DYNAMICS OF MACHINE
Tuesday, March 15, 2016
PROELL GOVERNOR
18.6.
Proell Governor
The Proell governor has the balls fixed at B and C
to the extension of the links DF and EG, as
shown in Fig.The arms FP and GQ are pivoted at P and Q
respectively.
Consider the
equilibrium of the forces on one-half of the governor as shown in Fig. The instantaneous centre (I) lies on the intersection of the line PF
produced and the line from D drawn perpendicualr to the spindle axis.
The prependicular B M is drawn on ID.
PORTER GOVERNOR
18.5.
Porter Governor
The Porter governor is a modification of a Watt’s governor, with
central load attached to the sleeve as shown in Fig. The load moves up and down the central spindle. This additional downward force
increases the speed of revolution required to enable the balls to rise to any
pre-determined level.
Consider the forces
acting on one-half of the governor as shown in Fig.
Let
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m = Mass of each ball in kg,
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w = Weight of each ball in newtons =
m.g,
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M = Mass of the central load in kg,
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W = Weight
of the central load in newtons = M.g,
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r = Radius
of rotation in metres,
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h
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=
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Height of
governor in metres ,
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N =
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Speed of
the balls in r.p.m .,
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ω =
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Angular
speed of the balls in rad/s
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= 2 π N/60 rad/s,
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FC
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=
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Centrifugal force acting on the ball
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in newtons
= m . ω 2.r,
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T1
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=
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Force in
the arm in newtons,
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T2
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=
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Force in
the link in newtons,
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α = Angle of inclination of the arm (or upper
link) to the vertical, and
= Angle of inclination of the link
(or lower link) to
the vertical.
Though there are several ways of
determining the relation between the height of the governor (h) and the
angular speed of the balls (ω), yet the
following two methods are important from the subject point of view :
1. Method of
resolution of forces
; and 2. Instantaneous centre
method.
1.
Method of resolution of forces
Considering the equilibrium of the forces acting
at D, we have
T
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cos β =
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W
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=
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M . g
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2
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2
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2
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T2
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=
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M . g
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or
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2 cos
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β
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A big hydel generator.
Governors are used to control the supply of working fluid (water in hydel
generators).
Note : This picture is
given as additional information and is not a direct example of the current
chapter.
. . . (i)
Again, considering
the equilibrium of the forces acting on B. The point B is in
equilibrium
under the action of the following forces, as shown in Fig.
18.3 (b).
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(i)
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The weight
of ball (w = m .g),
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(ii)
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The centrifugal force (FC),
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(iii)
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The
tension in the arm (T1), and
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(iv)
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The
tension in the link (T2).
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Resolving
the forces vertically,
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T1 cos α = T2
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cos β + w =
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M . g
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+ m.g
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. . . (ii)
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2
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T2
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cos β =
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M . g
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. . . 3
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Resolving
the forces horizontally,
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T1 sin α + T2 sin β = FC
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T1
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sin α +
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M . g
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⋅ sin
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β = FC
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=
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M
.g
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. . . 3 T2
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2 cos
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β
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2 cos β
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T
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sin α +
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M . g
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⋅ tan
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β = F
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1
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2
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C
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M . g
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∴
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T1 sin α = FC –
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⋅ tan β
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. . . (iii)
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2
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Dividing
equation (iii) by equation (ii),
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T1 sin α
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F –
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M . g
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⋅ tan β
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=
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C
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2
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T1 cos α
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M
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. g
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+ m . g
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2
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M . g
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M
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. g
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+ m . g
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tan α = FC
–
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⋅ tan β
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2
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2
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M . g
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+ m . g
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=
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FC
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–
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M . g
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⋅
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tan β
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2
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tan α
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2
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tan α
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Substituting
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tan β
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= q , and tan α =
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r
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, we have
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tan α
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h
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M . g
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+ m . g = m . ω2 . r ⋅
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h
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–
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M . g
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⋅ q
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. . . (∴ F = m.ω2.r)
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2
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r
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2
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C
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m . ω2 . h = m . g +
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M . g
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(1 + q)
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2
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m +
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M
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(1 + q)
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M . g
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1
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g
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∴
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h
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=
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m . g
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+
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(1 + q)
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=
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⋅
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2
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m
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2
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m .ω
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ω
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. . . (iv)
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m +
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M
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(1 + q)
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Mg
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1
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g
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2
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2
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ω =
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m . g
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+
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(1 + q)
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=
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⋅
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2
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m . h
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m
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h
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2 π N
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2
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m +
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M
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(1 + q)
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=
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⋅
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m
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h
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60
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m +
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(1 + q )
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2
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m +
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(1 + q)
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g
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60
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895
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∴
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N
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2
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=
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2
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⋅
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=
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2
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⋅
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m
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h
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2π
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. . . (v)
. . . (Taking g = 9.81
m/s2)
Notes
: 1. When the length of arms are equal to the length of links and the
points P
and D
lie on the same vertical line,
then
tan α = tan β
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orq
=
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tan α / tan β = 1
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Therefore,
the equation (v ) becomes
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N 2 =
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( m
+ M )
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⋅
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895
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. . . (vi)
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h
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m
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2. When
the loaded sleeve moves up and down the spindle, the frictional force acts on
it in a direction opposite to
that of the motion of sleeve.
