Governor

Chapter: Governor
Subtopics
Terms Used in Governors
Height of a governor, Equilibrium speed, Sleeve lift
Performance of Governors
Sensitiveness , Stability, Hunting, Isochronism, Governor Effort, Power
Types, Application, Function
With sketch explain construction working of
1.watt Governor
2.Porter governor
3.Proell Governor
4.Harnell Governor
Numerical Problems
Comparison of Flywheel and Governor
MCQs

Terms Used in Governors
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.
5. Sleeve lift – It is the vertical distance which the sleeve
travels due to change in equilibrium speed.
Note: There can be many equilibrium speeds between the mean
and the maximum and the mean and the minimum
equilibrium speeds

Performance of Governors :
1. Sensitiveness:
The bigger the displacement of the sleeve for a given fractional
change of speed, the more sensitive is the governor.
Sensitiveness is more correctly defined as the ratio of the
difference between the maximum and minimum equilibrium
speeds to the mean equilibrium speed.
A too sensitive governor changes the fuel supply by a large amount
when a small change in the speed of rotation takes place. This
causes wide fluctuations in the engine speed, resulting in the
hunting of the governor.

If,
Nmax = maximum equilibrium speed
Nmin = minimum equilibrium speed
Nmean = mean equilibrium speed
Nmean = ( Nmax +Nmin ) / 2
Speed Range : Nmax – Nmin
Then, Sensitiveness = (Nmax – Nmin) / Nmean
Putting Nmean Value in Above Equation we get,
Sensitiveness = 2( Nmax – Nmin) / ( Nmax +Nmin )

2. Stability:
A governor is said to be stable when for each speed within the working
range, there is only one radius of the governor balls at which the
governor is in equilibrium. For a stable governor, if the equilibrium
speed increases, the radius of the governor balls must also increase.
3. Isochronism :
A governor is said to be isochronous, when the equilibrium speed is
constant for all radii of rotation of the balls, within the working
range. An isochronous governor will be infinitely sensitive.

4. Hunting:
It is a condition in which the speed of the engine controlled by
the governor fluctuates continuously above and below the
mean speed. It is caused by a governor which is too sensitive.
5. Governor effort:
The effort of a governor is the force it can exert at the sleeve on
the mechanism, which controls the supply of fuel to the
engine. The mean force exerted during the given change of
speed is termed as effort. Generally efforts are defined for 1%
change of speed.

6. Power:
The power of a governor is defined as the work done at the
sleeve for a given percentage change of speed.
Power = efforts X displacement of sleeve

Function Of Governor
The function of the governor is to increase the supply of
working fluid going to the prime mover when the load on the
prime-mover increases and to decrease the supply when the
load decreases so as to keep the speed of the prime-mover
almost constant at different loads.
Governor is controlling the RPM of an engine by regulating the fuel supply

Watt Governor
• Figure
• Construction
• Working

Watt governor is
the simplest form of
centrifugal
governors with no
central load
Figure

Working Principle
Centrifugal Governor is shown above in the figure which maintains the
mean speed of the engine when there is variations in the load.
A centrifugal governor is a specific type of governor with a feedback system
that controls the speed of an engine by regulating the flow of fuel
or working fluid, so as to maintain a near-constant speed. It uses
the principle of proportional control.

i) When the engine speed increases :-
When the speed of engine increases , the load on the engine
decreases and the speed of rotation of spindle increases. The
centrifugal force on balls increases and the balls move upwards
and hence the sleeve moves upward. As the sleeve moves
upward . The upward movement of sleeve causes the throttle
valve at the end of the ball crank lever to decrease the fuel
supply. The power output is reduced.

ii) When the engine speed decreases :-
When the engine speed decreases, the load on engine
increases and speed of rotation of spindle decrease. The
centrifugal force on balls decreases and the balls moves
downwards. As the balls move downwards, hence the sleeve
moves downward which causes the throttle valve to increase
the fuel supply and the power output is increased.

