Handbook to ssc je mechanical
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This document provides information about the recruitment exam guide and handbook for junior engineers. It includes the following sections: engineering mechanics and strength of materials, theory of machines and machine design, thermal engineering, fluid mechanics and machinery, and production engineering. It also provides information on different engineering mechanics concepts like forces, force systems, equilibrium, moments, friction, kinematics, dynamics, and Newton's laws of motion. Additionally, it covers topics like stress, strain, stress-strain relationships, shear stress, and strain tensor. Examples of stress analysis for different structural components are also given.
![Z =Z Z =Z Z =Zxy yx' xz zx' ~ y z
Plane Stress problems are those in which the stress acting in one
of the mutual perpendicular directions is assumed to be zero
.. cr=0 Z =0 Z =0z xz yz
[cr]=[crx crxy]
cryx cry
For a given stress tensor
Z face
o ,o .o are normal stressesx y z
Remaining are shear stress.
STRESS TENSOR
Torsional shear stress
I IAxi I B di Direct shear stress
~mg
Tensile Compressive
Shear Stress
(acting parallel to
corresponding plane)
I
Normal Stress
(acting perpendicular to
corresponding plane)
I
Type of Stresses
I
When deformation or strain occurs freely in a direction,
stress developed in that direction is zero.
Whendeformationisrestrictedcompletely,orpartiallystress
is developed. Hence strain is the cause of stress.
•
NOIE
MKS : kgf/cm? 1 MPa = 106 N/m2
1GPa= 109N/m2
1 Pa = 1 N/m2 1kg;:::::0.1 MPa
cm
SI : Pa, MPa, GPa
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
Units of Stress
L
,
· ,••• p ---- ------ ------ -----_ •• _•• _... _.-
· ,
· ., ,, .. ., .. ,
/
F
d cr=-
A
p
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.
Centripetal and Centrifugal Force
Essential force for a circular motion acting radially inwards is
called centripetal forcewhich 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 ofmotion, the bodymust exert a force
radially outwards ofequal.
ANGULAR ACCELERATION
The rate of change of angular velocity is called angular
acceleration. It is expressed in radls2
dO)
a=cit
ANGULAR VELOCITY
The rate of change ofangular displacementofa bodywith respect
to time is called angular velocity.
de
0)=-
dt
ifa bodyisrotatingat Nr.p.m.then correspondingangularvelocity
21tN
0)= 60 rad/s
If the body is rotating 0)rad/s along a circular path of radius r,
then its linear velocity (v) is given by
v = (l}r
1
and Foc2
R
On combining the above two expressions,
Foc MIM2
R2
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 displacementis a vectorquantity.Angular
displacementO canbemeasuredin radians,degreesor revolutions.
1revolution = 21tradians = 360degrees
A-5Engineering Mechanics and Strength of Materials
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![Strain tensor is used to define the state of a strain at a
point
c :normal strain y: shear strain
•
STRAIN TENSOR
b
Shear strain = Shear angle (<l»
BJ...
L
B B'p
It is defined as the change in initial right angle between two line
elements which are parallel to x and y axes respectively.
V=/bt
'bV 0/ ob ot
Cv =--:Y-=T+-';+t
For a sphere,
oD
Cv = - D : diameter of sphere
D
SHEAR STRESS
'bV
Cv =v:s, +cy +cz
Another example of rectangular block is considered
p+-~~~----~-~----------------------------------------------¥------~-!--+p
I I
I I
: La :
I I
: L, :
~'---------------------+'
Compressive Tensile
Strain Strain
Consider a rod oflength La subjected to load P
Volumetric
Strain (s.)
ILateral
Strain (Slateral)
I
I
Longitudinal
Strain (Slong)
Shear StrainNormal Strain
Strain
I
Strain is defined as the ratio of change in dimension to original
dimension.
STRAIN
L2
Elongation of a conical bar under its self weight = ~E
DIL: : 8'1. 1
y = selfweight per unit volume
2
Elongation of a prismatic bar under its self weight = yL
2E
p
Elongation of a tapered bar subjected toaxial load P
'b=~
AE
Oll
Zxz = 10(i.e., shear stress acting onx face along Z direction)
Z = 0zy
o = 100 o = 50 o = 25x 'y 'z
Elongation of a bar Subjected toaxial load P
Units: MPa
[
100 120 10]
rr = 20 50 0
10 0 25
Engineering Mechanics and Strength ofMaterialsA-6
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![oct
IfJ.! = 0 => - = 0
do
J.!=
I
---- f--------------- --- ~dr do
"
1..-
Lo ...,
.. Lr ...
"
,
=> 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.
I
-lateral strain I
J.!= longitudinal strain
K=ooFor
HYDROSTATICSTATEOF STRESS
(NO DISTORTION, ONLY
VOLUME CHANGES)
(J
I
1
1
1
........
",...",..,;
- -------------~".",..
--+-~(J
-----------,... " 1
~_+--____,,_,,"''',... 1
1
1
1
1
1
1
(J
K=
Bulk Modulus (K)
Normal stress o
1
for a given 't, G ex. -.
Y
Shear Modulus or Modulus of Rigidity
As per Hooke's law,
Shear stress ex. shear strain
1't=Gyl
Engineering Stress (o)= 0" I .
ngma x section area
Instantaneous load
True stress = .
Instantaneous x section
Load
'E' is the slope of o vis ~ Ediagram
EL ---------------------- B
PL ------------- ~
Yxy=Yyx
Strain Tensor in 3D
[
Ex Yxy/2 YXZ/2]
[EhD = YyxJ2 Ey Yyz/2
Y2xJ2 Yzy/2 Ez
Relationship Between Elastic Constants
E=2G(1 +J.!)
E=3K(1-2J.!)
9KG
E= 3K+G
E 1
G=-x--
2 1+J.!
E 1
K=-x--
3 1-2J.!
Value of any Ee ~ 0
Note: J.!cork= 0
Young's Modulus or Modulus of Elasticity
As per Hooke's law upto proportional limit normal 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 = 200 GPa
AI 3
.. (oI)MS <(ol)CI <(0)Ai
Elastic Limit: Maximum value of stress upto which a material
can be completely elastic.
ProportionalLimit
It is the maximum value of stress upto which materials obey
Hooke's Law.
Shear strain like complementary shear stress are equal in
magnitude but opposite in direction.
[EhD = [Ex YXY/2]
Yyx/2 Ey
•
A-7Engineering Mechanics and Strength of Materials
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![Let, L = Length of the bar
F
where,
E = Young's modulus of elasticity
A= Area of cross - section
L = length ofwire
I= increase in length ofwire
Extension ofa tapered bar:
Let us consider a circular bar whose diameters are DJ and D2as
shownin figure. Let'F'be thetensile loadwhichis appliedaxially.
Uy = _1_(Stress)2
2E
or
EAI
where F=---, L
Energy stored/unit volume ofwire
u, =..!.E(Strain)2
2
1
Energy stored in wire, (U) = "2FI
_ h + J.l E2) _ (E2+ J.l Ed
al - E 2' a2 - E 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
El = E [al - J.l(a2 + (3)]
E2 = ~ [a2 - J.l(a3 + ad]
1
E3 = E [a3 - J.l(a1 + a2 )]
.. for biaxial state of stress/plane stress problems a3 = 0 but
E3 *- 0
al = E(El) + J.l a2
a2 = E E2 + J.l a 1
or
Relationship Between Principal Stress and
Principal Strain
(a) 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.
(b) 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.
(c) Ductility: It is the property of the material due to which
if can be drawn into thin wires. The length ofdeformation
is very large in a ductile materiaL
(d) 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.
(e) 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.
(f) 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.
(g) Hardness: It is the property of the material due to which
it resists cutting, scractehing, pinetration or inditation.
PROPERTIES OF MATERIALS
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.
Eng. stress vis Eng. strain curve MS under tension test
~--------------------------------~~
Engineering Mechanics and Strength ofMaterials
F
o
A-8
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![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.
THIN CYLINDERS
(e) Continuous beams: In continuous beams, there are more
than two supports.
II
(d) Fixed beam: In fixed beam, both of its ends are rigidly fixed
into the supporting walls.
I
(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.
Simply supported beam: A simply supported beam has
both of its ends are supported.
(b)
L
~
~
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.
8L = FILl + F2L2 + F3L3
Al EI A2E2 A3E3
Ifthe bars are of same material, then EI = E2= E3= E,
then,
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 = crILl +cr2L2 +cr3L3
Now, E E E
1 2 3
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 = FI= F2= F3
Let, 8L = total change in length
IE
F~ A" E, IA2, E,
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.
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
2
8L = wL = Pg L2
crE 6E
8L = wL
2AE
Elongation of a bar due to self weight:
Case (i) For uniform cross - section:
=> Ifbar is of uniform diameter 'D', then,
~
E = Young's modulus of elasticity
Extension of tapered bar (8L),
8L = 4FL
1t E DID2
A-9Engineering Mechanics and Strength of Materials
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![Beams
I
Statically Statically
Determinate Beam Indeterminate Beam
I
~J J t J lCantilever Simply Over Fixed Propped Continuous
Supported Hanging Beam Cantilever Beam
Beam Beam Beam
X
I
I
Pil
I
ILp i
1+ve shear force0
X
I
I
I
Ii P 1i-ve shear force
pLI GI
I
I
Bending Moment Sign Convention
X
I
C] D Ci)I
I I
I
concave I
X upwards
+ve bending
(+ve)
moment
SAGGING
BENDING
X
I
CJ D Ci)I
I
I I
X HOGGING
convex -ve bending
upwards
-ve
HOGGING
BENDING
SHEAR FORCE AND BENDING MOMENT DIAGRAMS
~ 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
Moreover,it can be seen from expressionsof Ehoopand Elongthat
Elong < Ehoop
:. The chances offailure ofthin cylinderismore longitudinally.
pd pd
along = -4--' ahoop =
t 11eJ 2t 11u
pd pd
.. a 1 = 21' a2= 4t
al pd
Absolute ~max = 2= 4t
_ 8d _ pd (2 _ )
Ehoop - d - 4tE ~
- 6L _ ~(1 - 2 )
Elong - L - 4tE ~
6V pd
Ev = -y= 4tE(5-4~)
STATEOF STRESS ATA POINT
IN THIN CYLINDER
pd pd
along = 4t' ahoop = 2t
Sometimes11 ofcircumferentialjointand longitudinaljointaregiven.
In that case,
-----If---~ (jlong
z
,1 __ -
y
)-x
Example of Thin Cylinder:
HydraulicCylinder.
Example of Thick Cylinder:
LPG Cylinder, Steam Pipes.
Assumptions for Thin Cylindrical Vessels
~ Stresses are assumed to be uniformly distributed as
thickness 't' is small.
~ Radial stresses are neglected.
Spherical
Pressure
Vessel
Cylindrical
Pressure
Vessel
on the basis of
shape of shell
Thin d : diameter
d t: thickness
-> 20
t
Thick
d
-~ 20
t
Engineering Mechanicsand Strength ofMaterials
Pressure vessel
A-tO
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![PAy
't=---
INA' b
Shear Stresses in Beams
D ~ diameter of log (given)
·. final dimensions of strongest rectangular cross-section
are
d
bx = b[~]
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
Zl1 = Z22 = Zxx
Ml1 = M22 = Mxx
(crb)ll = (crbb = (crb)xx
·. (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.
d = d Ixx ~L
depth should be varied parabolically.
(ii) width 'b' is varied
(+ve) 11b
x
Beams of Uniform Strength
A beam is said to be a beam of uniform strength when bending
stress developed at each and every cross-section is same.
(crb)max Y
(crb) = y .
max
A beam offering higher moment of resistance is stronger.
I-section beams are strongest as they have high section
modulus.
Fora giveneross-sectionalareaandmaterialsquarecross-section
beam is strongerthan circularcross-sectionbeam as
Zgquare > Zcircle
Thisfibre is
subjected
to tension
NeutralAxisis
neitherin tension
nor in compression
This fibreis
subjected
to compression
M
(crb)max = ±--
ZNA
. . ~A t => (crb)max "l.. => chances of failure "l..
For a given beam, (crb) ex y
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
NeutralAxis.
From bending eq",
crb = i ~to be used when 'R' is known.
As BM = const
above beam is under pure bending.
Bending Equation
I -ve
ML..--------l
P-0~~I I
MA=:P(CD) M ~ P(CD)
I I
I
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-l1Engineering Mechanics and Strength of Materials
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![1
f= a2
~ (end fixity coefficient)
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]
2
1t E Imin
.. Pe = 2
Le
Pe : Euler's buckling load.
Imin : min [Ixx and Iyy].
L, : effective length of column.
L : actual length of column.
L =aL
e 4length fixity coefficient
are more.
Euler's Formulae
Short Columns
(fail due to crushing)
Long Columns
(fail due to buckling)
Medium Columns
(fail due to buckling
as well as crushing)
~ As the length of structure, chances of it failing by buckling
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.
THEORY OF COLUMNS
= -3.6.
T1 = TA' T2 = TA- T.
TA+Tc=T.
91 + 92 = <1>0
=> 91 = (-92)
3T
=> TA= 4·
G1 J1 = G2 J2
1. Net TM = T (anti-clock)
Rxn = T (clock)
1'.
CD
T=T1+T2
9
=> T = (G1 J1 + G2 J2) L
91 = 92 = 9
Shafts with Both Ends Fixed
CD
T
Shafts in Parallel
Shafts in Series
~ Torsional Stiffness (q) 2.
T GJ 3.
q=-=-
9 L 4.
~ Torsion of a Tapered Shaft
A-15Engineering Mechanics and Strength of Materials
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![1
61 = E [ al - Y (a2 + (3)]
1
62 = E [ a2 - Y (al +(3)]
1
63 = E [a3 -y (al +(2)]
Now,According to failure creterion,
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 = maximumprincipal stress
a2 = minimum principal stress
a = yield stressy
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~
Let, 6)' 62, 63 = three principle strains
6 = strain at yieldingy
61,2,3::;6y
Now,
A-IS Engineering Mechanics and Strength of Materials
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![42. 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
(c)
(b) Lx Al.
2L
(d) ~L
L
~L
~L
L
(a)
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 :
Dr-D~
(d) 4Dl(c)
(a)
IfDI = external diameter ofshort column
D2= internal diameter of short column
F = external load applied
The highest value of eccentricity will be equal to :
Df-D~
(b) 8D1
39.
(c) PP (l-!':.) (d) PP(I-!':. )2tE 2 3tE 2
38. If the both the ends of the column made of mild steel are
hinged, then Rankin's constant value will be equal to :
1 1
(a) -- (b) --
7500 6500
1 1
(c) -- (d) --
5500 4500
(c) ~('L~)2 +('LFv )2+2('LFH)('LFv)
(d) ~('LFH)2 +('LFy )2 -2('L~)('LFv)
36. In the above question, the direction or angled of the
resultant will be give by:
LFy(a) tana=--
Lftl
(b) tana=Lftl
LFy
(c) tan a= 2)H x2)v
(d) tana=2 x LFH x~:)v
37. IfD = diameter ofthin cylindrical shell
L=length
t = thickness
~ = internal pressure, J..l= Poisson's ratio
then Hoop strain will be given by:
(a) PP(2-!) (b) PiD(1_!:)
2m 3J..l 3m 3
~
(b) ~YF:
33. A cyelindrical elastic bodysubjected 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. When number of forces are acting on the body and 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;
Longitudinal strain Lateral strain
(a) Lateral strain (b)
Longitudinal strain
stress strain
(c) strain (d) stress
32. Poisson's ratio is described as the ratio of:
29. If oI' cr2,cr3are three principal stresses, J..l= Poisson's ratio
and E = Young'smodulus of elasticity,then the expression
for strain energy/volume is given by
(a) ![crt + cr~+ cr~+ J..l(crlcr2+ cr2cr3+ cr3crd]
(b) _!_[crt + cr~+ cr~- J..l(crlcr2+ cr2cr3+ cr3crt)]
E
(c) 2~ [crt + cr~+ cr~- 2J..l(crlcr2+ cr2cr3+ cr3crd]
(d) _1_[crt +cr~ +crj +2J..l(crlcr2+cr2cr3+cr3crd]
2E
30. Impact strength ofa material represents:
(a) Hardness (b) Resilience
(c) Ductility (d) Toughness
31. 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:
I
(a) IE ="2 (b) IE=21
I
(d) IE="4
the body is given as :
'[2 '[
(a) -xV (b) -xV
2G 2G
'[2 r
(c) -xV (d) -xV
G G
A-21Engineering Mechanics and Strength of Materials
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![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
4ML2
(d) EI(c)
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 cross-
section about the neutral axis is I and the Young's modulus
is E. The magnitude ofthe maximum deflection is
ML2 ML2
(a) 2EI (b) EI
(c)
2a(LlT)E
(l-2v)
a(LlT)E
3(l-2v)
(b)
(d)
a(LlT)E
(l-2v)
3a(LlT)E
(1- 2v)
(a)
194. A solid circular shaft of diameter d is subjected to a
combined bending moment M and torque T. The material
property to be used for designing the shaft using the relation
.!.i_.JM2 + T2 is
nd3
(a) ultimate tensile strength (Su)
(b) tensile yield strength (Sy)
(c) torsional yield strength (Ssy)
(d) endurance strength (Se)
195. 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
196. 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
197. 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
198. 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
199. 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
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
P P
1+-[; t 2L .; L--+I
L ~p2J3 2p2L3
(a) (b) --
3EI 3EI
4p2J3 8p2JJ
(c) -- (d)
3EI 3EI
i q q
illllllllll]l M(illllllilit
1II1II L IJIII R, R2
5qL 3qL qL2
(a) s, = -8-' R2= -8-' M= 8
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
191. The strain energy stored in the beam with flexural rigidity EI
and loaded as shown in the figure is
(c) tid' (d) nd'
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) so
(c) 100 (d) 110
190. A uniformly loaded propped cantilever beam and its free
body diagrams are shown below. The reactions are
8T16T
(b)(a)
188. For a circular shaft of diameter d subjected to torque T, the
maximum value of the shear stress is
A-29Engineering Mechanics and Strength of Materials
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![(b) Yxy/Ji
(d) 2yxy
(a) Yxy
(c) Yxy/2
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) o = EEl (b)
E
o = --2 [El +J.lE2]
I-J.l
(c)
E
(d) o = E[EI - J.lE2]o = --2 [El- J.lE2]
1- J.l
o,
where, k = 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 o1
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.
(d)
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
(c)
]_~M2 +T2
2
~IM+~M2 +T2l
(a)
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.
(d)(c)
(b)
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 :
p2L3
(a) 3EI
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
(c)
9KG
E= K+G
9KG
E = 3K+G
(b)
(d)
KG
E = 9K+G
9KG
E = K+3G
(a)
215. A gun metal sleeveis fixed securelyto a steel shaft and the
compound shaft is subjected to a torque. If the torque on
the sleeveis twicethat onthe shaft, findthe ratio ofexternal
diameter of sleeveto diameter ofshaft [GivenNs= 2.5 NG]
(a) 2.8 (b) 1.6
(c) 0.8 (d) 3.2
216. A column sectionas 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/ m2and 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) 12t/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?
A-31Engineering Mechanics and Strength of Materials
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![= 100-40 = 60 = 30MPa
2 2
a -a
Maximum shear stress ('t"max) = x y
2
I .
where 2 = -_ = section modulus
, Ymax
M = ab(max) x Z
M
ab(max) = Z
1
ab(max) o: Z
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]= 1000X 10-6
62=-600 X 10-6
Maximum shear strain (ymax) = 6] - 62
= (1000 x 10-6)-(-600 x 10-6)
= 1000x 1Q-6+ 600 x 1Q-6
= 1600x lQ-6
As we know that,
E=2C (1+ u)
~=2(1+Ji)
Given: Principal stresses,
a = 100MPa, a =40 MPax y
For plane stress system,
QJA 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.
LlL
Hence,6=T
Fatigue is a fracturephenomenon in which material
is failed due to cyclic or repeated stresses usually
at low values of stress.
As we known that,
I
M = abc (max)x--
Ymax
66. (a)
65. (b)
64. (a)
61. (a)
53. (a)
50. (a)
49. (b)
41. (c)
33. (c)
30. (d)
.c=_wL2 _ wLNow,Total elongation - -- - -- - --
o E 2E 2AE
Here, w = WAL.
14. (a) Considering Principal of superposition,
8L=8L]+8L2 +8L3
PLI PL2 PL3
=--+--+--
EAI EA2 EA3
= !(.!i_+~+ L3JE Al A2 A3
Toughness of the material is defined as the
maximumamountofenergystoredin a materialupto
fracture under the application of impact loads.
8L= wy8y
E
Let, w = specific weight of bar
F=wAy
elongation(8L) = _FL_= ....:....(w_A_y.....:...)_.d_y
AE AE
AE L
P 8L
S
. () Stress(a)
tram 6 =---
E
I
. PL
E ongation 8L = -
AE
13. (d) Let a bar be hanging of weight 'w'
Load P
Stress(a) = --------
Area of cross- section A
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,
. 8L
Stram=-
L
U = '!_E(8L)2
v 2 L
where, E = Young's modulus of elasticity.
12. (b) As, we know that,
...,HINTS & EXPLANATIONSI···~
A-33Engineering Mechanics and Strength of Materials
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![ByFBD ofBook I,
LFy=O=> NI =mg
So,frictionalforce= flN 1 = fl mg
ByFBD of Book2,
LFy=0=>N2=N1 +mg=2mg
LFx=O
mg
FBD of Book 1
181. (d)
3.14[(10)4 -(5)4]x2
32 x10
= 183.98cm3
= 184cm3
175. (c) Considering equation,
T _ Ge _ Q
J L R
T=628 kg-em
~N/m
IE 1m )1
M
· b di (0.4)(1)2
axlmum en mg moment = ---
8
=0.05N-m
174. (a) Givenexternal diameter(D) = 10em
Internal diameter(D) = 5em
. n[(Do)4 -(Di)4Jx2
Sectional modulus (z) = 32
-.o,
(PC)Ll 4n2EI 4L2 16
--=--x--=-
(pc )L2 L2 n2EI 1
(PC)L1 : (PC)L2=16:1
172. (a)
pf'f;lOKN~: 6~ JR 9m R
p Q
RP + RQ= 10 x 6= 60 KN
Takingmoment about 'P',
~ x 0+ 9 x R = 180
RQ=20KN
~=60-20=40KN
173. (a)
For long column, (One end fixed and other end is free)
2
(p) _n EI
C L2 - 4L2
(SF)R= 6KN=w x 1.5
w=_i_=4KN
1.5
171. (d) For long column, (fixed at both ends)
2
(p) = 4n EI
C LJ L2
10mm 10mm
lOT~lOT
lO-3=7T~7T
9T~T
8L = lOx 10 8L = 7 x 10 8L = 9 x 10
1 AE' 2 AE' 3 AE
8L=8L1+8L2+8L3
100 70 90
=-+-+-
AE AE AE
260 26 x 10
=-=--mm
AE AE
167. (c) Given:A=3 em-
P= 12000N
Esteel= 3 x EAJ
Considering the relations,
P=P1+P2
P=P1A1 + P2~
weget, Psteel= 30N/mm2
170. (c) 1.5 m R
IE 10mm )IE )IE )1
= ~ x 16x 10-6 = 4n x 10---{jm'
4
Load = weight (w)= mg
= 100x 9.8= 980N
w 980
Stress(cr) = - = 6
A 4n x 10-
=78.03 X 106
= 78 MPa (approx.)
157. (d) Given,E= 200GPa
G=80Gpa
fl=?
Weknow that,
E=2G(1+fl)
200=2 x 80(1 + u)
(1+ ) = 200
fl 160
u= 1.25-1 =0.25
166. (a) Given,
lOT" I.____ __L...~ __ ~__L... __ __,I ,9T
L fTITIl !
A-35Engineering Mechanics and Strength ofMaterials
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![3
.. A = 1x 10 = 346.4
2.886
P=5MPa,d= 1000mm,t= 10mm
We know,
ex = _!_[crx- v(cry + crz)]
E
20 x(10)3 = 2.886
12x 10x 20
'Tmax=
cry = 100 MPa
0" = - 200 MPa
t"y= 100 MPa
Ifmaximumandminimumprincipal stressesaregiven,
the maximum shear stress is given by
'T = crl-cr2 = 100-40 = 30MPa
max 2 2
Factor of safety
Torsional yield strength
or
8
For safe designing 'T::;~
n
16T 16~M2 + T2
Induced shear stress is 'T- - - -----
- nd3 - nd3
Fora shaft subjectedtobending moment M and torque
T, the equivalent torque is
T =~M2+T2e
F = transverse shear load
'T= transverse shear stress
Here, shear stress 'Tis variable and is maximum at the
neutral axis.
198. (b)
199. (a)
196. (c)
195. (c)
194. (c)
= ~(-150)2 +(100)2 = 180.27MPa
L ( H'197 (b: 81 d ti 'I l':1=AK2,K = A)• 'J en erness ra 10 I, = K
Moment of'Inertia forrectangular section
bd3
1=-
12
Then K= {I= Jbd3
fA 12xA
3 3 F
where, 'Tmax= "'2'Tmean="'2b x h
FAy
'T=--
I.b
t~t-------ty
t- ------- -------
hl2
t L---_
8 = PL = 4PL
AE nD2E
For, finding the change in diameter (transverse
direction), Poisson's ratio v is needed.
But modulus ofelasticity E is also needed.
Now, E= 2G(1 +v) fromwhich 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
p
TL
1
.. RB=P
BMx=P x X
The total strain energy stored is given by
f
L(Px)2 xdx (pL)2 x2 JL(Px)2 x dx
u= + +
o 2EI 2EI 0 2EI
4p2L3
u=--3EI
192. (d) When the load P is applied in axial or longitudinal
direction, increase in length
4PL
P> 3L+ pX-RA x4L=0 => RA = --= P
4L
RA+ RB=2P
:EMB=O
PL PL
o~o
Bending moment diagram
D~L=tRa =P
p
2L--i B
191. (c)
A-37Engineering Mechanics and Strength of Materials
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![«(')}(')2Plane)and «(')}-(')2) = diameter of Mohr's
circle.
(')} - (')2
Maximum shear stress (r ) =max 2
1
From the above expression, r a T and r a -3max r max d
(d)Tr· -
T. = Tr.R =__ 2_
max J ( 4)
l1[~j
225. (a)
p2 [x3]L p2 L3
= 2E1 3 a = 2EI x]
p2L3
---
6E1
224. (d) From torsion equation,
Tr GQ= (')max
J L R
J
L(Px)2 p2 fL
= --dx = - x2dx
a 2E1 2E1 a
2
Strain energy due to bending (U) = fL M dx
o 2EI
. 1
Stram energy of bar = "2P8
= ~ x 30 x103 x 0.05 x10-3 = 0.75 N-m
= 700 X 10-6 - (-400 - 10-6)
= (700 + 400) X 10-6 = 1100 x 10-6
EIn2
Maximum! Critical load (F) = -2-
Le
So, critical/maximum load a Moment of inertia of
section
While, we know that M.O'! of thin hollow circular
sectionis maximum.
223. (b)
222. (a)
221. (b)
(')1 = (')2 = r = 'tl = 't2
So, it is the state of pure shear
219. (a) Maximum shear strain, (r) = EI - EI
= 2 x 1.5651 x 1/5 =
~(D4 _d4)
32 =5
~d4
32
D = 1.5651
d
f,g ('T)g T, D J,
r,:-= (T;)s =Tg·d·T.
0.62604
f
~=1.6
fss
Stressat comer B= 15+5- 8= 12t 1m2 (compressive)
As from the condition given,
216. (b)
217. (c)
Since both the steel shaft and gun metal slave are
e
securely fixed, L is the same for both.
l=l
JsNs JgNg
Jg Tg Ns
-=-·-=2x2.5=5
r, Ts Ng
L IN
U2 p2X2 •dx [2Xp2 •x3 ]LI2
:. W;= 2 [ 2x4EI = 2x4EIx3
Since,P = 1unit
r:w=--1 96EI
215. (b) Let us use suffix Sfor steel and suffixg for gun metal.
e T
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 surfaceforming
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 = J where M = - X
1 2EI' 2o
A-39Engineering Mechanics and Strength of Materials
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![B
~
P~ Q
I I
cr =1
On arranging the above equns.
E(El+~E2) E(E2+~El)
1_~2 ,cr2 = 1_~2
1
Are we know, El = E[cr1 -1·.ta2]
1
E2 = E[cr2 - ~crd
230. (b)'fsolid
THollow
~(D4 -D~)
32 0 I
THollow
228. (b)
227. (b)
Maximum principal strain = Yxy / 2
Engineering Mechanics and Strength ofMaterials
229. (c). . Tr GQ tmax
From torrsion equation, T = L = R
tmax should be equal for both shafter.
(;) solid = (;) Hollow
A-40
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![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
e = angle of contact in radians
Material used for flat belt is generally leather of various
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.
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.
I
Driving
PulleyI
Driver
Pulley
Fig.: Belt drive-open system
Driving
Pulley
Slack Side T2
1
T = "2x 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.
T - 2 w[ri -d]--XJ.! ---
3 rf-d
while 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
R = r1+ r2 is the mean radius of the bearing.
2
(iv) Frictional torque transmitted in a flat collar bearing is given
by
where
a = semi angle of the cone
(iii) Frictional torque trnsmitted in a trapezoidal or truncated
conical pivot bearing is given by
T = ~ x J.!W [ r ~ - r ~] cosec a
3 r1 -r2
while considering uniform pressure.
And in case ofuniform wear
I
T= - x J.!WR cosec a
2
2
T= - x J.!WR cosec a
3
while considering uniform pressure
And in case of uniform wear
1
T= -xJ.!WR
2
where J.!= Coefficient offriction
W = Load transmitted to the bearing
R = Radius of the shaft
(ii) Frictional torque transmitted in a Conical Pivot bearing is
given by
T = ~ x J.!WR while considering uniform pressure
3
And in case of uniform wear
Theory ofMachines and Machine DesignA-44
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![0)2 = N2 = dl + t (1- ~J0)) N) d2 + t 100
where dJ, d2 = diameters of driver and driven pulleys
0)1, ffi2 = angular velocities ofdriver and driven pulleys
NI,N2 = rotational speeds of driver and driven pulleys
expressedin revoluationsper 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
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.
Types of Gears
Commonly used gear are as below:
(a) Spur gear:Acylindricalgearwhosetoothtracesare straight
lines parallel to the gear axis. These are used for
where rl and r2are 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
L = 1t (rI + r2)+ I 2 + 2x
x
When the belt passesfromthe slack sideto 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.
Centrifugal Tension
The centrifugal tension (Tc)is given by
T, = mV2
where m = Mass per unit length of the belt
V = Linear velocityofthe 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 equaltothevalueofpowertransmitted givenbyeffective
turning force (TI - T2), that is the centrifugal tension has no
effecton 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 isthree
times the centrifugal tension.
T = 3Tc
=> T = 3mV2
Sovelocityforthe maximum powertransmitted is givenby
v > )3:
Velocity Ratio
The velocityratio of speeds ofdriver and driven pulleys is given
by
I
Fig. Circular Belt
The ratio of driving tensions in round belts and rope drive
is sameas V-beltdrive.
Length of Belt
In an open belt drive system the length of the belt is given by
(rl - r2)2
L = 1t (r. + r2)+ + 2x
x
Fig. V-belt
The ratio of driving tension forthe V-beltdrive is given by
.!L = e(f,.LCOSeC 13)e
T2
c> 2.3 log GJ= 11-9- cosec13
where J3 = Semi-angle ofthe groove
e = Angle of contact in radians
V-beltsare usually made ofcotton fabric, cards and rubber.
(c) Circular belts:The crosssectionof 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 ofthe clothing. Round belts are made ofleather,
canvas and rubber.
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.
(b) V-belts: Fig. shows the cross section of the V-belts. V-
belts are available in fivesections designedA, 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
ofbelts break, the entire set should be replaced at the same
time. The groove angle in the pulley forrunning the belt is
between 400to 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.
A-45Theory of Machines and Machine Design
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![... (2)
2E
I
1 2
E = "2ICOmean
... (1)Lllimax = lco~eanc,
also energy stored is a flywheel is given by
= 1(,) (comax- comin)x co
v-mean mean
COmean
Relation between maximum fluctution of energy .1Emaxand
coefficient of fluctuation of speed.
Lllimax = lCOmean(COmax - COmiu)
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 = .1comax
s
COmean
where
C - LlEmax
energy-
Wpercyc1e
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. Itabsorbs
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
FLYWHEEL
Fig. : Epicyclic gear train
. . NB TA
Velocity RatIo - = 1+-
Nc TB
ArmC
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.
4. 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.
D1 + D2 = D3 + D4
2 2
=> D1 + D2 = D3 + D4
. . N1 T2 x T4
Velocity rano = - =--=-_....:....
N4 T1x T3
Fig. : Compound gear train
N N N N T xT xT
Velocity ratio = _I = _1 X _3 x _5 = 2 4 6
N6 N2 N4 N6 T1x T3 x T5
3. Reverted gear train: Areverted gear train manifests when
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
Driven or
follower
Train value = ~ =IL
NI T4
2. Compound gear train: A compound gear train includes
two gears mounted on the same shaft as shown in Fig.
Driver
Theory ofMachines and Machine DesignA-48
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![where
where
=>
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
t = tl +t2
P = (tl + t2) x I x at
I = length of weld = width of plates
t1 = throat thickness at the top
(ii)
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
pp
r
)
(
>
)
(
1'] >
)
1
?) >
)
:>
Double parallel fillet weld
(b) 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.
pp
.--fI.~f~s_-r---lrp
P..-.1 -.J sf.-
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
pp
Ptransverse= 1.414 s x I x at
where s = Leg or size of weld
I = Length of the weld
at = Allowable tensile stress
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
where,
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.
Minimum of Pc' Pt and Ps
11=
P x t x at
P = Pitch of the rivets
t = Plate thickness
A-55Theory of Machines and Machine Design
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![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
where
_!_ (M + ~M2 + T2) = nd3 O"b(max)
2 32
nd3
Me= --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 byusing support bearings.
If gear is mounted on the shaft then
Now,
According to maximum normal stress theory (Rankine's
theory the maximum normal stress in the shaft is given by
1 1 ~ 2 2
O"b(max)= - O"b + - (O"b) + 4~
2 2
,,~_)~ ::3[~(M+ M + T2)]
=>
~
nd3
M2+T2 = __ ~
16 max
nd3
r, = 16~max
where T, =equivalent twisting moment = ~ M2+ T2
32M
where do = outer diameter of shaft in m
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
n 3
T - -x~xdo
- 16=>
T = torsional moment in N-m
For a hollow shaft
~= 16Txdo N/m2
( 4 4) ,n do - dj
d = shaft diameter in mwhere
SHAFTS
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 sizebyturning, 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.
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
r = 16; N/m2for solid shaft
nd
w=~
2~
w = width of the key
t = thickness of the key
O"c= permissible crushing stress
r = permissible shearing stress
where
Combined loading: When a shaft is subjectedto combined
twisting moment and bending moment, then the shaft is
designed on the basis of maximum normal stress theory
and maximum shear stress theroyand 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
~max = - J (o.,) + 4~
2
= _!_ 32M + 4 x (16 TJ2
2 nd3 nd3
r = ~ 'M2+T2
max nd3 j
(iii)
A-57Theory ofMachines and Machine Design
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![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 = mrw-
m = 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 w-
16 16
T = ~ /-!W [rf -d ]cosec a
3 rf-d
(b) Cone clutch: Total torque transmitted in the cone clutch is
given by
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
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= W
n(r?-d)
dr
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.
(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
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 driveforexamplejaw or clawclutch.
CLUTCHES
[
6 ]1/3P = Cx _!.Q_
60NL
or
IfN is r.p.m. the Life in hours is given by
(
CJ3 106
L= - x-- hours
P 60N
Variation of coefficientoffriction with the
bearing characteristic number (z:J
Rolling Contact Bearings
Bearing which operate on the basis ofprinciple ofrolling, 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%
failureis known asdynamicloadrating orbasicdynamiccapacity
or specific dynamic capacity.
Rating Life L ~ (~ J'
where P = load
C = dynamic basic load rating
(
IJ1/3P= C -
L
Theory ofMachines and Machine DesignA-60
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![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 :
Scooters
Rolling mills
Trucks
Mopeds
1.
2.
3.
4.
List - II
2flW(R3 +r3)
(d) 3sina(R2+r2)
2sina
flw(R+r)
(b)
Codes:
A B C D
(a) 3 4 2
(b) 1 3 2 4
(c) 3 2 4
(d) 3 4 2
2flW(R3 _r3)
(c) 3sina(R 2 _r2)
2cosa
234. Match List- I and List - II and select the correct answer
using the codes given below:
List - I
A Single plate clutch
B. Multi plate clutch
C. Centrifugal clutch
D. Jaw clutch
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:
4 uw [R~ -Rf] 3 uw [R~ -Rf]
(a) 3 cosu R~ -R? (b) 4 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 :
1 [R~-R?]() -flW
a 2 R2 -RI
2 [R~-R?] 2 [R~+R?]
(c) 3flW R2-RI (d) 3flW R2-RI
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 assuminguniform
pressure theory?
flw(R-r)
(a)
4 cote4 tan O
2 sine
IflWR
(d)
IflWR
(b)
IflWR
(a) 4 cosO
IflWR
(c)
229. If e = semi-cone angle, then in a conical pivot bearing
under uniform wear, the frctional torque transmitted will
be:
(d) 2e-<j)
2
K - Kr t;
(a) d - J2 (b) Kd = /2Kr
(c) Kd = 2~ (d) Kd = 4 x,
227. An external gear with 60 teeth meshes with a pinion of20
teeth, having module being 6mm, 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= <j)+e A <j)-e
(a) p (b) p=-
2 2
e-<j)
(c) ~=-
2
(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:
TI+2Tc + T2 TI+4Tc + T2
(c) 4 (d) 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
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 T1 = 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
Theory ofMachines and Machine DesignA-72
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![mh2
.. m1 = h + h ...(iv)
I 2
From equations (iii) and (iv)
mk2 mh2
hi(hi+h.) (hi +h.)
k2 = h1h2
m. = h,(h, +h2) ...(iii)
From the equations (i) and (ii) we get
mh
m+-I_I=m
I h
2
N2 = 300 rpm
N2 = T, =.!.
N, T2 3
..:5_ = .!.T = 24
T 3 I
2
Pitch line velocity = W1r1or W2r2
d2 8x72
= 21tN2- = 21tx300x--
2 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
C - (wmax - wmin)
S - (ro~ ;ro~")
or, Cs Wmax + CSWmin= 2 Wmax - 2wmin
wmax = 2+ Cs
wmin 2-Cs
28. (c) Static balance is a balance of forces due to the action
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.
29. (c) For dynamically equivalent
m1 + m2 = m (i)
m.h, = m2h2 (ii)
m.h] + m2h; = mk ' (iii)
From the equations (ii) and (iii); we get
m,h~ +(m,h,)xh2 =mk2
mk2
25. (c) T2 = 72
1
VR= -
3
21tN 2 x 1tx 160
W = -- = 16.75 rad/s
60 60
maximum velocity of sliding = W x d
= 16.75 x 0.018 = 0.302 mls
24. (a)
Arc of contact
So, No. of pair of teeth in contact = C· I . h
ircu ar pitc
31.4
= ----s;- = 2 pairs.
fp = rw2 (COS8+1CO:28)
At IDC 8 = 0
.. r. ~rCO>(l+~)
At ODC 8 = 1800
fp ~ -rCO> (l-~)
22. (c) 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
23. (c) Arc of contact = 31.4 mm
Module (m) = 5
Circular pitch = 1tm = 51t
21. (a)
W2 = WI+ at = O. + 5.59 x 12= 67.08 rad/s
1 1
KE = -mk2w2 = -x800x(0.36)2 X (67.08)2
2 2
= 233270N = 233.3 kJ
20. (a)
19. (c)
18. (d) Fp = Primary unbalanced force = mror'cos"
Fs = Secondary unbalanced force
mrci' (I)=-n-cos28 n =~
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.
a = _!__ = 580 = 5.59 rad/s '
mk" 800 x (0.36)2
Theory of Machines and Machine DesignA-76
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![1 5
Speed ratio = T· 1 3
ram va ue
44. (d) The centre of gravity of the coupler link in a 4 bar
mechanism would experience both linear and angular
accelerations.
45. (a) 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
46. (a) 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.
47. (b) 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
CD 45 3 3
Vc = VBx-= V =-= Vx-=-V
AB 30 2 2
. 3
:. Velocity of C = '2 V .
48. (c) The transmission angle is maximum when crank angle
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°.
49. (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
Tfirst 30 3
43. (a) Train values = T= 50 = 5"last
1= 0.1 kg-rrr'
42. (a) We know that
llE = ~ l( co~- con => 400 = ~ x [(210)2 -(190)2 ]
=> 400 = 4001
tan 30°
=26mm
15
Ijj
15.. Circular pitch
AXIal pitch = tan 300
.. 2k
=> 9 + -9 = 0
3m
OJ" = 2'"J!.40. (d) In toggle position, for a four bar linkage, the
mechanical advantage will be infinity.
41. (d) <I> = 30°
Normal circular pitch = circular pitch x cos<I>
= 15 x cos 30° = 13 mm
(
1 2 2J=> '2mr -i mr 9 + k (9r2) = 0
.. kr'
=> 9+--9=0
3 2
-mr
2
Taking moments about instantaneous centre' A'
I, e+ (kx) r = 0
=> (10 + m2)e+ kx (Or) r = 0
39. (c)
oicos O
If n is large COer= --
n
Angular velocity is maximum at 9 = 0, 180°
Angular velocity is zero at 9 = 90°
37. (c)
1
= '2l( COmax + comin) (comax- comin) = 1coay x Cs X coay
emax = lco;y Cs
From the figure, it shows that the value of'P' required
will be minimum when it is at 452 to the horizontal.
This can be solved by resolution of forces.
mT 40x12
R=-=-- = 240mm
2 2
mt 20x12
r=-=--
2 2
. r sin <I> = IR2_ R 2cos" '" - R sin '".. 2 'V a 'I' 'I'
120x;in20 ~R; -(240cos20)2 -240sin20
R, = 248 mm
addendum = 248 - 240 = 8 mm
ro cos O
36. (c)
35. (d)
30. (a) Since cylinder falls freely under effect of gravity, it
follows basic law of motion and
v2 = 2gh and v = ~2gh
31. (a) Ratio of angular speeds of F to A
TA. Tc .TE 20x30x 25
TB·To ·TF 60x80x75 24
32. (a) Point P being rigidly connected to point 3, will trace
same path as point 3, i.e. ellipse.
33. (c) A system of masses rotating in different parallel
planes is in dynamic balance ifthe resultant force and
the resultant couple are both to zero. This is known
as dynamic balancing.
34. (c) A bicycle remains stable in running through a bond
because of centrifugal action.
A-77Theory of Machines and Machine Design
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![6(6 -1)
NIC= 2 =3x5=15.
N(N -1)
Number of instantaneous centres, (N1C) = 2
where, N = Number oflinks
Hence,
(l = -r.I)
113. (c) Considering the following relation,
Given (N) = 6
v = rro[ sin00 + Sin;: 0) ] = roix O= °
v=O
(l = rro2[cos0" + cos:(O)]
.. 2 [ cos 2a]
Acceleration of piston (a) = rro cos a + -n-
Now, at inner dead centre, a = 0°, then
[
. sin 2a]
velocity of piston (v) = rto Sma + -2-
(a) Elliptical trammel
(b) Scotch - Yoke mechanism
(c) Oldham's coupling
109 . (d) Considering the following equations for a slider - crank
mechanism when the piston at inner dead centre,
hO
0.27= 1- 0.03 ho=0.0219mm
::::0.022mm
101. (b) Examples of inversion of double slider crank chain are
Let d = diameter of solid shaft
. T= Px60
.. 21tN
20xl03 x60 =955N-m
21tx 200
From torsion theory, we have torque transmitted by
solid shaft (T).
955x103 =~x~xd3
16
1t 3
=16x45xd d=47.6::::48mm
Given data
Eccentricityratio = 0.27 = 1- hO
c
99. (d)
C 50
orP= --1 =-=8.85K.N
(180)3 5.65
77. (c) In case of selflocking brake, no external force is required
for the braking action. This is not desirable condition
in normal application.
79. (b) It is proved by using theory of elasticity that the
theoretical stress concentration factor at the edge of
hole is given by I +2( ~)
87. (a) Allowable shear stress
~ = 2!L= 360 = 45 N/mm2
fos 8
6000 x 60 x 500
= 180:.(~r
L= 106 (C)360n P
Here, n = 500 r.p.m,
L= 6000 hours, C = 50 KN
73. (b)
Snap head rivet is used for boiler plates.
70. (d) Load stress factor
K = cr~ sin <P cos <P (_1_ +_1_J
1.4 Ep Eg
snap head
67. (a)
2x 16T = 140 ~ d= 15.4mm= 16mm
1td3
~
16T
Working shear stress = -3
1td
Allowable shear stress
Factor of safety = W ki h tror mg s ear s ess
61. (b)
A-79Theory ofMachines and Machine Design
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![di I . h (p) Number of teeth T
iametera PItC d = =-
diameter d
d
T
M d I ()
diameter
143 (a) 0 u e m =. Number of teeth
dp 100
Module ofthe gear (m) = Tp = 20= 5
dp + dg 200 => dp + dg = 400
2
dp + 3dp = 400 => 4dp = 400
dp = 400 = 100
4
...(i)
=.!.Q =0.2
50
140. (a) Given: distance between the axis = 200 mm
gear ratio = 3 : 1
Number ofteeth on pinion (Tp) = 20
Let dg = 3: 1
, dp
dg = ~ => dg = 3dp
dp 1
= 2x20 = 0.05
800
126. (a) Heightofthewalt governor (H)= lOCm=O.1 m
g= 10m/s2
H = g 2 => (Angular speed)2 =]_
(Angular speed) H
Angular speed = {10 = JlOO = 10 rad / s
~OJ
128. (a) Given: weight ofthe ball = 50 N
friction at the sluve = ION
. . frictio at sluve
coefficient of detention = -----
weight of ball
2(410-390)
(410+390)
_ 2 (112- 111)
- (112+ 111)
119. (d) Given: Mass moment of Inertia (Is) = 2500 kg-m-
Angular acceleration (a) = 0.6 rad/s-
Torque required (r)= Is x a
=2500 x 0.6= 1500N-m
120. (c) Given: Speed range of an engine,
111= 390 rpm, 112=410rpm
Coefficient of fluctuation = 112-111 = (112-111)
11m 112+ 111
2
_ (Nl + N2)
- 2
(Nl - N2)
_ (N1 -N2)
- 2(Nl - N2)
. . N m
then, coefficient of steadmess = ----=..:..:'--
N1-N2
w = 6.7981 = 6.7981 = 1.37 rad/s
e .J24.75 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
N, +N2
mean speed of flywheel (Nm) = ___:_-2--=-
7.85 x 0.866
.J25-0.25
= 7 85 cos30
th we . x
en, ~(5)2 _ sin2 30
h
r., 0.75
were, h=--=--=5
Lerank 0.15
= 7.85 rad/s
Considering the following formula,
3.14x2xO.15x500
60
3.14 x (2 x Lerank) x 500
60
ndN
Angular speed of crank (wemnk) = 60
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
116. (c)
Theory ofMachines and Machine DesignA-SO
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![w -4_c_=_l
0-4
wc-4=4
wc=4+4= 8radls
200. (a) Given: Moment of inertia (L)= 9.8 kg-m/
fluctuation speed (Nt) = 30 rpm
fluctuation energy (E) = 1936 j
D = 1-0.25 = 0.75
P
190. (d) 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,
Lf = (~rx 106
K = Constant = 3.3
72xl06 =e·2~OOOJ.3xl06
On solving, we get, w= 609.8 N (approx.)
197. (c) Given: Number oflinks (L) = 5
Number ofjoints (1)= 5
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 (TB)
Now, using the following relation,
wc-wA =_TB=_l
wB-wA r,
50 1 2
-=-=> 12 = 200N Imm
12 4
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,
P-D D
then TIt=0.25=--=1--, 'I P P
D = 1-0.35 = 0.65
P
177. (b) Given:Shearstrengthofrivet(11)=50N/mm2
diamenter of rivet (initial) (D1) = D
Final diamenter ofrivet (after doubling) (D2) = 2D
As we know that,
Shear strength of rivet IX (Diameterj-
1 IX (D)2
0.35 = P - D = 1- D
P P
44
22.5x1000=(T1-T2)X3
22.5 x 1000x 3 (
(T} - T2) = 44 1534.1N == 1540N approx)
168. (a) Inner radius (R)= 50 mm
Outer radius (RJ= 100 mm
axid force(w) =4 KN
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
11t=
PxtxFt
d T
mXPd =-x-= 1
T d
157. (c) Given: Average tension (Tavg.)l = 700 N
and (Tavg.h = 400 N
Linearvelocity(v) = 5 m1s
Power transmitted (P) = [(Tavg.)l - (Tavg.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-sl +S2JVelocity ratio - d2 100
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
=-m/s=-m/s
60 3
A-81Theory ofMachines and Machine Design
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![Availableand unavailable
energy in a cycle.
For a given T}, 11rev. will increase with the decrease of Tj. The
lowestpracticabletemperatureofheatrejectionis the temperature
of the surroundings, To.
u.E. = Ql- Wmax
T
For the given r. and T2, 11rev. = 1- T~
AE.+U.E.
AE. = Ql - U.E.or
Available Energy Referred to a Cycle.
The maximum workoutputobtainablefroma certainheat input in
a cyclicheat 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 ofthe energy
supplied.
High Grade Energy Low Grade Energy
(1) Mechanicalwork (1) Heat or thermalenergy
(2) Electricalenergy
(2) Heat derivedfromnuclear
fissionor :fusion
(3) Waterpower
(3) Heat derivedfromcombustion
of fossilfuels
(4) Windpower
(5) Kineticenergyof a jet
(6) Tidalpower
Available Energy
The sources of energy can be divided into two groups
(l) High grade energy
(2) Lowgrade 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.
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 ofsalt in water.
(iii) Rusting of iron.
(iv) Sudden expansion or contraction ofa gas.
AVAILABILITY AND REVERSIBILITY
Reversible Process
Anyprocesswhichcanbemade toproceedin thereversedirection
by variation in its conditions such that any change occurring in
anypart ofthedirectprocessisexactlyreversedinthecorresponding
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 approximatelyreversible.
(iii) Aworkingsubstancetakenalongthe completeCarnot's cycle.
(iv) The process of electrolysis is reversible if the resistance
offeredby the electrolyte is negligibly small.
A complete reversible process is an idealised concept as it
Dryness fraction (X) = My + ML .
where, My= mass of steam or vapour
ML= mass ofLiquid (saturated)
Drynessfractionis utilizedto calculatethe quantityofliquid
or vapour phase within the mixture. The value of dryness
fraction varies between 0 and 1. It also determines the
quality ofthe steam.
REVERSIBLE AND IRREVERSIBLE PROCESSES
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- TI = 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 ofmixture (water in saturated form in liquid
region as well as vapour region).
My
r, - T2 _ 1 T2---- --
11 11
Enthalpy: Enthalpy is regarded as the total energy of a
thermodynamic system.It is defined as the sum ofinternal
energy and product of pressure and volume. It is an
intensive property of the system. It is describe as:
H=U+PV
As
nR[ Tjfn( ~ )-T2fn(~)]
nRTjfn(~: )
A-87Thermal Engineerging
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0' r:--- r:--- ."f ~ ."f 00 0 ."f 0 ~ 0 0i '.c: ('f"] ....-: ~ ~ M 00
;;;.. 0 N 00 0 ~ '<:!'" M N ,......., ,......., ,......., t.n M M ......-< ,......., ,......., ,.......,00 0' ,.......,
N ,......., oq ~ ~
0 0 0
0 0 ,......., ,......., N M '<:!'" ~ 00 0 '<:!'" r:--- 0 N 0 0 r:--- M 00 M r:---
"0 0 0 0 0 0 0 0 0 0 ......-< ,......., ,.......,N N N M M '<:!'" '<:!'" ~ ~ ,......., '<:!'" r:---
...: 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
==
... ,......., ,......., ,.......,,......., ,......., ,.......,,......., ,......., ,.......,......-< ......-< ,......., ,.......,......-< ,......., ,......., ,......., ,......., ,.......,,......., ,.......,0 0 0
00 oS' ;;.- 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ,......., ,.......,,.......,
,_;;:j 0 ~ ~ ~ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ~ ~ ~ 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ~ ~ ~
0 0 0
S M 0 00 00 0 00 0' ......-< r:--- 0 r:---
S r:--- M r:--- N
0' r:--- 0
U 8 00
~ ~ ~ 0 0: 00 N 00
0: ~ 0' ~ rr: ~ ~ ~ r<"] ~
~ ~ 0: ~ ~ ,......., 0 M 0 M
~
0
0 0
M r:--- ......-< '<:!'" 00 N 0 ~ M 0 '<:!'" ~ ......-< ......-< 0'
0 ,......., ,.......,N ~ ~,......., ,......., N N N M '<:!'" '<:!'" ~ 0 0 0 r:--- 00 0' 0' ,......., ,......., ,.......,
0 '<:!'" r:---
Sj Sj Sj
~.. M
0 ~ 0 ~::I
==
,......., 0 0 0 0 0 0 0 0
III
=- =-
,......., ~ l£"! ~ l£"! ~ ~ ~ l£"! 0 0 0 0 0
0 N ~ r:---
III ~ ~ ~......-< ,......., ,......., ,.......,
~ e 0 ,......., ,.......,N N M '<:!'" ~ r:--- ......-< ,......., N N M '<:!'" ~ r:--- 0 0 0 0 0 ~ 0
.. 0 N ~
=-
0 N N N
0 0 0
~
~ ~
:s ......
