Governor

MECHANIS OF MATERIAL & MACHINES B 1

UNIT # GOVERNOR

TYPES OF

8

GOVERNOR

1-Centrefugal GOVERNOR It consists of two balls of mass which are attached to the mass as shown in figure. ARM

SPINDL E

FLY-BALL

FLY-BALL

S

LINK

SLEEVE

S

BAVEL GEAR

THROTTLE VALVE

These balls are known as governor balls and fly balls. The balls revolve with a spindle which is driven by the engine through bevel gears. The upper end of the arm are pivoted to the spindle, so that the balls may rise up or fall down as the revolve about the vertical axis. The arms connected by the links to a sleeve which is keyed to the spindle. This sleeve revolves with the

MECHANIS OF MATERIAL & MACHINES 2 spindle but can slide up and down. The balls and sleeve rises when the spindle speed increases and falls when the speed decreases. In order to limit the travel of a sleeve in up word and down word directions, two stops s,s are provided on the spindle. The sleeve is connected by a ball crank lever to a throttle valve. The supply of working fluids decreases when the sleeve rises and increases when falls. When the load on the engine increases the engine and governor speed decreases. This results in the decreases of centrifugal forces on the balls. Hence the ball move in words and the sleeve moves down words. The down word movement of the sleeve operates a throttle valve at the other end of the ball crank lever to increase the supply of working fluid and thus the engine speed is increased. In this case the extra power output is provided to balance the increased load. When the load on the engine decreases, the engine of the g speed increase which results in the increase of centrifugal force on the balls. Thus the balls move outward and the sleeve rise upwards. This upwards movement of the sleeve reduces the supply of working fluid and hence the speed is decreased. In this case the power output is reduced. WATT GOVERNOR The simplest form of centrifugal governor is Watt g is shown in figure.

It is basically a conical pendulum with links attached to a sleeve of negligible mass. The arms of governor may be connected to the spindle is following for three ways. 1. The pivot ‘P’ may be on spindle axis as shown in the figure (a). 2. The pivot ‘P’ may be offset from the spindle axis and the arms when produced intersect at ‘O’ as in figure ‘b’. The point ‘P’ may be offset but the arms cross the axis at ‘o’ as shown in figure ‘c’.

MECHANIS OF MATERIAL & MACHINES 3 Let , m = Mass of ball W = Weight of ball T = Tension in arm r = Radius fc = Centrifugal force h = Height of governor w = Angular velocity N = Spindle r.p.m. It is the assumed that the weight of the arms, links and the sleeve are negligible as compared to the weight of the balls. Now the balls in equilibrium under the action of 1. Centrifugal force (fc) 2. Tension of arm (T) 3. Weight of ball (w) Taking moment about “O” Fc × h = w × r m w r × h = mg × r w ×h=g h=g/w [w = 2πN/ 60] [g = 9.81 m/s ] h = 9.81/(2πN/ 60) [9.81/(2πN/ 60) ] h = 895 / N m [9.81 / (4 × 9.8696 N/ 3600)] N = Spindle r.p.m. QUESTION # 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. h1= 895 / N h1=895 / 60 h1= 0.248 mm h2= 895 / (61) h2= 0.240 mm Change vertical height = h1- h2 = 0.248 – 0.240 = 0.008 m Ans.

MECHANIS OF MATERIAL & MACHINES 4 PORTER GOVERNOR The porter governor is a modification of a watt governor with central load attached to the sleeve as shown below.

The load moves up and down the central spindle. Thos additional down word force increases the speed of revolution required to enable the ball to rise to any predetermined level . Consider the force acting on one – half of the governor as shown below.

Let m = Mass of each ball in kg. w or mg = Weight of each ball in Newton = mg M = Weight of central load in kg. = Mg r = Radius of rotation in meters. h = Height of governor in meters. w = Angular speed of balls in rad. / sec. = 2πN/ 60 rad. / sec Fc = Centrifugal force acting on each ball in Newton = mw r T1 = Tension in the arm in Newton .

