resin and allowed to get hardened at room temperature.

(v)

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sand casting process for most ferrous and non-ferrous metal in which un-bonded sand is held in the flask with a vacuum. The pattern is specially vented so that a vacuum can be pulled through it. A heat softened thin sheet (0.003 to 0.008 Inch) of plastic film is draped over the pattern and a vacuum is drawn (27-53 KPA). A special vacuum forming flask is placed over the pattern and is filled with a free-flowing sand. The sand is vibrated to compact the sand and a sprue and pouring cup are formed in the cope. Another sheet of plastic is placed over the top of the sand in the flask and a vacuum is drawn through the special flask, this hardens and strengthens the un-bonded sand. The vacuum is then released on the pattern and the cope is removed. The drag is made in same way then molten metal is poured, white cope and drag are kept under a vacuum because plastic vaporises but the vacuum keeps the shape of sand till the metal gets solidified. After which vacuum is turned off and the sand runs freely, releasing the casting.

permeability. No - Moisture generated defect.

S. Better life of pattern because sand did not touch the pattern surface.

Disadvantage: 1. Lowers, the production rate. 2.

Patterns may be classified according to the following factors: (a) (b) (c) (d)

(i)

Shape and size and casting Number of casting to be made Method of moulding to be used Parameters involved in the moulding operations

Solidpattern :-

Solid patterns are made in single piece having simple geometrical dimensions, it is easy to fabricate having separately defined parting line, runner and Gate etc.

(ii)

Split pattern :-

Vacuum Moulding:- (V-Process) is a variation ofthe

Advantage of Vaccum Moulding Process: 1. Produced very Good Surface finish. 2. Cost of bonding material is eliminated. 3. No- Production of toxic fumes and provide excellent 4.

A-179 etc. These patterns are made slightly over size, for over weight, material so that extra metal can be used for matching etc. Most commonly used patterns are listed below.

Takes more time hence increases production cost.

(B) Pattern: Pattern may be defined as a solid hollow shaped item used to make cavity in the mould or we can say the replica of shape what we desire to cast patterns are made by various metals and non-metals depending upon the requirement like, wood, wax, aluminium, ferrous and ceramics

)

()

()

f

,...,.....

:':;c

pattern

~~."

pattern

~g

When model have difficult geometrical dimensions then patterns are made in two parts that meet along the parting line of mould using two separate pieces allows the mould cavities in the cope and drag to be made separately and the parting line already determined.

(iii) Match Plate Pattern :A match plate pattern is similar to a split pattern except that each half of the pattern is attached to opposite sides ofa single plate. The plate is usually made up of wooden material. This pattern design ensures proper alignment of mold cavities in cope and drag and the runner system can be included on the match plate. Match plate patterns are used for larger production.

r-

Cope Pattern

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(iii) Machining Allowance: It is also known as finishing

Fig. (cope and drag pattern) (iv) Cope and drag pattern :A cope and drag pattern is similar to a match plate pattern, except that each half ofthe pattern is attached to a separate plate and the mould halves are made independently just as with match plate pattern. This match plate helps in proper alignments of mould cavities in the cope, drag and runner, etc. Match plate patterns are used for larger production and often used when the process is automated. Cope pattern

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allowance. After casting process every casting needs some machining or finishing operations in which a considerable amount of material needs to be removed from casting surface to compensate the loss of material from the surface of casting, some additional amount of material is provided in addition of draft allowance, this percentage of extra material over casting surface is known as machining allowance. This allowance is provided both inside walls and out side walls of castings. (iv) Shake Allowance: Before withdrawal of pattern from mould, the pattern is wrapped all around the faces to enlarge the mould cavity slightly which facilitates its safe removal and causes the enlargement in mould size. So it is desirable that the original pattern dimensions should be reduced to account for this increase in dimensions or we can say that shake allowance is provided in (-ve) to the original size of pattern. (v) Distortion Allowance: The tendency of distortion is not common in all castings. Only castings which have an irregular shape and some such design that the construction is not uniform through out will distort during cooling on account of setting up of thermal stresses in them. Such an effect can be easily seen in some dome shaped or 'U' shaped castings. To eliminate this defect an opposite distortion is provided in the pattern, so that the effect is neutralised and the desired casting can be achieved.

Colour Coding in Pattern

._______ Drag Pattern Fig. Match Plate Pattern

Design of Pattern Pattern as we know very well a master/ shape used to make cavities in mould of desired shape and size. During pattern designing we have to keep the following parameter in mind as given under, like material selection for pattern making. C patterns are made from wood, aluminium, plastic, rubber, ceramics and Iron etc. In general, pattern making process involves drawing making of desired object, to be made by casting along with addition of various allowance measurements with the dimensions. Most of the dimensional allowances to be added in pattern making are listed below: (i) Shrinkage Allowance: Shrinkage on solidification is the reduction in volume caused when metal loses temperature after casting. The shrinkage allowance is provided to compensate the reduction in volumetric dimensions. Aluminium permissible shrinkage allowance is 0.013 mm- 0.01 mm. (ii) Draft Allowance: At the time of withdrawing the pattern from the sand mould. It may damage the edge etc. so for making withdrawn easy, all patterns are given a slight taper on all vertical surface i.e. the surfaces parallel to the direction of their withdrawal from the mould. The taper is known as draft allowance.

Although colour coding is not accepted but the most commonly used coding are given below. (i) Red ~ machining surface (ii) Black ~ un-machining surface (iii) Yell ow ~ core prints (iv) Red strips on yellow base ~ Seats for loose pieces. (v) Black strips on yellow base ~ Stop ofts. (vi) No - colour ~ parting surface. (C) Core: Core is generally made up of sand having bonding resin in proper quantity these core's are used for making hollow section inside the casting. A good core must have following properties. (a) It should have good permeability, so that gas can easily escape during casting process. (b) It should be made good refractory material so that it can withstand the high temperature and pressure of flow of molten metal. (c) It should have high collapsibility i.e. it should be able to disintegrate quickly after solidification of casting metal. (d) The binding material or core material should not produce additional gases during casting process.

Classification ofCore:(i) (ii) (iii)

Horizontal Core Vertical core Balanced core

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Production Engineering (iv) (v)

hanging over core Wire core Core molding: Cores are made separately in a core box made up of wood or metal. cores are made by two ways (i) manually by hand and (ii) by using core making machines.

Characteristics of cores: (i) Permeability: Cores are made more permeable than the mold to achieve, good permeability. Coarse sand & fine sand in a specific quantity are mixed with molasses. (ii) Collapsibility:- Core should possess good collapsibility so that it can be easily removed from the casting after solidification without making any damage to the casting. (iii) Strength:- Core should possess enough strength so that it should not be de-shaped during placing in mold or during the molten metal pouring. (iv) Thermal Stability:Core material should have good thermal stability so that it can withstand the high temperature during casting process.

SOLIDIFICATION AND COOLING In this process molten metal loses heat to the surrounding atmosphere and changes its state from liquid to solid, if conductivity of mould is higher it acts as the center of nucleation and crystal growth commences from the mold and extends towards the center. We can say, solidification occurs by nucleation of minute crystals or grains, which then grow under the influence of crystallographic and thermal conditions. The size ofthese grains get affected by the composition of alloy and its cooling rate. During solidification heat is being extracted from the molten metal as soon as it enters the mold. This heat is called super heat. The latent heat of fusion is also evolved during solidification and it must be transferred to the surrounding mold before complete solidifications can be achieved. Thus there are three stages of cooling i.e. liquid-solid and solid

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Solidification (i)

A-1Sl can be controlled by providing risers or heads. These are attached to the casting at the right location so that they can continuously supply hot molten metal to the shrinking casting untill it is completely solidified. Delivery of molten metal is mostly accomplished by Gating System. Where as reserve metal is supplied by risers or heads. Both functions can be served by either of two. Hence no clear - cut distraction can be made.

Classification: (i)

Parting Gate

(ii) Branch Gating (iii) Step Gate (iv) Horizontal Gating

Design of gating system: The following formulas should be kept in mind while designing of gating system.

(a) Bernoulli's equation p

v2

pg

2g

- = - + h = constant Where, p = pressure, v = velocity of liquid h = head, p = density of liquid (b) ContinuityLaw: Flowrate Qr=A, VI =A2 V2 where, A = Area of cross - section V = Velocity ofliquid (molter n metal)

Time taken for pouring: volume of mould cavity Ag ~2gH

Pouring time (t) =

Where, A_g_ = area of gate

Design or Sprees: Area of ratio (R)= ~

= ~:

Properties

Fluidity: The ability of filling all parts of mold cavity is known as fluidity.

(ii) Hot cracking: During cooling process a part of casting may be placed under tension and these tensile stresses are greater when the metal is weak and thus ultimately metal gets cracks. Ifthere is a relatively large reduction in temperature during subsequent solidifications, thermal contraction may cause cracking. (iii) Effect of Inocculation: It is a process in which the properties and structures of casting are enchanced by adding another material (metal or non-metal) to the molten metal before pounng.

RISER AND GATING DESIGN Riser is a cavity made in mold to compensate the shrinkage arises

where, A3 = area of sprees at bottom Al = area of sprees at tope

Some most important formulas used: (a) Time taken to pour (t )

Volume of mould cavity Ag~2gH

in casting and acts as a reservoir of molten metal.

Gating: Gating design must control the phenomenon in such a way that no part of the casting is isolated from active feed channels during the entire freezing cycle, it is reffered to as a directional solidification. The degree ofprogressive solidification

A3 (b) Aspiration effect: A2 (c) Solidification time:

~

= ~h;

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where, c = constant, V = volume, A = surface area (d) Relative freezing time (Rp) R) (e) Volumeratio ( V

(A/V) casting = (A IV)river

Vriver .

=V

castmg

(f) Caini's formula: R p = --

a

RY-b

+c

where, a = Freezing characteristic constant b = Contraction ration from liquid to solid c = Relative freezing rate of river and casting CLASSIFICA nON OF CASTING (a) Sand Casting: In this process a cavity is made in a sand mold by using desired pattern and then after molten metal poured into mould. Which is after solidification known as casting. There are two main types of sand used for moulding Green Sand and dry sand. In green sand un-burned sand mixed with proper amount of clay as it binds and moistens and when the sand is mixed with binding material other than clay and moisture is known as Dry Sand. Application of Sand Casting: 1. It is mostly used for cheapest casting process to maintain low production cost. 2. Complex geometrical shape can be easily made by the process. 3. Sand casting method is used for producing very heavy parts like fly wheel of power press, Railway wheel etc. 4. Many large structures are produced by this method like engine blocks, engine manifolds cylinder heads and transmission cases etc.

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Steps involved in sand casting: (i) Mould making: - In the process expendable sand is packed around the pattern, which is a replica of the external shape of the casting when the pattern is removed, the cavity that will form is used for casting. Any internal feature of casting that cannot be made by pattern that is made by separate cores. (ii) Clamping: - Once the mould has been made, it must be prepared for the pouring of molten metal. So the surface of the mould cavity is first lubricated to facilitate the removal of the casting, then the cores are positioned and the mould halves are closed and securely clamped together. It is essential that the mould halves remains securely closed to prevent the loss of any material. (iii) Pouring: This process involves pouring of molten metal in to mould in such a way that all section of mould fills properly. This can be checked by rising level of molten metal in the risers. (iv) Cooling: This process involves cooling of molten metal

inside the mould, often pouring after cooling process when molten metal gets solidified casting comes out after breaking the mold. Trimming: During casting process some extra material remains attached with casting, this excess material removed from casting is known as trimming. (b) Die casting: - In this process cope and drag are replaced with metal die. Molten metal is poured into cavity, made in metal dies. (c) Pressure - Die Casting: - In this process molten metal poured in metal die along with a specific pressure. This pressure application enhance casting finishing and increase production rate. (d) Slush Casting: - In this method molten metal is poured into the mould and began to solidify at the cavity surface. When the amount of solidified material is equal to the desired wall thickness, the remaining slush is poured out the mould. As a result slush casting is used to produce hollow part without usmg core. (e) Plaster Mold Casting - In this method sand is replaced with plaster of paris is rest the process is similar to sand casting method. (f) Investment Casting: - In this method a mould is made of ceramic by using a wax pattern. When molten metal is poured into mould wax get melted and replaced by molten metal. It is mostly used for casting of (S.S), Aluminium alloy and magnesium alloys etc. (g) Centrifugal Casting: - In this process mold kept rotating at high speed and molten metal poured from centre of axis of mould. Then molten metal due to its moment of inertia moves towards inner wall of moving mould and due to light weight of impurities present in molten metal segregated and collected near the axis of rotation, which enables to make more pure casting having higher accuracy and lowest impurities. (h) Continuous Casting: - In continuous casting process molten metal is poured from a specific height in a vertical mould. This vertical mould kept cooling facilities so that the casting continuously cooled down. This process is mostly used for casting pipes, rod and sheet of brass, bronze copper, aluminium and Iron etc. (i) Shell mould casting: - This process is similar to sand casting method except the molten metal is poured into an mold having thin walled shell created from applying a sand resin mixture around a pattern. The pattern used in this method can be reuse to make many mold. This process is mostly used for casting carbon steel, alloy steel etc. CASTING DEFECfS: (i) Un - filled section: - This happens due to insufficient metal pouring at lower temperature than required. (ii) Blow holes or porosity: - This defect happens, if molting temperature is too high and non -uniform cooling on the permeability of molding sand is low. (iii) Shrinkage: - Some time after solidification the casting gets reduced in size at surface or internally which is known as shrinkage defect. Normally it happens due to improper

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cooling rate, improper gating, rising and type of material also. (iv) Hot tears: - Too much shrinkage mostly causes cracks internally and on external surface known as hot tears. It happens due to improper cooling, and over ramming of molding sand, etc. (v) Mis-Run: - When molten metal fails to reach at every section of mold then some sections remains un-filled known as misrun. (vi) Cold shut: - When molten metal comes from two or more paths into the mould and during meeting these different flow if not fuse together properly is known as cold shut. (vii) Inclusions: - Any un-wanted metallic / non-metallic waste present in casting is known as inclusion thus inclusions may be slag of sand oxides or gases etc. (viii)Cuts and washes: - These defects occurs due to erosion of sand from the mould or core surface by molten metal. (ix) Shot metal: - This defect appears in the form of small metal shots embedded in the casting which are exposed on the fractured surface of the latter. It happens when the molten metal is poured into mould particularly when its temperature is relatively lower. It may splash the small particle separated from the main stream during the spray and thrown ahead and solidified quickly to form the shots.

HEAT FLOW RATE DURING Badboys2

(i)

SOLIDIFICATION

Heat flows from the hoter portion to cooler portion ofthe casting.

Rate of heat flow per unit Area ~

C = Specific heat of molten metal 8p = Molten metal pouring temperature 8f= Cooling temperature of metal 80 = Initial temperature of mould

1

=- k

( :)

(a)

Hot forging: Hot forging may be defined as a process in which metal is heated up to its plastic state and then a suitable external pressure is applied to achieve desired shape and size. The deformation of shape of metal depends on the type of force applied on it. If the force is applied along its length the cross-section will increase on the cost ofreduction of its length. Similarly if the force is applied against its length the length will increase and the cross-sectional area will decrease. Forging may be used to bend the work piece. Without change its length along with using suitable dieand punch etc. In forging process external force may be applied by hand hammer, power operated hammers, and presses etc. If the force applied bymannually by hammers this process is known as smithy process. Classification of forging: Forging may be classified into following types (a) Upsetting (b) Drawing out or drawing down (c) Bending (d) Setting down (e) Forge Welding Up setting: In this process cross-section of work piece is increased with corresponding reduction in its length.

ky/ hrm' ~~~ force application

Where k = Thermal conductivity in KJlhrmk°. dt dx = thermal gradient in units of temperature

(T) and

distance (x). if metal is cooling against a large mold wall and heat flow is normal to the mold surface thickness (x) of solid metal deposited will be proportional to the square root oftime (t) or x =

K}

.Jt

Solidification time

o:

Volume ( S ~ Ar

unace

2 ea J

.: K = Constant

Where Pm = metal density P =densityofmolten metal L = latent heat ofliquid metal.

a = Thermal diffusityofmould Cm = Specific heat of mould

K = pc

Metal attains plastic state when an external force is applied along its length accross of its cross-section. Which results in increase in its dimensions at right angle to the direction of applied force with a corresponding reduction in its length, parallel to the direction of applied force. Normally bar stocks are used for being jumped by a desired amount so that this part can be given a desired shape through the jumped further operations. The jumping operation can be performed in any localised area i.e., the particular part in the bar shape, where said increase in cross-section is desired is heated till it acquires a plastic state. Than the length which do not required to be jumped cooled abruptly by quenching in water, and the hot portion is placed under suitable load. This operation may carried out manually ifthe work peice is small enough to handle and when heavy force is required (such as in large work peice) heavy hammer is used called as sledge hammer. The objects of cooling the bar length, which is not to be jumped out is two fold. Firstly to localised the reduction in length to the desired extent and secondly to prevent the bar from bending during up setting due to heavy blows.

