Chapter – 5
Governors
GOVERNORS
Governors are used to maintain constant mean speed of crankshaft, when load of engine vary.
When load on engine increases, the speed of engine decreases. In such condition, governor
increase the supply of fuel.
When load open engine decreases, the speed of engine increases, then governor decreases the
fuel supply.
Thus governor maintains the constant speed.
Types of governors
There are two types of governors:
i) Centrifugal governor
ii) Inertia governor
i) Centrifugal governor
In centrifugal governors, centrifugal force is produced by revolving the balls at constant speed which
actuates the sleeve.
This force is equal to controlling force which is provided by dead weight or spring or combination of two.
ii) Inertia governor
In inertia governors, balls are arranged in such away that inertia forces may changes their
position. This is done by using suitable linkages and springs.
In centrifugal governor, balls are operated by the actual change of engine speed.
In case of inertia governor, balls are operated by rate of change of engine speed.
Therefore, the response of inertia governors is faster than that of centrifugal governors.
Construction and working of centrifugal governors
Figure shows a centrifugal governor, which consist of two balls, two arms, two links, two stops, a sleeve and bell
crank lever.
Balls: there are two balls of equal masses which are attached to two arms
Arms: Arms are pivoted to a spindle.
Spindle: Spindle is driven by engine through bevel gears.
Links: The lower arms or links are connected to sleeve.
Sleeve: Sleeves rotate with spindle and moves up and down. The upward and downwards movement of sleeve is
restricted by stops S1 and S2.
Bell crank lever: Bell crank lever is connected to the throttle, which control the supply of fuel.
Working
When load on engine increases, speed decreases.
Thus, centrifugal force on balls decreases.
As a result, balls move inward and sleeve move downward.
The downward movement of sleeve increases the fuel supply and thus engine speed is increased.
When load on engine decrease speed increases.
Thus, centrifugal force on balls increases
As a result, balls move outwards and sleeve moves upward.
The upward movement of sleeve decreases the fuel supply and thus engine speed id decreased.
Watt governor
This is simplest form of centrifugal governor. This governor was used by prof. Watt in his earlier steam engines.
Now it is totally obsolete.
It consist of two one pair of balls.
Balls are attached with arms and arms are
pinned with spindle at point O.
Spindle is driven by engine though bevel
gears.
Lower arms or link are connected to sleeve.
As spindle rotates, balls take up position.
The position of ball depends upon speed of
engine.
Let
m = mass of each ball
w = weight of each ball = m.g
T = tension in the arm
ω = angular velocity of the balls, arms and sleeve
r = radial distance of the balls from spindle axis
FC = Centrifugal force acting on the ball = mrω2
h = height of governor i.e. the vertical distance
form the centre of ball to the point of intersection
of the upper arms along the axis of the spindle.
Height of governor decreases as speed of engine increases while height of governor increases as speed of
engine decreases.
Assuming that weight of the arms, links, and sleeve are negligible.
Each ball will be in equilibrium under the action of following forces:
The centrifugal force, FC acting on the ball where FC =mrω2
The weight of ball, w where w = m.g
The tension T in the upper arm
Resolving the forces acting on the ball in horizontal direction
T sin   FC  m  r   2 .......(1)
Resolving forces in vertical direction
T cos  w  m  g...............(2)
Dividing equation (i) and (ii), we get
T sin  m  r   2  2  r


T cos
m g
g
tan  
2  r
g
.......................(3)
From figure
r
tan   .........................(4)
h
Equating the two values of tan θ, given by equation (iii) and (iv),
2  r
g
h

r
h
rg
g

r 2 2
If g is taken in m/s2 and ω rad/sec, then h will be in meters. If N is the speed in r.p.m. then
2
1
 60 


