Types of Governors in Automobiles

types of governors used to control engine speed and types of governor used to control engine speed, types and role of governor in diesel engine pdf free download
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 Chapter 18 : Governors 653 18 Features Governors 1. Introduction. 2. Types of Governors. 3. Centrifugal Governors. 4. Terms Used in Governors. 18.1. Introduction 5. Watt Governor. The function of a governor is to regulate the mean 6. Porter Governor. speed of an engine, when there are variations in the load e.g. 7. Proell Governor. when the load on an engine increases, its speed decreases, 8. Hartnell Governor. therefore it becomes necessary to increase the supply of work- 9. Hartung Governor. ing fluid. On the other hand, when the load on the engine 10. Wilson-Hartnell Governor. decreases, its speed increases and thus less working fluid is 11. Pickering Governor. required. The governor automatically controls the supply of 12. Sensitiveness of Governors. working fluid to the engine with the varying load conditions 13. Stability of Governors. and keeps the mean speed within certain limits. 14. Isochronous Governor. A little consideration will show, that when the load 15. Hunting. increases, the configuration of the governor changes and a 16. Effort and Power of a valve is moved to increase the supply of the working fluid ; Governor. conversely, when the load decreases, the engine speed in- 17. Effort and Power of a Porter Governor. creases and the governor decreases the supply of working 18. Controlling Force. fluid. 19. Controlling Force Diagram Note : We have discussed in Chapter 16 (Art. 16.8) that the func- for a Porter Governor. tion of a flywheel in an engine is entirely different from that of a 20. Controlling Force Diagram governor. It controls the speed variation caused by the fluctuations for a Spring-controlled of the engine turning moment during each cycle of operation. It Governor. does not control the speed variations caused by a varying load. The varying demand for power is met by the governor regulating the 21. Coefficient of supply of working fluid. Insensitiveness. 18.2. Types of Governors The governors may, broadly, be classified as 1. Centrifugal governors, and 2. Inertia governors. 653 654 Theory of Machines The centrifugal governors, may further be classified as follows : 18.3. 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 Spring steel strip vertical axis. The arms are connected by the links to a sleeve, which is keyed to the spindle. This sleeve re- volves 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 Spindle to limit the travel of the sleeve in upward and down- controls fuel ward directions, two stops S, S are provided on the supply spindle. The sleeve is connected by a bell crank lever to a throttle valve. The supply of the working fluid de- creases when the sleeve rises and increases when it falls. When the load on the engine increases, the en- gine and the governor speed decreases. This results in Rotating the decrease of centrifugal force on the balls. Hence weight the balls move inwards and the sleeve moves down- wards. The downward movement of the sleeve oper- ates 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 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 in- A governor controls engine speed. As it rotates, the weights swing outwards, pulling crease of centrifugal force on the balls. Thus the balls down a spindle that reduces the fuel supply move outwards and the sleeve rises upwards. This up- at high speed. ward 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. The controlling force is provided either by the action of gravity as in Watt governor or by a spring as in case of Hartnell governor. Chapter 18 : Governors 655 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. 