Lecture 13 HYDRAULIC ACTUATORS[CONTINUED]

Transcription

1 Lecture 1 HYDRAULIC ACTUATORS[CONTINUED] 1.5Acceleration and Deceleration of Cylinder Loads Cylinders are subjected to acceleration and deceleration during their oeration. Cylinders are decelerated to rovide cushioning and cylinders are accelerated to reduce the cycle time of the oeration Acceleration To calculate the acceleration of cylinder loads, the equations of motion must be understood.let u be the initial velocity, v the velocity after a time t,s the distance moved during the time t anda the acceleration during the time t.the standard equations of motion are as follows: and with an acceleration a is given by v u at v u as 1 s ut at 1 s ( u v ) t The force F to accelerate a weight W horizontally Force = Mass Acceleration F = W a g where g is the acceleration due to gravity and is 9.81 m/s. The force P required to overcome friction is given by P = µw, where µ is the coefficient of friction. Note: Dynamic cylinder thrust In dynamic alications, the load inertia, seal friction, load friction, etc., must be allowed for calculating the dynamic thrust.as a first aroximation, the dynamic thrust can be taken as 0.9 times the static thrust. Cylinder seal friction varies with seal and cylinder design. The ressure required to overcome seal friction is not readily available from the majority of cylinder manufacturers. The seal friction breakout ressure can be taken as 5 bar for calculation uroses. It reduces when the iston starts to move. The ressure required to overcome seal friction reduces as the cylinder bore size increases and varies according to the seal design. Examle 1. A cylinder is required to move a 10 kn load 150 mm in 0.5 s. What is the outut ower? Solution: The velocity is given by Power is given by d 0.15 v =0. m/s t 0.5 P F v = kw Examle 1. A cylinder is required to extend at a minimum seed of 0.75 m/s in a system with a flow rate of 60 LPM. What cylinder size is required? 1

2 Solution: Let us first convert the LPM to : Q 60 LPM m / min 10 m / s Now we know flow rate and velocity, so we can calculate the diameter, Q A v Therefore, D D mm 0.75 Examle 1.5 An 8 cm diameter hydraulic cylinder has a cm diameter rod. If the cylinder receives flow at 100 LPM and 1 MPa, find the (a) extension and retraction seeds and (b) extension and retraction load carrying caacities. Solution: Let us first convert the flow in LPM to m /s before we calculate forward velocityq in =100 LPM = 100/( ) =1/600 m /s Now D C = Diameter of cylinder = 8 cm = 8 10 m d r = Diameter of iston rod = cm = 10 m = 1 MPa = N/m or Pa (a) Forward velocity is given by Q v ext = A = 1/ 600 = 0.15 m/s d / in Return velocity is given by 1 Q v in ret = m/s ( A Ar) π( dc dr ) (b) Force during extension is given by 6 (810 ) Fext a N Force during retraction is given by

3 F ( A A ) ret r 6 10 [(810 ) (10 ) ] N 5. kn Examle 1.6 A um sulies oil at m /s to a 0 mm diameter double-acting hydraulic cylinder. If the load is 5000 N (extending and retracting) and the rod diameter is 0 mm, find the (a) Hydraulic ressure during the extending stroke. (b) Piston velocity during the extending stroke. (c) Cylinder kw ower during the extending stroke. (d) Hydraulic ressure during the retracting stroke, (e) Piston velocity during the retracting stroke. (f ) Cylinder kw ower during the retracting stroke. Solution: We have Q in = m /s, F ext = F ret = 5000 N, d C = 0 mm = 0.0 m, d r = 0 mm = 0.0 m. (a) Hydraulic ressure during the extending stroke is Fext ext kpa A d C (0.0) (b) Piston velocity during the extending stroke is Qin Qin v ext 1.7 m/s A d C (0.0) (c) Cylinder kw ower during the extending stroke is Pext Fext vext kw (d) Hydraulic ressure during the retracting stroke is Fret ret = = kpa A Ar ( dc dr ) [(0.0) (0.0) ] (e) Piston velocity during the retracting stroke is Qin vret =1.697 m/s A Ar ( ) (f) Cylinder kw ower during the retracting stroke is Pret Fret vret kw Examle 1.7 A hydraulic cylinder has a rod diameter equal to one half the iston diameter. Determine the difference in load-carrying caacity between extension and retraction stroke if ressure is constant. What would haen if the ressure were alied to both sides of the cylinder at the same time? Solution: Forward or extending stroke is F A ext