~ i
oS~
== ~r:Il
r:Il
~
=-I.
~
-=~
~"'0
~
~
-=I.
=
- ~=00
~
=~
-;
e
~
t)IJ
~
e
as'
e
=
-=..C.J
~
°0
~
c.
00
Thermal EngineergingA-92
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![(a)
We+---
BRAYTON CYCLE WITH REGENERATOR AND
REHEATER
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.
...(iii)or 11 - 1 1
Brayton - (rp iy-l )/y
The efficiency of the Brayton cycle, therefore, depends upon
either the compression ratio or the pressure ratio.
...(ii)or
1
11Brayton = 1- y-l
rk
If rp = pressure ratio = P2/P 1the efficiency may be expressed in
the following form also
( ,(y-l)/y
11= l-l:~)
( J
Y-l
11= 1- :~
T4 -1= T3 -1
T} T2
T T T ( (Y-l)/y ( y-l
4 - 1 1 PI I v2 I
T3 - T2 = T2 = lpJ = l~)
If rk = compression ratio = V /v2 the efficiency becomes
[from Eq. (i)]
or
Now
As
...(i)
pressure.
Efficiency of Brayton Cycle:
11= 1- Q2 = 1- T4 - Tl
Q} T3- T2
Q 1= heat supplied = mcp (T3- T2)
Q2 = heat rejected = mcp (T4 - T 1)
T
(
J
(Y-l)/Y T
2 P2 3.
T;= Pt = T4 (Since PI = P3' and P4 = PI)
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
Heat
exchanger
TurbineCompressor
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.
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.
RANKINE CYCLEWITH REGENERATOR AND REHEAT
A-95Thermal Engineerging
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![AIR STANDARD CYCLES
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:
1. The working fluid is taken as air for the cycle and assumed
to be a perfect gas.
2. The engine is assumed to working in a closed cycle. There
is no change of mass of the working medium.
3. All the processes that conclude the cycle is of reversible
nature.
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 T6 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.
(c)
~v
f
(b)
.. s
(a)
-1 T} y-1/y
11- --y
T4 p
For a fixed ratio of (T /T 4) the cycle efficiency drops with
increasing pressure ratio.
Effect of Reheat on Brayton Cycle:
Effectiveness of the regenerator:
t3 - t2 actual temperature rise of air
E = --- =------=----------
Ts - T1 maximum possible rise of temperature
Efficiency of Brayton cycle with regenerator.
Q} T6 - T} T} [(T2 lTd-I]
11= 1- Q2 = 1- T4 - T3 = 1- T4 1- (Ts I T4)
T} T2 [1-(T} IT2)]
= 1- T4·r; 1-(Ts IT4)
_____. S
(b)
4
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![(b) (:) t
(d) (:
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) (!
(c) (:)p
1 1
(a) 11= 1- (r)y-I (b) 11= 1- (r)y+1
1 1
(c) 11= 1+ (r)y-I (d) 11=1--- y-I
(r)-
Y
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 :
--~"'S
constant
volume
2
--~S
constant
volume
constant
volume
T
(d) 1
"-
2""
T
1 1[' v
(a)
4
./
3"'
S
51. In thermodynamic cycles, the otto cycle is represented by
which of the following T - S diagram?
(d) <l>dQ~ 0
(b) <I> dQ < 0
T
(a) <I> dQ = 0
T
(c) <I> dQ > 0
T
44. 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,
[ Y-I]_Y_P1V1 (P2)Y -1
P2
(a) (b) mRTI/n-
Y-1 PI PI
T
(c) MCp (T2- Tl) (d) mR(l- ~~) (c) 1
T
(b) 1
(a) dQ=dU+qdV (b) Tds=dU+PdV
(c) dQ=Tds (d) dQ=Tds+dU
43. 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-t09Thermal Engineerging
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![VNcosa
(d) 2
VNsina
(b) 2(a) VN sin a
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 ofthe systemwill be:
1
(a) - (b) 2
4
(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 cyclebetweenthe same temperature limits.
Which of the following is correct?
(a) COP of refrigerator = COP of heat pump
(b) COP ofrefrigerator = COP ofheat pump + 1
(c) COP ofrefrigerator = COP of heat pump - 1
(d) COP ofrefrigerator = Inverse of COP ofheat pump
210. A solar energy based heat engine which receives 80 KJ of
heatat 100°Candrejects70KJ ofheattothe ambientat 30°C
is tobe designed.Then the thermal efficiencyofheat engine
willbe:
(a) 12.5% (b) 24.5%
(c) 40% (d) 70%
211. For an ideal gas, the expression [T(:;) p - T(:;) v ] is
always equal to :
(a) zero (b) R
Cp
(c) (d) Rf
Cv
212. During a phase change ofa pure substance:
(a) dG=O (b) dP= 0
(c) dH = 0 (d) dU= 0
213. At the triple point ofpure substance,the number ofdegrees
offreedom is :
(a) 2 (b) 1
(c) 0 (d) 4
214. A vessel ofvolume 1m' contains a mixture ofliquid water
and steam in equilibrium at 1.0bar. Given that 90% ofthe
volumeis occupiedbythe steam,then the fractionofmixture
will be equalto: Ifat 1bar,Vf= 0.001mvkg, V = 1.7mvkg
(a) 5.27 x 10-3 (b) 7.27 x 10-4 g
(c) 8.29X 10-3 (d) 9.23 x 10-3
215. 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°, Bladeexit angle = 25°,
Itblades friction is neglected, the work done/kg steams is :
(a) 124KJ (b) 164KJ
(c) 174KJ (d) 184KJ
216. if VN and a are the nozzle exit velocity and angle in an
impulseturbine, then the optimum blade efficiencyis given
by:
(c) fTds
204. The first Law ofthermodynamics takes the form w = -Ll D
when applied to:
(a) A closed system undergoing a reversible adiabatic
process
(b) An open systemundergoing an adiabatic process with
negligiblechangesinkineticenergyandpotentialenergy
(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 of60° C. Itisdropped 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
(a) fpdv (b) f vdP
ABC D
(a) 4 5 2 1
(b) 3 4 1 2
(c) 1 2 4 6
(d) 4 5 3
3. S.1.Engine
4. C.IEngine
5. Cooling towers
6. Heat exchangers
(D) dh = CpdT, even when
pressure, varies
1.
2.
(A)
(B)
List-II
Ideal gas
vander walls gas
cetane number
Approach and range
(C) (OT) :;t 0
oP h
201. In pressure lubrication systemofan engine, the maximum
oil pressure is controlled by :
(a) oilpump (b) oil pressure gauge
(c) oil pressure reliefvalue (d) oilfilter
202. A body of weight 100 N falls freely a vertical distance of
50m. The atmospheric drag forceis O.5N.For the body,the
work interaction is :
(a) -25J (b) +25J
(c) + 500J (d) - 500J
203. Match List - I and List - II and answer according to the
codes given below:
List- I
(a) Petrol (b) Splash
(c) Pressure (d) Dry sump
200. 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
(d) fsdT
Thermal EngineergingA-116
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![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
2
KN/m
A
5 1'---"","
2x314 2x314
v ------
opt. - cos 20° - 0.9396
= 668.4 mls ::::::668 mls
218. (a) Given
( )
2ub
Now, optimum velocity of steam vopt. =--
cos o;
1 a a a
=> P2 =10x2+ 1x2 - 22 = 5+4
Asa>O, p> 5
227. (d) 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.
229. (d) 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.
230. (b) Heat generated by bulb
= 100 x 24 x 60 x 60 J
= 8.64 x 106J
. . 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
131. (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
1and 2, clockwise (1-2-3) will be selected.
232. (c) 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
3
.____.___-........__--+ V (m )
0.01 0.03
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°
ndN n x 2 x 3000
Blade velocity (Ub) = 60 = 60
=314m/s
=n+4n
=5n
226. (b) T = constant
( a '1 ( a '1
Thus, lp, + vfjVt =lp2 + vljV2
=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,
VW1
IE >I< VW2 )1
~~
0.5294+100
Now, Dryness fraction = +
ms mw
0.03
0.02
3
2
= 1.5 KN/m2
= 1.5 kPa
220. (b) Given: Bore (d) = 2 unit, stroke (L) = 2 unit
Total area for heat loss = ~d2 + ndl. = ~(2)2 + n (2)(2)
4 4
0.5294
( ) v: 0.1
water mass mw = - = -- = 100 kg
Vf 0.001
=0.03KN-m
. () work done
Mean effectIve pressure MEP = ----
volume
( )
Vs 0.9
steam mass ms = - = - = 0.5294 kg
Vg 1.7
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![HI =2800kJ/kg
H2= 1800kJ/kg
Work done = HI - H2= (2800-1800) kJ/kg
= 1000kJ/kg
Then, specific steam consumptionm
3600
= 1000 =3.60kg/kW-h
242.(d) d=80mm
Strokelength L= 2 x Crank radian
=2 x60= 120mm
Then, swept volume Vs= A x L
= ~d2 xL
4
241. (a) H1
1
PemxVs xNx-
950xl03 = 2
60
950 xl 03 x 60 x 2
Pem = 0.0259 x 2200
=2 x 106 Pa=2MPa
1
For 4-stroke diesel engine, K = "2
P=950kW=950 x io'w
P = PemxALxNxK
60
T2=396.30K
W R(TI-T2)
m y-l
8.314x(300- 396.30)
1.67-1
W = 1194__!!_
m kmol
= 1194J/mol
Imol=Mg=40 g
For40 g,W=1194
1194
For 1kg W= --xl000
, 40
=29.7 kJ/kg
239. (d) In every case, entropy of universe is always positive.
(LS)universe~ 0
~Ssystem+ ~Ssurrounding~ 0
240. (a) AL= Vs(Sweftvolume)= 0.0259m3, Pem =?
N=2200rpm
0.67
=> 300 = (~)1.67 = (0.5)0.4012
T2 2
- x x 6 0.030
-0.8 0.015 10 en 0.015
= 8.31kJ
236. (a) Due to internal friction produced in irreversible
process, entropy of the system increases.
237. (b) From Clausiusinequality,
Cyclicintegral of di < 0 for irreversibleprocess
Ql + Q2 + Q3 + ....+ Qn < 0
Tl T2 T3 Tn
For process I,
2500 _ 2500 = -1.042
1200 800
For process II,
2000 _ 2000 = -1.5
800 500
Process II is more irreversible than process I.
238. (a) In adiabatic process,
W = PIVI- P2V2
y-l
y-l
_]_= (RL]rT2 P2
235. (a)
233. (c) Enthalpy of balloon is given by
H=U+Pr
Initially balloonkept 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 volumeis suchthat their product remains
constant.
Hence, h also remains constant.
234. (b) dH=dU +d(pVO)
dH= dU+ pdV + Vdp
f dH = f dq + f Vdp
~=Q+fVdp
Q-~=-fVdp
~kf;+~p.E+ W =-fVdp
2
W=-fVdp
1
Work is done on the system.
V2
Wisothermal= PIVIen-VI
Thermal EngineergingA-130
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![W W
ll+n = - = -----
Qs mcv(T3-T2)
h _ W
+n - mR~T
--
y-1
1
r= (~ )r-I =9.39.
W
Pm=-
Vs
TI =300k
T3 =6
TI
T3 = 1800k
we know that for maximum work output
T2T4=TIT3
T2 = ~TtT3
T3 = JTIT3
T4 = JTtT3
T2 = .)1800 x 300 = 734.84 k
t
~=(T2Jr-t
v2 TI
V3- v2 = 0.05 (VI - v2)
(:~ -I)=O.OS[ :~ -I]
rc-1 = 0.05 [12.26]
rc = 1.61
~diesel =1-(r )~-Il~(ct_~;l
1 1(l.61)l.4_1]
lldiesel = 1-(13.26)OAll.4 x 0.01 = 60.5
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.
3
303. (b)
302. (d)
301. (a)
r = 1+ ~ = 1+ 1177.5
= 1+5.99 ~7
Vc 196.3
W 1
1l=-=1---
Qs (r)y-l
~=1- 1
1800 (7iA-t
W = 973.44 kJ/kg
Vs = ~(25)2 (37.5) = 18398.43 em3
4
vc= v2 = 1500 em3
r = 1+~ = 1+ 18398.43 = 13.26
Vc 1500
297. (b)
1t 2 1t 2 3
Vs =-d L = -(10) x15=1177.5cm
4 4
295. (c)
TL = Lower temperature
TH = Higher temperature
RE 2
Power= --=- =lkW
COP 2
2
200
300-200
293. (c)
COP = Refrigeration effect = ~ = 2.33
Work done 1.5
288. (b)
.. VI
Compression ratio, r = 10 = V2
Now, mean effective pressure is given by
Work done
Pmean = -----
Swept volume
V4 Vt
Now -=-=10
'V3 V2
=> VI = 10V2 ...(i)
Also swept volume
VS=VI-V2
=> VS=0.9VI
Initiallyfor air
PIVl =nRTI
" _ nRT _lxO.287x300
.. vt - ~ - 100 =0.861 m3/kg
.. Vs=0.9x 0.861 =0.7749m3/kg
Work done in cycle
W = Heat supplied - Heat rejected = Qs - QR
= 1500 - 700 = 800 kJ/kg
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.
285. (c)
A-133Thermal Engineerging
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![= 1-[ 0.001 ]1.4-1 =58.2%
0.001+~xO.2002 xO.250
312. (c) Given PI = 1bar
P3 =40bar
r=5
vc=O.OOI m3
7t 2 3
Ys = -x 0.200 x 0.250m
4
311. (c)
= 1
1
11=1---
(r)Y-I
1
--1-:-4-:---:-1 = 0.46 = 46%
(4.68) .
V2 =1
v2' = 1 +0.7 (r-l)
=0.7r +0.3
v2' = 1 +0.3 (r-l)
=0.3r+0.7
lIx .'.
~=(_!i_J =(2.6)1.3 =1.7
v2 P2 1.3
0.7r+0.3
1.7
0.3r+0.7
r=4.68
30% V---+70%
I
I
I
I
I
--4--------1
P >I
11 v,1
V' 11
I 2
310. (c)
6-5
= (1.4-1)x-5-x100 =0.08 x 100=8%
M
= (r-1)-x 100
r
.1n M
-xl00= -(I-r)-xl00
n r
11= 1__ 1_= l-(r)y-1
(r)Y-I308. (d)
more efficient.
:. 11DieseI>11DuaI> 110tto
S
Thermal efficiency= 1 Qrejected = 1 Constant
Qsup plied Qsupplied
Thus the cycle with greater heat addition Qsuppliedis
V
4
P
32
1-(r)~-l=1-(r)~-I[r(:=~)]
sr-1 = r(s - 1)
sr-I - r( s - 1) = 0
306. (d) Efficiency of ideal regenerative cycle is exactly equal
to that of the corresponding Camot cycle. Hence it is
maximum,
307. (b) 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.
304. (a) 11otto= 11diesel
Vs =(r-l)vc
11+n(.1p)vc
Pm = -...;,...:=........:...__~-=----
(Y-1)(r-1)vc
11+n(M»
= (y-l)(r-l)
w = 11+nmR.1T = 11+nM>vc
y-1 (Y-1)
Thermal EngineergingA-134
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![Wc) = 0.582or58.2%
Cooling and dehumidification-temperature decreases 325. (d) Relative humidity of the air decreases.
and w also decreases
[ ]
1.4-1
11-1- 0.001
0.001+~(0.2)2 x(0.25)
11-1-( Vc JY-1
VC+VS
Substituting values,
1t 2 1t 2
Vs =-d L=-(200) x250cc
4 4
d=200mm
Vc=0.00Im3
Air is cooledat constant pressure to make unsaturated
air to saturated one,
Given,
StrokeL= 250mm
Diameter,
Clearancevolume
Now, swept volume
324. (c)
s
Tdew
T
323. (c)
30°
321. (a) On a psychrometric chart
Constant relative humidity lines are uphill curve not
straight, to the right.
Constant WBT lines straight downhill to the right.
Constantspecificvolumedownhillstraightto theright.
Constant enthalpy lines are not coincident to WBT.
322. (b) P=2u(v-u)(1 +cos<j»x flowrate
= 2 x 10(25-10) (1 + cos 120°)x 0.1
= 20 x 15x 0.5 x 0.1= 15kW
0.025
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
320. (b) On a psychrometric chart
75% RH
w (kg/kg)
313 () H· ..c. I ( ) QI QI
. c mt: usmg rormu a cop = W = Q2-Ql
=~
T2-Tl
319. (b) Chemicaldehumidification- temperatureincreases,w
decreases,
i.e., 0-2
Sensible heating - straight horizontal line towards
right,
i.e.0-1
1+~=5
Vc
Vs = 4vc
W
11=-
Qs
0.4746= W
76.225vc
W = 36.17vc
Pm = W = 36.17vc = 9.04 bar
Vs 4vc
P2 =Pl( :J=1.(5)14 =9.51 bar
1 1
11=1--- = 1--- =0.4746
(rr' (5)°.4
R(T3-T2)
Qs = Cv (T3 - TI)= _____..:.._--=----~
r-1
v2 (P3- P2) (40 - 9.51)xvc
= = 76.255vc
r-1 1.4-1
A-135Thermal Engineerging
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![13, d31
......_ -' I
121 ~ :
I
Pipes in Series
• In series, discharge (Q) remains same but head is divided.
FLOW THROUGH BRANCHED PIPES
KV2
h---f- 2g
K = Constant which depends upon angle of bend and its
radius of curvature.
BEND LOSSES
O.5V2
h---f- 2g
h.= vi [_1 _1]22g Cc
A2
CC=A
3
If'C, is not given, hj= 0.5 Vll2g
Head loss occurs after Venacontracta as boundry layer
separation occurs.
(iv) Entrance Losses
CD
V12
h.= 2g
(iii) Losses Due to Sudden contraction
,~
~:
N_CD I
®
(i) Losses Due to Sudden Enlargement
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.
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.
pVD
Re=--
J..l
V = velocity
D = diameter
m = dynamic Viscosity
Minor Losses
• Bernoulli's Equation, momentum Eq" are used to
determine these losses.
• The magnitude of minor losses is very loss.
...(1)
• Forced vortex flows are rotational flows and hence,
Bernoulli's equation cannot be applied.
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 = w:t I
• A pipe is a closed contour which carries fluid under
pressure.
• When fluid flowsthrough pipe, it encounters losses. These
losses can be broadly categorized into
(i) Major losses
(ii) Minor losses
Major Losses
• These losses are due to friction. The losses are evaluated
by Darcy-Weishback Equation.
fLV2
2gd
(ii) Exit Losses
Fluid Mechanics and MachineryA-144
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![--r-bj d~t-'--::I;;?::--u
Uy 1 (ap) 2
u=--- - (by-y )
b 2M ax
Case II : When both plates are at rest
u = - 2~ (:) (by - y2)(Poiseuille flow)
LAMINAR FLOW BETWEEN TWO PARALLEL PLATES
Case I : One plate is moving with a velocity of 'U' while the
other is stationary.
Umax
u=--
2
at r = R/.fi,= U = U i.e. average velocity equals the local
velocity.
Pressure drop (P, - P2) in a given finite length 'L'
32MUL
P1 - P2 = D2
t=(-:H
u = - _1_ (ap) (R 2 _ r2)
4M ax
from above expression of 'u', we can conclude that velocity is
varying parabolically.
1 ( ap) 2 ( 1t R2
'1
Umax = - 4M ax R ,Q = l-2-) U max
1" : shear stress
R: radius of pipe
M : dynamic viscosity of fluid
ap
ax :pressure gradient
u : velocity at a distance
'r' from cente
! ~
--! -~----
~P+.?Pdx
"[: Ox
• Shear between fluid layers e = Il du/dy(x-dir.)
Entrance Length
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
for maximum efficiency 1he = ~ 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.
Features of Laminar Flow
• No slip at boundary
• Flow is rotational
• No mixing of fluid layers
Ptheoretical = pQgH
Pactual = pQ (H - hf)
where hf are the head losses in pipe.
pQ (H - hf)
11= pQgH
H
POWER TRANSMISSION THROUGH PIPE
J
•
•
•
Siphon is a long bend pipe used in carrying water from a
reservoir at higher level to another reservoir at lower level.
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.
•
SYPHON
Q=Q1 +Q2
(hf)1 = (hfh
9+~---t-~-
PIPES IN PARALLEL
• In parallel arrangement, head losses remain same but
discharge gets divided.
=> h=~[LIV? + L2Vj + L3V}]
2g d1 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.
Q=Q] =Q2=Q3
h-= (hd] +(hf)2 +(hfh
A-145Fluid Mechanics and Machinery
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![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
Pshaft
l1mech. = Q H
P W W
The value of mechanical efficiency varies between 0.97
to 0.99.
(c) 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.
(d) Overall efficiency : It is defined as the product of
hydraulic efficiency (l1H)' mechanical efficiency (l1mech)
and volumemetric efficiency (11vol).
110 = l1H x l1mech x 11vol.
Francis Turbine
Francis turbines are by far the most common type in present-
day 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
1+ coso
Maximum hydraulic efficiency (l1max.) = 2
(b) Mechanical efficiency : It is described as the ratio of
power available at the shaft and the power produced by
the wheel.
Vw = velocity of whirl at outlet
2
u = peripheral velocity
Vr1 = relative velocity at inlet
Vr2 = relative velocity at outlet
VI = absolute velocity at inlet
V2 = absolute velocity at outlet
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.
2U(VWI ± VW2 )
l1H = -....:....._-:----....:....
V?
u
~ work done / weight = (VWl ± VW2 ).g
where, Vw = velocity of whirl at inlet
1
Gross head = HG = difference between head race and tail
race
Net head = Hnet = ~ - hf- h
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
VU1
Some important formulas :
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 :
u Vu
IE 2 * 2 )1
p
Specific Speed
where A = area ofthe jet
g = acceleration due to gravity.
The input power to the turbine
P = 1000 x Q x g x H = 1000x A~2gH x g x H
(.: Q= A~2gH )
Fluid Mechanics and Machinery
Q= A~(2gH)
A-148
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![Fig. : A Propeller or axial-flow turbine
Main parts ofa kaplan turbine:
(a) Scroll casing: It is the casing in which guiding the water
and controlling of passage of water takes palce.
(b) 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.
(c) Stay ring: The stay ring guides the water from scroll casing
to the guide vanes.
(d) Runner blades: Runner blades are connected to the hub
and there is an axial flow ofwater through the runner.
(e) Draft tube : It is utilized for the purpose of connecting
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.
or
110= 11H x 11mech. x 11vol.
VWIUl±VW2·U2
11H = gH
~ Mechanical efficiency : It is defined as the ratio of shaft
power to the power developed by the runner. It is denoted
as 11mech.
~ Volumetric efficiency (11vol):It is defined as the actual
quantity of fluid working on the runner to the total quantity
of fluid supplied.
Actual fluid quantity
11vol = T Ifl ·d .. ota U1 quantity
~ 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.
shaft power
110 = Input power
~ Work done per second: (W.D/s)
W.D/s = pQd[VW1Ul ± VW2U2]
• 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.
•
•
•
. P v2
At the exit of draft tube H = -- - Zl --
, P 2g
At the exit of pen stock when the position of draft
tube is at the tail race.
p v2
H=-+z.+-
P 2g
At the exit of draft tube when the position of draft
tube is at the tail race.
v2
H=--
2g
Ifthe velocity at the exit of draft tube is negligible,
then Net head, at the exit of penstock.
H=(~+Z<J
•
P v2
At the exit of penstock, H =-+zl +-
p 2g•
Some important formulas :
~ Net head (H) :
h f ifi ( ~ )T e range 0 speer IC speeds, lNs = 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
water from dams reaches to the inlet of the turbine.
(b) 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.
(c) 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.
(d) Runner: It is connected to the shaft of the turbine.
(e) Draft tube: Itis defined as tube which expands gradually
and it discharges water passing through the runner to the
tail race.
Velocity triangle of Francis turbine :
rotation.
A-149Fluid Mechanics and Machinery
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![efficiency.
~ 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.
•
•
(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
•
Fig. (3)
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 :
(e)
where, QA = Actual discharge, QL = rate of leakage
Hydraulic efficiency : It is defined as the ratio of
manometric head to the theoretical head.
Hm
11N=-
HT
Characteristic Curves
(d)
wVw2u2g
11mech.=
Ps
(c) Overall efficiency (110) : It is defined as the ratio of
power output to power input to the pump or shaft.
wHm
110 =-p-
Volumetric efficiency (l1vol) : It is defined as the ratio
of actual discharge to the sum of actual discharge and rate
of leakage.
QA
(VW2U2)/ g
Mechanical efficiency (l1mech)
It is defined as the ratio ofpower delivered bythe impeller
with the power input to shaft.
(b)
11mano.
Work Done by Impeller on the Water
[VW2.u2 - VWlul ]
WD = .:::...._--=------=---=
g
where, W.D = work done
VWl = velocity of whirl at inlet
VW2 = 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: u2
then work done = __ 2_
g
Heads in Centrifugal Pumps
(a) Suction head: Itis defined asthe 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 ofwater
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 asthe 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,
Manometric head (Hm )
11mano.= (VW2U2)/ g
(Back-ward
facing vanes)
(Radial Vanes)(Forward facing
Vanes)
A-153Fluid Mechanics and Machinery
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![11. A pump handling a liquid raises its pressure from 1bar 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
(b) 3
(d) 5
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
(c) 4
(b) n
(d) n + k
10.
(a) k
(c) n-k
9. A phenomenon is modelled using n dimensional variables
withk primary dimensions.The number ofnon-dimensional
variable is
(b) 354
(d) 707
8.
(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 1000kW power for a head of
40 m. If the head is reduced to 20 m, the power developed
(in kW) is
(a) 177
(c) 500
coefficient
z.
Match number
Skin friction
y.
W. Weber number
X. Froude number
U. Reynolds number
V. Nusselt number
(b) Irrotational flow
(d) Incompressible flow
7.
6. Forthe continuityequationgiven V.;= 0 tobevalid,when
; is the velocity vector, which one of the following is a
necessary condition?
(a) Steady flow
(c) Invescid flow
Match the following:
P. Compressive flow
Q. Free surface flow
R. Boundary layer flow
S. Pipe flow
T. Heat convection
For a Newtonian fluid
(a) shear stress is proportional to shear strain
(b) rate of sheer stress is proportional to shear strain
(c) shear stress is proportional to rate of shear strain
(d) rate of shear stress is proportional to rate of shear
strain
5.
au au
(d) v-+u-
Ox ay
Ov au
(c) u-+v-
ax ay
4.
A two-dimensional flow field has velocities along x and y
directions given by u = x2t and v = - 2xyt respectively,
where t is time. The equation of streamlines is
(a) x2y = 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 auOv
(a) u-+v- (b) u-+v-
ax ay Ox ay
3.
/-!uOL
(a) D2
8/-!uoL
(c) D2
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
2.
Ov
ay(d)
Ov
(c) ay
Ov
Ox
(b)
The velocity components in the x and y direction of a two-
dimensional potential flow are u and v respectively, then
au.
aX IS equal to
Ov
(a) Ox
1.
...,EXERCISEIIIIII~
Rate of flow (Lifi/s)-----7
(Muschel curves)
a'._/
--g
(])
...s:::
~
E§ I---.+--l-_
~ Muschel or constant efficiency curve
~ Discharge (Qd)
(Operating characteristic curves)
lip power
efficiency
Fluid Mechanics and MachineryA-154
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![A
(b) 2
(d) 2
C
1
3
B
3
1
B C
2 3
3 2
Codes
A
(a) 1
(c) 1
47.
44. 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'
45. 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
46. 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.
L~t I L~t II
(Loss) (Parameter responsible)
A Leakage Loss 1. Zero at design point
B. Friction Loss 2. Proportional to head
C. Entrance Loss 3. Proportional to half of
relative velocity square.
y2
(c) 2g
(b) nd2• 2g
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 v2 4f L y2
(a) -.-
d 2g
43.
l6000J..l2
(c)
40. When pressure p, flow rate Q, diameter D, and density d, a
dimensionless group is represented by
pQ2 __ P__
(a) dD4 (b) dQ2D4
pD4d pD4
(c) Q2 (d) dQ2
41. Maximum wall shear stress for laminar flow in tube of
diameter D with fluid properties J..land p will be
32000J..l2 6400J..l2
(a) pD3 (b) pD3
8000J..l2
(d) pD3
(
5 )115
(d) d= ~L(
3 )1/5
(c) d= ~L
_ (D5)1I4(b) d- -
8L
_ (D3)1I4(a) d- -
8L
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
38.
(c)
O'J..lY
~P
(a)
37.
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
36.
35.
34.
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
ax Oy
33.
(a)
ou+av +ow =0
(b)
Ou=av =Ow =0
ox Oy oz ox Oy oz
au ov Ow au av Ow
(c) -+-+-=1 (d) -+-+-=u.v.w
ax Oy oz ax Oy oz
42.
30. 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
31. 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
32. The general equation of continuity for three-dimensional
flow of a incompressible fluid for steady flow is
Fluid Mechanics and MachineryA-156
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![2
leg cot o
(d)
A'x' A'x'
where, Q = inclination of plane area.
x = distance ofc.g of plane area from free liquid surface.
2
leg tan e
(c)
88.
87.
86.
85.
84.
83.
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
(a) 2 3
(b) 2 3
(c) 3 2
(d) 2 1 3
Find the buoyantforceacting on the aluminium cubewhich
is suspended and immersed in ajar 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
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
The vertical distance of centre of pressure below the e.g
of the inclined plane area (submerged in liquid) is :
leg sin2 e leg cos2 o
(a) (b)
A'x' A'x'
82.
Fluid Mechanics and Machinery
Codes:
A B C D
(a) 4 1 3 2
(b) 2 3 4
(c) 4 3 2
(d) 2 4 1 3
A capilarity 1. cavitation
B. vapour pressure 2. Density of water
C. viscosity 3. shear forces
D. specific gravity 4. surface tension
70. Viscosity has the following dimensions:
(a) [ML T2] (b) [ML-1 T-1]
(c) [ML T-1] (d) [ML2T]
71. A piezometer tube is used forthe purpose ofmeasurement
of:
(a) Low pressure (b) High pressure
(c) Moderate pressure (d) Vacuumpressure
72. Atmospheric pressure is 1.03kg/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
73. 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
74. 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
75. 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
76. 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'
77. The volumetric change ofthe fluid caused by a resistance
is known as:
(a) volumetric strain (b) volumetric index
(c) compressibility (d) stress
78. 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
79. 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
80. A liquid compressed in a cylinder has initially a volume
of 20 m3 at a pressure of 100Pa. 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
81. Match List - I and List - II and select the correct answer
using the codes given below:
List - I List - II
A-15S
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![v-u (y)27. (c) ~ = 5.75 10glO k + 3.75
for v= u
5.75 + 10glO(f) + 3.75 = 0
.1_ = 0.223 => y = 0.223 R
R
r
So, - is independ at then (t = r where (is on stat)
E
At center r= 0, t =c x 0 =0
Zr wr
t=--
D
~ =~x twxr => TW = ~PD
L r D 4L
21. (d) u=x+2y+2
v=4-y
au Ov
-+- = I-I =0
ax Oy
As it satisfy the continuity equation for
incompressible flow so this is incompressible
Wz= H:-:]= ~[O-21 =-1
and since rotational component is not zero so flow is
not irrotational.
24. (a) Stagnation head = 20 m
Statichead = 5m
Dynamichead=20-5= 15m
u2
Now - = 15, 2g
u=17.15m/s
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
~P Zr
---
L 6L
Neither P nor I depend as on r.
Pl~
D~
x
12. (b) A streamline and equipotential line in a flowfield are
perpendicular to each other (because, when (slope)1
and (slope), are multiplied and we get, (slope), x
(slope), = -I
13. (a) Assumption:
a
(i) Flow is steady y (i.e.) at D = 0
(ii) Fully developed the ~ D- 0 ; properties are not
at
changing in the direction of the flow.
2.93 kl/kg
990xl000m p
m
-dp
p
Specific work done W = Vdp =
W = dp (30-I)xI05
11. (d)
... (i)Th
Powe~ Powe~
us, 1 1
(HeadI)·5 (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
In case of two fixed parallel plates, when the flow is
fully developed, the ratio vrnax/ vavgis given by
vrnax =~
vavg 2
Thus, vavg= (2/3) x vrnax.= 2 x 6/3 = 4 m/s
p] = I bar, P2 = 30 bar
p = 9900 kg/nr'.
10. (c)
Heat convection Nusselt number
8. (b) The relation between head and power is given by
Pu = Power/Head 1.5
Pipe flow
Froude number
Skin friction
coefficient
Reynolds number
Free surface flow
Boundary layer flow
dy
Now tOC-
dy
dx
oc--
dt dy
[dx / dy]
oc"":""'_-----'-
dt
(dx/dy)
where, = rate of change of shear strain.
dt
6. (d) Continuity equation
v· v=O
au Ov Ow
=> -+-+-=0
Ox Oy az
Multiplying by density on both sides, we get,
p(au + Ov + OwJ= 0
Ox Oy az
This is the equation for compressible flow.
7. (b) Compressive flow Weber number
1
rl 1------+1 V + dv ~ fluid flow
y -y I------------W
5. (c) ForNewtonian fluid,
dy
Shear stress a: dy
dy . .
where, dy = velocity gradient
Fluid Mechanics and MachineryA-168
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![74. (b)
72. (d)
70. (b)
=~=0.6
1000
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
From Newton's Law of viscosity, (J.!)
Shear stress (r) = J.!x velocity gradient
du
't = J.!x-
dy
't N
J.!= du =-2
_ m
dy S
J.!= J:£_s = [MLr2][T]
m2 [L2]
J.!= [ML- 1 T- 1]
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/ern-
(fauge pressure (Pg) = 21 bar
Atmospheric pressure (PatIn)= 1.013bar
Absolute pressure (Pab)= 't
Pab= Pg + Patm
= 21+ 1.013
=22.013 bar
69. (d)
density of wood piece
Specific gravity of wood = d . f
ensity 0 water
62. (b) Let H = height of water column,
H _ _E__13.6x103 x60x10-2 x9.81
then, - pg - 1000x9.81
=8.16m
= 816 em
65. (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 closedto 63765 Pa'
67. (b) Weight of wood piece (m) = 5 kg
Weight of wood piece = weight of liquid displaced
S= 60% ofvolume(v) x 1000
=~xl000
100
S = 600 v
513
v=-=-m
600 120
density of wooden piece (p] = Mass(m)
Volume(v)
5 3
=-=600kg/m
1
120
0.14 x 9.81x 10000
700 x9.81
1400
= 700 = 2 m of oil.
58. (b)
57. (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
Given: specific gravity of oil = 0.7
Pressure (p) = 0.14 kgf I em-
Density of oil (Poil) = 0.7 x 1000 = 700 kg/m'
H=_E_= 0.14x9.81
pg 700x9.81
2
On integrating, f ajl = -2x =>jI = - 2x + C (y)
ax z
ajl
- = 0 + C' (y)
ay
f c'(y) = f 2y
2y2 _2x2 2y2
C (y) = -2 +cl> lthen, 'I' = --+-+Cl
2 2
jI = y2 _ x2 + c1 or y2 - x2 + constant
49. (a) If flow in 2D, continuity equation becomes,
ay av
-+- =0
ax ay
So, for (i), u = x2 cos y, V = - 2x sin y
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
R, = 424.4,means Reynoldsnumber ofthis flowisless
then2000(424.4<2000)
Hencethe flowis "Laminar"
48. (a) Given here, <I> = 2xy, considering the following
relation,
-a<l> ajl
=>4= -=--
ax ay
ajl (a<l> ajl'1 _ a<l>
=> - - I -- -) --
ax -" ay' ax ax
a<l>= ~(2xy) = 2y = _ ajl
ax ax ay
Similarly,
a<l>= 2x = _ ajl
ay ax
Fluid Mechanics and MachineryA-170
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![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'
fluidviscosityru) = 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,
U =2V =2x5=10m/smax. avg.
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 ofpipe (L)= 1000m
Considering the following formula for head loss,
H _ 4fLv2
L - 2gd
for discharge (Qd) =Area x velocity
2
2000= vxl0x10 =?V= 2000xO.25 =0.5m/s
0.25 IOx102
96. (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
Re= vD = vD
Il v
P
1
= +-
2
94. (b) Given: kinematic viscosity (v) = 0.25 stokes
diameter ofpipe (D) = 10em
for a critical velocity, Reynold's number should be
between2000and 4000 .
(
PvD
Reynold'snumber Re) = --
Il
Vavg.
Vmax.
ap (Rf -R~)
93. (a) Average velocity (Vavg) = ax 81l
ap(Rf -R~)
Maximum velocity (Vmax) =
ax 41l
= dP =~=50Pa
dv 20
vI 20
82. (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
86. (b) Given: Side ofthe cube~a)= 5 em = 5 x 10-2m
Volumeofcube (v) = (a)
=(5 x 10-2)3
= 125x l0-6m3
Buoyant force acting on the cube
=v.pg
= 125x 10---6 x1000x 10 (Assumingg = 10m/s2)
= 125x 10-2
= 1.25 N
75. (a) Given: diameter ofpipe (D)= 20 em
kinematic viscosity (v) = 0.0101 stoke
Reynold'snumber (Re)= 2320
Let v = velocity of flowing water,
Re = vD = vx20
v 0.0101
2320 = v x 20 =? v = 2320 x 0.0101
0.0101 20
v = 1.1716 em/s ~ 1.117 cm/s
76. (a) Given: Mass ofliquid = 5 tonnes = 5 x 103 kg
volume= 10m3
. .. mass of liquid
mass density of liquid (p) = ---~-
volume
= 5x 103 = 5 x 102
10
= 500 kg/m'
78. (b) Given: diameter of glass tube (d) = 3 mm
surface tension (crT)= 0.0736 N/m
contact angle for water (a) = 0°
4crTcosa
Then capilary rise (H) = wd
4x 0.0736xcos 0° 4 x 0.0736 x 1
kg-f xd 9.81x 10-6 x 3
= 10.4 mm (approx.)
80. (c) Initial volume(vI) = 20 m3
Initial presssure (P1) = 100Pa
Final volume(v ) = 40 m3
Final pressure (p2)= 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 lumetri t .oume fIC s ram
A-171Fluid Mechanics and Machinery
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![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
....(i)
20
u=-nd
3
ndN rtd x 400
Speed of wheel (Il) = -6-0 = -6-0-
8
8 1 [7817]8
[Y]o- 81/7 S"y 0
1 [7 8/7] 7=8- 81/7 S"x8 =8-S"8
Given:
height (H) = 10m
flow rate (Q) = 11m3Is
Losses due to friction and others (HLf) = 5m
Pumping power (p)= PgQ(H+HLf )kw
1000
= 1000 x 9.8 x O.l (10 + 5) kw
1000
P = 14.7 kw
For maximum efficiency,
jet velocity = 2 x wheel speed
=2v
Given: Head (H) = 405 m
speed (N) = 400 rpm
~ = 0.45
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 x2h= wnd h
4 2
FA 2whnd2 = 4
FB wnd2h
2
Given: Velocity distribution is given as :
~=(x..JI/7
UO 8
displacement thickness (8*) is given as :
228. (a)
226. (a)
224. (a)
204. (c)
y2
pgH = E..___!!_+pg(H - h)
2
y2
pgH = E..___!!_+ pgH - pgh
2
y2
pgh = E..___!!_
2
2gh =v~
YB =figh
199. (c) Case I : When container placed with its large diameter
down ward: then force (FA)will be :
3n 3
=0 0003nm3/s = --m Is
. 10000
195. (b) Considering Bernoulli's equation, between section (A)
and section (B),
PYA
2
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
Area (A) = ~d2 = ~(0.04)2
4 4
=0.0004nm2
discharge (Qd) = A x vavg.
=O.0004nx 0.75
190. (b)
3 4
Given u = ')..xy3 - x2 y, v= xy2 -"4Y
For the case of incompressible flow,
au+av=o
ax ay
au 3 au 3
ax = ')..y -2xy, ay = 2yx = 3y
')..y3- 2xj + 2yx - 3y3 = 0
')..y3 - 3y = 0
y5 (')..- 3) = 0
')..-3=0
')..=3
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
vrnax 1.5
vavg. =-2-=T=0.75m/s
187. (d)
=dxgxHxA
=dgHA
181. (a) Given: kinematic viscosity of air = 1.6 x 10-5 m2/s
considering the following relation,
Ila: (T)1I2 ...(i)
1
p o: T ...(ii)
VOC (T)3/2
we get, kinematic viscosity at 70° C
= 2.2 x 10-5 m2/s
A-173Fluid Mechanics and Machinery
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![OR
Now considering, ilORP, coso, = OP
Let, fr is feed rate / (Feed / rev.)
F, = Feed / tooth
F -~t -
Nt
where, N, = Number of teeth
S= Ftsin<p
when, <p=<Pc'then, S = Smax.(maximum)
when, <p= 0, then, S = 0
Smax= r,sin <Pc
s = F~d (1- d)m t D D
Volume of removed metal, (VR) :~ 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: Itis 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: Itis 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.
D-2d
-2- D-2d 2
=---=--x-
D 2 D
2
S = Smax
m 2
MECHANICS OF MILLING OPERATIONS
=> Mean chip thickess :
O+Smax
Mean chip thickness (S~ = 2
(For conventional face milling)
sin o, = ~~( 1-~)
sino == 1_[D2+4d2-2.D.2d]
c D2
On solving, we get,where, A = Apporach distance
A = ~w c (D - w c) (for partial face milling)
A=D
2
. ?2d)
2
sin o, = ~l-ll-D)
.. . ( ) I+A
Machining time tm =--
Fr
where, tm= machining time (min.)
A= Approach distance
=)d(D-d)
In case of face milling:
1+2A
t =--
m F
r
5.
4.
D-2d (2d)
cos<Pc=-D-=I- D
As we know that, sin2<pe+ cos2<Pe= 1
sin2 <Pc= 1-cos2 <Pc
sin <Pc= ~1- cos2 <Pc
Material Removal Rate (M.R.R) = we x de x Fr
where, de = depth of cut (mm)
we = width ofcut(mm)
In case of slab milling:
3.