18.4. 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 mini- mum equilibrium speeds respectively. Note : There can be many equilibrium speeds between the mean and the maximum and the mean and the mini- mum equilibrium speeds. 5. Sleeve lift. It is the vertical distance which Centrifugal governor the sleeve travels due to change in equilibrium speed. 656 Theory of Machines 18.5. 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). 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, 2 F = Centrifugal force acting on the ball in newtons = m. ω .r, and C 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 (F ) acting on the ball, 2. the tension (T) in the arm, and 3. the weight C (w) of the ball. Taking moments about point O, we have F × h = w × r = m.g.r C 2 2 or m. .r.h = m.g.r or h = g / . . . (i) ω ω 2 When g is expressed in m/s and ω in rad/s, then h is in metres. If N is the speed in r.p.m., then ω =2 π N/60 9.81 895 2 h== metres . . . (∵ g = 9.81 m/s ) . . . (ii) ∴ 22 (2πNN / 60) 2 Note : We see from the above expression that the height of a governor h, is inversely proportional to N . 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. Chapter 18 : Governors 657 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 : N = 60 r.p.m. ; N = 61 r.p.m. 1 2 Initial height We know that initial height, 895 895 h== = 0.248 m 1 22 (N ) (60) 1 Change in vertical height We know that final height, 895 895 h== = 0.24 m 2 22 () N (61) 2 Change in vertical height ∴ = h – h = 0.248 – 0.24 = 0.008 m = 8 mm Ans. 1 2 18.6. 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 pre- determined level. Consider the forces acting on one-half of the governor as shown in Fig. 18.3 (b). (a)(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, 658 Theory of Machines 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 π N/60 rad/s, F = Centrifugal force acting on the ball C 2 in newtons = m. ω .r, T = Force in the arm in newtons, 1 T = Force in the link in newtons, 2 = Angle of inclination of the arm (or α 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. A big hydel generator. Governors are used to control the supply of working 1. Method of resolution of forces fluid (water in hydel generators). Considering the equilibrium of the forces acting Note : This picture is given as additional at D, we have information and is not a direct example of WM .g the current chapter. T cosβ= = 2 22 Mg . T = or 2 . . . (i) 2cos β 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), (ii) The centrifugal force (F ), C (iii) The tension in the arm (T ), and 1 (iv) The tension in the link (T ). 2 Resolving the forces vertically, Mg . TT cos α= cosβ +w = +m.g . . . (ii) 12 2 Mg  . ...  T cosβ=  2  2 Resolving the forces horizontally, T sin α + T sin β = F 1 2 C Mg .  Mg . TF sin α+ × sin β = ...  T = 1C  2 2cos β 2cos β  Mg . TF sin α+ × tan β = 1C 2 . Mg TF sin α= – × tan β ∴ . . . (iii) 1C 2 Chapter 18 : Governors 659 Dividing equation (iii) by equation (ii), Mg . F–t×βan C T sin α 2 1 = Mg . T cos α 1 . +mg 2 Mg.. Mg  +α mg . tan=F – × tanβ or  C 22  Mg.. F Mgtan β C += mg.– × 2 tanαα 2 tan tan β r = q, Substituting and tan α= , we have tan α h Mg.. h M g 2 2 mg.. m .r – q += ω × × . . . (∴ F = m.ω .r) C 22 r Mg . 2 mh ..ω=m.g+ (1+q) or 2 M mq ++ (1 )  Mg.1 g 2 ∴ hm=+.( g 1+q) = × 22 2 m  m.ωω . . . (iv) M mq ++ (1 )  Mg 1 g 2 2 or ω= mg.