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(b)

(c)

Drawing out or drawing down: This process is exactly a reverse process to that of up setting orjumping in the sense that, contrary to the latter, it is employed when a reduction in thickness or width on is desired with a corresponding increase in its length. In this process specific shape oftools also required to achieve the desire shape, known as pair of sewageand fullerthe selectionof the abovetools is governed by the shape of the cross-section of the stock, the Rod or bar heated up to pre-determined length to the plastic state followedby the cooling of the unwanted length for drawing by sudden quenching in water. If the bar is of rectangular or square cross section it is laid flat on the anvit face and hammered by the peen of cross-peen hammers by the limit. Ifthe reduction is to be done both in width as well as in thickness the operation is repeated by turning the bar at 90°. The desired result can be more quickly achieved by keeping the bar on the edge or hom of the anvit and then drawing. Bending: Bending of bars, flats and other simillar stock material is usually done in smithy shop, this can be done to produce different types of bent shapes such as angle, ovals and circles etc. Any desired angle or curvature can be made through this operation. For making a right angle bend that particular portion of stock, which is to be subjected to bending, is heated and jumped on the outer surface. This provides an extra material at that particular place which compensates for the elongation ofthe outer surface due to hammering during bending. This operationis carried out on the edge of a rectangular block. After bending, the outside bulging is finished by means of a flatter and the inside by means of a set hammer, this process can be made by mannually or by using forging. Machine along with jigs and fixture. Setting down: In the operationthrough which the rounding ofa comer is removed, to make it square, by means ofa set hammer. By putting the face of the set hammer over the round portion, formed by fullering or bending ofthe comer and hammering it at the top reduction in thickness takes place resulting in a sharp and square comer. Finishing is the operation through which the un-evenness of a flat surfaceis removed bymeans of a flatter or a set hammer and round stems are smoothened to the correct shape and required size by means of sewage after the job has been shaped roughly to the finished size through other operations. Forge welding: In this process two peice of simillar metals are heated properly up to sufficientwelding heat andjoined together by application of external heat, two important considerations are always made in order to get a sound weldedjoint. (i) Proper end preparation of the metal peices to be joint (ii) Rising the temperature of the prepared ends to the correctweldingheat. The surfacesto bejoined together should be quite clean i.e., they should be free from scale, dirt or ash. Otherwise this presence will lead to the failureof thejoint. In additionto this a fluxis applied on the hot metal which helps in over coming the above difficulties. This flux usually stand for wrought iron

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(d)

(e)

and borax for mild steelmelts at high temperature and form slag containing the iron oxide, ash etc. This slag forms a layer over the hot surface thus preventing it from comming in further contact with air. Which the result, further oxidation of iron is checked. There are three types ofweldedjoints in common which are • Butt weld: When two bare are made tojoin end to end bywelding, such that joint formation is at right angles to the lengths of the work piece it is known" as butt weld. • Scarf weld: This is alsoknown as lap weld. It is known so for the reason that the ends of the metal pieces to be joined are made to overlap each other and then hammered. Thus the weldformationis at an inclination with the top and bottom faces of the joined pieces. Also due to the distinct end preparation it is easy to apply correct pressure by hand hammering in proper direction. • "V" weld: It is also known with so many names e.g. split, splice or fork etc. It is employedwhere a highly strong weldedjoint is neededparticularlyin heavyworkwhere the greater thickness ofthe job enables the formation of =v: easily, to ensure perfect joining of metals the scarf of one piece should be made rough byproviding steps on it.

(i) b..tt VIA9Id

(ii) scerf VIA9Id

(iii) V-\I\9Id

Differentweldingjoin HOT EXTRUSION The process of extrusion consists of corresponding a metal inside a chamber to force it out through a small opening called die.Any plastic material can be extrudedsuccessfully. Most of the process used for extruding metals are hydraulically operated horizontal presses. A large number of extruded shapes are in common use, such as tubes, rods, structural shapes and lead covered cables. The principle of operation are the same for both hot and cold extrusion, and choice of one ofthese is governed by factors like the metal to be extruded, thickness of extruded section size of raw material being used, capacity of press, and type of product etc.billets of 125mm to 175mm in diameter and 300 to 675 mm length are in general used as raw material forextrusions of steel needs adequate lubrication around the billet. This is done by providing a coating of fine glass powder over the surface of hot billet. The process of extrusion suits best to the non ferrous metals and alloys although some steel alloys like stain less steel are also extruded. The extrusion process can be classified as follows (1) Direct or forward extrusion (2) Indirect or backward extrusion

Direction Extrusion: As shownin the figurebelow,In this process billetofraw metal to be extrudedis heatedto its forgingtemperature and forced in the machine chamber, this force push forward the billet and billet passed through the die. The length of extruded

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Production Engineering part will depend upon the size of the billet and cross-section of the die. The extruded part is then cut to the required length. The over hanging extruded length is fed is to a long support called the run out table.

A-1S5 The rod is fed up to stops through straightening rolls, cut to size and pushed into the header die. The rod is gripped in the die and punch operates on the projected part to apply pressure and form the head. The bending operation may be completed in a single or two strokes. Automatic machines for producing bolts and screw are also available in which all the operations like cutting stock to size, shank extrusion, heading trimming and threading etc. are performed simultaneously to produce finished components. The processis also successfullyadoptedfor producing rivets and nails. COLD EXTRUSION

Fig. Forwarded extrusion process It is a usual practice to leavethe last nearly 10% length of billet as un-extruded. This portion is known as discard which contains the surface impurities of billet. Indirect Extrusion: As shown in followingfigure ram or plunger used is hollow type, and as it pressed the billet against the back wall of the close chamber, the metal is extruded back in to the plunger. As the billet does not move inside the chamber, there is not friction between them. As such, less force is needed in this method in compressionto the directextrusion.Amore complicated type of equipment is required because plunger becomes weak due to the reduction in the effective area of cross-section and difficulty is exprienced in supporting the over heating extruded part.

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The most common cold extrusions process is impact extrusion, in which soft and ductile metal is used to formed various product like tubes for tooth paste, lotions, shaving cream, paints and condenser cans etc. The raw material used is in slug form having been either turned from a bar or punched out of a strip. The operation is performed with the help of a punch and a die. The prepared slug is in the die and struck from top by the punch operating at high pressure and speed. The metal flows up along the surface ofthe punch, forming a cup shaped component.When the punch moves up. Compressed air is used to separate the component from the punch. In the mean while a fresh slug is fed into the die. The production rate is quite high about 60 components per minute. Mostlywall thickness produced from 0.7 mm to 0.1 mm but only soft and ductile material can produced bythis processlike lead,tin, aluminium,zinc, and respectivealloys etc. Uniform diamensions, low scrap production and high production capacity is main advantage ofthis process. Although Die and punch are used in like drawing process but its high production rate, and tolerance of the order of ± 0.762 mm up to 12.7 mm diameter and± 0.127 mm upto 25 mm dia can be easily obtained. WELDING

Fig. BackwardExtrusions Advantage and limitations of hot extrusion 1. Due to application of higher pressure a very dense structure is produced. 2. Better surface finish is produced having higher dimensional tolerence. 3. Low tool cost involves and fast in production rate. 4. Most suitable for production of parts having uniform crosssection having fine surface finish and high dimensional accuracy. 5. Excessesive length object is creak problem in handling the extruded rod during extrusion.

Welding is a process ofjoining two or more than two similar or dissimilar metals together with or without use of pressure, and filler materials. Without use of external heat we get success in welding of Gold and Silver only till today but in future the use of temperature in welding may be reduced considerably. Welding process may be classified as follows: (i)

Homogeneous welding: In this method two similar metals are joined together by welding and use of filler of same material ifrequired. For example, mild steel with mild steel welding this process is also known as autogenous welding.

(ii)

Hetrogeneous welding: In this method, welding is done with two dissimilar metals and the filler metal used in this processis usuallykept, oflow meltingpoint than the parental metals. For example copper and brass, mild steel and cast iron etc.

COLD FORGING This is a cold up-setting process adopted for large scale production of small cold up set parts from wire stock. A few examples of such parts are small bolts rivets, screws, pins nails and small machine parts. Small balls for ball bearings are also made by this method. The machine, tool, and dies are almost simillar as in hot forging.

Classification Pressure (a)

fo Welding According

to Application

of

Non-pressure welding/fusion welding: In this process of weldingthe temperature ofjoining edge of metals are heated

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A-186

up tomeltingpoint andwhen it startsto meltthe fillermaterial is filled between joints. For example-Gas welding, Arc welding, Electric beam welding and Thermit welding etc. (b)

Pressure welding: In this process of welding two edge to be joined are heated up to their plastic state and then sufficientpressure is applied till the weld is performed. But no-fillermaterial is used commonlyin there weldingprocess. For example-Forge welding, Resistance welding etc.

Classification

of Welding on the Basis of Heat Source

(a)

Chemical welding: According to chemical method, heat is produced by oxidation or may be burning of coal and gas etc. Heat is also produced by chemical reaction of two or more salts together. For example iron oxide and aluminium powder produced heat by chemical reaction. This method of heat generation are employed in forge welding, gas welding and thermit welding.

(b)

Electric welding: These proceses use electrical energy to produce heat required to melt the work piece. Electrical energy based joining process may further classified as follows: (i)

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Electric arc welding: In an open circuit when resistance of air gap between two terminals of conductors is less than the quantity of current/voltage carrying acrossthem, the electrons willjump from one terminal to another. This is calledjumping of electrons and due to this arc a high temperature generates at bothterminals which is about 3700°C-4000°C.

(ii) Resistance welding: In this method heat is produced when sufficient quantity of current is passed through a conductor having proper resistances. For example spot welding, projection welding etc. (iii) Induction welding: In this method heat is producedby

use of high frequency current to produce sufficient eddy current in the workpiece to be weld. (c)

Mechanical method: This method is rarely used in modem practice because the heat production in this method is very low as compared to energy applied as heat produced by friction or heavy blow/impact load etc.

CLASSIFICATION OF RESISTANCE WELDING (A) Spot welding: Welding machine used in this type of welding consists of two cylindrical pointed electrodes, out of them one electrode is kept fixed and other electrode is movable. Movable +-- Electrode

Spot welding

The work piece is to be weld placed between these electrodes and a high ampere current is passed through them for a limited time till the metal get fused at the place of welding and then applied sufficientpressurewhich make completethe weldingjoint. This process is mostly used in thin metal sheet welding like domestic utensils, cabinets like structure etc. Important

Factors Related to Spot Welding

1.

Welding pressure control: For good welding a sufficient pressure application for enough time is very important and this pressure applied for welding is known as welding pressure. This pressure should be applied on job on accurate time of plastic state of metal. Time for pressure application depends upon the thickness and properties of metals to be welded. 2. Time management: It is the total time consumed while completing different stages of welding and it is known as cycle time also. This cycle time must be adjusted in such a manner so that the metal should acquire sufficient plastic stage required for good welding and cut the supply automatically after completing the heating stage. Time control may be managed by different methods like das pot circuit breaker, electronic circuit etc. 3. Surface penetration: The surface to be welded should be free from all/dust or any un-wanted materials. So that the penetration of welding joint should be max, min and weld can make proper. 4. Electrodes: The electrodes must possess mainly these characteristics i.e. high electrical and thermal conductivity and it should have sufficient mechanical strength to with stand high pressure to which they are subjected.These are made water cooled. The surface of electrodes must be easily cleanable so that the resistance between the surface of electrodes and work metal should be kept minimum. Electrodes are mainly made up of copper alloys with molybdenum and tungsten. Spot welding process may further be classified into following types depending on their application. (a) Rocker arm type: The machineusedin spotweldingprocess consists of one fixed electrode and other movable electrode which is mounted on a rocker arm and moved in up and down direction by mechanical arrangements. In some machine mechanicalarrangement is poweredwith hydraulic system to make more automated system. (b) Press type: In this types machines are used in heavy or thick sheet welding and movable electrode is operated electrically or by compressed air. (c) Portable Guns: In many places it is not feasibleto transport hence for that purpose a portable machine is required. This Portable Gun carries two electrodes and the transformer is supported generally on over head rails. Mainly it is used in automobile industry. (B) Seam Welding

Fig. Spot welding

Itis series of closely spaced or single line spot welding. The weld shape for individual spot may be of any shape like round or rectangular. In this process, two circular disc shaped electrodes

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Production Engineering

A-1S7

are used out of which one is kept moving and other kept only movable but not joined with any moving arrangement. The work piece is placed between these electrodes and current is passed through these electrodes. This current carrying electrode when rotates the work piece moved forward and a continuous line of spot welding performed.

In resistance butt welding the metal to be weld should have equal cross section and properly faced at their respective ends. For welding wires and rods up to 12.7 mm diameter the machine may be used as spring operated and for larger diameter high pressure is mostly applied hydrauliceley or pneumatically. Resistance butt welding may be used to increase length of pipes, rodes, wires and bars of highly conductive material like copper, brass and aluminium etc.

(E) Resistance Flash Welding It is almost similar to resistance butt welding except that it is operated comparatively on less current. In this method, current is switched on before abutting the ends of bar etc. and then the movable clamp is transported towards the fixed clamp containing another metal piece maintaing small gap between both mating ends.

Continuous line of welding joint

Fig.Line diagram of Seam welding The machine used in seam welding is almost similar to spot welding except it contains circular disc shaped electrodes attached with revolving mechanism between two circular disc like electrodes one powered by rotating force is known as drive and another kept movable is known as driven. The pressure applied on driving wheel electrode by hydraulically or phenumatically. The seam welding is mostly used for metal having sufficient electrical resistivity. For example mild steel, tin plates and many dissimilar metals like steel with brass and bronze.

Badboys2 Seam welding

may be further classified as circular longitudinaltype, universaltypeandportabletype.

type,

Work Piece Fixed clamp

Movable clamp

Fig. Resistance flash welding

(C) Projection Welding This welding is almost similar to spot welding except of having any projection on both faces of electrodes. So it is most effectively used in mass production of multi point spot welding in single stroke as desired projection.

Due to this small gap, a flash developed between the ends which produce a high heat at both ends and metal at both ends gets melted, after this melting sufficient pressure is applied on movable clamp and both ends get fused and welding joint gets completed.

(D) Resistance Butt Welding

The flash developed at the ends of work piece only on a small part of it, so comparatively less electric current consumed. It is more faster process then the resistance butt welding and no facing at ends of metal required in this method. During welding, slug and remaining molten metal comes out from the weld joint, so weld joint made by this method is more stronger than resistance butt welding joint.

Resistance butt welding has similar working principle of welding as in spot welding except that electrodes are in clamp shape in which one clamp is fixed type and another is movable type. The job to be weld are normally bars, pipes, wires etc. One piece to be weld kept in fixed clamp and other clamped in movable clamp. Both metal pieces are faced (finished at ends) properly. Then movable clamp containing working metal (steel pipe) is so adjusted that both ends meet together which are to be welded. After properly meeting ends of metal , the current is switched to till corresponding ends of metals are reached to the fusion point.

These resistance butt and flash welding processes are limited on the capacity of clamping size of welding machine and the material coming out from the joints need extra machining etc. which may increase production cost of welding.

PERCUSSION WELDING This method of welding involves the use of stored electrical energy either in reactors, capacitors or storage batteries etc.

i

-:

Movable clamp

Work Piece

/

Fixed clamp

Fig. Resistance Butt welding

In percussion welding the heat for welding is secured simultaneously over the complete area of abutting surface from an arc produced by rapid discharge of stored electrical energy followed immediately by application of pressure. Percussion welding permits welding harden steels without affecting heat treatment and dissimilar metals can be weld successfully like steel with Mg etc.

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A-ISS

MUL TI IMPULSE WELDING This method is also known as pulsating welding, it is applicable to spot welding, seam welding and butt welding processes. etc. Pulsating welding process consists of applying the current in a series of impulses which may be a fraction of cycle or no. of cycles. This process has certain advantage, for example more thicker materials can be easily welded with same equipments increase electrode life and spettering of welding reduced considerably.

welding due to development of magnetic field around. It generally happens in D.C. arc welding due to having fixed polarity. Arc blow generally occurs in three directions forward, backward and side.

7.

Arc crater: It may be defined as the penerat ions of arc in base metal, it depends upon arc length, electrode width and thickness of base metal.

8.

Spatter: Molten metal dispersed around the welding beads in small drops form is known as spatters.

ELECfRIC ARC WELDING

9.

In this method no external pressure is applied, only the metal to be welded are heated up to welding temperature and a pool of molten metal fills the gap in between the joints, then these joints allow to cool in air and weld get completed.

Chipping: Removing the spatters and slage etc. formed on welding bead on metal surface during welding is known as chipping. The slage is formed as a by-product due to use of coated electrode in welding process.

10.

Edge preparation: For making different types ofjoint, some

In some types of electric arc welding an additional filler material is applied known as electrode and heat is produced by electric Arc about 3400°C. At initial stage electrode requires potential about 60 - 100 volt and in running condition when a regular arc is produced, it requires only 15 - 45 volt normally to maintain the welding operation.

Important 1.

side of work metal has to be grinded in specific shape and size. The grinding at edge/side of work piece is known as edge preparation.

11. Weaving of electrodes: This term is related with forward motion of welding electrode on the surface of welding plane. Weaving means tilting of electrode simultaneously along with forward motion of electrode. This is used for increasing width of deposition of molten metal over weld.

Terms Related with Electric Arc Welding

Open circuit voltage: This voltage may be called the voltage at electrode when no Arc is formed and machine is in switched on condition generally it remainsant 60 -100 volt.

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Motion of Electrode

Arc voltage: This voltage may be defined as the voltage of an electrode on electrode when regular Arc is formed during welding operation. Arc voltage i.e.

3.

= Cathode drop + volumn drop + Anod drop Fig. Different weaving styles

V=V epa +V +V

Duty cycle: Duty cycle is the time duration up to which that specific machine can supply a specific current and voltage for a specific time duration without making any hazard to a welding machine.