MECHANIS OF MATERIAL & MACHINES 5 T2 = Tension in the link in Newton . α = Angle of inclination of the arm to the vertical. β = Angle of inclination of the link to the vertical. There are several ways of determining the relation between the height of governor (h) and the angular speed of the balls (w). The following two methods are important from subject point of view . 1. Method of relation of forces. 2. Instantaneous centre method. 1. Method forces:

of

relation

of Considering the equilibrium of

forces acting at ‘D’ we have, T2 cos β = W / 2 T2 cos β = Mg / 2 T2 = Mg / 2 cos β (1) Again considering the equilibrium of forces acting on ‘B’. The ‘B’is in equilibrium under the action of the following forces. I. Weight of the ball. II. Centrifugal force. III. Tension in arms. IV. Tension in the link. Resolving the forces vertically, T1 cosα = T2 cos β + W T1 cosα = mg + Mg / 2 (2) Resolving the forces horizontally , T1 sinα + T2 sin β = Fc T1 sinα = Mg / 2 cos β . sin β = Fc T1 sinα + Mg / 2 .tan β = Fc T1 sinα = Fc - Mg / 2 .tan β (3) Dividing eg (3) by eg (2), T1 sinα = Fc - Mg / 2 .tan β T1 cosα = mg + Mg / 2 (mg + Mg / 2) tanα = Fc – Mg / 2. tan β mg + Mg / 2 = Fc / tanα - Mg / 2 . tan β / tanα Substituting.

MECHANIS OF MATERIAL & MACHINES 6 tan β / tanα = q , tanα = r / h mg + Mg / 2 = m .w .r.h / r - Mg / 2 . q mg + Mg / 2 = m .w .r.h - Mg / 2 . q mg + Mg / 2 + Mg / 2 . q = m .w .h mg + Mg / 2(1 + q) = m .w .h [m + M / 2 (1 + 2)g] / m .w = m .w .h / m .w h = [m + M / 2 (1 + q)] / m = g / w (4) w = [m + M / 2 (1 + q)] / m × g / h (2πN/ 60) = [m + M / 2 (1 + q)] / m × g / h 39.44 N / 3600 = [m + M / 2 (1 + q)] / m × 9.81 / h 3600 × 39.44 N / 3600 = [m + M / 2 (1 + q)] / m × 9.81 / h × 3600 39.44 N = [m + M / 2 (1 + q)] / m × 35316 / h 39.44 N / 29.44 = [m + M / 2 (1 + q)] / m × 35316 / h × 39.44 N = [m + M / 2 (1 + q)] / m × 895 / h N = [m + M / 2] / m × 895 / h (5) When the length of arms are equal to the length and the points ‘P’ and ‘D’ lie on the same vertical line then , tanα = tan β q = tanα / tan β = 1 There equation (5) be comes, N = [m + M ] / m × 895 / h (6) QUESTION # 2 A porter governor has equal arms each 250 mm long and pivoted on the axis of rotation. Each ball has a mass of 5kg and the mass of load on the sleeve is 15kg. the radius of rotation of the ball is 150 mm. When the governor being to lift and 200 mm when the governor is at maximum speed. Find the minimum and maximum speed and range of speed of the governor.

MECHANIS OF MATERIAL & MACHINES 7

BP = BD = 250 mm =0.25 m m = 5kg M = 15 kg r1 = 150 mm = 0.15m r2 = 200 mm = 0.2m For minimum speed when r1 = BG = 0.15 m (N1) = m + M / m × 895 / h1 h1 = √ (250) - (150) h1 = 200 mm h1 = 0.2 m (N1) = m + M / m × 895 / h1 (N1) = 5 + 15 / 5 × 895 / 0.2 (N1) = 4 × 4475 N1 = √ 17900 N1 = 133.79 r.p.m. For maximum speed (N2) = m + M / m × 895 / h2 h2 = √ (250) - (200) h2 = 150 mm h2 = 0.15 m (N2) = m + M / m × 895 / h2 (N2) = 5 + 15 / 5 × 895 / 0.15 (N2) = 4 × 5966.67 N2 = √ 23866.67 N2 = 154.5 r.p.m. Range of increase of speed = N2 - N1 = 154.5 – 133.79

MECHANIS OF MATERIAL & MACHINES 8 Question # 3

=

20.71 r.p.m.