( + 1+q) = ×  2. mh m h  M 2 mq ++ (1 ) 2πNg  2 or =×  60 mh  MM mq ++ (1 ) 2mq + (1+ ) g 60 895 222 ∴ N=× =×  mh 2π m h  . . . (v) 2 . . . (Taking g = 9.81 m/s ) 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 = tan α / tan β = 1 Therefore, the equation (v) becomes () mM + 895 2 N=× . . . (vi) mh 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 M . g ± F  m.(gq ++1)  2 895 2  . . . (vii) N=× m.g h mg .+± (M .g F) 895 =× . . . (When q = 1) . . . (viii) mg . h 660 Theory of Machines The + sign is used when the sleeve moves upwards or the governor speed increases and negative sign is used when the sleeve moves downwards or the governor speed decreases. 3. On comparing the equation (vi) with equation (ii) of Watt’s governor (Art. 18.5), we find that the mM + mass of the central load (M) increases the height of governor in the ratio . 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, W FB×= M w×IM+ ×ID C 2 Mg . =× m.g IM+ × ID 2 IM M . g ID Fm=× .g + × ∴ C BM 2 BM IM M .g IM + MD  =× mg . +  BM 2 BM  IM M .g IM MD  =× mg . + + Fig. 18.4. Instantaneous centre  BM 2 BM BM  method. Mg . =α mg . tan+ (tanα+ tanβ) 2 IM MD  ...  =α tan , and = tanβ  BM BM  Dividing throughout by tan α, F Mg.tan β Mg.  tan β C .1 . (1) =+ mg + =m g+ +q ...  q =   tanαα 2 tan 2 tan α   r 2 We know that F = m.ω .r, and tanα= C h hM.g 2 mr ..ω× =m.g+ (1+q) ∴ r 2 Mg . M mg.(++1 q) m+(1+q) 1 g 22 or h=×= × 22 mm ωω . . . (Same as before) When tan α = tan β or q = 1, then mM + g h=× 2 m ω Chapter 18 : Governors 661 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 ; r = 150 mm 1 = 0.15m; r = 200 mm = 0.2 m 2 Fig. 18.5 The minimum and maximum positions of the governor are shown in Fig. 18.5 (a) and (b) respectively. Minimum speed when r = BG = 0.15 m 1 Let N = Minimum speed. 1 From Fig. 18.5 (a), we find that height of the governor, 22 2 2 hP==G (PB) – (BG)= (0.25) – (0.15)= 0.2 m 1 We know that mM++ 895 5 15 895 2 ( ) 17 900 N=× = × = 1 mh 50.2 1 ∴ N = 133.8 r.p.m. Ans. 1 Maximum speed when r = BG = 0.2 m 2 Let N = Maximum speed. 2 From Fig. 18.5 (b), we find that height of the governor, 22 2 2 hP==G (PB) – (BG)= (0.25) – (0.2)= 0.15 m 2 We know that mM++ 895 5 15 895 2 (N )=× = × = 23 867 2 mh 50.15 2 ∴ N = 154.5 r.p.m. Ans. 2  662 Theory of Machines Range of speed We know that range of speed = N – N = 154.4 – 133.8 = 20.7 r.p.m. Ans. 2 1 Example 18.3. 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, determine how the speed range is modified. Solution. Given : BP = BD = 250 mm ; m = 5 kg ; M = 30 kg ; r = 150 mm ; r = 200 mm 1 2 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 N = Minimum speed when r = BG = 150 mm, and 1 1 N = Maximum speed when r = BG = 200 mm. 2 2 Fig. 18.6 Speed range of the governor From Fig. 18.6 (a), we find that height of the governor, 22 2 2 hP==G (PB) – (BG)= (250) – (150)= 200 mm= 0.2 m 1 We know that mM++ 895 5 30 895 2 (N )=× = × = 31 325 1 mh 50.2 1 ∴ N = 177 r.p.m. 1 From Fig. 18.6 (b), we find that height of the governor, 22 2 2 hP==G (PB) – (BG)= (250) – (200)= 150 mm= 0.15 m 2 We know that mM++ 895 5 30 895 2 () N=× = × =41767 2 mh 50.15 2 ∴ N = 204.4 r.p.m. 2  Chapter 18 : Governors 663 We know that speed range of the governor = N – N = 204.4 – 177 = 27.4 r.p.m. Ans. 2 1 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 mg . + (M .g – F ) 895 2 () N=× 1 mg . h 1 5×+ 9.81 (30× 9.81 – 20) 895 =×= 29500 59 × .81 0.2 ∴ N = 172 r.p.m. 1 We also know that when the sleeve moves upwards, the frictional force (F) acts downwards and the maximum speed is given by mg .++ (M .g F) 895 A series of hydel generators. 2 () N=× 2 Note : This picture is given as additional mg . h 2 information and is not a direct example of the current chapter. 