Governors Dr. L. S.

Al-Ansari

Governors
Introduction:
The resultant
The function of a governor is to regulate the mean speed of an engine, when there
are variations in the load e.g. when the load on an engine increases, its speed
decreases, therefore it becomes necessary to increase the supply of working fluid. On
the other hand, when the load on the engine decreases, its speed increases and thus
less working fluid is required. The governor automatically controls the supply of
working fluid to the engine with the varying load conditions and keeps the mean
speed within certain limits.
A little consideration will show, that when the load increases, the configuration of
the governor changes and a valve is moved to increase the supply of the working fluid
; conversely, when the load decreases, the engine speed increases and the governor
decreases the supply of working fluid.
Types of Governors:
The governors may, broadly, be classified as
1. Centrifugal governors, and 2. Inertia governors.
The centrifugal governors, may further be classified as follows :

Centrifugal Governors:
The centrifugal governors are based on the balancing of centrifugal force on the
rotating balls by an equal and opposite radial force, known as the controlling force*.It
consists of two balls of equal mass, which are attached to the arms as shown in Fig.
18.1. These balls are known as governor balls or fly balls. The balls revolve with a
spindle, which is driven by the engine through bevel gears. The upper ends of the
arms are pivoted to the spindle, so that the balls may rise up or fall down as they
revolve about the vertical axis. The arms are connected by the links to a sleeve, which
is keyed to the spindle. This sleeve revolves with the spindle ; but can slide up and
down. The balls and the sleeve rises when the spindle speed increases, and falls when
the speed decreases. In order to limit the travel of the sleeve in upward and downward
directions, two stops S, S are provided on the spindle. The sleeve is connected by a
bell crank lever to a throttle valve. The supply of the working fluid decreases when
the sleeve rises and increases when it falls.
When the load on the engine increases, the engine and the governor speed
decreases. This results in the decrease of centrifugal force on the balls. Hence the
balls move inwards and the sleeve moves downwards. The downward movement of
the sleeve operates a throttle valve at the other end of the bell crank lever to increase
the supply of working fluid and thus the engine speed is increased. In this case, the

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extra power output is provided to balance the increased load. When the load on the
engine decreases, the engine and the governor speed increases, which results in the
increase of centrifugal force on the balls. Thus the balls move outwards and the sleeve
rises upwards. This upward movement of the sleeve reduces the supply of the working
fluid and hence the speed is decreased. In this case, the power output is reduced.
Note : When the balls rotate at uniform speed, controlling force is equal to the
centrifugal force and they balance each other.

Fig. 18.1. Centrifugal governor.

Terms Used in Governors:
The following terms used in governors are important from the subject point of view;
1. Height of a governor. It is the vertical distance from the centre of the ball to a
point where the axes of the arms (or arms produced) intersect on the spindle axis. It is
usually denoted by h.
2. Equilibrium speed. It is the speed at which the governor balls, arms etc., are in
complete equilibrium and the sleeve does not tend to move upwards or downwards.
3. Mean equilibrium speed. It is the speed at the mean position of the balls or the
sleeve.
4. Maximum and minimum equilibrium speeds. The speeds at the maximum and
minimum radius of rotation of the balls, without tending to move either way are
known as maximum and minimum equilibrium speeds respectively.
Note : There can be many equilibrium speeds between the mean and the maximum
and the mean and the minimum equilibrium speeds.
5. Sleeve lift. It is the vertical distance which the sleeve travels due to change in
equilibrium speed.
Watt Governor:
The simplest form of a centrifugal governor is a Watt governor, as shown in Fig.
18.2. It is basically a conical pendulum with links attached to a sleeve of negligible
mass. The arms of the governor may be connected to the spindle in the following
three ways :
1. The pivot P, may be on the spindle axis as shown in Fig. 18.2 (a).
2. The pivot P, may be offset from the spindle axis and the arms when produced
intersect at O, as shown in Fig. 18.2 (b).

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3. The pivot P, may be offset, but the arms cross the axis at O, as shown in Fig. 18.2
(c).

Fig. 18.2. Watt governor.