4 Retracting stroke is Also, given that Therefore Now Therefore, Again F ( A A ) ret r d = d r A = A r F A ext F ( A A ) ret r A A A A A 0.5A r F F ext ret ext ret F F A ( A 0.5 A ) A A A A F F ext ret Therefore, Pr essure Piston area Difference = If the ressure were alied to both sides of the cylinder at the same time, there would be a net force to extend the cylinder. This net force will be the same as obtained above: Pr essure Piston area F net-extending = Examle 1.8 A cylinder with a bore of 150 mm and a iston rod diameter of 105 mm, has to extend with a seed of 7 m/s, ressure alied is 150 bar. Calculate (a) The flow rate in LPM of oil to extend the cylinder (b) The flow rate in LPM from annulus side to extend the cylinder. (c) The retract seed in m/min using (a). (d) The flow rate from full bore end on retract. Solution: Area of iston is given by A D (0.15) m Area of rod is Ar Dr (0.105) m

5 Hence, A Ar m. Given, v ext = 7 m/min and = 150 bar. (a) Flow rate of oil to extend in LPM: ( Q ) A v ext ext (b) Flow rate of oil to extend from annulus in LPM: ( Q ) ( A A ) v m /min 1.69 LPM ext a r (c) Retract seed in m/min using (a): Q ( Q ) ( A A ) v ext m /min 6 LP ret ext r ret v ret v ret 1.78 m/min (d) Flow rate from full bore end on retract: Q A v ret ret M m /min LPM 1.6Various Methods of Alying Linear Motion Using Hydraulic Cylinders A cylinder must roduce a force equal to the load the cylinder is required to overcome. A cylinder may be laced with its axis vertical, horizontal or inclined deending on the load to be actuated. 1. Vertical cylinder: Ina vertical cylinder, the load to be actuated is in the vertical direction as shown in Fig Then the cylinder load F is equal to the weight W of the object, acting in the vertical direction. Load = Cylinder force Figure 1.1Cylinder load vertical cylinder. Horizontal cylinder:the schematic diagram of horizontal cylinder is shown in Fig Ina horizontal cylinder, the cylinder load is theoretically zero,because no comonent of the object s weight acts along the axis of the cylinder. However, when the object slides across the horizontal surface, the cylinder must overcome the frictional force created between the object and the horizontal surface. 5

6 Load Figure 1.1Cylinder load horizontal cylinder. Inclined cylinder:in an inclined cylinder as shown in Fig. 1.1, the cylinder load equals the comonent of the object s weight acting along the axis of the cylinder and frictional force = W = Cylinder force Figure 1.1Cylinder load inclined cylinder For an inclined cylinder, the load the cylinder must overcome is less than the weight of the object to be moved if the object does not slide on an inclined surface. The cylinder loads calculated as above are based on moving an object at a constant velocity. But when the object has to be accelerated from zero velocity to a steady-state velocity, an additional force called inertia force must be added to the weight comonent and any frictional force involved. LetF load = W = weight or load acting vertically downward, F cyl = load acting on the cylinder, F bear = force on the bearings and α= angle between the load W and the axis of the cylinder.then, F cyl = F load cos(a) F bear = F load sin(a) Examle 1.9 Find the cylinder force required to move a 6000 N weight along a horizontal surface at a constant velocity (Fig. 1.15). The coefficient of friction between the weight and horizontal suort surface is

7 Load Figure 1.15 Solution: The cylinder force is given by F Frictional force ( F ) W cyl F cyl N f Examle 1.10 Find the cylinder force required to lift a 6000 N weight along a direction that is 0 from the horizontal direction as shown in Fig The weight is moved at a constant velocity. = Cylinder force Figure 1.16 Solution: Let F load = W = weight or load acting vertically downward F cyl = load acting on the cylinder F bear = force on the bearings = angle between the load W and the axis of the cylinder Then F cyl = F load cos F bear = F load sin Hence, 7

8 Fcyl 6000 cos N Examle 1.11 A 6000 N weight is to be lifted uward in a vertical direction for the system shown in Fig Find the cylinder force required to (a) Move the weight at a constant velocity of 1.75 m/s. (b) Accelerate the weight from zero velocity to 1.75 m/s in 0.5 s. Load = Cylinder force Figure 1.17 Solution: (a) For a constant velocity, the cylinder force to move weight at a constant velocity of 1.75 m/s Fcyl Fa W 6000 N (b) Force required to accelerate the weight: First we shall calculate acceleration a.5 m/s 0.5 Force required to accelerate the weight is Fa ma N 9.81 The cylinder force F cyl required is equal to the sum of the weight and acceleration force Fcyl N Examle 1.1 A N weight is to be lowered by a vertical cylinder as shown in Fig The cylinder has a 75 mm diameter iston and 50 mm diameter rod. The weight is to decelerate from 100 m/min to a sto in 0.5 s. Determine the required ressure in the rod end of the cylinder during the deceleration motion. 8