The arms of porter governor are each 250 mm long and pivoted on the governor axis. The mass of each ball is 5kg and mass of the central sleeve is 30kg. The radius of rotation of ball is 150 mm when the sleeve being to rise and reaches a value of 200 mm maximum speed. Determine the speed of governor. If the friction of the sleeve is equivalent of 20N load of the sleeve. Determine how the speed range is modified.

(N1) = m + M / m × 895 / h1 h1 = √ (250) - (150) h1 = 200 mm h1 = 0.2 m (N1) = 5 + 30 / 5 × 895 / 0.2 (N1) = 7 × 4475 N1 = √ 31325 N1 = 176.98 r.p.m. (N2) = m + M / m × 895 / h2 h2 = √ (250) - (200) h2 = 150 mm h2 = 0.15 m (N2) = m + M / m × 895 / h2 (N2) = 5 + 30 / 5 × 895 / 0.15 (N2) = 7 × 5966.67 N2 =√ 41766.69

MECHANIS OF MATERIAL & MACHINES 9 N2 = 204.36 r.p.m. Range increase speed = N2 - N1 = 204.36 – 176.98 = 27.38 r.p.m. When , F = 20 N (N1) = mg + (Mg – F) / mg × 895 / h1 [ For minimum position Friction show be Minus] (N1) = 5 × 9.81 + (30 × 9.81 - 20) / 5 × 9.81 × 895 / 0.2 (N1) = 49.05 + 274.3 / 49.05 × 4475 (N1) (N1) N1 N1 (N2)

= 49.05 + 274.3 / 49.05 × 4475 = 6.6 × 4475 = √ 29535 = 171.8 r.p.m. = mg + (Mg + F) / mg × 895 / h2 ∴[ For maximum position Friction show be positive] (N2) = 5 × 9.81 + (30 × 9.81 + 20) / 5 × 9.81 × 895 / 0.15 (N2) = 49.05 + 314.3 / 49.05 × 5966.7 (N2) = 7.4 × 5966.7 N2 = √44153.58 N2 = 210.12 r.p.m. Modified range of speed of governor = N2 - N1 = 210.12 – 171.8 = 38.37 r.p.m. Question # 4 In an engine governor of porter type, the upper and lower arms are 200 mm and 250 mm respectively and pivoted on the axis of rotation. The mass of the central load is 15kg, the arms of each ball is 2kg and friction of the sleeve together with resistance of the operating gear is equal to a load of 24N at the sleeve. If the linking inclination of the upper arms to the vertical to the 30 degree and 40 degree. Find range of speed of governor, taking friction in to account.

MECHANIS OF MATERIAL & MACHINES 10

BP = 200 mm = 0.2 m BD = 250 mm =0.25 m M = 15kg m = 2kg F = 24 N α1 = 30° α2 = 40° Range of speed of governor = n – n = ? For minimum position (N1) = mg + (Mg – F / 2)(1 + q) / mg × 895 / h1 r1 = BG = BP sin30° r1 = 200 × 0.5 r1 =100 mm r1 = 0.1 m h1 = BP cos 30° h1 = 200 × h1 = 0.1732 m q1 = tanβ1 / tanα1 tanα1 = tan 30° = 0.5774 tanβ1 = ? (BD)² = (BG)² + (DG)² (DG)² = (BD)² - (BG)² (DG)² = (0.250)² - (0.1)² √(DG)² = √0.0625 – 0.01 DG =√ 0.0525 DG = 0.2291 m tanβ1 = r1 / DG