5×+ 9.81 (30× 9.81+ 20) 895 =×= 44200 59 × .81 0.15 ∴ N = 210 r.p.m. 2 We know that speed range of the governor = N – N = 210 – 172 = 38 r.p.m. Ans. 2 1 Example 18.4. In an engine governor of the 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 15 kg, the mass of each ball is 2 kg and friction of the sleeve together with the resistance of the operating gear is equal to a load of 25 N at the sleeve. If the limiting inclinations of the upper arms to the vertical are 30° and 40°, find, taking friction into account, range of speed of the governor. Solution . Given : BP = 200 mm = 0.2 m ; BD = 250 mm = 0.25 m ; M = 15 kg ; m = 2 kg ; F = 25 N ; α = 30°; α = 40° 1 2 First of all, let us find the minimum and maximum speed of the governor. The minimum and maximum position of the governor is shown Fig. 18.7 (a) and (b) respectively. Let N = Minimum speed, and 1 N = Maximum speed. 2 From Fig. 18.7 (a), we find that minimum radius of rotation, r = BG = BP sin 30° = 0.2 × 0.5 = 0.1 m 1 Height of the governor, h = PG = BP cos 30° = 0.2 × 0.866 = 0.1732 m 1  664 Theory of Machines 22 2 2 and m DG== (BD) – (BG) (0.25) – (0.1)= 0.23 ∴ tan β = BG/DG = 0.1/0.23 = 0.4348 1 and tan α = tan 30° = 0.5774 1 tan β 0.4348 1 q== = 0.753 1 ∴ tan α 0.5774 1 Fig. 18.7 We know that when the sleeve moves downwards, the frictional force (F) acts upwards and the minimum speed is given by  Mg.–F mg.(++1 q)  1 2 895 2  () N=× 1 mg . h 1 15 × 9.81 – 24  29×+ .81 (1+0.753)  2 895  =×= 33596 2 × 9.81 0.1732 ∴ N = 183.3 r.p.m. 1 Now from Fig. 18.7 (b),we find that maximum radius of rotation, r = BG = BP sin 40° = 0.2 × 0.643 = 0.1268 m 2 Height of the governor, h = PG = BP cos 40° = 0.2 × 0.766 = 0.1532 m 2 22 2 2 and DG== (BD) – (BG) (0.25) – (0.1268)= 0.2154 m ∴ tan β = BG/DG = 0.1268 / 0.2154 = 0.59 2 and tan α = tan 40° = 0.839 2 tan β 0.59 2 q== = 0.703 2 ∴ tan α 0.839 2 Chapter 18 : Governors 665 We know that when the sleeve moves upwards, the frictional force (F) acts downwards and the maximum speed is given by mg . + F  mg.(++1 q)  2 2 895 2  () N=× 2 mg . h 2  15×+ 9.81 24 2×+ 9.81 (1+ 0.703)  895 2  =×= 49 236 2 × 9.81 0.1532 ∴ N = 222 r.p.m. 2 We know that range of speed = N – N = 222 – 183.3 = 38.7 r.p.m. Ans. 2 1 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 ; r = 150 mm ; r = 200 mm 1 2 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. Fig. 18.8 Let N = Minimum speed when r = BG = 150 mm ; and 1 1 N = Maximum speed when r = BG = 200 mm. 2 2 From Fig. 18.8 (a), we find that height of the governor, 22 22 hP==G (BP) – (BG)= (250) – (150)= 200 mm= 0.2 m 1 BF = BG – FG = 150 – 30 = 120 mm . . . ( FG = DH) 666 Theory of Machines 22 2 2 and DF== (DB) – (BF) (250) – (120)= 219 mm ∴ tan α = BG/PG = 150 / 200 = 0.75 1 and tan β = BF/DF = 120/219 = 0.548 1 tan β 0.548 1 q== = 0.731 1 ∴ tan α 0.75 1 M 50 mq ++ (1 ) 5+ (1+ 0.731) 1 895 895 2 22 We know that (N )=×= ×= 43206 1 mh 50.2 1 ∴ N = 208 r.p.m. 1 From Fig. 18.8(b), we find that height of the governor, 22 2 2 hP==G (BP)− (BG)= (250)− (200)=150 mm= 0.15 m 2 BF = BG – FG = 200 – 30 = 170 mm 22 2 2 and DF=− (DB) (BF )= (250)−(170)=183mm ∴ tan α = BG/PG = 200/150 = 1.333 2 and tan β = BF/DF = 170/183 = 0.93 2 tan β 0.93 2 ∴= q== 0.7 2 tan α 1.333 2 We know that M 50 mq ++ (1 ) 5+ (1+ 0.7) 2 895 895 222 (N )=×= ×=56 683 2 mh 50.15 2 ∴ N = 238 r.p.m. 2 We know that range of speed = N – N = 238 – 208 = 30 r.p.m. Ans. 2 1 Example 18.6. The arms of a Porter governor are 300 mm long. The upper arms are pivoted on the axis of rotation. The lower arms are attached to a sleeve at a distance of 40 mm from the axis of rotation. The mass of the load on the sleeve is 70 kg and the mass of each ball is 10 kg. Determine