4.

As per requirement of joint, there are different weaving styles as shown in above figure. 12.

electrode tip from work metal during welding process. Actually it is better known by practices about the correct length of Arc. The distance between the electrode tip and work metal depends upon the voltage and current used for various welding process. Normally about 3 mm distance is assumed as correct distance, less than 2 mm is counted as short Arc length and above 3 mm upto 6 mm is known as long Arc length.

Power factor: It is the relation in between the current used and total current supplied to machine. Power factor =

5.

Current used

kW

Current supplied

kVA

Polarity: This term is mainly associated with D.C. arc welding because, D.C. current has fixed polarity i.e. + ve and - ve terminal and for the A.C. they interchanged at every cycle. It may be classified as follows.

13.

Straight polarity: Work piece made positive terminal and the electrode is made negative terminal, it is used for more thick plates etc.

Blow holes: It is a type of defect formed during welding process due to presence of any impurity or air bubbles or any space remains un-filled by molten metal during welding process.

14.

(ii) NegativepolaritylReverse polarity: Work piece is made negative terminal and electrode is made positive terminal, it is used for thin plates welding.

Buckling: It is also a type of defect. When work metal is twisted or deshaped in un-wanted direction during welding the process is known as buckling.

15.

Hard facing: Hard facing may be defined as the process of

(i)

The polarity have a considerable effect in welding because heat generated at positive terminal is much more than the negative terminal. Heat generated at positive terminal is about 2/3rd higher than negative terminal. 6.

Arc length: Arc length may be defined as the distance of

Arc blow: It may be defined as the deviation of arc during

hardening the surface by welding process. 16.

Heat affected zone: During welding process some time weld metal looks separated from work metal, it happens due to improper heating. This effect is called Heat affected zone or we may say the place had been effected by improper welding heat.

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Production Engineering 17. Padding: This is the process of making number oflayers of metal on a used part of metal to increase its dimensions.

18. Penetration: It may be known as depth of fusion during welding process.

19. Slag: When a flux coated electrode is used in welding process then a layer of flux material is collected over welding bead which contains the impurities of weld material. This layer is known as slag. It is removed by chipping of weld.

Classification of Electric Arc Welding (A) Metal arc welding: In this welding process the arc is made between work metal and electrode (may be bare or coated electrode). Base electrode is made up of same material but using it having certain disadvantages such as welded surface may be subjected to oxidation. To prevent the oxidation of welding surface, coated electrodes are used.

(B)

Carbon arc welding: This process is mainly employed with D.C. supply only due to having specified polarity in D.C. supply. A carbon electrode with negative polarity produce arc when close to work metal connected with positive polarity current. Straight polarity connections are made to prevent carrying over of carbon contents over metal surface during carbon electrode fusion. Otherwise deposition of carbon contents may result in a brittle and bad weld.

A-189 Metal Arc Welding The basic principle of metal arc welding is the development of electric arc between the metal electrode and work metal. The metal electrode (bare or coated) having sufficiently high ampere current when kept at proper distance to job an electric arc is developed or we can say the high ampere current value over come to the resistance offered by air gap between the electrode and job having different polarity of current. And a certain amount of electrons jump over the work metal surface from electrode which produced a high temperature near about 3400°C. This high temperature is utilised in melting the work metal up to molten stage at joining points and the electrode also melts simultaneously. Melting of work metal at joining make a pool of molten and alongwith in the molten filler metal cover this pool of metal. This covering of molten electrode over pool of molten metal is known as " welding bead".

Electrical Energy: Both A.c. and D.C. electrical energy are widely used in arc welding process. Both have some advantages and disadvantages which regulate the use of particular electrical energy for a specific welding. Use of electrical energy also depends on the material of work metal properties of material to be weld like thickness of metal etc.

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Carbon arc welding is mostly for steel sheet and casting holders consist of magnetic coil which guide the Arc. This welding process is operated manually or by machine or both.

Properties

of A.C. and D.C. Arc Welding

Properties

A.C. welding process

D.C. welding process

1.

Installation cost

Less initial cost investment.

Higher initial investment.

2.

Maintenance

Economical and easier.

Critical and costiler.

3.

Current value

Mostly suitable with higher current value

Better suitable with lower current value.

4.

Arc

In some cases it is comparatively difficult

Comparatively easier to develop an arc it

to develop an arc and maintaing of arc

consists almost every time problem of Arc

is little difficult than D.C. arc welding. It

blow etc. In D.C. welding process

consists very rarely problem of arc like

maintaining of arc is comparatively

arc blow etc.

easier.

S.No.

5.

Power supply

It is most preferred withA.C. mains supply

It is easily used with A.C. and any D.C. power supply also.

6.

Polarity

Its polarity is interchanged with every change

It has fixed polarity.

of cycle of power,

7.

8.

9.

Electrode

Bare electrodes are not suitable, so only flux

In this process bare and coated both types

coated electrodes are mostly used.

of electrodes, can be used easily.

Maintaining small arc is difficult, only

Maintaining of small arc is easier than

iron powder electrodes are exceptional.

A. C. Arc welding.

Welding

By this process welding of thin sheet is

Thin sheet can be easily welded by this

capabilities

difficult. Welding capability is limited up to

process. It has distinct polarities so it is

Arc length

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Production Engineering

A-190

to.

Welding distance

only ferrous metals generally due to change

easier to weld different metals also other

in polarity in every cycle.

than ferrous metal.

Voltagedrops are less as compared to D.C.

It has relatively more voltage drops so

supply at a distance from main supply. So for distance welding from power mains supply

welding is preferred to do at nearest to the D.C. mains supply.

A.C. welding is mostly preferred. GAS WELDING

It may also be considered under non-pressure fusion welding. The source of heat required for fusion of metal is achieved by flame of suitable gas combustion. It consists of a flow of any suitable gas under specific pressure which gives a flame after burning in presence of oxygen etc. Tools and Equipments In gas welding process different tools and equipments are used. Some of the mainly used are mentioned below: Welding torch - or blow pipe may be defined as the equipment designed for mixing oxygen and combustible gas (acetylene etc.) in required proportion and injecting for combustion and making flame or we may say that with this equipment we can acquire an adequate mixed proportion of oxygen and acetylene (in oxyacetylene gas welding) to develop a suitable flame for welding

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Classification: (a) According to pressure of acetylene gas (i) High pressure welding torch (ii) Low pressure welding torch (b) According to number of tips used with torch (i) Single tip welding torch (ii) Multiple tips welding torch (c) According to fuel used (i) Acetylene welding torch (ii) Hydrogen welding torch (d) According to application (i) Mannual welding torch (ii) Automatic welding torch Hose pipe: Itis used for supply gases from pressure regulator to welding torch. These are made up of rubber coating overthreaded net pipe. It should have sufficient strength, light in weight, economical, and non-reactive with gas which they tend to carry. These are fixed with welding torch with the hose pipe clamp. Pressure regulator: Itis a pressure controlling devices used for supply of desired pressure of gas to loose pipe connected with welding torch. Itis mounted directly over gas cylinders. Classification: (a)

(b)

Single stage regulator: Itregulates pressure of gas at one stage only. It has to be regulated from time to time as the internal pressure inside cylinder varies. Two stage regulator: It is desired to regulate pressure of gases at two stages. One is auto-controlled and other is

controlledbyextremelymounted screw.It consistsofa small storage chamber due to which the out going pressure is out of effect of pressure variation inside the cylinder. This type of regulator consists of pressure gauges mounted on regulator which shows the pressure of gas inside the cylinder and out going gas pressure. Acetylene gas purifier: These are used in low pressure acetylene gas generators. It is used for detecting impurities like sulphides and phosphomines etc from the acetylene gas to improve the properties of acetylene gas. Water seal or hydraulic back pressure value: It is used in low pressure acetylene generator system. It is mounted between welding torch and acetylene generating cylinders/tank. Important

Applications:

Itreduces the back fire hazards It works as non-return value against atmospheric air and oxygen when the pressure of acetylene gas is reduced than the atmospheric pressure inside the tank. Safety valve: It is a safety device used to provide safety against high pressure of gas than the recommended range. Welding table: It is used for placing jobs during welding operations. It is made up of mild steel and top is made by some refractorymaterial/refractory brisk etc. Welding torch lights: Itis an instrument which produces spark used for lightening weldingtorch. In practice, electronicgas lights are commonly used other gas welding equipments are welding goggles, apron, gloves, and wire brush etc. (i) (ii)

Gases used in welding process: 1.

2.

Oxygen (02) It does not go through combustion itself but very helpful in combustion process with different gases. It is storedin metalliccylindersat about 120 kg!em?in liquefied state. It is prepared by following two methods mainly (a) Byliquefication of air (b) By electrolysis of water Acetylene (C2H2): It is highly inflamable gas and produces about 3600°C temperature.

Production

method:

(a)

Combination of carbon and hydrogen: In this processtwo carbon electrodes are used to produce arc in presence of hydrogen gas which make C2H2 in which a little amount of methane and ethane gases are found.

(b)

Natural gas de-composition: It is most popular method

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Production Engineering

A-191 5.

producting acetylene gas in modern life. In this method natural gas is treated by electric arc which produces acetylene and hydrogen. (c)

By calcium carbide: "In this method calcium carbide is reacted with water as resultant acetylene gas and lime are produced. CaC2+ Calcium Acetylene

2~O Water

~

Ca(OH)2 + C2~ Lime

carbide The reactor vessels used for producing acetylene are called generator. According to pressure of generated gas, the generator may be classified as under (i) Low pressure Generator: Containing gas pressure of about 0.1 kg/cm-. (ii) Medium Pressure Generator: Containing gas pressure of about 0.1 to 1.5 kg/cm-. (iii) High Pressure Generator: Containing gas pressure of more than 1.5 kg/cm-. Properties of Acetylene: (i) It is colourless gas and lighter than air. (ii) It explodes at about 300°C itself in presence of oxygen. (iii) It has mild smell and having no harmful action to being but in more than 40% cases it creates problems in respiratory system. (iv) It can be converted, into liquid state at about 1°C temperature and 49 kg/ern? pressure. Properties of Hydrogen Gas: (i) It is highly inflamable gas and produces about 2400°C temperature. (ii) It is a colourless, odourless and tasteless gas. (iii) It is generally used for cutting and welding soft metal likealuminium, magnesium and lead etc. (iv) Retort gas: It is a mixture of number of in flam able gases produced by decomposition of oil at about 740°C in a retort. Natural Gas: It is a colourless and odourless gas which is a mixture of hydrocarbons and achieved from oil mines. Propane and butane: These are produced from oil refineries. Some other gases also used in gas welding process. For example coke oven gas, petrol or kerosene gas, argon and helium etc. Filler material or Electrode: Filler material may be defined as the material rod required to fill the gap between the metal in molten state. Dry various metal electrodes are used with different welding processes. Welding Rods

Applications Mild steel etc.

2. 3.

Low carbon steel (copper coated) High carbon steel Stainless steel

4.

Aluminium

L

For making hard weld etc. Stain less steel goods welding Aluminium goods welding

7.

Cast iron

Welding of copper made articles Mainly in gas welding and brazing etc. For cast iron welding

Properties: 1. It should be economical and easily available. 2. It should have low melting point than the filler metal. 3. It should have sufficient quality of dissolving impurities of molten metal and light inweight so that it can float above the welding metal in molten condition. 4. It should be easily removable after welding. 5. It should not produce any deflect in weld.

Flux

Application

1.

Borax (Na2B407)

2.

Cast iron flux

3.

Brazing flux

4.

Alumina

Used with mild steel and low carbon steel etc. Used with cast iron, high carbon steel, ferrous silicon and silver steel etc. Used with copper, brass and bronze etc. Aluminium and its alloys etc.

S.No.

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S.No.

6.

Copper silver alloy Brass

FLAME It is produced by combustion of gases and due to oxidation,

different temperature are achieved. A flame can be adjusted for different temperature range. So these different flames have a distinct role in gas welding process. Middle

:.:J- ~c:n_e_......~Outer ~~~:,-----~~) I, , Zone

--Inner Zo~~"'--

./

J

-i-----T-

./

- ----

....

Fig. National flame

Middle Zone

Outer Zone

Fig. Carburising flame

Inner Zone

r-----J< -----~~~ vy-----i

Middle Zone Outer Zone

Fig. Oxidising flame Classification

of Flame

1.

Neutral flame: It is achieved when acetylene and oxygen are used in equal quantity. It consists of only two specified parts of flame, one is inner and outer envelop. It is most widely used in gas welding, it produces above 3200°C temperature.

2.

Carburising flame: This flame can be achieved by increasing acetylene gas quantity in flame it consists three distinct flames and acetylene feather can be easily detected in this flame, it is generally used in hard facing, nickel, and monel welding etc.

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Production Engineering

A-192 3.

Oxidising flame: It can be achieved

by increasing percentage of oxygen in natural flame. It is generally used with brass welding.

Common Difficulties in Flame Formation 1.

2.

3.

4.

Breaking offlame: Looks like burning gas with maintaining some distance from tip of welding torch. It can be rectified by reducing pressure of gas etc. Flickering offlame: In this fault, flame shows flickering. It happens due to increase in moisture contents in acetylene and it can be removed by removing moisture contents from acetylene gas. Popping: In this fault as usual sound like pit-pit comes from welding torch. It can be rectified by regulating the pressure of gas. Back fire: In this fault flame disappear suddenly with an abnormal sound, it happens due to following reasons. (a) Using welding torch less than its recommended pressure (b) When tip of welding torch get two close to job (c) Over heating of tip etc.

WELDING METHODS 1.

Leftward welding: In this process most ofheat is absorbed by filler material rod so it is preferred in welding thin upto 6 mm thick sheet.

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SOLDERING AND BRAZING In this method of joining metals, particularly in the shape of sheet thin wire form, or thin wire with thin sheet like electronic part with PCB. In this method a low melting filler material is used and no fusion takes place in work piece. These filler metal used in this process is known as solder. These are made in various composition depending upon the application and requirement of strength of joint. Some important compositions are as follows:

1.

Tin 67% : lead 33%

2.

Tin 50% : lead 50%

3.

Tin 30% : lead 67%

In some process of soldering alloy of copper and zinc to which silver is also added sometimes is known as hard solder. Germal silver, used as a hard solder for steel in an alloy of copper, zinc and nickel, in general the classification of solder in the above two catagories is according to their melting point. Soft solders usually melt at a temperature below 350°C and hard solder above 600°C the operation performed by using a soft solder is known as soft soldering and when using a hard solder is known as hard soldering. In this process work piece is cleaned properly and than a solder ion tool is used in heated condition, which melts the solder and then a suitable flux is applied to joining point. This flux works to prevent the formation of oxidation. Normally zinc chloride is used as soldering flux. The soldering tool is made up in two types one is total iron made which is used by heated in furnace and another is copper tiped placed between electrical elements and the tip is heated electrically.

Brazing: Brazing is almost similar to the joining process of +- Work

Piece

e\O

/~\o~

o'~

~\~ev

2.

Fig. Leftward welding Rightward welding: In this technique most of heat offlame is absorbed by base metal so it is preferred in welding thick sheet generally 6 mm to 25 mm thick. Rest of flat, vertical, horizontal and overhead welding methods are similar as described in electric arc welding method. Welding

~

torch

.-V':'ork piece

Fig. Rightward welding

soldering except hard solder material is used in place of soft solder and work piece is heated up to red hot in brazing but in soldering process work piece remains cools only soldering material is melted and spreaded over the work piece to make soldering. But in brazing process work piece is heated up to red hot condition and then after hard solder material is allowed to melt with flux over the joint to be weld. So that solder material get melted and filled the small gap between the joint of work piece to be brazed.

SPECIAL WELDING TECHNIQUIES (a)

Some of the special welding techniques are given as follows : TIG (Tungsten Inert Gas welding): It is also known as Gas Tungsten Arc welding (GTAw). This process utilizes a non - consumable tungsten electrode that provides a very intense current to the welding arc. This welding arc provides the required heat to melt the metal. This electric arc is struck between a non consumable electrode and the metal work piece. The tungsten and weld puddle arc given a protective enviroment and also cooled with the help of an inert gas (eg. argon). A welding rod is also ted at joints alongwith filler material and melted with the base metal.

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A-193

(iv) It requires minimum post weld cleaning Applications: It can be applied for deep groove welding of plates and castings. All commercial metals can be welded by this process. It also finds its application in automotive repair. MIG can also be incorporated into robotics. Some more applications are rebuilding equipment, overlay of wear resistant coating, welding pipes, reinforcement of the surface of a worn out rail road tracks. WELDING DEFECTS:

Copper backing bar

Gas Tungsten are (TIG) welding (GTAW) Advantages : 1. It produces, perfect, precise welds with suitable selection of proper welding rods and wires. 2. It has the capability to weld various metals. Most of the common metals or alloyslike mild steel, Stainless steel, titanium, aluminium and copper. 3. It uses a lesser amount of amperage as compared with other processes. 4. It is a clear welding process and does not leave any deposite over weld pead. 5. It has a high value of controlability 6. TIG welds are strong, ductile and resistant to corrosion. (b) MIG (Metal Inert Gas welding) : It is generally regarded as a high deposition rate welding process. In this process, consumable electrodes are used, which is generally in the form of coiled wire fed by a motor drive to argon shielded arc. Wire is consistently fed from a spool. A high value of current densities arc utilized. The diameter of wire is kept generallywithin the range of 0.80 mm to 2.30 mm. The consumable electrode in this process serves two purposes (i) its acts as a source for the arc column (ii) It also acts as the supply for the filler material. The shielding gas in this process, forms the arc plasma, stabilizes the arc on the metal being welded, shields the arc and molten weld pool, and allows smooth transfer of metal from the weld wire to molten weld pool.