MECHANIS OF MATERIAL & MACHINES 11 tanβ1 = 0.1 / 0.2291 tanβ1 = 0.4364 q1 = tanβ1 / tanα1 q1 = 0.4364 / 0.5774 q1 = 0.753 (N1) = [mg + {Mg – F} / 2 (1 + q)] / mg × 895 / h1 (N1) = [2 × 9.81 {15 × 9.81 – 24}/ 2 (1 + 0.753)]/ 2 × 9.81 × 895 / 0.1732 (N1) = [19.6 + {61.5} (1.753)] / 19.6 × 5167.43 (N1) = [19.6 + 107.8] / 19.6 × 5167.43 (N1) = 127.4 / 19.6 × 5167.43 (N1) = 6.5 × 5167.43 (N1) = 33588.3 √(N1) =√ 33588.3 N1 = 183.3 r.p.m. For maximum position (N2) = [mg + {Mg + F }/ 2(1 + q2)] / mg × 895 / h2 r2 = BG = BP sin40° r2 = 200 × 0.64 r2 = 0.2 × 0.64 r2 = 0.128 m h2 = BP cos 40° h2= 200 × 0.766 h2= 0.2 × 0.766 h2= 0.153 m (BD)² = (BG)² + (DG)² (DG)² = (BD)² - (BG)² (DG)² = (0.250)² - (0.128)² √(DG)² = √0.0625 – 0.01638 DG =√ 0.04612 DG = 0.214 m q2 = tanβ2 / tanα2 tanα2 = tan 40° = 0.839 tanβ2 = r2 / DG tanβ2 = 0.128 / 0.2146 tanβ2 = 0.596 q2 = tanβ2 / tanα2 q2 = 0.596 / 0.839

MECHANIS OF MATERIAL & MACHINES 12 q2 = 0.7105 (N2) = [mg + {Mg + F }/ 2(1 + q2)] / mg × 895 / h2 (N2) = [2 × 9.81 {15 × 9.81 + 24} / 2(1 + 0.7105)] / 2 × 9.81 × 895 / 0.153 (N2) = [19.6 + {85.57}(1.753)] / 19.6 × 5849.6732 (N2) = [19.6 + 150] / 19.6 × 5849.6732 (N2) = 169.6 / 19.6 × 5849.6732 (N2) = 8.65 × 5849.6732 (N2) = 50599.673 √(N2) =√ 50599.673 N2 = 224.94 r.p.m. Range of speed of governor = N2 - N1 = 224.94 – 183.3 = 41.69 r.p.m. 39.18 r.p.m.

Problem # 05:A porter Governor has four arms 250mm long. The upper arms are attached on the axis of rotation and lower arms attached to the sleeve at distance of 30mm from the axis. The mass of each ball is 5kg and the sleeve has mass of 50kg. The external radiuses of rotation are 150mm and 200mm. Determine the range of speed of the governor. Given Data:BP = 250mm, BG = 150mm m = 5kg, M = 50kg Solution:First we find h1 for minimum speed, ( PG )

2

= BP ( )− BG ( 2

h1 =

( 0.250−)

2

2

)

0.150 (

= 0.0625 − 0.0225 = 0.04 h1 = 0.2m

Now find tan α1 BG 0.15 = = 0.75 h1 0.2 Now find tan β1 tan α1 =

)

2

MECHANIS OF MATERIAL & MACHINES 13 ( FD )

2

= ( BD ) − ( BF ) 2

=

( 0.250 )

2

2

− ( 0.12 )

2

= 0.0625 − 0.0144 = 0.0481 = 0.2193 mm Put the value in Eq. BF 0.12 tan β1 = = = 0.547 FD 0.2193 Putting the values of tan α1 & tan β1 in q1 q1 =

tan β1 0.547 = = 0.729 tan α1 0.750

PuttingthevaluesinEq . M m+ + (1q )1 895 2 2 × ( N1 ) = m h 1 50 5 + (1+ 0.729) 2 = × 5 48.225 895 = × 5 0.2 = 207.75 . r.pm