Production EngineeringA-202
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Handbook to ssc je mechanical
- 1. o Theory with Exercises o Practice Question Bank Recruitment Exam Guide Handbook to sse Junior Engineer <-«. ~ disha Badboys2Badboys2 Badboys2
- 2. o Theory with Exercises o Practice Question Bank Recruitment Exam Guide Handbook to sse Junior Engineer Badboys2Badboys2 Badboys2
- 3. 1. Engineering Mechanics and Strength of Materials A-I - A-40 2. Theory of Machines & Machine Desig A-4I - A-82 3. Thermal Engineering A-83 - A-135 4. Fluid Mechanics and Machinery A-136 - A-173 5. Production Engineering A-174 - A-220 Badboys2Badboys2 Badboys2
- 4. (x-y plane) F3 .' F2 ./ Fl •.•»" -: / (Complete classification of force system) 1. Coplanar collinear : In this case, all the forces act in the same plane and also have a common line of action. Non-concurrent Non-parallel ParallelConcurrent Non-concurrent Non-parallel lParallelConcurrent LCollinear Non-coplanar JCoplanar I Forcelsystem Statics deals with forces in terms of their distribution and effect on a body at absolute or relative rest. Dynamicsdeals with the studyofbodiesin motion. Dynamics is further dividedinto kinematics and kinetics. Kinematics is concerned with the bodies in motion without taking into account the forceswhich 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) KineticsKinematics Dynamics I Statics 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 ENGINEERING MECHANICS I~NfJINI~I~IIINfJ)11~(~II1INI(~S lINI) srl'111~NfJrl'IIf)lf )11Irl'I~III1II..S SECTION A : MECHANICAL ENGINEERING Badboys2Badboys2 Badboys2
- 5. -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 distance of the point from the line of action of the force. a, ~,y = Angles included between three forces P, Q and R then, Q 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 ( Psin a J -1( P sin a Jtan a = :::::>a=tan Q+pcosa Q+pcosa or, 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, Resultant 'R' = ~p2 + Q2 + 2PQ cos a Psina Angle made by resultant 'R' = Q + P cos a L?7Q p 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. z F} (Non-coplanar concurrent forces) (xy plane) y 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. ---~F2 ---~Fl (x y plane) Coplanar parallel force: All the forces act in a plane and parallel with each other irrespective of direction. 3. 2. Coplanar concurrent : In this case all the forces act in the same plane and meet or intersect at a common point. (xy plane) Engineering Mechanics and Strength ofMaterialsA-2 Badboys2Badboys2 Badboys2
- 6. 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 flowor 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. Cone of friction 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. Inclined plane with horizontal w=mg From fig : tan <I> = : = Il =><I> = 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. by the resultant of normal reaction with the limiting force of friction with the normal reaction. 21RF I 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. F u > -=>F=IlR R Angle of friction (<1»: Itis defined as the angle subtended Horizontal surface F (Frictional for~ ~ 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 Surface/ Support R Action and Reaction: From Newton's third law,for every action there is a equal and opposite reaction. d 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-3Engineering Mechanics and Strength ofMaterials Badboys2Badboys2 Badboys2
- 7. Gravitational law is also known as universal law of gravitation. According to this law,Every substance or bodyhas 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 GRAVITATIONAL LAW where IF=m~=mal In=mass of the body v = velocity of the body F =Force acting on the body a = acceleration produced in the body. 3. Newton's third law of motion: This law states that there is always an equal and opposite reaction to every action. d(mv) - --=F dt There are three laws of motion known as Newton's laws of motion. 1. Newton's first law ofmotion: This law statesthat ifa bodyis in the state ofrest it remains in the state ofrest 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 secondlawofmotion:Itstatesthattherateofchange ofmomentum is directlyproportional to the impressedforce, and take place in the samedirection, in which the force acts. Momentum = mv NEWTON'S LAWS OF MOTION 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 irrespectiveofits direction,thereforeit is a scalarquantity. Velocity: The rate ofchange ofposition of a bodywith respect to time is called velocity.Velocityis a vector quantity. In other way wecan sayvelocityisthe speedofa 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 calledretardation. Beam is subjected to following set of forces after the beam is detached from the supports. (a) Weightofthe beamW 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 = 0 LF = 0 LM = 0x ' y , => Equilibrium equations for concurrent forces LF = 0, LF = 0 (only two conditions are required)x y w ~M The free body diagram (FBD) of the above system can be drawn as in Fig. FREE BODY DIAGRAMS A free body diagram (FBD) is a simplified representation of particle or rigid bodythat 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 bodymust be considered and equally important. Anyforcenot directlyapplied on the body must be excluded. Let us consider a system of a beam loaded and supported as shown in Fig. Engineering Mechanicsand Strength ofMaterialsA-4 Badboys2Badboys2 Badboys2
- 8. Z =Z Z =Z Z =Zxy yx' xz zx' ~ y z Plane Stress problems are those in which the stress acting in one of the mutual perpendicular directions is assumed to be zero .. cr=0 Z =0 Z =0z xz yz [cr]=[crx crxy] cryx cry For a given stress tensor Z face o ,o .o are normal stressesx y z Remaining are shear stress. STRESS TENSOR Torsional shear stress I IAxi I B di Direct shear stress ~mg Tensile Compressive Shear Stress (acting parallel to corresponding plane) I Normal Stress (acting perpendicular to corresponding plane) I Type of Stresses I When deformation or strain occurs freely in a direction, stress developed in that direction is zero. Whendeformationisrestrictedcompletely,orpartiallystress is developed. Hence strain is the cause of stress. • NOIE MKS : kgf/cm? 1 MPa = 106 N/m2 1GPa= 109N/m2 1 Pa = 1 N/m2 1kg;:::::0.1 MPa cm SI : Pa, MPa, GPa 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 Units of Stress L , · ,••• p ---- ------ ------ -----_ •• _•• _... _.- · , · ., ,, .. ., .. , / F d cr=- A p 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. Centripetal and Centrifugal Force Essential force for a circular motion acting radially inwards is called centripetal forcewhich 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 ofmotion, the bodymust exert a force radially outwards ofequal. ANGULAR ACCELERATION The rate of change of angular velocity is called angular acceleration. It is expressed in radls2 dO) a=cit ANGULAR VELOCITY The rate of change ofangular displacementofa bodywith respect to time is called angular velocity. de 0)=- dt ifa bodyisrotatingat Nr.p.m.then correspondingangularvelocity 21tN 0)= 60 rad/s If the body is rotating 0)rad/s along a circular path of radius r, then its linear velocity (v) is given by v = (l}r 1 and Foc2 R On combining the above two expressions, Foc MIM2 R2 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 displacementis a vectorquantity.Angular displacementO canbemeasuredin radians,degreesor revolutions. 1revolution = 21tradians = 360degrees A-5Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 9. Strain tensor is used to define the state of a strain at a point c :normal strain y: shear strain • STRAIN TENSOR b Shear strain = Shear angle (<l» BJ... L B B'p It is defined as the change in initial right angle between two line elements which are parallel to x and y axes respectively. V=/bt 'bV 0/ ob ot Cv =--:Y-=T+-';+t For a sphere, oD Cv = - D : diameter of sphere D SHEAR STRESS 'bV Cv =v:s, +cy +cz Another example of rectangular block is considered p+-~~~----~-~----------------------------------------------¥------~-!--+p I I I I : La : I I : L, : ~'---------------------+' Compressive Tensile Strain Strain Consider a rod oflength La subjected to load P Volumetric Strain (s.) ILateral Strain (Slateral) I I Longitudinal Strain (Slong) Shear StrainNormal Strain Strain I Strain is defined as the ratio of change in dimension to original dimension. STRAIN L2 Elongation of a conical bar under its self weight = ~E DIL: : 8'1. 1 y = selfweight per unit volume 2 Elongation of a prismatic bar under its self weight = yL 2E p Elongation of a tapered bar subjected toaxial load P 'b=~ AE Oll Zxz = 10(i.e., shear stress acting onx face along Z direction) Z = 0zy o = 100 o = 50 o = 25x 'y 'z Elongation of a bar Subjected toaxial load P Units: MPa [ 100 120 10] rr = 20 50 0 10 0 25 Engineering Mechanics and Strength ofMaterialsA-6 Badboys2Badboys2 Badboys2
- 10. oct IfJ.! = 0 => - = 0 do J.!= I ---- f--------------- --- ~dr do " 1..- Lo ..., .. Lr ... " , => 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. I -lateral strain I J.!= longitudinal strain K=ooFor HYDROSTATICSTATEOF STRESS (NO DISTORTION, ONLY VOLUME CHANGES) (J I 1 1 1 ........ ",...",..,; - -------------~".",.. --+-~(J -----------,... " 1 ~_+--____,,_,,"''',... 1 1 1 1 1 1 1 (J K= Bulk Modulus (K) Normal stress o 1 for a given 't, G ex. -. Y Shear Modulus or Modulus of Rigidity As per Hooke's law, Shear stress ex. shear strain 1't=Gyl Engineering Stress (o)= 0" I . ngma x section area Instantaneous load True stress = . Instantaneous x section Load 'E' is the slope of o vis ~ Ediagram EL ---------------------- B PL ------------- ~ Yxy=Yyx Strain Tensor in 3D [ Ex Yxy/2 YXZ/2] [EhD = YyxJ2 Ey Yyz/2 Y2xJ2 Yzy/2 Ez Relationship Between Elastic Constants E=2G(1 +J.!) E=3K(1-2J.!) 9KG E= 3K+G E 1 G=-x-- 2 1+J.! E 1 K=-x-- 3 1-2J.! Value of any Ee ~ 0 Note: J.!cork= 0 Young's Modulus or Modulus of Elasticity As per Hooke's law upto proportional limit normal 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 = 200 GPa AI 3 .. (oI)MS <(ol)CI <(0)Ai Elastic Limit: Maximum value of stress upto which a material can be completely elastic. ProportionalLimit It is the maximum value of stress upto which materials obey Hooke's Law. Shear strain like complementary shear stress are equal in magnitude but opposite in direction. [EhD = [Ex YXY/2] Yyx/2 Ey • A-7Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 11. Let, L = Length of the bar F where, E = Young's modulus of elasticity A= Area of cross - section L = length ofwire I= increase in length ofwire Extension ofa tapered bar: Let us consider a circular bar whose diameters are DJ and D2as shownin figure. Let'F'be thetensile loadwhichis appliedaxially. Uy = _1_(Stress)2 2E or EAI where F=---, L Energy stored/unit volume ofwire u, =..!.E(Strain)2 2 1 Energy stored in wire, (U) = "2FI _ h + J.l E2) _ (E2+ J.l Ed al - E 2' a2 - E 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 El = E [al - J.l(a2 + (3)] E2 = ~ [a2 - J.l(a3 + ad] 1 E3 = E [a3 - J.l(a1 + a2 )] .. for biaxial state of stress/plane stress problems a3 = 0 but E3 *- 0 al = E(El) + J.l a2 a2 = E E2 + J.l a 1 or Relationship Between Principal Stress and Principal Strain (a) 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. (b) 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. (c) Ductility: It is the property of the material due to which if can be drawn into thin wires. The length ofdeformation is very large in a ductile materiaL (d) 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. (e) 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. (f) 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. (g) Hardness: It is the property of the material due to which it resists cutting, scractehing, pinetration or inditation. PROPERTIES OF MATERIALS 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. Eng. stress vis Eng. strain curve MS under tension test ~--------------------------------~~ Engineering Mechanics and Strength ofMaterials F o A-8 Badboys2Badboys2 Badboys2
- 12. 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. THIN CYLINDERS (e) Continuous beams: In continuous beams, there are more than two supports. II (d) Fixed beam: In fixed beam, both of its ends are rigidly fixed into the supporting walls. I (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. Simply supported beam: A simply supported beam has both of its ends are supported. (b) L ~ ~ 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. 8L = FILl + F2L2 + F3L3 Al EI A2E2 A3E3 Ifthe bars are of same material, then EI = E2= E3= E, then, 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 = crILl +cr2L2 +cr3L3 Now, E E E 1 2 3 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 = FI= F2= F3 Let, 8L = total change in length IE F~ A" E, IA2, E, 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. 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 2 8L = wL = Pg L2 crE 6E 8L = wL 2AE Elongation of a bar due to self weight: Case (i) For uniform cross - section: => Ifbar is of uniform diameter 'D', then, ~ E = Young's modulus of elasticity Extension of tapered bar (8L), 8L = 4FL 1t E DID2 A-9Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 13. Beams I Statically Statically Determinate Beam Indeterminate Beam I ~J J t J lCantilever Simply Over Fixed Propped Continuous Supported Hanging Beam Cantilever Beam Beam Beam Beam X I I Pil I ILp i 1+ve shear force0 X I I I Ii P 1i-ve shear force pLI GI I I Bending Moment Sign Convention X I C] D Ci)I I I I concave I X upwards +ve bending (+ve) moment SAGGING BENDING X I CJ D Ci)I I I I X HOGGING convex -ve bending upwards -ve HOGGING BENDING SHEAR FORCE AND BENDING MOMENT DIAGRAMS ~ 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 Moreover,it can be seen from expressionsof Ehoopand Elongthat Elong < Ehoop :. The chances offailure ofthin cylinderismore longitudinally. pd pd along = -4--' ahoop = t 11eJ 2t 11u pd pd .. a 1 = 21' a2= 4t al pd Absolute ~max = 2= 4t _ 8d _ pd (2 _ ) Ehoop - d - 4tE ~ - 6L _ ~(1 - 2 ) Elong - L - 4tE ~ 6V pd Ev = -y= 4tE(5-4~) STATEOF STRESS ATA POINT IN THIN CYLINDER pd pd along = 4t' ahoop = 2t Sometimes11 ofcircumferentialjointand longitudinaljointaregiven. In that case, -----If---~ (jlong z ,1 __ - y )-x Example of Thin Cylinder: HydraulicCylinder. Example of Thick Cylinder: LPG Cylinder, Steam Pipes. Assumptions for Thin Cylindrical Vessels ~ Stresses are assumed to be uniformly distributed as thickness 't' is small. ~ Radial stresses are neglected. Spherical Pressure Vessel Cylindrical Pressure Vessel on the basis of shape of shell Thin d : diameter d t: thickness -> 20 t Thick d -~ 20 t Engineering Mechanicsand Strength ofMaterials Pressure vessel A-tO Badboys2Badboys2 Badboys2
- 14. PAy 't=--- INA' b Shear Stresses in Beams D ~ diameter of log (given) ·. final dimensions of strongest rectangular cross-section are d bx = b[~] 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 Zl1 = Z22 = Zxx Ml1 = M22 = Mxx (crb)ll = (crbb = (crb)xx ·. (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. d = d Ixx ~L depth should be varied parabolically. (ii) width 'b' is varied (+ve) 11b x Beams of Uniform Strength A beam is said to be a beam of uniform strength when bending stress developed at each and every cross-section is same. (crb)max Y (crb) = y . max A beam offering higher moment of resistance is stronger. I-section beams are strongest as they have high section modulus. Fora giveneross-sectionalareaandmaterialsquarecross-section beam is strongerthan circularcross-sectionbeam as Zgquare > Zcircle Thisfibre is subjected to tension NeutralAxisis neitherin tension nor in compression This fibreis subjected to compression M (crb)max = ±-- ZNA . . ~A t => (crb)max "l.. => chances of failure "l.. For a given beam, (crb) ex y 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 NeutralAxis. From bending eq", crb = i ~to be used when 'R' is known. As BM = const above beam is under pure bending. Bending Equation I -ve ML..--------l P-0~~I I MA=:P(CD) M ~ P(CD) I I I 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-l1Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 15. ---)-e ~ +ve---)-Deflection upwards (+ve) e = dy dx JMxx+C, = EI(:) --> slope equation is obtianed If Mxx + C1x+ C2 = EI(y) ---)-deflection eq" Sign Convention 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 e :slope Y : deflection R : radius of curvature d2y Mxx -- - -- dx2 R sr., DEFLECTION OF BEAMS 9 I----~ rmax = "8 tavg ~ Squarewith DiagonalsVertical: ~ T section: Shear stress Distribution 1'max = "2 1'avg' 1'NA = "3 1'avg 1'max 9 --=- = 1.125. 1'NA 8 43 4 in a circular cross-section 1'max = "3 1'avg· ---)-For square, circle, rectangle, 1'NA is the maximum shear stress. But in triangular cross-section, it isn't so. In triangular cross-section, 4 K=- 3 For circle, A= bd 3 K= - 2 A = a2 3 K= - 2 P where, 1'avg = A· Expression for Maximum Shear Stress Across Various Cross-Sections .. r a: y2 (parabolic variation) As r o: f(y2) :. As 's' t t ~ at extreme fibres l' = 0 1'= b By using the above formulae, we get P shear force on plane of cross-section. A area. y distance of hatched portion from neutral axis. INA: moment of inertia of entire cross-section about neutral axis. b width. Consider a Beam of Rectangular Cross-Section ~ I section: Engineering Mechanicsand Strength ofMaterialsA-12 Badboys2Badboys2 Badboys2
- 16. WL WL3 Mmax = 4; Ymax = Yc = 48EI WL2 9 =9 =- max A, B 16EI Case III: Simply supported beam subjected to uniformly distributed load L ML ML2 9max=9BA= 2EI; Ymax=YC=--·, 8ill Case II: Simply supported beam subjected to concentrated point load 'W' at mid-span. W 1 I 1 Arl ~*~C------------~1B 4$ ~Ll2 GAMr-----: -------'ltF-----C -------::I~ ~ 4$ :G_)1 Ll2 >1 Ll2 1 IE '" WL2 5 WL3 9B = 9c = 9max = 8EI' Ymax= Yc = 48 EI Case VI: Cantilever beam subjected to uniformly distributed load over half its length from fixed end. Aro~o~WN/m DC 7 WL4 WL3 Yc = Ymax = 384 ill' 9max= 9B = 9c = 48EI Expressions for Deflections in Simply Supported Beam Case I: Simply supported beam subjected to pure bending. Alr(---L-/2--~!_B------~OC ML ML2 9max = 9B = EI' Ymax= YB = 2EI Case V: Cantilever beam of length 'L' subjected to point load 'W' at its mid-span. J.I-----" -L -DB)~~.~------------------------~>M Case IV: Cantilever beam subjected to concentrated moment 'M' at free end. WL4 Ymax = YB = 30EI WN/m A L IB Case III: Cantilever beam subjected to uniformly varying load WL4 Ymax=YB= 8EI W N/m A For cantilever, y = Ymaxat x = 0 _WL3 .. Ymax=~. Case II: Cantilever beam subjected to uniformly distributed load L EI d4y __,.4 times integration to / Wxx = dx4----rJ? obtain deflection 'y' load intensity Expression for Deflection in Cantilever Beams Case I: Cantilever beam subjected to point load W at free end X W EI d3Y ~ 3 times integration to JC Fxx = dx3 obtain deflection 'y' shear force A-13 9 (-veDeflection downwards (-ve) Also, Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 17. TL 9 = GJ GJ i => 9 ,j.. => <P,j.. => r ,j.. => chances of failure ,j.. ~ A shaft offering higher value of Tr' has more strength. Shafts with high value of polar section modulus are preferred. ~ Torsional Rigidity GJ : Torsional rigidity d K= - D 9A' = 9B, = 9C' = 9A = 9B =ge <PA= <PB= <Pe= <P <PA'= <PB'= <PC'« <p) T 'tmax = - Zp Zp : polar section modulus . 7t 3 For sohd shafts, Zp = 16 d For hollow shafts, Zp = 1: D3 (1 - K4) D : outer diameter d : inner diameter tA = 'tB = 'te = 'tmax 'tA' = 'tB' = 'te' = 'tmax I C Cross-section of a shaft at free end -B Moreover, <pex rand 9 ex L A I 9 : maximum angle of twist. <P:maximum shear angle. J : polar moment of inertia. Tr : Twisting moment R9 Now,<p= L RLJ Torsion Equation e: angle of twist <1>: shear angle L : distance of cross-section from fixed end ---~~::~_~_~~~~~----~L------~ Cm~ P Shear Stress Distribution Pure Torsion A member of a shaft is subjected to pure torsion when the magnitude oftwisting moment remains constant throughout the length of shaft. TORSION 5 WL3 Y =---W=W Be 48 EI' e A 5 WL3 YeB= 48 ill' WB=W We YeB = WB YBe A Deflection at C due to load at B Deflection at B due to load at C ,?ILoad at C = We YeB ~ ~ Stiffness of beam = Max. deflection Higher flexural rigidity is an indicative of higher stiffness of beam but lower deflection and slope. ~ Maxwell's Reciprocal Theorem (valid for beams under point load and having same L, E and I) Load /Yc doesn't give max. deflection o = Wb (a2 - ab) s/ c 3EIL doesn't give max. slope b 5 WL4 WI} Ymax = Yc = 384 ill ;9max = 9B = 9A = 24EI Case IV: Simply supported beam subjected to a concentrated point load acting not at mid-span W 1 Engineering Mechanicsand Strength ofMaterialsA-14 Badboys2Badboys2 Badboys2
- 18. 1 f= a2 ~ (end fixity coefficient) 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] 2 1t E Imin .. Pe = 2 Le Pe : Euler's buckling load. Imin : min [Ixx and Iyy]. L, : effective length of column. L : actual length of column. L =aL e 4length fixity coefficient are more. Euler's Formulae Short Columns (fail due to crushing) Long Columns (fail due to buckling) Medium Columns (fail due to buckling as well as crushing) ~ As the length of structure, chances of it failing by buckling 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. THEORY OF COLUMNS = -3.6. T1 = TA' T2 = TA- T. TA+Tc=T. 91 + 92 = <1>0 => 91 = (-92) 3T => TA= 4· G1 J1 = G2 J2 1. Net TM = T (anti-clock) Rxn = T (clock) 1'. CD T=T1+T2 9 => T = (G1 J1 + G2 J2) L 91 = 92 = 9 Shafts with Both Ends Fixed CD T Shafts in Parallel Shafts in Series ~ Torsional Stiffness (q) 2. T GJ 3. q=-=- 9 L 4. ~ Torsion of a Tapered Shaft A-15Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 19. 2 2 SE = _!_ T9 =_!_ T L =2._(AL). 2 2 GJ 4G where T : twisting moment. Zp : polar section modulus for circular x section. ~ = C~)d3 T t=- Z 'p SE of bar = work done by load P 1 P2L cr2 crEAL Strain energy of bar =- Po =-- =- x AL =--. 2 2AE 2E 2 ~ Strain energy of solid circular shaft subjected to torsion P 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. STRAIN ENERGY METHODS oe : buckling stress n2E cr =-- e 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 I I 1 -=-+- RR PE Pc where, PR = Rankine's Load PE = Crippling loadbyEuler's formula Pc= crushing load h cry.A were PR = _--=-__ 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. a4 nr4 na4 a4 I = - I = -=--=- 1 12' 2 4 n2(4) 4n (Pe)l = 4n 072 = 4n(0.72) _ (Pe)2 12· 12 -0.513 . . (2) is stronger. Rankines formula: It is a combination of Euler and crushing load. It is also known as Rankine Gordon formula. (2)(1) If remaining all other parameters are same, (Pe)BF > (Pe)FH > (Pe)BH > (Pe)FF Which of the following column is stronger? Engineering Mechanics and Strength ofMaterialsA-16 ~ Both Ends Both Ends Fixed and Fixed and Hinged Fixed Hinged Free (BH) (BF) (F &H) (FF) 1 1 a 1 - J2 2 2 1 1 11=- 1 4 2 - 0.2 4 Badboys2Badboys2 Badboys2
- 20. Now, Let o.(maximum principal stress) and o,ora3 (minimun principal stress) and a by the yield stress.y For design creterion, maximum principal stress must not exceed the working stress (aw) al,2 ~ away For considering yield creterion, a1 = ± ay or a2 = ± ay This theory is utilized for brittle meterials. (b) 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. 0"3 J 0"2 ~,. 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. Wb Wa e e In this case using the above relation, we get Wa2b2 U=--- 6EIf Theories of failure: ( b A Mxx : moment at section x-x X I I r- ~; B~~ ~1(------7) M X x ~x =-M L M2 M2L U ~ !2Eldx~ 2EI W 2 U =.!_ P8 = 2P L A 2 nd2E UB = UI + U2 2p2 (~) p2 (~) nd2 E + 2nd2 E = 0.5 UA" STRAIN ENERGY DUE TO BENDING b (M )2 u=J xx dx 2EIxxa U : strain energy P (B) P (A) Modulus of Resilience ~ Two bars A and B are as shown:- Modulus of Toughness EL PL ~2 T SE = -- (AL) (1 + K2), where t = 4G 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. 0" d K= - D K = 0 for solid K<l ~ Strain energy of hollow circular x section shaft. d : Inner diameter. D : Outer diameter. A-17Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 21. 1 61 = E [ al - Y (a2 + (3)] 1 62 = E [ a2 - Y (al +(3)] 1 63 = E [a3 -y (al +(2)] Now,According to failure creterion, 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 = maximumprincipal stress a2 = minimum principal stress a = yield stressy 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~ Let, 6)' 62, 63 = three principle strains 6 = strain at yieldingy 61,2,3::;6y Now, A-IS Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 22. PL PL (a) -- (b) - 2AE AE (c) PL2 (d) PL2 -- -- AE 2AE 13. In the above question, if w be the total weight of the bar hanging fixedat one end, then elongation(8L)willbe equal to: (a) 8L = wL (b) 8L = 2wL AE AE 2 8L = wL(c) 8L = wL (d) AE 2AE 12. IfP = axial loadapplied,A = cross- sectionalare ofuniform circular bar, L = length of the bar, E = Young'smodulus of elasticity, then elongation of the bar will be equal to : (b) Uu =..!..E(strain) 2 1 (d) Uu = - E(stress) 2 (a) U; = ..!..E(strain)2 2 (c) U,= ..!..E(stress)2 2 (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 ofthe 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: CA B A B C (a) -- = -- = -- cosu cos J3 cosy A B C (b) --=--=-- sm u sin J3 sin y C A 4. 3. Iftwo co-planar forces 'PIand '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 : (a) R = Jp2 +Q2 +2PQsinQ (b) R = Jp2 +Q2 -2PQsin9 (c) R = ~p2 +Q2 +2PQcos9 (d) R = Jp2 +Q2 -2PQcos9 In the question number 2, the direction or angle mode by the resultant will be equal to: (a) u = tan-I (Q:~::se) (b) a = tan-1 ( Pcos9 ) Q+Psin9 (c) a = tan-1 ( Psin9 ) Q-Pcos9 (d) a = tan-I ( Pcos 9 ) Q-Psin9 Which of the following expressions represents Lami's theorem,ifA, B, Carethree are in equilibriumand as shown in figure. 2. (a) F=G MIM2 (b) (MIM2)2 F=G R2 R2 (c) MIM2 (d) (MIM2)2 F=G-- F=G R R If M1 and M2 are two masses of two bodies, 'R' is the distance betweentheir centers then which ofthe following expression represents gravitational law 'G' is universal gravitational constant. B 1. ...,EXERCISEI···.. A-19Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 23. wL4 wL4 (d) 8max = 8B = --, Ymax = YB =-- 2EI 3EI 28. If t = shear stress, G = shear modulus and v = volumn ofthe body then the expression of strain energy stored withing (a) wL2 wL2 8max = 8B = --, Ymax =YB =-- 3EI 5EI (b) wL3 wL4 8max = 8B = --, Ymax =YB =-- 6EI 8EI (c) wL6 wL6 8max = 8B = --, Ymax =YB =-- 6EI 4EI 27. Ifcantilever beam is subjected to uniformly distributed load (UdI), then expressions for deflection are given by: l= G8 =!_(d) "["max L R T, = G8 = "["max L J R (c) "["maxL J G8 RTr = G8 = "["max (a) J L R r (c) v = - (d) y= 0)2r (J) 24. 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 25. 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 26. Torsion equation is given as: if8 = maximum angle of twist, J =Polar moment of inertia, Tr = Twisting moment, R8 <I> = maximum shear angle = L 23. 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 21tN (b) -rad/s 360 21tN (d) -rad/s 180 21tN (a) --rad/s 120 21tN (c) --rad/s 60 (b) Rate of change of momentum is inversely proportional 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 22. If a body is rotating at N rpm, the corresponding angular velocity will be equal to : 20. 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 21. 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 (b) "v"(')1-(')2 2 2 (d) o y = o 1 - o 2 19. 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 2 2 (c) (')y= (')1 + (')2 1 1 1 1 1 1 (a) -=-+- (b) -=-+- PE PR Pc Pc PE PR 1 1 1 (c) - = --- (d) -=-+- PR PE Pc PR PE Pc P (Ll L2 L31 (a) 8L = E Al + A2 + A3) 8L = PE(.!1_+~+ L31) (b) Al A2 A3 P (c) 8L = p(LIAI + L2A2 + L3A3) (d) 8L = PE (LIAI + L2A2 + L3A3) 15. 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 16. 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 17. A beam whose both ends are supported is known as : (a) simply supported beam (b) fixed beam (c) overhanging beam (d) continuous beam 18. IfPR = Rankin's Load, PE = crippling load by Euler's formula ~nd.Pc = crushing load, then Rankin's formula for columns IS given as : P~ A"E I A"E A3,E ~P IE ~IE ~IE ~I L1 L2 L3 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 Engineering Mechanics and Strength ofMaterialsA-20 Badboys2Badboys2 Badboys2
- 24. 42. 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 (c) (b) Lx Al. 2L (d) ~L L ~L ~L L (a) 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 : Dr-D~ (d) 4Dl(c) (a) IfDI = external diameter ofshort column D2= internal diameter of short column F = external load applied The highest value of eccentricity will be equal to : Df-D~ (b) 8D1 39. (c) PP (l-!':.) (d) PP(I-!':. )2tE 2 3tE 2 38. If the both the ends of the column made of mild steel are hinged, then Rankin's constant value will be equal to : 1 1 (a) -- (b) -- 7500 6500 1 1 (c) -- (d) -- 5500 4500 (c) ~('L~)2 +('LFv )2+2('LFH)('LFv) (d) ~('LFH)2 +('LFy )2 -2('L~)('LFv) 36. In the above question, the direction or angled of the resultant will be give by: LFy(a) tana=-- Lftl (b) tana=Lftl LFy (c) tan a= 2)H x2)v (d) tana=2 x LFH x~:)v 37. IfD = diameter ofthin cylindrical shell L=length t = thickness ~ = internal pressure, J..l= Poisson's ratio then Hoop strain will be given by: (a) PP(2-!) (b) PiD(1_!:) 2m 3J..l 3m 3 ~ (b) ~YF: 33. A cyelindrical elastic bodysubjected 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. When number of forces are acting on the body and 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; Longitudinal strain Lateral strain (a) Lateral strain (b) Longitudinal strain stress strain (c) strain (d) stress 32. Poisson's ratio is described as the ratio of: 29. If oI' cr2,cr3are three principal stresses, J..l= Poisson's ratio and E = Young'smodulus of elasticity,then the expression for strain energy/volume is given by (a) ![crt + cr~+ cr~+ J..l(crlcr2+ cr2cr3+ cr3crd] (b) _!_[crt + cr~+ cr~- J..l(crlcr2+ cr2cr3+ cr3crt)] E (c) 2~ [crt + cr~+ cr~- 2J..l(crlcr2+ cr2cr3+ cr3crd] (d) _1_[crt +cr~ +crj +2J..l(crlcr2+cr2cr3+cr3crd] 2E 30. Impact strength ofa material represents: (a) Hardness (b) Resilience (c) Ductility (d) Toughness 31. 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: I (a) IE ="2 (b) IE=21 I (d) IE="4 the body is given as : '[2 '[ (a) -xV (b) -xV 2G 2G '[2 r (c) -xV (d) -xV G G A-21Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 25. (a) 1600x 10--6 (b) 400 x 10-6 (c) 800 x 10-6 (d) 300 x 10-6 65. In terms ofPoisson'sratio (u),the ratio of young'smodulas ofelasticity(E) to shear modulus (C) of elastic material is : (a) 2 (1- u) (b) 2 (1+ u) (c) (1- Jl) (d) (1+ Jl) 2 2 (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~ssO"ei~ its perpendicular direction. O"tand 0"earenot equalInmagnitude,thenontheplaneofmaximum 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, thenthe maximumshearstrainwillbe equalto 20emO 1 0 em 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 valueofPoisson'sratio foraluminium 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"w<O"u (c) O"w>O"u (d) None of these 60. The point of contraflexure is found to be in which of the followingbeam? (a) cantilever beam (b) Simplesupportedbeam (c) overhanging beam (d) None of these 61. A largeplate (uniform)consistingofa rivethole is subjected touniformuniaxialtensilestressof 100MPa.Themaximum stress in the plate will be equal to : 1 (c) 0" ocz2 (d) O"boc- b z2 51. 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 52. 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 53. Stiffness is measured in which of the following: (a) Modulus of elasticity (b) Toughness (c) density (d) ultimate strength 54. Percentage elongation is associated with which of the following terms during tensile test? (a) Malleability (b) creep (c) Hardness (d) ductility 1 (a) O"boc- z 50. 49. 48. 47. 46. 45. 44. 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 givenmaximum 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) ductilematerials (c) brittlematerials (d) None of these If llwefficiency of weldedjoint, llR = efficiencyofriveted joint, then which of the followingrelation is true: (a) llw>llR (b) llw<llR (c) llw=llR (d) llw~llR When the cyclic or repeated stresses are applied to the material, then its behaviour is termed as : (a) creep (b) fatigue (c) stiffness (d) endurance I( O"b= stressin a beam,z = sectionmodulus,then, which of the following expression represents the relation between them: 43. Engineering Mechanicsand Strength ofMaterialsA-22 Badboys2Badboys2 Badboys2
- 26. (d) Y "C aL (c) Y (a) ac (c) -= E aT Free Bodydiagram shows: (a) No forces are acting of the body (b) All the internal forces acting on the body (c) All the internal and external forcesacting onthe body (d) None of these During tensile testing for cast iron specimen, the stress- strain curve shows: (a) No yield point (b) upper yield point only (c) lower yield point only (d) Both upper and lower yield points 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 IfG = Modulusof rigidity,"C = shear stress, Y= shear strain, aL = Longitudinal stress, E YL = Longitudinal strain, then the expression for 'G' will be given by: 75. 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 76. The unit of stress in SI system is given as : (a) 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 n2EI n2EI (a) 4L2 (b) 2L2 n2EI n2EI (c) 8L2 (d) 16L2 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 : 1 .. 1 .. (a) - x ongmal value (b) - x ongmal value 2 8 1 .. 1 .. 1 (c) - x ongmal value (d) - x ongmal va ue 4 16 73. Ifthe length of the column is doubled, the value of critical load becomes : -8F (d) nd2(c) Fi 2FI (a) (b) - 4 9 78. FI FI (c) - (d) - 9 3 71. The secondmoment ofa circular area aboutthe diameter is given by if'd'is the diameter: 79. (a) nd4 (b) nd4 64 32 (c) nd4 (d) nd4 -- -- 16 8 80. 72. A circular rod ofdimeter '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 81. 1~) (~ F 6F -12F (a) nd2 (b) nd2 (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 I span l at a distance of"3 from the left - end. The bending moment at the point of application ofload is given by : 66. If the principal stresses in a plane stress systems are 100 MPa and 40 MPa, then maximum shear stresswill be equal to: (a) 30 (b) 40 (c) 100 (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 = 502MPa, 0'2 = 62 MPa (c) 0'1 =2018,MPa,a2=46MPa (d) 0'1 = 702 MPa, 0'2 = 88MPa 68. In case of simply supported beam on two end support, the valueofbending moment ismaximum will be : (a) On the supports (b) at mid - span (c) where there is no shear force (d) where the deflectionismaximum 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: )P nnlnmllLm A-23Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 27. (b) S< 32 (d) S~32 97. 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 valueof's' for shortcolumn should be in the range of : (a) S=32 (c) S>32 (c) (b) T x J (d) T x o T J T e (a) 95. 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) a; (d) None of these 96. If T = Torque transmitted, O = angle of twist, J = Polar moment ofinertia, then torsionalrigidity ofthe shaftwillbe equal to: (d) 16(c) 2 1 8 (b) 4(a) (c) 3ae (l-v) (d) 3ae (l+v) 2E 2E Ifwe round a thin cylinderwitha 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 PR PR (a) - (b) 2H H PR PR (c) (d) - 4H 8H 92. In a thin spherical pressure vessel, the volumetric strain is given by: (a) 3ae (l-v) (b) 3ae (1+ v) E E 91. IfH = wall thickness, P = pressure, R = mean radius, then maximum shear stress in case ofthin cylindrical pressure vessel will be : (a) ah = 2 (b) ~=4 at at (c) ah = 8 (d) ah = 16 at at 90. (a) Hyperbolic (b) linear (c) Parabolic (d) None of these When the concentrated load is applied, then the nature of variation ofbending 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 89. 88. 87. 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 forceand bending moment diagram, ifthe 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 ofvariation ofthe bending moment diagram will be 86. (c) Wl2 (b) 3(a) wl2 84. 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 : 93. wL3 (b) wL2 (a) 8 4 wL2 wL2 94. (c) 8 (d) 16 85. 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 : wL 2 (d) wL 8 (c) wL 4 (b) wL 3 (a) 82. 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 83. 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 : Engineering Mechanicsand Strength ofMaterialsA-24 Badboys2Badboys2 Badboys2
- 28. F (d) 2(c) F (b) FL FL 2 (a) (a) Jlk (b) &_ R R (c) Jl~ (d) Jl~ 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: 1 (a) <I>=a (b) ¢=- a (c) <1>=a2 (d) a= 2<1> 118. If <I>= angle of friction, u= coefficient offriction, then which ofthe following relation is true? (a) <I>=COC1(Jl) (b) <I>=tan-1(Jl) (c) <I>=sin-1(Jl) (d) <I>=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 e (b) w cos e (c) wtan e (d) w cot e 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 : 116. If <I>= angle of friction, Jl = coefficient of friction, then which of the following relation is true? (a) Jl= cot <I> (b) Jl= sin <I> (c) Jl= tan <I> (d) Jl = cos <I> 117. If <I>= angle of friction, a = angle of repose, then which of the following relation is true? R (a) Jl=- (b) Jl=RxF F F Jl=F2R(c) Jl=- (d) R 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: (b) Normal reaction and frictional force (c) Force on the body and normal reaction (d) None of these 112. 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 and normal reaction 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: D D (a) - (b)- 2 4 (c) 2D (d) 4D 104. 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 105. 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 106. 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 107. 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 108. For aquiring equilibrium condition, How many are minimum number of coplaner and non - collinear forces required? (a) 1 (b) 5 (c) 3 (d) 4 109. Ifthree 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 110. 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 111. Angle offrictiion is defined as the angle between (a) normal reaction and the resultant of frictional force A-25Engineering Mechanics and Strength ofMaterials Badboys2Badboys2 Badboys2
- 29. 141. Tangent of angle offriction is equal to: (a) kinetic friction (b) Limitingfriction (c) Frictional force (d) coefficientof friction 142. The coefficient ofrolling 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 600N. The forcerequired to roll the wheel will be equal to: (a) 90N (b) 180N (c) 45N (d) 270N 143. The ratio oflinear stress to linear strain is given by : (a) Modulus of elasticity (b) Modulus of rigidity (c) Bulkmodulus (d) Poisson's ratio 144. the value of Poisson's ratio is always: (a) less than one (b) greater than one (c) equal to one (d) None of these 145. Young's modulus of elasticity for a perfectly rigid bodies 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 cylindricalshellofdiameter200 mm andwallthicknessof 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 Nzmm- 148. The bulk moduless of elasticity: (a) does not increase with pressure (b) increases with pressure (c) independent of pressure (d) None of these (c) (b) 5 2 (d) 5 10 3 3 10 (a) 135. The unit of moment is: (a) N-m (b) N/m (c) N/m2 (d) N/m4 136. The quantity,whichis equaltorate ofchange 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) td<ts (b) td>ts (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 valueofslidingfriction willbe: (a) less (b) more (c) equal (d) double 140. Abodyofweight30N rest ona horizontal floor.Agradually 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 : 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 ofparallelogram offorces gives the resultant of: (a) Parallel forces (b) Likeparallel forces (c) two coplanar concurrent forces (d) Non coplaner concurrent forces F (d) 4(c) J3F 130. Which of the following is a vector quantity (a) Energy (b) Mass (c) Angle (d) Force 131. Twoforcesof equalmagnitude 'F' act an angle of 1200 with each other. Then their resultant will be equal to: (a) 2F (b) F T r M T (a) - = - (b) - = -- J R o Ymax T G9 r G9 (c) - = - (d) - = - J L R L 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 associatedfordesigning of shaft base on strength is given by: wz2 (d) 16 2 wz2(b) 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 : wz2 (a) w z2 (b) 8 w/2 w/2 (c) 4 (d) 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 : wz2 (a) 8 Engineering Mechanics and Strength ofMaterialsA-26 Badboys2Badboys2 Badboys2
- 30. 170. Uniformalydistributedload'WI isacting overperunit length ofa cantilever beam of Sm length. If the shear force at the midpointofbeamis6KN.thenth valueofwwillbeequalto : (a) 2 KN/m (b) 3 KN/m (c) 4KN/m (d) 6KN/m wi 6 (d) wi 8 (c) 3wl 8 (b) wI 4 (a) 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. IfEsteel= 3 xEAI' then the stress in steel will be equal to: (a) 10N/mm2 (b) 20N/mm2 (c) 30N/mm2 (d) 40N/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 : lOT~ ~ ~ r9T IE 10mm "IE "IE "I 10mm lOmm 26xlO 16x 10 (a) ---mm (b) --mm AE AE 6xl0 32xlO (c) --mm (d) ---mm AE AE Strain (a) Brittlematerial (b) Hardmaterial (c) Softmaterial (d) ductilematerial 163. Necking pheneomenon in stress-strain curve is observed for: (a) ductilematerial (b) brittlematerial (c) (a) and (b) both (d) None of them 164. 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)2.0 165. 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 166. The total extension ofthe bar loaded as shown in figure is : 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) elasticlimit 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 100kg lamp is supported by a single cable of diameter 4 mm. The stress carried bythe cable will be equal to : (a) 40 MPa (b) 78MPa (c) 48 MPa (d) 88MPa 157. The modulus ofelasticity and rigidity ofa 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 compositebar ofcopper 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 deformationofmetals 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 (b) 0.4 (d) OJ plane of maximum shear stresswill be : (a) 2KN/mm2 (b) 4KN/mm2 (c) 8KN/mm2 (d) 3KN/mm2 151. Ifelasticmodulus(E)= 12GPa, shearmodulus(G)= 50GPa, then the value of Poisson's ration for the material will be eqaul to: (a) 0.1 (c) 0.2 149. To represent, stress - strain relations for a livearlyelastic homogeneous and isotropicmaterial, minimum number of material constants required are: (a) 2 (b) 3 (c) 1 (d) 4 150. A tension memberwith a cross- sectionalarea of30 - mm- resists a load of 60 KN. The normal stress induced on the 162. The stress strain curve below represents for: A-27Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 31. E1 n2E1 (a) Pcr=~ (b) p =-- nL c 3L2 (c) rrEI n2E1 Pcr=7 (d) p =-- c L2 00 Q~ ~ Q~ (c) 0.98 (d) 1.37 186. A ballweighing 0.01kg.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 2s 2s (c) gsin9(tan9-1-1) (d) gsin9(tan9+ 1-1) 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 forceis 08±L(a) zero (b) 0.8N (c) 0.89N (d) 1.2N 185.A blockR ofmass 100kg is placed on a block S ofmass 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 surfaceis 0.4, the minimum forceF (in kN) needed to move the block S is g cos9(tan 9+ 1-1)g cos 9(tan9 -1-1) (b)(a) 2s2s (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, shownin the figure below. The coefficientof 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 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 supportedbeamPQ oflength 9 m, carries a Udlof 10 KN/m fora distance of 6m from end P.What will bethe 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 Udlof0.4N/m. The maximumbendingmomentoccuringin the beam will be: (a) O.OSN-m (b) 1N-m (c) 2N-m (d) 4N-m 174. A hollow shafthas external and internal diametersof 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) 628Kg-cm (d) 324Kg-cm 176. For designing a connecting rod, which of the following formulais utilized? (a) Rankin'sformula (b) Euler'sformula (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 ofmass 1kg 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 forceF. 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 ofvirtual work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be Engineering Mechanicsand Strength ofMaterialsA-28 Badboys2Badboys2 Badboys2
- 32. 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 4ML2 (d) EI(c) 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 cross- section about the neutral axis is I and the Young's modulus is E. The magnitude ofthe maximum deflection is ML2 ML2 (a) 2EI (b) EI (c) 2a(LlT)E (l-2v) a(LlT)E 3(l-2v) (b) (d) a(LlT)E (l-2v) 3a(LlT)E (1- 2v) (a) 194. A solid circular shaft of diameter d is subjected to a combined bending moment M and torque T. The material property to be used for designing the shaft using the relation .!.i_.JM2 + T2 is nd3 (a) ultimate tensile strength (Su) (b) tensile yield strength (Sy) (c) torsional yield strength (Ssy) (d) endurance strength (Se) 195. 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) 20 196. 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) 223.6 197. 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 198. 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 199. 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 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 P P 1+-[; t 2L .; L--+I L ~p2J3 2p2L3 (a) (b) -- 3EI 3EI 4p2J3 8p2JJ (c) -- (d) 3EI 3EI i q q illllllllll]l M(illllllilit 1II1II L IJIII R, R2 5qL 3qL qL2 (a) s, = -8-' R2= -8-' M= 8 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 191. The strain energy stored in the beam with flexural rigidity EI and loaded as shown in the figure is (c) tid' (d) nd' 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) so (c) 100 (d) 110 190. A uniformly loaded propped cantilever beam and its free body diagrams are shown below. The reactions are 8T16T (b)(a) 188. For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is A-29Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 33. where, fc= calculated average axial compression stress fb = maximum allowable bending compressive stress on the extreme fibre, and fe' = calculated bending stress in the extreme fibre. (a) fe' + f; <1 (b) f:_h' <1 Ie h Ie h (c) fe' + f; > 1 (d) f:_h' >1 Ie h Ie h 214. The design ofa eccentrically loaded column needs revision when C (d) 16EI C 96EI (c) r! (b) 192EI C 48EI (a) 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 I: c (B L .j4 L ----.j (a) pC 2PL (b) pC PL 3EI' 3EI' (c) 8PC 2PL (d) 8Pr! PL 3EI' 3EI' 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 n2EI 2n2EI (a) L2 (b) L2 3n2EI 4n2EI (c) L2 (d) L2 1.5 tim (d) Afooaf4t/m 1.5 tim Ai.__ ~i~10_t c~f~~uouou~uounuo~ou!3t(c) A f (a) 4t ~4m""""I---- 9t .------.14t 208. For the shear force diagram shown in figure, the loaded beam will be K-G (b) J.l= 2G+6K K-G (d) u= G+3K 3K-G (a) J.l= 2G+6K 3K-2G (c) u= 2G + 6K 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 ina 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 209. When a column is fixed at both ends, corresponding Euler's critical load is Engineering Mechanicsand Strength ofMaterialsA-30 Badboys2Badboys2 Badboys2
- 34. (b) Yxy/Ji (d) 2yxy (a) Yxy (c) Yxy/2 229. 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) o = EEl (b) E o = --2 [El +J.lE2] I-J.l (c) E (d) o = E[EI - J.lE2]o = --2 [El- J.lE2] 1- J.l o, where, k = 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 o1 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. (d) 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 (c) ]_~M2 +T2 2 ~IM+~M2 +T2l (a) 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. (d)(c) (b) 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 : p2L3 (a) 3EI 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 (c) 9KG E= K+G 9KG E = 3K+G (b) (d) KG E = 9K+G 9KG E = K+3G (a) 215. A gun metal sleeveis fixed securelyto a steel shaft and the compound shaft is subjected to a torque. If the torque on the sleeveis twicethat onthe shaft, findthe ratio ofexternal diameter of sleeveto diameter ofshaft [GivenNs= 2.5 NG] (a) 2.8 (b) 1.6 (c) 0.8 (d) 3.2 216. A column sectionas 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/ m2and 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) 12t/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? A-31Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 35. ABC (b) 4 1 3 (d) 2 4 3 C 1 2 B 2 1 Codes: A (a) 3 (c) 4 3. Riveted joints 4. Continuous beam C. Perry's formular Deflection of beam Eccentrically loaded column A. Clopeyroh'sn theorem 1. B. Maculayr's method 2. C 231. Match list I with list II and select the correct answer using the codes given below the lists : List I List-D SFD (d) 230. The SFD for a beam in shown in the fig. The BMD is shown by : (c)c A-32 Engineering Mechanics and Strength ofMaterials Badboys2Badboys2 Badboys2
- 36. = 100-40 = 60 = 30MPa 2 2 a -a Maximum shear stress ('t"max) = x y 2 I . where 2 = -_ = section modulus , Ymax M = ab(max) x Z M ab(max) = Z 1 ab(max) o: Z 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]= 1000X 10-6 62=-600 X 10-6 Maximum shear strain (ymax) = 6] - 62 = (1000 x 10-6)-(-600 x 10-6) = 1000x 1Q-6+ 600 x 1Q-6 = 1600x lQ-6 As we know that, E=2C (1+ u) ~=2(1+Ji) Given: Principal stresses, a = 100MPa, a =40 MPax y For plane stress system, QJA 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. LlL Hence,6=T Fatigue is a fracturephenomenon in which material is failed due to cyclic or repeated stresses usually at low values of stress. As we known that, I M = abc (max)x-- Ymax 66. (a) 65. (b) 64. (a) 61. (a) 53. (a) 50. (a) 49. (b) 41. (c) 33. (c) 30. (d) .c=_wL2 _ wLNow,Total elongation - -- - -- - -- o E 2E 2AE Here, w = WAL. 14. (a) Considering Principal of superposition, 8L=8L]+8L2 +8L3 PLI PL2 PL3 =--+--+-- EAI EA2 EA3 = !(.!i_+~+ L3JE Al A2 A3 Toughness of the material is defined as the maximumamountofenergystoredin a materialupto fracture under the application of impact loads. 8L= wy8y E Let, w = specific weight of bar F=wAy elongation(8L) = _FL_= ....:....(w_A_y.....:...)_.d_y AE AE AE L P 8L S . () Stress(a) tram 6 =--- E I . PL E ongation 8L = - AE 13. (d) Let a bar be hanging of weight 'w' Load P Stress(a) = -------- Area of cross- section A 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, . 8L Stram=- L U = '!_E(8L)2 v 2 L where, E = Young's modulus of elasticity. 12. (b) As, we know that, ...,HINTS & EXPLANATIONSI···~ A-33Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 37. . n 2 Area of cross-section (A) = 4d =~(4xl0-3)2 4 w=30N N=w=30N when the bodyjust starts moving, P=9N and also, P = F = 9N we know that, F = Il R or IlN F 9 3 Il= N = 30 =10 142. (b) Given: coefficientoffriction (u) = 0.3 Load or Reaction (R)= 600N force required to roll the wheel =f..lR =0.3 x 600 =180N 147. (b) Given: diameterofshell(D)= 200mm wall thickness (t) = 5mm Internal fluid pressure(P)= 10Nrmm' Maximum shearing stress (tmro) PD lOx200 1 4t 4x5 2 =100N/mm2 150. (a) Given: Area ofcross - section (A) = 30 mm- Load(P)= 60KN 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 =~-1=~-1= 120_1 Il 2G 2 x 50 100 u= 1.2-1 =0.2 156. (b) Given: mass(m)= 100kg diameter(d)= 4mm = 4 x IO-3m Stress (c)=? F~ N 140. (c) = ~ F2 + F2 +2 cos 120 = ~2F2 -2F2 cos300 (0: P = Q = F) R=# R=F 2 2 pI = n EI = n EI (2L)2 4L2 pI L2 - - p- 4L2-"4 pI = _!_x P 4 74. (a) Given: length of column = L flexural rigidity = EI effective length = 2 x L n2EI n2EI n2EI P--------- - L~ - (2L)2 - 4L2 79. (a) During tensile testing of cast iron specimen, the stress - strain curve shows no yield point because the length of deformation is very little. 131. (b) Given; P = Q = F, 9 = 120°, R = ~p2 +Q2 +2PQcos9 nd2 nd2 nd2 nd2 4 16 -12F nd2 73. (c) For the case of column, 2 P = n EI L2 Now, If length is doubled, V=2L 2F F RA =3,RB =3 Now,at point If, the bending moment, (BM)c = 2F x!....=2Fl 3 3 9 72. (b) Considering the followingrelations, o = compressive stress (o) + tensile stress (crt) F F 4F 16F 70. (b) ;}~A B f tn ~ RA RB Pdi crN=crC=- 2t crl' cr2= 1098MPa,41 MPa Using the following relations, For thin cylinders, nd3t J=- 4 67. (a) Engineering Mechanicsand Strength ofMaterialsA-34 Badboys2Badboys2 Badboys2
- 38. ByFBD ofBook I, LFy=O=> NI =mg So,frictionalforce= flN 1 = fl mg ByFBD of Book2, LFy=0=>N2=N1 +mg=2mg LFx=O mg FBD of Book 1 181. (d) 3.14[(10)4 -(5)4]x2 32 x10 = 183.98cm3 = 184cm3 175. (c) Considering equation, T _ Ge _ Q J L R T=628 kg-em ~N/m IE 1m )1 M · b di (0.4)(1)2 axlmum en mg moment = --- 8 =0.05N-m 174. (a) Givenexternal diameter(D) = 10em Internal diameter(D) = 5em . n[(Do)4 -(Di)4Jx2 Sectional modulus (z) = 32 -.o, (PC)Ll 4n2EI 4L2 16 --=--x--=- (pc )L2 L2 n2EI 1 (PC)L1 : (PC)L2=16:1 172. (a) pf'f;lOKN~: 6~ JR 9m R p Q RP + RQ= 10 x 6= 60 KN Takingmoment about 'P', ~ x 0+ 9 x R = 180 RQ=20KN ~=60-20=40KN 173. (a) For long column, (One end fixed and other end is free) 2 (p) _n EI C L2 - 4L2 (SF)R= 6KN=w x 1.5 w=_i_=4KN 1.5 171. (d) For long column, (fixed at both ends) 2 (p) = 4n EI C LJ L2 10mm 10mm lOT~lOT lO-3=7T~7T 9T~T 8L = lOx 10 8L = 7 x 10 8L = 9 x 10 1 AE' 2 AE' 3 AE 8L=8L1+8L2+8L3 100 70 90 =-+-+- AE AE AE 260 26 x 10 =-=--mm AE AE 167. (c) Given:A=3 em- P= 12000N Esteel= 3 x EAJ Considering the relations, P=P1+P2 P=P1A1 + P2~ weget, Psteel= 30N/mm2 170. (c) 1.5 m R IE 10mm )IE )IE )1 = ~ x 16x 10-6 = 4n x 10---{jm' 4 Load = weight (w)= mg = 100x 9.8= 980N w 980 Stress(cr) = - = 6 A 4n x 10- =78.03 X 106 = 78 MPa (approx.) 157. (d) Given,E= 200GPa G=80Gpa fl=? Weknow that, E=2G(1+fl) 200=2 x 80(1 + u) (1+ ) = 200 fl 160 u= 1.25-1 =0.25 166. (a) Given, lOT" I.____ __L...~ __ ~__L... __ __,I ,9T L fTITIl ! A-35Engineering Mechanics and Strength ofMaterials Badboys2Badboys2 Badboys2
- 39. = ~q x L2 _.9.L2 = _ qL2 8 2 8 5 So, finally R1 = gqL, 3 qL2 R2 = gqL, M = -8- 3 5 R1 =qL-R2 = qxL-gqL=gqL Also, moment M = R2 x L - q x L x (~) qxL4 3 => R2 =-qxL 8EI 8 qL4 For a cantilever with UDL, 81= 8EI For cantilever with load R2 at end R xL3 8 - ---=-2__ 2 - 3EI => :EFy= 0 Rl + R2 = q x L Also, 81=~ I q/length I~ k==:::======J 190. (a) By using the relation, tN =3a(I_2V) V E 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. 189. (b) 16T t=-3 =t max nd T r 8 T r By-=-=-=> ---- J r L ~d4 d 32 2 188. (c) 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 n2 xEI PO' = 2 L ~F I I I I I I ~R2+~----~----~I------~ I it (100 + 150) = 250 kg 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 (Fs), F <Fs-so that the friction developed will equal to the applied force F = 0.8 N. 185. (d) Hint: Given FBD, For block'S' ~g F=08;:i~184. (b) gcos8(tan8-1-l) 2s => t = 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, s = ut +"!"at2 2 => s = 0 + ..!..g cos 8 (tan 8 - ~ )t2 2 183. (a) => 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 182. (b) 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. Engineering Mechanicsand Strength ofMaterialsA-36 Badboys2Badboys2 Badboys2
- 40. 3 .. A = 1x 10 = 346.4 2.886 P=5MPa,d= 1000mm,t= 10mm We know, ex = _!_[crx- v(cry + crz)] E 20 x(10)3 = 2.886 12x 10x 20 'Tmax= cry = 100 MPa 0" = - 200 MPa t"y= 100 MPa Ifmaximumandminimumprincipal stressesaregiven, the maximum shear stress is given by 'T = crl-cr2 = 100-40 = 30MPa max 2 2 Factor of safety Torsional yield strength or 8 For safe designing 'T::;~ n 16T 16~M2 + T2 Induced shear stress is 'T- - - ----- - nd3 - nd3 Fora shaft subjectedtobending moment M and torque T, the equivalent torque is T =~M2+T2e F = transverse shear load 'T= transverse shear stress Here, shear stress 'Tis variable and is maximum at the neutral axis. 198. (b) 199. (a) 196. (c) 195. (c) 194. (c) = ~(-150)2 +(100)2 = 180.27MPa L ( H'197 (b: 81 d ti 'I l':1=AK2,K = A)• 'J en erness ra 10 I, = K Moment of'Inertia forrectangular section bd3 1=- 12 Then K= {I= Jbd3 fA 12xA 3 3 F where, 'Tmax= "'2'Tmean="'2b x h FAy 'T=-- I.b t~t-------ty t- ------- ------- hl2 t L---_ 8 = PL = 4PL AE nD2E For, finding the change in diameter (transverse direction), Poisson's ratio v is needed. But modulus ofelasticity E is also needed. Now, E= 2G(1 +v) fromwhich 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 p TL 1 .. RB=P BMx=P x X The total strain energy stored is given by f L(Px)2 xdx (pL)2 x2 JL(Px)2 x dx u= + + o 2EI 2EI 0 2EI 4p2L3 u=--3EI 192. (d) When the load P is applied in axial or longitudinal direction, increase in length 4PL P> 3L+ pX-RA x4L=0 => RA = --= P 4L RA+ RB=2P :EMB=O PL PL o~o Bending moment diagram D~L=tRa =P p 2L--i B 191. (c) A-37Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 41. Given a = 15kN/cm2 ax = 9kN/cm2 tY =4kN/cm2xy :. a',2 =12±~(3)2+(4)2 =12±5kN/cm2 a} = 17kN/cm2 a2 = 7kN/cm2 Both of these are tensile in nature. 212. (c) Ductile materials generally fail inshear, therefore, when subjected to torsion, a specimen made of a ductile But the bending moment we depend up on rigidity or flexibility of the beam . B.Mat A=2PL. 211. (b) Principal stresses, 209. (d) Euler's critical load, P = 1[2EI (Leff )2 where Leff= effective length ofthe column. When both ends are fixed, Le.ff=0.5L 1[2El 1[2El 41[2 El . P - ---or-- •• cr - (O.5L)2 - 0.25L2 L2 210. (a) PartAC ofthe beam isrigid. Hence C will act as fixed end. PI! thus Os = 3EI 207. (c) 206. (a) Per could be non linearly related to slenderness ratio so better to avoid choice (d). a = Pd _ 700x103 x2xO.5 _ x 6- h 2t - 2 x 25 x 10-3 - 14 10 - 14 MPa. We know that E=2G(1 +)l) E=3K(1-2)l) .. 2G(1+)l)=3K(1-2)l) 2G+ 2)lG=3 K-6)lK )l(2G+6K)=3 K-2G 3K-2G )l = 2G + 6K 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. (slenderness ratio )2 203. (a) I ratio and -; ratio as per I.S. Code 800 : 1984. Therefore, y D 202. (a) Since allowable compressive stress depends upon T (atx=L) ML2 Ymax = 2EI Mx2 Again,EIy= -2-+C2 Atx = 0, y= 0 (fixed end) So, C2 =0 M 2 s= 2EI x dy At x = 0, dx = 0 as fixed at end .. C}=O dy or EI-=Mx+C1 dx d2y 201. (a) We know, EI dx2 = M 4 1 (pcr )ctamped (Per)hinged Ea~T So, athermal = (1-2v) 200. (c) We know critical buckling load 2 p = 1[ EI cr 2 Le For both ends hinged, L, = L L For both ends clamped, L,= "2 ~T= a (1- 2v) E Ea~T a=....,----...,... (1-2v) As it will be compressive stress. e=~(1-2v) E We know e= a~T Let thermal stress is a and for the symmetrical system, ax=ay=az=a 1 ex = ey = ez = - (a - )l 2a) E it will not depend upon the modulus of elasticity. Engineering Mechanicsand Strength ofMaterialsA-38 Badboys2Badboys2 Badboys2