W=weight
Fc= Centrifugal force
FC × h= w × r = m.g.r
Or
FC × h = m. ω 2.r.h
Where
g is expressed in m/s2
ω in rad/s,
h is in metres.
N is the speed in r.p.m.,
WATT GOVERNOR Formulae

Problem 1.
A Watt governor is mounted on an engine and runs at 70
rpm,
its speed is reduced to 68 rpm when load on the engine
increases.
Find
(I) the height of the governor at 70 rpm and
(II) also find the change in the height of the governor.
Numerical Problems on Watt Governor

We know that,
h1 = 895/ N12
= 895/ 702
= 895/ 4900
= 0.18266 meter ----Answer 1
h2 = 895/ N22
= 895/ 682
= 895/ 4624
= 0.19355 meter
Solution:
Given Data,
N1 = 70 RPM
N2 = 68 RPM
Find
h1 = ?
h2 - h1 = ?

Change in height of the governor
= h2 - h1
= 0.19355- 0.18266 = 0.01089 meter ----Answer 2

Problem 2.
The height of a governor is 300 mm at sometime. The load
on the engine is increased and the height is increased 340
mm.
Find the percentage change in the speed of the governor.
Solution:
Given data,
h1 = 300 mm = 0.3 meter
h2 = 340 mm = 0.34 meter
Find:
Percentage Change in speed = (N1 – N2 / N1) x 100 = ?

We know that,
h1 = 895/ N12 ,
0.3 = 895/ N12 ,
N12 = 895/ 0.3 = 2983.33,
N1 = 54.61 RPM
h2 = 895/ N22
0.34 = 895/ N22 ,
N22 = 895/ 0.34 = 2632.35,
N2 = 51.306 RPM
Change in speed,
N1 – N2 = 54.61 – 51.306 = 3.304 RPM
Percentage change in speed of the governor
= (N1 – N2 / N1) x 100
= (54.61 – 51.306 / 54.61) x 100
= (3.304 / 54.61) x 100
= 6.05 % ----Answer

Problem 3.
In Figure A and Figure B, the positions of the arms of the open
arm and crossed governors are shown with respect to spindle,
find out the heights of the both governors.

Figure A Figure B
Figure A and Figure B are given in
data.

In figure A,
Height, h1 = 120 cos300 + 20 cot 300
= 120 x 0.866 + 20 x 1.732
= 103.92 +34.643
= 138.561 mm ----Answer 1
In figure B,
Height, h2 = 200 cos 300 - 20 cot 300
= 200 x 0.866 – 20 x 1.732
= 173.20 – 34.641
= 138.599 mm ----Answer 2

Problem 4.
A crossed arm type Watt governor is shown in figure C. The length of each
arm is 300 mm.
The angle of inclination of each arm from the spindle axis is 350.
If each arm is pivoted at a distance of 30 mm from the axis of the spindle as
shown in figure C,
find out the speed of the Governor.

Solution:
The height of the governor,
h = 300 cos 350 - 30 cot 350
= 300 x 0.866 – 30 x 1.427
= 259.8 – 42.81
= 216.99 mm = 0.21699 meter
h= 0.21699 m
We know that,
h = 895/ N2 ,
N2 = 895 / h,
N2 = 895 / 0.21699 = 4124.61
N = 64.223 RPM ----Answer

With sketch explain construction working of
1.Porter governor
2.Proell Governor
3.Harnell Governor

Question:
With sketch explain construction working of porter governor

Construction :-
•Porter governors has two fly balls which are attached to the arms
of the porter governor .
•These two arms are pivoted to the top of the spindle. This spindle
is driven by the engine.
•The parts of the arms just above the fly balls are connected to the
central sleeve.

•This sleeve moves up and down according to the movement of
the balls.
•There are stoppers place in the spindle to limit the vertical
movement of the spindle.
•This sleeve is carries a heavy central load.
•The movement of the sleeve controls opening and closing of the
throttle valve.