895 0.2

Now, find the maximum speed, where r2=0.2 m

MECHANIS OF MATERIAL & MACHINES 14 ( PG )

2

= ( BP ) − ( BF ) = 2

2

( 0.25 )

2

− ( 0.2 )

2

= 0.0625 − 0.04 = 0.0225 h2 = 0.15 m Now find tan α 2 tan α 2 =

r2 0.2 = = 1.333 h2 0.15

putting the values for tan β 2

( FD )

2

= ( BD ) − ( r2 ) 2

=

( 0.25 )

2

2

− ( 0.17 ) = 0.0625 − 0.0289 2

= 0.0336 FD = 0.1833 m r 0.17 tan β 2 = 2 = = 0.928 FD 0.1833 Putting the values of tan α 2 & tan β 2 in q2 q2 =

tan β 2 0.928 = = 0.69 tan α 2 1.333

Putting the values for find the maximum speed,

( N2 )

2

=

m+

M (1 + q2 ) 895 2 × m h2

50 (1 + 0.69) 895 47.42 895 2 = × = × = 56564.03 5 0.15 5 0.15 N 2 = 237.83 r. p.m 5+

Now, find range of speed , N 2 − N1 = 237.83 − 207.75 N 2 − N1 = 30.08 r. p.m

Problem # 06:The arms of a porter governor are 300mm long. The upper arms are pivoted on the axis of the rotation. The lower arms are attached to sleeve at distance of 40mm from axis of rotation. The mass of the load on the sleeve is 70kg and mass of each ball is 10kg. Determine the equilibrium speed when the radius of the rotation of the

MECHANIS OF MATERIAL & MACHINES 15 balls is 200mm. If the friction is equitant to a load of 20Nat the sleeve, what will be the range of speed for this position. Given Data:BP = BD = 300mm = 0.3m 20N m = 10kg, Solution:-

BG = 200mm = 0.2m, M = 70kg

Find the value of h

( GP )

2

= ( BP ) − ( BG ) = 2

2

( 0.3)

2

− ( 0.2 )

2

− ( 0.16 )

2

= 0.05 = 0.2236 m BG 0.2 tan α = = = 0.8944 PG 0.2236

( FD )

2

= ( BP ) − ( BF ) = 2

2

( 0.3)

2

= 0.0644 = 0.253 m BF 0.16 tan β = = = 0.630 BD 0.253 Putting the values of tan α & tan β in q tan β 0.630 q= = = 0.704 tan α 0.8944

Now, find average speed of governor N = 2

m+

M (1 + q ) 895 2 × m h

70 ( 1 + 0.704 ) 895 69.64 895 2 = × = × 10 0.2236 10 0.2236 = 27874.686 = 166.95 r. p.m 10 +

Now, find minimum speed,

F

=

MECHANIS OF MATERIAL & MACHINES 16 ( N1 )

2

 Mg  mg +  − F  ( 1+ q) 895  2  = × mg h

 70 × 9.81  10 × 9.81 +  − 10  ( 1 + 0.704 ) 895 2   = × 10 × 9.81 0.2236 98.1 + (333)(1.704) 895 895 = × = 6.79 × 98.1 0.2236 0.2236 N1 = 164.7 r. p.m

Now, find maximum speed,

( N2 )

2

=

mg + ( Mg ± F ) mg

×

895 h

 70 × 9.81  10 × 9.81 +  + 10  ( 1 + 0.704 ) 895 2   = × 10 × 9.81 0.2236 98.1 + (353) × (1.704) 895 895 = × = 7.137 × 98.1 0.2236 0.2236 N 2 = 169 r. p.m Now, find range of speed N 2 − N1 = 169 − 164.7 N 2 − N1 = 4.25 r. p.m