9.81


 
2
 2  N 2
 2 N 


 60 
9.81
h
895
N2
From above equation, it is clear that height of a watt governor is inversely proportional to the square of speed.
Therefore, at high speed the value of h is very small.
Porter governor
Figure shows the diagram of a porter governor. In case of porter governor, a central heavy load is attached to the
sleeve. The central load and sleeve moves up and down the spindle.
Let
M = mass of central load
W = weight of central load = M X g
w = weight of each ball = m x g
m = mass of each ball
h = height of governor
r = radius of rotation
FC = Centrifugal force on the ball = mrω2
ω = angular speed of ball =2πN/60
N = speed of ball in r.p.m
T1 = tension in upper arm
T2 = tension in lower arm
α = angle of inclination of the upper arm to
the vertical
β = angle of inclination of the lower link to the
vertical
Ff = Force of friction between sleeve and
spindle
Here lower arm AC is considered in equilibrium to analyze the porter governor. The forces acting on the lower
arm AC are:
Centrifugal force FC
Weight of ball
Half weight of sleeve
Now find out the instantaneous centre of arm AC.
The direction of velocity of point A is perpendicular to OA and
direction of velocity of point C is parallel to the axis of rotation of
spindle.
Now, draw the perpendiculars from points A and C.
These perpendiculars will intersect at point ‘I’.
Here ‘I’ is instantaneous centre of arm AC.
Now taking the moments of forces acting on arm AC about point I,
FC  AD   m  g   ID 
FC   m  g  
 m  g  tan  
W
 IC
2
ID W IC
 
AD 2 AD
W  ID  CD 


2  AD 
 m  g  tan  

IC  ID  DC 
W  ID CD 



2  AD AD 
M g
 m  g  tan  
  tan   tan  
2
 m  g  tan  
M g
 tan  
 tan  1 

2
 tan  



ID
CD

 tan  ;
 tan  ;W  M  g 
AD
AD

 m  g  tan  
M g
tan  1  k 
2
where
M g

FC  tan   m  g 
1  k 
2


But from triangle OAB,
r
tan   ; FC  mr 2
h
Substitute the these values in the above equation, we get
r
M g
mr 2   m  g 
1  k 
h
2

2 
m g 
M g
1  k 
2
mh
if k=1 which is true when tan α =tan β, then above equation becomes as
2 
m g  M  g m  M  g

m h
m h
Put

2 N
60
k
tan 
tan 
 2 N   m  M   g

 
60
mh


2
m  M  g

h
 2 N 
m

 60 
2
m  M   g  3600


m  4 2  N 2

m  M 9.81 3600

m
4 2  N 2
h
m  M 894.56

m
N2
In above equation, h is in meter.
Proell Governor
Figure shows the line diagram of Proell governor.
It is similar to the porter governor because it heavy central load on sleeve.
But, it is differ from porter governor in arrangement of balls.
Here, balls are attached at extension of lower arms instead of conjunction of upper and lower arms.
The action of this governor is similar to the watts governor. As speed increases, radius of rotation
increased and sleeve lifts. As a result, fuel supply decreased.
The proell governor is analyzed by considering the lower arm AC in equilibrium state.
Now find out the instantaneous centre of arm AC.
The direction of velocity of point A is perpendicular to OA and direction of velocity of point C is parallel to the axis
of rotation of spindle.
Now, draw the perpendiculars from points A and C. these perpendiculars will intersect at point ‘I’.
Here ‘I’ is instantaneous centre of arm AC.
Now take the moments of all forces about point ‘I’ and assume AG is extension of lower arms on which balls are
attached.
FC  GD  m  g  ID 
M g
 IC
2
Dividing both sides by AD, we get
FC 
GD
ID M  g IC
 m g 


AD
AD
2
AD
 m  g  tan  
M  g  ID  DC 


2
 AD 
M  g  ID DC 



2  AD AD 
M g
 m  g  tan  
 tan   tan  
2
GD
M g
 tan  
FC 
 m  g  tan  
 tan  1 

AD
2
 tan  
AD
M g

FC 
 tan   m  g 
1  k 
GD
2


 m  g  tan  
But from triangle OAB, tanα = r/h and FC = mrω2
Substituting these values in above equation
mr 2 
AD r 
M g
 m  g 
1  k 
GD h 
2

ID


IC

ID

DC
;
 tan  

AD


2 
AD r 
M g
1
 m  g 
1  k  
GD h 
2
 m r
M g


m

g

1

k



AD 
2
2 

GD 
mh




if k=1 which is true when tan α=tan β, then the above equation becomes as
2 
AD  m  g  M  g 