Working :-
•When the load of engine decreases, there will be sudden increase in speed of
engine and spindle speed will also increase.
•As the spindle speed increases two fly balls also start rotating around the spindle
fast.
•The centrifugal force will push the balls outward making the balls move in
upward direction.
•Since the ball moves upward , the arms also moves upward and the sleeve
connected to the lower part of the arm also moves upward.
•Upward movement of sleeve actuates the throttle valve via a mechanism
connected to the sleeve to decrease the fuel supply to the engine.
•The decreases in fuel supply decreases the speed.
Hence speed is maintained.

•In the other case, when the load of the engine increases speed of the engine
decreases.
•As the engine speed decreases , speed of the spindle also decreases and
centrifugal force in the balls.
•Hence the balls come down with the arms.
•As the arms comes down , the sleeve connected to the arm also comes down
and it actuates the throttle valve which increase the fuel supply.
•Due to increase in fuel supply, speed of engine also increases.

Difference between watt and porter governor
The porter governor is a modification of a watts governor; with
central load attached to the sleeve.
The only difference between the watt and porter governor is the
inclusion of a dead weight

Question:
With sketch explain construction working of proell governor
Proell Governor -Construction And Working
•Proell Governor is a different type of governor in which the fly balls are
connected to the spindle using an extended arm.
•The fly balls are mounted on this additional arm. Like the porter governor,
proell governor also has central weight which increases the speed of rotation.

Construction :-
•Proell Governor has following parts :- spindle, central weight, arms,
extended arms, fly balls and sleeve. The spindle is connected to the engine
and its rotation speed increases and decreases with speed of the engine.
• Spindle has a central load to increase the speed of rotation. Two arms are
pivoted at the top of the spindle and these two arms are connected to the
extended arms which are connected to the fly balls.
•The other ends of the two arms which are pivoted to the top of the spindle is
connected to the sleeve and moves the sleeve up and down . The sleeve
actuates a mechanism which open and close the throttle valve

Working :-
•When the load on the engine decreases, the speed of engine increases
suddenly and also spindle rotation speed is increased as the spindle is
connected to the engine.
•As the rotation of spindle becomes fast, the arms pivoted to top of
spindle also rotates with high speed and the balls move outward due to
increased centrifugal forced on the balls.
• When the balls move outward, the sleeve connected to the arms moves
up and actuates a mechanism which closes the throttle valve and
decreases the fuel supply which decreases the engine speed. Hence the
engine speed is maintained.

On the other hand, when the load on the engine increases, speed of the engine
decreases. Since the engine speed decreases, speed of rotation of spindle also
decreases and hence the ball rotates at low speed and moves inward due to
decrease in centrifugal force. As the balls moves inward, the sleeve moves in
downward direction which actuates a mechanism which opens the throttle
valve and increases the fuels supply and hence the engine speed increases.
In this way speed is maintained by this governor in both the cases.

Difference between porter and proell governor
• The porter governor is a modification of a Watt's governor, with central
load attached to the sleeve. The load moves up and down the central
spindle.
• The Proell governor has the balls fixed at the extension of the links

Question:
With sketch explain construction working of Hartnell governor
Hartnell governor is a spring controlled centrifugal governor, in which
a spring controls the movement of the ball and hence the sleeve

Construction:-
•Hartnell Governor has a spindle which is connected the engine and rotates
with the same speed as that of engine.
•This spindle is connected with two fly balls which rotates and experience
centrifugal force when spindle rotates. These fly balls are connected to sleeve
via a bell crank lever which translate side ways motions of balls to up-down
motion of sleeve using rollers.
•The upper side of sleeve has a spring which restricts the upward movement
of sleeve. This spring has a nut which is used to adjust the force of the spring.
•The sleeve is connected to throttle valve through a link mechanism which
controls the fuel supply to the engine.