Let:
m = Mass of the ball in kg,
w = Weight of the ball in Newtons = m.g,
T = Tension in the arm in Newtons,
ω = Angular velocity of the arm and ball about the spindle axis in rad/s,
r = Radius of the path of rotation of the ball i.e. horizontal distance from the centre
of the ball to the spindle axis in metres,
FC = Centrifugal force acting on the ball in Newtons = m.ω2.r, and
h = Height of the governor in metres.
It is assumed that the weight of the arms, links and the sleeve are negligible as
compared to the weight of the balls. Now, the ball is in equilibrium under the action
of
1. the centrifugal force (FC) acting on the ball,
2. the tension (T) in the arm, and
3. the weight (w) of the ball.
Taking moments about point O, we have
FC × h = w × r = m.g.r or m.ω2.r.h = m.g.r or h = g /ω2 . . . (i)
When g is expressed in m/s2 and ω in rad/s, then h is in metres. If N is the speed in
r.p.m., then ω = 2π N/60
9.81 895
ℎ = (2𝜋𝑁/60)2 = (𝑁)2 . . . (∵ g = 9.81 m/s2) . . . (ii)
Note : We see from the above expression that the height of a governor h, is inversely
proportional to N2. Therefore at high speeds, the value of h is small. At such speeds,
the change in the value of h corresponding to a small change in speed is insufficient
to enable a governor of this type to operate the mechanism to give the necessary
change in the fuel supply. This governor may only work satisfactorily at relatively low
speeds i.e. from 60 to 80 r.p.m.
Example 18.1. Calculate the vertical height of a Watt governor when it rotates at
60 r.p.m. Also find the change in vertical height when its speed increases to 61 r.p.m.
Solution. Given : N1 = 60 r.p.m. ; N2 = 61 r.p.m.
Initial height
We know that initial height,
895 895
ℎ1 = = = 0.248 𝑚
(𝑁1 )2 (60)2
Change in vertical height
We know that final height,

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895 895
ℎ2 = 2
= = 0.24 𝑚
(𝑁2 ) (61)2

 Change in vertical height= h1 – h2 = 0.248 – 0.24 = 0.008 m = 8 mm Ans.

Porter Governor:
The Porter governor is a modification of a Watt’s governor, with central load
attached to the sleeve as shown in Fig. 18.3 (a). 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 predetermined level.
Consider the forces acting on one-half of the governor as shown in Fig. 18.3 (b).

Fig. 18.3. Porter governor.

Let m = Mass of each ball in kg,
w = Weight of each ball in Newtons = m.g,
M = Mass of the central load in kg,
W = Weight of the central load in Newtons = M.g,
r = Radius of rotation in metres,
h = Height of governor in metres ,
N = Speed of the balls in r.p.m .,
ω = Angular speed of the balls in rad/s = 2 πN/60 rad/s,
FC = Centrifugal force acting on the ball in Newtons = m.ω2.r,
T1 = Force in the arm in Newtons,
T2 = Force in the link in Newtons,
α = Angle of inclination of the arm (or upper link) to the vertical, and
β = 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
𝑊 𝑀. 𝑔
𝑇2 𝐶𝑜𝑠(𝛽) = =
2 2
𝑀.𝑔
or 𝑇2 = 2𝐶𝑜𝑠(𝛽) . . . (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).
(i) The weight of ball (w = m.g),

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(ii) The centrifugal force (FC),

(iii) The tension in the arm (T1), and
(iv) The tension in the link (T2).
Resolving the forces vertically,
𝑀.𝑔
𝑇1 𝐶𝑜𝑠(𝛽) = 𝑇2 𝐶𝑜𝑠(𝛽) + 𝑤 = 2 + 𝑚. 𝑔 . . . (ii)

Resolving the forces horizontally,

𝑇1 𝑆𝑖𝑛(𝛼) + 𝑇2 𝑆𝑖𝑛(𝛽) = 𝐹𝐶
𝑀.𝑔
𝑇1 𝑆𝑖𝑛(𝛼) + 2𝐶𝑜𝑠(𝛽) 𝑆𝑖𝑛(𝛽) = 𝐹𝐶
𝑀.𝑔
𝑇1 𝑆𝑖𝑛(𝛼) + 𝑇𝑎𝑛(𝛽) = 𝐹𝐶
2
𝑀.𝑔
𝑇1 𝑆𝑖𝑛(𝛼) = 𝐹𝐶 − 2 𝑇𝑎𝑛(𝛽) . . . (iii)
Dividing equation (iii) by equation (ii),
𝑀.𝑔
𝑇1 𝑆𝑖𝑛(𝛼) 𝐹𝐶 − 2 𝑇𝑎𝑛(𝛽)
= 𝑀.𝑔
𝑇1 𝐶𝑜𝑠(𝛼) + 𝑚. 𝑔
2