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Metal Inert Gas (MIG) welding Adavantages : (i) It can work in all positions according to the need. (ii) it has a high deposition rate (iii) It requires less shilled labour

There are various types of welding defects which are given as follows:

Weld defects ~ (a) Undercut

(e) Lack of fusion

(b) Cracks

(f) Lack of penetration

(a) Slag inclusions: Various types of oxides, fluxes and electrode material are trapped in the welding zone. Dueto this trapping, inclusions are produced. These inclusion can be removed by grinding process or any other suitable mechanical process. (b) Under-cut: It can be defined as the notch which is formed due to the melting away the base/parent metal at the toe ofthe weld. It generally increases the stress and also reduces the fatigue strength of the material. It can be prevented by cleaning the metal before welding. It can be repaired with smaller electrode. (c) Porosity: Porosity is devloped when gas bubbles are entraped during cooling of weld pool. It is also devloped due to chemical reactions happened during welding. It can be controlling the welding speed. (d) Incomplete fusion: It is developed when the insufficient heat is provided and the travelling speed of weld torch or electrode is very fast. It is developed due to low amperage, steep electrode angle short arc gap, lach of pre-heat etc. It can be repaired by removing and rewelding. (e) Overlap: Overlaping in welding is caused due to improper welding technique, steep electrode angle and fast travel speed. It can be prevented by using a proper welding technique. (f) Underfill: It is developed when joint is not completely filled by with weld metal. It is caused by improper welding technique. It can prevented by applying proper welding technique for the weld type and position.

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A-194 (g) Spatter: It is developed due to high power arc, magnetic arc blow and damp electrodes. It can be prevented by reducing arc power, arc length and by using dry electrodes. (h) Incomplete penetration: It occurs due to low amperage, low preheat, tight root opening, short arc length and fast tra vel speed. (i) cracks: The development of cracks results in the premature failure of the parts when they are subjected to dynamic loading conditions. There arc many types of cracks, some ofthem are given as: (a) Longitudinal cracks (b) transverse cracks (c) crater cracks (d) under bead cracks (e) toe cracks These cracks occur when the joint is at elevated temperature or after the solidification of weld metal. These can be prevented by altering the design in joint, altering the parameters, procedures, preheating the component etc.

(b) Magnetic particle testing: It is used to defect surface discontinuities in materials like iron, cobalt, nickel and their alloys. A magnetic field is produced into the component to be tested. The magnitization of the component can be done directly or indirectly. It the defects are present in the component after magnetization, then the defects will create a leakage field. After magnetization, iron particles are applied to the surface of the component. The particles will be attracted and aggregate near leakage fields, thus giving an indication of defect. It is used in gas pipe welding. (c) Ultrasonic testing: (UT) In this testing, ultrasonic waves are propagated in the component to be tested. The very short ultrasonic wave of frequencies ranging from 0.1- 15 MHz and upto 50 MHz are used for the purpose of defection of internal flows or cracks. In ultrasonic testing, electrical pulses are converted into mechenical vibrations and the returned mechanical vibrations arc converted into electrical pulses. Adevice called transducer converts electrical energy into mechanical vibrations. In this testing,a propr (Connected to ultrosonic machine) is passed over the surface of the component to be tested. As the wave travels through the materical, from the defective location, the wave get reflected. The transducer picks up the signals and CRT (cathode Ray Tube) screen records the pulse - height pattern. The spacing between pulses and height of pulses are interpreted for the purpose of finding the correct location of cracks in the component. (d) Radiographic testing: (RT) In this testing, the hidden flows are defected by using the ability of short wave length electromagnetic radiation to penetrate various materials. Radiographic Testing method reveals the surface and sub-surface defects.

(NOn NON - DESTRUCTIVE TESTING (FORWELDING) It is defined as the process of testing the welded components for discontineities, cracks, inclusions, spatters penetrations, undercuts, porosity etc. In this type of test, the component is not destructed and after testing the component, it can be further used. Some important kinds ofNDT (non-destructive testing) are given as : (a) liquid penetrant test: It is also known as Dye penetrant test or penetrant test. It is utilized for the purpose of detecting the surface detects, porosities, cracks etc in welding components. In this test, the material (component) is first cleaned and coating is applied with a fluorscent dye solutions. The excess solution after some time (dwell time) is removed. The bleedout is easily detected in visible dyes while fluorescent dyes are view with an ultrovoilet lamp.

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S. No.

Name of Machine

MACHINING Machining may be defined as a process of removing extra material from the work piece to achieve a desired shape and dimensions by using any cutting tool. Metal may be removed either in chips form or in fine powder form like metal removed form is tabulated as under:

Narne of operation to be carried out

Removed Metal form (Either Chip / Powder)

1

Lathe

Turning, Drilling, Inner turning, Threading and Taper turning, etc.

Metal removed in form of chips

2

Drill Machine

Drilling, Tapping, etc.

Chips

3

Shaper

Shaping

Chips

4

Milling Machine

5

Planer

6

Milling and Boaring, etc. Chips Planning, Turning, etc.

Chips

Broaching Machine

Broaching

Chips

7

Grinding Machine

Grinding

Powder

8

Polishing Machine

Polishing

Very fme powder

9

Buffing Machine

Buffing and Polishing, etc.

Very fme powder

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The common features of machining process are listed below:1. The material of tool should be harder than the work piece to be machined. 2. The tool should be strong enough and hold rigidity on a proper support so that it can withstand the heavy pressure during machinery. 3. The shape of cutting tool should be designed in such a manner that cutting edge produce maximum pressure on work piece. 4. There is always a relative motion oftool with regard to the work or that of the work with regard to the tool or both in relation to each other. Basic Elements of Machine Tool All machine tools do one similar work that of removal of material from work piece and all these machine tools have some common elements as given below:1. Frame Structure. 2. Slides and Guideways. 3. Spindles and Spindle bearing, etc. 4. Machine Tool Drive.

MACHINE TOOL CONTROLS On observing machine tools, we find that it contains many levers, hand wheels, stop switches, drivers etc. All of which are known as the control of machine tool which performs a specific function in every machine tool. All their controls specified are of the following types: 1. Mannual control. 2. Semi-automatic control. 3. Automatic control. 4. Numerical control.

Importantfactors requiredin today's scenarioasfollowing: (a) (b) (c)

Quick metal removal. High class surface finish with economic tooling cost. Minimum idle time of machining at lower power consumption. Cutting Action For cutting action, a relative motion between the tool and work piece is necessary. The relation motion between tool and work piece can be maintained either by keeping work piece stationary and moving to tool or by keeping tool stationary and moving work piece. The cutting action can be classified into following types:1. Orthogonal cutting and 2. Oblique cutting.

Work Piece ....---------.._

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)-

--, Movement

1-

-..- _.-.-.-.-.-.-.-- _o_._._._._.-c-.- _.-.-

---·-0-·-

\

--.

Work Piece Movement

-c-·

~ Cutting Tool Movement (a) Orthogonal Cutting

(b) Oblique Cutting

Turning on Lathe in Cutting Process

As shown in above figure, two types of tool shapes are used in orthogonal cutting process. We see that the cutting edge is rectangular and the turning face of work piece is made flat. This type of cutting is known as two-dimensional cutting. while in oblique cutting process, the tool's cutting edge is made like triangular / inclined. This process is known as three-dimensional cutting. CUTTING TOOLS Cutting tools may be defined as the tools required for cutting. The cutting tools used in power operated machines are commonly harder and having more red hot hardness than manually operated tools. These tools are designed to acquire more useful cutting using minimum power consumption. Properties of Good Cutting Tool Material 1. It should be tough enough and having good strength. 2. It should have good resistance against shock, wear, corrosion, cracking and creep, etc. 3. It should have good response for hardening, tempering and annealing, etc. 4. It should be economical and easily available. 5. It should have capability to retain these physical and mechanical properties at elevated temperature during

6.

machining operations. This property may be known as red hot hardness. It should be easily fabricated into tool shape.

Classification of Cutting Tools Cutting tools may be classified as follows on the basis of having number of cutting point / edges:-

1. Single Point Cutting Tools: These cutting tools contain only one cutting edge/point. For example, turning, parting and grooving tools for lathe machine, shaper tools and planer tools, etc.

2. Multi Point Cutting Tools: These cutting tools contain more than one cutting edge / points. For example, drill bit, broach and milling cutter, etc. On the basis of motion cutting may be broadly classified as follows:-

1. Linear or Reciprocating Motion Tools: For example, shaper tools, lathe tools and planer tools, etc.

2. RotaryMotion Tools:For example, drill bit, milling cutter, grinder wheels and honning tool, etc. Common Cutting ToolMaterials Depending upon their physical, chemical and mechanical properties, etc. some metal and alloys in common use are

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bronze and cast iron, etc. It can be employed for two times more speed than common High Speed Steel tools. 5. Cemented Carbide:These are generally used in sintered tips form made up of powder metrology process. These are directly manufactured into desired shape and size and mounted on suitable holders (either by brazing or by clamping, etc.). These holders are normally made by medium carbon steel. It gives better results than satellite and high speed steel. It can be used with four times more cutting speed than high speed steel tools and can retain its hardness up to 1200°C temperature. 6. Ceramics or Cemented Oxides: These are made by applying sintering process with aluminium oxides and boron nitride in powder form. It is also made up in readymade tips form. Which is used after mounted on a suitable tool holder (either by brazing or by fastening). These can easily retain their hardness up to 1200°C temperature and can work 2-3 times faster than tungsten carbide tips. Sometimes these ceramics give more satisfactory results in finishing, etc. than tungsten carbide, etc.

mentioned below:1. High Carbon Steel: High carbon steel shows different hardness with different percentage of carbon contents. It shows BHN hardness from 400-750 with different percentage of carbon. It contains carbon percentage 0.6%1.5%normally. But high carbon steel start losing its hardness above 200°C. So, its application is limited in slow moving / operating tools, hand tools and wood working machine tools, etc. For example, hammers, cold chisels, files, anvil, saws, screw drivers, center punch and razors, etc. 2. Diamond: Diamond is the hardest and brittle material but its use is limited due to its high cost. It consists great wear resistance but low shock resistance. So, it is used in slow speed cutting of hard materials like glass cutting tool, grinder wheel, dressing tool and other cutting tools, etc. 3. High SpeedSteel: It is most commonly known cutting tool material. It contains 18W, 4Cr, 1% V. In some tools, additional cobalt with 2%-15% is also added to increase its hardness up to 600°C. It contains sound ability to bear impact loading and perform intermittent cutting. 4. Stellite:It contains 40%-50% cobalt, 15%-35%chromium + 12.25%vanadium + 1%-4% carbon normally and it consists good shock resistance, wear resistance and hardness. Normally, it retains its hardness up to 920°C temperature and it is used for comparatively harder materials like hard

Cutting ToolGeometry The different angles provided in cutting tool also plays a significant role in machining process along with the material of tools. Here we give a sketch of single point cutting tool designed for different turning processes.

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~nd

Nose Radius ~ Side cutting ~ angle

cutting angle

Face _....3....._

Shank ---1

___;;""-L-

Top View

Side Rake angle

-:r

Top rake angle

~ Side Clearance angle

End relif angle

Front View Front clearing angle Side View Cutting Tool Angles

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Rake angle: The angle between face of tool and a plane parallel to its base. If this inclination is towards the shank, it is known as back rake angle or top rake angle and if measured along with side is known as side rake angle. These angles reduce the strength of tool's cutting edge. But along with reducing the strength, these angles also through away the chip from the cutting edge, which causes reduction of pressure on cutting edge of tool. Negative rake: When these angles are made in reverse direction to the above are known as negative rake angle. Obviously these angles strengthen the tools but reduce the keenness of cutting edge but these angles are used for extra hard surfaces and hardened steel parts, etc. and used generally carbide tips, etc. Lip angle: Lip angle may be defined as the angle between face and the flank of tool. As the lip angle increases, cutting edge will go stronger. It would be observed that since the clearance angle kept constant, this angle varies inverse to the rake angle. So, when the strong cutting edge is required like for harder material, rake angle is reduced and lip angle increased. Clearance angle: As the name resembles, this angle is made in tool to provide clearance between job and cutting edge of tool. If the angle is provided in side of cutting edge, it is known as side clearance angle and if this angle is given at front of tool it is known as front clearance angle. Relief angle: This angle formed between the flank of tool and a perpendicular line drawn from the cutting point to the base of the tool. Cutting angle: The total cutting angle of the tool is the angle formed between the tool face and a line through the point which is a tangent to the machined surface of the work at that point. Obviously, its correct value will depend upon the position of tool in which it is held in relation to the axis of the job.

The grains of metal in front of cutting edge of tool start elongation the line AB and continue to do so until they are completely deformed along CD. The region between ABCD is known as shear zone. Types of Chips Chips may be classified as given under:1. Discontinuous or Segmental chip. 2. Continuous chip. 3. Continuous chip with built-up edge. 1. Discontinuous Chip

Work Piece

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Discontinuous Chip Formation

CHIP FORMA nON Chip may be defined as a thin strip of metal removed from the work piece as the tool progressed into work piece. Like in lathe machine, where job is kept moving and a study tool advanced into it, the metal's thin strip removed from work piece due to its plastic deformation but as the length of chip increase a stress compress the chip and after a limit, this chip gets fractured and removed from work piece. The shearing of metal chip formation does not, however, occurs sharply along a straight line.

2.

These type of chips formed in small pieces as shown in figure. This type of chips are produced during machining of brittle material like cast iron and bronze, etc. In machining of brittle materials, shear plane gradually reduce until the value of compressive stress acting on the shear plane becomes too low to prevent rupture along with as the tool advance formed in work piece. At this stage, any further advancement of tool results in the fracture of metal ahead of it, that's why it results in production of segmented chips. In this type of chip formation, excessive load has to withstand by tool which results in poor surface finish of work piece. Continuous Chip Formation

Continuous Chip Work Piece Work Piece B

Continuous Chip Formation Chip Formation

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As shown in figure, the chip formed in a continuous ribbon form and breaks after a certain length. It happens when ductile material is machined. In this chip formation, minimum load forced on the tool's cutting edge. So, that a better finish is achieved and minimum wear and tear occur in tool edge.

Built-up Edge

3. Continuous Chip with built-up Edge This type of chip is generally formed during machining ductile material and a high friction exists at the chip tool interface. Due to high friction, a high temperature generates at melting point of chip and cutting edge of tool. Due to generation of high temperature, chip formed at high temperature. As the cutting proceeds, the chip flows over this edge and up along the face of tool. Periodically, a small amount of the built-up edge separates and leaves with the chip or embedded in the turned surface. Due to this, chip formed is not smooth. When the tool is operating with a built-up edge a short distance, back from the cutting edge, the wear takes the form of cratering of tool face caused by the extreme abrasion of chip. This type of chip formation may be reduced by using proper coolant.

Work Piece

Showing Built-up edge Due to built-up edge chip formation, surface finish achieved is rough and chance of production in crater on the surface of work piece. CUITING FORCE Cutting force is a very important factor in tool designing like we consider a lathe turning tool, it is a single point cutting tool. The force acting on the tool is the vector sum of three component cutting force mutually at right angle. The resultant cutting force is denoted by (R).

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Work Piece

-: _._._._._._._._._._

..

I

Tool where, Fn = force normal to machine surface Ff = force acting parallel to the axis of work piece F, = tangential force along work piece Out of these three components, force Ft is the largest and Fn the smallest. In case of orthogonal cutting, only two component force come into play since the value of Fn is zero in that case. In single point cutting turning process, the component Fn- Ff and F, can be easily determined with the help of suitable force dynometer. Thus resultant R can then be calculated from the following relationship:R

= '\jIF n2 + F f2 + F t2

_._._.-._._._._._._._

..

~_.

R

and in case of orthogonal cutting process, as stated that Fn is almost zero. So, value of R=

IF2f

'\j

+ F2 t

According to A.S.M.E. cutting manual, tangential cutting force will be as given below:Pt

=K --p K a TC Ld

where, P, = tangential cutting force ~ = constant depending upon the material Ka = constant depending upon the true rake angle of tool T = average chip thickness L = length of cutting edge in active engagement

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c and d are exponents depending upon the material being out. The variable T and L are introduced in order to embrace the nose angle. Nose radius feed per revolution and depth of cut.

(Ft cos $ + Fc sin $) sin $ bx t

Stress in Metal Cutting

Shear Strain

As we know that when tool applied a force on work piece and resulting chip formation, the chip production occurs due to stress and strain development. To compute the stress and strain developed on chip, we consider a single point cutting tool as given below:-

It has been defined as the deformation per unit length. In metal

cutting, the diagram for measuring shear strain is taken from a shear plane, we have AB

Shear Strain, y = CD =

AD+DB CD

= tan ($ - a) + cos $

= -----

cosu

sin $ cos ($ -

a) .