Working :-
Working at low load :-
•When the load on the engine decreases, the speed of the engine will
increase suddenly and hence the spindle speed will also increases. With the
spindle rotates the balls also starts rotating, then due to centrifugal force they
are pulled outward. As the balls move outward, the sleeve moves upward as
they are connected through bell crank lever.
•Bell crank lever change the angle of motion to 90 degrees. The sleeves are
connected to the bell crank lever via roller.

•As the balls move outward, these roller takes the ball upward.
•As the sleeve move upwards the spring present above the sleeve gets
compressed. This spring limits the movement of sleeve in upward direction.
As the sleeve move upward, the throttle valve which connected to the sleeve
through a link mechanism starts closing and the supply of working fuel
through the throttle valve gets reduces. As the fuel supply reduces, the speed
of the engine also get reduced.

Working at high load :-
When the load on the engine increases, the speed of engine will decrease and
requires more fuel to increase its speed. As the speed of the engine decreases,
the rotational speed of spindle which is connected to the engine also gets
decreased. Due to decrease in spindle speed, the speed of rotation of balls also
decreases and the centrifugal force on the balls also decreases.
So the balls comes closer to each other and since the sleeve is connected to the
balls via a bell crank lever , so the sleeve comes down .
The spring which is located above the sleeve also pushes the sleeve downward.
When the sleeve comes down, the throttle valve connected to the sleeve via a
link mechanism starts opening and the supply of working fuel to the engine
increases and hence the speed of engine is maintained

The height of the Porter Governor can be determined by the following
formula
Mass of each ball =m kg
Mass of the dead load on the sleeve = M Kg
Initial radius of rotation = r1
Final radius of rotation = r2

Examples of:
1.Porter governor
2.Theory and Equation Derivation  Hartnell Governor

Example 4: (Friction at sleeve is not considered, k=1)
The arms of a Porter governor are each 240 mm long and
pivoted on the governor axis. The mass of each ball is 4 kg and
the mass of the central sleeve is 16 kg. The radius of rotation of
the balls is 150 mm when the sleeve is in condition to rise, the
diameter of rotation of ball is 280 mm and at maximum speed
is 400 mm.
Find out:
i. Rage of speed
ii. Maximum lift of the sleeve

Solution:
Given Data
i. The length of each arm = 240 mm
ii. Mass of each ball =4 kg
iii. Mass of the dead load on the sleeve = 16 Kg
iv. Initial radius of rotation, r1 = 140 mm
v. Final radius of rotation, r2 = 200 mm
As per the question, each arm is pivoted on the axis of the spindle, so α =
β
Tan α = Tan β, value of K = 1

We know that,
------(1)
by formula
i. Ininitial position
h12 = (240)2- (140)2 = 57600 – 19600
h1 = 194.9 mm = 0.1949 m

From formula
-------(2)
On putting the values of h1, m , M and K in the above formula
N12 = 5 x 4592.09 = 22960.492
N1 = 151.527 RPM (Ans)

ii. final position
…….(3)
h22 = (240)2- (200)2 = 57600 – 40000 = 17600
h2 = 132.67 mm = 0.13267 m
On putting the values of h2, m , M and K in the equation 3
N22 = 5 x 6746.06 = 33730.308
N2 = 183.658 RPM

Range of speed
= N2 – N1
= 183.658 – 151.527
= 32.131 RPM …….Answer
Lift of the sleeve
= 2( h1 – h2)
= 2 ( 194.90 – 132.67) = 2x 62.23
= 124.46 mm …….Answer

Example 5: (Friction at sleeve is not considered, k=0)
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 15 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

The minimum and maximum positions of the governor are shown in Fig.