GD 
m h
2 
AD   m  M   g 
GD  m  h 
By comparing the final equation of porter and proell governor, we find that
Angular speed of proell governor reduces for a given values of m, M, and h.
In other words, in proell governor, the balls of smaller masses can be used for same angular speed.
Hartnell Governor
It is a spring loaded governor.
Two bell crank levers are pivoted to the frame at points O and O’.
The frame is attached to the governor spindle.
Rollers fit into a groove in the sleeve.
A helical spring is also fitted to provide the downward force on the sleeve through a collar.
Working:
When speed increases, the radius of rotation of balls increases and balls moves away from the spindle axis.
As balls move away from the spindle axis, the rollers lift the sleeve against spring force.
If the speed decreases, the sleeve move downwards.
The movement of sleeve is transferred to throttle of engine which controls the fuel supply.
Let
r1 = minimum radius of rotation of ball centre from spindle axis,
r2 = maximum radius of rotation of ball centre from spindle axis,
S1 = spring force exerted on sleeve at minimum radius,
S2 = spring force exerted on sleeve at maximum radius,
m = mass of each ball,
M = mass of sleeve,
N1 = minimum speed of governor at minimum radius,
N2 = Maximum speed of at maximum radius,
ω1 and ω2 = Corresponding minimum and maximum angular velocities
(FC)1 = Centrifugal force corresponding to minimum speed = m x ω2 x r1
(FC)2 = Centrifugal force corresponding to maximum speed = m x ω2 x r2
s = stiffness of spring or the force required to compress the spring by one mm,
r = radius of rotation when governor is in mid-position,
a = length of ball arm of bell crank lever i.e. distance OA
b = length of sleeve arm of bell crank lever i.e. distance OC
h = compression of the spring when radius of rotation changes form r1 to r2. This is also known as lift of sleeve.
Figure shows the forces acting on the bell crank lever in two positions i.e. at minimum radius position and
maximum radius position.
I)Position of minimum radius
The position of bell crank lever at the minimum radius is shown by AOC whereas
The position of bell crank lever when governor is in-mid position is shown by dotted line A1OC1
Let h1 = lift of sleeve i.e. vertical CC1.
θ1 = angle turned by bell-crank lever between mid position and minimum position. This means the angle
between OA and OA1 is same as between OC and OC1.
CC1 AA1

OC OA
CC1 AA1

OC OA
1 
Arc
CC1 




 1

Radius OC 

h1  r  r1 
[ AA1  r  r1; OA  a; OC  b]

b
a
b
h1   r  r1  .........................(1)
a
ii) Position of maximum radius
The position of the bell crank lever at the maximum radius shown by AOC whereas the position of bell crank when
governor is in mid-position is shown by dotted line A2OC2.
Let
h2 = Lift of sleeve form mid-position i.e. vertical C2C.
θ2 = angle turned by bell crank lever between mid-position and maximum radius position i.e.
 C2OC =  A2OA=θ2
h
AA
2  2  2
OC OA
h2  r2  r 
 AA2  r2  r 

b
a
b
h2   r2  r  .........................(2)
a
Adding equation (i) and (ii)
b
b
 r  r1    r2  r 
a
a
b
b
b
b
b
 r  r1  r2  r   r2  r1 
a
a
a
a
a
h1  h2 
h
b
 r2  r1 .....................(3)
a

h  h1  h2  .total.lift
iii) Position of minimum radius
Taking moments about fulcrum O, we get
 M  g  S1   b
 m  g  AA1   FC 1  a1....................( A)
1
2
 M  g  S1   b
1
2
 M  g  S1  
  FC 1  a1  m  g  AA1
2
 FC 1  a1  m  g  AA1 
b1 
 M  g  S1  
2
 FC 1  a1  m  g   r  r1   .............(4)
b1 

AA1  r  r1 
IV) Position of maximum radius
 M  g  S2   b
2
2
  FC 2  a2  m  g  AA2 ...............( B )
  FC 2  a2  m  g   r2  r 
 M  g  S2  