- 42. «(')}(')2Plane)and «(')}-(')2) = diameter of Mohr's circle. (')} - (')2 Maximum shear stress (r ) =max 2 1 From the above expression, r a T and r a -3max r max d (d)Tr· - T. = Tr.R =__ 2_ max J ( 4) l1[~j 225. (a) p2 [x3]L p2 L3 = 2E1 3 a = 2EI x] p2L3 --- 6E1 224. (d) From torsion equation, Tr GQ= (')max J L R J L(Px)2 p2 fL = --dx = - x2dx a 2E1 2E1 a 2 Strain energy due to bending (U) = fL M dx o 2EI . 1 Stram energy of bar = "2P8 = ~ x 30 x103 x 0.05 x10-3 = 0.75 N-m = 700 X 10-6 - (-400 - 10-6) = (700 + 400) X 10-6 = 1100 x 10-6 EIn2 Maximum! Critical load (F) = -2- Le So, critical/maximum load a Moment of inertia of section While, we know that M.O'! of thin hollow circular sectionis maximum. 223. (b) 222. (a) 221. (b) (')1 = (')2 = r = 'tl = 't2 So, it is the state of pure shear 219. (a) Maximum shear strain, (r) = EI - EI = 2 x 1.5651 x 1/5 = ~(D4 _d4) 32 =5 ~d4 32 D = 1.5651 d f,g ('T)g T, D J, r,:-= (T;)s =Tg·d·T. 0.62604 f ~=1.6 fss Stressat comer B= 15+5- 8= 12t 1m2 (compressive) As from the condition given, 216. (b) 217. (c) Since both the steel shaft and gun metal slave are e securely fixed, L is the same for both. l=l JsNs JgNg Jg Tg Ns -=-·-=2x2.5=5 r, Ts Ng L IN U2 p2X2 •dx [2Xp2 •x3 ]LI2 :. W;= 2 [ 2x4EI = 2x4EIx3 Since,P = 1unit r:w=--1 96EI 215. (b) Let us use suffix Sfor steel and suffixg for gun metal. e T 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 surfaceforming 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 = J where M = - X 1 2EI' 2o A-39Engineering Mechanics and Strength of Materials Badboys2Badboys2 Badboys2
- 43. B ~ P~ Q I I cr =1 On arranging the above equns. E(El+~E2) E(E2+~El) 1_~2 ,cr2 = 1_~2 1 Are we know, El = E[cr1 -1·.ta2] 1 E2 = E[cr2 - ~crd 230. (b)'fsolid THollow ~(D4 -D~) 32 0 I THollow 228. (b) 227. (b) Maximum principal strain = Yxy / 2 Engineering Mechanics and Strength ofMaterials 229. (c). . Tr GQ tmax From torrsion equation, T = L = R tmax should be equal for both shafter. (;) solid = (;) Hollow A-40 Badboys2Badboys2 Badboys2
- 44. where j = number ofjoints To determine the nature of chain we use equation . h 3/ J+-=- -2 2 2 h = No. of higher pairs where . 3 J = -/- 2 2 also 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. 2. 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. 3. 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: Itconsists 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. Ithas three degrees of freedom. DEFINITION OF KINEMATIC CHAIN Combination of kinematic pairs joined in such a way that the last link isjoined 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 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 ofkinematic 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 THEORY OF MACHINES 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. rl'III~f)11Yf)l~ )11I(;IIINI~S lINI) )11I(;IIINI~1)I~SIfJN Badboys2Badboys2 Badboys2
- 45. 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 II I I I Types: (i) Beam engine (ii) Locomotive coupling rod (iii) Watts indicator ©©©©mechanism (b) Single slider crank chain Types : (i) Pendulum pump (ii) Oscillating cylinder engine (iii) Rotary internal combustion engine (iv) Crank and slotted liver quick return mechanism (v) Whitworth quick return mechanism (c) Double Slider crank chain Types: (i) Elliptical trammel (ii) Scotch-yoke mechanism (iii) 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) The instantaneous centre I lies at a distance Va along the (J) perpendicular to the direction of velocity Va at point A on a rigid body shown in Fig. IA = Va (J) 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. 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 MECHANISM Fig. Linkage shown in Fig. 1 is Grashoftype if s+l<p+q 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. GRASHOF'S CRITERIA 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. 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 Theory ofMachines and Machine DesignA-42 Badboys2Badboys2 Badboys2
- 46. The pressure over the rubbing surfaces is uniformly distributed through out the bearing surface. The wear is uniform throughout the bearing surface. Frictional torque transmitted in a flat bearing is given by 2. (i) Pivot and Collar bearings are used to take axial thrust of the rotating shaft. While studying the friction in bearing it is assumed that 1. c = Distance between the pivots ofthe front axles b = Wheel base a = Angle of inclination ofthe links to the vertical FRICTIONAL TORQUE IN PIVOT AND COLLAR BEARING where Fig. where v is the velocity ofthe particle C with respect to coincident pointC. ACKERMAN STEERING GEAR MECHANISM 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 <I> - cot e= c/b where c = Distance between the pivots of the front axles b = Wheel base <I> and e 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. DAVIS STEERING GEAR MECHANISM According to Davis Steering gear the condition for the correct steering is given by tan a = c/2b a c = at = 2r..rvcc cc UI 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 at a = ~ which is perpendicular to the link PQ PQ and the tangential component ofthe linear acceleration ofQ with respect to P is given by at =axPQQP ar = co2 X PQ = ( VQP J2 X PQ = V~P QP PQ PQ Fig. Radial component of the linear acceleration of Q with respect to P is given by t Clap Fig. Now ifthe point Q moves with respect to P with an angular velocity coand 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. Va = co xIA v, = 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. Number of Instantaneous Centres in a Constrained Kinematic Chain If n are the number oflinks in a constrained kinematic chain, then the number of instantaneous centre (N) is given by N= n(n-l) 2 VELOCITY AND ACCELERATION OF MECHANISMS 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. A-43Theory ofMachines and Machine Design Badboys2Badboys2 Badboys2
- 47. 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 e = angle of contact in radians Material used for flat belt is generally leather of various 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. 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. I Driving PulleyI Driver Pulley Fig.: Belt drive-open system Driving Pulley Slack Side T2 1 T = "2x 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. T - 2 w[ri -d]--XJ.! --- 3 rf-d while 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 R = r1+ r2 is the mean radius of the bearing. 2 (iv) Frictional torque transmitted in a flat collar bearing is given by where a = semi angle of the cone (iii) Frictional torque trnsmitted in a trapezoidal or truncated conical pivot bearing is given by T = ~ x J.!W [ r ~ - r ~] cosec a 3 r1 -r2 while considering uniform pressure. And in case ofuniform wear I T= - x J.!WR cosec a 2 2 T= - x J.!WR cosec a 3 while considering uniform pressure And in case of uniform wear 1 T= -xJ.!WR 2 where J.!= Coefficient offriction W = Load transmitted to the bearing R = Radius of the shaft (ii) Frictional torque transmitted in a Conical Pivot bearing is given by T = ~ x J.!WR while considering uniform pressure 3 And in case of uniform wear Theory ofMachines and Machine DesignA-44 Badboys2Badboys2 Badboys2
- 48. 0)2 = N2 = dl + t (1- ~J0)) N) d2 + t 100 where dJ, d2 = diameters of driver and driven pulleys 0)1, ffi2 = angular velocities ofdriver and driven pulleys NI,N2 = rotational speeds of driver and driven pulleys expressedin revoluationsper 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 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. Types of Gears Commonly used gear are as below: (a) Spur gear:Acylindricalgearwhosetoothtracesare straight lines parallel to the gear axis. These are used for where rl and r2are 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 L = 1t (rI + r2)+ I 2 + 2x x When the belt passesfromthe slack sideto 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. Centrifugal Tension The centrifugal tension (Tc)is given by T, = mV2 where m = Mass per unit length of the belt V = Linear velocityofthe 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 equaltothevalueofpowertransmitted givenbyeffective turning force (TI - T2), that is the centrifugal tension has no effecton 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 isthree times the centrifugal tension. T = 3Tc => T = 3mV2 Sovelocityforthe maximum powertransmitted is givenby v > )3: Velocity Ratio The velocityratio of speeds ofdriver and driven pulleys is given by I Fig. Circular Belt The ratio of driving tensions in round belts and rope drive is sameas V-beltdrive. Length of Belt In an open belt drive system the length of the belt is given by (rl - r2)2 L = 1t (r. + r2)+ + 2x x Fig. V-belt The ratio of driving tension forthe V-beltdrive is given by .!L = e(f,.LCOSeC 13)e T2 c> 2.3 log GJ= 11-9- cosec13 where J3 = Semi-angle ofthe groove e = Angle of contact in radians V-beltsare usually made ofcotton fabric, cards and rubber. (c) Circular belts:The crosssectionof 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 ofthe clothing. Round belts are made ofleather, canvas and rubber. 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. (b) V-belts: Fig. shows the cross section of the V-belts. V- belts are available in fivesections designedA, 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 ofbelts break, the entire set should be replaced at the same time. The groove angle in the pulley forrunning the belt is between 400to 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. A-45Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 49. P = 1tD c T D = Pitch circle diameter T = Number of teeth on the wheel. where Fig. 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 <I> where <I> 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 Root or dedendum circle Working depth element Addendum 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 ofobliquity: 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 1° angles are 14- and 20°. 2 Addendum: Radial distance of a tooth from the pitch circle to the top of the tooth. Fig. (c)(b) Pinion (a) _a:0n ..-- Rack (g) Internal and external gearing: Two gears on parallel shaft may gear either externally or internally as shown in Fig. Fig. : Rack and pinion Fig. : Worm gear (f) 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. Fig. : Bevel gear (d) Spiral gear: These are identical to helical 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. (e) 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. Itis used for connecting two non-parallel, non-intersecting shafts which are usually at right angles. transmitting motion between two shafts whose axis are parallel and coplanar. (b) 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. (c) 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. Theory ofMachines and Machine DesignA-46 Badboys2Badboys2 Badboys2
- 50. no. of teeth on the driven wheel no. of teeth on the driving wheel Fig. : Simple gear train Speed of the driving wheel Velocity ratio = --=---------==----- Speed of the driven wheel I I I i""Il'§""""'~'II"'~III"II"II'§"""lIi Driven or follower Idle gears where <l> 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. L h fAr f Length of path of contact engt 0 co contact = ---=-----==------- cos <l> from the begining to the end of engagement. Length ofthe path ofcontact: Length ofthe common normal cut- offby the addendum circles of the wheel and pinion. Arc of contact: The path traced by a point on the pitch circle from 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 R · Length of arc of contact ontact abo = --=-------- Circular Pitch Contact ratio is the number pairs of teeth in contact. Length of Arc of contact: Length of the arc of contact can be calculated as Working depth: Radial distance from the addendum circle to the clearance circle. Working depth = Addendum of first gear + Addendum of second gear 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 m D 1 m=-=- T Pd :::::> 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 +dedendum 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 . . T 1t DIameter pitch Pd = - = - D Pc :::::> PcxPd=1t Module: It represents the ratio of pitch circle diameter (in mm) to the number of teeth. N2 = IL NI T2 Diametral pitch: It represents the number of teeth on a wheel per unit of its diameter. ~ = 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 For two gears to mesh correctly their circular pitch should be same A-47Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 51. ... (2) 2E I 1 2 E = "2ICOmean ... (1)Lllimax = lco~eanc, also energy stored is a flywheel is given by = 1(,) (comax- comin)x co v-mean mean COmean Relation between maximum fluctution of energy .1Emaxand coefficient of fluctuation of speed. Lllimax = lCOmean(COmax - COmiu) 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 = .1comax s COmean where C - LlEmax energy- Wpercyc1e 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. Itabsorbs 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 FLYWHEEL Fig. : Epicyclic gear train . . NB TA Velocity RatIo - = 1+- Nc TB ArmC 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. 4. 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. D1 + D2 = D3 + D4 2 2 => D1 + D2 = D3 + D4 . . N1 T2 x T4 Velocity rano = - =--=-_....:.... N4 T1x T3 Fig. : Compound gear train N N N N T xT xT Velocity ratio = _I = _1 X _3 x _5 = 2 4 6 N6 N2 N4 N6 T1x T3 x T5 3. Reverted gear train: Areverted gear train manifests when 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 Driven or follower Train value = ~ =IL NI T4 2. Compound gear train: A compound gear train includes two gears mounted on the same shaft as shown in Fig. Driver Theory ofMachines and Machine DesignA-48 Badboys2Badboys2 Badboys2
- 52. 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. 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) Circular cam 2. Cylindrical Cam: A cam in which the follower reciprocates or oscillates in a direction parallel to the cam axis. CAMS Is1eeve= Xcompression= (r2 - r.) y_ x where r1 = 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 S = S2 - S1 h where S1 = Spring force at minimum radius of rotation S2 = Spring force at maximum radius of rotation Ifhp is the height of porter governor (when length of arms and links are equal). and h., is height of watt's governor then ~=m+M 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 from (1) and (2) Lllimax = 2ECs where C, = coefficient of fluctuation of speed. GOVERNORS 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) SleeveLift: 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 ofrotation 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 . . 2 (N1 - N2) Sensitiveness = ___;'----.!_----=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 = -- metres N2 where N = speed of the arm and ball about the spindle axis. (2) Porter governor: Itis 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. A-49Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 53. BASIS OF LIMIT SYSTEM 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. Standard tolerances The system oflimits and fits comprise of 18grades 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, ITOand IT 1 as below For ITOl, i (microns) = 0.3 + 0.008D For ITO, i(microns)=0.5+0.012D Malleable cast iron (Cast iron alloy which solidifies in the as-cast condition in a graphite free structure) 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) Grey cast iron (3-3.5% carbon) Dead mild steel (upto 0.15% carbon) 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) 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. 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 MACHINE DESIGN VELOCITY AND ACCELERATION OF THE FOLLOWER (a) 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 sec2e and acceleration of the follower is given by af = (1)2(r, + r2) (2 - cos' e) sec2 e where r1= 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 (b) 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 7. 6. 5. 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. 8. Lift or Stroke:The maximum travel offollower from its lowest position to the topmost position is called life or stroke. 9. Angle ofAscent: It is the angle moved by cam from the time the follower begins to rise till it reaches the highest point. 10. Angle of Descent: Angle during which follower returns to its initial position. 11. Angle ofAction: It is the total angle moved by cam from the beginning of ascent to finish of descent. 12. Under Cutting: The situation ofa Cam Profile which has an inadequate curvature to provide correct follower movement, is known as under cutting. Cam profile , "- "- ".......... _----".,. Pitch point Maximum pressure angle Pitch "_--l.~~~-+-+-e point Theory ofMachines and Machine DesignA-50 Badboys2Badboys2 Badboys2
- 54. "['torsion = torsional shear stress a distance r in Nzmm? T = applied torque r = radial distance where The torsional shear stress induced at a distance r from the centre is given by Txr "['torsion = -1- p I t~--(~{;~~~~~~;~)~ T e Pc"['=_s 2A If <I>is the deformation produced due to shear stress r then r o: <I> "['=C<I> 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 e radians as shown in Fig. T 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 IS where (b) The direct shear stress induced in the rivet is given as r = Ps A r = direct shear stress in N/mm2 I I I ~ d (a) 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. a = stress e = strain E = Young's modulus or modulus of elasticity a P x l E= -=-- e Ax 8/ or where STATIC LOADING AND DYNAMIC LOADING (a) Static loading: Atype 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:Atype ofloading which varies in magnitude as well as direction, very frequently, such type ofloading is called dynamic loading or fluctuating or alternating loads. 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. 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 Pt at= - A where Pt = Axial tensile force in N A = Area of cross-section of the body in mm? at = Tensile stress in N/mm2 the strain produced can be calculated as 8/ e= - / 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. Pc ac= - A where ac = compressive stress in Nzmm? Pc = compressi ve force A = area of cross-section of the body in mm- Compressive strain is given by 8/ e= - / where 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 a = Ee A-51 ForlTl, i(microns)=O.8.0.020D where D is the size or diameter in mm. Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 55. K - (Jrnax t--- (Jo where K, = stress concentration factor (Jrmx = maximum stress at the discontinuity (Jo = nominal stress at the same point Stress concentration factor is also known as theoretical or form stress concentration factor. Stress Concentration 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. I The stress at the surface of contact between the rivet and the plate is given by P (Jb or (Jc = -- d-t-n where d = diameter of the rivet t = thickness of the plate d- t = projected area of the rivet n = 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 P P Pb= -=- 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 I M= (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. y From the equation, we have M (J Neutral axis R o (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 where Y I R (J M E r Ip where C = modulus of rigidity e = 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 Ip = polar 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 1 T co d ....!,!.- Theory ofMachines and Machine DesignA-52 Badboys2Badboys2 Badboys2
- 56. (F.S.X:: J+S.X ::J=1Elliptic Method Soderberg Method Goodman Method valid for ductile material _1_ = (am J2 x F.S. +~ F.S. au ae Gerber Method Mathematical RelationMethod Name Inference corresponding to each line are shown in the table. Table cry Tensite-----. Meanstress (crm) 10 1 1 Compressive-+---1 1+--- S-Ndiagram 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. 10 1d 102 103 104 105 106 1rJ 108 Numberof stresscyclesN - Sut en ~ 11~ S~t----t-------'lf--------- s; Low H ighcycle ~-c-y-c~le_'~--~~-----__'~ Infinitelife _______.. Endurance limit FS.fatigueloading= --------- Design or working stress 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 Ultimate strength FS.brittlemateria!s= For static loading Design or working stress 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 Material Empirical Relation 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 Table 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. Increase in actul stress K, -lover nominal value q = -- =---------- K, -1 Increase in theoretical stress over nominal value amax,actual= Actual maximum stress at notch or discontinuity Kf<Kt Notch sensitivity: Notch sensitivity is calculated by using the relation where amax.actual In practical the actual effectof 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-53Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 57. Doublerivetellap joint (zig-zigriveting) 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 distancebetweenthe 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 Rivetedjoints 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: Ajoint 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 wherePt= 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 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. Lapjoint with single riveted 2. 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 aboveor 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. Cross- sectional view of the double riveted joint is shown in Fig. c -cr· cr = variable stress = max mm v 2 o y = yield point stress FS. = Factor of safety RIVET JOINTS A rivet is made ofa shortcylindricalbar with ahead 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: Ifoneplate overlapsthe otherand the twoplates are riveted together, then this type of joint is called. Lap joint, Fig. shows a cross sectional view of a lap joint. cr +cr· cr = mean stress = max mm m 2 cru = ultimate stress cre = endurable limit for reverse loading stress where Theory ofMachines and Machine DesignA-54 Badboys2Badboys2 Badboys2
- 58. where where => 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 t = tl +t2 P = (tl + t2) x I x at I = length of weld = width of plates t1 = throat thickness at the top (ii) 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 pp r ) ( > ) ( 1'] > ) 1 ?) > ) :> Double parallel fillet weld (b) 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. pp .--fI.~f~s_-r---lrp P..-.1 -.J sf.- 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 pp Ptransverse= 1.414 s x I x at where s = Leg or size of weld I = Length of the weld at = Allowable tensile stress 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 where, 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. Minimum of Pc' Pt and Ps 11= P x t x at P = Pitch of the rivets t = Plate thickness A-55Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 59. Tangentkey (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. 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 cross- sectional view of a tangent key is shown in Fig. where d = diameter of shaft w=_Q_ 4 t=2w=_Q_ 3 6 I=1.5d Rectangular sunk key (iii) Gib-head key: Cross-sectional view of a Gib-head sunk key is shown in Fig. d w=T t=_!' 12 1= 1.5 d (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 I Square sunk key Shaft cross-section d w=t=T length of the key ( = 1.5 d 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 ofdiameter, coarse threads have large pitch and lead as compared to fine threads. Coarse threads are more in strength and chances ofthread shearing and crushing is veryless. They are preferred for vibration free applications as they offer lessresistance to unscrewing. Finethreads givebetter 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 weuse the valuesof 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. KEYS Toprevent the relative motionofthe shaft and the machinerypart 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: Asquaresunkkeyis showninFig. If d is the diameter of the shaft width of the square sunk keyis equalto d/4and the thicknessis same as width. 1+-'fJ""+I -f--- - --- t I J_~~ _t. _L~2 pi -j~ - ---Inn-t-:tpt Double V-Butt joint t2 = throat thickness at the bottom crt = allowabletensile stress for weldment inN/mm2 Theory ofMachines and Machine DesignA-56 Badboys2Badboys2 Badboys2
- 60. 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 where _!_ (M + ~M2 + T2) = nd3 O"b(max) 2 32 nd3 Me= --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 byusing support bearings. If gear is mounted on the shaft then Now, According to maximum normal stress theory (Rankine's theory the maximum normal stress in the shaft is given by 1 1 ~ 2 2 O"b(max)= - O"b + - (O"b) + 4~ 2 2 ,,~_)~ ::3[~(M+ M + T2)] => ~ nd3 M2+T2 = __ ~ 16 max nd3 r, = 16~max where T, =equivalent twisting moment = ~ M2+ T2 32M where do = outer diameter of shaft in m 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 n 3 T - -x~xdo - 16=> T = torsional moment in N-m For a hollow shaft ~= 16Txdo N/m2 ( 4 4) ,n do - dj d = shaft diameter in mwhere SHAFTS 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 sizebyturning, 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. 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 r = 16; N/m2for solid shaft nd w=~ 2~ w = width of the key t = thickness of the key O"c= permissible crushing stress r = permissible shearing stress where Combined loading: When a shaft is subjectedto combined twisting moment and bending moment, then the shaft is designed on the basis of maximum normal stress theory and maximum shear stress theroyand 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 ~max = - J (o.,) + 4~ 2 = _!_ 32M + 4 x (16 TJ2 2 nd3 nd3 r = ~ 'M2+T2 max nd3 j (iii) A-57Theory ofMachines and Machine Design Badboys2Badboys2 Badboys2
- 61. Static Tooth Load 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 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 Sliding Contact Bearings 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. velocities upto 12.5 mls velocities upto 12.5 mls 3 c. di .-- lor or mary cut gears operatmg at 3+v C =v 0"0 X C, velocity factor 4.5 full .--- for care y cut gears operatmg at 4.5+v O"w= C =vwhere 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 = 0.124 _ 0.684 T . 0.912 Y for 20° full depth mvolute system = 0.154 - -- T where K2 y=_l_ 6K2 Ft = o"wb PcY Lewis Equation y = form factor called Lewis form factor b = width of gear face Y for 14!.: composite and full depth involute system 2 Let O"wbK~ Pc = O"wbPc K~ 6K2 6K2 F = O"wX bt3 = O"wbt2 t 12 x ~ 6h 2 Now if circular pitch is P, then we can represent t, and h in terms of Peas . t Now for y for beam of height t = - 2 from (1) ... (2) ... (1) . . My Maximum value ofbendmg stress = O"w= - I where M is maximum bending moment (i.e. at BC) M= Ftxh M Ft= - h M = O"wI Y -__rTangent to the - - - base circle 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. Theory ofMachines and Machine DesignA-58 Badboys2Badboys2 Badboys2
- 62. ZN P K Partial lubrication (5 c "~-+---~-- ~ o 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 = -, W = Load on the Journal [·d The variation of coefficientoffriction with respect to the bearing characteristic number is shown in Fig. r-:------,_ Thin film or boundary lubrication t 1 (unstable) 3 c o ""B E (ix) where K is a factor for end leakages for 0.75 < !:_ < 2.8, K = 0.002 d 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 [ - = 1 bearing is called square bearing d [ - > 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 ofpressure fedbearings 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 D-d C1 C2=R-r=--=- 2 2 (iii) Diametral clearance ratio: Ratio between diametral clearnace to journal diameter . . C1 Diametral clearance ratio = - 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 = (z;)(~J where N = Journal speed in r.p.m., Z = lubricant viscosity, P = bearing pressurenormally we take its valueas 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 P - Z N ( d J2 ( [J NI 2 - 4.75 x 106 c; [+ d mm where N = Journal speed in r.p.m. Z = Absolute viscosity of the lubricant (viii)Coefficient of friction: Coefficient of friction can be expressed as I I I I I Hydrodynamicjournal bearing Diameter of the bearing = D = 2R Diameter of the journal = d = 2r Length of the bearing = l Terminologies associated with a hydrodynamicjournal 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 andjournal Journal Hydrodynamic Journal Bearing Terminology Cross-sectionalviewofa hydrodynamicjournal bearing is shown in Fig. A-59Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 63. 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 = mrw- m = 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 w- 16 16 T = ~ /-!W [rf -d ]cosec a 3 rf-d (b) Cone clutch: Total torque transmitted in the cone clutch is given by 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 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= W n(r?-d) dr 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. (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 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 driveforexamplejaw or clawclutch. CLUTCHES [ 6 ]1/3P = Cx _!.Q_ 60NL or IfN is r.p.m. the Life in hours is given by ( CJ3 106 L= - x-- hours P 60N Variation of coefficientoffriction with the bearing characteristic number (z:J Rolling Contact Bearings Bearing which operate on the basis ofprinciple ofrolling, 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% failureis known asdynamicloadrating orbasicdynamiccapacity or specific dynamic capacity. Rating Life L ~ (~ J' where P = load C = dynamic basic load rating ( IJ1/3P= C - L Theory ofMachines and Machine DesignA-60 Badboys2Badboys2 Badboys2
- 64. (b) 16 (d) 32 15. Maximum fluctuation of energy is the (a) Ratio of maximum and minimum energies (b) sum ofmaximum and minimum energies (c) Difference ofmaximum and minimum energies (d) Difference of maximum and minimum energies from mean value 16. In full depth 114degree involute system, the smallest number of teeth in a pinion which meshes with rack without interference is (a) 12 (c) 25 (a) -1 (b) zero (c) 1 (d) 2 12. 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 13. 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 14. 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 (c) Geneva mechanism is an intermittent motion device (d) 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 (d) ?:2 11. A double-parallelogram mechanism is shown in the figure. Note that PQ is a single link. The mobility ofthe mechanism is P Q (a) 50 mm (b) 120 mm (c) 150mm (d) 280mm 2. The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is (a) 1 (b) 2 (c) 3 (d) 4 3. Match the items in Column I and Column II. ColumnI ColumnII P. 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 4. Match the items in Column I and Column II. ColumnI ColumnII P. Addendum 1. Cam Q. Instantaneous centre 2. Beam of velocity R Section modulus 3. Linkage S. Prime circle 4. Gear (a) P-4, Q-2, R-3, S-l (b) P-4, Q-3, R-2, S-l (c) P-3, Q-2, R-1, S-4 (d) P-3, Q-4, R-1, S-2 5. The number of inversions for a slider crank mechanism is (a) 6 (b) 5 (c) 4 (d) 3 6. For a four-bar linkage in toggle-position, the value of mechanical advantage is (a) zero (b) 0.5 (c) 1.0 (d) infinite 7. 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 8. 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 9. 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 1. 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? ...,EXERCISEI···.. A-61Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 65. (c) (d) 12 24 4 15 1 (b) 8(a) 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 1 31. (b) Jib {gh3h(d) /3 (a) .J2 gh l3gh (c) 30. A cord is wrapped around a cylinder of radius 'r' and mass 'm' as shown in the given figure. Ifthe cylinder is releasd from rest, velocity of the cylinder, after it has moved through a distance 'h' will be (a) dynamically balanced (b) statically balanced (c) statically and dynamically balanced (d) not balanced 29. 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) h}+h2=k (b) hT+h;=k2 28. A rotor supported at A and B, carries two masses as shown in the given figure. The rotor is 2+Cf (d) 2-Cf l+Cf (c) 1-Cf 2-Cf (b) 26. 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) 27. If Cf is the coefficient of speed fluctuation of a flywheel then the ratio of O)~O)min will be 1-2Cf (a) 1+2Cf (b) 6.12 mls (d) 12.13 mls 25. 24. 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 (c) 9.04 mls 23. 21. 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 22. 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 (b) 349.8kJ (d) None of these 20. (a) 2 degrees of freedom (b) 3 degrees of freedom (c) 4 degrees of freedom (d) 6 degrees of freedom 18. 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 19. 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 (c) 487.5 kJ o x ) 17. The two-link system, shown in the figure, is constrained to move with planer motion. It possesses y Theory ofMachines and Machine DesignA-62 Badboys2Badboys2 Badboys2
- 66. c 47. 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 46. (d) S + P > L+ Q(c) S+PSL+Q 45. 44. 43. 42. 41. 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 W S+LSP+Q ~ S+L>P+Q 40. (a) horizontal (b) vertical (c) at45° to the horizontal (d) perpendicular to the line CO 37. Two gear 20 and 40 teeth respectively are in mesh. Pressure 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) 8mm (d) lOmm 38. In a slider-bar mechanism, when does the connecting rod have zero angular velocity? (a) When crank angle = 0° (b) When crank angle = 90° (c) When crank angle = 45° (d) Never 39. 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 32. 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 ill(a) ellipse 2 3 4 (b) parabola (c) approximately straight line (d) circle p 33. 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. 34. 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 35. 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 2 (C) lIKes (1) av (d) lKes(1)2av (where I= Mass moment of inertia of the flywheel (1)av = Average rotational speed K;= coefficient of fluctuation of speed) 36. 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 0 ""'"("I') ("I') I A-63 o, C) (a) I~ _I fJ2n: m (b) 2n: m (c) I~ I~ 2n: 3m (d) 2n: 2m Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 67. (b) 30° (d) 60° 64. 63. 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° (c) 45° 62. (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 61. cr2 3. cr1 - cry - cry R. Maximum-shear stress criterion cry cry cr ". Y - cry Q. Maximum-distortion- 2. energy criterion 1.P. Maximum normal- stress criterion The tangential force transmitted (in newton) is (a) 3552 (b) 2611 (c) 1776 (d) 1305 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) 81116 (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 o2 Theory ofMachines and Machine Design IS (a) 8000 (b) 6000 (c) 4000 (d) 1000 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 56. 55. 54. 53. 52. 51. 50. A link OP is 0.5 m long and rotate about point O. It has a slideratpermit B.Centripetal accelerationofP relativeto 0 is 8m/sec'. The slidingvelocityofsliderrelativeto P is 2 mI sec. The magnitude of Coriolis component of acceleration IS (a) 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 hydrodynamicjournal 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 37- tooth 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 forceon 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 o (b) 20cm (d) 50cm (a) 10cm (c) 30cm A cB 49. The transmission angle is maximum when the crank angle withthe fixedlink is (a) ff (b) 90° (c) 180° (d) 270° In the given figure, ABCD is a four-bar mechanism. At the instant shown,ABand CDareverticaland BC ishorizontaL AB is shorter than CD by 30 cm, AB is rotating at 5 radls and CD is rotating at 2 rad/s. The length ofAB is 48. A-64 Iy 57. (a) y (b) 2 (c) 2_y (d) 3.y 58. 4 3 Badboys2Badboys2 Badboys2
- 68. o 4 8 12 16 20 24 Module of spur gears (mm) (a) AandB (b) Band C (c) AandC (d) A,BandC 0.08 Error (e) (in mm) 0.04 where Il = co-efficient of friction W = load over bearing R = radius of bearing 75. 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) 76. 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 77. In case of self locking brake the value of actuating force is (a) Positive (b) Negative (c) Zero (d) None of these 78. I.S. specifies which ofthe following total number of grades of tolerances? (a) 18 (b) 16 (c) 20 (d) 22 79. The theoretical stress concentration factor at the edge of hole is given by (a) l+(~) (b) 1+2m (c) 1+3(~) (d) 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 80. 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 81. 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 82. 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 83. Which of the following graph is correctly represent? 0.12 r-------------, (c) 3 (b) "4IlWR 1 (d) "2IlWR (a) IlWR 2 -IlWR 3 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 (c) (a) 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-65Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 69. 94. In a belt-drive if the pulley diameter is doubled keeping the tension and belt width constant, then it will be necessaryto (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 ofminimum 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 84. For self-locking which of the following condition is satisfied? (a) <I> ~ 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) Taperedbearing (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) differencebetween 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 onlyonetypeofstress whereas the beam has both tensile and compressive stress. Theory ofMachines and Machine DesignA-66 Badboys2Badboys2 Badboys2
- 70. 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) N}-N2 Nl +N2 (c) Nl+N2 (d) Nl-N2 119. The flywheel of a steam engine has mass - moment of inertia2500 kg - m2.Ifthe angular accelerationis0.6radls2, the starting torque required will be equal to: (a) 3500 NM (b) 3700 NM (c) 1800 NM (d) 1500 NM 120. 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) 0.09 121. The safe rim velocity of a flywheel is influenced by: (a) Mass of the flywheel (b) energy fluctuation (c) centrifugal stresses (d) speed fluctuation 122. 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 123. Porter governor is a : (a) Pendulum type governor (b) Dead weight type governor (c) Spring loaded type governor (d) Inertia type governor 124. In case of an isochronous governor, the value of sensitivity is : (a) infinity (b) zero (c) one (d) None of these 125. 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 (c) Hex:- (d) Hex:2 co co 126. 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 127. Which one of the following governors cannot be isochronous? (a) Hartnell (b) Porter (c) Watt (d) Hartung 128. 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) 0.5 129. 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) 2vco vco (d) 2 117. 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 106. Oldham'scoupling is an inversionofwhichofthe 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) Processof obtaining byreversing 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 followingpair, a ball and socketjoint forms? (a) Spherical pair (b) Rolling pair (c) Turningpair (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 : (a) 2.y (b) ly 2 2 (c) iy (d) 2y 3 3 113. 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) 6 114. Ifthe number oflinks ina mechanism is 8,then the number of pairs will be equal to : (a) 6 (b) 12 (c) 3 (d) 18 115. 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 116. 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 A-67Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 71. 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 (d) I::!L= (T2)2 N2 Tl 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 T1 = Number of teeth on driving gear T2= Number of teeth on driven gear ~ ~ ~ ~ (a) -=- (b) -=- N2 T2 N2 Tl (d) m= circular Pitch(c) m = diameteral pitch 1t circle pitch (b) m=-~- 1t 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 (a) m = number of teeth 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/d93in 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 followermoves 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 rmmmum (c) Point on cam pitch curve at which pressure angle is maximum (d) Any point on the pitch circle 138. 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 139. Which pair of gears usually has higher frictional losses: (a) Helical gears (b) Spur gears (c) Bevel gears (d) Worm and Worm wheel 140. 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 141. The outer circle of spur gear is called: (a) pitch circle (b) base circle (c) addendum circle (d) dedeundum circle 142. 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 143. The product of module and diameteral pitch is equal to : (a) 1 (b) 4 (c) 3 (d) 0 144. Axial thrust is minimum in case of: (a) Helical gear (single) (b) Helical gear (double) (c) Bevel gear (d) Spur gear Theory ofMachines and Machine DesignA-68 Badboys2Badboys2 Badboys2
- 72. (b) cycloidal teeth (d) All of these 168. The inner and outer radius of friction surface of a plate clutch are 50 mm and 100mm 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 ofthe rivetis 50Nrnm", 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 Nzmm- 178. The thickness of the boiler plate is 16 mm, then diameter of rivet used in the boiler joint 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 <I> (b) (R + r) cos <I> (c) (R'+rj tan o (d) (R+r) cot <I> 182. Which type of teeth are normally used and satisfy law of gearing? (a) conjugate teeth (c) involute teeth (a) Si~ oc (b) c~ oc (c) Jl sin o: (d) Jl cos o: 161. Velocityof belt for maximum power transmission by the belt and pulley arrangement will be equal to : (a) ~ (b) t;: (c) f:F (d) 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 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, N1 = Number of discs on driving shaft N2 = Number of discs on driving shaft then, number of pairs of contact surfaces will be : (a) N1 - N2+ 1 (b) N1 - N2-1 (c) N1+N2+1 (d) N1+N2-1 160. IfJl = actual coefficient of friction in a belt moving in a grooved pulley a: = groove angle. then virtual coefficient of friction will be : A-69Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 73. ~c B (a) 4 rad/s (b) 8 radls (c) 10 rad/s (d) 15 rad/s 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 : ABC D (a) 2 3 1 4 (b) 4 1 3 2 (c) 4 3 2 1 (d) 3 2 4 (D) Spur gears Codes: (C) Bevel gears (a) 3 (b) 4 (c) 2 (d) 5 198. Match the following items in List 1 and List 2 List 1 List 2 (A) Worm gears 1. Parallel shafts (B) Cross-helical gears 2. Non parallel, intersecting shafts 3. Non - parallel, non intersecting shafts 4. Large speed ratios 193. Which of the following is not the part of roller bearing (a) shaft (b) inner race (c) outer race (d) cage 194. Which of the following assumption regarding the lubricant film is made in Petroffs equation? (a) Itis converging (b) Itis diverging (c) Itis converging or diverging (d) It is uniform 195. 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 196. The life of bearing is expressed in : (a) Lacs of revolutions (b) billions of revolutions (c) thousands of revolutions (d) None of these 197. The number of degrees of freedom of a five link plane mechanism with five revolute pairs will be: 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 4R 3 (d) 2R 3 (c) 189. When the intensity of pressure is uniform in a flat pivot bearing of radius 'R', then the frictional force is : (a) R (b) 2R 1 (d) 3JlwR 4 (c) 3JlwR 188. The friction torque transmitted in case of flat pivot bearing for uniform ratio of wear will be equal to : 2 (a) Jlw R (b) 3JlwR cos(8+<1»-1 (d) cos (8 - <1»+1 cos(8+ <1»+1 (c) cos (8 - <1»+ 1 1- sin <I> 1+ sin <I> (a) llmax = 1 . A- (b) llmax = 1 . A- +~n~ -~n~ 1- cos <I> 1+ cos <I> (c) llmax = (d) llmax = 1 A- 1+ cos <I> -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 to : (a) 10 (b) 20 (c) 30 (d) 40 185. If '<1>' is the face angle of a bevel gear, then which of the following relation is correct? (a) <I> = pitch angle + addendum angle (b) <I> = pitch angle - addendum angle (c) <I> = axial pitch (d) <I> = pitch angle 186. If '8' is the root angle of a bevel gear, then which of the following relation is correct? (a) 8 = pitch angle + addendum angle (b) 8 = pitch angle - addendum angle (c) 8 = pitch angle + dedendum angle (d) 8 = pitch angle - dedendum angle 187. In case of spiral gears, maximum efficiency is given by : cos (8 - <1»+ 1 cos (8 + <1»-1 (a) cos(8 + <1»+ 1 (b) cos(8 + <1»-1 183. The maximum efficiency of worm and worm wheel system will be : Theory ofMachines and Machine DesignA-70 Badboys2Badboys2 Badboys2
- 74. (b) 1.8 (d) 2.55 219. The stress concentration factor(k) given in Lewisequation has the values for most of the designing purposes is given by: (a) 0.8 (c) 1.55 (a) y = 0.154- 0.912 (b) y= 0.154+ 0.912 T T (c) Y = 0.254_ 0.952 (d) y = 0.254+ 9.52 T T 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) 1and 2 (b) 1and 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 : (d) ZNP (a) ZN P ZN2 (c) P 215. If Z = absolute Viscosity of lubricant P = bearing pressure N = journal speed then the bearing characteristic number is given by : (a) FhP o: Dj 3 (c) ~p o: Dj 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 : (c) P.d tFt 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 Pdt (d) - Ft T (d) 6(c) 2T 200. A flywheel ofmoment of inertia 9.8 Kg - m2 fluctuates by 30 rpm fora fluctuationinenergyof 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 : T (a) 4 (b) T A-71Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 75. 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 : Scooters Rolling mills Trucks Mopeds 1. 2. 3. 4. List - II 2flW(R3 +r3) (d) 3sina(R2+r2) 2sina flw(R+r) (b) Codes: A B C D (a) 3 4 2 (b) 1 3 2 4 (c) 3 2 4 (d) 3 4 2 2flW(R3 _r3) (c) 3sina(R 2 _r2) 2cosa 234. Match List- I and List - II and select the correct answer using the codes given below: List - I A Single plate clutch B. Multi plate clutch C. Centrifugal clutch D. Jaw clutch 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: 4 uw [R~ -Rf] 3 uw [R~ -Rf] (a) 3 cosu R~ -R? (b) 4 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 : 1 [R~-R?]() -flW a 2 R2 -RI 2 [R~-R?] 2 [R~+R?] (c) 3flW R2-RI (d) 3flW R2-RI 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 assuminguniform pressure theory? flw(R-r) (a) 4 cote4 tan O 2 sine IflWR (d) IflWR (b) IflWR (a) 4 cosO IflWR (c) 229. If e = semi-cone angle, then in a conical pivot bearing under uniform wear, the frctional torque transmitted will be: (d) 2e-<j) 2 K - Kr t; (a) d - J2 (b) Kd = /2Kr (c) Kd = 2~ (d) Kd = 4 x, 227. An external gear with 60 teeth meshes with a pinion of20 teeth, having module being 6mm, 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= <j)+e A <j)-e (a) p (b) p=- 2 2 e-<j) (c) ~=- 2 (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: TI+2Tc + T2 TI+4Tc + T2 (c) 4 (d) 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 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 T1 = 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 Theory ofMachines and Machine DesignA-72 Badboys2Badboys2 Badboys2
- 76. ANSWER KEY 1 (c) 26 (d) 51 (a) 76 (c) 101 (b) 2 (c) 27 (d) 52 (c) 77 (c) 102 (a) 3 (d) 28 (c) 53 (a) 78 (a) 103 (d) 4 (b) 29 (c) 54 (c) 79 (b) 104 (b) 5 (c) 30 (a) 55 (d) 80 (b) 105 (a) 6 (d) 31 (a) 56 (b) 81 (c) 106 (b) 7 (a) 32 (a) 57 (a) 82 (a) 107 (c) 8 (b) 33 (c) 58 (d) 83 (d) 108 (a) 9 (a) 34 (c) 59 (b) 84 (a) 109 (d) 10 (d) 35 (d) 60 (c) 85 (a) 110 (b) 11 (c) 36 (c) 61 (b) 86 (b) 111 (a) 12 (a) 37 (c) 62 (a) 87 (a) 112 (a) 13 (d) 38 (b) 63 (a) 88 (a) 113 (c) 14 (a) 39 (c) 64 (c) 89 (b) 114 (a) 15 (c) 40 (d) 65 (a) 90 (d) 115 (a) 16 (d) 41 (d) 66 (c) 91 (c) 116 (c) 17 (a) 42 (a) 67 (a) 92 (a) 117 (a) 18 (d) 43 (a) 68 (a) 93 (d) 118 (b) 19 (c) 44 (d) 69 (a) 94 (c) 119 (d) 20 (a) 45 (a) 70 (d) 95 (d) 120 (c) 21 (a) 46 (a) 71 (b) 96 (b) 121 (c) 22 (c) 47 (b) 72 (b) 97 (a) 122 (b) 23 (c) 48 (c) 73 (b) 98 (d) 123 (b) 24 (a) 49 (b) 74 (c) 99 (d) 124 (a) 25 (c) 50 (a) 75 (b) 100 (a) 125 (d) K (d) 10 K (c) 8 (c) H = m - M x 895 m 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 followershouldmovewith 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: K K (a) - (b) - 4 6 238. The height of the porter governor is defined as : (a) H=m+Mx895 (b) H=m-Mx995 m N2 m N2 T2 +r9 (c) -= e Tl 237. In case of flat belts having negligible centrifugal tension, then the ratio of driving tensions is given by: _T1 Jl Tl r9 (a) (b) -= e T2 o T2 (d) None of these 236. In cam design, the rise motion is given by the SHM s = %( I - cos ~9)where h is the total rise, 9is cam shaft angle and ~ is the total angle of rise interval. Then the jerk is given by : (a) %(I-COS ~9) (b) ~-%cos( ~9) (b) 3H(n;Y (d) ~( n;r (a) ~(';y ( C) 2H (';'f A-73Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 77. 2 2 (0.025(02) + (0.030 (02) +2 x 0.025 x 0.030 (04x (-0.866) = 0.015033 (02. Now 0.1r(02 = 0.015033 (02. => r=0.150m => r= 150mm 2. (c) Accordingto Grubler's criterion,thenumber ofdegrees of freedom of a mechanism is given by F = 3(n-1) - 2j- h = 3(8-1) -2 x 9- 0 = 21-18 = 3 5. (c) For a 4-bar chain/mechanism like slider-crank mechanism, there are as number of inversions as the number oflinks orbars. These different inversionsare obtained by fixing different links one at a item for one inversion. r = mror' = 0.1 r (02N From the above force polygon, R = ~Fl + Fi + 2F}F2cos 1500 I I I I I I Let (0be angular velocity of disc . . F1= mlr1(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 ofbalancing mass 0.1 kg, so -----------------~ I' 1 " 1 , 1 , 1 ' IR ,,, 1. (c) Since, all the masses lie in the single plane of the disc. So, we have a force polygon. ...,HINTS & EXPLANATIONS 126 (a) 151 (d) 176 (b) 201 (b) 226 (a) 127 (b) 152 (c) 177 (b) 202 (b) 227 (c) 128 (a) 153 (d) 178 (d) 203 (a) 228 (a) 129 (c) 154 (b) 179 (c) 204 (d) 229 (b) 130 (a) 155 (a) 180 (a) 205 (b) 230 (c) 131 (b) 156 (b) 181 (b) 206 (b) 231 (a) 132 (a) 157 (c) 182 (c) 207 (d) 232 (d) 133 (b) 158 (a) 183 (a) 208 (a) 233 (c) 134 (b) 159 (d) 184 (b) 209 (c) 234 (d) 135 (a) 160 (a) 185 (a) 210 (b) 235 (a) 136 (c) 161 (b) 186 (d) 211 (a) 236 (c) 137 (c) 162 (a) 187 (d) 212 (b) 237 (b) 138 (a) 163 (d) 188 (c) 213 (a) 238 (a) 139 (b) 164 (c) 189 (c) 214 (c) 239 (d) 140 (a) 165 (a) 190 (d) 215 (a) 240 (b) 141 (c) 166 (c) 191 (a) 216 (c) 241 (c) 142 (b) 167 (b) 192 (b) 217 (c) 242 (a) 143 (a) 168 (a) 193 (a) 218 (a) 144 (b) 169 (b) 194 (d) 219 (c) 145 (d) 170 (c) 195 (a) 220 (c) 146 (a) 171 (b) 196 (d) 221 (d) 147 (d) 172 (a) 197 (c) 222 (a) 148 (c) 173 (c) 198 (c) 223 (b) 149 (b) 174 (a) 199 (b) 224 (d) 150 (a) 175 (a) 200 (a) 225 (b) l 1 l 1 l l 1 I A-74 Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 78. Hence, it possesses 2 degree of freedom. ~-----------------+xa Kutzbach criterion for movabilityof a mechanism, Number of degree of freedom = 3 (1- 1) - 2j - h = 3(2 - 1)- 2 x 0 - 1 =3-1=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. Type of instantaneous centres: (a) Fixed instantaneous centres (b) Permanent instantaneous centres (c) Neither fixed nor permanent instantaneous centres Fixed instantaneous centre: They remains in the same place for all configuration of the mechanism. Permanent instantaneous centres: They move when the mechanism move, but the joints are of permanent mature. Neither fixed nor permanent instantaneous centre:- They vary with the configuration of the Mechanism. 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 isknown as maximum fluctuation ofenergy. .. AE = Maximum energy - Minimum energy 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 t . =--=32 nun sin! <j) y A-75 1 Similarly, for 6-bar or more chains, F > 2 Hence, for a statically indeterminate structures, Mobility ~ 2 Theory of Machines and Machine Design Hence, number of inversions for a slider-crank 11. (c) mechanism will be four. Load to be lifted 6. (d) Mechanical advantage = Effort applied Output force Input force For a four bar linkage in toggle position Effort = 0 .. Mechanical advantage = 00 7. (a) For flywheel which controls the fluctuations in speed 12. (a) during a cycle at constant output load, 1 (2 2)~E ="21 (02 -(01 ~ 400 = ±.I(2102 - 1902 ) 14. (a) ~ 1= 0.1 kg-m-. 8. (b) The critical or whirling speed of centrally loaded shaft between two bearings (a) "'e ="'0 =t; =~ "'e =J 9.81 =73.82 rad/ s (b) 0.0018 2nNc = 73.82 (c) 60 ~ N, = 704.96 ::::705 rpm 9. (a) According to Grsashoff's rule for a planar crank- rocker four bar mechanism, the sum of lengths of 15. (c) 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. 10. (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 16. (d) 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 17. (a) h = numbers of higher pairs Now, a 5-bar chain is the simplest statically indeterminate structure in which link 1is 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. Badboys2Badboys2 Badboys2
- 79. mh2 .. m1 = h + h ...(iv) I 2 From equations (iii) and (iv) mk2 mh2 hi(hi+h.) (hi +h.) k2 = h1h2 m. = h,(h, +h2) ...(iii) From the equations (i) and (ii) we get mh m+-I_I=m I h 2 N2 = 300 rpm N2 = T, =.!. N, T2 3 ..:5_ = .!.T = 24 T 3 I 2 Pitch line velocity = W1r1or W2r2 d2 8x72 = 21tN2- = 21tx300x-- 2 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 C - (wmax - wmin) S - (ro~ ;ro~") or, Cs Wmax + CSWmin= 2 Wmax - 2wmin wmax = 2+ Cs wmin 2-Cs 28. (c) Static balance is a balance of forces due to the action 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. 29. (c) For dynamically equivalent m1 + m2 = m (i) m.h, = m2h2 (ii) m.h] + m2h; = mk ' (iii) From the equations (ii) and (iii); we get m,h~ +(m,h,)xh2 =mk2 mk2 25. (c) T2 = 72 1 VR= - 3 21tN 2 x 1tx 160 W = -- = 16.75 rad/s 60 60 maximum velocity of sliding = W x d = 16.75 x 0.018 = 0.302 mls 24. (a) Arc of contact So, No. of pair of teeth in contact = C· I . h ircu ar pitc 31.4 = ----s;- = 2 pairs. fp = rw2 (COS8+1CO:28) At IDC 8 = 0 .. r. ~rCO>(l+~) At ODC 8 = 1800 fp ~ -rCO> (l-~) 22. (c) 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 23. (c) Arc of contact = 31.4 mm Module (m) = 5 Circular pitch = 1tm = 51t 21. (a) W2 = WI+ at = O. + 5.59 x 12= 67.08 rad/s 1 1 KE = -mk2w2 = -x800x(0.36)2 X (67.08)2 2 2 = 233270N = 233.3 kJ 20. (a) 19. (c) 18. (d) Fp = Primary unbalanced force = mror'cos" Fs = Secondary unbalanced force mrci' (I)=-n-cos28 n =~ 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. a = _!__ = 580 = 5.59 rad/s ' mk" 800 x (0.36)2 Theory of Machines and Machine DesignA-76 Badboys2Badboys2 Badboys2