Minimum speed
when r1 = BG = 0.15 m
N1 = Minimum speed
Referring Fig.
height of the governor,

Minimum speed of the governor is given by

Referring Fig. ,
height of the governor,
Maximum speed
when r2 = BG = 0.2 m

Maximum speed of the governor is given by
Range of speed
…..ANSWER

Example: Porter governor
Theory and Equation Derivation  Hartnell Governor

Example 6: (Friction at sleeve is considered)
The arms of a Porter governor are each 250 mm long and pivoted on the
governor axis.
The mass of each ball is 5 kg and the mass of the central sleeve is 30 kg.
The radius of rotation of the balls is 150 mm when the sleeve begins to rise
and reaches a value of 200 mm for maximum speed.
Determine the speed range of the governor, If the friction at the sleeve is
equivalent of 20 N of load at the sleeve.

Solution:
Given : BP = BD = 250 mm ;
m = 5 kg ;
M = 30 kg ;
r1 = 150 mm ;
r2 = 200 mm
Minimum and maximum speed of the governor
N1 = Minimum speed when r1 = BG = 150 mm, and
N2 = Maximum speed when r2 = BG = 200 mm.
The minimum and maximum position of the governor is shown
in Fig.

Referring Fig. , height of the governor,
h1
150
250

Referring Fig. , height of the governor,
h2

Speed range when friction at the sleeve is equivalent of 20 N of load (i.e.
when F = 20 N)
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

Hartnell governor
is spring controlled
centrifugal
governor,
in which
a spring controls
the movement of
the ball and hence
the sleeve

A Hartnell governor is a spring loaded governor as shown in Fig. It consists of
two bell crank levers pivoted at the points O,O to the frame. The frame is
attached to the governor spindle and therefore rotates with it.
A helical spring in compression provides equal downward forces on the two
rollers through a collar on the sleeve.

Consider the forces acting at one bell crank lever. The minimum and
maximum position is shown in Fig.
Let h = the compression of the spring when the radius of rotation changes
from r1 and r2.
as shown in Fig. (a), the compression of the spring or the lift of sleeve ℎ1 is
given by
…….1

Similarly, for the maximum position i.e. when the radius of rotation
changes from r to r2,
as shown in Fig. (b), the compression of the spring or lift of sleeve ℎ2 is
given by
……..2
……3

Example: Hartnell Governor
Example
to find out:
Range of speed,
Sleeve lift,
Governor effort,
Power of the governor

Problems on Hartnell Governor
Problem 7:
A Hartnell governor having a central sleeve spring and two right-angled bell
crank levers moves between 290 rpm. and 310 rpm. for a sleeve lift of 15 mm.
The sleeve arms and the ball arms are 80 mm and 120 mm respectively. The
levers are pivoted at 120 mm from the governor axis and mass of each ball is
2.5 kg. The ball arms are parallel to the governor axis at the lowest equilibrium
speed. Determine
1. Loads on the spring at the lowest and the highest equilibrium speeds, and
2. Stiffness of the spring.

Problem 8:
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
1. the range of speed,
2. sleeve lift,
3. governor effort and
4. power of the governor in the following cases :
1. When the friction at the sleeve is neglected, and
2. When the friction at the sleeve is equivalent to 10 N.

Given : BP = BD = 250 mm ; m = 5 kg ; M = 25 kg ; r1 =150 mm ; r2 = 200 mm ;
F = 10 N
1. When the friction at the sleeve is neglected
First of all, let us find the minimum and maximum speed of rotation. The
minimum and maximum position of the governor is shown in Fig. (a) and (b)
respectively.
Let N1 = Minimum speed, and
N2 = Maximum speed.