Problem # 08:All arms of a porter governor are 178mm long and are hinged at a distance of 38mm from the axis of rotation. The mass of each ball is 1.15kg and mass of sleeve is 20kg. The governor sleeve begins to rise at 280 r.p.m, when the links are at angle of 30o to the vertical. Assuming the friction force to be constant. Determine the minimum and maximum speed of rotation, when the inclination of the arms to the vertical is 45o. Given Data:M = 20kg, m = 1.15kg, BP = BD = 178mm, Solution:-

N=280 r.p.m α = β = 300

MECHANIS OF MATERIAL & MACHINES 17 r r ⇒ sin 30o = BP 0.178 o r = 0.178 × sin sin 30 = 0.089 r = 0.089 + 0.038 ⇒ r = 0.127 m r 0.127 tan α = ⇒ tan 30o = h h 0.127 h= ⇒ h = 0.22m tan 30o mg + ( Mg ± F ) 895 N2 = × mg h

sin α =

±F =

N 2 × mg × h − ( mg + Mg ) 895

( 280 ) × ( 1.15 × 9.81) × 0.22 −  1.15 × 9.81 + 20 × 9.81  = ) ( ) ( 2

895 = 217.41 − 207.48 ± F = 9.99 N

α = β = 45o r 0.127 0.127 tan α = ⇒ tan 45o = ⇒ h1 = = 0.127 m h h1 tan 45o

( N1 ) =

2

=

mg + ( Mg − F ) mg

×

895 h1

( 1.15 × 9.81) + ( 20 × 9.81 − 9.99 ) × 895 0.127 ( 1.15 × 9.81)

= 17.51×

895 ⇒ N1 = 123405.01 = 351.29 r. p.m 0.127

Find r, r r ⇒ sin 45o = BP 0.178 o r = sin 45 × 0.178 = 0.125m + 0.038m = 0.163m = h

sin α =

MECHANIS OF MATERIAL & MACHINES 18 Find minimum speed,

( N1 )

2

= =

mg + ( Mg − F ) mg

×

895 h

( 1.15 × 9.81) + ( 20 × 9.81 − 9.99 ) × 895 0.163 ( 1.15 × 9.81)

895 = 310.07 r. p.m 0.163 Find maximum speed, N1 = 17.51×

( N2 )

2

= =

mg + ( Mg + F ) mg

×

895 h

( 1.15 × 9.81) + ( 20 × 9.81 + 9.99 ) × 895 0.163 ( 1.15 × 9.81)

895 = 325.29 r. p.m 0.163 Now, find range of speed , N 2 − N1 = 325.29 − 310.07 N 2 = 19.27 ×

N 2 − N1 = 15.24 r. p.m

Problem # 07:A loaded porter governor has four links each 250mm long. Two revolving masses each of 3kg and a central dead weight of mass 20kg. All the links are attached to respective sleeve at radial distances of 40mm from the axis of rotation. The mass revolves at a radius of 150mm at minimum speed and at a radius of 200 mm at minimum speed. Determine the range of speed. Given Data: r1 = 150mm,

r2 = 200mm

m = 3kg ,

M = 20kg

Solution:

MECHANIS OF MATERIAL & MACHINES 19 find h1 & α1 ,

( FP )

2

= ( BO ) − ( BF ) 2

=

( 0.25)

2

2

− ( 0.11) = 0.0504 = 0.224m 2

BF 0.11 = = 0.491 PF 0.224 α1 = tan −1 0.491 = 26.15o tan α1 =

tan α1 =

0.15 0.15 ⇒ h1 = = 0.305m h1 tan 26.15o

find minimum speed, m + M 895 3 + 20 895 2 × = × ( N1 ) = m h1 3 0.305 = 7.67 × 2934.46 = 22507.04 N1 = 149.99 r. p.m

find h2 & α 2 ,

( FP )

2

= ( BO ) − ( BF ) 2

=

( 0.25)