AA2  r2  r 
2
 FC 2  a2  m  g   r2  r   .............(5)
b2
Subtracting equation (4) from equation (5), we get
S2  S1 
2
2
 FC 2  a2  m  g   r2  r     FC 1  a1  m  g   r  r1 
b2
b1
But spring stiffness is given by
s
S2  S1 S2  S1

Total.lift
h
Put the value of h in above equation
h
b
 r2  r1 
a
from equation (3)
S2  S1
a  S2  S1 
s
 

b
b
r

r


2
1


 r2  r1 
a
In working range of governors, θ is usually small the obliquity effects of the arms of the bell-crank levers may be
neglected. In that case
 b1  b2  b; a1  a2  a 
and
If moments due to weight of balls are also neglected, then
 mg  AA1  0; mg  AA2  0 
Substitute the above values in equation (A) and (B)
 M  g  S1   b
1
 m  g  AA1   FC 1  a1
2
 M  g  S1   b  0  F  a.....................(6)
 C 1
2
 M  g  S2   b
2
2
 M  g  S2   b 
2
  FC 2  a2  m  g  AA2
 FC 2  a  0....................(7)
Substituting equation (6) from equation (7)
S2  S1 
2a
 FC 2   FC 1  .....................(8)
b
But spring stiffness is given by
s
S2  S1 S2  S1

Total.lift
h
2a
 FC 2   FC 1 
b
s
b
 r2  r1 
a
2
 a   FC 2   FC 1 
s  2 
 r2  r1 
b
where
h
b
 r2  r1 
a
Wilson-Hartnell Governor
Bell crank Lever: It is a spring loaded governor. In this governor, bell crank lever is pivoted to arms which rotate
with spindle.
Balls: Balls are attached to bell crank lever and also connected with a spring.
Spring: Horizontal arms of bell crank lever carry two rollers which are fitted in the sleeve.
Sleeve: Sleeve is connected with an adjustable auxiliary spring through a lever.
Lever: One end of lever is connected with spring while other end of lever is connected with sleeve. It is also
pivoted at fulcrum.
Auxiliary spring: Usually main spring is not adjustable. So second adjustable auxiliary spring is provided
Working:
When governors speed increases, balls moves in outwards direction and spring pull the balls in inwards
direction.
Sleeve is also lifted due to increase in speed.
At same time, lever also exerts the pressure in downward direction due to tension of spring.
Auxiliary spring tends to keep the sleeve down.
Let
m = mass of each ball
M= mass of sleeve
W= Weight of sleeve = M x g
P= Tension in main spring
S = Tension in auxiliary spring
FC = Centrifugal force on each ball
r = Radius of rotation of balls
s = stiffness of each ball spring
s* = Stiffness of auxiliary spring
FC1 = Centrifugal force corresponding to minimum speed = m x ω2 x r1
FC2 = Centrifugal force corresponding to maximum speed = m x ω2 x r2
F  X  S Y
F
Y
S
X
Total downward force on sleeve= Weight of sleeve + Force at end “A” of lever due to auxiliary spring
W  F
Y

W    S 
X

Y

 M g  S
X

Taking moments about point O, we get

Y

M

g


S



X

   b.......................(1)
 FC  P   a  
2




 values of centrifugal
 force, tension in main spring, tension in auxiliary spring
Let FC1, P1, and S1 are corresponding
for minimum speed. And
Let FC2, P2, and S2 are corresponding values of centrifugal force, tension in main spring, tension in auxiliary spring
for maximum speed.
Substitute these values in equation (1),
We have for minimum speed


Y

M

g


S


1

X

   b.................(2)
FC1  P1  a  
2







Similarly, for maximum speed


Y

M

g


S

2

X

   b.....................(3)
FC2  P2  a  
2







Subtracting equation (2) from equation (3), we have


 FC  FC   P2  P1   a   S2  S1   y  b ..................(4)
1
 2

x 2
P2-P1 = Net pull (or tension) in two main spring when radius increases from r2 to r1.
Note: Here, we consider that there are two main spring of half length.
= 2 x [Force exerted by each main spring]
= 2 x [Stiffness of main spring x extension of ball springs]
 2   s   d 2  d1 
 2   s  2  r2  r1  
 4s  r2  r1 
The net force in auxiliary spring is given by
S2 –S1 = Stiffness of auxiliary spring x extension of auxiliary spring
S2  S1  s*  h*...........................(5)
h h*