- 80. 1 5 Speed ratio = T· 1 3 ram va ue 44. (d) The centre of gravity of the coupler link in a 4 bar mechanism would experience both linear and angular accelerations. 45. (a) 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 46. (a) 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. 47. (b) 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 CD 45 3 3 Vc = VBx-= V =-= Vx-=-V AB 30 2 2 . 3 :. Velocity of C = '2 V . 48. (c) The transmission angle is maximum when crank angle 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°. 49. (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 Tfirst 30 3 43. (a) Train values = T= 50 = 5"last 1= 0.1 kg-rrr' 42. (a) We know that llE = ~ l( co~- con => 400 = ~ x [(210)2 -(190)2 ] => 400 = 4001 tan 30° =26mm 15 Ijj 15.. Circular pitch AXIal pitch = tan 300 .. 2k => 9 + -9 = 0 3m OJ" = 2'"J!.40. (d) In toggle position, for a four bar linkage, the mechanical advantage will be infinity. 41. (d) <I> = 30° Normal circular pitch = circular pitch x cos<I> = 15 x cos 30° = 13 mm ( 1 2 2J=> '2mr -i mr 9 + k (9r2) = 0 .. kr' => 9+--9=0 3 2 -mr 2 Taking moments about instantaneous centre' A' I, e+ (kx) r = 0 => (10 + m2)e+ kx (Or) r = 0 39. (c) oicos O If n is large COer= -- n Angular velocity is maximum at 9 = 0, 180° Angular velocity is zero at 9 = 90° 37. (c) 1 = '2l( COmax + comin) (comax- comin) = 1coay x Cs X coay emax = lco;y Cs From the figure, it shows that the value of'P' required will be minimum when it is at 452 to the horizontal. This can be solved by resolution of forces. mT 40x12 R=-=-- = 240mm 2 2 mt 20x12 r=-=-- 2 2 . r sin <I> = IR2_ R 2cos" '" - R sin '".. 2 'V a 'I' 'I' 120x;in20 ~R; -(240cos20)2 -240sin20 R, = 248 mm addendum = 248 - 240 = 8 mm ro cos O 36. (c) 35. (d) 30. (a) Since cylinder falls freely under effect of gravity, it follows basic law of motion and v2 = 2gh and v = ~2gh 31. (a) Ratio of angular speeds of F to A TA. Tc .TE 20x30x 25 TB·To ·TF 60x80x75 24 32. (a) Point P being rigidly connected to point 3, will trace same path as point 3, i.e. ellipse. 33. (c) A system of masses rotating in different parallel planes is in dynamic balance ifthe resultant force and the resultant couple are both to zero. This is known as dynamic balancing. 34. (c) A bicycle remains stable in running through a bond because of centrifugal action. A-77Theory of Machines and Machine Design Badboys2Badboys2 Badboys2
- 81. 57. (a) T T - 60P - 60 x 15000 1492077 N m orque - 2nN - 2n x 960 . - Pitch circle diameter of gear D = m x Z = 4 x 21 = 84 mm T . . Tangential force Ft = D12 = 149207.7 = 3552.56 N 84/4 58. (d) Interference is a phenomenon in which the addendum 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. 59. (b) Pp = 30 kN PQ = 45 kN As we know that, L = (~) a , a = 3 for ball bearing Life of bearing P = Lp = (PQ J3 Life of bearing Q LQ Pp 56. (b) Ft x factor = <Ja x bny x m => <Ja = 3552.56x1.5 = 166.52 MPa 25x0.32x4 106100 = 2122 N (5 x 20) 12 Resultant (normal) force FN = Ft 2122 2258.1 N cos 20° cos 20° 55. (d) For a ball bearing, the life-load relationship is L= (C)3 => Loc= _1F F3 forceTangential T = 60x 20000 106.10 N-m 2n(60x30) 54. (c) :~tr~~:~:~::5m[Zp;Zg1 = 5('9~37) 56 = 5x- = 140mm 2 P = 2nNT => T = 60 P 60 2n N 53. (a) For the shaft subjected to alternating stress of 100 MPa y= 10glO 100 = 2 .. 2 = 3.535 - 0.28167 => x = 5.44964 .. 10giON = 5.44964 N = 105.44964 = 281604.53 cycle X3 6 X-axis 10glON According to 2-point form, the equation of straight line connecting (6, 10giO70) and (3, 10giO490) is 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 2.000 = loglO100 1.845 = 10glO70 I I I ---1--------I I ---r-------T--I I I I I I I I 2.69 = loglO490 51. (a) The Sommerfield number defmed as Z;(~ris 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. 52. (c) 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. 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 Theory ofMachines and Machine DesignA-78 Badboys2Badboys2 Badboys2
- 82. 6(6 -1) NIC= 2 =3x5=15. N(N -1) Number of instantaneous centres, (N1C) = 2 where, N = Number oflinks Hence, (l = -r.I) 113. (c) Considering the following relation, Given (N) = 6 v = rro[ sin00 + Sin;: 0) ] = roix O= ° v=O (l = rro2[cos0" + cos:(O)] .. 2 [ cos 2a] Acceleration of piston (a) = rro cos a + -n- Now, at inner dead centre, a = 0°, then [ . sin 2a] velocity of piston (v) = rto Sma + -2- (a) Elliptical trammel (b) Scotch - Yoke mechanism (c) Oldham's coupling 109 . (d) Considering the following equations for a slider - crank mechanism when the piston at inner dead centre, hO 0.27= 1- 0.03 ho=0.0219mm ::::0.022mm 101. (b) Examples of inversion of double slider crank chain are Let d = diameter of solid shaft . T= Px60 .. 21tN 20xl03 x60 =955N-m 21tx 200 From torsion theory, we have torque transmitted by solid shaft (T). 955x103 =~x~xd3 16 1t 3 =16x45xd d=47.6::::48mm Given data Eccentricityratio = 0.27 = 1- hO c 99. (d) C 50 orP= --1 =-=8.85K.N (180)3 5.65 77. (c) In case of selflocking brake, no external force is required for the braking action. This is not desirable condition in normal application. 79. (b) It is proved by using theory of elasticity that the theoretical stress concentration factor at the edge of hole is given by I +2( ~) 87. (a) Allowable shear stress ~ = 2!L= 360 = 45 N/mm2 fos 8 6000 x 60 x 500 = 180:.(~r L= 106 (C)360n P Here, n = 500 r.p.m, L= 6000 hours, C = 50 KN 73. (b) Snap head rivet is used for boiler plates. 70. (d) Load stress factor K = cr~ sin <P cos <P (_1_ +_1_J 1.4 Ep Eg snap head 67. (a) 2x 16T = 140 ~ d= 15.4mm= 16mm 1td3 ~ 16T Working shear stress = -3 1td Allowable shear stress Factor of safety = W ki h tror mg s ear s ess 61. (b) A-79Theory ofMachines and Machine Design Badboys2Badboys2 Badboys2
- 83. di I . h (p) Number of teeth T iametera PItC d = =- diameter d d T M d I () diameter 143 (a) 0 u e m =. Number of teeth dp 100 Module ofthe gear (m) = Tp = 20= 5 dp + dg 200 => dp + dg = 400 2 dp + 3dp = 400 => 4dp = 400 dp = 400 = 100 4 ...(i) =.!.Q =0.2 50 140. (a) Given: distance between the axis = 200 mm gear ratio = 3 : 1 Number ofteeth on pinion (Tp) = 20 Let dg = 3: 1 , dp dg = ~ => dg = 3dp dp 1 = 2x20 = 0.05 800 126. (a) Heightofthewalt governor (H)= lOCm=O.1 m g= 10m/s2 H = g 2 => (Angular speed)2 =]_ (Angular speed) H Angular speed = {10 = JlOO = 10 rad / s ~OJ 128. (a) Given: weight ofthe ball = 50 N friction at the sluve = ION . . frictio at sluve coefficient of detention = ----- weight of ball 2(410-390) (410+390) _ 2 (112- 111) - (112+ 111) 119. (d) Given: Mass moment of Inertia (Is) = 2500 kg-m- Angular acceleration (a) = 0.6 rad/s- Torque required (r)= Is x a =2500 x 0.6= 1500N-m 120. (c) Given: Speed range of an engine, 111= 390 rpm, 112=410rpm Coefficient of fluctuation = 112-111 = (112-111) 11m 112+ 111 2 _ (Nl + N2) - 2 (Nl - N2) _ (N1 -N2) - 2(Nl - N2) . . N m then, coefficient of steadmess = ----=..:..:'-- N1-N2 w = 6.7981 = 6.7981 = 1.37 rad/s e .J24.75 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 N, +N2 mean speed of flywheel (Nm) = ___:_-2--=- 7.85 x 0.866 .J25-0.25 = 7 85 cos30 th we . x en, ~(5)2 _ sin2 30 h r., 0.75 were, h=--=--=5 Lerank 0.15 = 7.85 rad/s Considering the following formula, 3.14x2xO.15x500 60 3.14 x (2 x Lerank) x 500 60 ndN Angular speed of crank (wemnk) = 60 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 116. (c) Theory ofMachines and Machine DesignA-SO Badboys2Badboys2 Badboys2
- 84. w -4_c_=_l 0-4 wc-4=4 wc=4+4= 8radls 200. (a) Given: Moment of inertia (L)= 9.8 kg-m/ fluctuation speed (Nt) = 30 rpm fluctuation energy (E) = 1936 j D = 1-0.25 = 0.75 P 190. (d) 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, Lf = (~rx 106 K = Constant = 3.3 72xl06 =e·2~OOOJ.3xl06 On solving, we get, w= 609.8 N (approx.) 197. (c) Given: Number oflinks (L) = 5 Number ofjoints (1)= 5 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 (TB) Now, using the following relation, wc-wA =_TB=_l wB-wA r, 50 1 2 -=-=> 12 = 200N Imm 12 4 178. (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, P-D D then TIt=0.25=--=1--, 'I P P D = 1-0.35 = 0.65 P 177. (b) Given:Shearstrengthofrivet(11)=50N/mm2 diamenter of rivet (initial) (D1) = D Final diamenter ofrivet (after doubling) (D2) = 2D As we know that, Shear strength of rivet IX (Diameterj- 1 IX (D)2 0.35 = P - D = 1- D P P 44 22.5x1000=(T1-T2)X3 22.5 x 1000x 3 ( (T} - T2) = 44 1534.1N == 1540N approx) 168. (a) Inner radius (R)= 50 mm Outer radius (RJ= 100 mm axid force(w) =4 KN 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 11t= PxtxFt d T mXPd =-x-= 1 T d 157. (c) Given: Average tension (Tavg.)l = 700 N and (Tavg.h = 400 N Linearvelocity(v) = 5 m1s Power transmitted (P) = [(Tavg.)l - (Tavg.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-sl +S2JVelocity ratio - d2 100 166. (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 =-m/s=-m/s 60 3 A-81Theory ofMachines and Machine Design Badboys2Badboys2 Badboys2
- 85. 6x80 480 =--=-=240mm 2 2 1000 x 60 x 3000 = (4900)3 =.!. 2000 x 60 x L2 9800 8 1500 1 ~ ="8 => L2 = 1500 x 8 = 12000 hours 203. (a) Given, Initially, wI = w Life = LI 25.2xl06 =_7_xl06 xu2 1000 2 25.2 xl000 u =---- 7 u= ..)3600 = 60 m / s 227. (c) Given: Number ofteeth (T) = 60 Module (m) = 6 mm Number ofteeth on pinion (Tp) = 20 Centre distance (D) = m(T + Tp ) = 6 (60 + 20) 2 2 w After half of load, w 2 =- 2 Life=L, Considering the relation, Lw3=c _ 112-111 2( 112-111) 111+ 112 111+ 112 2 2(510-430) 160 = 510+ 430 940 = 0.17 225. (b) Given: Safe stress (as) = 25.2 MN/m2 density (p) = 7g/cm3 Safe stress (aJ = p x maximum periphered velocity (v2) as = p x v2 . . 112-111 224. (d) Coefficient of fluctuation of speed = ......:..:....___:~ 11m . ffici () P-D Dteanng e ICleny 11t = _- = 1-- P P = 1-0.25 = 0.75 202. (b) Given: Load (wI) = 9800N Speed(N1)= 100rpm (Life) (LI) time duration = 3000 hours Now, ifload (w2) = 4900 N Speed (N2) = 2000 rpm then Life (L2) = ? Considering the following, L(w)3 = Constant =>..s. = (W2J3 L2 wI W mean = 629 rpm == 600 rpm (Approx.) 201. (b) Given: diamenter of rivet hole = D Pitch of the rivet = P D -= 0.25 P L1 = L2 8 L2 = 8L1 =>8 times. 207. (d) Given: dynamic load capacity = 22 x 103N Speed (N) = 600 rev/min. (time duration) Life (L) = 2000 hours Using the following relation, L = (:rX 106 rev we get w = 5.29 KN E 1936 wmean= Tfxw =9.8x1t =629rpm(Approx) . 21tNt 2 x 1tX 30 Now, Angular velocity (w) = _- = --- 60 60 = n rad/s Now, considering the following relation, E =Lx w x wmean A-82 Theory ofMachines and Machine Design Badboys2Badboys2 Badboys2
- 86. 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. (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) Surrounding ~system boundary (Thermodynamic system) Typesofthermodynamic systems: There are three types ofthermodynamic systems: (a) 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 lnputmass stem boundary Input mass Exit mass Surroundings Exit energy (Open system) THERMODYNAMIC SYSTEM 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. THERMODYNAMICS 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. Some Thermodynamical Terms in brief (i) Thermodynamic system: A thermodynamical system is an assembly oflarge number ofparticles 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. rrIII~ll)ll'l~ I~Nf.INI~I~IIINf. Badboys2Badboys2 Badboys2
- 87. --~V (Total work done) I = (Total heat)eye e 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 errthalpywith respect to temperature keeping the pressure as constant. C =(dH) p dT v THERMODYNAMICAL PROCESSES Any process may have own equation of state, but each thermodynamical process must obeyPV = nRT. 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. Someterms like specificvolume, specificenergy etc. come under the category of specific extensive properties. Thermodynamic equilibrium A systemis saidtobe in equilibriumwhenthereis 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) Mechanicalequilibrium (b) Thermalequilibrium (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. IfdU 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 fromstate'A' to state'B' and the systemundergoes a cyclic process. A f ~B System 1 System 2 If SI and S2are the entropies ofthe system I and 2 respectivelyat 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 ~Q M= T for constant T For a systemwith variable T, we have sf dQ ~S = Sf - s, = f T 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. 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. Differentialformofthe 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 efficiencyof any heat engine always be less than 100%. (ii) 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. ENTROPY Entropy is the another thermodynamical variable which many times veryuseful to understand the system.Entropy is related to the disorder orrandomness in the system.Tounderstand this, let us consider two systems as shown in Fig. Thermal EngineergingA-84 Badboys2Badboys2 Badboys2
- 88. p (vi) First law ofthermodynamics in isothermal process. As ~T 0, :. ~U=O Q ~U+W=O+W or Q W AdiabaticProcess: An adiabatic process is one in which pressure, volume and temperatureofthe systemchangebutheat willnot exchange between system and surroundings. p 4. C = (v) Specificheat at constant temperature: As~T=O, -P -P V or tan O = V dP dV or (i) An isothermal process obeys Boyle's law PV = Constant. (ii) The wall ofthe containermust 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 sufficienttime forthe exchange ofheat. (iv) Slope of P - Vcurve: For isothermal process PV = Constant After differentiating w.r.t. volume,we get P+V dP 0 dV p p 3. Isothermal Process: A thermodynamical process in which pressure and volume of the system change at constant temperature, is called isothermal process. »c.sr (v) Work done: W = P~V= 0 (vi) First law ofthermodynamics in ischoric process Q ~U+W=~U+O or Q ~U dP (ii) Slope of P - Vcurve, dV = 00 (iii) Specific heat at constant volume 3R 5R Cv = 2formonoatomicand Cv = 2fordiatomic (iv) Bulk modulus of elasticity :As Vis constant, ~V= 0 An isochoric process obeysGay - Lussac'sLaw, P oc T(i) '-----------·vo W=O j T (v) (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:Boilingofwaterand freezingofwater atconstant pressure etc. 2. Isochoricor Isometric Process: A thermodynamical process in which volume ofthe system remain constant, is called isochoric process. and M B (-A:f Work done: W = P~V=nR~T (i) Isobaric process obeys Charle's law, VOC T dP (ii) Slopeof P ~ V curve, dV = O. (iii) Specific heat at constant pressure 5R 7R C;= 2 for monoatomic and Cs= Tfordiatomic (iv) Bulk modulus of elasticity: As P is constant, M = 0 Compression '---~----~- ....v V Expansion W = +Pav p .........:-~ ........-...........,p ......------to--~ p 'W=-P~V P 1. Isobaric Process: Ifa thermodynamic system undergoes physical change at constant pressure, then the process is called isobaric. M (-~Vr00 B A-85Thermal Engineerging Badboys2Badboys2 Badboys2
- 89. 11 Heat absorbedby enginefrom source({1) Efficiency ofcarnot engine Workdonebyengine(W) ...(iv) V3 V4 or VIV3 V2V4 Also V2 V3 VI V4 From equations (ii) and (iii) , wehave V2 5_ ...(iii)or Similarlyin the adiabaticcompressionD ---+ A T Vy-1 1',v,y-l 2 4 1 1 ...(ii)or In the adiabatic expansionB ---+ C 1',v:y-1 T Vy-1 1 2 2 3 ~ -nRT2fn(~J Adiabatic compression: IfW4 is the work done during the adiabatic compression, then nR(lj - Tf) nR(T2 -Ii) -nR (Ii- T2) W4 y-1 y-I y-I Net work done in the whole cycle W =Wj+W2+W3+W4 - R'T'II (V2J nR(Ii -T2) RT II(V3J (Ii-T2)- n -LJ~n - + n 2~n - -nR__:__:_----='-'- Vi y-I V4 y-I 4. 3. 2. o-------------+v p ... 1. Isobaric Process 3. Adiabatic Process CARNOT CYCLE Camot cyclehas four operations. Thermodynamic coordinates after each operation are shown in Fig. Initially at A coordinates are r;Vj,Tj. 500K 400K 300K L------------___.v 2. Isothermal Process 4. Isochoric Process ...p Q 0 (v) Specific heat:C = nflT = 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 k RY = another constantor k V RT P Also or RT PV=RT, or P= V Substituting in PVf = k, we get (Rnvy k k R = new constant (i) Adiabatic process must be sudden, so that heat does not get time to exchange between system and surroundings. (ii) The walls ofthe container must be perfectlyinsulated. (iii) Adiabatic relation betweenP and V PVY =k (iv) Adiabatic relation between Vand T & P and T For one mole of gas Isothermal expansion: If Qj is the heat absorbed from the source and WI is the work done, then, QI ~ WI ~nR1Jfn(~ ) (As AU ~0) nR1Jfn(~ J Adiabatic expansion: If W2 is the work done during the adiabatic expansion, then nR(lj - Tf ) nR(Ii - T2) W2 = y-I y-I Isothermal compression: If Q2 is the heat rejectto the sink and W3 is the work done during the process, then Q2 ~ W3~nRT2fn(~) ~ nRT2fn(~:J (As flU = 0) 1. Thermal EngineergingA-86 Badboys2Badboys2 Badboys2
- 90. Availableand unavailable energy in a cycle. For a given T}, 11rev. will increase with the decrease of Tj. The lowestpracticabletemperatureofheatrejectionis the temperature of the surroundings, To. u.E. = Ql- Wmax T For the given r. and T2, 11rev. = 1- T~ AE.+U.E. AE. = Ql - U.E.or Available Energy Referred to a Cycle. The maximum workoutputobtainablefroma certainheat input in a cyclicheat 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 ofthe energy supplied. High Grade Energy Low Grade Energy (1) Mechanicalwork (1) Heat or thermalenergy (2) Electricalenergy (2) Heat derivedfromnuclear fissionor :fusion (3) Waterpower (3) Heat derivedfromcombustion of fossilfuels (4) Windpower (5) Kineticenergyof a jet (6) Tidalpower Available Energy The sources of energy can be divided into two groups (l) High grade energy (2) Lowgrade 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. 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 ofsalt in water. (iii) Rusting of iron. (iv) Sudden expansion or contraction ofa gas. AVAILABILITY AND REVERSIBILITY Reversible Process Anyprocesswhichcanbemade toproceedin thereversedirection by variation in its conditions such that any change occurring in anypart ofthedirectprocessisexactlyreversedinthecorresponding 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 approximatelyreversible. (iii) Aworkingsubstancetakenalongthe completeCarnot's cycle. (iv) The process of electrolysis is reversible if the resistance offeredby the electrolyte is negligibly small. A complete reversible process is an idealised concept as it Dryness fraction (X) = My + ML . where, My= mass of steam or vapour ML= mass ofLiquid (saturated) Drynessfractionis utilizedto calculatethe quantityofliquid or vapour phase within the mixture. The value of dryness fraction varies between 0 and 1. It also determines the quality ofthe steam. REVERSIBLE AND IRREVERSIBLE PROCESSES 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- TI = 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 ofmixture (water in saturated form in liquid region as well as vapour region). My r, - T2 _ 1 T2---- -- 11 11 Enthalpy: Enthalpy is regarded as the total energy of a thermodynamic system.It is defined as the sum ofinternal energy and product of pressure and volume. It is an intensive property of the system. It is describe as: H=U+PV As nR[ Tjfn( ~ )-T2fn(~)] nRTjfn(~: ) A-87Thermal Engineerging Badboys2Badboys2 Badboys2
- 91. (p+:2)tV -nb) = nRT Here a and b are Constant called Vander waal's constant. ANALYSIS OF THERMODYNAMIC CYCLES RELATED TO ENERGY CONVERSION 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) bythe systemagainst the surroundings. i.e., 1Q= ~U + 1W Heat(~Q) andworkdone(~W) arethe path functionsbut internal energy (~U) is the point function. Work Let us consider a gas or liquid contained in a cylinder equipped with a movablepiston, 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. total volume of the gas. Therefore volume of the gas is equal to volume of the vessel. (4) The moleculesofgases are in a stateofrandom motion, i.e., theyare constantlymoving with all possiblevelocitieslying 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 moleculesdo not posses any potential energy.Their energy iswhollykinetic. (8) The collisions are instantaneous i.e., the time spent by a moleculein a collisionis verysmallas comparedto the time elapsed between two consecutive collisions. (9) Thoughthemoleculesareconstantlymovingfromoneplace toanother,the averagenumberofmoleculesperunit 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 wholevesselremains uniform. (11) The mass ofa molecule is negligibly small and the speed is very large, there is no effectof gravity on the motion ofthe 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 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 closelyat lowpressure and high temperature. 4. Equation of state for real gases is given by Vander waal's equation Thermal Engineerging 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. 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. LetU, S, andVbethe initial valuesofthe internal energy,entropy, and volumeofa systemand Uo,So,and Votheir [mal valueswhen 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 <I>is the availabilityfunction and is a compositepropertyof both the system and its environment, with U, S, and V being properties of the system at someequilibrium state, and Toand Po the temperature and pressure of the environment. (In the Gibbs function, G = U- TS + PV,T, and p refer to the system). The decrease in the availabilityfunction in a process in which the systemcomes to equilibrium with its environment is <I>-<I>o=(U- ToS+ PoV)-(Uo- ToSo+PoVo) .. W::;;<I>-<I>o Thus the useful work is equal to or less than the decrease in the availability function. Irreversibility of the Process The actualwork donebya systemis alwayslessthan the idealized reversible work, and the differencebetween the two is called the irreversibility ofthe process. 1= Wmax- W This isalsosometimesreferredtoas 'degradation' or 'dissipation'. For a non-flow processbetween the equilibrium states, when the system exchangs heat only with the environment .. I~O I= To[(1S)system+ (~S)SUlT.J Similarly, for steady flowprocess, I = To(1Ssystem+ ~SSUlT) The same expression for irreversibility applies to both flow and non-tlowprocesses. The quantityTo(~Ssystem+ 1SSillT)represents an increase in unavailable energy (or energy). BEHAVIOUR OF IDEAL AND REAL GASES 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 moleculesarerigid andperfectlyelastic spheres ofvery smalldiameter. (3) Gas moleculesoccupyverysmall space.The actual volume occupied by the molecule is very small compared to the A-88 hrrnx TO and1-- Tl Wmax (1-~~)QI .. Wmax AE. Badboys2Badboys2 Badboys2
- 92. -T I .F..U's3:m,a-.n-e- ',2.,3__1 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 If the heating of ice at - 10°C to stream at 250°C at the constant pressure of 1atm is considered 1-2is solid(ice)heating, 2-3 ismelting ofice at O°C,3-4 is the liquid heating, 4-5 isthe 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 1upon cooling. The curve passing through the 2, 3 points is called the fusion curve and the curve passing through the 4, 5points (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 ofCO2 isat 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. Fig.(i) Fig.(ii) PROPERTIES OF PURE SUBSTANCES 1. It is a single substance and has a uniform composition. It has constant chemical composition through its mass. 2. It has a same colour, taste and texture. 3. It has a fixed melting point and boiling point. W= - area A RCT>F.FA Fig.(iii) W= area A RCA '-----!-f-;-' -------;-;,D-----t> V JiV = area ARC[)F.A {JL___~A _ _.v p!' (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. 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. The work done in this expansion Wx = +areaAXBCDA Now gas returns to its initial stateB via path BYA. Work done during this compression Wy -area BYADCB The net work done W Wx+ Wy areaAXBCDA-areaBYADCB +areaAXBYA o'----,:r:c-)-------'-c'--.v p Work done in Cyclic Process Suppose gas expands from initial stateA to final state B via the pathAXB. (b) Non-cyclic process(a) cyclic process L------- ... vL----~-- ..v pp W = P(Vf -VJ =Pi1V Cyclic Process and Non-cyclicProcess 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. VI W = f PdV V; If the pressure remain constant while volume changes, then the work done 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 F=P~ A-89 I-dx Thermal Engineerging Badboys2Badboys2 Badboys2
- 93. 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. --7 (s) Entropy 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. Fig. (a) p-v- T surface for water which expands a freezing 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 (:~) p ~T 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. 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. 373 = 1 x 4.187t'n 273 = 1.305 kJ/-K (iv) Entropy increase of water as it is vaporized at 100°C, absorbing the latent heat of vaporization (2257 kJ/kg) ilS4 = S5- S4 mL 2257 =T= 273 =6.05kJ/kg-K ... (wherem= 1kg) (v) Entropy increase of vapour as it is heated from 100°C to 250°C at 1atm. ... (wherem = lkg) (iii) Entropy increase of water as it is heated from O°C to 100°C (cPwater = 4.187 kJ/kg-K) T3 ilS3 = S4- S3= m cp en T2 273 273 = mcp en 268 = 1 x 2.093 en 268 = 0.0398 kJ/-K (ii) Entropy 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 = T = -----ri3 = 1.232 kJ/-K (i) 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 T2=273 me dT =j-= j _p T T1=268 T fl00'C - -------------- 2500C 6_ T-s diagram for a pure substance Consider heating of the system of 1kg 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. Thermal EngineergingA-90 Badboys2Badboys2 Badboys2
- 94. Temp. Pressure Sat Sat Sat Sat. Sat Sat Sat Sat °C kPa,MPa Liquid Vapour Liquid Evap. Vapour Liquid Evap. Vapour Liquid Evap. Vapour T P Vr Vg Or Org ug hr hrg hg Sf Srg Sg 0.01 0.6113 0.001000 206.132 0.00 2375.3 2375.3 0.00 2501.3 2501.3 0.0000 9.1562 9.1562 5 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 EntropyKJIKgKEnthalpy KJIKgInternal Energy KJIKGSpecificvolume, m3/kg In steam table, properties of water are arranged as a function of pressure and temperature. Saturates steam: Temperature table STEAM TABLES A-91Thermal Engineerging Badboys2Badboys2 Badboys2
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- 96. A-93 P=10 kPa. (45.81) P=50 kPa.(81.33) P =100 kPa. (99.62) 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 2682.57.6947 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 3131.53488.1 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 <XX> 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 P=200 kPa (120.23) P =300 kPa (133.55) P=400 kPa (143.65) 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 1.341363300.8 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 Thermal Engineerging Badboys2Badboys2 Badboys2
- 97. QI = h, -h6s + h3 - h2s; Q2 = h4s - hs WT = h, - h2s + h3 - h4s; Wp = h6s - hs 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. 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, Pump Wp The flow diagram for the ideal Rankine cycle with reheat is shown in fig. where, Q1 = heat transferred to the working fluid Q2 = heat rejected from the working fluid WT = work transferred from the working fluid Wp = work transferred into the working fluid RANKINE CYCLE WITH REHEATER (hI -h4)-(h2 -h3) (h, - h4) (c) ... s (a) 1" Simple steam power plant Process 1 : Reversible constant pressure heating process of water to form steam insteam 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 High pressure water Process ~, r I : I Boiler I I I I I I I I I I I I Air _J T and Combus- fuel tion products Process 1 High pressure, high temperature steam 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 The efficiency of the Rankine cycle is given by Thermal EngineergingA-94 Badboys2Badboys2 Badboys2
- 98. (a) We+--- BRAYTON CYCLE WITH REGENERATOR AND REHEATER 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. ...(iii)or 11 - 1 1 Brayton - (rp iy-l )/y The efficiency of the Brayton cycle, therefore, depends upon either the compression ratio or the pressure ratio. ...(ii)or 1 11Brayton = 1- y-l rk If rp = pressure ratio = P2/P 1the efficiency may be expressed in the following form also ( ,(y-l)/y 11= l-l:~) ( J Y-l 11= 1- :~ T4 -1= T3 -1 T} T2 T T T ( (Y-l)/y ( y-l 4 - 1 1 PI I v2 I T3 - T2 = T2 = lpJ = l~) If rk = compression ratio = V /v2 the efficiency becomes [from Eq. (i)] or Now As ...(i) pressure. Efficiency of Brayton Cycle: 11= 1- Q2 = 1- T4 - Tl Q} T3- T2 Q 1= heat supplied = mcp (T3- T2) Q2 = heat rejected = mcp (T4 - T 1) T ( J (Y-l)/Y T 2 P2 3. T;= Pt = T4 (Since PI = P3' and P4 = PI) 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 Heat exchanger TurbineCompressor 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. 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. RANKINE CYCLEWITH REGENERATOR AND REHEAT A-95Thermal Engineerging Badboys2Badboys2 Badboys2
- 99. AIR STANDARD CYCLES 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: 1. The working fluid is taken as air for the cycle and assumed to be a perfect gas. 2. The engine is assumed to working in a closed cycle. There is no change of mass of the working medium. 3. All the processes that conclude the cycle is of reversible nature. 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 T6 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. (c) ~v f (b) .. s (a) -1 T} y-1/y 11- --y T4 p For a fixed ratio of (T /T 4) the cycle efficiency drops with increasing pressure ratio. Effect of Reheat on Brayton Cycle: Effectiveness of the regenerator: t3 - t2 actual temperature rise of air E = --- =------=---------- Ts - T1 maximum possible rise of temperature Efficiency of Brayton cycle with regenerator. Q} T6 - T} T} [(T2 lTd-I] 11= 1- Q2 = 1- T4 - T3 = 1- T4 1- (Ts I T4) T} T2 [1-(T} IT2)] = 1- T4·r; 1-(Ts IT4) _____. S (b) 4 Thermal EngineergingA-96 Badboys2Badboys2 Badboys2
- 100. Process 3-4, ( J Y-1 T2 = ~ T1 v2 Process 1-2, 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 efficiencyof air-standard otto cycle 11= 1- Q2 = 1 mcy (T4 - T}) = 1 (T4 - T}) Q1 mCy(T3 - T2) (T3 - T2) ...(i) (d)(c) o 4. Exahust EVO ~It rl~ The air-Standard Otto Cycle of spark-ignition (SI) engine. It is named after N.A.Otto a German engineer who first built a four- stroke 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. 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 ODC---.~ V1 DC 2,6 Indicator diagram: (b)(a) BDC or outer _____. dead centre (ODC) TDC or inner dead centre (IDC) The schematic diagram of each stroke is shown in fig. 4. Heat is considered to be provided from a constant High temperature source and not be chemical combustion. 5. Heat loses are negligible. 6. The value of specific heats of workings substance remain constant throughout the cycle. 7. Kinetic energy and Potential energy remain constant throughout the process. 1. Induction 2. Compression 3. Expansion IVO Both value closed Jl~ Spark 0 0 0 AIR-STANDARD OTTO CYCLE A-97Thermal Engineerging Badboys2Badboys2 Badboys2
- 101. Process 2-3 Process 3-4 Cut-off ratio, It is see that Expansion ratio, VI vI rk =-=- V2 v2 V4 v4[ -_-- e - V3 - v3 V3 v3[ -_-- c - V2 - v2 rk=re·rc T4 =(~JY-I T3 v4 rrl y-l rc T4 =T3--l Y-rk T2 = P2v2 =~=_ T3 P3v3 v3 rc Compression ratio, The efficiency of diesel engine: 11= 1- 21._ = 1- mcv (T4 - T1) = 1 (T4 - TI) QI mCp(T3 - T2) Y (T3 - T2) Here, 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 cut- offratio. 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. --_'.~ V 3 4 r 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. W = PIVI(r -1) (£Y-l_l) net 1 P ky- El_ = P4 = rp (say) P2 PI Wnet = PlVl (P3V3 _ P4V4 _P2V2 +IJ y-l PlVl PlVl PlVl Now, Volume at the beginning of compression VI = ~ rk = Volume at the end of compression 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 or ...(ii) 1 11otto = 1- y-l rk whe-e rk is called the compression ratio and given by ( J Y-l . V2 Fromeq. (1) 11= 1- ~ or The indicator diagram ofdiesel cycle Thermal EngineergingA-98 Badboys2Badboys2 Badboys2
- 102. =1- Q2 =1- mCp(Tc-TB)+mcp(TD-TC) Q1 mCv(TE- TA) 11=1 (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 forOttocycleandminimum fordieselcycle. Also for same heat input, efficiency of Otto cycle is maximum whilethat ofDieselcycleisminimum. (b) For samemaximum pressure and sameheat input, the efficiencyof Diesel cycleis more than Dual cycle. (c) For same pressure and temperature, the efficiency of Diesel cycle is more than dual cycle. Steam boilers Steamboiler isbasicallya closedvesselinto 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) Steamboilers are also used forheating 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 ofwater (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) Lowpressure boiler: It is described as a boiler which developespressure of the steam whose value ofbelow 80 bar.ExamplesareCochran,Lancashire,Locomotive boilers etc. (b) High pressure boiler : It is defined as the boiler in which steamis developedat morethan 80 barpressure. Examplesare Babcock and Wilcox,Benson,Lamount etc. Weknow that, efficiency(11) ~ S Heat supplied, Q, = me (T,- T + me (T - T )P B P D C Heatrejected, Q2 = mcv (TE - TA) Q1-Q2 Q1 T IB constant volume 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 1C~D Isentropic ~ process ()) B ~ EtZl £ 1 1(ry -IJAs rc > 1, = Y ~-1 is also greater than unity. Therefore, the efficiencyof the Diesel cycleis less than ofthe Otto cyclefor the 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 1 rJ-l 11Diese1= 1--. y-1 . --1 Y rk re- y-1 T l__ T3 _1_ 3 . ry-1 re ry-1 k k 11= 1 1 T3 1 T1=T2 .--=-.-- "y-1 r "y-1 "k e J.k Substituting the values of Tl' T2 and T4 in the expression of efficiency. Process 1-2 A-99Thermal Engineerging Badboys2Badboys2 Badboys2
- 103. 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. Itis 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 Fire Tube Boiler Water tube boiler Hot gas es are inside the Water is available inside tubes while water out side the tube while hot gases the tubes outside the tubes. Firing is done internally Firing is done externally. Operatinal pressure is upto Operational pres sure is 16 bar high as upto 100 bar Less amount of steam is High amount of steam is generated generated It requires large floor area for It requires les s floor area g rven power for the same power It is not generally used in It is used in large power large power plants plants Boiler shell diameter is large Boiler shell diameter is for given power small for the same amount of given power Parts are not accessible Parts are easily accessible easily for the purpose of for the purpose of maintenance maintenance Efficiency is less Efficiency is high Initial cost is less Initial co st is high It has a large ratio ofwater It has a comparatively small content to s team capacity ratio ofwater content to steam capacity Slow in evaporation Quick in evaporation (viii)Passage of gas: 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. Difference between fire tube boiler and water tube boiler: (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. (iii) Tube contents : 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. (iv) Circulation ofwater: 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, steam boilers may be classified follows (a) Horizontal (b) Vertical (c) Inclined Thermal EngineergingA-tOO Badboys2Badboys2 Badboys2
- 104. (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 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). where, C,= specific heat at constant pressure TB = Temperature of heated feed water TA = initial feed water temperature H= Enthalpy H = Sensible heat of waterw (d) Air Pre-heater : Itis 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 ~ 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. Cp(TB-TA) % saving in fuel consumption H _ H x 100 w x.toiThermal Engineerging Badboys2Badboys2 Badboys2
- 105. VW2 Velocity triangle of steam turbine )1 Steam flow through turbine blades: (some important formulas) P3 -.................... --Entropy (~) ~ I+-dH(,+ll+-dH2-+1 Vr2>Vrj Fixed (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. Velocitycompounding (d) Pressure velocity compounded turbine: Itis the combinationarrangement ofpressurecompounding 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 ofa smaller turbine for similar value of pressure drop. The efficiency is very low and it is very rarey used nowadays. Pressure compounding (c) Simple velocitycompounded impulse turbine: This typeofturbinesconsistsofa setofnozzlesand moving blades rows which are connected the shaft whereas the fixedblade rows are connected to casing. Ithas in general, moving and fixedblades.The steam at a veryhigh 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. pressure absorbed is equally (nearly) divided in stages which reduces the steam velocity entering into moving blade. Thermal EngineergingA-l02 Badboys2Badboys2 Badboys2
- 106. Classification ofI.C. Engine: An I.e. Engine can be classified as follows: (i) According to the type of fuel used Steam Engine LC. Engine In steam engine, In r.c Engine, Combustion combustion takes place takes place inside the engine. outside the engine. Steam engines are Operation one. Engines operated at temperature done at temperature values of values of around 600° C about 2400°C It does not require It requires cooling due to cooling 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 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% 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. Comparison between I.C. Engine and Steam angine : Impulse turbine Reaction turbine Only kinetic energy is Kinetic energy and utilized for the purpose Pressure energy both are of rotation of turbine utilized for the purpose of rotation of turbine Water flows over the Guide mechanism directs nozzle followed by the water for the purpose striking the blades of flowing over turbine Pres sure is decreas ed Pressure is decreased in in nozzle and not in fixed blades (nozzles) and moving blades also in moving blades The type of the blades The type of the blade is is profile aerofoil Low power is High power is produced Low efficiency High efficiency For same power For same power generations ifrequires generations, it requires less space more space 11. Maximum efficiency of reaction turbine; 2cos2 a 11max = 2 ls-cos U Comparison between Impulse and reaction turbine: impulse turbine 9. If blade speed = u = vcosa, then efficiency of reaction turbine is maximum. 10. Maximum efficiency of impulse turbine, cos2 a ( vr1cos<l> ' 11max = -l +1)2 vr2 cosn vI CoSU If blade speed = u = 2 .fhen efficiency of8. Axial force on wheel = w (v fl - Vf2 ) g 7. (VWl-VW2)u (11) = g x ~h x J 11= blade efficiency x nozzle efficiency work done on blade (11) = Total energy supplied per stage 6. 2u(vW1 -VW2) Blade efficiency = 2 vI Stage efficiency 5. v2 Energy supplied to blades/kg of steam = 2~4. 3. W u(VW1-VW2) Horse power = g 752. w 1. Work done on blades/sec. = -(VWl - VW2)X u g Vrl = relative velocity at inlet vr2 = relative velocity at output VI = Absolutive velocity of steam at inlet v2 = Absolute velocity of steam at output u = nozzle angle J3 = Angle with which discharged steam makes with tangeal of wheel e = Inlet angle of moving blade <I> = exit angle of moving blade co = steam weight which flowes through blade Qv = volume of steam flowing through blades D = diameter of blade drum Let us consider the following: u = velocity (Linear) of blade vWI= velocity of whirl at inlet VW2= velocity of whirl at outlet vfl = flow velocity at inlet vf2 = flow velocity of output A-I03Thermal Engineerging Badboys2Badboys2 Badboys2
- 107. B.D.C. But actually exhaust value begines to open when about 85% ofworking stroke is completed. A pressure (4- 5bar) forcesabout60%ofburnt gasesintoexhaustmaxifold 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 opensand inertia offlywheeland 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 coveredby it and if also consists of the values. (c) Piston: The other end ofthe working space ofcylinder is coveredbythe piston. Powerproduced dueto combustion products is transmitted to the crank shaft. (d) Connecting rod: It changes and transmits the translatorymotion ofthe 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) Rockerarms (P) Value - springs (q) Crank case : It holds the cylinder piston and crank shaft. (r) Flywheel (s) Bed plate (t) Cooling waterjackets 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 air- fuelmixtures are given below: Upperlimit =>20 : 1 Lowerlimit => 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 (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 cycleengine (b) Oho cycle engine (c) Dual cycleengine (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 bycausing an electric spark by spark plug. During combustionprocess, 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,pressuredecreasesdue 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 Thermal EngineergingA-l04 Badboys2Badboys2 Badboys2
- 108. Four stroke cycle Two stroke cycle Cycle is completive in Cycle is combusted in four strokes two strokes Heavy flywheel is Light flywheel is required required Engine is heavy for Engine is light for same same power output power output Less cooling and High cooling and lubrication is needed lubrication is needed High initial cost Low initial cos t High thermal efficiency Lower thermal efficiency Volumetric efficiency is Volumetric efficiency is more less n 2 where, A=-d 4 d = borediameter (m) L = length of stroke(m) (xiv)Calorificvalue 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: Specificfuel 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: Specificoutput = ~(kw /m3) AxL R 1 · fu I . .(lL.) Actual fuel- air ratio e ative e - aIr ratio .._K = -------- Stoichiometric fuel air ratio (xi) Relativefuel air ratio: l1lTH l1Rel. =-- 11cycle (x) Fuel- Air ratio: F I Ai . Mass of consumed fuel ue - r ratio = -------- Mass of air taken inside during the same period of time (vii) Volumetricefficiency: It is defined asthe ratio ofmass 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. B.P IP B.P BP+FP Brake power Indicated Power11mech. B.P mfxC (vi) Mechanical efficiency: mf xCL where, m,= mass of fuel supplied (kg/s) CL = Lower calorificvalue ofthe fuel (v) Brake thermal efficiency: Brake Power 11 =------ BTH Input fuel energy LP 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. 2nNT Brake Power (BP) = 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: IndicatedPower 11 =------ IlH Input FuelEnergy PmxLxAxNxKxh Indicated Power (lP) = 60 kw where, Pm= mean effectivepressure (KPa) L= length of stroke (m) A = Piston area (m/) N = Number ofrevolution ofcrank shaft (rpm) 1 K = 2"for4- strokeengine and 1for2 - strokeengine n = number of cylinders (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. A-105Thermal Engineerging Badboys2Badboys2 Badboys2
- 109. (b) Dry sump lubrication system In this system, oil supply is carried in a separats tank. Scavengerpumps 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 lowother 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 ofair flowingthrough cooling tins, tin surface area and thermal conductivity of material of the fins used. Applications: Small engines, Industrial and agricultural engines. (ii) Wateror liquid cooling: In this system,water or liquid ismade to circulatearoundthe cylindersand 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. 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 filmmade 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, flashpoint 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 S.I Engine C.I. Engine Ifperforms on OHO cycle Ifperforms on diesel cycle Compression ratio ranges Compression ratio from 5 to 10 ranges from 13 to 27. Carburetor supplies fuel Fuel injector supplies fuel Maintenance cost is low Maintenance cost is but running cost is high high but running cost is low Spark plug is used No Spark plug is used Thermal efficiency is low Thermal efficiency is high Thermal Engineerging Difference between SI and CI Engine: A-l06 Badboys2Badboys2 Badboys2
- 110. 13. 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 14. According towhich law,all perfectgases change in volume by II 273 rd oftheir original volume at 0° C forevery 1° C change in temperature while pressure is kept constant. (a) Joule's Law (b) Boyle'slike (c) Charles's law (d) Gay - Lussca Law 15. Properties of substances like pressure, temperature and density in thermodynamic co-ordinates are: (a) Path function (b) Point function (c) Cyclicfunction (d) Real function 16. 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 17. 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 ofwork transter (c) amount of heat rejected (d) amount of mass transfer 18. Heat is being supplied to air in a cylinder fiiled with frictionless piston held by a constant weight. The process willbe: (a) Isochoric (b) Isothermal (c) Isobaric (d) Adiabatic 19. During an adiabaticprocess,the pressure P of a fixedmass of a ideal gas changes by ~P and its volume v changes by ~V ~V. THe value of V is given by : 5 (b) C=-(F+32) 9 9 (d) C=-(F-32) 5 5 (a) C= -(F -32) 9 5 (c) F=-(F+32) 9 10. The fixed points for celcius temperature scale are: (a) Ice point as 0° C (b) Steampoint as 100°C (c) Both ice and steam points as O°C and 100°C respectively (d) Triple point ofwater as 0.Q1° C 11. For the calculation ofreal 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) 9. 8. 7. 6. 5. 4. Stirling and Ericsson cycles are: (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 ofperpetual 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 3. (c) 2 (d) 1 4 3 2 3 1 3 4 2 (a) (b) 4 ABC D 234 D. Cribbs function Codes: 1. Point function 2. Path function 3. Second Law of thermodynamics 4. F=C-P+2 A. Heat B. Energy C. Entropy 2. Which ofthe following isnot a property ofthermodynamic 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 1. ...,EXERCISEI···.. A-t07Thermal Engineerging Badboys2Badboys2 Badboys2
- 111. 34. 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 35. 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 36. Actual expansion process in a throttling device is : (a) Reversible adiabatic expansion (b) Isenthalpic expansion (c) Isothermal expansion (d) Irreversible 37. 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 38. The first Law of thermodynamics was given by : (a) Obert (b) Keenan (c) Joule (d) Newton 39. 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 40. 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 41. 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) + 10kcal (b) -10 kcal (c) 0 (d) +5kcal 42. 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? y-1 y-1 (a) (:~F (b) (~~F y-1 y (c) (~~)2y (d) (~~) y-1 32. Which of the following in an irreversible process? (a) Isothermal process (b) Isentrobic process (c) Isoparic process (d) Isenthalpic process 33. In a reversible adiabatic process, the ratio (T 1 : T2) will be equal to: y -1 ) (b) -mR(T1-T2 2 31. 30. 29. 28. 27. 26. 25. 24. 23. 22. 21. ~P ~v (c)v- (d) v.- P 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) T3/2 (d) TI/2 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 Anideal 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 : 20. 1 ~P (b) v P 1 P v ~P (a) Thermal EngineergingA-lOS Badboys2Badboys2 Badboys2
- 112. (b) (:) t (d) (: 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) (! (c) (:)p 1 1 (a) 11= 1- (r)y-I (b) 11= 1- (r)y+1 1 1 (c) 11= 1+ (r)y-I (d) 11=1--- y-I (r)- Y 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 : --~"'S constant volume 2 --~S constant volume constant volume T (d) 1 "- 2"" T 1 1[' v (a) 4 ./ 3"' S 51. In thermodynamic cycles, the otto cycle is represented by which of the following T - S diagram? (d) <l>dQ~ 0 (b) <I> dQ < 0 T (a) <I> dQ = 0 T (c) <I> dQ > 0 T 44. 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, [ Y-I]_Y_P1V1 (P2)Y -1 P2 (a) (b) mRTI/n- Y-1 PI PI T (c) MCp (T2- Tl) (d) mR(l- ~~) (c) 1 T (b) 1 (a) dQ=dU+qdV (b) Tds=dU+PdV (c) dQ=Tds (d) dQ=Tds+dU 43. 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-t09Thermal Engineerging Badboys2Badboys2 Badboys2
- 113. 68. The critical point forwater is : (a) 374°C (b) 373°C (c) 273°C (d) 323°C 69. Theamountofheatrequiredtoraisethe temperaturevolume is known as: (a) specific heat at constant pressure (b) specific heat at constant volume (c) kilo -joule (d) None of these 70. Critical pressure is the pressure of steam at (a) exitofsteamnozzle (b) either at inlet or at outlet of steam nozzle (c) inlet ofsteam nozzle (d) throat of steam nozzle 71. As the pressure increases, the saturation temperature of vapour: (a) increases (b) decreases (c) increases first then decreases (d) decreases first then increases 72. 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 73. Expression for the specific entropy ofwet steam is: L (a) Sg+Xsf (b) hf+XT L ~ ~+X~ ~ ~+XT 74. The latent heat ofvapourization ofa fluid at lookK is 2560 KJ/kg. Then change of entropy associated with the evaporation will be equal to: (a) 6.86KJ/kgOK (b) 9.86KJ/kgOK (c) 25.6KJ/kgOK (d) -25.6 x 103KJ/kgOK 75. 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 76. Critical pressure for steam is equal to : W nObM ~ nl~ (c) 163bar (d) 184bar 77. Throttling calorimeter is used forthe 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) Specificvolume 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 (b) H +wH (d) H:w-H: 67. 66. 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 Sensible heat is the needed to (a) vaporise water into steam (b) change the temperature ofa liquid or vapour (c) convert water into steam and super heat it (d) measure dew point temperature If H = enthalpy of dry air, H,= enthalpy of water vapour w= ~pecifichumidity,then the enthalpy ofmoist air will be equal to: (a) Ha+n, (c) Ha-wHy 65. Mass of water vapour in suspension Mass of water vapour in suspension + Mass of dry steam (d) Mass of dry steam + Mass of water vapour Mass of dry steam (c) Massof watervapourinsuspension (b) Mass of water vapour in suspension Mass of dry steam (a) 64. 63. 62. 61. 60. 59. 58. 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 losephase 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 : Massof dry steam 57. Thermal Engineergingx.uo Badboys2Badboys2 Badboys2
- 114. (c) Feed check value (d) Fusible plug 92. Steam in boiler drum is always (a) wet (b) dry (c) superheated (d) Both (a) and (b) 93. 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 94. In a Lancashire boiler, the economiser is located: (a) beforeair pre-heater (b) After air preheater (c) Between feed pump and boiler (d) Not used 95. Locomotive boiler produces steam at : (a) Medium rate (b) Lowrate (c) Veryhigh rate (d) Verylow rate 96. 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 97. Ifcirculation ofwater in a boiler ismade bypump, then it is known as: (a) Forced circulation boiler (b) Natural circulation boiler (c) Internally firedboiler (d) Externallyfiredboiler 98. 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 99. Maximum heat is lost in boiler due to: (a) Unburnt carbon (b) Flue gases (c) Incomplete combustion (d) Moisture in fuel 100. What salts of calcium and magnisium cause tempeorary hardness of boiler feed water? (a) Sulphides (b) Carbides (c) Nitrates (d) Bi - carbonates 101. 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 102. In parson's steam turbine, steam expands in: (a) partly in nozzle and blade (b) blades only (c) nozzles only (d) None of these 103. 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 104. Blade efficiency of steam turbine is equal to: (a) V (Vw,- Vw2) / 2g (b) 2V (Vw,- Vw2) / V/ (c) equal enthalpy of saturated steam (d) None of these 80. The enthalpy of vapour at lower pressure depends (a) Temperature (b) Volume (c) Temperature and volume (d) Neither temperature nor volume 81. For high boiler efficiency,the feed water is heated by: (a) regenerator (b) convective heater (c) super heater (d) economiser 82. 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 83. Lancashire boiler is a : (a) water tube boiler (b) fire tube boiler (c) locomotiveboiler (d) high pressure boiler 84. 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 85. 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 ofwater convertedinto steam from 1000 C to 1100 C in one hour. 86. 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 87. Boiler rating is generally defined in terms of: (a) maximum temperatureofsteam (b) heat transfer rate KJ/hr (c) heating surface (d) heating output kg/hr 88. Boiler mountings are necessary for: (a) operation and safety of a boiler (b) increasing the efficiency ofboiler (c) Both (a) and (b) (d) None of these 89. Bab cockand wilcox boiler is (a) water tube type (b) fire tube type (c) Both (a) and (b) (d) None of these 90. Range of high pressure boilers are: (a) blow 80 bar (b) above 80 bar (c) 40 - 80bar (d) 60- 80bar 91. 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 A-11IThermal Engineerging Badboys2Badboys2 Badboys2
- 115. (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. Ifthe compressor ratio is increased, then the volumetric efficiency of a compressor : 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 tity 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. Anaxial 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 11m (c) 11BT =~ IT 118. n (c) P2 = (_2_)n+1 PI n-i-I Multi-stage turbines are: (a) Reaction type (b) Pressure compounded (c) Velocity compounded (d) All of these If11BT = Brake thermal efficiency 111T= Indicated thermal efficiency 11m= Mechanical efficiency then, which ofthe following relation is correct? 117. n (b) P2 = (_2_)n-I PI n-l n (a) ~ =C!Jn-1 (c) V(Vw,+Vwz)/g (d) 2Vrvw,+ Vw2)/V,2 105. Steam turbines are governed by the following methods: (a) Throttle governing (b) Nozzle control governing (c) By - Pass governing (d) All of these 106. For Parson's reaction turbine, degree of reaction is: (a) 50% (b) 100% (c) 75% (d) 25% 107. 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 108. 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 109. 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 110. Stage efficiency of steam turbine is equal to : Blade efficiency Nozzle efficiency (a) Nozzle efficiency (b) 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: cosa sin a (a) -- (b) 2 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: Thermal Engineergingx.nz Badboys2Badboys2 Badboys2
- 116. (a) 7.75 (b) 8.75 (c) 9.75 (d) 10.75 139. In a refrigeration system,the refrigerant gain heat: (a) Compressor (b) Condenser (c) Expansion value (d) Evaporator 140. The Bell- columan refrigeration cycleuses: (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 141. 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= 375Kl/kg. Thenthe COPofrefrigerationis: (a) 5 (b) 10 (c) 6 (d) 9 142. A camot refrigeration cycle has a COP of 4. The ratio of higher temperature to lowertemperature will be equal to : (a) 2.5 (b) 2 (c) 2.8 (d) 1.25 143. The refrigeration systemworks on: (a) First Law ofthermodynamics (b) Second Law ofthermodynamics (c) Zeroth Law of therodynomics (d) None of these 144. One ton ofrefrigeration is equal to: (a) 210KJ/min (b) 21KJ/min (c) 420KJ/min (d) 20KJ/min 145. If a heat pump cycle operates between the condensar temperatureof+27°Cand evaporatortemperatureof-23°C, then the COP of camot will be equal to : (a) 6 (b) 12 (c) 5 (d) 1.2 146. Which of the followinghas minimum freezingpoint? (a) Freon-12 (b) Freon-22 (c) Ammonia (d) Carbon di-oxides 147. 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( Py J is associated with: Pt -Py (a) Relativehumidity (b) Specifichumidity (c) Degree of saturation (d) Partial pressure [where, Py= Vapourpressure, P, = total pressure = Pa +PyJ 149. In the absorption refrigeration cycle, the compressor of vapour compression refrigeration cycle is replace by : (a) Liquidpump (b) Generator (c) absorber and generator (d) Absorber, Liquid pump and generator 150. When waterLithiumbromideisused in a vapourabsorption refrigeration system, then: (a) they together act as refrigerant (b) water is the refrigerant (c) Lithiumbromide isrefrigerant (d) None of these (c) Compression ratio (d) Index of compressor performance 129. 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 130. 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) 6bar (b) 12bar (c) 18bar (d) 24 bar 131. An engine operators betweentemperature 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 132. A heat engine develops60 kw ofwork having an efficiency of 60%, then the amount of heat rejected will be equal to: (a) 400kw (b) 40kw (c) 20kw (d) 200kw 133. In carnot cycle,addition and rejectionofheattakesplaceat : (a) constant pressure (b) constant temperature (c) constant volume (d) constant speed 134. A heat engine receives 1120KJ ofheat 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 135. 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 136. A heat pump working an reversed camot cyclehas COP of 5. Ifit work as refrigerator taking 1kw of work input the refrigerating effectwill be : (a) 1kw (b) 2kw (c) 3kw (d) 4kw 137. Tworeversiblerefrigerator's arearranged in seriesand their COP are 4 and 5respectively. Then the COP of composite refrigeration systemwill be : (a) 1.5 (b) 2. (c) 4 (d) 3.7 138. An ideal refrigerator is operating between a condenser temperature of -37°C and an evaporator temperature of -3 ° C. Ifthe machine isfunctioningas a heatpump, its COP will be equal to: displacement (b) Airactually delivered Amount of piston Stroke volume (a) Clearance Volume (a) decreases (b) increases (c) constant (d) None of these 128. Volumetricefficiencyof air compressor is defined as : A-l13Thermal Engineerging Badboys2Badboys2 Badboys2