If F = Frictional force
acting on the sleeve in newtons, then the equations (v) and (vi) may be
written as
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M . g ± F
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m . g +
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(1
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+ q )
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N 2
=
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2
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⋅
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895
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. . . (vii)
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m.g
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h
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= m . g + (M . g ± F ) ⋅
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895
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. . . (When q = 1) . . . (viii)
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m . g
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h
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The + sign is used
when the sleeve moves upwards or the governor speed increases and negative sign
is used when the sleeve moves downwards or the governor speed decreases.
3. On
comparing the equation (vi) with
equation (ii) of Watt’s
governor (Art. 18.5), we find that the m
+ M
mass of the central load (M) increases the height of
governor in the ratio
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m
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2. Instantaneous centre method
In this method,
equilibrium of the forces acting on the link BD are considered. The
instantaneous centre I lies at the point of intersection of PB
produced and a line through D perpendicular to the spindle axis, as
shown in Fig. 18.4. Taking moments about the point I,
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F ⋅ BM = w ⋅ IM +
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W
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⋅ ID
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C
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2
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= m .g ⋅ IM +
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M . g
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⋅ ID
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2
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∴
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FC
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= m .g ⋅
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IM
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+
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M . g
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⋅
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ID
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BM
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BM
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2
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||||||||
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= m .g ⋅
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IM
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+
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M . g
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IM + MD
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BM
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2
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BM
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= m . g ⋅
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IM
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+
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M . g
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IM
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+
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MD
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||||||||||||
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BM
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2
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BM
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BM
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= m . g tan
α + M . g (tan α + tan β) 2
Fig. 18.4. Instantaneous
centre method.
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IM
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MD
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. . . 3
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= tan α, and
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= tan β
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||||||||||||||||||||||||||
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BM
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BM
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||||||||||||||||||||||||
Dividing throughout by tan α,
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|||||||||||||||||||||||||||
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FC
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= m . g +
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M . g
|
|
+
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tan β
|
= m . g +
|
M . g
|
+ q)
|
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=
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tan β
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||||||||||||||||||||||||||||||
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1
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(1
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. . . 3q
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|||||||||||||||||||||||||||||||
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tan α
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|||||||||||||||||||||||||||||||||||||
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2
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tan α
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2
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tan α
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|||||||||||||||||||||||||||
|
We know that F = m.ω2.r,
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and
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tan α =
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r
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|||||||||||||||||||||||||||||
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C
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h
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||||||||||||||||||||||
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||
|
∴ m.ω2 . r ⋅
|
h
|
= m . g +
|
M .g
|
(1+ q)
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||||||||
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|||||||||
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r
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2
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|||||
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m . g +
|
M . g
|
(1 + q )
|
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|
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m +
|
M
|
(1 + q)
|
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||||||||
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1
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g
|
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|||||||||||||
or
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2
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2
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||||||
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|||||
|
h =
|
|
m
|
|
⋅ ω2 =
|
|
|
⋅ ω2
|
|
|||||||||||||||||
|
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|
m
|
|
||||||||||||||||||||
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|
|
. . (Same as before)
|
|
|
When tan α = tan β or q
= 1, then
|
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|||||||||||||
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|
|
h =
|
m + M
|
⋅
|
g
|
|
|
|
|
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|
||||||
|
|
|
ω2
|
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|||||||||
|
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|
m
|
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