𝑀. 𝑔 𝑀. 𝑔
( + 𝑚. 𝑔) 𝑇𝑎𝑛(𝛼) = 𝐹𝐶 − 𝑇𝑎𝑛(𝛽)
2 2
𝑀. 𝑔 𝐹𝐶 𝑀. 𝑔 𝑇𝑎𝑛(𝛽)
+ 𝑚. 𝑔 = −
2 𝑇𝑎𝑛(𝛼) 2 𝑇𝑎𝑛(𝛼)
𝑇𝑎𝑛(𝛽) 𝑟
Substituting 𝑇𝑎𝑛(𝛼) = 𝑞 and 𝑇𝑎𝑛(𝛼) = ℎ we have
𝑀. 𝑔 ℎ 𝑀. 𝑔
+ 𝑚. 𝑔 = 𝑚. 𝜔2 . 𝑟. − .𝑞
2 𝑟 2

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) or q=1
Therefore, the equation (v) becomes

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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

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, we find
that the mass of the central load (M) increases the height of governor in the
ratio((m+M)/m).
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,

Fig. 18.4. Instantaneous centre method.

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Example 18.2. A Porter governor has equal arms each 250 mm long and pivoted on
the axis of rotation. Each ball has a mass of 5 kg and the mass of the central load on
the sleeve is 25 kg. The radius of rotation of the ball is 150 mm when the governor
begins to lift and 200 mm when the governor is at maximum speed. Find the minimum
and maximum speeds and range of speed of the governor.
Solution. Given : BP = BD = 250 mm = 0.25 m ; m = 5 kg ; M = 15 kg ; r1 = 150
mm = 0.15m; r2 = 200 mm = 0.2 m
First of all, let us find the minimum and maximum speed of the governor. The
minimum and maximum position of the governor is shown in Fig. 18.6 (a) and (b)
respectively.
Let N1 = Minimum speed when r1 = BG = 150 mm, and
N2 = Maximum speed when r2 = BG = 200 mm.

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(a) Minimum Position. (b) Maximum Position.

Fig. 18.6

We know that speed range of the governor = N2 – N1 = 204.4 – 177 = 27.4 r.p.m.
Ans.
Speed range when friction at the sleeve is equivalent of 20 N of load (i.e. when F
= 20 N)
We know that when the sleeve moves downwards, the friction force (F) acts
upwards and the minimum speed is given by

We also know that when the sleeve moves upwards, the frictional force (F) acts
downwards and the maximum speed is given by

We know that speed range of the governor = N2 – N1 = 210 – 172 = 38 r.p.m. Ans.

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Example 18.5. A Porter governor has all four arms 250 mm long. The upper arms
are attached on the axis of rotation and the lower arms are attached to the sleeve at a
distance of 30 mm from the axis. The mass of each ball is 5 kg and the sleeve has a
mass of 50 kg. The extreme radii of rotation are 150 mm and 200 mm. Determine the
range of speed of the governor.
Solution. Given : BP = BD = 250 mm ; DH = 30 mm ; m = 5 kg ; M = 50 kg ;
r1 = 150 mm ; r2 = 200 mm
First of all, let us find the minimum and maximum speed of the governor. The
minimum and maximum position of the governor is shown in Fig. 18.8 (a) and (b)
respectively.

(a) Minimum Position. (b) Maximum Position.

Fig. 18.8
Let N1 = Minimum speed when r1 = BG = 150 mm ; and
N2 = Maximum speed when r2 = BG = 200 mm.
From Fig. 18.8 (a), we find that height of the governor,

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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. 18.12 (a). 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.
18.12 (b). The instantaneous centre (I) lies on the intersection of the line PF produced
and the line from D drawn perpendicular to the spindle axis. The perpendicular BM is
drawn on ID.

Fig. 18.12. Proell governor.

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Notes : 1. The equation (i) may be applied to any given configuration of the
governor.
2. Comparing equation (iii) with the equation (v) of the Porter governor (Art. 18.6),
we see that the equilibrium speed reduces for the given values of m, M and h. Hence
in order to have the same equilibrium speed for the given values of m, M and h, balls
of smaller masses are used in the Proell governor than in the Porter governor.
3. When α = β, then q = 1. Therefore equation (iii) may be written as

Example 18.9. A Proell governor has equal arms of length 300 mm. The upper and
lower ends of the arms are pivoted on the axis of the governor. The extension arms of
the lower links are each 80 mm long and parallel to the axis when the radii of
rotation of the balls are 150 mm and 200 mm. The mass of each ball is 10 kg and the
mass of the central load is 100 kg. Determine the range of speed of the governor.
Solution. Given : PF = DF = 300 mm ; BF = 80 mm ; m = 10 kg ; M = 100 kg ;
r1 = 150 mm; r2 = 200 mm
First of all, let us find the minimum and maximum speed of the governor. The
minimum and maximum position of the governor is shown in Fig. 18.13.
Let N1 = Minimum speed when radius of rotation, r1 = FG = 150 mm ; and
N2 = Maximum speed when radius of rotation , r2 = FG = 200 mm.
From Fig. 18.13 (a), we find that height of the governor,
h1 = PG == 260 mm = 0.26 m

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(a) Minimum Position. (b) Maximum Position.