Work Done in Cutting

A Strain in Cutting The values are calculated for the conditions at the shear plane where the two normal force Fs and Ns are existing. Let, Fs = force across the shear plane As = area of shear plane $ = shear angle b = width of chip t = thickness of chip Fc = cutting force Ft = tangential force Fn = force normal to shear plane

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(Z) =

F AS s

(kg F/mm2)

and (as> (mean normal stress) =

F

__!!_

As

2

(kg F/mm )

where, Fs = Fc cos $ - Ft sin $ Fn = Ft cos $ + Fc sin $ Ao A = -.(where Ao = area of chip before removed) s sin o So, mean shear stress (t) =

Fs

A=

Fc cos$ - Ft sin $ bxt

s

The work done in cutting process may be calculated by adding work done in shearing and work done in overcoming friction arise. If W = total work done Ws = work done in shearing Wf = work done in overcome friction Wm = (work done in cutting + work spent in feeding) Ao = (cross-sectional area of chip before removal) Now, assuming that there is no work loss, then total work done must be equal to the work supplied, then total work done, we have W = Ws + Wf ... (1) Now, we assume that total work supplied is used in cutting but partly used in feeding the tool, then we have Wm = work consumed in cutting + work spent in feeding Wm = Fc x x Ft x feed velocity

v,

Now, assuming that the Ft is very minor in comparison of Fc. So, neglecting the feeding work, we have Wm = Fc x Vc ... (2) Assuming that there is no work loss, we have Wm = W ... (3) So, putting value in equation in (3), we have FcxVc=Ws+Wf ... (4) as we know, Ws = Fs x Vs (shear force x shear velocity) Wf = F x Vf (friction force x velocity of chip flow) then, FcxVc=FsxVs+FxVf ... (5) if the forces are taken in kg and velocity in metre per minute, the work done will be in kgf m/min. Then, W=

Total work done in cutting per unit time ... Volume of the metal removed m unit time

sine (Fc cos $ - Ft sin $) sin $ bxt and mean normal stress,

So, we have

Iw

=

Fe Ao

I

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Horse Power Calculation . . Work done in cutting / minute H.P. required for cuttmg = 4500 Power =

Fe x Ve H.P. 4500

... (1)

Fe x Vc kw 4500 x 1.36

... (2)

Source of Heat in Metal Cutting

4.

Engineering

face and therefore the chip does not get hardened. The chip separates from work piece at the shear plane. Accounting all above Lee and Shaffer's had developed a slip-line field for stress zone, in which no deformation would occur even if it is stressed to its field point. From all these, both of them had derived the following relationship: 1t

= - + a - 1 = 45° + a - 1 4 or we can say,

1<1>+1-0,=45°1

... (1)

BASIC PRINCIPLES OF MACHINING

3 WorkPiece

Area (1) = Primary deformation area

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Area (3) = Tool work piece interface Assuming that all work done is converted into heat, then the heat generated we have (Q), where Wm = Fe x then we have,

v,

(i) Drilling: Drilling is the process or operation used for manufacturing circular holes. These holes are produced by a specific type of end cutting rotating tool which is generally termed as drill. The machine used for the purpose of drilling is known as drill machine. The operationsperformedby drill machine in addition to producing holes are tapping, reaming, boring, counter boring, spot facing etc. • Working principle: A large amount of force is exerted by the rotating edge of the drill on the workpiece and then the hole is produced. During driling operation, the metal is removed by shearing and extresion. ~-001

Tool (Drill) feed motion

IQ= Fe x Vel· EARNST-MERCHANT THEORY It is based on the principle of minimum energy consumption. It states that during cutting the metal, shear should occur in the direction in which the energy requirement for shearing is minimum. The other assumption made by them includes= 1. The behaviour of metal being machined is like that of an ideal plastic. 2. At the shear plane the shear stress is maximum is constant and independent of shear angle (<1». They deduced the following relationship:

I~=%-~+II LEE AND SHAFFER'S THEORY It is a theory about analysation the process of orthogonal metal cutting by applying theory of plasticity for an ideal rigid plastic material. The principal assumptions made for this include: 1. The work piece material ahead of the cutting tool behaves like an ideal plastic material. 2. The deformation of metal occurs on a single shear plane. 3. There is a stress field within the produced chip which transmits the cutting force from the shear plane to the tool

LIP Working Principle (Drilling) • Typesof drilling machine: (a) Based on construction -e Portable ~ bench drilling machine ~Radial -e upright ~ Multi-spindle ~Automatic ~ Turret -s-Deep hole (b) Based on feed • Hand and power driven portable drilling machine • Geometryof drill

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Lip angle + Lip relief angle + Helix angle = 90°

nDN Cutting speed (Vc ) = 1000 rpm

F. Ft = sm

"2

(ii) Milling: Milling is a maching process in which rotary cutters arc utilized for the purpose of removing material from the work piece. In this process, the workppiece is feeded in a direction at an angle with the fool axis.

2<1> = po int angle F = feed = mmlrev. Flute

(1)

Helix angle

Drill axis

L_:~:~m~di~;!: ~

-r-,,__..~~

;r-.....

/

PRINCIPLE OF MILLING Tip

Rotational direction

Margin Cutting lip

~I 1'-

Milling cutter

Chisel edge

-,

/ •

'.

VI V'I

Q)

c

::~.'

(2)

~~ ~ I

\

..0

Chisel edge angle

:;

Point)' angle \-

Work piece

Q)

Work Piece

\ <,

(Drill Geometry) Rahe angle: - It is the angle formed between the axis of drill and leading edge ofland. Point angle :- It is also termed as cutting angle. It is the angle formed between the lips which are opposite in nature of a drill calculated in a plane containing the axis of drill and lips. Feed angle :- The angle produced by cutting edge which tries to strike the cutting edge for the purpose of breaking

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it.

C BC F/2 F tan=-=-=AB nr 2m Clearance angle: this angle is formed between the flank and a plane which is perpendicular to the axis ofthe drilL Clearance angle for ductile material ranges from 8 - 12° and that for brittle material ranges from 6 - 9° . Machining time and cutting speed:Machinetime

down milling

Face

L (Tm) = N x F

where, L = length of drill's axial travel (mm) N = speed of drill (rpm) F = Feed/rev. (mm) Tm = machining time (min.) L=t+A t = thickness of work-piece A = drill approach = 0.30 D = drill diameter

~ Working principle: The working principle of milling is on the based of rotating motion. During milling operation, a milling cutter spins about an axis and the workpiece is feeded. While the feeding of workpiece, cutter blades remove the material in each pass. Various operations can be performed such as face milling. End milling, keyway cutting, dovetail cutting, T-slot cutting, circular slat cutting, up-milling and down milling peripheral milling, slab, slotting, side & straddle, milling etc. • Types of milling machines: => Horizontal milling machine (a) Horizontal spindle: It is utilized for peripheral milling operations. => vertical milling machine (a) verticle spindle: It is used for face milling operations => column and knee milling machine => Turret type milling machine => Universal type milling machine => Bed type milling machine => Planar type milling machine => CNC milling machines Cutting parameters in milling:

L

Rotational speed (N R )

=

Cutting speed (V) nD

where, D = diameter oftool NR = Rotational speed (rev.! min) V = cutting speed (m/min.) 2. Rotational speed in milling can also be related with the desired cutting speed at work piece surface. NR=--

Fr

nt x fc

where, F r = feed rate (mm/min.) nt = Number of teeth on cutter fc = chip load (mmltooth)

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Production

A-202

3.

4.

Material Removal Rate (M.R.R) = we x de x F r where, de = depth of cut (mm) we = width ofcut(mm) In case of slab milling: .. . ( tm ) =-I+A Machining time Fr where, tm= machining time (min.) A= Approach distance

(2d)

D

As we know that, sin2
. sin

=)d(D-d)

5.

D-2d cos
Engineering

?2d) o, = ~l-ll-D)

In case of face milling: t

m

1+2A F

sin o

=--

1_[D2+4d2-2.D.2d]

==

D2

c

r

where, A = Apporach distance

On solving, we get,

A = ~w c (D - w c) (for partial face milling)

sin o, A=D

(For conventional face milling)

2

MECHANICS OF MILLING OPERATIONS

= ~~( 1- ~)

=> Mean chip thickess : Mean chip thickness (S~ = S

= Smax

2

m

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sm = Let, fr is feed rate / (Feed / rev.) F, = Feed / tooth Ft --~ Nt where, N, = Number of teeth S= Ftsin

OR ilORP, coso, = OP

D-2d -2-

D-2d

2

D

2

D

=---=--x-

2

O+Smax 2

F ~dD(1- Dd) t

Volume of removed metal, (V R) : ~ x Vc where, Am = mean chip area Ve = cutting velocity => Some important milling operations in brief: (a) Slab milling: During operation, the width of cutter extends beyond the work piece on both of the sides. (b) Slot milling: In this type, the width of the work piece is more than the width of the cutter, by creating a slot. (c) Side milling: In this operation, the milling cutters provide machining along the side ofthe work piece. (d) Straddle milling: In this operation, the milling cutters provide machining along both sides of the work piece. (e) Up - milling: It is also known as conventional milling. In this type, the wheel rotates in the opposite direction of feeding. In starting, the chips produced by cutter tooth are very thin and then increases its thickness. Tool life is short chip length is relatively longer. (f) Down - milling: It is also known as down milling. In this type, the wheel rotates parallel to the feeding. The chips are thick in the starting and leaves out thin. The length of the chip is relatively shorter. Tool life is relatively longer. (g) Face milling: During operation, the axis ofthe cutter makes a 90° angle with developed surface. The surface is generated is due to combined result of operations of cutter teeth located on both periphery and the cutter face.

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A-203

=> Conventional face milling: In this type, the diameter of the tool is kept larger than the width of the work piece. => Partial face milling: In this type, the milling cutter is in overhanging position from one side of the work-piece. => End milling: In this type, the diameter of milling cutter is less than the width of the work piece. => Profile milling: In this type, outside periphery of the flat part of work-piece is cut.

SHAPING: (WORKING PRINCIPLE) It is described as a process in which metal is removed from metal work piece surface in horizontal, vertical and angular planes. In these operations, a single point cutting tool is utilized, which is held on the ram that provides a reciprocating motion to the tool. A single point cutting tool is clamped in the tool post which is mounted on the machine's ram. The motion ofthe ram is the reciprocating TO and FRO, which resulting the tool cuts the material in the forward stroke. There is no cutting during return or bachward stroke. Shaping operations are generally used for producing slots, grooves and keyways. It also produces contour of can cave or conven or a combination of these.

k = Return stroke time Cutting stroke time

As we know that,

Return stroke = K

x

cutting stroke time

ke 1000V Time taken to complete one double stroke, (T

=>kxT

c

=--

e

=--+-1000V T

_

2J

ke 1000V

e + ke

_ e(k + I)

2s - 1000V - 1000V

1000V

1

Now, N = _1_=

e(k + 1) 1000V

T2s

e(k

+ I)

Machining time: (T.J As we know that, Time taken to complete one double stroke (T 2s)' T

_ e(k+ 1) 1000V

2s -

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Let, b = breadth of work piece, (mm), f= feed rate (mm/double stroke) Now, Total number of double strokes needed to complete

b

(vertical surface)

(Horizontal surface)

(Inclined surface)

(Working principle) Classification of shaping machine: (i) Horizontal type (ii) Vertical type (iii) crank type (iv) Hydraulic type (v) Universal type Mechanisms used in shaping machines: (i) crank and slotled lever mechanism (ii) Hydraulic shaper mechanism (iii) Whitworth quick return mechanism Cutting speed: In is defined as the ratio oflength of cutting stroke to the time required by the cutting stroke. Let, V = cutting speed, m/min. N = Number of douple strokes ofthe ram/min. K = ratio of return time to cutting time 1= length of cutting stroke Time required by cutting stroke (T c) cutting stroke length (m) cutting speed (m / min)

T = c

e

the work =f"

Hence, Time taken to complete the cut b)

eb(k+l)

=> T2s x ( f" = 1000Vf (iv) Lathe machine (working principle) A Lathe is defmed as a machine tool on which work piece is rotated on its own axis for the purpose ofperforming various operations like cutting, knurling, turning, facing etc. In a lathe machine, the work piece is helded between the chucks which revolve. The tool post consists of a cutting tool which is fed against the work piece for required depth and also in required direction. The material from the work - piece is removed in the form of chips and the required shape is obtained. Some parts of a lathe: (a) Bed (b) Legs (c) Head stock (d) Tail stock (e) Gear- box (t) carriage

Workpiece \ ( ( \

Tool

I

/'"

V x 1000 Working Principle of Lathe machine

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A-204

Some operations performed on Lathe in brief: => Turning: In this operation, straight, curved and conical workpieces are produced. => Facing: In this operation, the flat surface is developed at the end of the work piece. => Boring: In this operation, a hole or a cylindrical cavity is entarged which are manufactured by another process => Threading: In this operations, threads are produced internally or externally => Knurling: In this operation, a regurlarly shaped roughness is developed on cylindrical surfaces. Machining properties / cutting parameters: => Feed: It is defined as the distance through which the cutting tool advances between two consecutive cuts. => Depth of cut : It is defined as the advancement of cutting tool into the job in a transverse direction => Cutting speed : It is defined as the speed through which the spindle rotates. (a)

. ( ) 7tDN Cuttmg speed V =--

1000

where, D = diameter of workpiece (rum) N = rotational speed (rpm) (b)

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Machining time: (T)

= _L_

FxN where, L = length of work-piece F = feed rate (mmlrev.) N = rotational speed (rpm) (D-d)

(c)

Depth of cut : (tc)

= -2-

where, d = diameter of work piece after machining D = diameter of work - piece before machining (d) Metal Removal Rate (MRP) = 7tD tc FN Types of Lathes : (a) Centre or Engine lathe (b) Bench lathe (c) Speed lathe (d) Tool room lathe (e) Automatic lathe (t) Turret lathe (g) Capstan lathe (h) Computer - controlled lathe Grinding: Grinding is a machining purpose used for the purpose of removal of the metal with the help of applying abrasives which are bonded to form a rotating wheel. It is generally utilized for good surface finishing, grinding of craks and burns etc. It can be utilized for flat, conical and cylindrical surfaces.

Engineering

Types of grinding machine/operations: The following are the grending machines: (a) Surface grinding (b) cylindrical or External grinding or centre - type grinding (c) Internal cylindrical grinding (d) centerless grinding (e) Form and profile grinding (t) Plange cut grinding => In surface grinding. It utilizes a rotating abrasive wheel for the purpose of removing material and thus resulting in a flat surface => In cylindrical grinding, It is utilized for the purpose of grinding cylindrical surfaces and work-piece shoulders. => In internal cylindrical grinding, It is used for the purpose of grinding the internal diameter ofthe work piece and also tapered holes => In form and profile grinding, the grinding wheel does not transverse the work-piece and having the exact shape as of the finished product. => In plunge - cut grinding, It is used to grind the work pieces having projections, multiple diameters or other irregular shapes. Various types of grinding wheel:

n

Wheel thickness

Grinding faces

IE

wheel diameter

I

I

I

)!

.--r---------r---,

Type 1 (straight)

'" L; r--? 1= Type 2 Rrecessed t (one side straight)

Diameter of Recessed

~

c

I I ~

Type 3 Recessed (Both sides straight) Grinding face

) Type 4 (Tapered face straight wheel)

Grinding face

Thickness

J,

t

LI...........__ _-----'--'II

O'----------r----r----O

Type (cylindrical or wheel ring)

~II~

Type 6 (straight cup wheel)

Grinding face

Type 7 (Flaring cup wheel)

Type 8 : Saucer wheel

Type 9 : Dish wheel WorkTable Grinding Principle

=> Type 1, Type 2 and Type 3 are utilized for cylindrical, internal centreless and surface grinding.

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=> Type 4 is usually used for thread grinding of a gear teeth.

=> Type 5 is utilized for producing flat surfaces. => Type 6 is utilized for grinding flat surfaces by applying grinding wheel. => Type 7 is utilized for the purpose of grinding tools. => Type 8 is utilized for the purpose of sharpening of circular or band saw. => Type 9 is utilized for the purpose of grinding various kinds of tools in the tool room. Characteristics of grinding wheel The performance of a grinding wheel depends on the following factors: (a) Abrasives: Abrasives are used due to its two main mechanical properties i.e. hardness and toughness. It also has a sharp edges. Some ofthe properties of abrasives are indentation, fracture r resistance, wear resistance etc. There are generally two types of abrasives which are as : => Natural abrasives: These are sand stone, corundum diamond and gasnet etc. => Synthetic abrasives : These are manufactured and have well defined properties of roughness and hardness. Eg : silicon carbide and aluminium oxide. (b) Bond: It has the property of adhesiveness. Due to this property, the abrasive grains are cemented together for the purpose offormation of grinding wheel. As per the demand, it serves the imparting of hardness or softness properties to the grinding wheel. Some bonds are given as follows: => vitrified bond => silicate bond c> shellac bond => Rubber bond => Oxy chloride bond c> Resinoid bond (c) Grit: It is also termed as grain size. After passing the materials through screens, the size of the grain grit is determined with the number of meshes / linear inch. It influences the stock removal rate and surface finish. Grain size selection depends upon the type of grinding, type of material; material removal rates (MRR) and required surface finish. (d) Wheel grade: The wheel grade is measured by the strength ofthe bonding material. These are generally two kinds of wheels used which are hard wheel (Strong bond and abrasive grains can with stand with larger forces) and soft wheels (ifthe material to be grinded is hard then the abrasives grains are wear out and resulting losing of sharp edges for cutting is lost, this process is known as glazing.) Selection of grinding wheel: The grinding wheels are selected depending upon the following given factors. (a) Material's properties (b) Required quality of surface finish (c) Accuracy in dimensions (d) Method ofgriding i.e. either dry or wet (e) Rigidity, size and machine type (t) Speed and feed of wheel

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A-205 (g) Type of grinding to be done Grinding wheel parameters : => Depth of cut : Itis defined as a thickness of the material removed through grinding wheel in a single transverse stroke. Depth of cut(Tc) d, d2

= (dl

~d2)

= diameter ofthe work-piece before grinding = diameter of the work-piece after grinding

=> Feed: Feed is described as the motion of the workpiece longitdinally per revolution in cyclindrical grinding. Feed (f) = k..> A (where A is constant) where, A = face width of wheel in mm = 0.4 to 0.6 (finish grinding) = 0.6 to 0.9 (Rough grinding) fxN

=> Work travel : work travel = --m/ .