9 10000 N Figure 1.18 Solution: As er Newton s law of motion, we have F m a where m 1 min 100 min 60 s 1.67 m/s a. m/s 0.5 s 0.5 s Summing forces on the N weight, we have N ( A Ar ) N. m/s 9.81 m/s (N/m ) ( ) m N 08 N Solving we get (N/m ) Pa 550 kpa Examle 1.1 A 7000 N weight is being ushed u on an inclined surface at a constant seed by a cylinder, as shown in Fig The coefficient of friction between the weight and the inclined surface equals (a) Determine the required cylinder iston diameter for the ressure of 689 kpa, (b) Determine the required cylinder iston diameter, if the weight is to accelerate from a 0 mm/s to a 15 mm/s in 0.5 s. 9

10 7000 N Figure 1.19 Solution: (a) Cylinder iston diameter for the ressure of 689 kpa The comonent of the weight W acting along the axis of the cylinder is W sin0. The comonent of weight W acting along normal to the incline surface is W cos0. The frictional force equals the coefficient of friction times the force normal to sliding surfaces. Therefore, the frictional force f acting along the axis of the cylinder is f Wcos cos0 507 N Total force on the cylinder is frictional force and vertical comonent: Fcylinder f W sin sin N F A =17007 N cylinder Diameter to resist N is given by Fcylinder D =17007 N D = m = 56 mm (b) Cylinder iston diameter if the weight is to accelerate from a 0 mm/s to a 15 mm/s in 0.5 s As er Newton s law of motion, we have calculate the acceleration F m a where 10

11 15mm/s a 08 mm/s.08 m/s 0.5 s Summing forces on the 7000 N weight using values determined in (a), we have A D D D = m = 68.5 mm Examle 1.1 A hydraulic cylinder has a bore of 00 mm and a iston rod diameter of 10 mm. For an extend seed of 5 m/min, calculate (a) The suly flow rate. (b) The flow rate from the annulus side on extend. (c) The retract seed using Q E. (d) The flow rate from the full bore end on retract. Also, if the maximum ressure alied to the cylinder is 100 bar, calculate the (e) dynamic extend thrust and the (f) dynamic retract thrust assuming that dynamic thrust = 0.9 static thrust. Moreover, the hydraulic cylinder having a bore of 00 mm diameter and a rod of 10 mm diameter are connected regeneratively. (g) If the same flow rate of 157 L/min is used, calculate the extend seed. (h) If the maximum system ressure is 100 bar, calculate the dynamic extend thrust. Solution: (a) Flow rate of oil to extend cylinder at 5 m/min: Q E = Area of iston Velocity = (00/1000) 5 60 =0.006 m /min = = 157 L/min (b) Flow of oil leaving cylinder Q E is given by Q E = Annulus area Velocity = [(00/1000) (10/1000) ] 5 60 = 80 L/min (c) The same fluid flow rate used to extend the cylinder (157 LPM) is used to retract the cylinder. The retract cylinder velocity v is given by Now Q E v A a Q E = 157 L/min = m (A a) = Annulus area= m 11

12 v= = 0.61 m/s = 9.8 m/min (d) Flow from the full bore end of cylinder Q R is given by Q = A v= = m /s = 09 LPM (e) We have 0. Full bore area = = 0.01 m Dynamic extend thrust = 0.9 Pressure Full bore area 5 N = m m = 8 kn ( ) (f) Annulus area = = m 5 N Dynamic retract thrust = m m = 1 kn 0.1 (g) Piston rod area = = m Flow rate Extend seed = Piston rod area L m = min L m = 10. m/min This comares with 5 m/min when connected conventionally. (h) For a regenerative system Dynamic extend thrust = N m m = 18.6 kn As the area of the annulus is almost equal to that of the rod, the regenerative extend and conventional retract thrusts and seeds are almost the same. Examle 1.15 A mass of 000 kg is to be accelerated horizontally u to a velocity of 1 m/s from the rest over a distance of 50 mm (Fig. 1.0). The coefficient of friction between the load and guide is Calculate the bore of the cylinder required to accelerate this load if the maximum allowable ressure at the full bore end is 100 bar (take seal friction to be equivalent to a ressure dro of 5 bar). Assume that the back ressure at the annulus end of the cylinder is zero. 1

13 Load P 1 =100 bar P = 0 Figure 1.0 Solution: In this case, u = 0, v = 1 m/s, s = 0.05 m and a is unknown. Using the equation v u as We have, 1 0 a 0.05 a 10 m/s The force to accelerate the load is given by w F a g N 9.81 The force P to overcome the load friction is given by P W N The total force to accelerate the load and overcome friction is F + P = = 9 N The cylinder area required for a given thrust is calculated from Thrust = Force Area The ressure available is ressure at the full bore end of the cylinder minus the equivalent seal break-out ressure. Pressure available = = 95 bar = bar Area is given by 9 Area m 15 mm Now area is also given by D Area where D is the diameter. Comaring the two equations, we get D = 55. mm. The cylinder diameter is thus 55. mm. This neglects the effect of any back ressure. The nearest standard cylinder above has a 6 mm diameter bore. 1