Fig. 18.13

Example 18.11. The following particulars refer to a Proell governor with open
arms :
Length of all arms = 200 mm ; distance of pivot of arms from the axis of rotation =
40mm ; length of extension of lower arms to which each ball is attached = 100mm ;
mass of each ball = 6 kg and mass of the central load = 150 kg. If the radius of
rotation of the balls is 180 mm when the arms are inclined at an angle of 40° to the
axis of rotation, find the equilibrium speed for the above configuration.

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Solution. Given : PF = DF = 200 mm ; PQ = DK = HG = 40 mm ; BF = 100 mm ;

m = 6 kg; M = 150 kg ; r = JG = 180 mm = 0.18 m ; α = β = 40°
Let N = Equilibrium speed.

Fig. 18.15. All dimensions in mm.

From the equilibrium position of the governor, as shown in Fig. 18.15, we find that

Example 18.12. A Proell governor has all four arms of length 305 mm. The upper
arms are pivoted on the axis of rotation and the lower arms are attached to a sleeve
at a distance of 38 mm from the axis. The mass of each ball is 4.8 kg and are attached
to the extension of the lower arms which are 102 mm long. The mass on the sleeve is
45 kg. The minimum and maximum radii of governor are 165 mm and 216mm.

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Assuming that the extensions of the lower arms are parallel to the governor axis at
the minimum radius, find the corresponding equilibrium speeds.
Solution. Given : PF = DF = 305 mm ; DH = 38 mm ; BF = 102 mm ; m = 4.8 kg ;
M = 54 kg
Equilibrium speed at the minimum radius of governor
The radius of the governor is the distance of the point of intersection of the upper
and lower arms from the governor axis. When the extensions of the lower arms are
parallel to the governor axis, then the radius of the governor (FG) is equal to the
radius of rotation (r1).
The governor configuration at the minimum radius (i.e. when FG = 165 mm) is
shown in Fig. 18.16.
Let N1 = Equilibrium speed at the minimum radius i.e. when FG = r1 = 165 mm.
From Fig. 18.16, we find that

Fig. 18.16

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Note : The valve of N1 may also be obtained by drawing the governor configuration
to some suitable scale and measuring the distances BM, IM and ID. Now taking
moments about point I,

Equilibrium speed at the maximum radius of governor

Let N2 = Equilibrium speed at the maximum radius of governor, i.e. when F1G1 = r2
= 216 mm.

The governor configuration at the maximum radius of F1G1 = 216 mm is shown in

Fig. 18.17. From the geometry of the figure,

Fig. 18.17
Since the extension is rigidly connected to the lower arm (i.e. DFB or D1F1B1 is
one continuous link) therefore B1D1 and angle B1D1F1 do not change. In other
words,
B1D1 = BD = 400 mm and γ – β = γ1 – β1 or γ1 = γ – β + β1= 18.5° – 24.6° + 35.7° =
29.6°
∴ Radius of rotation, r2 = M1D1 + D1H1 = B1D1 × sin γ1 + 38 mm
= 400 sin 29.6° + 38 = 235.6 mm = 0.2356 m
From Fig. 18.17, we find that
B1M1 = B1D1 × cos γ1 = 400 × cos 29.6° = 348 mm = 0.348 m
F1N1 = F1D1 × cos β1 = 305 × cos 35.7° = 248 mm = 0.248 m

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I1N1 = F1N1 × tan α1 = 0.248 × tan 45.1° = 0.249 m

N1D1 = F1D1 × sin β1 = 305 × sin 35.7 = 178 mm = 0.178 m
∴ I1D1 = I1N1 + N1D1 = 0.249 + 0.178 = 0.427 m
M1D1 = B1D1 sin γ1 = 400 sin 29.6° = 198 mm = 0.198 m
∴ I1M1 = I1D1 − M1D1 = 0.427 – 0.198 = 0.229 m
We know that centrifugal force,

Note : The value of N2 may also be obtained by drawing the governor configuration
to some suitable scale and measuring the distances B1M1, I1M1 and I1D1.

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