1000

. mm.

where, N = Rotational speed (m/min).

MANUFACTURING PROCESSES IN BRIEF Manufacturing process is defined as the conversion raw material into finished or find product. Classification of manufacturing processes: (i) Primary shaping processes: ~ casting ~ Powder metallurgy ~ Plastic technology (ii) Forming processes: ~ Forging ~ Extresion ~ Rolling ~ Sheet metal working ~ Rotary swaging ~ Explosive forming ~ Electromagnetic forming (iii) Machining Processes : ~ Turning

~

of

Drilling

~ Milling ~ Grinding ~ Shaping and Planning ~ Non - Traditional machining such as : ultra sonic machining, Electro-chemical maching etc. (iv) Joining Processes ~ Pressure welding ~ Resistance welding ~ Diffusion welding ~ Soldering ~ Brozing (v) Surface finishing processes ~ Honing ~ Lapping ~ Electro-plating ~ Plastic coating ~ Metallic coating

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A-206

~ ~

Sanding Tumbling

COMMONLY USED MACHINES AND TOOLS: (i) Lathe machine: ~ Cylindrical turning : It involves the reduction of diameter of work-piecebyremoving material along the axis of work - piece from the cylindrical job's surface. ~ Taper turning: In this type, material is removed at an angle to the work-piece axis. And thus diameter of the workpiece is increased or decreased. ~ Eccentric turning: In this type, the axis of work-piece does not coincide with the main axis. ~ Knurling : In this type, a diamond shaped impression is embossed on the work piece. ~ Facing: In this type, flat surface is developed by machining the ends ofthe work-piece. ~ Parting - off: In this type, the work piece is cut after obtaining required shape and size. ~ Chamfering: In this type, the end ofthe work - piece is bevelled. (ii) Milling Machine: ~ Plain milling: In this type, a flat, horizontal surface is made paraller to the axis of rotation of plain milling cutter ~ Side milling : In this type, a flat vertical surface is developed on the side of work-piece with the help ofa side milling cutter. ~ Facemilling: In this type,facemilling cutter is utilized with rotating motion about a perpendicular axis to the work -piece. ~ End milling: In this type, a flat surface is developed. The developed flat surface may be horizontal, vertical or at an angle with the table. ~ Thread milling: In this type, threads are produced by utilizing a single or multiple thread milling cutter. ~ Form milling: In this type, irregular contours are generated with the help ofa form cutter. (iii) Drilling machine : ~ Drilling: In this type, a cylindrical hole is developed

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with a drill which is a cutting tool having cutting edges ~ Boring: In this type, the hole (pre-existing is enlarged by using drilling operation. ~ Reaming: In this type, a preexisting hole produced by drilling or boring is finished and sized. ~ Counter-boring: In this type, the pre existing drilled hole is enlarged cylindrically at the end of the hole. (iv) Shaper machine: ~ Horizontal surfaces : In this type, a flat surface is generated on a workpiece by holding it in a vise. ~ Vertical surfaces: In this type, the end of a workpiece, squaring up a component are produced. ~ Angular surfaces : In this type, an angular cut at an angle other than 90° with the horizontal or vertical plane. (v) Planer machine: ~ Horizontal surfaces: In this type, the tool is feeded crosswise for the purpose of completing the cut, while the work piece is provided a reciprocating motion along with the table. ~ Vertical surfaces: In this type, the tool is feeded down ward for the purpose of completion of the cut, while the work piece is provided reciprocating motion along with the table. ~ Angular surfaces : In this type, the tool is feeded at an angle for the purpose of completion ofthe cut, while the work - piece is provided reciprocating motion along with the table. (vi) Grindingmachine: ~ Cylindrical surfaces: In this type, cylindrical surfaces of a work piece are fmishedby utilizing cylindricalgrinders. ~ Tapered surfaces : In this type, tapered surfaces of a work piece are finished by using cylindrical grinders ~ Horizontal surfaces : In this type, the horizontal surfaces of work pieces are finished by utilizing the surface grinders. ~ Threaded surfaces: In this type, threads are produced by utilizing a thread grinding machine along with single or multiple rib wheels.

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EXERCISE 1.

For TIG welding, which ofthe following gases are used? (a) Hydrogen and carbon dioxide (b) Argon and helium (c) Argon and Neon (d) Hydrogen and oxygen 2. The pre-heating of parts to be welded and slow cooling of the welded structure will lead to reduction in : (a) residual stresses and incomplete penetration (b) cracking and incomplete fusion (c) cracking and residual stress (d) cracking and underfill Which one ofthe following is a solid state joining process? 3. (a) Gas-Tungsten arc welding (b) Resistance spot welding (c) Friction welding (d) Submerged arc welding 4. Arc stability is better with: (a) ACwelding (b) DC welding (c) Both (a) and (b) (d) None of these In which type ofwelding, molten metal is poured forjoining 5. the metals? (a) Arc welding (b) Thermit welding (c) MIG (d) llG The gases used in tungsten inert gas welding are: 6. (a) argon and helium (b) neon and helium (c) neon and argon (d) ozone and neon 7. Amount of current required in electric resistance welding is regulated by changing the: (a) Input supply (b) Primary turns of the trasnformers (c) Seondary turns of the transformers (d) All ofthese The material used for coating the electrode: 8. (a) Protective layer (b) Blinder (c) De- oxidiser (d) Flux 9. The electric resistance welding operates with: (a) Low current and high voltage (b) High current and low voltage (c) Low current and Low voltage (d) High current and High voltage 10. Fluxes are used in welding in order to protect the molten metal and the surfaces to be joined from: (a) oxidation (b) carburizing (c) unequal temperature distribution (d) distortion and warping 11. Twostainless steelfoilsof 0.1 mm thickness are to bejoined. Which of the following processes would be best suited? (a) Gas welding (b) TIGwelding (c) MIG welding (d) Plasma arc welding

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12. In oxy-acetylene welding: (a) Pressure is applied (b) Filler metal is applied (c) Both Pressure and filler metal arc applied (d) Neither pressure, nor filler metal is applied 13. What should be the size of weld in case of buss welded joint? (a) Twice the throat of weld (b) Half ofthe throat (c) Equal to the throat of weld (d) None of these. 14. Welding process in which two pieces to be joined are over - llaped and placed between two electrodes in known as : (a) percussion welding (b) spot welding (c) seam welding (d) projection welding 15. The abbriviation ERW in ERW pipes stands for: (a) electrically resistance welded (b) elastic reinforced with wire (c) extra reinforcement welded (d) electrically reinforced and welded 16. T - joint weld is used: (a) where longitudinal shear is present (b) where sever loading is encountered and the upper surface of both piece must be the same plane (c) To joint two pieces of metal in the same manner as rivet joint metals (d) Tojoin two pieces perpendicularly 17. Half corner weld is used: (a) where longitudinal shear is present (b) where sever loading is encountered (c) tojoin two pieces of metal in the same manner as rivet joint metals (d) none of these 18. The range of optimum pressure applied in electric resistance welding is given by : (a) 0-5MPa (b) 5-lOMPa (c) 10-25 MPa (d) 25 - 50 MPa 19. Electronic components are often joined by : (a) soldering (b) brazing (c) welding (d) adhesive 20. The method ofjoining two similar or dissimilarmetals using a special fussible alloy is : (a) Soldering (b) brazing (c) Arc welding (d) All of these 21. The taper provided on pattern for its easy and clean withdrawl from the mould is known as : (a) Taper allowance (b) Distortion allowance (c) Pattern allowance (d) draft allowance 22. Sand are graded according to their: (a) clay content (b) gram SIze (c) clay content and grain size (d) None of these

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Production Engineering

A-20S 23.

24.

25.

26.

27.

Sweep pattern is used for moulding parts having: (a) Triangular shape (b) Elliptical shape (c) Uniform symmetrical shape (d) Complicated shapes having intricate details In foundaries, a square pan fitted with a wooden handle is known as: (a) Bellow (b) Slick (c) Shovel (d) Riddle An aluminium cube of 20 em side has to be cast along a cylinderical riser. If the volume shrinkage during solidification is 6%, then shrinkage volume of cube after solidification will be : (a) 400cm3 (b) 480cm3 (c) 500cm3 (d) 540cm3 With a solidification factor of 0.97 x 106 s/m-', the solidification time in (seconds) for spherical casting of200 mm diameter is : (a) 539 (b) 4311 (c) 1078 (d) 918 Hot chamber die-casting machines are used for alloys with

35.

(a) Law melting temperature (b) High melting temperature (c) Low thermal conductivity (d) low electric resistance Which of the following processes is commonly used to manufacture powder coated steel central heating radiators? (a) sand casting (b) Bending (c) Shaping (d) Press work In an orthogonal cutting process, the cutting force and thrust force are 1200 Nand 600 N respectively. It the rake angle of the tool is zero, then what will be the coefficient of friction in fool- chip interface?

40.

36.

37.

38.

39.

28. Badboys2

29.

(a) 2 (c)

30.

31.

32.

33.

34.

112

(b)

-Ii u Ji

(d) Which one of the following cutting tool bits are made by powder metallurgy process? (a) carbon steel tool bits (b) Stellite tool bits (c) less tool bits (d) Tungsten carbide tool bits Which one of the following is a single point cuting tool? (a) hacksawblade (b) millingcutler (c) pasting tool (d) grinding wheel The lip angle of a single point cutting tool is : (a) 10°-30° (b) 300t060° (c) 50°-60° (d) 60°-80° A milling machine has a metal removal rate 25 cm3/min. for a steel work piece. The depth of cut is 4.5 mm and width of cut is 90 mm. Then the required table feed will be : (a) 61.7 mmlmin. (b) 51.7 mmlmin. (c) 65.4mm1min (d) 48.8mm1min. For cutting tool material, which is correct order of increasing hot hardness (a) H.S.S, carbide, diamond (b) Carbide, H. S. S, diamond (c) Diamond, carbide, H.S.S (d) Carbide, diamond, H.S.S

41.

42.

43.

44.

45.

46.

47.

Which ofthe following among the given options is a single point cutting tool? (a) Milling cutter (b) Hack saw blade (c) Turning tool (d) Grinding wheel Which process involves increasing ofthe cross - sectional area by pressing or hammering in a direction parallel to the original ingot axis? (a) up setting (b) Peening (d) Setting down (c) Swaging Which of the following is not a type of industrial forging? (a) Drop forging (b) Roll forging (c) Blast forging (d) upset forging Which of the following statement is correct? (a) Hot rolling produces a stronger shaft than cold rolling (b) Cold rolling produces a stronger shaft than hot rolling (c) Shafts are not made by rolling process (d) Angle of twist of shaft is inversely proportional to shaft diameter Which ofthe following is commonly used die material? (b) Molybdenum (a) Tungsten (c) Cast iron (d) Hot work tool steel Reaming operation can be performed on : (a) Drilling and milling machine (b) Lathe and drilling machine (c) Shaper and drilling machine (d) Shaper and milling machine In a drilling machine the metal is removed by : (a) shearing and extrusion (b) Extrusion (c) Shearing (d) shearing and compression Which is not the part of drilling machine (a) Spindle (b) Tool holder (c) Table (d) Cross-slide Lathe beds arc produced by which of the following production processes? (a) Rolling (b) casting (c) Drawing (d) Forging When work piece is fed in the same direction and that of the cutter tooth at the point of contact, that type of milling is known as: (a) Down milling (b) upmilling (c) slot milling (d) slab milling Disign of jigs and fixtures need careful attention to: (a) Idle time reduction (b) Disign for safety (c) Swarf clearance (d) All of these TheA.P.F (atomic Packing Factor) for BCC structure is: (a) 0.52 (b) 0.68 (c) 0.74 (d) 0.84 Which of the following surface hardening processes needs quenching? (a) Induction hardening (b) Flame hardining (c) Nitriding (d) case carburizing

o, C)

I

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("I') ("I')

Production Engineering 48.

49.

50.

51.

In iron-carbonequilibriumdiagram,the x-axisis represented by: (a) carbon percentage (b) Temperature (c) Nickel percentage (d) None of these In annealing heat treatment process, the hypocutectoid steel is: (a) Heated from 40° C to 50° C above the critical temperature and then cooled slowly in the tumace. (b) Heat from 40° C to 50° C above the upper critical temperature and then cooled suddenly in a suitable coolingmedium (c) Heated from 40° C and 50° C below the critical temperature and then cooled in still air (d) Heated below or close to the lower critical tempeature and then cooled slowly. 18 : 8 stainless steel consists of: (a) 18%vanadium, 8% chromium (b) 18%chromium,8.1nickel (c) 18%tungsten, 8% nickel (d) 18%tungsten, 8% chromium On high rate of cooling, austenite converts into: (b) Ferrite (a) martensite (d) Pearlite (c) Ledeburite Which ofthe followingis correctfornormalazing operation? (a) It relieves internal stresses (b) It produces a uniform structure (c) After heating, the material is allowed to cool in atmosphere (d) The rate of cooling is slow The crystal structure of austenite is : (a) Simplecubic (SC) (b) Body centred cubic (BCC) (c) Face centered cubic (FCC) (d) Hexagonal closed packed (HCP) Austenite decomposes into territe and cementite at a temperature of: (a) 1148°C (b) 727°C (c) 1495°C (d) 1539°C Alloy steel as compared carbon steel is more (a) strong (b) tough (c) fatigue resistant (d) All of these Shock resistance of steel is increased by adding (a) Aluminium (b) Cobalt (c) Nickelchromium (d) Carbon Carbon steel is (a) produced by adding carbon in steel (b) an alloy of iron and carbon with varying quantities of phosphorus and sulpher (c) purer than the cast iron (d) None of these The raise yield point oflow carbon steel (a) Phosphorus is added (b) Silicon is added (c) Carbon is added (d) Sulphur is added Stress-concentration occurs when a body is subjected to (a) Extensive stress (b) reverse stress (c) fluctuating stress (d) non-uniform stress

52. Badboys2

53.

54.

55.

56.

57.

58.

59.

A-209

60. Diamondweight is expressedin terms of carats. One carat is equal to (a) 20mg (b) 200mg (c) 400mg (d) 1mg 61. When a body recovers its original dimensions on removing the load then it is called (a) plastic (b) brittle (c) elastic (d) None of these 62. Abilityofmaterial to under go large permanent deformations in tension is called (a) plasticity (b) stiffness (c) toughness (d) hardness 63. Shock resistance steel should have (a) high wear resistance (b) low wear resistance (c) toughness (d) low hardness 64. Essential gradient of any hardened steel is (a) carbon (b) pearlite (c) martensite (d) cementite 65. Steel containing 18% chromium and 8% nickel is called (a) austinitic stainless steel (b) ferritic stainless steel (c) martensitic stainless steel (d) None of these 66. Steel having combination 88.7% ferrite and 13%cementite is known as (a) martensite (b) austenite (c) pearlite (d) All of these 67. A metal which is brittle in tension can become ductile (a) in presence of notches (b) in presence of emprillement agents such as hydrogen (c) under hydrostatic condition (d) None of these 68. Etching solution used for medium and high carbon steel, pearlite steel and cast iron is (a) Nital- 2% RN03 is ethyl alcohol (b) 1% hydrofluoric acid in water (c) 50% NH2, OH and 50% water (d) picral- 5% pieric acid and ethyl alcohol 69. Steelcontaining 15to 20% nickel and 0.1% carbon is called (a) ferritic stainless steel (b) austenitic stainless steel (c) martensitic stainless steel (d) None of these 70. Chrome steel is widely used for (a) connecting rod (b) cutting tool (c) handtool (d) motor car crank shaft 71. Carbon steel castings are (a) easily weldable (b) tough and ductile (c) brittle (d) All of these 72. Vandium when added to steel it (a) decreases tensile strength (b) increases tensile strength (c) remain constant (d) None of these 73. High speed steel should have (a) wear resistance (b) hardness (c) toughness (d) both (a) and (b) 74. Alloy steel containing 36% Nickel is known as (a) Stainless steel (b) High speed steel (c) Die steel (d) HS.S.

o, C)

I

Badboys2

Production Engineering

A-210

75.

76.

77.

78.

79.

80.

81.