2

2

− ( 0.16 ) = 0.0369 = 0.192m 2

BF 0.16 = = 0.8329 PF 0.192 α 2 = tan −1 0.8329 = 39.79o tan α 2 =

tan α 2 =

0.2 0.2 ⇒ h2 = = 0.240m h2 tan 39.79o

find maximum speed, m + M 895 3 + 20 895 2 × = × ( N2 ) = m h2 3 0.240 = 7.67 × 3729.167 = 28602.708 N 2 = 169.123 r. p.m Now, find range of speed, N 2 − N1 = 169.123 − 149.99 N 2 − N1 = 19.13 r. p.m

MECHANIS OF MATERIAL & MACHINES 20 PROELL GOVERNOR

W × ID 2 Mg × BM = mg × IM + 2 × (IM + MD) MD  Mg  IM IM Fc = mg × BM + 2  BM + BM    Mg  IM IM MD  FM  Fc = FM mg × BM ÷ 2  BM ÷ BM    

Fc × BM = w × IM + Fc

Fc = Fc = Fc = Fc = Fc = mw²r q =

Mg  IM IM MD  FM  mg × ÷ ÷    FM 2  FM FM  BM  Mg FM  mg × tan α ÷ ( tan α ÷ tan β )   2 BM    Mg  tan β  FM mg ÷ 1 ÷  ×tanα  2  tan α  BM  Mg FM  (1 ÷ q )  mg ÷ ×tanα  2 BM   r , tanα = h

tan β tan α

mw²r = mw² =

FM BM FM BM

w² =

FM BM

w =

2πN 60

Mg   mg ÷ 2 (1 ÷ q )    Mg 1  (1 ÷ q )  mg ÷  2 h   r

×h × ×

M   mg ÷ 2 (1 ÷ q )    m      

g h

MECHANIS OF MATERIAL & MACHINES 21  2πN   ²  60 

=

FM BM

N2 =

FM BM

N2 =

FM BM

M   m ÷ 2 (1 ÷q )  g  × m   h     M   m + 2 (1 ÷q )  9.81 × 3600  × m   h × 4 × 9.86     M   m + 2 (1 ÷q )  895  × m h      

(

(

) )

QUESTION # 1 A proell governor has equal arms of length 300 mm. The upper and lower ends of the arms are pivoted on the axis of governor. The extension arms of 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 central load is 100 kg. Determine the range of the speed of the governor.

r1 = 150 mm = .15 m r2 = 200 mm = 0.2 m m =10 kg M =100 kg BF = 80 mm = 0.08 m ( FP ) 2 = ( PG ) 2 + FG 2 (300 ) 2 = h 2 + 150 2 h12 = (300 ) 2 - 150 2 h1 = √ (300 ) 2 - 150 2 h1 = √(0.3)2 – (0.15)2 h1 = PG = 0.26 m

MECHANIS OF MATERIAL & MACHINES 22 FM = PG = GD = 0.26 m BM = BF + FM BM = 0.08 + 0.26 BM = 0.34 m N12 = N12 = N12 = N12 = N1 = N1 = For maximum position

FM  m + M  895 ×  m  h BM   0.26 10 +100  895 ×  0.34   10  0.26

0.764 × 11 × 3442.3 28929 √ 28929 170 r.p.m.

( FP ) 2 = ( PG ) 2 + FG (300 ) 2 = h 22 + 200 2 h 22 = (300 ) 2 - 200 2 h 22 = √ (300 ) 2 - 200 2 h 22 = √(0.3)2 – (0.2)2 h 2 = PG = 0.224 m h 2 = PG = FM = 0.224 m BM = h 2 + BF BM = 0.224 + 0.08 BM = 0.304 m

2

N 22 = FM ×  m + M  895  m  h2 BM   10 +100

  N 22 =   × 0.304  10  0.224 2 N 2 = 0.736 × 11 × 3995.535 0.24

895

N 22 = 32347.851 N2

Question # 2

=√ 32347.851 N 2 =180 r.p.m. RANGE OF THE SPEED = N2 – N1 = 180 - 170 = 10 r.p.m.