- 117. 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 ofgoverning isemployedis: (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 flowrate (d) Fueltemperature 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 fromthe compressedair that is supplied for the combustion (d) airfuelmixing (a) Air only (b) Liquid fuel only (c) Both (a) and (b) (d) Solidfueland 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 vaporisethe fueland 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) Smokepoint (d) Ignition temperature 169. In four cylinder petrol engine, the standard firing order is: W 1-3-3-4 (b) 1-4-3-3 ~ 1-3-3-4 ~) 1-3-4-3 170. Which ofthe followingis not a part ofMPFI petrol engine? (a) Fuel injector (b) Carburetor (c) MAP sensor (d) Electronic control unit 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 byvolume 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 fuelinjection system ofan IC engine means injection of: (b) 3: 1 (d) 1: 2 IS given as : (a) 1: 1 (c) 2: 1 156. 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) combustionoffueltakes placewithout sparkprovided 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. Forthe samecompressionratioandheat 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 stroketo the crank radius in an LC. engine (d) 1- Vc Vs 1+ Vc (c) Vs 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 ofan IC engine is given as : ifV c = cylindervolume, Vs= sweptvolume 1+ Vs 1- Vs (a) Vc (b) Vc Thermal EngineergingA-114 Badboys2Badboys2 Badboys2
- 118. (d) Length of base of indicator diagram 189. Most important properly of an IC engine lubricant is: (a) density (b) viscosity (c) thermal conductivity (d) none of these 190. The process of scavenging is associated with: (a) two stroke engine (b) four stroke engine (c) gas turbine (d) compressor 191. The function oflubrication in engine are: (a) Lubrication and cooling (b) cleaning, sealing and noise reduction (c) efficiency increase (d) Both (a) and (b) 192. 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) calorificvalue (b) pour point (c) specific gravity (d) carbon content 193. Control ofmaximum oil pressure in the lubrication system is attected by : (a) oilfilter (b) Pressure switch (c) Pressure relief value (d) Pump motor 194. What techniqueis adoptedforthe lubricationofthe cylinder of a scooter engine? (a) splash lubrication (b) Forced feed lubrication (c) Gravity feedlubrication (d) Petrol lubrication 195. Using lubricants on engine parts reduces: (a) motion (b) force (c) Acceleration (d) Friction 196. 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 197. 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) Boiliningpoint temperatureofthe coolantis decreased by its use 198. Radiator tubes are generally made up of: (a) Brass (b) Steel (c) Cast iron (d) None of these 199. Which of the following lubrication system is used in a car engine generally? Area of indicator diagram Area of indicator diagram (c) (b) Length of base of indicator diagram x spring scale Length of base of indicator diagram x spring scale (c) 70em3 (d) 84em3 187. Unit ofbrake specific fuel consumption is: (a) kg-hr-kw (b) kglkw-hr (c) kw - hr / kg (d) kg - hr / kw 188. Indicated mean effectivepressure is : Area of indicator diagram x spring scale (a) Length of base of indicator diagram Area of indicator diagram 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 Pm·LAN (d) 2 Pm·LAN 4 (c) Pm·LA (b) N(a) Pm.LAN (a) Efficiency (b) Specific fuel consumption (c) Air fuel ratio (d) Total fuel consumption 178. Value overlapping happens: (a) completelybeforeT.D.C (b) completelyafter T.D.C (c) partially beforeT.D.Cand partially afterT.D.C (d) completelybeforeB.D.C 179. Morse test is used for multicylinder spark ignition engine to determine: (a) Thermal efficiency (b) Mechanical efficiency (c) Volumetricefficiency (d) Relative efficiency 180. 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 181. Flash point for diesel fuel should be: (a) Minimum49°C (b) Maximum49°C (c) Minimum99°C (d) Maximum99°C 182. Theratio ofbrakepowerto indicatedpowerofan I.C. engine is called: (a) Thermal efficiency (b) Mechanical efficiency (c) Volumetricefficiency (d) Relative efficiency 183. The primary winding of ignition coil consist of: (a) fewturns of thin wire (b) many turns of thin wire (c) fewturns of thick wire (d) many turns of thick wire 184. How many power stroke/second will take place in a four strokepetrol engine rotating at 3000 r.p.m? (a) 25 (b) 50 (c) 100 (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: --+------~xo Power Y 177. The curve shown in the given figure is characteristic diesel engines: Then 'Y' axis shows: A-115Thermal Engineerging Badboys2Badboys2 Badboys2
- 119. VNcosa (d) 2 VNsina (b) 2(a) VN sin a 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 ofthe systemwill be: 1 (a) - (b) 2 4 (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 cyclebetweenthe same temperature limits. Which of the following is correct? (a) COP of refrigerator = COP of heat pump (b) COP ofrefrigerator = COP ofheat pump + 1 (c) COP ofrefrigerator = COP of heat pump - 1 (d) COP ofrefrigerator = Inverse of COP ofheat pump 210. A solar energy based heat engine which receives 80 KJ of heatat 100°Candrejects70KJ ofheattothe ambientat 30°C is tobe designed.Then the thermal efficiencyofheat engine willbe: (a) 12.5% (b) 24.5% (c) 40% (d) 70% 211. For an ideal gas, the expression [T(:;) p - T(:;) v ] is always equal to : (a) zero (b) R Cp (c) (d) Rf Cv 212. During a phase change ofa pure substance: (a) dG=O (b) dP= 0 (c) dH = 0 (d) dU= 0 213. At the triple point ofpure substance,the number ofdegrees offreedom is : (a) 2 (b) 1 (c) 0 (d) 4 214. A vessel ofvolume 1m' contains a mixture ofliquid water and steam in equilibrium at 1.0bar. Given that 90% ofthe volumeis occupiedbythe steam,then the fractionofmixture will be equalto: Ifat 1bar,Vf= 0.001mvkg, V = 1.7mvkg (a) 5.27 x 10-3 (b) 7.27 x 10-4 g (c) 8.29X 10-3 (d) 9.23 x 10-3 215. 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°, Bladeexit angle = 25°, Itblades friction is neglected, the work done/kg steams is : (a) 124KJ (b) 164KJ (c) 174KJ (d) 184KJ 216. if VN and a are the nozzle exit velocity and angle in an impulseturbine, then the optimum blade efficiencyis given by: (c) fTds 204. The first Law ofthermodynamics takes the form w = -Ll D when applied to: (a) A closed system undergoing a reversible adiabatic process (b) An open systemundergoing an adiabatic process with negligiblechangesinkineticenergyandpotentialenergy (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 of60° C. Itisdropped 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 (a) fpdv (b) f vdP ABC D (a) 4 5 2 1 (b) 3 4 1 2 (c) 1 2 4 6 (d) 4 5 3 3. S.1.Engine 4. C.IEngine 5. Cooling towers 6. Heat exchangers (D) dh = CpdT, even when pressure, varies 1. 2. (A) (B) List-II Ideal gas vander walls gas cetane number Approach and range (C) (OT) :;t 0 oP h 201. In pressure lubrication systemofan engine, the maximum oil pressure is controlled by : (a) oilpump (b) oil pressure gauge (c) oil pressure reliefvalue (d) oilfilter 202. A body of weight 100 N falls freely a vertical distance of 50m. The atmospheric drag forceis O.5N.For the body,the work interaction is : (a) -25J (b) +25J (c) + 500J (d) - 500J 203. Match List - I and List - II and answer according to the codes given below: List- I (a) Petrol (b) Splash (c) Pressure (d) Dry sump 200. 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 (d) fsdT Thermal EngineergingA-116 Badboys2Badboys2 Badboys2
- 120. r Group I Group II Group In Group IV Group V I When Differential Function Phenomenon added to the system, is EHeat G Positive I Exact KPath M Transient F Work H Negative J Inexact L Point N Boundary P T • •Fig. 1 V Fig. 2 S P T • •Fig. 3 V Fig. 4 S According to the first law of thermodynamics, equal areas are enclosed by (a) Figures 1and 2 (b) Figures 1and 3 (c) Figures 1and 4 (b) Figures 2 and 3 229. Match items from groupsI, II, III, IV and V. 228. The following four figures have been drawn to represent a fictitious thermodynamic cycle,on the p-Vand T-S planes. (a) will be slightly less than 5 bar (b) will be slightlymore than 5bar (c) will be exactly5bar (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 reciprocatingcompressor.Which ofthe followingrepresents that diagram? (a) p (b) P Pout =2 Pout I~Pin Pin Vc V V (C) P (d) P Pout -n Pout I~Pin Pin I I I I Vc V Vc V The final pressure :---~ I I I I I I I I V(mj 0.01 0.03 (a) 1.5kPa (b) 3kPa (c) 4.5kPa (d) 6kPa 219. Inordertoburn 1kg ofCH4 completely,theminimumnumber ofkg of oxygenneeded is (take atomic weights ofH, C and oas 1, 12and 16respectively) (a) 3 (b) 4 (c) 5 (d) 6 220. 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 221. Atmosphericair from40° 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 222. The use of Refrigerant R- 22 fortemperature below-30°C is not recommended due to its: (a) goodmiscibility with lubricating oil (b) Poormiscibilitywith lubricating oil (c) lowevaporating temperature (d) None of these 223. IfAir-Fuelratio ofthemixture in petrol engine is morethan 15: 1,then, (a) NO is reduced (b) CO2 is reduced (c) HCxisreduced (d) CO is reduced 224. 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 isothermallyto a final volume of2 m3. The p-V-T relation is (p+:2Jv= RT, wherea>O. 217. For a single stageimpulseturbine withrotordiameterof2 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 figurebelow shows a thermodynamic cycleundergone by a certain system.Then the mean effectivepressure in NI m'; A-117Thermal Engineerging Badboys2Badboys2 Badboys2
- 121. (b) 0.36 (d) 0.01 234. 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 2 (a) f pdV (b) f Vdp 1 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 W &TI ~ ~OO (c) 554.67 (d) 8320.00 236. 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 237. Consider the following two processes: I. A heat source at 1200K loses2500 kJ ofheat to sink at 800 K. Il. A heat source at 800 K loses2000 kJ ofheat 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 238. 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 239. 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 240. 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) 0.1 241. The values of enthalpy of steam at the inlet and outlet of a steam turbine ina 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 (c) 0.06 Thermal Engineerging . . b W fOutlet dVincorrect; It must e = Inlet p 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 mass flow rate is given by W = _fOlutlet Vdp ,where V is the In et 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) 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 3 400 kPa y pV = constant (2007, 2m) 100 kPa 2 3 VI1m V (a) T (b) T 3 312~2 S S (c) T (d) T 3 2//2£J 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 p A-11S (a) 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 Badboys2Badboys2 Badboys2
- 122. 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 in0.00028 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. 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 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. (c) (d) zero (a) 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 TI 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-119Thermal Engineerging Badboys2Badboys2 Badboys2
- 123. y-I T2=(P2JYTI PI then the system consists of 270. An ideal gas expands isothermally from volume vIto v2 and then compressed to original volume vI adiabatically initial pressure is PI and final pressure is P3. The total work done by gas is w, then (a) P3>P1,w>0 (b) P3<P1,w<0 (c) P3 > PI' w < 0 (d) P3 = PI' w = 0 271. If during a process, the temperature and pressure of system are related by E (d) p=- 3 2 p=-E 3 (c) Cp (a) zero (b) Cv (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 1 3E (a) p=-E (b) p=- 3 2 [T(;)P-T(;)vN!;)p -T(;l.s always equal to ( a ) (c) lP+ y2) (Y - b)= RT ( a ) (d) lP- y2) (Y - 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 TL'°K 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 (b) (p+ ;2 }V+b)=RT OOK TH = 300 K ~ ~ • 0 0 U U (a) (b) TL,oK TL,oK TH = 300 K TH=300K ~ ~ J0 0 U U (c) (d) 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, pv»= 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 TL for different values of COP for a constant value ofT H = 300 K (say)? (c) y y (Truax tY+1) (b) (Tmin ) 2(Y-l) Tmm lTmaJ y-l y-l (Truax y (d) (Tmm yTmln Tmax (a) 259. The compression ratio of a gas power plant cycle corresponding to maximum work output for the given temperature limits ofT minand Tmaxwill be Thermal EngineergingA-120 Badboys2Badboys2 Badboys2
- 124. (d) 1 (y-l)/y rp 1 1-- [I/y P (c) 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 powerplant P. always increases the thermal efficiencyofthe 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) Intakeand exhaustprocessesare constantvolumeheat rejection processes (c) The combustion process is a constant volume heat addition process (d) The workingfluidis an idealgas with constantspecific 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. Specificgas constant forair = 0.287kJ/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_ 1-_!_ (a) [y-I (b) rY p p Temperature (0C) - 15 -10 -5 0.01 5 10 15 20 Saturation pressure (kPa) 0.1 0.26 0.4 0.61 0.87 1.23 1.71 2.34 Specificenthalpy ofwater 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 ofwater in field is formed after a farmer has a wateredit. The ambientair conditionsare:temperature20°C and relative humidity 5%. An extract of steam tables is given below. Specific volume (m3/kg) Enthalpy (kJ/kg) Temperature Psat (bar) Saturated Saturated Saturated Saturated(0C) liquid vapour liquid vapour 45 0.09593 0.001010 15.26 188.45 2394.8 342.24 150 0.001658 0.001658 1610.5 2610.5 (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. Givenbelowis an extract from steam tables. 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) P-S-5,R-U-3,P-S-l, Q-T-2 (b) P-S-l, R-U-3,P-S-4,Q-T-2 276. 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 GroupI, II and III. (a) T-~ (b) T _ 27R.b 1- 27R.b 1- 8a (c) T _ 2R.b (d) T = 2aI- I R.b8a 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 (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 enterthe turbine at20 bar, 1500K exits at 1bar. The isoentropicefficiencyof 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 pI and P2 are as shown in the figure. It can be concluded that A-121Thermal Engineerging Badboys2Badboys2 Badboys2
- 125. td3 - tdl (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 (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 1kW (b) 1 and2kW (c) 2andlkW (d) 2and2kW 294. The bypass factor, in case of sensible cooling of air, is given by tdl - td3 td2 - td3 (a) td2 -dd3 (b) tdl -dd3 (c) T T i (d) 1 (b)(a) 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 (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 (b) more (d) none of these 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 (c) equal 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 ~8Q(a) ~8Q > 0 and T<O ~8Q(b) ~8Q <0 and T<O ~8Q(c) ~8Q>0 and T>O ~8Q(d) ~8Q<0 and T>O 283. Which one of the following pairs of equations describes an irreversible heat engine? Thermal EngineergingA-122 Badboys2Badboys2 Badboys2
- 126. (b) 5kW (d) None of these 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 (c) 6kW 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 con- vective 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. TI and T3,where T3is themaximum temperatureand Tl is themini- mum temperature? 307. Forthe samemaximum pressure and temperature (a) Otto cycleis more efficient than diesel cycle. (b) Diesel cycleis more efficientthan Otto cycle. (c) Dual cycleis more efficientthan Otto and dieselcycles (d) Dual cycleis lessefficientthan Otto and Dieselcycles. 308. Ifcompression ratio of an engine working on otto cycle is increasedfrom5to6,itsairstandardefficiencywillincreaseby (a) 1% (b) 20% (c) 16.67% (d) 8% 309. An air standard diesel cycleat 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 offratio (c) thermal efficiencyremains the samewethe increasein 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 strokeare 1.3bar and 2.6 bar respectively. Assuming that compression follows the lawPV1.3 = constant,whatwillbe the air standardefficiency of cycle (a) 36% (b) 42% (c) 46% (d) 48% 311. The stroke and bore ofa four stroke spark ingition engine are250mm and200mm respectively.Theclearancevolume is0.001m3. Ifthe specificheat ratioy = 1.4,the air-standard cycle efficiencyofthe engine is (a) 46.40% (b) 56.10% (c) 58.20% (d) 62.80% 312. An engine working on otto cyclehaving compression ratio of5. The maximum andminimum pressureduring the cycle are 40 bar and 1 bar respectively. The mean effective pressure of cyclewill be (a) 7bar (b) 7.89bar (c) 9.04bar (d) 11.79bar 304. The diesel engine and otto engine has same compression ratio. The cut offratio ofdieselengineis S.The air standard efficiencyof these cycleswill 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 (c) (d) (Y-I)(r-l) (~).llth (y-l)(r-l) (b) (~).llth (Y-l)(r-l) (~).llth (Y-l)r (a) (c) diesel fuel is balance with petrol (d) none of these 297. A engine isworkingonair standarddieselcycle.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 efficiencyofengine will be (a) 53.25% (b) 60.5% (c) 64.89% (d) 67.75% 298. Number of processes in a Rankine cycles are (a) 3 (b) 4 (c) 5 (d) 6 299. The comfort condition in air conditioning are at (a) OODBTandO%RH. (b) 20°CDBTand60%RH (c) 30°CDBTand 80%RH. (d) 40°CDBTand90%RH. 300. 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 301. The air standard diesel cycleis less efficient than the Otto cycle for the (a) same compression ratio and heat addition (b) same pressure and heat addition (c) samerpm and cylinderdimensions (d) same pressure and compression ratio 302. An otto cycletakes in air at 300k. The ratio ofmaximum to minimum temperature is 6 for maximum work output the optimum pressureratio willbe (a) 7.48 (b) 8.37 (c) 8.93 (d) 9.39 303. The mean effective pressure of an Otto cycle can be expressedas where (~P =Pressureriseduring heat addition) A-123Thermal Engineerging Badboys2Badboys2 Badboys2
- 127. 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 are250 mm and 200 mmrespectively.The clearancevolume is 0.001 m'. Ifthe specificheat ratio y = 1.4,the air-standard cycle efficiencyof the engine is (a) 46.40% (b) 56.10% (c) 58.20% (d) 62.80% 325. Ifa mass of moist air in an airtight vessel is heated to a higher temperature, then (a) specifichumidity of air increases (b) specifichumidity of air decreases (c) relative humidity of air increases (d) relative humidity of air decreases (b) 15.0kW (d) 37.5kW 321. The statements concern psychrometric chart. 1. Constantrelativehumiditylinesareuphill 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 speedis 10m/s,the waterjet velocityis 25 mls and volumetricflowrate ofthejet is 0.1 m3/s. Ifthejet defletion angle is 120°and the flowis ideal, the power developed is (a) 7.5kW (c) 22.5kW (a) 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 ofmoisture in kg per kg ofdry air will be approximately? (a) 0.002 (b) 0.027 (c) 0.25 (d) 0.75 0 ('I') ('I') ('I') I Thermal Engineerging o, C) Process in figure Name of the process P. 0-1 i Chemical dehumidification Q. 0-2 ii. Sensibleheating R 0-3 ill. Coolingand dehumidification S. 0-4 IV. Humidificationwithsteam injection T. 0-5 v. Humidificationwithwater injection w (kg/kg) 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 followingadvantages? 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. D. Kaplan turbine C. Propeller turbine fixed runners vanes 2. Specificspeed from 10 to 50 +tangential flow 3. Specificspeed from 60 to 300+mixedflow 4. Specificspeedfrom300 to 1000+axialflowwith adjustablerunner vanes B. Prancis turbine 1. List II Specificspeedfrom300 to 1000+axialflowwith Pelton turbineA. Isentropic Isenthalpic Isobaric Isothermal 1. 2. 3. 4. A. Compression B. Heat rejection C. Expansion D. Heat absorption 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 316. Match list I (processer with) list II (Type)for Bell coleman or Joule or Reverse Brayton cycle for gas cycle refrigera- tion and select the correct answer using the codes given below the lists. List! ListII A-124 Y y-l (a) fp=(~t (b) fp=(~r y 2(y-l) (c) fp=(~ tY-1) (d) fp=(~ J-y- Badboys2Badboys2 Badboys2
- 128. 5. (b) Considring the followingdiagram, 12. (a) As we know that, C F-32 l' 1 - = -- ~ ~2 100 180 C Q) C= 100(F_32)=~(F_32)S!ZJ 180 9!ZJ Q) 1-0 P-. 5 Volume(V) ~ C =-(F -32) 9 hence, (QA -wA) = (QB -wB) 13. (b) Let, C = specific heat at constant pressurep ...,HINTS & EXPLANATIONS 1 (c) 26 (b) 51 (c) 76 (b) 101 (b) 126 (b) 151 (d) 176 (c) 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) 14 (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) 16 (c) 41 (c) 66 (b) 91 (d) 116 (a) 141 (a) 166 (a) 191 (d) 216 (d) 241 (a) 266 (b,d) 291 (a) 316 (b) 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 (a) 325 (d) • •• , ANSWER KEYI···.. A-125Thermal Engineerging Badboys2Badboys2 Badboys2
- 129. swept volume (v) = ~ D2L = ~ x (0.2)2 x 0.25 =0.00785m3 Now using the following relation, Volume of air induced Volumetric efficiency (11vot)= 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 -1- T2 112- T2 - 400 - 400 = 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 ( Tf-Ti) thermal efficiency (11thermal ) = ~ x 100 dQv change of entropy (ds) = 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 0.0344= 150 QA Q =~=4360KJ A 0.0344 74. (c) Given Latent heat ofvapourization (dQ) = 250 Temperature (T) = 100 K 11 11= - = 0.0344 320 We also know that, . () work Produced (w p) efficiency 11 = ---------,-----'--,-:::--:- Heat added (QA) T2=47°C=47+ 273 =320K work produced (wp) = 150 KJ T2- T1 (320- 309) Now, efficiency (11)=T= 320 Q2 = 42 =~ T2 200 100 Q1 * Q2 Here, T; ~ So, Engine is not possible. 35. (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 41. (c) Given: Initial volume (v) = 4m3 Final volume (vf) = 2 m' Pressure (P) = 4.2 kg/em- work done (w) = Pdvp =4.2 x 104 x (4-2) =4.2x2xl04 = 8.4 X 104joule As the system undergoes a reversible process, then, Heat (Q)H= 20 x 4.2 X 104 = 8.4 x 104joules change in internal energy = wD - QH =8.4x 104-8.4 x 104=0 55. (c) Given, TI = 36° C = 36 + 273 =309K C, = specific heat at-constant volume then, Cp - C, = R (gas constant) where, R = 287 Jzkg" K 19. (b) As we knonw that, during an adiabatic process, PvY = constant PIVIY=P2 V/=C Here, PI =P,P2=M>, V1=Y, V2=tN Hence tN = _!_ ( dP) 'V y P 28. (d) Let v = mean square molecular velocity p = density, 1 VI ~2thenvoc-~-= - .JP V2 PI 30. (c) Given: T = 0 Now considering kinetic energy equation of gases, (kinetic 3 energy) K.E = "2KT Here, T =0 Hence, K.E = 0 34. (b) Given: Q1 + 105 MJ, Q2 = 42 MJ T1=400K, T2=200K Q1 105 21 Now, T;= 400 = 80 Thermal EngineergingA-126 Badboys2Badboys2 Badboys2
- 130. 375 -125 250 425-375 50 (COP). =5Refrigerator 142. (d) Given: coefficient of performance (COP) = 4 Let T, = Higher temperature T2= Lower temperature () Hl-H4 COP Refrigerator= H H 2 - 1 Throttle (COp) Refrigerator= (COp) Refrigerator+ 1 =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, 310- 270 270 = 6.75 40 270 (COp) Refrigerator= TE Tc-TE 136. (d) Given: coefficient of performance (COP)p = 5 work input (W,/p)= 1kw (COP)p=~=Qa=5 WI/P 1 Q =5kw Now, (COP)p-(COP)R =w= 1 5 -(COP)R = 1 (COP)R= 5-1 =4 (COP) = 4 = Qar = Qar Now, R W 1 liP Q =4x 1=4kw 137. (b) COP of refrigerator' l' = (COP), = 4 COP of refrigerator '2' = (COP)2 = 5 Now, COP of composite refrigeration system (COP)c; (COp) = (COP)l x (coph = 4x5 c 1+ (COP)l + (coph 1+4+5 20 (COP)c =-= 2 10 (COP)c = 2 138. (a) Given: condensortemperature (T)= 37°C =37+273=31OK Evaporator temperature (TE) = - 3°C =-3 +273 =270K B QA-B = f T.ds = T.(SB - SA) A dQ = T.ds Hence, Area under T - S diagram shows heat transfer for reversible process. Now, we know that, LQ = fT.ds T2 280 Ql *- Q2 Hence, -T T1 2 Hence, engine operates on Irreversible cycle 135. (a) Considering the following Temperature (T) - Entropy(s) diagram: 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, Ql = 1120 = 14 = ~ Tl 560 7 Q2 840 3 -=-=- w 11=- QG 0.6 = 60 QG Now, 11,= 112 1- 900 =1- T2 T2 400 T22=360000 T2 = '-'360000 = 600K 132. (b) Given: work developed / produced (W) = 60 kw Efficiency (11)= 60% or 0.6 considering the following formula, Effi . () work Produced (w ) rciency 11 = ( ) Amount of Heat Generated QG A-127Thermal Engineerging Badboys2Badboys2 Badboys2
- 131. = 80-70 xlOO 80 = !Qxl00 80 =12.5% 214. (a) Given volume of vessel (v) = 1m' steam volume (Vs) = 0.9 m' water volume (Vw) = 0.1 m' At 1bar, Vf= 0.001 mvkg V = 1.7m3/kgg = 120 = 4 30 210. (a) Given: Heat received (Q) = 80 KJ Temperature (TJ) = 100°C Heat rejected (Q2) = 70 KT Temperature (T2)= 30° C ( ) Ql -Q2 100 Thermal efficiency 11th. = Ql x 307.72 7Vc = 307.72 => Vc = -7- V =43.96cm3=44cm3 202. (a) Weight of body = 100N,dragforce(Fo)=0.5N 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) = 1kg 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) =O.4(T -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 4.58 207. (c) () Heat rejected COP system = Power Consumed =~x(7)2x8 4 =307.72cm3 considering the following relation, Vs -v,C.R = ____::__-=- Vc Where, Vc= clearanace volume 7t 2 Swept volume (V) = "4D L 11thermal = 0.5 or 50% 184. (a) Given: speed (N)= 3000 rpm 3000 3000 Power stroke / s = __ = _- = 25 2 x 60 120 186. (a) Given: Bore diameter (D)= 7 em Stroke length (L)= 8 em compression ratio (C.R) = 8 1 1 11thermal = 1-(4)0.5 = 1-(4)112 = 1__ 1_= I-.!.= 1-0.5 .J4 2 = 1 1 (4)1.5-1 I thermal efficiency of cycle (11thermal ) = 1- ( )r-l C.R 400 Comression ratio (C.R) = 100 = 4 ( ) _ ____:!i_ _ 300 = 300 = 6 COP cornot - Tl - T2 - 300 - 250 50 166. (a) Given: Total volume = 400 cc compressed volume = 100 cc r = 1.5 ..!L= ~= 1.25 T2 4 145. (a) Given: TJ = 27° C TJ =27+273=300k T2=-23°C T =-23+273=250k2 COP = ____.:!l__ T1-T2 _1_ = Tl - T2 =..!L-l COP T2 T2 Thermal EngineergingA-12S Badboys2Badboys2 Badboys2
- 132. 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 2 KN/m A 5 1'---""," 2x314 2x314 v ------ opt. - cos 20° - 0.9396 = 668.4 mls ::::::668 mls 218. (a) Given ( ) 2ub Now, optimum velocity of steam vopt. =-- cos o; 1 a a a => P2 =10x2+ 1x2 - 22 = 5+4 Asa>O, p> 5 227. (d) 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. 229. (d) 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. 230. (b) Heat generated by bulb = 100 x 24 x 60 x 60 J = 8.64 x 106J . . 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 131. (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 1and 2, clockwise (1-2-3) will be selected. 232. (c) 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 3 .____.___-........__--+ V (m ) 0.01 0.03 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° ndN n x 2 x 3000 Blade velocity (Ub) = 60 = 60 =314m/s =n+4n =5n 226. (b) T = constant ( a '1 ( a '1 Thus, lp, + vfjVt =lp2 + vljV2 =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, VW1 IE >I< VW2 )1 ~~ 0.5294+100 Now, Dryness fraction = + ms mw 0.03 0.02 3 2 = 1.5 KN/m2 = 1.5 kPa 220. (b) Given: Bore (d) = 2 unit, stroke (L) = 2 unit Total area for heat loss = ~d2 + ndl. = ~(2)2 + n (2)(2) 4 4 0.5294 ( ) v: 0.1 water mass mw = - = -- = 100 kg Vf 0.001 =0.03KN-m . () work done Mean effectIve pressure MEP = ---- volume ( ) Vs 0.9 steam mass ms = - = - = 0.5294 kg Vg 1.7 A-129Thermal Engineerging Badboys2Badboys2 Badboys2
- 133. HI =2800kJ/kg H2= 1800kJ/kg Work done = HI - H2= (2800-1800) kJ/kg = 1000kJ/kg Then, specific steam consumptionm 3600 = 1000 =3.60kg/kW-h 242.(d) d=80mm Strokelength L= 2 x Crank radian =2 x60= 120mm Then, swept volume Vs= A x L = ~d2 xL 4 241. (a) H1 1 PemxVs xNx- 950xl03 = 2 60 950 xl 03 x 60 x 2 Pem = 0.0259 x 2200 =2 x 106 Pa=2MPa 1 For 4-stroke diesel engine, K = "2 P=950kW=950 x io'w P = PemxALxNxK 60 T2=396.30K W R(TI-T2) m y-l 8.314x(300- 396.30) 1.67-1 W = 1194__!!_ m kmol = 1194J/mol Imol=Mg=40 g For40 g,W=1194 1194 For 1kg W= --xl000 , 40 =29.7 kJ/kg 239. (d) In every case, entropy of universe is always positive. (LS)universe~ 0 ~Ssystem+ ~Ssurrounding~ 0 240. (a) AL= Vs(Sweftvolume)= 0.0259m3, Pem =? N=2200rpm 0.67 => 300 = (~)1.67 = (0.5)0.4012 T2 2 - x x 6 0.030 -0.8 0.015 10 en 0.015 = 8.31kJ 236. (a) Due to internal friction produced in irreversible process, entropy of the system increases. 237. (b) From Clausiusinequality, Cyclicintegral of di < 0 for irreversibleprocess Ql + Q2 + Q3 + ....+ Qn < 0 Tl T2 T3 Tn For process I, 2500 _ 2500 = -1.042 1200 800 For process II, 2000 _ 2000 = -1.5 800 500 Process II is more irreversible than process I. 238. (a) In adiabatic process, W = PIVI- P2V2 y-l y-l _]_= (RL]rT2 P2 235. (a) 233. (c) Enthalpy of balloon is given by H=U+Pr Initially balloonkept 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 volumeis suchthat their product remains constant. Hence, h also remains constant. 234. (b) dH=dU +d(pVO) dH= dU+ pdV + Vdp f dH = f dq + f Vdp ~=Q+fVdp Q-~=-fVdp ~kf;+~p.E+ W =-fVdp 2 W=-fVdp 1 Work is done on the system. V2 Wisothermal= PIVIen-VI Thermal EngineergingA-130 Badboys2Badboys2 Badboys2
- 134. p 1 3 l~ i :2, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, , 270. (c) or n2 x2 = = 0.625 nl +n2 ~S = - R(0.375 £n 0.375 + 0.625 £n 0.625) = 0.66 R = 5.49Jik 1 221 2 pV =-mnc = -.-mnc 332 or 3p = _!_ mnc2 = E 2 2 V 2E p=- 3 269. (c) 11T=0.96 266. (b,d) Signs of work for the four cases are given below (a) 0 (b) '-' ve (c) 0 (d) '-' ve 267. (b) ~S = - R (nl£nxl + n2 £n X2) 8.4 0nl =--= .375 22.4 14 n2 =-=0.625 22.4 (2732-335) 0.64 = 11T --'----3-6-08---'- 264. (d) 258. (d) According to zeroth law of thermodynamics, "when two systems which are equal in temperature to a third system, they are equal in temperature to each other '. Accordingly when 50 cc of water at 25°C are mixed with 150cc ofwater at 25°C, the resulting temperature ofthe mixture will be 25°C, Same analogy applies to situationsin(b) and (c). However,this argument isnot valid when water and sulphuric acid, initially at the sametemperature,aremixed.Heretemperaturewillrise dueto chemical reaction - the change is oftenviolent. 263. (b) 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 differencebetweenoperating temperatures. 11= WTurbine QSuppIied L....-------------+S 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 ofheat addition to the cycle. 254. (d) du = 8Q- 8W Since du is the property and it is exact differential so 8Q- 8Wisthe exactdifferential. 256. (d) 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 = 0.0045x 0.075x 80 x 103 =27 joule. 257. (a) In throttling process enthalpy remains constant. h, =h2 1000= 800+x(2800- 800) x=O.1 T 3 252. (c) and 8=-5 KW m= 1kg/s In thermodynamics, energy or available energy of a systemin the maximum usefulwork possible during a process that brings a system into equilibrium with surroundings (heat reservoir). 247. (c) 246. (a) 245. (a) dS = 8+mR dV V (as isothermal process, dT = 8) fdS=mRfdV or I I V ~S=mR£n V2 VI ( PI' ~S=mR£nlp2) (asPIVI =P2V2)or = ~ x (80)2 x 120 =603cm3 244. (b) We have dQ = dU + pdU mRTdV or T dS = mCvdT + --V-- A-131Thermal Engineerging Badboys2Badboys2 Badboys2
- 135. 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 v p 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. ~----;------------------;--~v BDCTDC 3p L...-----1f----------+----.v Intake Exhaust valve open I p End of combustion ~ Patm Incorporation ofreheater in a steam power plant always increases dryness fraction of steam at condenser inlet and always increases specific work output. 281. (d) 280. (b) 279. (b) 11-1 0.28305 T4 = (P4 , ----;-=> l = (20) 1.28305 T3 lp3) 1500 1 T4= 29047434k. 11= T3 - T4' => 0.94 = 1500- T4 T3- T4 1500-2904.7434 T'4=2820.45 Work w= Cp (T3-T'4) w= 0.98(1500 - 2820.45) w= 1294.049 kJ/kg. 274. (c) ~suniverse= ~Ssyst+ DSsurrounding ~ssurrounding = 0 (Throtteled) ~Su = ~Ssys I .P2 I 0.1 =R ogJ-=8.314 og- PI 0.5 ~suniverse= 13.38 kJ/k 275. (d) Gases become cool during Joule Thomson's expansion only if they are below a certain temperature called inversion temperature TI: The inversion temperature is the characteristic of each gas. Itis related to the Van der Waals' constants 'a' and 'b' by the relation T _ 2a 1 - R.b 276. (a) 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. Volume constant heat addition and volume constant heat removal are Otto cycles. 277. (d) For a given saturation pressure, iftemperature is lower than the saturation temperature then it is subcooled liquid or compressed 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. P3 > PI 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. 272. (a) Cp = 0.98, c,= 0.7638 PI =20bar, T3= 1500k P2 = lbar, 11=0.94 11= Cp =~=1.28305 c, 0.7638 Thermal EngineergingA-132 Badboys2Badboys2 Badboys2
- 136. W W ll+n = - = ----- Qs mcv(T3-T2) h _ W +n - mR~T -- y-1 1 r= (~ )r-I =9.39. W Pm=- Vs TI =300k T3 =6 TI T3 = 1800k we know that for maximum work output T2T4=TIT3 T2 = ~TtT3 T3 = JTIT3 T4 = JTtT3 T2 = .)1800 x 300 = 734.84 k t ~=(T2Jr-t v2 TI V3- v2 = 0.05 (VI - v2) (:~ -I)=O.OS[ :~ -I] rc-1 = 0.05 [12.26] rc = 1.61 ~diesel =1-(r )~-Il~(ct_~;l 1 1(l.61)l.4_1] lldiesel = 1-(13.26)OAll.4 x 0.01 = 60.5 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. 3 303. (b) 302. (d) 301. (a) r = 1+ ~ = 1+ 1177.5 = 1+5.99 ~7 Vc 196.3 W 1 1l=-=1--- Qs (r)y-l ~=1- 1 1800 (7iA-t W = 973.44 kJ/kg Vs = ~(25)2 (37.5) = 18398.43 em3 4 vc= v2 = 1500 em3 r = 1+~ = 1+ 18398.43 = 13.26 Vc 1500 297. (b) 1t 2 1t 2 3 Vs =-d L = -(10) x15=1177.5cm 4 4 295. (c) TL = Lower temperature TH = Higher temperature RE 2 Power= --=- =lkW COP 2 2 200 300-200 293. (c) COP = Refrigeration effect = ~ = 2.33 Work done 1.5 288. (b) .. VI Compression ratio, r = 10 = V2 Now, mean effective pressure is given by Work done Pmean = ----- Swept volume V4 Vt Now -=-=10 'V3 V2 => VI = 10V2 ...(i) Also swept volume VS=VI-V2 => VS=0.9VI Initiallyfor air PIVl =nRTI " _ nRT _lxO.287x300 .. vt - ~ - 100 =0.861 m3/kg .. Vs=0.9x 0.861 =0.7749m3/kg Work done in cycle W = Heat supplied - Heat rejected = Qs - QR = 1500 - 700 = 800 kJ/kg 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. 285. (c) A-133Thermal Engineerging Badboys2Badboys2 Badboys2
- 137. = 1-[ 0.001 ]1.4-1 =58.2% 0.001+~xO.2002 xO.250 312. (c) Given PI = 1bar P3 =40bar r=5 vc=O.OOI m3 7t 2 3 Ys = -x 0.200 x 0.250m 4 311. (c) = 1 1 11=1--- (r)Y-I 1 --1-:-4-:---:-1 = 0.46 = 46% (4.68) . V2 =1 v2' = 1 +0.7 (r-l) =0.7r +0.3 v2' = 1 +0.3 (r-l) =0.3r+0.7 lIx .'. ~=(_!i_J =(2.6)1.3 =1.7 v2 P2 1.3 0.7r+0.3 1.7 0.3r+0.7 r=4.68 30% V---+70% I I I I I --4--------1 P >I 11 v,1 V' 11 I 2 310. (c) 6-5 = (1.4-1)x-5-x100 =0.08 x 100=8% M = (r-1)-x 100 r .1n M -xl00= -(I-r)-xl00 n r 11= 1__ 1_= l-(r)y-1 (r)Y-I308. (d) more efficient. :. 11DieseI>11DuaI> 110tto S Thermal efficiency= 1 Qrejected = 1 Constant Qsup plied Qsupplied Thus the cycle with greater heat addition Qsuppliedis V 4 P 32 1-(r)~-l=1-(r)~-I[r(:=~)] sr-1 = r(s - 1) sr-I - r( s - 1) = 0 306. (d) Efficiency of ideal regenerative cycle is exactly equal to that of the corresponding Camot cycle. Hence it is maximum, 307. (b) 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. 304. (a) 11otto= 11diesel Vs =(r-l)vc 11+n(.1p)vc Pm = -...;,...:=........:...__~-=---- (Y-1)(r-1)vc 11+n(M» = (y-l)(r-l) w = 11+nmR.1T = 11+nM>vc y-1 (Y-1) Thermal EngineergingA-134 Badboys2Badboys2 Badboys2
- 138. Wc) = 0.582or58.2% Cooling and dehumidification-temperature decreases 325. (d) Relative humidity of the air decreases. and w also decreases [ ] 1.4-1 11-1- 0.001 0.001+~(0.2)2 x(0.25) 11-1-( Vc JY-1 VC+VS Substituting values, 1t 2 1t 2 Vs =-d L=-(200) x250cc 4 4 d=200mm Vc=0.00Im3 Air is cooledat constant pressure to make unsaturated air to saturated one, Given, StrokeL= 250mm Diameter, Clearancevolume Now, swept volume 324. (c) s Tdew T 323. (c) 30° 321. (a) On a psychrometric chart Constant relative humidity lines are uphill curve not straight, to the right. Constant WBT lines straight downhill to the right. Constantspecificvolumedownhillstraightto theright. Constant enthalpy lines are not coincident to WBT. 322. (b) P=2u(v-u)(1 +cos<j»x flowrate = 2 x 10(25-10) (1 + cos 120°)x 0.1 = 20 x 15x 0.5 x 0.1= 15kW 0.025 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., 04 320. (b) On a psychrometric chart 75% RH w (kg/kg) 313 () H· ..c. I ( ) QI QI . c mt: usmg rormu a cop = W = Q2-Ql =~ T2-Tl 319. (b) Chemicaldehumidification- temperatureincreases,w decreases, i.e., 0-2 Sensible heating - straight horizontal line towards right, i.e.0-1 1+~=5 Vc Vs = 4vc W 11=- Qs 0.4746= W 76.225vc W = 36.17vc Pm = W = 36.17vc = 9.04 bar Vs 4vc P2 =Pl( :J=1.(5)14 =9.51 bar 1 1 11=1--- = 1--- =0.4746 (rr' (5)°.4 R(T3-T2) Qs = Cv (T3 - TI)= _____..:.._--=----~ r-1 v2 (P3- P2) (40 - 9.51)xvc = = 76.255vc r-1 1.4-1 A-135Thermal Engineerging Badboys2Badboys2 Badboys2
- 139. The viscous shear stress between two layers at a distance 'y' du from the surface can be written as: r ex - dy Newton's Law of Viscosity stresser. 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 VISCOSITY Liquids are highly incompressible :. dp = 0 dp Gases are highly compressible as P ex p. 5. Bulk Modulusof Elasticity(K) It is defined as reciprocal of compressibility. dV 13 = _____:y_ = _!_ dp dp P dp . P Mathematically, pressure head (h) = - pg 4. Compressibility(13) Hydrostatic law: It states that rate ofincrease ofpressure in a vertical direction is equal to weight density of fluid at that point. V I v=-=- m P Sg = 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. S _ weight of fluid g - weight of standard fluid Density of Fluid 3. Relative densitySpecific 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. mg - =pg V co= p = V Specific Weight(co) : It is defined as weight per unit volume of substance. m FLUID PROPERTIES 1. Density (p) : It is defined as mass per unit volume of substance. Fluid: Fluid is a substance which has the property tendency to flowunder the action of shear and tangential forces. Liquids and gases both are fluids. 2. 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 fluidis describedas a fluidwhich 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 fluidssare defined as fluids those obeyNewton's law of viscosity.The density ofthese fluids may be constant or variable. The viscosity is calculated according to Newton's law of viscosity as : du t=f..l- dy 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 ofpolymers, 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: Incompressiblefluids 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. I~I..(JI)) )11~(~IlllNI(~S llN)) )lll(~IIINI~11Y Badboys2Badboys2 Badboys2
- 140. P, = py = pz· PRESSURE MEASUREMENT DEVICES L BAROMETER It is a device made by Torricelli and is used to measure local atmospheric pressure. II. PIEZOMETER • It is a device used for measurements of moderate pressure (gauge) of liquids only. • Piezometer cannot measure the pressure of gas. 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 h = 2 cr cos 8 pg(r2 - r1) Foran annular capillaryhaving externalradiusr2and innerradius r., pgr 2 c cos 8 h=--- 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 levelin 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 CAPILLARITY • For pure water 8 = 0°. For Mercury-glass, 8 = 130° to 140°. • • • 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. 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. SURFACE TENSION (0') Cohesive and Adhesive forces: Cohesive forces are intermolecular attraction of forever be- tween molecular of same liquid/fluid. Examples • Newtonian: Water, air • Dilatent : Butter, starch solution • Psuedo plastic : Paints • Bingham plastic: Gel, cream • Thixotropic: Printer's ink and enamel (~~) ~ Ideal fluid r Newtonian '"'" ~~---....0::: <I.) ,.<:1 r:.rJ. Rheopactic Bingham plastic RHEOLOGY It is the branch of science in which we study about different types of fluids 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 in- crease with increase in temperature. • In ordinary situations, effect of pressure on viscos- ity is not so significant but in case of some oils, vis- cosity increase with increase in pressure. • du If Il is low => velocitygradient dy is high => easy to flow fluid. Kinematic Viscosity (v) • • • du as r = Il- dy '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 ofviscosity are known as Newtonian fluids. If Ilis high =>velocitygradient du is less=>highlyviscous dy fluid. • A-137Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 141. leg: moment of inertia of the plane surface about c.g F = pghA - - leg sirr' 8 hcp = h + ---=-,=--- hA '" y"" cp : Centre of pressure cg : Centre of gravity F : hydrostatic force acting on the plane surface inclined to free surface. . h hcp h SIn e = == =--= -y ycp y 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 Atmospheric pressure t w 1 r, Gauge pressure I' Gauge pressure, I' absolutepressure ~ II iPressure 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 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. 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 Vacuum pressure is defined as the pressure which is below the atmospheric pressure. Absolute pressure 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. pAHg Pressure (P) = ---;::- =pHg hence from the above expression, pap and p aH -+ The pressure of fluid is equal in all directions at any (Inciined tube manometor) 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. +-- DIFFERENTIAL U TUBE MANOMETER PI + pgy - Pn gh = 0 -I-h- -- -- - --- -- SIMPLE U TUBE MANOMETERy p m MANOMEIER • used for measurement of high pressure. • It makes the use of a manometric fluid. Fluid Mechanics and MachineryA-138 Badboys2Badboys2 Badboys2
- 142. d d h 2 leg = ~d4 64 A = ~d2 4 (iv) :l. l+-~11a ':" ~ I a h 2 leg = a4 12 A = a2 (iii) h 3 A= bd 2 leg bd2 -- 36 STABILITY OF SUBMERGED BODY • 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. 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. Metacentric Height (GM) 1 GM =- -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 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 d (ii) F : Hydrostatic force acting on the curved portion FH: Horizontal component of F Fe : Vertical component of F FH = pghA Fy = pgV F=~~ +F~ V = Volume till the free surface h d/2 leg = bd3 - 12 A= bd (i) Consider a curved surface as shown in the figure. v HYDROSTATIC FORCES ON CURVED SURFACESFor a horizontal surface, O= 0° => 11ep = 11 For a vertical surface, O = 90° - - leg => h =h+-ep hA Vertical Surfaces (9 = 90°) A-139Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 143. 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) _1 _ H a=~a2x+a2y+a2z For uniform flow, convective acceleration = 0 For steady flow au au au au a =u-+v-+w-+- x ax ay az at av av av ava =u-+v-+w-+- Y ax ay az at aw aw aw awa =u-+v-+w-+- z ax ay az at ~ ~ ~ ~ ~ ev av ev ev ava=- =u-+v-+w- + at ax ay az at J.. J..L-...J Convective acceleration temporal or local acceleration ~ A A A V = u i-i v j-r w k Acceleration of A Fluid Particle ~ (pu) + ~ (pv) + ~ (pw) + ap = 0 ax ay az at for incompressible fluid, A1V1= A2V2• Continuity equation in cartesian - co-ordinates 2 2 Continuity equation: Ifstates 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 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 flowi.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. 2. Uniformflow Ifthe properties (velocityat any giventime) isnot changing with respect to space, such a flow is known as uniform flow. 3. Incompressibleflow If the density of the fluid doesn't change with respect to pressure, the flow is known as incompressible flow. 4. Rotationaland Irrotationalflow Ifthe fluid particles are rotating about their centre ofmass, 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 5. 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. density of liquid 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 solid NOTE: 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 Fluid Mechanics and MachineryA-140 Badboys2Badboys2 Badboys2
- 144. ( dY) x ( dY) = _ ~ x ~ = (- 1) dx tjl=constant dx jI=constant V U :. Equistream and Equipotential lines are orthogonal to each other. Cauchy-Riemann EqD In irrotational flows, u= _B<I> =_ a.v~ IB<I> =a.v1 ...(I) ax ay ax ay v=-:=~ ~ 1:=-~1 ...(2) Equations (1) and (2) areknown 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 (a) 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 Slope of equistream line dy =~ dx u • There is no boundation on jJ as it satisfies continuiting equation. Equistream Line Itis a line obtainedbyjoining points having same streamfunction values. ajJU= - - ay ajJ V=- ax where dy= IdY = -~Idx dx v Slope of equipotential line Stream Function (w) • It is defined only for 2D flows and is a function of space and time. 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. 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 aq,u =-- ax aq, v =-- ay aq, w =-- az CIRCULATION m It is defined as the line integral of velocityvector along a closed loop. VORTICITY It is defined as double of angular velocity.(Circulation per unit of enclosed area) Vorticity = 20) 1 (av au1 w =-l---)z 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 PATH LINE It is the actual path traced by a fluid particle. STREAK LINE Itis the locusofall fluidparticles at a moment which have passed through a given point. Rotational components in flow V = ui = v} Equation of streamline in differential form Stream Line It isan imaginaryline drawn in sucha waythat the tangent drawn at any point on this line gives the direction of velocity vector of the fluid particle at that point. A-141Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 145. QTH = A2Al A2 _ A2 1 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 ofan orifice meter is same as that ofventurimeter. 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 Diverging sectionConverging section 2gh 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 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: v=~2gH (f) Flow sensors Flow Measurement Devices Venturimeter It is a highly accurate deviceused for measurement ofdischarge. CD CDThroat (minimum cross-sectional area) P V2 TheBernoulli's Eq=in sucha casecanbewrittenas-1 + -21 + Zt pg g BERNOULLI'S EQUATION FOR REAL FLUID In real fluids, viscous shear stresses are present due to which energy is not conserved. 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 everypoint in a flow. P V2 - + - + Z= constant (Head form) pg 2g 1 P + - pV2 + pgz = constant 2 mgH mg Potential head = H (d) Pressure Head: It is defined as the fluid pressure per unit specific weight. P h d fluid pressure P ressure ea = = - specificweight pg (e) Total head: It is defined as the sum of kinetic head, potential head and pressure head. Totalhead = kinetic head + potential head + Pressure head v2 P HT=-+H+- 2g pg Euler's Equation of Motion The Euler's equation considers the following assumptions • Flow is irrotational • Flow is laminar • Flow is invicid. I~ + V dv + g dz = 0I~ Euler's Eqn for steady flow Integrating the above equation. We obtain Bernoulli's equation 2 kinetic Head = ~ 2g (c) Potential head : It is defined as the potential energy per unit weight. P . Ih d Potential energyotentia ea = ------=..::...- weight of the body mg 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. ki . h d kinetic energymetre ea = -------=::..::...._- weight of the body Fluid Mechanics and MachineryA-142 Badboys2Badboys2 Badboys2
- 146. Free vortex flows are irrotational flows and thus, Bernoulli's equation can be applied. • NOTE: FORCED VORTEX • External torque is required to maintain its angular velocity at a constant value. w= constant V o: r 1 V: velocity Voc- r r: radius 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. P1 A1 Fx,Fyare 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 = IF2 + F2j x Y Fluid Fluid (system) Working principle: Apitot 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 ishorizontal. Hence,there wouldbe no effect of weight. Consider a pipe bend as shown, 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. 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: Cd = (coefficient of discharge) or Q-CdAo~2gH where, Orifice Meter Discharge is given as : Cc·~d2J2gH Q= ~ Differential manometer Direction of flow -+-1---+ 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 orificemeter. 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 flowoffluid in pipeshaving fluids ofsinglephase. ~2 Venaconfracta I' :rl~~e A-143Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 147. 13, d31 ......_ -' I 121 ~ : I Pipes in Series • In series, discharge (Q) remains same but head is divided. FLOW THROUGH BRANCHED PIPES KV2 h---f- 2g K = Constant which depends upon angle of bend and its radius of curvature. BEND LOSSES O.5V2 h---f- 2g h.= vi [_1 _1]22g Cc A2 CC=A 3 If'C, is not given, hj= 0.5 Vll2g Head loss occurs after Venacontracta as boundry layer separation occurs. (iv) Entrance Losses CD V12 h.= 2g (iii) Losses Due to Sudden contraction ,~ ~: N_CD I ® (i) Losses Due to Sudden Enlargement 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. 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. pVD Re=-- J..l V = velocity D = diameter m = dynamic Viscosity Minor Losses • Bernoulli's Equation, momentum Eq" are used to determine these losses. • The magnitude of minor losses is very loss. ...(1) • Forced vortex flows are rotational flows and hence, Bernoulli's equation cannot be applied. 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 = w:t I • A pipe is a closed contour which carries fluid under pressure. • When fluid flowsthrough pipe, it encounters losses. These losses can be broadly categorized into (i) Major losses (ii) Minor losses Major Losses • These losses are due to friction. The losses are evaluated by Darcy-Weishback Equation. fLV2 2gd (ii) Exit Losses Fluid Mechanics and MachineryA-144 Badboys2Badboys2 Badboys2
- 148. --r-bj d~t-'--::I;;?::--u Uy 1 (ap) 2 u=--- - (by-y ) b 2M ax Case II : When both plates are at rest u = - 2~ (:) (by - y2)(Poiseuille flow) LAMINAR FLOW BETWEEN TWO PARALLEL PLATES Case I : One plate is moving with a velocity of 'U' while the other is stationary. Umax u=-- 2 at r = R/.fi,= U = U i.e. average velocity equals the local velocity. Pressure drop (P, - P2) in a given finite length 'L' 32MUL P1 - P2 = D2 t=(-:H u = - _1_ (ap) (R 2 _ r2) 4M ax from above expression of 'u', we can conclude that velocity is varying parabolically. 1 ( ap) 2 ( 1t R2 '1 Umax = - 4M ax R ,Q = l-2-) U max 1" : shear stress R: radius of pipe M : dynamic viscosity of fluid ap ax :pressure gradient u : velocity at a distance 'r' from cente ! ~ --! -~---- ~P+.?Pdx "[: Ox • Shear between fluid layers e = Il du/dy(x-dir.) Entrance Length 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 for maximum efficiency 1he = ~ 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. Features of Laminar Flow • No slip at boundary • Flow is rotational • No mixing of fluid layers Ptheoretical = pQgH Pactual = pQ (H - hf) where hf are the head losses in pipe. pQ (H - hf) 11= pQgH H POWER TRANSMISSION THROUGH PIPE J • • • Siphon is a long bend pipe used in carrying water from a reservoir at higher level to another reservoir at lower level. 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. • SYPHON Q=Q1 +Q2 (hf)1 = (hfh 9+~---t-~- PIPES IN PARALLEL • In parallel arrangement, head losses remain same but discharge gets divided. => h=~[LIV? + L2Vj + L3V}] 2g d1 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. Q=Q] =Q2=Q3 h-= (hd] +(hf)2 +(hfh A-145Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 149. where 8 : momentum thickness 1:0 : plate shear stress p : density UOC): free stream velocity Drag force (FD) ~ d8 pU~ dx dP -=0 dx • Flow is 2D, incompressible and steady• 0* Shape factor (H) =e Von Karman's Momentum Integral Equation Assumptions • • ap For separation of boundary layer, ax > O. 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 Displacement thickness (0) 0* =1(ll- ~ ') dy o UOC) Momentum thickness (8) 8 = 1~(ll- ~)' dy o UOC) UOC) Energy thickness (OE) f 8 u ( u 2 , OE = -U II--2 j dy o OC) Uoc> • • 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. • y=O u = 0.99Uo au =0 ay a2u-=0 ay2 Conditions u=O Boundry at y= 0 y=o y=o Boundary layer 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 K 6 .. 0.25 <- < => transition 0' Hydrodynamically Rough and Smooth Boundaries From Nikuradsee's experiment, K 87 < 0.25 => smooth boundary K 87 > 6 => rough boundary ReR > 100 => rough pipe ReR < 4 => smooth pipe 4 <ReR < 100 => transition • In turbulent flow in pipes, average velocity equals local velocity at y = 0.223 R. Thickness of Laminar Sublayer (0') 11.6v 0' =-- V* v • u ( y, V* = 5.75 10glO ly') y' = 0' I 107 (for smooth pipes) y' = Kl30 (for rough pipes) Reynold's condition for rough & smooth pipes • V. : Shear velocity V. =t Umax - U ---=..:..:,:=..:....___ = 5.75 10glO (R/y) V* • 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. • • 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: = pu'v' = P I ldy) TURBULENT FLOWS u = __ 1 (ap)b2 max 8J.l ax _ 2 U =3Umax large velocity gradients existing in it. Velocity within the boundary layer increases from zero to main stream velocity asumptically. • Fluid Mechanics and MachineryA-146 Badboys2Badboys2 Badboys2
- 150. 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 (b):) (e);:: ~) v=Q (8) 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. Impulse turbine Reaction turbine (i) In this, the conversion (i) A part of energy offluid of potential energy into is converted into kinetic kinetic energy takes energy before entering the place by nozzle before fluid into turbine. entering to turbine (ii) There are no losses in (ii) There are losses in flow flow regulations regulations (iii) The whole unit is (iii)The whole unit is placed above the tailrace 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 (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) 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 flowturbine : 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 flowturbines : In this type, the flowing of water is through the runner along the direction parallel to the rotational axis of the runner. ~ Mixed flowturbine : 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. ComparisonbetweenImpulseTurbineand ReactionTurbine 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 ajet, depending on the type of turbine. The efficiency of any turbomachine 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 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. 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 4.64 x 8 = ~Rex 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 toC _-- fx _ 1 2 -pU 2 00 Itis 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. A-147Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 151. 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 Pshaft l1mech. = Q H P W W The value of mechanical efficiency varies between 0.97 to 0.99. (c) 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. (d) Overall efficiency : It is defined as the product of hydraulic efficiency (l1H)' mechanical efficiency (l1mech) and volumemetric efficiency (11vol). 110 = l1H x l1mech x 11vol. Francis Turbine Francis turbines are by far the most common type in present- day 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 1+ coso Maximum hydraulic efficiency (l1max.) = 2 (b) Mechanical efficiency : It is described as the ratio of power available at the shaft and the power produced by the wheel. Vw = velocity of whirl at outlet 2 u = peripheral velocity Vr1 = relative velocity at inlet Vr2 = relative velocity at outlet VI = absolute velocity at inlet V2 = absolute velocity at outlet 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. 2U(VWI ± VW2 ) l1H = -....:....._-:----....:.... V? u ~ work done / weight = (VWl ± VW2 ).g where, Vw = velocity of whirl at inlet 1 Gross head = HG = difference between head race and tail race Net head = Hnet = ~ - hf- h 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 VU1 Some important formulas : 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 : u Vu IE 2 * 2 )1 p Specific Speed where A = area ofthe jet g = acceleration due to gravity. The input power to the turbine P = 1000 x Q x g x H = 1000x A~2gH x g x H (.: Q= A~2gH ) Fluid Mechanics and Machinery Q= A~(2gH) A-148 Badboys2Badboys2 Badboys2