x y
h*  h 
Put
h
h* 
y
x
b
 r2  r1 
a
from equation (3) [Hartnell governor (Previous article]
b
y
 r2  r1  
a
x
Put the value of h* in equation (5)
b y
S2  S1  s*. r2  r1   
a x
Now, Substituting the values of (P2-P1) and (S2 – S1) in equation (4)
 FC  FC  4s  r2  r1   a   s*. r2  r1   b  y   y  b
1
 2

a x x 2



 FC
2
 FC
2

b y y b 1
 FC1  4s  r2  r1   s*. r2  r1      
a x x 2 a
 FC1

s*
b y
 4s  r2  r1   . r2  r1    
2
a x
Dividing by (r2 – r1) to both sides, we get
FC2  FC1
r2  r1
s*  b y 
 4s  .  
2 a x
FC2  FC1
r2  r1
2
s*  b y 
 4s  .  
2 a x
2
If auxiliary spring is not used, then s*=0, then
FC2  FC1
r2  r1
 4.s
2
Some important definitions
a) Sensitivity: A governor is said to be sensitive, if displacement of sleeve is bigger for small change in speed.
Sensitivity is also defined as the ratio of ‘difference between the maximum and minimum equilibrium speeds’
to the ‘mean equilibrium speed’.
Let N1= minimum equilibrium speed corresponding to full load condition
N2 = Maximum equilibrium speed corresponding to zero load condition
N= Mean equilibrium speed  N1  N 2
2
Sensitivity of governor =(Difference of maximum and minimum equilibrium speeds)/(Mean equilibrium speed)

N 2  N1 N 2  N1 2( N 2  N1 )


N1  N 2
N
N1  N 2
2
b) Stability: A governor is said to be stable when for each speed there is only one radius of rotation of
governor balls at which the governor is in equilibrium. The speed should be within working range of governor.
c) Isochronisms: A governor is said to be isochronous if the equilibrium speed is constant for all radii of rotation of
the balls within the working range. This means that when radius of rotation changes from minimum radius to
maximum radius, the equilibrium speed remains constant.
c) Hunting: If the speed of engine controlled by the governor fluctuates continuously above and below the mean
speed, the governor is said to be hunting. This is caused by too sensitive governor which changes the fuel supply
by a large amount when a small change in speed of rotation takes place.
Governor effort
•The effort of governor is the force exerted by the governor at the sleeve and sleeve tends to move.
•When speed is constant, force exerted to the sleeve is zero as sleeve does not tends to move.
•When speed changes, sleeve tends to move to its new equilibrium position and hence a force exerted on the
sleeve. This force gradually diminishes to zero as sleeve moves to new equilibrium position
The mean force exerted on the sleeve during a given change of speed, is known as effort of the governor.
Governor power
The power of governor is defined as the work done at the sleeve for a given percentage change of speed. Hence
the power of governor is the product of the governor effort and the displacement of the sleeve.
Power of governor = (Governor effort) X (Displacement of sleeve)
Method of determining the effort and power of a governor
The effort and power of a governor may be determined by following method.
Let us apply the method for porter governor. The same method can be used for other types of governor.
Let
N= Equilibrium speed corresponding
W= Weight of sleeve=Mxg where M is mass of sleeve
h = height of governor corresponding to speed N
c.N= Increase of speed
c= A factor which when multiplied to equilibrium speed, gives the increase in speed
Increased speed= Equilibrium speed + increase of speed
 N  c.N  (1  c) N [example 205(inreased speed) =200 (Equilibrium speed)+5 (Increase of speed)]
h1 =Height of governor corresponding to increased speed (1+c)N
The equilibrium position of the governor for increased speed is shown in figure (b)
In order to prevent the sleeve from rising when the increase of speed takes place, a downward force will have to
be exerted on the sleeve.
Let W1 = Weight of sleeve (which is more than W) so that rising of sleeve is prevented when the speed is (1+c)N.
This means that W1 is the weight of sleeve when height of governor is h.
Also
W1  M1  g
Downward force to be applied when the rising of sleeve is
to be prevented when speed increases from N to (1+c)N= W1  W
When speed is N r.p.m., and angles α and β are equal so that k=1, the height h is given by equation
h
m  M 894.56