MECHANIS OF MATERIAL & MACHINES 23 A governor of proell type has each arm 250 mm long. The pivots of the upper and the lower arms are 25 mm from the axis. The central load acting on the sleeve has a mass of 25 kg and each rotating ball has a mass of 3.2 kg. when the governor sleeve is in mid position the extension link of the lower arm is vertical and the radius of the path of rotation of the masses is 175 mm. the vertical height of the governor is 200 mm. if the governor speed is 160 r.p.m. when is mid position, Find 1. length of the extension link. 2. Tension of the upper arm.

m = 3.2 kg M = 25 kg R = 175 mm = 0.175 m h = 200 mm = 0.2 m N = 160 r.p.m BF = ? BM = FM + BF FM  m + M  895 ×  m  h FM + BF   0.2  3.2 + 25  895 2 (160 ) =  × 0.2 + BF  3.2  0.2 (160 ) 2 = 0.2 ( 8.812 ) × 4475 0.2 + BF (160 ) 2 = 0.2 × 39433 .7 0.2 + BF 2 (160 ) = 0.2 39433 .7 0.2 + BF 0.2 0.649 = 0.2 + BF 0.649 ( 0.2 + BF ) = 0.2

N12 =

MECHANIS OF MATERIAL & MACHINES 24 0.2 + BF =

0.2 0.649

BF = 0.308 – 0.2 BF = 0.108 m

T1cosα = mg +

Mg 2

0.2 ( 3.2 × 9.81) +  25 × 9.81  2   T1 0.25 = T1 0.8 = 31 .392 +122 .625 T1 0.8 = 154 154 T1 = 0.8

T1 = 192.5

Sensitivity of Governor Consider two governors ‘A’ and ‘B’ running at a same speed. When the speed is increases or decreases by a certain amount, the life of the sleeve of governor ‘A’ is grater than the life of sleeve of governor ‘B’. it is then said that the governor ‘A’ is more sensitivity than the governor ‘B’. In general the greater the life of the sleeve corresponding to a given frictional change in speed, the greater is the sensitiveness of the governor. It may also be stated in another way that for a given life of the sleeve, the sensitiveness of the governor increases as the speed range decreases. This definition of sensitiveness may be quite satisfactory when the governor is considered as an independent mechanism. But when the governor is fitted to an engine, the practical requirement is simply that the change of equilibrium speed from the full load to the low load position of the sleeve should be as small a fraction as possible of the mean equilibrium speed. The sensitiveness is define as the ratio of difference between the maximum and the minimum equilibrium speed to the mean equilibrium speed. Let N1 = Minimum equilibrium speed. N2 = Maximum equilibrium speed. N1 − N 2 2 N 2 − N1 N

N = mean equilibrium speed. =

∴sensitiveness of governor =

MECHANIS OF MATERIAL & MACHINES 25 =

2( N 2 − N 1 ) N1 + N 2

Huntin g A governor is set to be hunt if the speed of the engine fluctuates continuously above an below the mean speed. This is cussed a too sensitive governor which changes the fuel supply by a large amount when a small change in the speed of rotation takes place . for example when the load on the engine increases the engine speed decreases and if the governor is very sensitive , the governor sleeve immediately falls to its lowest position. This will results in the opening of the control ball wide which will supply the fuel to the engine in excess of its requirement so that the engine speed rapidly increases again and the governor speed rises to its highest position. Due to this movement of the sleeves the control value will cut off the fuel supply the engine and thus the engine seeped begin to fall once again. This cycle is repeated indefinitely. Such a governor may admit either the maximum or minimum amount of fuel. The effect this will be to cause wide flocculation in the engine speed or in other words the engine will hunt. Stability of Governor A governor is said to be stable when for every speed within the working range there is definite configuration i,e, there is one radius of rotation of the governor balls at which the governor is in equilibrium. For stable governor if the equilibrium speed increase the radius of governor ball must also increased.

MECHANIS OF MATERIAL & MACHINES 26

MECHANIS OF MATERIAL & MACHINES 27