Case hardening process is (a) carburizing (b) cynidity (c) nitridity (d) All of these Normalising of steel is done to (a) remove strains caused by cold working (b) refine grain structure (c) remove dislocation causedin the internal structure due to hot working. (d) All of these Steel containing pearlite and ferrite is (a) ductile (b) soft (c) hard (d) tough Percentage of carbon in carbon steel is (a) 0-1% (b) 0.1-1.5% (c) 1.5-4.2% (d) 1- 3% Cutting tools are manufactured by (a) High speed steel (b) Nickel steel (c) Chormesteel (d) None of these Silicon Steel is widely used in (a) electrical industry (b) connecting rod (c) cutting tool (d) All of these Steel containing 11- 14% chromium and 0.35% carbon is called (a) ferritic stainless steel (b) martensitic stainless steel (c) austenitic stainless steel (d) All of these Nitriding is a process for (a) softening (b) hardening (c) tempering (d) All of these Temperature at which the first tiny new grains appears is called (a) melting temperature (b) criticaltemperature (c) pointing temperature (d) recrystallinetemperature Annealing of steel is done to (a) improvemachinability (b) softeners of metal (c) release internal stress (d) All of these Machining properties of steel are improved by adding (a) Carbon (b) Chromimum (c) Silicon (d) Sulphur, lead and phosphorus To make low carbon steel tougher and harder (a) Carbon is added (b) Carbon reduced (c) Silicon added (d) Aluminium added Chilling heat treatment and alloy adding (a) decreases machinability (b) increase machinability (c) increase carbon percentage (d) None of these Ifsteel is cooled in still air, the structure obtained is called (a) sorbite (b) pearlite (c) toorsite (d) mortensite Heat treatment process used for castings is (a) hardnening (b) normalising (c) annealing (d) tempering

Badboys2 82.

83.

84.

85.

86.

87.

88.

89.

90. Temperature at which the change starts on heating the steel is called (a) uppr critical temperature (b) point of recalesense (c) lower critical temperature (d) All of these 91. Heat treatment process used to soften the hardened steel is (a) annealing (b) hardening (c) tempering (d) quenching 92. Eutectoid based composition of carbon steel at room temperature is called (a) martensite (b) ferrite (c) comontite (d) pearlite 93. In steel, main alloy causing corrosion resistance is called (a) cobalt (b) vandium (c) carbon (d) chromium 94. Hardness of Steel depends on (a) Carbon percentage (b) Silicon percentage (c) Shape and distribution of carbide in iron (d) None of these 95. Advantage of austempering is (a) mere uniform microstructure is obtained (b) quenching eracts are avoided (c) None of these (d) All of the above 96. Delta iron exists in the temperature range of (a) 1400°C-1530°C (b) 768°Cto 900°C (c) 1400°C-1550°C (d) 350-786°C 97. Induction hardening have high (a) carbon percentage (b) cemiteteformation (c) power factor (d) frequency 98. Sorbite is obtained by (a) quenching of steel in oil (b) heating above its critical temperature (c) reduction of silicon percentage (d) annealing of steel 99. Temperature at which the changes end on heating the steel is called (a) uppercriticallimit (b) lowercriticallimit (c) melting point (d) point ofrecalesence 100. Heat treatment process is (a) hardening by quenching(b) annealing (c) tempering (d) All of these 101. Ifsteel is slowlycooled in furnace, the structure obtained is called (a) ferrite (b) sorbite (c) martensite (d) pearlite 102. Steel having combination of6.67% carbon and 93.33% of iron is known as (a) austenite (b) pearlite (c) cementite (d) martensite 103. By normalising of steel, its (a) ductility decrease (b) ultimate tensile strength increase (c) field point increases (d) All of the above

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Production Engineering 104. An alloy steel contains (a) more than 0.5% Mn and 0.5% Si (b) more than 0.15% Mn and 0.5% Si (c) less than 0.5% Mn and 0.15% Si (d) more than I%MnandO.05 Si 105. In carbon steel castings the percentage of (a) carbon between 1.5 - 2.5% (b) carbon below 1.7% (c) various carbon between 0.5 - 1.5% (d) more than 1.5% carbon 106. In steel as the percentage of carbon increase the following has decrease (a) ductility (b) tensile strength (c) hardness (d) toughness 107. Silicon steel is widely used in (a) chemical industry (b) mechanical parts making (c) electrical industry (d) die and puncher 108. Weld decay is the phenomenon found with (a) mild steel (b) wrought iron (c) cast iron (d) stainless steel 109. Annealing of white cast iron results in the production of (a) nodulariron (b) cementite (c) malleable iron (d) cast iron 11O. Solder is an alloy of (a) copper and tin (b) lead and copper (c) lead with zinc (d) lead and tin 111. The manufacturing process in which metal change its state from liquid to solid. (a) Casting (b) Machining (c) Forging (d) Turning 112. In which casting consumable pattern is used. (a) Sand casting (b) die-casting (c) PD.C (d) Investment casting 113. In case of Investment casting (a) wax pattern used (b) wooden pattern used (c) metallic pattern used (d) any of these can be used 114. The casting process by which hollow casting produced without using core is known as (a) Sand casting (b) Die casting (c) Centrifugal casting (d) Slush casting 115. For non sysmetric shape suitable casting method is (a) Sand casting (b) Slush casting (c) investment casting (d) all of these 116. The purpose of adding wood flour to foundry sand is to improve (a) collapsibility (b) strength (c) mouldability (d) all of these 117. Surface finish of casting depends upon (a) mold degassing (b) pattern fmish (c) casting process (d) all of these 118. Cores are used to make casting (a) Hollow (b) moresolid (c) more economic (d) moreweak

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A-211

119. Wood for pattern is considered dry when moisture content is (a) 5% (b) zero (c)

less than 15%

(d)

less than 30%

120. For steel casting following type of sand is better. (a) coarse grain (b) fine grain (c) medium grain (d) None of these 121. Trowel is (a) pointed tool (b) wooden hammer (c) tool used to repair corner (d) long, flat metal plate fitted with a wooden handle 122. Shrinkage allowance is made by providing (a) cores (b) taper in casting (c) addition in dimension of pattern (d) all of above 123. Casting process in which molten metal poured into mould under pressure is known as (a) sand casting (b) slush casting (c) vacuum casting (d) pressure die casting 124. Casting process in which mould kept revolving is known as (a) slush casting (b) vacuum casting (c) centrifugal casting (d) die casting 125. Facing sand used in foundary work comprises of (a) Silica and Clay (b) Clay, sand and water (c) Clay and abumina (d) Silica and aluminium 126. Accuracy of shell moulding is of the order of (a) O.oInvm (b) 0.1 nvm (c) 0.003m1mtoO.005m1m (d) None of these 127. Mark the most suitable material for die casting in the following (a) copper (b) Nickel (c) Steel (d) Cast iron 128. In general, the draft on casting is of the order of (a) 10-15m1m (b) 10-5m1m (c) 20-10mlm (d) 1-IOmlm 129. The purpose of riser in a casting process (a) act as feeding way in mould (b) act as reservoires (c) feed molten metal from basis to gate (d) None 130. Match plate pattern is (a) Green sand moulding (b) Pitmoulding (c) machining moulding (d) Pit moulding 131. For making ornaments and toys casting process used is (a) die casting (b) Investment casting (c) sand casting (d) slush casting 132. True centrifugal casting is used to get (a) chilled casting (b) accurate casting (c) dynamically balanced casting (d) Solid casting 133. Draft on pattern for casting is providing for (a) Sapteremoval from mould (b) adding shrinkage allowance (c) providing better finishing in casting (d) for machining allowance

Badboys2

Production Engineering

A-212

134. The gate is provided in mould to (a) provide a reservoires (b) constant flow (c) feed mould according to rate of cooling (d) all of above 135. Sand slinger gives (a) better packing of sand (b) uniform sand density (c) better packing of sand near flask (d) none of above 136. As the size of casting increases, it is often better to use increasingly (a) Coarse grain (b) fine grain (c) mediumgrain (d) none of these 137. Black colour marking in pattern is used to indicate (a) machined surface (b) un-machined surface (c) parting surface (d) None 138. Loam Sand comprises of percentage of sand and mould (a) 10: 50 (b) 20 : 80 (c) 50 : 18 (d) 80:20 139. The ratio between the pattern shrinkage allowances of steel and iron is approx. (a) 2: 1 (b) 1: 1 (c) 1:2 (d) 1: 10 140. Sweep pattern is suitable for __ casting (a) small (b) medium (c) large (d) any of these 141. Fluidity is greatly influenced by the temperature of (a) tapping (b) melting (c) solidification (d) pouring 142. Chills are used in mould to (a) achieve directional solidification (b) reduce the possibility of blow holes (c) reduce freezing time (d) smoothens metal flow for reducing splatter 143. Which ofthe followingmaterialrequiresthe largestshrinkage allowance,while making a pattern for casting. (a) Aluminium (b) Brass (c) cast Iron (d) carbon steel 144. The height of the down - sprue is 175 mm and its cross sections area at the base is 200 mm-, the cross-sectional area ofthe horizontal runner is also 200 mm/. Assuming no losses the correct choice for the time (in second) required to fill a mould cavityof volume 106 mm-, will be (use g = 10m!

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S2)

(a) 2.67 (b) 8.45 (c) 26.72 (d) 84.50 145. Two castings of the same metal have the same surface are one casting is in the form of a sphere and the other is a cube. What is the ratio ofthe solidification time for the sphere to that of the cube. (a)

(c)

3 4 5 41t

6 (b)

1t

31t

(d)

8

146. Consumable patterns are made of (a) wax (b) polystyrene (c) ceramics (d) none of above 147. Limestone used in melting of cast iron acts as (a) flux (b) catalyst (c) alloying element (d) none of these 148. Electric indirect arc furnace is normally used for melting of (a) non-ferrous alloys (b) cast steel (c) ferrous alloys (d) all of these 149. The draw back with metallic patterns is (a) costly (b) heavy in weight (c) difficult to shape (d) all of these 150. There is no need of withdrawal of pattern from the mold if is used (a) solid pattern (b) split pattern (c) thermoplastic pattern (d) consumable pattern 151. Polystyrene used as consumable pattern material has a relative density of (a) 1.2-1.25 KN/m3 (b) 0.2-0.25 KN/m3. (c) 0.2-1.0 KN/m3 (d) all ofthese 152. In small castings which of the following allowance can be ignored (a) draft allowance (b) shrinkage allowance (c) matching allowance (d) rapping allowance 153. Small patterns are often used for (a) bends (b) pipework (c) drainage pelting (d) all ofthese 154. Permeability of sand decreases when (a) moisture percentage increases (b) compactness increases (c) bonding contents increases (d) all of above 155. Providing more than adequate machining allowance (a) increase machining cost (b) reduce machining cost (c) reduce casting weight (d) all of above 156. By compacting, sand density (a) increases (b) decreases (c) have no effect (d) None 157. Compacting of sand affects its (a) strength (b) permeability (c) density (d) all of these 158. The draft allowance to be provided on a pattern depends on (a) vertical length of pattern (b) intricacyofpattern (c) molding method (d) all of above 159. Contraction allowance in cast steel casting will be least for casting, having dimensions (a) upt0600mm (b) 600-1000mm (c) 1000-1800mm (d) above1800mm 160. Distortion in casting can be reduced by (a) modifying design (b) sufficientmachining allowance (c) improving foundary facility (d) all of above

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Production

Engineering

161. Clay content of green sand is usually (a) 5-10% (b) 18-30% (c) 5-30% (d) 10-50% 162. The water percentage in green sand is kept normally (a) 6-8% (b) 5-10010 (c) 10-20010 (d) 20-30% 163. Clay used for foundary sand should be (a) kaolinite (b) mont-morillonite (c) illite (d) all of these 165. Main contents of moulding sand are (a) Silica sand, clay and water (b) Silica sand, dust and carbon (c) Sand, coal powder and water (d) Green Sand and water 165. is used in magnesium moulding process. (a) boric sulphur (b) molasis (c) charcoal (d) all of these 166. Graphite is sprinkled on the surface of green sand mold to (a) exclude the burn out effect (b) minimize surface defects (c) improve surface finish (d) reduce the number of blow holes. 167. Hot tears in casting are caused due to (a) too much ramming of mold (b) grain size of sand (c) size of casting (d) rate of poring of molten metal 168. Rough surface may appears due to (a) large grain size sand (b) lowramming (c) high permeability (d) anyone of above 169. Scabs may be caused by (a) low permeability of sand (b) high moisture content of sand (c) intermittent running of molten metal over sand surface (d) all of the above 170. The advantage of shell moulding is (a) less sand requirement (b) dimensional accuracy (c) good finish (d) high productivity 171. Hardness of the mould is affected by (a) ramming of moulding sand (b) percentage of moisture (c) binder percentage (d) all of above 172. Blow holes in casting are due to (a) high moisture content of sand (b) low permeability of sand (c) excessive fine grains and gas producing ingredients (d) any of above

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A-213

173. Which of the following defect may occur due to improper design of gating system. (a) Cold sheets (b) mis-run (c) rough surface (d) all of these 174. Sprue are generally (a) uniform in size (b) tapered downwards (c) tapered upward (d) None 175. The design of gate should be able to (a) avoid erosion of cores and moulding cavity (b) prevent scum slag and eroded sand particles from entering the mould cavity (c) minimise turbulence and dross formation (d) all of above 176. In Magnesium alloy casting, normally solidification shrinkage is of (a) 1% (b) 2 % (c) 4 % (d) 10% 177. Solidification time for riser should be (a) less than that of casting (b) more than casting (c) same as casting (d) none of above 178. Forging of steel is done at a temperature of (a) 800°C (b) lO00°C (c) lO00°F (d) 1200°C 179. Process used for making Nut and Bolts is (a) hot piercing (b) upsetting (c) hot drawing (d) none of these 180. Process of shaping metal sheet by processing them against a desired shape is known as (a) upsetting (b) spinning (c) rolling (d) all of these 181. Porosity of metal is largely eliminated in _ (a) cold working (b) hot working (c) annealing (d) casting 182. Production of countours in flat blank is term as (a) piercing (b) punching (c) blanking (d) upsetting 183. Forging temperature used for plain carbon steel is (a) 800°C (b) lO00°C (c) 11OO°C (d) 1300°C 184. Gear shaping is related to (a) upsetting (b) hot (c) template (d) drawing 185. Mass production generally done by (a) Casting (b) Machining (c) Hobbing (d) All of these 186. Effect associated with cold forging is (a) shrinking (b) elongation (c) strain hardening (d) all of these 187. Crank shaft is made by (a) hot forming (b) coldforming (c) machining (d) casting

Badboys2

Production Engineering

A-214

188. For extrusion process (a)

199. Notching is the operation of

complex section are produced from bar stocks

(b) the strength of finished product is improved due to cold working (c)

Good surface finish and close tolerence is generated

(d)

all of these

189. Seam less tube can be produced by (a)

steam hammer forging (b)

piercing

(c)

casting

none of these

(d)

190. Process of extrusion is like (a)

a tooth paste coming from tube

(b)

air press from nozzle

(c)

both (a) and (b)

(d) none of these

(a)

removal of excess metal from the edge of strip to make it suitable for drawing without wrinkling

(b)

cutting in a single line across a part of the metal strip allow bending or forming in progressive die operation while the part remains attached to the strip

(c)

both (a) and (b)

(d)

none of these

200. Process consists of pushing the metal inside a chamber to force it out by high pressure through an orifice which is shaped to provide the desired form of the finished part, is called (a)

piercing

(b)

forging

(c)

extrusion

(d)

cold peening

201. Parts of circular cross section which are symmetrical about the axis of rotation are made by

191. Material good for extrusion is (a)

Low carbon steel

(b)

Cast iron

(a)

hot forging

(b)

(c)

S.S.

(d)

HS.S.