the equilibrium speed when the radius of rotation of the balls is 200 mm. If the friction is equivalent to a load of 20 N at the sleeve, what will be the range of speed for this position ? Solution. Given : BP = BD = 300 mm ; DH = 40 mm ; M = 70 kg ; m = 10 kg ; r = BG = 200 mm Equilibrium speed when the radius of rotation r = BG = 200 mm Let N = Equilibrium speed. The equilibrium position of the governor is shown in Fig. 18.9. From the figure, we find that height of the governor, 22 2 2 ( ) – ( ) (300) – (200) 224 mm hP== G BP BG = = = 0.224m Chapter 18 : Governors 667 ∴ BF = BG – FG = 200 – 40 = 160 . . . ( FG = DH)  22 2 2 and DF== (DB) – (BF) (300) – (160)= 254 mm ∴ tan α = BG/PG = 200 / 224 = 0.893 and tan β = BF/DF = 160 / 254 = 0.63 tan β 0.63 q== = 0.705 ∴ tan α 0.893 We know that M mq ++ (1 ) 895 2 N=× 2 mh 70 10++ (1 0.705) Fig. 18.9 895 2 =×= 27 840 10 0.224 ∴ N = 167 r.p.m. Ans. 2 Range of speed when friction is equivalent to load of 20 N at the sleeve ( i.e. when F = 20 N) Let N = Minimum equilibrium speed, and 1 N = Maximum equilibrium speed. 2 We know that when the sleeve moves downwards, the frictional force (F) acts upwards and the minimum equi- librium speed is given by  Mg.–F mg.(++1 q)  2 895 2  () N=× 1 mg . h An 18th century governor. 70 × 9.81 – 20  10×+ 9.81 (1+ 0.705)  2 895  =×= 27 144 10 × 9.81 0.224 ∴ N = 164.8 r.p.m. 1 We also know that when the sleeve moves upwards, the frictional force (F) acts downwards and the maximum equilibrium speed is given by Mg . +F  mg.(++1 q)  2 895 2  () N=× 2 mg . h 70×+ 9.81 20  10×+ 9.81 (1+ 0.705)  895 2  =×= 28 533 10 × 9.81 0.224 ∴ N = 169 r.p.m. 2 668 Theory of Machines We know that range of speed = N – N = 169 – 164.8 = 4.2 r.p.m. Ans. 2 1 Example 18.7. A loaded Porter governor has four links each 250 mm long, two revolving masses each of 3 kg and a central dead weight of mass 20 kg. All the links are attached to respective sleeves at radial distances of 40 mm from the axis of rotation. The masses revolve at a radius of 150 mm at minimum speed and at a radius of 200 mm at maximum speed. Determine the range of speed. Solution. Given : BP = BD = 250 mm ; m = 3 kg ; M = 20 kg ; PQ = DH = 40 mm ; r = 150 mm ; r = 200 mm 1 2 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. 18.10 (a) and (b) respectively. Let N = Minimum speed when r = BG = 150 mm, and 1 1 N = Minimum speed when r = BG = 200 mm. 2 2 From Fig. 18.10 (a), we find that BF = BG – FG = 150 – 40 = 110 mm and sin α = BF / BP = 110 / 250 = 0.44 or α = 26.1° 1 1 ∴ Height of the governor, h = OG = BG / tan α = 150 / tan 26.1° = 306 mm = 0.306 m 1 1 Fig. 18.10 Since all the links are attached to respective sleeves at equal distances (i.e.40 mm) from the axis of rotation, therefore tan α = tan β or q = 1 1 1 mM++ 895 3 20 895 2 (N )=× = × = 22424 We know that 1 mh 3 0.306 1 N = 150 r.p.m. 1  Chapter 18 : Governors 669 Now from Fig. 18.10 (b), we find that BF = BG – FG = 200 – 40 = 160 mm and sin α = BF/BP = 160 / 250 = 0.64 or β = 39.8° 2 2 ∴ Height of the governor, h = OG = BG / tan α = 200 / tan 39.8° = 240 mm = 0.24 m 2 2 In this case also, tan α = tan β or q = 1 2 2 mM++ 895 3 20 895 2 (N )=× = × = 28 590 2 We know that mh 30.24 2 ∴ N = 169 r.p.m. 2 We know that range of speed = N – N = 169 – 150 = 19 r.p.m. Ans. 2 1 Example 18.8. All the arms of a Porter governor are 178 mm long and are hinged at a distance of 38 mm from the axis of rotation. The mass of each ball is 1.15 kg and mass of the sleeve is 20 kg. The governor sleeve begins to rise at 280 r.p.m. when the links are at an angle of 30° to the vertical. Assuming the friction force to be constant, determine the