Range of speed
We know that range of speed = N2 – N1 = 189 – 164 = 25 r.p.m.
Sleeve lift
x = 2 (h1 – h2) = 2 (200 – 150) = 100 mm = 0.1 m
Governor effort
Let c = Fraction Percentage increase in speed.
increase in speed or range of speed,
c.N1 = N2 – N1 = 25 r.p.m.
c = 25/N1 = 25/164 = 0.152
Governor effort
P = c (m + M) g = 0.152 (5 + 25) 9.81 = 44.7 N
Power of the governor = P.x = 44.7 × 0.1 = 4.47 N-m

2. When the friction at the sleeve is taken into account

Range of speed
We know that range of speed = N2 – N1 = 192.4 – 161 = 31.4 r.p.m.
Sleeve lift
The sleeve lift (x) will be same as calculated above.
Sleeve lift, x = 100 mm = 0.1 m
Governor effort
Let c = Fraction Percentage increase in speed.
We know that increase in speed or range of speed,
c.N1 = N2 – N1 = 31.4 r.p.m.
c = 31.4/N1 = 31.4/161 = 0.195
We know that governor effort,
P = c (m.g + M.g + F) = 0.195 (5 × 9.81 + 25 × 9.81 + 10) N= 57.4 N
We know that power of the governor= P.x = 57.4 × 0.1 = 5.74 N-m

COMPARISON OF FLYWHEEL AND GOVERNOR
Example of Proell Governor

BASIS OF
COMPARISON
FLYWHEEL GOVERNOR
Description
A flywheel is a mechanical device
specifically designed and attached
to the crank shaft so as to control
the fluctuations of speed during
different rotational operations.
A governor is a mechanical device
used to govern the speed of
machines. It allows the engine to run
at the selected speed without any
effect of changing the load.
Influence on
Speed
Flywheel controls the variation of
speed in each cycle of the
operating engine.
Governor controls the mean speed
only when load varies or when it’s
required by controlling the fuel supply
to the engine.

BASIS OF
COMPARISON
FLYWHEEL GOVERNOR
Weight
Flywheel is a heavy machine
component, though it cannot
change the speed of the
crankshaft.
Governor is relatively light machine
component used to limit engine
speed.
Main Function
Flywheel stores the energy
during the power stroke when
requirement is less and transfer
to shaft when it is required
during other strokes of engine.
Governor on the other hand, works
only to keep the mean speed of
running machine constant by
regulating the fuel supply. When the
load increases the fuel supply
increases by opening of the throttle.
The reverse is also true.

Condition
The flywheel may not be used if the
cyclic fluctuations of energy output are
small or negligible (insignificant).
Governor is essential for all types
of engines to adjust the fuel
supply as per the demand.
Mandatory
Flywheel is mandatory for the
machines like bikes for the start of
operation.
Governor is mandatory for
devices where constant speed is
desired like in generator.
Moments of
inertia
Flywheel has large moments of inertia
because of the heavy mass of the
rotating wheel. The wheel is usually
made up of high density cast iron.
The governor has relatively less
moments of inertia when
compared to flywheel.

Under
Operation
The flywheel is under operation
when the engine is running
because it is attached to the
crankshaft.
The governor is under
operation only when the
engine is not running at its
mean speed.
Extra
Mechanism
Flywheel does not need any extra
mechanism to supply back energy.
Governor needs a piston
(plunger) mechanism.
Size
The size of flywheel may differ
from one machine type to another.
Governors come in different
variety to be used in different
prime movers.

Energy
Storage
Flywheel is an energy storing
device and prevents the changes
in energy during each cycle of
engine operation.
Governor is not an energy
storing component.
Connection
Flywheel is connected to the crank
shaft and runs it at a constant
speed, both the flywheel and
crankshaft rotates simultaneously.
The governor is not
connected to the crankshaft;
it can sometimes be
electronically operated.

The height of the Proell Governor can be determined by the
following formula
Where
h = Height of governor
N= Speed of the balls in r.p.m
m = Mass of the Balls
M = Mass of the central load

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 216 mm.
Assuming that the extensions of the lower arms are parallel to the governor
axis at the minimum radius, find the corresponding equilibrium speeds.
Example 9

Given data : 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

N1 = Equilibrium speed at the minimum radius i.e. when FG = r1 = 165 mm.