(c)

cold forging

(d) none of these

192. Upsetting or cold heading machine is a (a)

rolling process

(b)

extrusion process

(c)

forging process

(d)

none of these

Badboys2 193. The major problem in hot extrusion

is

(a)

design of punch

(b)

design of die

(c)

wear of punch

(d) wear and tear of die

194. Process of increasing the cross-section ofa bar and reducing its length is called (a)

drawing down

(b)

drifting

(c)

spinning

(d) upsetting

195. Cold working (a)

requires much higher pressure than hot working

(b)

increase hardness

(c)

distort grain structure

(d)

all of these

202. Mechanical properties of the metal improve in hot working due to (a)

grain growth

(b) recrystallisation

(c)

recovery of grains

(d) refmement of grain size

203. If there are bad effects of strain hardening on a cold formed part the part must be (a)

tempered

(b)

annealed

(c)

hardned

(d) normalised

204. A tooth paste tube can be produced by (a)

hollow backward extrusion

(b)

forging

(c)

solid forward extrusion

(d)

none of these

205. The true strain for a low carbon steel bar which is doubled in length by forging is

196. Cold working process can be applied on the component having diameters up to (a)

20mm

(b)

25mm

(c)

30mm

(d)

50mm

197. Which of the following is a gear finishing operation (a)

hobbing

(b)

milling

(c)

saving or burnishing

(d)

none of these

198. Roll forging (a)

causes a steadily applied pressure instead of impact force

(b)

is a forging method for reducing the making it longer

(c)

is used to force the end of a heated bar into a desired shape

(d) none of these

hot spinning

(a)

0.307

(b)

0.5

(c)

0.693

(d) 1

206. The process of hot extrusion is used to produce (a)

certain rods made of aluminium

(b)

steel pipes for domestic water supply

(c)

stainless steel tubes used in furniture

(d)

large size pi pes used in city water main s

207. Extrusion process can effectively Reduce the cost ofproduct through (a)

Saving in tooling cost

(b)

Saving in administrative

(c)

material saving

(d)

all of these

cost

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Production

Engineering

208. Hot press forging (a) causes a steadily pressure instead of impact force (b) is used to force the end of a heated bar into a desired shape (c) is a forging method for reducing the diameter of a bar and in the process making it layers (d) all of these 209. In hot working (a) annealing operation is not necessary (b) power repowerments are low (c) surface finish is good (d) grain refinement is possible 210. In a solid extrusion die, purpose of knock out pin is (a) shopping the part to extrude through the hose (b) ejecting the part after extrusion (c) allowing the job to have better surface finish (d) reducing the waste of material 211. In electric resistance welding, two copper electrodes used to cooled by (a) air (b) water (c) both (a) and (b) (d) None of these 212. An example of fusion welding is (a) Thermitwelding (b) Arc welding (c) Forge welding (d) Gas welding 213. Welding process in which flux is used in form of gannual is (a) D.C.Arc welding (b) Spot welding (c) Thermitwelding (d) SubmergedArc welding 214. In arc welding face shield used to protect eyes from (a) Spatters (b) Spark (c) Infra-red and ultraviolet rays (d) None of these 215. Gases used in tungsten gas welding are (a) Carbon dioxide and H2(b) CO2 and oxygen (c) Argon and helium (d) Acetylene and nitrogen 216. Open circuit voltage for Arc welding is of the order of (a) 20-40V (b) 10-20V (c) 40-50V (d) 40-95V 217. Welding of steel structure on site work of a building easily made by (a) Spot welding (b) Buttwelding (c) Arcwelding (d) Any of the above 218. Tig welding is preffered in followingmetal welding (a) Silver (b) Aluminium (c) Mild steel (d) All of these 219. In arc welding temperature generated is of the following order. (a) lOOO°C (b) 1800°C (c) 3500°C (d) More than 4000°C

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A-215 220. Temperature of oxy-hydrogen flame as compared to oxyacetylene flame is (a) less (b) more (c) same (d) depends on oxygen percentage 221. Oxidising flame is obtained by supplying (a) more oxygen and less volume of acetylene. (b) both oxygen and acetylene kept in same volume. (c) acetylene volume kept more than oxygen volume (d) None of these 222. Oxidising flameas comparedto neutral flame has inner core (a) shorter in size (b) less luminous (c) more luminous (d) both (a) and (b) 223. Maximum flametemperature occurs (a) at inner core of flame (b) at outer core of flame (c) attipofflame (d) next to the inner core 224. Maximum used flame in gas welding method is (a) oxidising (b) neutral (c) carburising (d) None of these 225. Strongest brazing joint is (a) Lapjoint (b) Buttwelding (c) Scrafwelding (d) None of these 226. Melting point of the filler metal in brazing should be above (a) 400°C (b) 420°C (c) 6(X)°C (d) 800°C 227. Seam welding is continuous (a) spot welding process (b) type of projection welding (c) multi-spot welding (d) None of these 228. Welding process preferred for cutting and welding for nonferrous metal is (a) MIGwelding (b) TIGwelding (c) Inert gas welding (d) None of these 229. The welding process in which electrode do not consumed is (a) MIG welding (b) TIGwelding (c) Argon welding (d) None of these 230. The welding process in which electrode get consumed is (a) MIGwelding (b) TIGwelding (c) Spot welding (d) None of these 231. Grey cast iron is usually welded by (a) resistance welding (b) gas welding (c) spot welding (d) arc welding 232. In arc welding using direct current amount of useful arc heat at the anode and cathode respectively are (a) two third of one third (b) One third and two third (c) equal (d) none of these 233. Multipoint welding process is (a) seam welding (b) spot welding (c) projection welding (d) percussion welding

Badboys2

Production Engineering

A-216

234. Amount of current required in electrical resistance welding regulated by changing the (a) polarity (b) input supply (c) by altering no. ofturns of primary winding (d) by changing no. of turns of secondary winding 235. Welding of chromium molybdenum steels cannot use (a) Oxygen acetylene welding (b) Thermitwelding (c) Soldering (d) Electric arc welding 236. Spot-welding, projection welding and seam welding are classification of (a) Thermitwelding (b) Resistance welding (c) Arc welding (d) Spot welding 237. An arc is produced between a bare metal electrode and workin (a) D.C.welding (b) Submerged arc welding (c) Spot welding (d) None of these 238. In arc welding, two low welding speed results in (a) Excessivepilling up of weld metal (b) Electrode waistage (c) Over hauling without penetration edge (d) All of these 239. Fillers material is essentially used in (a) Spot welding (b) Gas welding (c) Seamwelding (d) Projection welding 240. Rate of welding steel by carburising flame as compared to neutral flame is (a) less (b) same (c) more (d) all of the above 241. Carburising flame is used to weld (a) Brass and bronze (b) Steel, and copper (c) Hard surfacing materials such as satellite (d) Any of above 242. Filler material is used in (a) Spot welding (b) Butt welding (c) Seamwelding (d) None of these 243. Cleaning of metal in electrical resistance welding is (a) important (b) not important (c) have no effect (d) none of these 244. An example of fusion welding is (a) Spot welding (b) Gas welding (c) Projection welding (d) All of these 245. Welding process using a pool of molten metal is (a) TIGwelding (b) MIGwelding (c) Submergedarc welding(d) None of these 246. In spot welding the electrode tip diameter (d) should be equal to

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(a) Less than (c)

j15i

.[t

(b).[t (d)

247. In arc welding current used is (a) AC. current at low frequency (b) AC. current at high frequency (c) D.C. current (d) All of these 248. An arc is produced between a bare metal electrode and the work in welding process known as (a) Gas welding (b) Submerged arc welding (c) D.C.welding (d) None of these 249. Seam welding used for metal sheets having thickness in the range (a) below3 mm (b) 3-5mm (c) 3-6mm (d) 0.025-3mm 250. In Arc welding, range of temperature generated at arc is (a) IOOO°C - 2000°C (b) 2000°C- 4000°C (c) 4000°C-6000°C (d) None of these 251. Projection welding is a (a) type of arc welding (b) type of continuous spot welding (c) type of gas welding (d) none of these 252. In resistance welding voltage used for heating is (a) below 10V (b) 10V (c) higher than 10V (d) None of these 253. In arc welding, penetration is minimum for (a) DCSP (b) OCRP (c) AC. (d) None of these 254. In electrical resistance welding, pressure applied varies in the range (a) 50-100kgF/cm2 (b) 100-150kgF/cm2 (c) 150-200 kg F/cm2 (d) 250-550kgF/cm2 255. Which of the following current is preferred for welding of non-ferrous metal by arc welding? (a) DC (b) AC. at high frequency (c) AC. at low frequency (d) None of these 256. Main criterion for electrode diameter selection is (a) Thickness of work piece (b) Typeof work piece metal (c) Welding pressure applied (d) Welding process applied 257. In projectionwelding diameterofthe projectionas compared to thickness of the sheet is approximately (a) same (b) half (c) double (d) 1.5times 258. Number of zones of heat generation in spot welding are (a) 1 (b) 2 (c) 3 (d) None of these 259. In spot welding tip of electrode made up of (a) Sinteredmetal (b) Carbide (c) Copper (d) Brass

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Production Engineering 260. Material used for coating the electrode is called (a) binder (b) oxidiser (c) flux (d) slag 261. In arc welding, arc is created between work piece and electrodes due to (a) type of current (b) electronsjumping from electrode to workpiece (c) high resistivity due to presence of air (d) none of these 262. During Arc welding with increaseof thickness ofmaterial to be welded, welding current have to (a) decrease (b) increase (c) remain constant (d) none of these 263. In resistance welding pressure released (a) after welds gets cool (b) when work gets heated (c) just after the weld completed (d) none of these 264. Welding process used for joining round bars is (a) Thermitwelding (b) Projection welding (c) Seamwelding (d) Butt welding 265. Welding in which the metals to be joined are heated to a molten state are allowed to solidify in presence of a filler materials is called (a) Spot welding (b) Fusion welding (c) D.C.welding (d) None of these 266. In a welding process, flux is used to (a) Permit perfect cohesion of metal (b) remove oxides of metal formed at high temperature (c) both (a) and (b) (d) none of these 267. In electrical resistance welding (a) Voltagekept high and current also high (b) Voltagekept high and current kept low (c) Voltagekept low and current kept high (d) None of these 268. In forehand gas welding operation, the angle between the rod and work piece is kept around (a) 15° (b) 10-20° (c) 30° (d) 45° 269. Material best weldable with itselfis (a) copper (b) aluminium (c) mild steel (d) all of these 270. Arc length in electric Arc welding is the distance between tip of the electrode and (a) work piece (b) bottom of crates (c) centre of crates (d) none of these 271. Oxygen to acetylene ratio in case ofneutral flame is (a) 0: 1 (b) 1:2 (c) 0.8:2 (d) 2: 1

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A-217 272. In MIG welding helium or argon is used in order to (a) act as flux (b) act as shielding medium (c) providing cooling effect(d) all of these 273. Oxygen to acetylene ratio in carburising flame is (a) 0.5: 1 (b) 0.9: 1 (c) 1: 1 (d) 1: 2 274. A lathe machine special in (a) Diameter oflathe (b) Gross weight of machine (c) Speed of lathe (d) Swing oflathe 275. Lathe machine bed made up of (a) alloys (b) cast iron (c) mild steel (d) prg Iron 276. Shanks of tapes drills are provided standard tapes known as (a) tapes shank (b) morse tapes (c) chapman tapes (d) None of these 277. The length of a hacksaw blade is measured from (a) extreme end to extreme end (b) centre of hole at one end to the center of holes at the other end (c) the formula L = 16 x width (d) None of the above 278. A plug gauge is used for measuring (a) out side bore (b) cylindrical bores (c) spherical holes (d) tapes bores 279. Standard milling arbores size is (a)

1"

(b)

1_!_"

(c)

_!_"

(d)

Anyone of above

2

4

280. Standard taper generally used on milling machine spindles is (a) Morsetaper (b) Shellr'sistaper (c) Champman taper (d) None of these 281. Sintered and tungsten carbides can be machined by (a) Conventional process (b) Grinding only (c) E.DM. (d) None 282. The binding material used in cemented carbide tool is (a) chromium (b) cobalt (c) sulpher (d) nickel 283. Discontinous chips are formed during machining by (a) mild Steel (b) aluminium (c) cast Iron (d) brass 284. Continous chips are formed while machining of (a) cast iron (b) mild steel (c) aluminium (d) None of these 285. To prevent tool from rubbing the work, angle provided on tool is (a) reliefangle (b) rake angle (c) clearance angle (d) None of these

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A-21S 286. In metal machining due to burnishing friction, heat is generated in the (a) friction zone (b) friction less zone (c) work-tool contact zone (d) None of these 287. A single point tool has (a) rake angle (b) cutting angle (c) clearance angle (d) None of these 288. Angle on which the strength of the tool depends is (a) cutting angle (b) lip angle (c) rake angle (d) clearence angle 289. Velocity oftool relative to workpiece is called (a) average velocity (b) cutting velocity (c) shear velocity (d) chip velocity 290. The angle provided to prevent rubbing between workpiece and cutting tool is known as (a) relief angle (b) rake angle (c) lip angle (d) None of these 291. Cutting tool used in lathe, shaper and planer is (a) Multi point cutting tool (b) Two point cutting tool (c) Single point cutting tool (d) Multi point cutting tool 292. Angle between the tool face and the ground and surface of fank is called (a) rake angle (b) lip angle (c) clearance angle (d) cutting angle 293. Velocity of tool along the tool face is called (a) Chip velocity (b) Cutting velocity (c) Shear velocity (d) None of these 294. The depth of cut depends upon (a) tool material (b) cutting speed (c) regidityofmachining tool (d) All of these 295. The metal in machining operation is removed by (a) distortion of metal (b) shearing the metal across a zone (c) tearing chips (d) cutting the metal across a zone 296. Tool life is most affected by (a) tool geometry (b) cutting speed (c) feed and depth (d) None of these 297. Tool signature (a) description of tool shape (b) the plane of tool (c) design and description of various angles provide on tool (d) brandlmodle none of tool 298. Tool signature comprised of (a) property of tool (b) speed of cutting tool (c) 7-various elements (d) 6-elements 299. Depth of cut for roughing operation as companied to finishing operation is (a) same (b) more (c) less (d) None of these

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Production Engineering 300. Cutting speed should be kept low while machining (a) Soft material (b) Regular shape material (c) Casting (d) All of above 301. The type of chip produced when cutting ductile material is (a) continous (b) discontinous (c) built up edge (d) None of these 302. The average cutting speed for machining a cast iron by a high speed tool steel tool is (a) 10m/min (b) 20m/min (c) 30m/min (d) None of these 303. Relief angle on high speed tools generally vary in the range (a) 0-5° (b) 5°_10° (c) 10°-20° (d) 20° to 30° 304. In metal machining, due to friction between the moving chip and the tool face, heat is generated in the (a) Shear zone (b) Friction zone (c) Work-tool contact zone (d) None of the above 305. Material having lowest cutting speed is (a) Bronze (b) Aluminium (c) High carbon steel (d) Cast iron 306. Cutting tools used on milling machining machine is (a) Single point (b) Double point (c) Multi point (d) Any of above 307. The cutting edge of the tool is perpendicular to the direction of tool travel in (a) oblique cutting (b) orthogonal cutting (c) both (a) and (b) (d) None of these 308. Orthogonal cutting system is also called (a) Single-dimensional cutting system (b) Two-dimensional cutting system (c) Three dimensional cutting system (d) Any of above 309. In metal cutting operations, chips are formed due to (a) stress deformation (b) shear deformation (c) sharpness of cutting edge (d) linear transformation 310. With increase of cutting speed, the built up edge made (a) larger in size (b) smaller (c) remains same (d) None of these 311. Cutting ratio is the ratio of (a) Chip thickness to depth of cut (b) Chip velocity to cutting velocity (c) Both (a) and (b) (d) None of the above 312. Continous chips formed when machining speed is (a) lower (b) constant (c) higher (d) None of these 313. Which of the following tool material has highest cutting speed? (a) HS.S. (b) Carbon steel (c) Tool steel (d) Carbide tools

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Production Engineering

A-219

314. Tool cutting forces, with increase in cutting speed (a) increase linearly (b) decrease linearly (c) remains constant (d) None of these 315. Chip breakers are provided on cutting tool is (a) for operator's safety (b) better finish (c) permit short ships (d) forminimizing heat generation 316. Maximum cutting angles are used for machining (a) cast iron (b) mild steel (c) aluminiumalloys (d) None of these 317. When radial force in cutting is two large will cause (a) better finish (b) poor finish (c) decrease tool life (d) increase tool life 318. Segmentedchips are formedwhile machining (a) softmaterial (b) tough material (c) brittlematerial (d) high speed steel 319. As cutting speed increase the built up edge (a) reduced (b) increase (c) becomelarger (d) None of these 320. In tool signature, the largest nose radius is indicated (a) in starting (b) at the end (c) in middle (d) All of these 321. In equation YIn = C, value ofn depends on (a) Material of workpiece (b) Material of tool (c) Cutting position (d) All of these 322. The relationship betweentool life (T) and cutting speed (Y) m/min is given as

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yn

(a)

v r-c

(b)

-=C

(c)

r

(d)

vr=-c

Y

=C

T

323. Chips are broken effectively due to which ofthe following property (a) Elasticity (b) Toughness (c) Workhardening (d) Stress produced 224. Continous chips are formed when machining (a) brittle metal (b) ductilemetal (c) high speed (d) All of these 325. Finishing obtained on workpiece mostly affected by (a) Cutting speed (b) Feed rake (c) Lubricant used (d) Depth of cut 326. Machinability tends to decrease with (a) increase in strain-hardening (b) increase in tensile strength (c) increase in carbon contents (d) None of these 327. Machinability tends to increase with (a) increase in hardness (b) decrease with decrease hardness (c) remain same as hardness varies (d) proper stress releaving and proper heat treatment

328. With high speed steel tools, the maximum safe operating temperature is in order of (a) below200°C (b) above300°C (c) 200°C (d) None of these 329. Best method of increasing the rate of removaling metal is (a) increase feed rate (b) increase depth of cut (c) increase speed of tool (d) increase cutting angle 330. In a cutting operation, the largest force is (a) Radial force (b) Longitudinal force (c) Tangential force (d) Force along shear plane 331. Metal in machining operation is removed by (a) distortion of material (b) shearing of metal (c) fracturing of metal (d) any of above 332. When material is ductile and cutting speed is slow then chips formed are (a) Continuous (b) Discontinuous (c) Powder shape (d) None of these 333. During machining process when ductile metal is cutting at medium speed then chips formed are (a) Continuous (b) Discontinuous (c) Continuous with built up edge (d) Power shape 334. Chip formedwhen Ductile Metalmachined with high speed (a) Continuous chips (b) Discontinuous chips (c) Continuous chips with built up edge (d) Fragmented chips with built up edge 335. Material having hight cutting speed is (a) Bronze (b) Aluminium (c) Cast Iron (d) High carbon steel 336. An angle provided between tool face and line tangent to the machined surface at cutting points called as (a) rake angle (b) lip angle (c) cutting angle (d) clearance angle 337. Angle provided in a single point cutting tool to control chip flow is (a) Siderake angle (b) End relief angle (c) Backrake angle (d) Sliderelief angle 338. Velocityof chip relative to work-piece is acting (a) Along the shear plane (b) Normal to shear plane (c) Normal to tool place (d) Along the tool face 339. The coefficient of friction between chip and tool can be reduced by reducing the (a) loweringrake angle (b) feed of tool (c) width of tool (d) dept of cut 340. In metal machining due to plastic deformation of metal maximum heat is generated in the (a) Friction zone (b) Shear zone (c) Point of contact of cutting tip and work piece (d) All of above

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