- 152. Fig. : A Propeller or axial-flow turbine Main parts ofa kaplan turbine: (a) Scroll casing: It is the casing in which guiding the water and controlling of passage of water takes palce. (b) 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. (c) Stay ring: The stay ring guides the water from scroll casing to the guide vanes. (d) Runner blades: Runner blades are connected to the hub and there is an axial flow ofwater through the runner. (e) Draft tube : It is utilized for the purpose of connecting 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. or 110= 11H x 11mech. x 11vol. VWIUl±VW2·U2 11H = gH ~ Mechanical efficiency : It is defined as the ratio of shaft power to the power developed by the runner. It is denoted as 11mech. ~ Volumetric efficiency (11vol):It is defined as the actual quantity of fluid working on the runner to the total quantity of fluid supplied. Actual fluid quantity 11vol = T Ifl ·d .. ota U1 quantity ~ 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. shaft power 110 = Input power ~ Work done per second: (W.D/s) W.D/s = pQd[VW1Ul ± VW2U2] • 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. • • • . P v2 At the exit of draft tube H = -- - Zl -- , P 2g At the exit of pen stock when the position of draft tube is at the tail race. p v2 H=-+z.+- P 2g At the exit of draft tube when the position of draft tube is at the tail race. v2 H=-- 2g Ifthe velocity at the exit of draft tube is negligible, then Net head, at the exit of penstock. H=(~+Z<J • P v2 At the exit of penstock, H =-+zl +- p 2g• Some important formulas : ~ Net head (H) : h f ifi ( ~ )T e range 0 speer IC speeds, lNs = 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 water from dams reaches to the inlet of the turbine. (b) 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. (c) 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. (d) Runner: It is connected to the shaft of the turbine. (e) Draft tube: Itis defined as tube which expands gradually and it discharges water passing through the runner to the tail race. Velocity triangle of Francis turbine : rotation. A-149Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 153. T- I I I I I I I I I I I U2 I I -l---~~====~====~~I----Vwl ---+! :+-Vw2 ~---------~VW----------~ Here, a = guide blade angle or runner vane angle at inlet Fixed Blade Shift --*)~-- Disc Blade Fig. : Ranges of application of different types of turbine. Note the overlap at the boundaries VELOCITY DIAGRAMS flowrate 1m)$-1 10 20 kW I MW Francis and similar turbines 100 .¬ 1.c 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. e, <I> = vane tip angle a, ~ = angle which absolute velocity makes with tangential directions Degree ofreaction (R): It is defined as the ratio of change in static energy in the rotor to the total energy transfer. (Vr} - Vr{l (uf-ufl l 2g ) +l 2g ) R = -----=--------=-- H 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 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 given as : • Conical draft tube • Single elbow tube • Moody spreading tube • Elbow draft tube with circular inlet and rectangular outlet. 1000 Fluid Mechanics and MachineryA-150 Badboys2Badboys2 Badboys2
- 154. 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 Nu Muschelcurves ~.•. , 'Best performance curves ....... - 25% Full opening H = constant % full load (Operating characteristic curves) Constant efficiency or Muschel curves : In these curves, the data obtained from constant head and constant speed curves are drawn forthe purpose offinding the constant efficiency zone. (c) unit speed (NJ Unit speed vs unit power (b) 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 unit speed (Nu) Unit speed vsefficiency Unit speed (Nu) (Unit speed vs unit discharge) ~;:::i 100%CI'-' (l) 75%eo ~,J::I o 50%fZl :.a .~ 25% e= wheel vane angle <I> = vane angle at outlet Vf1 Vf1 V tan a = -- .tan e = ;tan <I> = ____f1_ Vwl ' Vwl -u 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= JH (b) Unit discharge (Q,) : The turbine discharge working under unit head is known as unit discharge. Q Qu= JH (c) 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 The performance of turbines should be studied for the purpose ofproviding 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 ofturbine is varied byvarying the flowrates 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 A-151Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 155. (g) (f) (e) (d) (c) Radial flow 10- 25 Mixed flow 70 - 135 Axial- flow,propeller 100 - 425 Based on types of casing : -+ Volutechamber 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 guidsvanes areplaced 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. 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 flowtakes place due to which flow area is enhanced. -+ Axial flow: In this type, high discharge and lowheads is used. Eg : Irrigation. 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 ofthe sides ofthe impeller. Due to which, discharge is increased. 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. 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 -+ 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. NJQspecific speed (N s) = ~ H Type of pump specific speed -+ casmg 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 : Fluid Mechanics and MachineryA-152 Badboys2Badboys2 Badboys2
- 156. efficiency. ~ 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. • • (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 • Fig. (3) 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 : (e) where, QA = Actual discharge, QL = rate of leakage Hydraulic efficiency : It is defined as the ratio of manometric head to the theoretical head. Hm 11N=- HT Characteristic Curves (d) wVw2u2g 11mech.= Ps (c) Overall efficiency (110) : It is defined as the ratio of power output to power input to the pump or shaft. wHm 110 =-p- Volumetric efficiency (l1vol) : It is defined as the ratio of actual discharge to the sum of actual discharge and rate of leakage. QA (VW2U2)/ g Mechanical efficiency (l1mech) It is defined as the ratio ofpower delivered bythe impeller with the power input to shaft. (b) 11mano. Work Done by Impeller on the Water [VW2.u2 - VWlul ] WD = .:::...._--=------=---= g where, W.D = work done VWl = velocity of whirl at inlet VW2 = 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: u2 then work done = __ 2_ g Heads in Centrifugal Pumps (a) Suction head: Itis defined asthe 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 ofwater 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 asthe 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, Manometric head (Hm ) 11mano.= (VW2U2)/ g (Back-ward facing vanes) (Radial Vanes)(Forward facing Vanes) A-153Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 157. 11. A pump handling a liquid raises its pressure from 1bar 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 (b) 3 (d) 5 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 (c) 4 (b) n (d) n + k 10. (a) k (c) n-k 9. A phenomenon is modelled using n dimensional variables withk primary dimensions.The number ofnon-dimensional variable is (b) 354 (d) 707 8. (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 1000kW power for a head of 40 m. If the head is reduced to 20 m, the power developed (in kW) is (a) 177 (c) 500 coefficient z. Match number Skin friction y. W. Weber number X. Froude number U. Reynolds number V. Nusselt number (b) Irrotational flow (d) Incompressible flow 7. 6. Forthe continuityequationgiven V.;= 0 tobevalid,when ; is the velocity vector, which one of the following is a necessary condition? (a) Steady flow (c) Invescid flow Match the following: P. Compressive flow Q. Free surface flow R. Boundary layer flow S. Pipe flow T. Heat convection For a Newtonian fluid (a) shear stress is proportional to shear strain (b) rate of sheer stress is proportional to shear strain (c) shear stress is proportional to rate of shear strain (d) rate of shear stress is proportional to rate of shear strain 5. au au (d) v-+u- Ox ay Ov au (c) u-+v- ax ay 4. A two-dimensional flow field has velocities along x and y directions given by u = x2t and v = - 2xyt respectively, where t is time. The equation of streamlines is (a) x2y = 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 auOv (a) u-+v- (b) u-+v- ax ay Ox ay 3. /-!uOL (a) D2 8/-!uoL (c) D2 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 2. Ov ay(d) Ov (c) ay Ov Ox (b) The velocity components in the x and y direction of a two- dimensional potential flow are u and v respectively, then au. aX IS equal to Ov (a) Ox 1. ...,EXERCISEIIIIII~ Rate of flow (Lifi/s)-----7 (Muschel curves) a'._/ --g (]) ...s::: ~ E§ I---.+--l-_ ~ Muschel or constant efficiency curve ~ Discharge (Qd) (Operating characteristic curves) lip power efficiency Fluid Mechanics and MachineryA-154 Badboys2Badboys2 Badboys2
- 158. J..l1td40) (a) 32h J..l1td30) (c) 32h 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 A type of flowin which the fluid particles while moving in the direction offlowrotate abouttheir mass centre, is called (a) steady flow (b) uniform flow (c) laminar flow (d) rotational flow A-2d flowhaving velocityV = (x +2y +2) i +(4 - y)jwill be (a) compressible and irrotational (b) compressible and not irrotational (c) incompressible and irrotational (d) incompressible and not irrotational 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 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. Ifthe 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 capillaryrise in anarrow 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' 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 Pressure in Pascals at a depth of 1m belowthe free surface of a body of water will be equal to (a) 1 Pa (b) 98.1 Pa (c) 981 Pa (d) 9810 Pa A circular disc of diameter' d' is slowlyrotated 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-155 (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 27. point on it shows the velocity vector at that point (ii) There is no flow across streamlines dx dy dz. h diff . I . f(iii) - = - = - IS tel erentra equation 0 a u v w 28. 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 29. 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 Group A Group B P: Biotnumber 1 Ratio of buoyancy to viscous force Q: Grashof number 2 Ratio of inertia force to viscous force R: Prand tl number 3 Ratio of momentum to thermal diffusivities s: Reynolds number 4 Ratio of internal thermal resistance to boundary layer thermal Fluid Mechanics and Machinery 12. A streamline and an equipotential line in a flowfield 19. (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 20.diameter D, neglecting gravity effects, the pressure drop dP over a length L and the wall shear stress 'tw are related by dPD ~PD2 21. (a) 't =-- (b) 'tw =--2- w 4L 4L dPD 4dPL (c) 't =-- (d) 't =-- w 2L w D 14. Biot number signifies the ratio of 22. (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 23. (b) an unsteady uniform flow (c) a steady non-uniform flow (d) an unsteady non-uniform flow 16. Match Group A with Group B: Badboys2Badboys2 Badboys2
- 159. A (b) 2 (d) 2 C 1 3 B 3 1 B C 2 3 3 2 Codes A (a) 1 (c) 1 47. 44. 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' 45. 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 46. 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. L~t I L~t II (Loss) (Parameter responsible) A Leakage Loss 1. Zero at design point B. Friction Loss 2. Proportional to head C. Entrance Loss 3. Proportional to half of relative velocity square. y2 (c) 2g (b) nd2• 2g 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 v2 4f L y2 (a) -.- d 2g 43. l6000J..l2 (c) 40. When pressure p, flow rate Q, diameter D, and density d, a dimensionless group is represented by pQ2 __ P__ (a) dD4 (b) dQ2D4 pD4d pD4 (c) Q2 (d) dQ2 41. Maximum wall shear stress for laminar flow in tube of diameter D with fluid properties J..land p will be 32000J..l2 6400J..l2 (a) pD3 (b) pD3 8000J..l2 (d) pD3 ( 5 )115 (d) d= ~L( 3 )1/5 (c) d= ~L _ (D5)1I4(b) d- - 8L _ (D3)1I4(a) d- - 8L 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 38. (c) O'J..lY ~P (a) 37. 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 36. 35. 34. 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 ax Oy 33. (a) ou+av +ow =0 (b) Ou=av =Ow =0 ox Oy oz ox Oy oz au ov Ow au av Ow (c) -+-+-=1 (d) -+-+-=u.v.w ax Oy oz ax Oy oz 42. 30. 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 31. 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 32. The general equation of continuity for three-dimensional flow of a incompressible fluid for steady flow is Fluid Mechanics and MachineryA-156 Badboys2Badboys2 Badboys2
- 160. 0 ""'"("I') ("I') I Fluid Mechanics and Machinery A-157 o, C) 48. The velocity potential in a flow field is <I> = 2xy. The 59. Compressibility (B) is equal to : corresponding value of stream function is (If k is bulk modulus) (a) (y2 _ x2) + constant (b) (x2 - y2) + constant (a) ~ = _!_ (b) ~=k 122 k (c) -(x - y ) + constant (d) 2 (x - y) + constant ~=_12 (c) (d) ~ = k2 49. Consider the following k2 The components of velocity u and v along X and Y 60. Kinematic viscosity is equal to : directions in a two dimensional flow problem of an (a) Dynamic viscosity x density incompressible fluid are Dynamic viscosity (i) u = x2 cos Y ; v = - 2x sin y (b) Density (ii) u = x + 2; v = 1 - 4 (iii) u = xyt ; v = x3 _ y2t12 Density (iv) In u = xty; v = xy - ylx (c) Dynamic viscosity Which of that will satisfy the continuity equation ? (a) 1,2 and 3 (b) 1,2 and 4 (d) (c) 2, 3 and 4 (d) 1,2, 3 and 4 Dynamic viscosity x Density 50. Consider the following statements regarding bernaulli's 61. Void ratio does not depend upon: theorom for fluid flow (a) Liquid limit (b) Volume 1. Conservation of energy (c) Bulk volume (d) Porosity 2. Steady flow 62. The height of the water column corresponding to a 3. Viscous flow pressure equivalent to 60 m of mercury column will be 4. In compressible flow equal to: Which of the above statements is/are correct? (a) 8160 em (b) 816 em (a) 1,2 and 4 (b) 1 only (c) 81.6 em (d) 7996.0 em (c) 2,3 and 4 (d) 1,2, 3 and 4 63. The difference of pressure between inside and outside of 51. An ideal fluid is defined as the fluid which: a liquid drop is (a) is compressible M>=!(b) is in compressible (a) ~P=Txr (b) (c) is in compressible and non-viscous r (d) has negligible surface tension M>=_!_ M>= 2T52. Which of the following is the unit ofkinematic viscosity? (c) (d) (a) N- s/m? (b) m2/s 2r r (c) kg/s m2 (d) m/kg-s 64. Falling drops of water become spherical in shape due to 53. Poise is the unit of: the property of : (a) dynamic viscosity (b) kinematic viscosity (a) adhesion (b) cohesion (c) mass density (d) velocity gradient (c) surface tension (d) viscosity 54. Surface tension is a phenomenon because of: 65. The pressure at a depth of 5 km below the surface of sea (a) viscous forces only water considering specific gravity of water to be 1.3, will (b) Adhesion between liquid and solid molecules be equal to : (c) Difference in magnitude between the forces due to (a) 63765 Pa (b) 637654 Pa adhesin and cohesion (c) 1.27 x108Pa (d) 1.48 x 108 Pa (d) cohesion only 66. Capilary action is due to : 55. In case of a static fluid: (a) viscosity of liquid (a) only normal stresses can exist (b) cohesion of liquid particles (b) linear deformation is small (c) surface tension (c) fluid pressure is zero (d) None of these (d) resistance to shear stress is small 67. A piece of wood having weight 5 kg floats in water with 56. Gauge pressure is equal to: 60% of its volume under the liquid. Then the specific (a) absolute pressure + atmospheric pressure gravity of the wood will be equal to: (b) absolute pressure - atmospheric pressure (a) 0.83 (b) 0.60 (c) atmospheric pressure - absolute pressure (c) 0.71 (d) 0.72 (d) atmospheric pressure - vaccum 68. Hydrostatic low states that the rate ofincrease of pressure 57. It pressure intensity is 1.006 MN/m2 and specific gravity in vertical direction is equal to : of sea water is 1.025, then the depth of a point below (a) fluid density (b) fluid specific weight water surface in sea will be equal to : (c) fluid weight (d) fluid specific gravity (a) 10 m (b) 1000 m 69. A rectangular tank of square cross-section (2m x 2m) (c) 100 m (d) 1m and height 4 m is completely filled up with a liquid. Then 58. An oil of specific gravity 0.7 and pressure 0.14 kgf I cm-. the ratio oftotal hydrostatic forceon any vertical wall to its Then the weight of the oil will be equal to : bottom is equal to: (a) 70 em of oil (b) 2 m of oil (a) 2.0 (b) 4.0 (c) 20 em of oil (d) 80 cm of oil (c) 6.0 (d) 1.0 Badboys2Badboys2 Badboys2
- 161. 2 leg cot o (d) A'x' A'x' where, Q = inclination of plane area. x = distance ofc.g of plane area from free liquid surface. 2 leg tan e (c) 88. 87. 86. 85. 84. 83. 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 (a) 2 3 (b) 2 3 (c) 3 2 (d) 2 1 3 Find the buoyantforceacting on the aluminium cubewhich is suspended and immersed in ajar 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 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 The vertical distance of centre of pressure below the e.g of the inclined plane area (submerged in liquid) is : leg sin2 e leg cos2 o (a) (b) A'x' A'x' 82. Fluid Mechanics and Machinery Codes: A B C D (a) 4 1 3 2 (b) 2 3 4 (c) 4 3 2 (d) 2 4 1 3 A capilarity 1. cavitation B. vapour pressure 2. Density of water C. viscosity 3. shear forces D. specific gravity 4. surface tension 70. Viscosity has the following dimensions: (a) [ML T2] (b) [ML-1 T-1] (c) [ML T-1] (d) [ML2T] 71. A piezometer tube is used forthe purpose ofmeasurement of: (a) Low pressure (b) High pressure (c) Moderate pressure (d) Vacuumpressure 72. Atmospheric pressure is 1.03kg/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 73. 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 74. 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 75. 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 76. 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' 77. The volumetric change ofthe fluid caused by a resistance is known as: (a) volumetric strain (b) volumetric index (c) compressibility (d) stress 78. 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 79. 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 80. A liquid compressed in a cylinder has initially a volume of 20 m3 at a pressure of 100Pa. 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 81. Match List - I and List - II and select the correct answer using the codes given below: List - I List - II A-15S Badboys2Badboys2 Badboys2
- 162. v2 their usual meanings then the term 2g represents: (a) kinetic energy (b) Pressure energy (c) kinetic energylunit weight (d) None of these (d) c, =~~~C = J4XH(c) v 2 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 (b) Cv=~ 99. 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 flowis said tobe subsonic ifthe value ofmach number IS : (a) Mach No. = I (b) Mach No. < 1 (c) Mach No. > 1 (d) Mach No. = 2 103. Ratio of inertia force to surface tension is termed as : (a) Mach number (b) Froude number (c) Reynold's Number (d) Weber's number 104. 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 105. 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 106. 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 107. The experimental determination of coefficient of velocity is given as : (a) c, =J?;;. 98. 97. 96. 95. An oil having kinematic viscosity of 0.25 stokes flows through a pipe of 10em diameter. The flowwill be critical at a velocity of about: (a) 1.5 mls (b) 0.5 m/s (c) 2.5 m/s (d) I mls 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 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 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 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 94. (b) 4 I (d) ~ (a) 2 3 (c) 4 93. 89. 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 90. 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 91. 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 92. 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 flowin a circular pipe the ratio between average velocity to maximum velocity will be equal to: A-159Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 163. (c) f=~ (d) f=~ Re Re 124. 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 125. 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 126. For maximum power,transmission through a pipeline, the frictional head loss equals: (a) HI 3 (b) HI 4 (c) HI 6 (d) HI 9 127. 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 128. 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 129. 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 121. Which of the following parameter is measured using orifices? (a) Velocity (b) Pressure (c) Flowrate (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: (d) ReO.25(c) ReO.33 (b) ReO.5(a) Re 119. Energy loss in flowthrough 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 : 111. A pivot tube is used for measuring: (a) pressure of flow (b) velocity of flow (c) flow rate (d) total energy 112. 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 113. 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 114. 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 115. Flow through a supersonic nozzle is an example of: (a) Isolated system (b) open system (c) closed system (d) insulated system 116. 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 117. 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 118. 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 5. p2U2 6. 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 3. 4. C. Weber's Number D. Euler's Number List - II pi pu2 U I (gd) U/Jid pLU2/cr P 1. 2. 11O. Match the List - I and List - II and selectthe correctanswer using the codes below: List - I A Froude number B. Mach Number Fluid Mechanics and MachineryA-160 Badboys2Badboys2 Badboys2
- 164. 2 V2 VI V _ 1 (c) Vb =-- (d) b --- cosa cos2 a 148. 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 149. 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 150. 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 151. 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 1 152. 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 (b) Vb = V? cos a 142. Ajet ofwater ofO.002m2 area movingwith 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 ofblades 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 efficiencyof reaction turbine is given by: (b) kaplan turbine (d) Propeller turbine l+cos2 a l+sina (c) 2 (d) 2 141. Which of the following turbine does not requir a draft tube? (a) Pelton turbine (c) Francis turbine l-cosa (b) 2 1+ cosa 2 (a) 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 : NHS/4 NJP (a) JP (b) HS/4 NJP NJP (c) PHS/4 (d) gHs/4 136. Ifthejet 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 : 130. Ifa draft tubeis usedwith 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 speedratio in kaplan turbine is generally kept A-161Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 165. 165. Air vessel is used in reciprocating pumps to obtain: (a) Rise in delivery head (b) Reduction of suction head (c) Increase in supply of water (d) continuous supply of water at uniform rate 166. 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 167. 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 168. For small discharge at high pressure, which of the following preferred? (a) Mixed flow (b) Reciprocating (c) Axial flow (d) Centrifugal 169. The discharge through a double acting reciprocating pump is where N is rpm : (a) Q = 2ALN (b) Q = ALN 60 (c) Q = ~N (d) Q= 2ALN 170. Centrifugal pump is started with its delivery value is : (a) kept 50% open (b) irrespective of any position (c) kept fully closed (d) kept fully open 171. 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)? 172. 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 173. 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 174. 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 175. 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 ALN (c) Qd = 120 (d) Qd = ALN 157. In a centrifugal pump, the liquid enters the pump: (a) at the top (b) at the bottom (c) at the centre (d) None of these 158. 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 159. 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% 160. 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 161. 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° 162. The vanes of a centrifugal pumps are usually: (a) curved forward (b) curved backward (c) radial (d) None of these 163. 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 164. 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) d - 60 153. 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: Fluid Mechanics and MachineryA-162 Badboys2Badboys2 Badboys2
- 166. (a) 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 velocityof 1.5mls is: 3n 3n (a) 50 (b) 10000 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 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 (x- y) 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 x y3- x2y, V = xi -~v". Then, the value of 'A' 4 for the possible flow field involving an incompressible fluid is : 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 (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 : 1r --:-:Vessel H ~_~_~ Liquid --------- -_.---------- .- -----------_.------------ Manometer 176. A centrifugal pump will start delivering water only when the pressure rise in the impeller is equal to or greater than the: (a) manometrichead (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= 5HP 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-163Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 167. 34 2 List - II ML2T-3 ML-l T-2 ML-lil ML2T-2 ML2T-l (c) 8 205. 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 A Dynamic viscosity 1. B. Moment of momentum 2. C. Power 3. D. Volume modulus 4. of elosticity 5. Codes: ABC D (a) 3 5 2 (b) (c) 1 2 3 4 (d) 2 4 5 1 (d) 28 8 (b) 7(a) 8 8 (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 (b) GM=_!_+BG V V I I (c) GM= BG-- (d) GM=-xBG V V 201. For irrotational and incompressible flow, and, if <I> = velocity potential, 'V = stream function, then which of the following is correct? (a) V2<1> = 0,V2'V = 0 (b) V2<1> *- 0,V2'V *- 0 (c) 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 effectbe 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 : 2d >1IE IE ;;.1 1 d (c) ~2g(H + h) (d) ~2g(H - h) 196. 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) 1and 2 (d) 3 and 4 197. 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 198. 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) 199. 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 : (b) fiijl(a) ~2gH H 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: Fluid Mechanics and MachineryA-164 Badboys2Badboys2 Badboys2
- 168. (b) 3v (d) 5v is : (a) 0.45 (b) 0.15 (c) 2.0 (d) 1.18 226. A series ofnormal flatvanes 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 (c) 4v 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) vanepump (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 specificspeed 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 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 415T 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) Tubularturbine (d) Turbo - impulse turbine 214. Consider the followingcomponents ofa 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) 1and 2 (c) 1, 2 and 4 (d) 1, 2 and 3 A-165Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 169. 18 16 13 8 6 3 ~Nu (a) kaplan turbine (b) Pelton turbine (c) Francis turbine (d) Propeller turbine 228. 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 229. 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 230. Specific energy is minimum at a depth of flow is known (a) critical depth (b) normal depth (c) sub-critical depth (d) alternate depth 231. 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 232. 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 - 12J.l2ii(a) 12J.lu (b) d2 d 12J.l2ii - (c) (d) 6J.lu d2 d2 - 236. It u = average velocity,L= length ofplate, J.l= coefficient of viscosity for the viscous low between two parallel plates, the pressure drop per unit length is equal to : Area (a) (wetted perimeter) 2 (b) wetted perimeter area (c) square root of area Area (d) wetted perimeter 233. If h, = atmospheric pressure head h,= vapour pressure head h,= suction head h = working head of turbine then cavitation factor is defined as : ha +hv +hs ha - hv - hs (a) h (b) 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 : A 227. Theunit discharge,Quand unit speed,Nucurvefora turbine is shown is shown in figure, then the curve 'A' shows: Fluid Mechanics and MachineryA-166 Badboys2Badboys2 Badboys2
- 170. au Since, az = 0 : for 2-dimensional flow. au au :. Convective acceleration = u ax + v ay . we get, -x2t = - x2t + f'(y) Since, f(y) = c => '" = -x2 yt +C c is a numerical constant taking it zero. '" =-x2 yt For the equations of streamlines, '" = constant :. -x2yt = constant For a particular instance, x2y= constant 4. (a) In a two-dimensional velocityfluid with velocitiesu, v along x and y directions. Acceleration along x direction ax = aconvective+3temporalorlocal au au au au=u-+v- + w-+v- ax ay az at'------v----' ~ convective acceleration temporal acceleration From equation(ii)putting value of alfl in equation(iv) ay Uo Here, uav = 2 ... (ii) From Eqs. (i) and (ii), 16J...luoL ~p=---=-- D2 3. (a) Given u x2t and v = - 2xyt We know, : = v = - 2xyt ...(i) a", 2 and - = -u = - x t ... (ii) ax Integrating Eq. (i), we get '" = x2yt + f(y) ...(iii) Differentiating Eq. (iii) w.r.t. y (iv) we get 8Jf =-x2t+f'(Y)ay ...(iv) 1. (d) For two-dimensional flow, continuity equation has to be satisfied. au Ov 00_ Ov ax +v ay = 0 => & - - ay 2. (d) Pressure drop across a straight pipe of length L is given by 32J...lvavL ~p = (i)D2 ... 1 ••• , HINTS & EXPLANATIONSIIIIII~ 121 (c) 141 (a) 161 (c) 181 (a) 201 (a) 221 (c) 122 (a) 142 (b) 162 (b) 182 (d) 202 (b) 222 (a) 123 (b) 143 (b) 163 (b) 183 (c) 203 (a) 223 (d) 124 (a) 144 (c) 164 (d) 184 (c) 204 (c) 224 (a) 125 (b) 145 (a) 165 (d) 185 (a) 205 (b) 225 (c) 126 (a) 146 (d) 166 (a) 186 (b) 206 (a) 226 (a) 127 (b) 147 (a) 167 (c) 187 (d) 207 (c) 227 (c) 128 (c) 148 (a) 168 (b) 188 (c) 208 (c) 228 (a) 129 (a) 149 (b) 169 (a) 189 (a) 209 (a) 229 (b) 130 (c) 150 (c) 170 (c) 190 (b) 210 (a) 230 (a) 131 (d) 151 (d) 171 (b) 191 (c) 211 (b) 231 (a) 132 (b) 152 (a) 172 (a) 192 (a) 212 (d) 232 (c) 133 (a) 153 (b) 173 (a) 193 (a) 213 (c) 233 (b) 134 (a) 154 (d) 174 (c) 194 (b) 214 (b) 234 (b) 135 (b) 155 (a) 175 (b) 195 (b) 215 (a) 235 (d) 136 (a) 156 (a) 176 (a) 196 (b) 216 (a) 236 (a) 137 (b) 157 (c) 177 (a) 197 (a) 217 (c) 138 (a) 158 (a) 178 (d) 198 (d) 218 (a) 139 (d) 159 (c) 179 (a) 199 (c) 219 (b) 140 (a) 160 (d) 180 (b) 200 (a) 220 (a) A-167Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 171. v-u (y)27. (c) ~ = 5.75 10glO k + 3.75 for v= u 5.75 + 10glO(f) + 3.75 = 0 .1_ = 0.223 => y = 0.223 R R r So, - is independ at then (t = r where (is on stat) E At center r= 0, t =c x 0 =0 Zr wr t=-- D ~ =~x twxr => TW = ~PD L r D 4L 21. (d) u=x+2y+2 v=4-y au Ov -+- = I-I =0 ax Oy As it satisfy the continuity equation for incompressible flow so this is incompressible Wz= H:-:]= ~[O-21 =-1 and since rotational component is not zero so flow is not irrotational. 24. (a) Stagnation head = 20 m Statichead = 5m Dynamichead=20-5= 15m u2 Now - = 15, 2g u=17.15m/s 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 ~P Zr --- L 6L Neither P nor I depend as on r. Pl~ D~ x 12. (b) A streamline and equipotential line in a flowfield are perpendicular to each other (because, when (slope)1 and (slope), are multiplied and we get, (slope), x (slope), = -I 13. (a) Assumption: a (i) Flow is steady y (i.e.) at D = 0 (ii) Fully developed the ~ D- 0 ; properties are not at changing in the direction of the flow. 2.93 kl/kg 990xl000m p m -dp p Specific work done W = Vdp = W = dp (30-I)xI05 11. (d) ... (i)Th Powe~ Powe~ us, 1 1 (HeadI)·5 (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 In case of two fixed parallel plates, when the flow is fully developed, the ratio vrnax/ vavgis given by vrnax =~ vavg 2 Thus, vavg= (2/3) x vrnax.= 2 x 6/3 = 4 m/s p] = I bar, P2 = 30 bar p = 9900 kg/nr'. 10. (c) Heat convection Nusselt number 8. (b) The relation between head and power is given by Pu = Power/Head 1.5 Pipe flow Froude number Skin friction coefficient Reynolds number Free surface flow Boundary layer flow dy Now tOC- dy dx oc-- dt dy [dx / dy] oc"":""'_-----'- dt (dx/dy) where, = rate of change of shear strain. dt 6. (d) Continuity equation v· v=O au Ov Ow => -+-+-=0 Ox Oy az Multiplying by density on both sides, we get, p(au + Ov + OwJ= 0 Ox Oy az This is the equation for compressible flow. 7. (b) Compressive flow Weber number 1 rl 1------+1 V + dv ~ fluid flow y -y I------------W 5. (c) ForNewtonian fluid, dy Shear stress a: dy dy . . where, dy = velocity gradient Fluid Mechanics and MachineryA-168 Badboys2Badboys2 Badboys2
- 172. 15000 So U = 47.75 em/sec. , (~)x(20)2 ud 20x47.75 Now Re= - = = 424.4 , v 2.25 vd Re=- v = 92.69 x 10-3 N 45. (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 2 1226(6)2 = 2 x 1 x 1 x 2.1 x 10-3 x --- ~p For 'to to be maximum Lshould be maximum so V should be maximum. In laminar flow, maximum velocity will be pVD attained when -- = 2000 M ~p 32MX 2000M2 64000M2 L pDoD2 pD3 16000M2 'to = pD3 VL 6xl 42. (b) R, = --;- = 0.15 X 10-4 4 x 105 Hence the boundary layer is laminar over the entire length of the plate. rx; lx 0.15 X 10-4 6= 5~lx0.15xl0-4 = 5 ..{~ = 5 6 = 7.91 x 10-3 m = 7.91 mm 6xO.5 Rex= 0.15 X 10-4 = 2 x 105 (for middle point of plate) - 1.328 _ 1.328 -2 1 10-3CD---- - x ~RcL ~4x 105 . 6V2 FD= 2 x 1 x 1 x Cf-- 2 dM -=0 d6 M= W 7td2V; g 4 from continuity equation = ~D2V =~d2V Q 4 '4 2 IfH is the head causing the flow, then V2 6LV2 H= _2 +--' 2g 2gD w 39. (b) Momentum of issuing jet is M = -QV2 g = 0.02036 kg/rrr' 129.3) d/22 4 T = J 7tMCOr3dr = M7tdco o h 32h I · di du m/Ve ocrty gra lent = - = 0.25 sec meter dy Kinematic Viscosity, U = 6.30 x 10-4 m2/s du du Shear stress = fl - = pv - dy dy = 129.3 x 6.30 x 10-4 x 0.25 (Asp = 31. (b) ~:" 1('",,:,,I,,:, :~ ,r,"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 dy assuming that gap h is small so that velocity distribution may be assumed linear du rco dy h rro 27tMCO dT= M- 2m2dr = -- r3dr h h dr 29. (a) Consider an element of disc at a distance r and having width dr. A-169Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 173. 74. (b) 72. (d) 70. (b) =~=0.6 1000 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 From Newton's Law of viscosity, (J.!) Shear stress (r) = J.!x velocity gradient du 't = J.!x- dy 't N J.!= du =-2 _ m dy S J.!= J:£_s = [MLr2][T] m2 [L2] J.!= [ML- 1 T- 1] 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/ern- (fauge pressure (Pg) = 21 bar Atmospheric pressure (PatIn)= 1.013bar Absolute pressure (Pab)= 't Pab= Pg + Patm = 21+ 1.013 =22.013 bar 69. (d) density of wood piece Specific gravity of wood = d . f ensity 0 water 62. (b) Let H = height of water column, H _ _E__13.6x103 x60x10-2 x9.81 then, - pg - 1000x9.81 =8.16m = 816 em 65. (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 closedto 63765 Pa' 67. (b) Weight of wood piece (m) = 5 kg Weight of wood piece = weight of liquid displaced S= 60% ofvolume(v) x 1000 =~xl000 100 S = 600 v 513 v=-=-m 600 120 density of wooden piece (p] = Mass(m) Volume(v) 5 3 =-=600kg/m 1 120 0.14 x 9.81x 10000 700 x9.81 1400 = 700 = 2 m of oil. 58. (b) 57. (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 Given: specific gravity of oil = 0.7 Pressure (p) = 0.14 kgf I em- Density of oil (Poil) = 0.7 x 1000 = 700 kg/m' H=_E_= 0.14x9.81 pg 700x9.81 2 On integrating, f ajl = -2x =>jI = - 2x + C (y) ax z ajl - = 0 + C' (y) ay f c'(y) = f 2y 2y2 _2x2 2y2 C (y) = -2 +cl> lthen, 'I' = --+-+Cl 2 2 jI = y2 _ x2 + c1 or y2 - x2 + constant 49. (a) If flow in 2D, continuity equation becomes, ay av -+- =0 ax ay So, for (i), u = x2 cos y, V = - 2x sin y 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 R, = 424.4,means Reynoldsnumber ofthis flowisless then2000(424.4<2000) Hencethe flowis "Laminar" 48. (a) Given here, <I> = 2xy, considering the following relation, -a<l> ajl =>4= -=-- ax ay ajl (a<l> ajl'1 _ a<l> => - - I -- -) -- ax -" ay' ax ax a<l>= ~(2xy) = 2y = _ ajl ax ax ay Similarly, a<l>= 2x = _ ajl ay ax Fluid Mechanics and MachineryA-170 Badboys2Badboys2 Badboys2
- 174. 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' fluidviscosityru) = 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, U =2V =2x5=10m/smax. avg. 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 ofpipe (L)= 1000m Considering the following formula for head loss, H _ 4fLv2 L - 2gd for discharge (Qd) =Area x velocity 2 2000= vxl0x10 =?V= 2000xO.25 =0.5m/s 0.25 IOx102 96. (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 Re= vD = vD Il v P 1 = +- 2 94. (b) Given: kinematic viscosity (v) = 0.25 stokes diameter ofpipe (D) = 10em for a critical velocity, Reynold's number should be between2000and 4000 . ( PvD Reynold'snumber Re) = -- Il Vavg. Vmax. ap (Rf -R~) 93. (a) Average velocity (Vavg) = ax 81l ap(Rf -R~) Maximum velocity (Vmax) = ax 41l = dP =~=50Pa dv 20 vI 20 82. (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 86. (b) Given: Side ofthe cube~a)= 5 em = 5 x 10-2m Volumeofcube (v) = (a) =(5 x 10-2)3 = 125x l0-6m3 Buoyant force acting on the cube =v.pg = 125x 10---6 x1000x 10 (Assumingg = 10m/s2) = 125x 10-2 = 1.25 N 75. (a) Given: diameter ofpipe (D)= 20 em kinematic viscosity (v) = 0.0101 stoke Reynold'snumber (Re)= 2320 Let v = velocity of flowing water, Re = vD = vx20 v 0.0101 2320 = v x 20 =? v = 2320 x 0.0101 0.0101 20 v = 1.1716 em/s ~ 1.117 cm/s 76. (a) Given: Mass ofliquid = 5 tonnes = 5 x 103 kg volume= 10m3 . .. mass of liquid mass density of liquid (p) = ---~- volume = 5x 103 = 5 x 102 10 = 500 kg/m' 78. (b) Given: diameter of glass tube (d) = 3 mm surface tension (crT)= 0.0736 N/m contact angle for water (a) = 0° 4crTcosa Then capilary rise (H) = wd 4x 0.0736xcos 0° 4 x 0.0736 x 1 kg-f xd 9.81x 10-6 x 3 = 10.4 mm (approx.) 80. (c) Initial volume(vI) = 20 m3 Initial presssure (P1) = 100Pa Final volume(v ) = 40 m3 Final pressure (p2)= 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 lumetri t .oume fIC s ram A-171Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 175. 180. (b) 179. (a) Difference of mercury level (H) = 10 mm = 10 x 1O-3m 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 = 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. 2 Pabs 4tLv HL (head Loss) = pg = 2gd 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 Ql Nl 1200 1000 -=-:::::>--=-- Q2 N2 Q2 1500 Q _ 1200 x 1500 2 - 1000 1800 m3/s = 0.679 u _ 1tDN where, - 60 k= 1tDN 60~2gH 0.44 = 3.14 x D x 600 60·J2 x 9.8 x 300 0.44 x 60.J2 x 9.8 x 300 D=-------- 3.14 x 600 = 1.07 m or 1.08 (approx.) 159. (c) 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 1000 x 9.8 x 0.013 x 32 1000x 6 149. (b) Given: wheel speed (N)= 600 rpm Speed ratio (k) = 0.44 net head (H) = 300 m Speed ratio (k) = b,,2gH P xl000 3000 x 1000 pgQdH 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 1 20 Number of buckets (Nb) = 15+2m = 15+2 = 15 + 10 = 25 136. (a) Given: jet ratio (m) = 18 Number of buckets = 15 + 0.5 m = 15 + 0.5 x 18 = 15 + 9 = 24 138. (a) Given: Power developed (P) = 3000 kw Head (H) = 5 m discharge (Qd) = 75 m3/s Overall efficiency of turbine ('10) = ( p ) pgQdH 1000 6 3 2x6 --=-:::::>vmean =--=4m/s vmean 2 m 134. (a) Given: mean diameter of runner (DQ1)= 200 mm Least diameter of jet (dL) = 1 em = 10 mm . . ( ) Dm 200 jet ratio m =-=-=20 dL 10 005=Axv= =~d2 xv. 4 0.05 = ~(0.2)2 xv 4 4 x 0.05 v= 1.59m/s 1tx (0.2)2 N H dl (H) 4xO.0025xl000x(I.59)2 ow, ea oss L = 2xl0x0.2 = 6.32 m (which is near to 6.44 m) 4fL V2 125. (b) Head loss (HL) = --- 2gd H ocv2L HLI vr HLI (v)2 1 --=-:::::>--=--=- HL2 v~ HL2 (2v)2 4 HL2 = 4 X HLI = 4 times 127. (b) Given: MaximuI? velocity (vmax) = 6 m/s We know that the Ratio Maximum velocity (vmax.) _ i Mean velocity (v mean) - 2 vmax. =i vmean 2 Fluid Mechanics and MachineryA-I72 Badboys2Badboys2 Badboys2
- 176. 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 ....(i) 20 u=-nd 3 ndN rtd x 400 Speed of wheel (Il) = -6-0 = -6-0- 8 8 1 [7817]8 [Y]o- 81/7 S"y 0 1 [7 8/7] 7=8- 81/7 S"x8 =8-S"8 Given: height (H) = 10m flow rate (Q) = 11m3Is Losses due to friction and others (HLf) = 5m Pumping power (p)= PgQ(H+HLf )kw 1000 = 1000 x 9.8 x O.l (10 + 5) kw 1000 P = 14.7 kw For maximum efficiency, jet velocity = 2 x wheel speed =2v Given: Head (H) = 405 m speed (N) = 400 rpm ~ = 0.45 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 x2h= wnd h 4 2 FA 2whnd2 = 4 FB wnd2h 2 Given: Velocity distribution is given as : ~=(x..JI/7 UO 8 displacement thickness (8*) is given as : 228. (a) 226. (a) 224. (a) 204. (c) y2 pgH = E..___!!_+pg(H - h) 2 y2 pgH = E..___!!_+ pgH - pgh 2 y2 pgh = E..___!!_ 2 2gh =v~ YB =figh 199. (c) Case I : When container placed with its large diameter down ward: then force (FA)will be : 3n 3 =0 0003nm3/s = --m Is . 10000 195. (b) Considering Bernoulli's equation, between section (A) and section (B), PYA 2 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 Area (A) = ~d2 = ~(0.04)2 4 4 =0.0004nm2 discharge (Qd) = A x vavg. =O.0004nx 0.75 190. (b) 3 4 Given u = ')..xy3 - x2 y, v= xy2 -"4Y For the case of incompressible flow, au+av=o ax ay au 3 au 3 ax = ')..y -2xy, ay = 2yx = 3y ')..y3- 2xj + 2yx - 3y3 = 0 ')..y3 - 3y = 0 y5 (')..- 3) = 0 ')..-3=0 ')..=3 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 vrnax 1.5 vavg. =-2-=T=0.75m/s 187. (d) =dxgxHxA =dgHA 181. (a) Given: kinematic viscosity of air = 1.6 x 10-5 m2/s considering the following relation, Ila: (T)1I2 ...(i) 1 p o: T ...(ii) VOC (T)3/2 we get, kinematic viscosity at 70° C = 2.2 x 10-5 m2/s A-173Fluid Mechanics and Machinery Badboys2Badboys2 Badboys2
- 177. Steel Alloys: Along with the carbon, all steels may be alloyed by mixing some other elements in various proportions to improve followingmost common properties of steel. Some of them are given below: (a) Toimprove 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. Chromium It promotes hardness, toughness and corrosion resistance. Silicon Improveselasticity,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. Effect on Steel ElementAlloying steel is slightly better than that of semi-killed steel. Effect ofAlloying Elements on SteelIron 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 saysteel is a mixture ofiron 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: (a) Dead Mild Steel - below 0.15% carbon. (b) Mild Steel (low carbon steel) - carbon from 0.15%-0.3%. (c) Medium Carbon Steel- 0.3%-0.8% carbon. (d) High Carbon Steel- 0.8%-1.5% carbon. 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 steelsare 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 de- oxidised. 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 1)llf)I)'Jf~rl'If)~ 1~~f.I~I~I~11I~f. Badboys2Badboys2 Badboys2
- 178. 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 USED IN ENGINEERING PRACnCE 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 ofheat 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 ofheat 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). (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. (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-175Production Engineering Badboys2Badboys2 Badboys2
- 179. 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 ofmetal. Italso 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 orjust below the A 1eutectoid 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 carbidesizeacceleratessperoidization, 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 slowlyto a temperature just belowthe 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 exactamount 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 (e) Toimprove surfaceproperties (surface hardening, corrosion resistance-stabilizing treatment and high temperature resistance-precipitation hardening, surfacetreatment). (t) To improve cutting properties of tool steels (hardening and tempering). (g) To improve magnetic and electrical properties (hardening, phase transformation). (h) Toimprovemachinability(fullannealing andnormalizing). The process of heat treatment may be classified into following types:- (a) Annealing: Annealing isbasicallyknown as metal softening process in which the metal heated upto its critical temperature or 30°-50°C aboveits 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) Toimprove machinability. (d) Tofacilitate further coldworking byrestoring 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 Production EngineeringA-176 Badboys2Badboys2 Badboys2
- 180. 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 conductivematerial 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 ofhardening 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 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°C- 350°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 temperingprocess, 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. (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. Itprovides equal advantage for both carbon-alloy steel and other alloys steel. It can be used for A-177Production Engineering Badboys2Badboys2 Badboys2
- 181. Metal casting is a process in which molten metal is poured (in liquidstate)into a mould.There moltenmetal 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 mostpopular casting method was "Sand Casting" in which desired shape article is pressed in to sand and when the article removed from sand it leavesan impression or cavityin the sand. Which is exactly according to the shape of article after removal of article molten metal is poured in this cavity formedin the sand. The article used to make cavity in sand is known as pattern and the cavitymade in sand isknown as mould. Advantage of Casting 1. Casting is a cheap,fastand economicalmethod ofproducing any shape ofmetal and Non-metals. 2. Large and heavy structures can be made easily by casting method. 3. Foridenticalmassproductioncastingisverysuitablemethod. 4. Duetoproductionofminimumscrap,wastageofrawmaterial isminimised. 5. Complex shape can be made easily by casting method with low production cost and in less time investment. 6. Casting is suitable formetal, non-metal and alloys. 7. Insertionof anyobjectofsamematerial ordissimilarmetal is easier in casting method. 8. Somemechanical propertiesachievedin casting processare distinct from any other manufacturing method. Someimportant terms (A) Mould. (B) Pattern (C) Core. (A) Mould: Itmay be defined as a shape made up of sand, Die Steel, Ceramic, and rubber etc. in which desired cavity is produced with the help ofsuitable pattern. According to the material used, in making cavity,the material can represent the mould's name like,ifsand is primematerialthen it willbe known as sand mould, and rubber mould if rubber is prime material in masking mould. Sand mould may further be classified in followingtypes : METAL CASTING part design, maintain proper hardness and doing tempering on proper timing. (c) 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. (d) 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. (e) Shape Distortion: This defect caused by non-uniform heating. This can be controlled by heating gradually up to specified temperature. (f) 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. 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 flameand 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 ofmetal. 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 lowingthe 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 super- saturated solid solution. This process results in increasing hardness. This one of the reason due to which this process is known as precipitation hardening also. 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 byimproving Production EngineeringA-178 Badboys2Badboys2 Badboys2
- 182. r-Cope Pattern 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. :':;c pattern ,...,..... ~g pattern f ~~."() ) ( ) 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) Splitpattern :- 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. Patterns may be classified according to the following factors: (a) Shape and size and casting (b) Number of casting to be made (c) Method of moulding to be used (d) Parameters involved in the moulding operations (i) Solidpattern :- (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:- Itis 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) Vacuum Moulding:- (V-Process) is a variation ofthe 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. 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 permeability. 4. No - Moisture generated defect. S. Better life of pattern because sand did not touch the pattern surface. Disadvantage: 1. Lowers, the production rate. 2. 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 A-179Production Engineering Badboys2Badboys2 Badboys2
- 183. (iii) Machining Allowance: Itis also known as finishing 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 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) Yellow ~ 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) Itshould 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) Horizontal Core (ii) Vertical core (iii) Balanced core 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 ofwithdrawing the pattern from the sand mould. Itmay 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. ._______ Drag Pattern Fig. (cope and drag pattern) (iv) Copeand 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 Production EngineeringA-ISO Badboys2Badboys2 Badboys2
- 184. where, A3 = area of sprees at bottom Al = area of sprees at tope Somemost important formulas used: (a) Time taken to pour (t ) Volume of mould cavity Ag~2gH A3 ~ (b) Aspiration effect: A2 = ~h; (c) Solidification time: 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 - = - +h = constant pg 2g 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 Pouring time (t) = Ag ~2gH Where, A_g_= area of gate Design or Sprees: Area of ratio (R)= ~ = ~: Riser is a cavity made in mold to compensate the shrinkage arises 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 RISER AND GATING DESIGN 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 Solidification Properties (i) Fluidity: The ability of filling all parts ofmold 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 ofInocculation: 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. SOLIDIFICATION AND COOLING (iv) hanging over core (v) 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. A-1SlProduction Engineering Badboys2Badboys2 Badboys2
- 185. 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 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 - DieCasting: - 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 ofimpurities 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 re- use to make many mold. This process is mostly used for casting carbon steel, alloy steel etc. 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 ofmould fills properly. This can be checked by rising level of molten metal in the risers. (iv) Cooling: This process involves cooling of molten metal 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 ofSand 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. a (f) Caini's formula: R p = -- +c RY-b where, a = Freezing characteristic constant b = Contraction ration from liquid to solid c = Relative freezing rate of river and casting where, c = constant, V = volume, A = surface area (A/V) casting (d) Relative freezing time (Rp) = (A IV)river (R) Vriver (e) Volumeratio V = V . castmg Production EngineeringA-182 Badboys2Badboys2 Badboys2
- 186. 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 bejumped 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. ~~~ force application C = Specific heat of molten metal 8p = Molten metal pouring temperature 8f= Cooling temperature of metal 80 = Initial temperature of mould (i) 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 die- and 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 (a) Up setting: In this process cross-section of work piece is increased with corresponding reduction in its length. Cm = Specific heat of mould K a = Thermal diffusityofmould = - pc Where Pm = metal density P =densityofmolten metal L = latent heat ofliquid metal. .: K = Constant ( J 2 Volume Solidification time o: S ~ Ar unace ea HEAT FLOW RATE DURING SOLIDIFICATION Heat flows from the hoter portion to cooler portion ofthe casting. Rate of heat flow per unit Area ~ 1=-k ( :) ky/ hrm' 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 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 mis- run. (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. Itmay splash the small particle separated from the main stream during the spray and thrown ahead and solidified quickly to form the shots. A-183Production Engineering Badboys2Badboys2 Badboys2
- 187. The process ofextrusion consists of corresponding a metal inside a chamber to force it out through a small opening calleddie.Anyplasticmaterial canbeextrudedsuccessfully. Most of the process used for extruding metals are hydraulically operated horizontal presses. A large number ofextruded shapes are in common use, such as tubes, rods, structural shapes and lead coveredcables. The principle of operation are the same forboth 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.billetsof 125mm to 175mm in diameter and 300to 675 mm lengthare in generalusedasrawmaterial forextrusions of steel needs adequate lubrication around the billet. This is done by providing a coating of fine glass powder over the surface ofhot billet. The process ofextrusion 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 forwardextrusion (2) Indirect or backwardextrusion Direction Extrusion: As shownin the figurebelow,Inthisprocess billetofrawmetaltobe extrudedisheatedtoitsforgingtemperature and forced in the machine chamber, this forcepush forward the billet and billet passed through the die. The length of extruded HOT EXTRUSION (i) b..tt VIA9Id (ii) scerf VIA9Id (iii) V-I9Id Differentweldingjoin and borax formild 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 ofiron is checked. There are three types ofweldedjoints in commonwhich 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 lapweld.Itisknown so for the reason that the ends of the metal pieces to be joined are made to overlap each other and then hammered.Thusthe weldformationis atan 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: Itis also known with somany names e.g. split, spliceor fork etc. Itis employedwhere a highly strong weldedjoint isneededparticularlyin 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. (b) Drawing out or drawing down: This process is exactly a reverse process to that ofup setting orjumping in the sense that, contrary to the latter, it is employedwhen a reduction in thickness or width on is desired with a corresponding increase in its length. In this process specificshape oftools also required to achieve the desire shape, known as pair of sewageand fullerthe selectionofthe abovetoolsisgoverned 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 coolingofthe unwanted length fordrawing by sudden quenching in water. Ifthe bar is ofrectangular or square cross section it is laid flat on the anvit face and hammered bythe peen of cross-peenhammers bythe 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. (c) Bending: Bending of bars, flats and other simillar stock material is usually done in smithy shop, this can be doneto produce different types of bent shapes such as angle, ovals and circles etc.Anydesired 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 carriedout 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. (d) Setting down: Inthe operationthrough which the rounding ofa comer is removed, to make it square, bymeans ofa set hammer. By putting the face of the set hammer over the round portion, formedby 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 surfaceisremovedbymeans ofa flatteror a sethammer 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. (e) Forge welding: In this processtwopeiceof simillar metals are heated properlyup 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 bejoint (ii) Rising the temperature of the prepared ends to the correctweldingheat. The surfacestobejoined together should be quite clean i.e., they should be free from scale, dirt or ash. Otherwise this presence will lead to thefailureofthejoint. Inadditiontothisa fluxisapplied on the hot metal which helps in overcoming the above difficulties. This flux usually stand for wrought iron Production EngineeringA-184 Badboys2Badboys2 Badboys2