mini- mum and maximum speed of rotation when the inclination of the arms to the vertical is 45°. Solution. Given : BP = BD = 178 mm ; PQ = DH = 38 mm ; m = 1.15 kg ; M = 20 kg ; N = 280 r.p.m. ; α = β = 30° First of all, let us find the friction force (F). The equilibrium position of the governor when the lines are at 30° to vertical, is shown in Fig. 18.11. From the figure, we find that radius of rotation, r = BG = BF + FG = BP × sin α + FG = 178 sin 30° + 38 = 127 mm and height of the governor, h = BG / tan α = 127 / tan 30° = 220 mm = 0.22 m We know that mg.(+± Mg F) 895 2 N=× mg . h . . . (∴ tan α = tan β or q = 1) Fig. 18.11 1.15×+ 9.81 20× 9.81± F 895 2 (280)=× 1.15 × 9.81 0.22 2 (280)×× 1.15 9.81× 0.22 ±= F – 1.15× 9.81 – 20× 9.81 or 895 = 217.5 – 11.3 – 196.2 = 10 N We know that radius of rotation when inclination of the arms to the vertical is 45 (i.e. when α = β = 45°), r = BG = BF + FG = BP × sin α + FG = 178 sin 45° + 38 = 164 mm 670 Theory of Machines and height of the governor, h = BG / tan α = 164 / tan 45° = 164 mm = 0.164 m Let N = Minimum speed of rotation, and 1 N = Maximum speed of rotation. 2 We know that mg . + (M .g – F ) 895 2 () N=× 1 mg . h 1.15×+ 9.81 (20× 9.81 – 10) 895 =×= 95 382 1.15 × 9.81 0.164 ∴ N = 309 r.p.m. Ans. 1 mg .++ (M .g F) 895 2 () N=× and 2 mg . h 1.15×+ 9.81 (20×+ 9.81 10) 895 =×= 105 040 1.15 × 9.81 0.164 N = 324 r.p.m. Ans. 2 18.7. 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 perpendicualr to the spindle axis. The prependicular BM is drawn on ID. Fig. 18.12. Proell governor. Taking moments about I, using the same notations as discussed in Art. 18.6 (Porter governor), WM .g F× BM= w×+ IM × ID= m. g×+ IM × ID . . . (i) C 22 IM M . g IM + MD  Fm=× .g + ∴ C  . . . ( ID = IM + MD)  BM 2 BM  Chapter 18 : Governors 671 Multiplying and dividing by FM, we have FM IM M . g IM MD  Fm=× .g + + C  BM FM 2 FM FM   FM M . g =× mg . tanα+ (tanα+ tanβ)  BM 2    FM M.t g an β =× tanα mg .+ 1+   BM 2tan α   tanβ r 2 q = tanα= We know that F = m.ω r ; and C tan α h FM r M . g  2 ∴ mr ..ω= × m.g+ (1+q)  BM h 2  M  mq ++ (1 )  FM g 2 2 ω= and  . . . (ii) BM m h    2 Substituting ω = 2π N/60, and g = 9.81 m/s , we get M  mq ++ (1 )  FM 895 2 2 N =  . . . (iii) BM m h    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 FM m + M 895 2  N = (h being in metres) ...(iv)  BM m h  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 ; r = 150 mm; r = 200 mm 1 2 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 N = Minimum speed when radius of rotation, r = FG = 150 mm ; and 1 1 N = Maximum speed when radius of rotation , r = FG = 200 mm. 2 2 From Fig. 18.13 (a), we find that height of the governor, 22 2 2 hP==G (PF) – (FG)= (300) – (150)= 260 mm= 0.26 m 1 672 Theory of Machines and FM = GD = PG = 260 mm = 0.26 m ∴ BM = BF + FM = 80 + 260 = 340 mm = 0.34 m FM m + M 895  2 We know that () N = . . . (∴ α = β or q = 1) 1  BM m h  1 0.26 10 + 100 895 == 28 956 or N = 170 r.p.m.  1 0.34 10 0.26  Fig. 18.13 Now from Fig. 18.13 (b), we find that height of the governor, 22 2 2 hP==G (PF) – (FG)= (300) – (200)= 224 mm= 0.224 m 2 and FM = GD = PG = 224 mm = 0.224 m ∴ BM = BF + FM = 80 + 224 = 304 mm = 0.304 m FM m + M 895  2 We know that () N = . . . ( α = β or q = 1)  2  BM m h  2 0.224 10 + 100 895  or N = 180 r.p.m. == 32 385  2 0.304 10 0.224  We know that range of speed = N – N = 180 – 170 = 10 r.p.m. Ans. 2 1 Note : The example may also be solved as discussed below : From Fig. 18.13 (a), we find that sin α = sin β = 150 / 300 = 0.5 or α = β = 30° and MD = FG = 150 mm = 0.15 m FM = FD cos β = 300 cos 30° = 260 mm = 0.26 m

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