From Fig, we find that height of the governor

MCQ Questions
1. The height of a Watt’s governor is
a) directly proportional to speed
b) directly proportional to (speed)2
c) inversely proportional to speed
d) inversely proportional to (speed)2
2. The height of a Watt’s governor is equal to
a) 8.95/N2
b) 89.5/N2
c) 895/N2
d) 8950/N2

3. A Watt’s governor can work satisfactorily at speeds from
a) 60 to 80 r.p.m
b) 80 to 100 r.p.m
c) 100 to 200 r.p.m
d) 200 to 300 r.p.m
Answer: a
Explanation: A watt’s governor may only work satisfactorily at low
speeds i.e. from 60 to 80 r.p.m.

4. A Watt’s governor is a spring loaded governor.
a) True
b) False
Answer: b
Explanation: A Hartnell governor is a spring loaded governor.
Watt’s governor is a pendulum type governor.

5. Which of the following is a pendulum type governor?
a) Watt’s governor
b) Porter governor
c) Hartnell governor
d) None of the mentioned
Answer: a
Explanation: Watt’s governor is a pendulum type governor.

6. The sensitiveness of a governor depends upon the lift of the sleeve.
a) True
b) False
Answer: a
Explanation: In general, the greater the lift of the sleeve corresponding to
a given fractional change in speed, the greater is the sensitiveness of the
governor.

7. For two governors A and B, the lift of sleeve of governor A is more than
that of governor B, for a given fractional change in speed. It indicates that
a) governor A is more sensitive than governor B
b) governor B is more sensitive than governor A
c) both governors A and B are equally sensitive
d) none of the mentioned
Answer: a
Explanation: In general, the greater the lift of the sleeve corresponding to a
given fractional change in speed, the greater is the sensitiveness of the
governor.

8. The sensitiveness of the governor ____________ as the speed range
decreases.
a) remains unaffected
b) decreases
c) increases
Answer: c
Explanation: For a given lift of the sleeve, the sensitiveness of the
governor increases as the speed range decreases.

9. A governor is said to be stable, if the
a) radius of rotation of balls increases as the equilibrium speed decreases
b) radius of rotation of balls decreases as the equilibrium speed decreases
c) radius of rotation of balls increases as the equilibrium speed increases
d) radius of rotation of balls decreases as the equilibrium speed increases
Answer: b,c
Explanation: A governor is said to be stable when for every speed within
the working range there is a definite configuration

10. When the radius of rotation of balls ______________ as the equilibrium
speed increases, the governor is said to be unstable.
a) remains constant
b) decreases
c) increases

11. A governor is said to be isochronous when range of speed is zero for all
radii of rotation of the balls within the working range, neglecting friction.
a) True
b) False
Answer: a
A governor is said to be isochronous when the equilibrium speed is constant
(i.e. range of speed is zero) for all radii of rotation of the balls within the
working range, neglecting friction. The isochronism is the stage of infinite
sensitivity.

12. When the speed of the engine fluctuates continuously above and below
the mean speed, the governor is said to be
a) stable
b) unstable
c) isochronous
d) hunt

13. A very sensitive governor will cause hunting.
a) True
b) False
Answer a
14. Effort of a governor is the
a) mean force exerted at the sleeve for a given percentage change of speed
b) work done at the sleeve for maximum equilibrium speed
c) mean force exerted at the sleeve for maximum equilibrium speed

15. Power of a governor is the
a) mean force exerted at the sleeve for a given percentage change of speed
b) work done at the sleeve for maximum equilibrium speed
c) mean force exerted at the sleeve for maximum equilibrium speed
Answer: d
Explanation: The power of a governor is the work done at the sleeve for a
given percentage change of speed. It is the product of the mean value of
the effort and the distance through which the sleeve moves.
Mathematically,
Power = Mean effort × lift of sleeve

Governor

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