- 188. Classification fo Welding According to Application of Pressure (a) Non-pressure welding/fusion welding: In this process of weldingthe temperatureofjoining edgeofmetalsare heated 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 ofGold and Silver onlytill 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 steelwith 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 meltingpointthan the parental metals. For example copper and brass, mild steel and cast iron etc. WELDING The most common coldextrusionsprocessis impactextrusion, 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 formhaving 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 surfaceofthe punch, forming a cup shapedcomponent.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 componentsper minute. Mostlywall thickness produced from 0.7 mm to 0.1 mm but onlysoftand ductilematerial canproduced bythis processlikelead,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 12.7mm diameter and± 0.127 mm upto 25 mm dia can be easily obtained. COLD EXTRUSION 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 projectedpart to apply pressure and form the head. The bending operation may be completedin a single or two strokes. Automatic machines for producing bolts and screw are also available in which all the operations like cutting stockto size, shank extrusion, heading trimming and threading etc. are performed simultaneouslyto produce finished components. The processis also successfullyadoptedforproducingrivetsand nails. 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. COLD FORGING Fig. BackwardExtrusions Advantage and limitations of hot extrusion 1. Dueto application ofhigher pressure a verydense structure is produced. 2. Bettersurfacefinish isproducedhaving higher dimensional tolerence. 3. Low tool cost involves and fast in production rate. 4. Most suitableforproduction ofparts having uniform cross- section having fine surface finish and high dimensional accuracy. 5. Excessesivelength object is creak problem in handling the extrudedrod during extrusion. Fig. Forwarded extrusion process It is a usual practiceto leavethe last nearly 10%length ofbilletas un-extruded. This portion is known as discard which contains the surfaceimpurities ofbillet. Indirect Extrusion: As shownin followingfigureram orplunger 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 methodin compressiontothe 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. 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 overhanging extruded length is fedis to a long support calledthe run out table. A-1S5Production Engineering Badboys2Badboys2 Badboys2
- 189. 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 appliedsufficientpressurewhichmakecompletethe 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 veryimportant 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 cycletime also. This cycletime 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 circuitbreaker,electroniccircuitetc. 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 canmake 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: Themachineusedin spotweldingprocess consists ofone fixedelectrode and other movableelectrode which is mounted on a rocker arm and moved in up and down direction by mechanical arrangements. In some machinemechanicalarrangementispoweredwithhydraulic systemto 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 compressedair. (c) Portable Guns: In manyplaces it is notfeasibletotransport hence forthat purpose a portable machine is required. This Portable Gun carries two electrodes and the transformer is supported generally on overhead rails. Mainly it is used in automobile industry. (B) Seam Welding Itis series ofcloselyspacedor single line spotwelding. The weld shape for individual spot may be of any shape like round or rectangular. In this process, two circular disc shaped electrodesFig. Spotwelding Movable +-- Electrode Spot 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. CLASSIFICATION OF RESISTANCE WELDING up tomeltingpointandwhenit startstomeltthefillermaterial is filled between joints. For example-Gas welding, Arc welding,Electricbeamwelding 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-fillermaterialisusedcommonlyinthereweldingprocess. For example-Forgewelding, Resistancewelding 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 saltstogether.For exampleiron oxide and aluminium powder produced heat by chemical reaction. This method of heat generation are employed in forge welding, gas weldingand 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) 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 electronswilljump fromone terminal to another.This is calledjumping ofelectrons and due to this arc a high temperature generates at bothterminals which is about3700°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 methodheat isproducedby use of high frequency current to produce sufficient eddy current in the workpiece to be weld. (c) Mechanical method: This methodis rarelyused inmodem practice because the heat production in this method is very low as compared to energy applied as heat produced by friction or heavyblow/impact load etc. Production EngineeringA-186 Badboys2Badboys2 Badboys2
- 190. This method of welding involves the use of stored electrical energy either in reactors, capacitors or storage batteries etc. 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. Fig.Resistance flash welding 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. 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. 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 Fixed clampMovable clamp Work Piece 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. Fig.Resistance Butt welding Fixed clamp Work Piece /-:Movable clamp i 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. Seam welding may be further classified as circular type, longitudinaltype, universaltypeandportabletype. (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. (D) Resistance Butt Welding 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 bewelded. After properly meeting ends of metal ,the current is switched to till corresponding ends of metals are reached to the fusion point. Continuous line of welding joint 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. A-1S7Production Engineering Badboys2Badboys2 Badboys2
- 191. 16. 15. 14. 13. Fig. Different weaving styles As per requirement of joint, there are different weaving styles as shown in above figure. Arc length: Arc length may be defined as the distance of 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. 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. 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. Hard facing: Hard facing may be defined as the process of hardening the surface by welding process. 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. 12. Motion of Electrode 11. 10. 9. 8. 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. Arc crater: Itmay be defined as the penerat ions of arc in base metal, it depends upon arc length, electrode width and thickness of base metal. Spatter: Molten metal dispersed around the welding beads in small drops form is known as spatters. 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. Edge preparation: For making different types ofjoint, some 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. 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. 7. 5. 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. (i) Straight polarity: Work piece made positive terminal and the electrode is made negative terminal, it is used for more thick plates etc. (ii) NegativepolaritylReverse polarity: Workpiece is made negative terminal and electrode is made positive terminal, it is used for thin plates welding. 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 blow: Itmay be defined as the deviation of arc during kW kVA Current used Current supplied Power factor = 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. 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 Terms Related with Electric Arc Welding 1. 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. 2. 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 = Cathode drop + volumn drop +Anod drop i.e. V=V +V +Vepa 3. 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. Power factor: Itis the relation in between the current used and total current supplied to machine. ELECfRIC ARC 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. MUL TI IMPULSE WELDING Production EngineeringA-ISS Badboys2Badboys2 Badboys2
- 192. S.No. 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. 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. 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. 8. Arc length Maintaining small arc is difficult, only Maintaining of small arc is easier than iron powder electrodes are exceptional. A.C. Arc welding. 9. 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 Properties of A.C. and D.C. Arc Welding 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. 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. Carbon arc welding is mostly for steel sheet and casting etc. Electrode holders consist of magnetic coil which guide the Arc. This welding process is operated manually or by machine or both. A-189Production Engineering Badboys2Badboys2 Badboys2
- 193. controlledbyextremelymountedscrew.It consistsofa small storage chamber due to which the out going pressure is out ofeffectofpressure 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 usedin lowpressureacetylene 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: Itis used in low pressure acetylene generator system. It is mounted between welding torch and acetylene generating cylinders/tank. Important Applications: (i) Itreduces the back fire hazards (ii) 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/refractorybrisk etc. Welding torch lights: Itis an instrument which produces spark usedforlighteningweldingtorch. Inpractice,electronicgas lights are commonly used other gas welding equipments are welding goggles, apron, gloves, and wire brush etc. Gases used in welding process: 1. Oxygen (02) Itdoes not go through combustion itself but very helpful in combustion process with different gases. It isstoredinmetalliccylindersatabout120 kg!em?in liquefied state. Itis prepared by following two methods mainly (a) Byliquefication ofair (b) By electrolysis of water 2. Acetylene (C2H2): It is highly inflamablegas and produces about3600°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 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 formixing oxygenand combustiblegas (acetylene etc.) in required proportion and injecting for combustion and making flame or wemay say that with this equipment we can acquire an adequate mixed proportion of oxygen and acetylene (in oxy- acetylene gas welding) to develop a suitable flame for welding Classification: (a) According to pressure of acetylene gas (i) High pressure welding torch (ii) Lowpressure welding torch (b) According to number oftips used with torch (i) Singletip 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 weldingtorch. These aremade up ofrubbercoating overthreaded net pipe. It should have sufficient strength, light in weight, economical, and non-reactive with gas which they tend to carry. These are fixedwith welding torch with the hose pipe clamp. Pressure regulator: Itis a pressure controlling devicesused for supply of desired pressure of gas to loose pipe connected with welding torch. Itis mounted directly over gas cylinders. Classification: (a) 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. (b) Two stage regulator: It is desired to regulate pressure of gases at two stages. One is auto-controlled and other is GAS WELDING Ithas relatively more voltage drops so welding is preferred to do at nearest to the D.C. mains supply. Voltagedrops are less as compared to D.C. supply at a distance from main supply. So for distance welding from power mains supply A.C. welding is mostlypreferred. Welding distanceto. easier to weld different metals also other than ferrousmetal. only ferrous metals generally due to change in polarity in every cycle. Production EngineeringA-190 Badboys2Badboys2 Badboys2
- 194. Inner Zone r-----J< -----~~~ vy-----iMiddle 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. Fig. Carburising flameFig.National flame ~~~:,-----~~) -i-----T- Middle Outer Zone Zone 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 ./ J --- ./ Inner Zo~~"'-- - ---- .... FLAME S.No. Flux Application 1. Borax (Na2B407) Used with mild steel and low carbon steel etc. 2. Cast iron flux Used with cast iron, high carbon steel, ferrous silicon and silver steel etc. 3. Brazing flux Used with copper, brass and bronze etc. 4. Alumina Aluminium and its alloys etc. Properties: 1. It should be economical and easily available. 2. Itshould 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. 7. Cast iron 6. Welding of copper made articles Mainly in gas welding and brazing etc. For cast iron welding S.No. Welding Rods Applications L Low carbon steel Mild steel etc. (copper coated) 2. High carbon steel For making hard weld etc. 3. Stainless steel Stain less steel goods welding 4. Aluminium Aluminium goods welding processes. 2~O ~ Ca(OH)2+ C2~ Water Lime CaC2+ Calcium Acetylene 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) Lowpressure 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) Ithas 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 like- aluminium, magnesium and lead etc. (iv) Retort gas: It is a mixture of number of inflam 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. Dryvarious metal electrodes are used with different welding 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. Copper silver alloy Brass 5. A-191Production Engineering Badboys2Badboys2 Badboys2
- 195. SPECIAL WELDING TECHNIQUIES Some of the special welding techniques are given as follows : (a) 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. In this method ofjoining 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 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. Fig. Rightward welding .-V':'ork piece eO o'~/~o~ ~~ev 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 2. +-Work Piece 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. 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. 2. 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. 3. Popping: In this fault as usual sound like pit-pit comes from welding torch. It can be rectified by regulating the pressure of gas. 4. 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. SOLDERING AND BRAZING Production EngineeringA-192 Badboys2Badboys2 Badboys2
- 196. (a) Slag inclusions: Various types of oxides, fluxes and electrode material are trappedin the weldingzone.Duetothis trapping, inclusions are produced. These inclusion can be removed bygrinding 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 stressand 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 devlopedwhen gas bubblesare 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 ofweld torch or electrode is very fast. It is developed due to lowamperage, 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, steepelectrodeangle and fast travel speed. It can be prevented byusing a proper welding technique. (f) Underfill: It is developedwhenjoint is not completely filled bywith weld metal. It is caused by improper welding technique. It can prevented by applying proper welding technique for the weld type and position. (e) Lack of fusion (f) Lack of penetration (a) Undercut (b) Cracks Weld defects ~ WELDING DEFECTS: There are various types ofwelding defects which are given as follows: (iv) Itrequires 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. 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 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 comparedwith 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 (MetalInert 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 coiledwire fedby a motor drive to argon shielded arc. Wire is consistently fed from a spool.Ahigh value of current densities arc utilized. The diameter ofwire is keptgenerallywithintherange of0.80mm to 2.30mm. 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. Copper backing bar Direction of travel A-193Production Engineering Badboys2Badboys2 Badboys2
- 197. S. Name of Narne of operation Removed Metal form No. Machine to be carried out (Either Chip / Powder) 1 Lathe Turning, Drilling, Inner Metal removed in form turning, Threading and of chips Taper turning, etc. 2 Drill Machine Drilling, Tapping, etc. Chips 3 Shaper Shaping Chips 4 Milling Machine Milling and Boaring, etc. Chips 5 Planer Planning, Turning, etc. Chips 6 Broaching Machine Broaching Chips 7 Grinding Machine Grinding Powder 8 Polishing Machine Polishing Very fme powder 9 Buffing Machine Buffing and Polishing, Very fme powder etc. 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: (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. Itthe 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- 15MHz 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. (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: Itoccurs 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 pre- mature 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. Production EngineeringA-194 Badboys2Badboys2 Badboys2
- 198. machining operations. This property may be known as red hot hardness. 6. 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 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 ToolMaterial 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 CUTTING TOOLS 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 ofwork 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 processis known as three-dimensional cutting. (b) Oblique Cutting Turning on Lathe in Cutting Process Work Piece )- --. Movement (a) Orthogonal Cutting ~ Cutting Tool Movement -..- _.-.-.-.-.-.-.- - _o_._._._._.-c-.- _.-.----·-0-·- -c-· Work Piece ....---------.._ 1- --, Movement Importantfactors requiredin today'sscenarioasfollowing: (a) Quickmetal removal. (b) High class surface finish with economic tooling cost. (c) 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 bykeeping workpiece 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. 2. Semi-automaticcontrol. 4. Numerical control. 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. 3. Automatic control. 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 A-195Production Engineering Badboys2Badboys2 Badboys2
- 199. Cutting ToolAngles Front clearing angle Side View Top rake angle Top View bronze and cast iron, etc. It can be employed for two times more speed than common High Speed Steel tools. 5. CementedCarbide: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. 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. Front View End relif angle~ Side Clearance angle Side Rake -:rangle 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:Itis 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 contains40%-50% cobalt, 15%-35%chromium+ 12.25%vanadium + 1%-4% carbonnormally 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 ~nd cutting angle Nose Radius ~ Face Shank Side cutting ~ angle _....3....._ ___;;""-L- ---1 Production EngineeringA-196 Badboys2Badboys2 Badboys2
- 200. Continuous Chip Formation Work Piece Continuous Chip 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. 2. Continuous Chip Formation Discontinuous Chip Formation Work Piece 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 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. Chip Formation B Work Piece Rake angle: The angle between face of tool and a plane parallel to its base. Ifthis 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. 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. A-197Production Engineering Badboys2Badboys2 Badboys2
- 201. 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 and in case of orthogonal cutting process, as stated that Fn is almost zero. So, value of R = IF2 + F2 'j f t According to A.S.M.E. cutting manual, tangential cutting force will be as given below:- P = K K TC Ldt --p a RI 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 followingrelationship:- R = IF 2 + F 2 + F 2 'j n f t _._._._._._._._._._ .. _._._.-._._._._._._._ .. ~_. Work Piece -: 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). CUITING FORCE 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. Work Piece Built-up Edge 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. 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. Dueto 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. Production EngineeringA-198 Badboys2Badboys2 Badboys2
- 202. So, we have Iw = Fe IAo 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, Total work done in cutting per unit time W= ... Volume of the metal removed m unit time Work Done in Cutting 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 v,x Ft x feed velocity 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) cosu = tan ($ - a) + cos $ = ----- sin$ cos ($ - a) . 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 AD+DB Shear Strain, y = CD = CD (Ft cos$ + Fcsin$) sin$ b x t and mean normal stress, (Fc cos$ - Ft sin$) sin$ bxt sine Fs Fc cos$ - Ft sin $ So, mean shear stress (t) = A = b x t s where, Fs = Fc cos $ - Ft sin $ Fn = Ft cos $ + Fc sin $ Ao A = -.- (where Ao = area of chip before removed) s sin o F 2 and (as> (mean normal stress) = __!!_ (kg F/mm ) As 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 F (Z) = AS (kg F/mm2) s A 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:- 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. Stress in Metal Cutting Shear Strain A-199Production Engineering Badboys2Badboys2 Badboys2
- 203. • Typesofdrilling 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 •Geometryofdrill Working Principle (Drilling) LIP ~-- Tool (Drill) 001 feed motion 1<1>+1-0,=45°1 ... (1) BASIC PRINCIPLES OF MACHINING (i) Drilling: Drilling is the process or operationused 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 usedforthe purposeofdrilling is known asdrill machine. The operationsperformedbydrill machine in addition toproducingholesaretapping, reaming, boring, counter boring, spot facing etc. • Workingprinciple: Alarge amount of forceis exerted bythe rotating edge ofthe drill on the workpiece and then the hole is produced. During driling operation, the metal is removed by shearing and extresion. or we can say, 1t <I> = - + a - 1 = 45° + a - 1 4 face and therefore the chip does not get hardened. 4. 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: 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 Area (1) = Primary deformation area Area (2) = Tool chip interface 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 v, then we have, 3 WorkPiece Source of Heat in Metal Cutting ...(2) Fe x Vc kw 4500 x 1.36 ...(1) Fe x Ve Power = H.P. 4500 . . Work done in cutting / minute H.P. required for cuttmg = 4500 Horse Power Calculation Production EngineeringA-200 Badboys2Badboys2 Badboys2
- 204. ~ 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: Itis 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: Cutting speed (V) L Rotational speed (N R ) = 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. Fr NR=-- nt x fc where, Fr = feed rate (mm/min.) nt = Number of teeth on cutter fc= chip load (mmltooth) down milling Work Piece Work piece Rotational Milling direction cutter PRINCIPLE OF MILLING nDN Cutting speed (Vc) = 1000 rpm (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. BC F/2 F tan<l>=-=-=- 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:- L Machinetime (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 C (Drill Geometry) Rahe angle: - Itis 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 it. Face Chisel edge angle <, Point)' angle - Margin Cutting lip ~I 1'- Chisel edge -, / VI '. • V'I Q) c ::~.' ~~ ~ I ..0 Q) :; (2) Lip angle + Lip relief angle + Helix angle = 90° F. Ft = "2sm<l> 2<1>= po int angle F = feed = mmlrev. Flute Helix angle Drill axis ~ -r- ,,__..~~ ;r-..... / (1) L_:~:~m~di~;!: Tip A-201Production Engineering Badboys2Badboys2 Badboys2
- 205. OR Now considering, ilORP, coso, = OP Let, fr is feed rate / (Feed / rev.) F, = Feed / tooth F -~t - Nt where, N, = Number of teeth S= Ftsin<p when, <p=<Pc'then, S = Smax.(maximum) when, <p= 0, then, S = 0 Smax= r,sin <Pc s = F~d (1- d)m t D D Volume of removed metal, (VR) :~ 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: Itis 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: Itis 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. D-2d -2- D-2d 2 =---=--x- D 2 D 2 S = Smax m 2 MECHANICS OF MILLING OPERATIONS => Mean chip thickess : O+Smax Mean chip thickness (S~ = 2 (For conventional face milling) sin o, = ~~( 1-~) sino == 1_[D2+4d2-2.D.2d] c D2 On solving, we get,where, A = Apporach distance A = ~w c (D - w c) (for partial face milling) A=D 2 . ?2d) 2 sin o, = ~l-ll-D) .. . ( ) I+A Machining time tm =-- Fr where, tm= machining time (min.) A= Approach distance =)d(D-d) In case of face milling: 1+2A t =-- m F r 5. 4. D-2d (2d) cos<Pc=-D-=I- D As we know that, sin2<pe+ cos2<Pe= 1 sin2 <Pc= 1-cos2 <Pc sin <Pc= ~1- cos2 <Pc Material Removal Rate (M.R.R) = we x de x Fr where, de = depth of cut (mm) we = width ofcut(mm) In case of slab milling: 3. Production EngineeringA-202 Badboys2Badboys2 Badboys2
- 206. Working Principle of Lathe machine Workpiece I( Tool( /'" (b) Legs (d) Tail stock (t) carriage (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 (c) Head stock (e) Gear- box Machining time: (T.J As we know that, Time taken to complete one double stroke (T2s)' T _ e(k+ 1) 2s - 1000V Let, b = breadth of work piece, (mm), f= feed rate (mm/double stroke) Now, Total number of double strokes needed to complete b the work =f" Hence, Time taken to complete the cut ( b) eb(k+l) => T2s x f" = 1000Vf 1000V e(k +I) Now, N = _1_= 1 T2s e(k +1) 1000V k = Return stroke time As we know that, Cutting stroke time Return stroke = K x cutting stroke time ke =>kxT =-- c 1000V Time taken to complete one double stroke, (T2J e ke =--+-- 1000V 1000V T _ e+ ke _ e(k + I) 2s - 1000V - 1000V T = e c V x 1000 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 (Tc) cutting stroke length (m) cutting speed (m / min) (Working principle) (vertical surface) (Horizontal surface) (Inclined surface) 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. => 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. A-203Production Engineering Badboys2Badboys2 Badboys2
- 207. Type 9 : Dish wheel => Type 1, Type 2 and Type 3 are utilized for cylindrical, internal centreless and surface grinding. Type 8 : Saucer wheelType 7 (Flaring cup wheel) Grinding face Type 6 (straight cup wheel)Type (cylindrical or wheel ring) Grinding face Grinding face Thickness J, t ~II~ LI...........___-----'--'II O'----------r----r----O Type 4 (Tapered face straight wheel) Type 3 Recessed (Both sides straight) c ) Diameter of Recessed ~ I I ~ Grinding faces wheel diameter nIE )! .--r---------r---, I I I Wheel thickness '"L;r--? 1=Type 2 Rrecessed t (one side straight) Type 1 (straight) 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: Grinding Principle WorkTable 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. . ( ) 7tDN (a) Cuttmg speed V =-- 1000 where, D = diameter of workpiece (rum) N = rotational speed (rpm) (b) 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 tcFN 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. Production EngineeringA-204 Badboys2Badboys2 Badboys2
- 208. Manufacturing process is defined as the conversion of 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 ~ 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 MANUFACTURING PROCESSES IN BRIEF Depth of cut(Tc) = (dl ~d2) d, = diameter ofthe work-piece before grinding d2 = diameter of the work-piece after grinding => Feed: Feed is described as the motion of the work- piece 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/ mm. . 1000 where, N = Rotational speed (m/min). (g) Type of grinding to be done Grinding wheel parameters : => Depth of cut : Itis defined as a thickness ofthe material removed through grinding wheel in a single transverse stroke. => 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. Italso 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: Itis 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 A-205Production Engineering Badboys2Badboys2 Badboys2
- 209. with a drill which is a cutting tool having cutting edges ~ Boring: In this type,the hole (pre-existing is enlarged byusing 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. ~ Verticalsurfaces: 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. ~ Verticalsurfaces: In this type, the tool is feededdown ward for the purpose of completion of the cut, while the work pieceis providedreciprocating motion along with the table. ~ Angular surfaces : In this type, the tool is feeded at an angle forthe 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 ofaworkpiecearefmishedbyutilizing 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 surfacesofwork pieces are finished byutilizing the surface grinders. ~ Threaded surfaces: In this type, threads are produced byutilizing a thread grinding machine along with single or multiple rib wheels. ~ Sanding ~ Tumbling COMMONLY USED MACHINES AND TOOLS: (i) Lathe machine: ~ Cylindrical turning : It involves the reduction of diameterofwork-piecebyremovingmaterial along the axis ofwork - piece from the cylindrical job's surface. ~ Taperturning: In this type, material is removed at an angle to the work-piece axis. And thus diameter of the workpiece is increased or decreased. ~ Eccentricturning: In this type,the axis ofwork-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 surfaceis made paraller to the axis ofrotation ofplain milling cutter ~ Side milling : In this type, a flat vertical surface is developed on the side ofwork-piece with the help ofa side milling cutter. ~ Facemilling: In this type,facemilling cutteris utilized with rotating motion abouta perpendicular axis to the work -piece. ~ End milling: In this type, a flat surface is developed. The developedflat surfacemay behorizontal, vertical or at an angle with the table. ~ Thread milling: In this type, threads are produced by utilizing a single or multiplethread 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 Production EngineeringA-206 Badboys2Badboys2 Badboys2
- 210. 1. For TIG welding, which ofthe following gases are used? 12. In oxy-acetylenewelding: (a) Hydrogen and carbon dioxide (a) Pressure is applied (b) Argon and helium (b) Fillermetal is applied (c) Argon and Neon (c) Both Pressure and filler metal arc applied (d) Hydrogen and oxygen (d) Neither pressure, nor filler metal is applied 2. The pre-heating of parts to be welded and slow cooling of 13. What should be the size of weld in case of buss welded the welded structure will lead to reduction in : joint? (a) residual stresses and incomplete penetration (a) Twicethe throat ofweld (b) cracking and incomplete fusion (b) Half ofthe throat (c) Equal to the throat ofweld (c) cracking and residual stress (d) None of these. (d) cracking and underfill 14. Welding process in which twopieces to bejoined are over 3. Which one ofthe followingis a solid statejoining process? - llaped and placed between two electrodes in known as : (a) Gas-Tungsten arc welding (a) percussion welding (b) spot welding (b) Resistance spot welding (c) seam welding (d) projection welding (c) Friction welding 15. The abbriviation ERW in ERWpipes stands for: (d) Submergedarc welding (a) electricallyresistance welded 4. Arc stability is better with: (b) elastic reinforced with wire (a) ACwelding (b) DC welding (c) extrareinforcement welded (c) Both (a) and (b) (d) None of these (d) electrically reinforced and welded 5. Inwhich typeofwelding,molten metal ispoured forjoining 16. T - joint weld is used: the metals? (a) where longitudinal shear is present (a) Arc welding (b) Thermitwelding (b) where sever loading is encountered and the upper (c) MIG (d) llG surface of both piece must be the same plane 6. The gases used in tungsten inert gas welding are: (c) Tojoint two pieces of metal in the same manner as (a) argon and helium (b) neon and helium rivetjoint metals (c) neon and argon (d) ozone and neon (d) Tojoin two pieces perpendicularly 17. Half corner weld is used: 7. Amount of current required in electric resistance welding (a) where longitudinal shear is present is regulated by changing the: (b) where sever loading is encountered (a) Input supply (c) tojoin twopieces ofmetal in the samemanner asrivet (b) Primary turns ofthe trasnformers joint metals (c) Seondary turns of the transformers (d) none of these (d) All ofthese 18. The range of optimum pressure applied in electric 8. The material used for coating the electrode: resistance welding is given by : (a) Protective layer (b) Blinder (a) 0-5MPa (b) 5-lOMPa (c) De- oxidiser (d) Flux (c) 10-25 MPa (d) 25 - 50 MPa 9. The electric resistance welding operates with: 19. Electronic components are oftenjoined by : (a) Low current and high voltage (a) soldering (b) brazing (b) High current and low voltage (c) welding (d) adhesive (c) Low current and Low voltage 20. Themethodofjoining twosimilaror dissimilarmetalsusing (d) High current and High voltage a special fussible alloy is : 10. Fluxes are used in welding in order to protect the molten (a) Soldering (b) brazing metal and the surfaces to bejoined from: (c) Arc welding (d) All of these (a) oxidation 21. The taper provided on pattern for its easy and clean withdrawl from the mould isknown as : (b) carburizing (a) Taper allowance (b) Distortion allowance (c) unequal temperature distribution (c) Pattern allowance (d) draft allowance (d) distortion and warping 22. Sand are graded according to their: 11. Twostainlesssteelfoilsof0.1mm thicknessareto bejoined. (a) clay content Which of the following processes would be best suited? (b) gram SIze (a) Gas welding (b) TIGwelding (c) clay content and grain size (c) MIGwelding (d) Plasma arc welding (d) None of these ...,EXERCISE Badboys2Badboys2 Badboys2
- 211. 30. 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 31. Which one of the following is a single point cuting tool? (a) hacksawblade (b) millingcutler (c) pasting tool (d) grinding wheel 32. The lip angle of a single point cutting tool is : (a) 10°-30° (b) 300t060° (c) 50°-60° (d) 60°-80° 33. 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) 61.7mmlmin. (b) 51.7mmlmin. (c) 65.4mm1min (d) 48.8mm1min. 34. 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 (d) uJi(c) 112 (a) Law melting temperature (b) High melting temperature (c) Low thermal conductivity (d) low electric resistance 28. 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 29. 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 bethe coefficient of friction in fool- chip interface? (a) 2 (b) -Ii 23. Sweep pattern is used for moulding parts having: (a) Triangular shape (b) Elliptical shape (c) Uniform symmetrical shape (d) Complicated shapes having intricate details 24. In foundaries, a square pan fitted with a wooden handle is known as: (a) Bellow (b) Slick (c) Shovel (d) Riddle 25. 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 26. 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 27. Hot chamber die-casting machines are used for alloys with 0 ('I') ('I') ('I') I Production Engineering o, C) 35. 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 36. 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 (c) Swaging (d) Setting down 37. Which of the following is not a type of industrial forging? (a) Drop forging (b) Roll forging (c) Blast forging (d) upset forging 38. 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 39. Which ofthe following is commonly used die material? (a) Tungsten (b) Molybdenum (c) Cast iron (d) Hot work tool steel 40. 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 41. In a drilling machine the metal is removed by : (a) shearing and extrusion (b) Extrusion (c) Shearing (d) shearing and compression 42. Which is not the part of drilling machine (a) Spindle (b) Tool holder (c) Table (d) Cross-slide 43. Lathe beds arc produced by which of the following production processes? (a) Rolling (b) casting (c) Drawing (d) Forging 44. 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 45. Disign of jigs and fixtures need careful attention to: (a) Idle time reduction (b) Disign for safety (c) Swarf clearance (d) All of these 46. TheA.P.F (atomic Packing Factor) for BCC structure is: (a) 0.52 (b) 0.68 (c) 0.74 (d) 0.84 47. Which of the following surface hardening processes needs quenching? (a) Induction hardening (b) Flame hardining (c) Nitriding (d) case carburizing A-20S Badboys2Badboys2 Badboys2
- 212. 0 ""'"("I') ("I') I Production Engineering A-209 o, C) 48. Iniron-carbonequilibriumdiagram,the x-axisisrepresented 60. Diamondweight is expressedin terms ofcarats. One carat is by: equal to (a) carbon percentage (a) 20mg (b) 200mg (b) Temperature (c) 400mg (d) 1mg (c) Nickel percentage 61. When a bodyrecoversits original dimensions on removing (d) None of these the load then it is called 49. In annealing heat treatment process, the hypocutectoid (a) plastic (b) brittle steel is: (c) elastic (d) None of these (a) Heated from 40° C to 50° C above the critical 62. Abilityofmaterialtoundergo largepermanentdeformations temperature and then cooled slowly in the tumace. in tension is called (b) Heat from 40° C to 50° C above the upper critical (a) plasticity (b) stiffness temperature and then cooled suddenly in a suitable (c) toughness (d) hardness coolingmedium 63. Shock resistance steel should have (c) Heated from 40° C and 50° C below the critical (a) high wear resistance (b) lowwear resistance temperature and then cooledin still air (c) toughness (d) low hardness (d) Heatedbelowor closeto the lower critical tempeature 64. Essential gradient of any hardened steel is and then cooled slowly. (a) carbon (b) pearlite 50. 18 : 8 stainless steel consists of: (c) martensite (d) cementite (a) 18%vanadium, 8%chromium 65. Steelcontaining 18%chromiumand 8%nickel iscalled (a) austinitic stainless steel (b) 18%chromium,8.1nickel (b) ferritic stainless steel (c) 18%tungsten, 8% nickel (c) martensitic stainless steel (d) 18%tungsten, 8% chromium (d) None of these 51. On high rate of cooling, austenite converts into: 66. Steelhaving combination 88.7% ferrite and 13%cementite (a) martensite (b) Ferrite is known as (c) Ledeburite (d) Pearlite (a) martensite (b) austenite 52. Whichofthe followingiscorrectfornormalazing operation? (c) pearlite (d) All of these (a) It relieves internal stresses 67. A metal which isbrittle in tension canbecome ductile (b) It produces a uniform structure (a) in presence of notches (c) After heating, the material is allowed to cool in (b) in presence of emprillement agents such as hydrogen atmosphere (c) under hydrostatic condition (d) The rate of cooling is slow (d) None of these 53. The crystal structure of austenite is : 68. Etching solution used for medium and high carbon steel, (a) Simplecubic(SC) pearlite steel and cast iron is (b) Bodycentred cubic (BCC) (a) Nital- 2% RN03 is ethylalcohol (c) Face centered cubic (FCC) (b) 1% hydrofluoric acid in water (d) Hexagonal closed packed (HCP) (c) 50%NH2, OHand 50%water 54. Austenite decomposes into territe and cementite at a (d) picral- 5%pieric acid and ethyl alcohol temperature of: 69. Steelcontaining 15to 20%nickel and 0.1% carbonis called (a) 1148°C (b) 727°C (a) ferritic stainless steel (c) 1495°C (d) 1539°C (b) austenitic stainless steel 55. Alloy steel as compared carbon steel is more (c) martensitic stainless steel (a) strong (b) tough (d) None of these (c) fatigue resistant (d) All of these 70. Chrome steel is widely used for 56. Shock resistance of steel is increased by adding (a) connecting rod (b) cutting tool (a) Aluminium (b) Cobalt (c) handtool (d) motor car crank shaft (c) Nickelchromium (d) Carbon 71. Carbon steel castings are 57. Carbon steel is (a) easilyweldable (b) tough and ductile (a) produced by adding carbon in steel (c) brittle (d) All of these 72. Vandium when added to steel it (b) an alloy of iron and carbon with varying quantities of (a) decreases tensile strength phosphorus and sulpher (b) increases tensile strength (c) purer than the cast iron (c) remain constant (d) None of these (d) None of these 58. The raise yield point oflow carbon steel 73. High speed steel should have (a) Phosphorus is added (b) Silicon is added (a) wear resistance (b) hardness (c) Carbon is added (d) Sulphur is added (c) toughness (d) both (a) and (b) 59. Stress-concentration occurs when a body is subjected to 74. Alloy steel containing 36% Nickel is known as (a) Extensive stress (b) reverse stress (a) Stainless steel (b) High speed steel (c) fluctuating stress (d) non-uniform stress (c) Die steel (d) HS.S. Badboys2Badboys2 Badboys2
- 213. (b) pearlite (d) martensite lowercriticallimit(b) (d) point ofrecalesence (a) uppercriticallimit (c) melting point 100. Heat treatment process is (a) hardening by quenching(b) annealing (c) tempering (d) All of these 101. Ifsteel is slowlycooledin 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 (c) cementite 103. Bynormalising of steel, its (a) ductility decrease (b) ultimate tensile strength increase (c) fieldpoint increases (d) All of the above 90. Temperatureat which the change starts on heating the steel is called (a) uppr critical temperature (b) point of recalesense (c) lowercriticaltemperature (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) Siliconpercentage (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 aboveits critical temperature (c) reduction of silicon percentage (d) annealing of steel 99. Temperature at which the changes end on heating the steel is called 75. Case hardening process is (a) carburizing (b) cynidity (c) nitridity (d) All of these 76. Normalising of steel is done to (a) remove strains caused by cold working (b) refine grain structure (c) removedislocationcausedin the internal structuredue to hot working. (d) All of these 77. Steelcontaining pearlite and ferrite is (a) ductile (b) soft (c) hard (d) tough 78. Percentage of carbon in carbon steel is (a) 0-1% (b) 0.1-1.5% (c) 1.5-4.2% (d) 1- 3% 79. Cutting tools are manufactured by (a) High speed steel (b) Nickel steel (c) Chormesteel (d) None of these 80. Silicon Steel is widelyused in (a) electrical industry (b) connecting rod (c) cutting tool (d) All of these 81. 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 82. Nitriding is a process for (a) softening (b) hardening (c) tempering (d) All of these 83. Temperature at which the first tiny new grains appears is called (a) melting temperature (b) criticaltemperature (c) pointing temperature (d) recrystallinetemperature 84. Annealing of steel is done to (a) improvemachinability (b) softeners of metal (c) release internal stress (d) All of these 85. Machining properties of steel are improved by adding (a) Carbon (b) Chromimum (c) Silicon (d) Sulphur, lead and phosphorus 86. Tomake low carbon steel tougher and harder (a) Carbon is added (b) Carbon reduced (c) Silicon added (d) Aluminiumadded 87. Chilling heat treatment and alloyadding (a) decreasesmachinability (b) increasemachinability (c) increase carbon percentage (d) None of these 88. Ifsteel is cooledin still air, the structure obtained is called (a) sorbite (b) pearlite (c) toorsite (d) mortensite 89. Heat treatment process used for castings is (a) hardnening (b) normalising (c) annealing (d) tempering Production EngineeringA-210 Badboys2Badboys2 Badboys2
- 214. 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) 0.1nvm (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 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 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 A-211Production Engineering Badboys2Badboys2 Badboys2
- 215. 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) alloyingelement (d) none of these 148. Electric indirect arc furnace is normallyused formelting 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 oftenused 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 ofpattern (b) intricacyofpattern (c) moldingmethod (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 above31t 8 (d)(c) 6 (b) 1t 3 4 5 41t (a) 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) Coarsegrain (b) finegrain (c) mediumgrain (d) none of these 137. Black colourmarking in pattern is used to indicate (a) machined surface (b) un-machined surface (c) parting surface (d) None 138. Loam Sand comprises ofpercentage 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) achievedirectional solidification (b) reduce the possibility of blow holes (c) reducefreezingtime (d) smoothens metal flow forreducing splatter 143. Whichofthe followingmaterialrequiresthe largestshrinkage allowance,while making a pattern forcasting. (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 lossesthe correct choicefor the time (in second) required to filla mould cavityofvolume 106 mm-, willbe (use g = 10m! 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 ofa sphere and the other is a cube. What is the ratio ofthe solidification time for the sphere to that of the cube. Production EngineeringA-212 Badboys2Badboys2 Badboys2
- 216. (b) coldforming (d) casting 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) coldworking (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 (c) machining (b) mis-run (d) all of these 173. Which of the following defect may occur due to improper design of gating system. (a) Cold sheets (c) rough surface 174. Sprue are generally (a) uniformin 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. Solidificationtime forriser shouldbe (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 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 surfacedefects (c) improve surface finish (d) reduce the number of blow holes. 167. Hot tears in casting are caused due to (a) toomuchramming ofmold (b) grain size of sand (c) size of casting (d) rate ofporing ofmolten 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) lowpermeability ofsand (b) high moisture content of sand (c) intermittent running ofmoltenmetal oversand 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 ofmoulding 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) lowpermeability ofsand (c) excessive fine grains and gas producing ingredients (d) any of above A-213Production Engineering Badboys2Badboys2 Badboys2
- 217. 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 pipes 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 cost (c) material saving (d) all of these (d) 1(c) 0.693 (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 (a) 0.307 (b) 0.5 (b) recrystallisation(a) grain growth 201. Parts of circular cross section which are symmetrical about the axis of rotation are made by (a) hot forging (b) hot spinning (c) cold forging (d) none of these 202. Mechanical properties of the metal improve in hot working due to (b) forging (d) cold peening (a) piercing (c) extrusion 199. Notching is the operation of (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 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 (b) 25mm (d) 50mm (a) 20mm (c) 30mm 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 196. Cold working process can be applied on the component having diameters up to (d) wear and tear of die(c) wear of punch 188. For extrusion process (a) 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 (d) none of these 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 191. Material good for extrusion is (a) Low carbon steel (b) Cast iron (c) S.S. (d) HS.S. 192. Upsetting or cold heading machine is a (a) rolling process (b) extrusion process (c) forging process (d) none of these 193. The major problem in hot extrusion is (a) design of punch (b) design of die Production EngineeringA-214 Badboys2Badboys2 Badboys2
- 218. 220. Temperature of oxy-hydrogen flame as compared to oxy- acetylene flame is (a) less (b) more (c) same (d) depends on oxygen percentage 221. Oxidising flame is obtained bysupplying (a) more oxygen and less volume of acetylene. (b) both oxygen and acetylene kept in same volume. (c) acetylenevolume kept more than oxygen volume (d) None of these 222. Oxidisingflameas comparedtoneutral flame hasinner core (a) shorter in size (b) less luminous (c) moreluminous (d) both (a) and (b) 223. Maximumflametemperatureoccurs (a) at inner core offlame (b) at outer core of flame (c) attipofflame (d) next to the inner core 224. Maximum used flamein gaswelding 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 ofthe 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-spotwelding (d) None of these 228. Weldingprocesspreferred for cutting and welding fornon- ferrousmetal 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) MIGwelding (b) TIGwelding (c) Argon welding (d) None of these 230. The welding process in which electrode get consumed is (a) MIGwelding (b) TIGwelding (c) Spotwelding (d) None of these 231. Grey cast iron is usually welded by (a) resistance welding (b) gas welding (c) spot welding (d) arcwelding 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) seamwelding (b) spot welding (c) projectionwelding (d) percussion welding (b) 1800°C (d) Morethan 4000°C 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 forArc welding is ofthe 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) Spotwelding (b) Buttwelding (c) Arcwelding (d) Any of the above 218. Tig welding isprefferedin 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 (c) 3500°C 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 forreducing 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) powerrepowerments 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 thejob to have better surface finish (d) reducing the waste ofmaterial 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 fusionwelding is (a) Thermitwelding (b) Arc welding (c) Forge welding (d) Gaswelding 213. Weldingprocess in which flux is used in form of gannual is (a) D.C.Arc welding (b) Spotwelding (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 A-215Production Engineering Badboys2Badboys2 Badboys2
- 219. 250. In Arc welding, range oftemperature generated at arc is (a) IOOO°C- 2000°C (b) 2000°C- 4000°C (c) 4000°C-6000°C (d) None of these 251. Projectionwelding 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) below10V (b) 10V (c) higher than 10V (d) None of these 253. In arcwelding,penetrationis 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 forelectrodediameter selectionis (a) Thickness of work piece (b) Typeofwork piece metal (c) Welding pressure applied (d) Welding process applied 257. In projectionweldingdiameteroftheprojectionas 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 (b) 3-5mm (d) 0.025-3mm 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) Gaswelding (b) Submergedarc welding (c) D.C.welding (d) None of these 249. Seamweldingused formetal sheets having thickness in the range (a) below3mm (c) 3-6mm (d)(c) j15i 238. In arc welding, two lowwelding speed results in (a) Excessivepilling up ofweldmetal (b) Electrode waistage (c) Over hauling without penetration edge (d) All of these 239. Fillers material is essentiallyused in (a) Spotwelding (b) Gaswelding (c) Seamwelding (d) Projectionwelding 240. Rate ofwelding 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. Fillermaterial is usedin (a) Spotwelding (b) Butt welding (c) Seamwelding (d) None of these 243. Cleaning ofmetal in electricalresistance welding is (a) important (b) not important (c) have no effect (d) none of these 244. An example of fusionwelding is (a) Spotwelding (b) Gaswelding (c) Projectionwelding (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 (a) Less than .[t (b).[t (b) Submergedarc welding (d) None of these 234. Amount ofcurrent 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 acetylenewelding (b) Thermitwelding (c) Soldering (d) Electricarcwelding 236. Spot-welding, projection welding and seam welding are classification of (a) Thermitwelding (b) Resistance welding (c) Arc welding (d) Spotwelding 237. An arc is produced between a bare metal electrode and workin (a) D.C.welding (c) Spotwelding Production EngineeringA-216 Badboys2Badboys2 Badboys2
- 220. (b) rake angle (d) None of these 280. Standard taper generally used on milling machine spindles is (a) Morsetaper (b) Shellr'sistaper (c) Champmantaper (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) mildSteel (b) aluminium (c) cast Iron (d) brass 284. Continous chips are formedwhile machining of (a) cast iron (b) mild steel (c) aluminium (d) None of these 285. Toprevent tool from rubbing the work, angle provided on tool is (a) reliefangle (c) clearance angle 2 (d) Anyone of above_!_"(c) 1_!_" 4 (b)(a) 1" 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. Oxygento acetyleneratio in carburising flame is (a) 0.5: 1 (b) 0.9: 1 (c) 1: 1 (d) 1: 2 274. A lathe machine specialin (a) Diameter oflathe (b) Grossweight ofmachine (c) Speed of lathe (d) Swingoflathe 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) extremeendto extremeend (b) centre of hole at one end to the center of holes at the other end (c) the formulaL = 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 sizeis 271. Oxygento acetyleneratio in case ofneutral flame is (a) 0: 1 (b) 1:2 (c) 0.8:2 (d) 2: 1 (b) bottom of crates (d) none of these (b) Fusion welding (d) None of these 266. In a welding process, flux is used to (a) Permit perfect cohesion ofmetal (b) remove oxidesofmetal formedat 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 lowand 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 (c) centre of crates 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 electrodeto workpiece (c) high resistivity due to presence of air (d) none of these 262. DuringArcweldingwith increaseofthickness 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 forjoining round bars is (a) Thermitwelding (b) Projectionwelding (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 materialsis called (a) Spotwelding (c) D.C.welding A-217Production Engineering Badboys2Badboys2 Badboys2
- 221. Carbon steel Carbide tools (b) (d) 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. (c) Tool steel 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 (b) more (d) None of these 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 (c) less (d) brandlmodle none of tool (b) cutting speed (d) None 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 (c) feed and depth 297. Tool signature (a) description of tool shape (b) the plane of tool (c) design and description of various angles provide on tool 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 Production EngineeringA-21S Badboys2Badboys2 Badboys2
- 222. 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 removedby (a) distortionofmaterial (b) shearing ofmetal (c) fracturingofmetal (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 speedthen chips formed are (a) Continuous (b) Discontinuous (c) Continuous with built up edge (d) Power shape 334. Chip formedwhen DuctileMetalmachinedwith 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 singlepoint cutting tool to control chip flowis (a) Siderake angle (b) End reliefangle (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 toolplace (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 generatedin the (a) Friction zone (b) Shear zone (c) Point of contact of cutting tip and work piece (d) All of above Y (c) r =C (d) vr=-c 323. Chips are broken effectivelydue to which ofthe following property (a) Elasticity (b) Toughness (c) Workhardening (d) Stress produced 224. Continous chips are formedwhen machining (a) brittlemetal (b) ductilemetal (c) high speed (d) All of these 325. Finishing obtained on workpiece mostly affected by (a) Cutting speed (b) Feedrake (c) Lubricant used (d) Depth of cut 326. Machinability tends to decrease with (a) increasein 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 (b) yn -=C T (a) v r-c 314. Toolcutting forces, with increase in cutting speed (a) increaselinearly (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 anglesare used formachining (a) cast iron (b) mild steel (c) aluminiumalloys (d) None of these 317. When radial forcein 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) toughmaterial (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) inmiddle (d) All of these 321. In equation YIn= C, value ofn depends on (a) Material ofworkpiece (b) Material oftool (c) Cutting position (d) All of these 322. The relationship betweentool life(T) and cutting speed(Y) m/min is given as A-219Production Engineering Badboys2Badboys2 Badboys2