.................(i)
m
N2
if speed increases to (1+c)N and height remains the same and mass of sleeve is M1, then we have
h
m  M1
894.56

.................(ii )
m
(1  c)2 N 2
Where M1 is the mass of sleeve corresponding to weight W1.
Equating the two values of h given by equations (i) and (ii), we get
m  M 894.56 m  M1
894.56



m
N2
m
(1  c)2 N 2
m  M  (m  M 1 ) 
1
(1  c)2
894.56


Cancelling
.
.to.both.sides 

2
mN


(1  c) 2 (m  M )  (m  M 1 )
M 1  (m  M )(1  c) 2  m
Subtracting the M from both sides
M 1  M  (m  M )(1  c) 2  m  M
M1  M  (m  M )(1  c)2   m  M 
M 1  M  (m  M ) (1  c) 2  1
Now
W1  W   M1  g   (M  g )
W1  W   M1  M   g
Put the value of (M1-M) in above equation
W1  W  (m  M ) (1  c) 2  1  g

M1  M  (m  M ) (1  c)2  1

The mean force P exerted on the sleeve during the change of speed from N to (1+c)N is given by:
2
W1  W (m  M ) (1  c)  1  g
P

2
2

(m  M ) 1  c 2  2c  1  g
2
 c 2  2c 
 (m  M )  g 

 2 
 c 2  2c 
 (m  M )  g 

 2 
 2c 
 (m  M )  g  
2
(Neglecting C 2 the which is very small)
 (m  M )  g  c
 c(m  M )  g
Governor effort:
P  c(m  M )  g
Power of governor
Power of governor is defined as the product of governor effort and displacement of sleeve
Let
x= displacement of sleeve when speed increases from N to (1+c)N
h= Height of governor when speed is N
h1 = height of governor when speed is (1+c)N. The mass of sleeve is not changed i.e.
mass of sleeve is M
Then
x  2(h  h1 ).....................( A)
h
m  M 894.56

m
N2
(when speed is N, height is h)
h1 
mM
894.56

m
(1  c)2 N 2
mM
894.56

h1
m
(1  c) 2 N 2

m  M 894.56
h

m
N2
h1
1

h (1  c) 2
h1 
h
(1  c) 2
Substituting the value of h1 in equation (A), we get

h 
x  2 h 
2 
(1

c
)


 (1  c)2  1 
x  2h 
2 
 (1  c) 
 (1  c 2  2c  1 
x  2h 

2
(1

c
)


 c 2  2c 
x  2h 

2
 1  c  2c 
(When speed is (1+c)N, height is h1)
x
2h  2c
(1  2c)
x  2h 
(Neglecting C 2 the which is very small)
2c
(1  2c)
Now Power of governor = Governor effort X displacement of sleeve
 P x
(Here, P= Governor effort and displacement of sleeve = x)
  c ( m  M )  g   2h 
2c
(1  2c)
4c 2
Governor power 
(m  M )  g.h
(1  2c)

2c 
 P   c(m  M )  g  and .x  2h 

(1

2
c
)


Controlling force
When the speed of rotation is uniform, the balls of a governor are subjected to outward centrifugal force, which
tend to move them outwards.
But the outward movement of balls is resisted by an equal and opposite force acting radially inwards. This inward
force is known as controlling force.
Hence controlling force is equal to centrifugal force but acting in the opposite direction. It is denoted by Fc.
Controlling force:
FC  mr 2
Controlling force is provided by:
(i)
Weight of rotating mass in a watt governor
(ii) Weight of rotating mass and weight of sleeve in a Porter governor
(iii) Compression spring in case of a Hartnell governor
Controlling force curve
Controlling force curve is a curve which shows the variation of the controlling force with radius of rotation.
The radius of rotation is taken along x-axis whereas controlling force is taken along y axis.
The curve is useful in finding the stability and sensitivity of a governor
The curve also shows the effect of friction.
In figure, Curve OAB represents the controlling force curve.