Lecture 5 HYDRAULIC CIRCUIT DESIGN AND ANALYSIS [CONTINUED] 1.1 Circuit for Fast Approach and Slow Die Closing A machine intended for high volume production has a high piston velocity. If not controlled, the highspeed platen approaching the job instead of making a smooth contact will bang on the job. This is not desirable. In all such cases, rapid traverse and feed circuits are employed. (5) (6) () () Figure 1.15 Rapid traverse and feed circuit. In the circuit shown in Fig. 1.15, pump delivery normally passes through FCV(). During fast approach,the solenoid-operated DCV () is energized. This diverts pump delivery to the cap end of the cylinder through valve (). Full flow is thus available for the actuator to advance at the rated speed. A few millimeters before the platen makes contact with the die, solenoid valve () is de-energized forcing the pump delivery to pass through FCV (). The platen now approaches the die at a controlled speed because the flow to cylinder (6) is now regulated. Directional valves () and (5), however, must be energized simultaneously for the approach phase to begin. Valves () to (5) are solenoid-controlled pilot-operated valves intended for handling large flows with minimum pressure drop. While valve (5) requires a.5 bar check valve (6) in the return line to develop the pilot pressure required to move the main spool, no such facility is required in the case of valve() because the back pressure generated by valve () would serve as the pilot pressure for this valve. 1.15Rapid Traverse and Feed, Alternate Circuit
In this circuit (Fig. 1.16), full flow from the pump is allowed to the cap end through the directional valve ()for fast approach and the rod end oil freely passes through the normally open deceleration valve back to the tank. Near about the end of the stroke, a cam depresses a roller attached to the deceleration valve spool and therefore the valve shifts blocking the flow from the rod end. The flow now has only one pathway back to the tank and that is through FCV (). The approach speed is now governed by the setting of this valve. During piston retraction stroke, full flow is allowed to the rod end through check valve (6). (5) () (6) () () (1) Figure 1.16 Rapid traverse and feed circuit alternate circuit. Example 1.1 A double-acting cylinder is hooked up in a regenerative circuit. The relief-valve setting is 105 bar. The piston area is 10 cm and the rod area is 65 cm. If the pump flow is 0.0016 m /s, find the cylinder speed and load-carrying capacity for the Solution: (a) Extending stroke. (b) Retracting stroke. (a) We have Qp 0.0016 ve xt 0.6 m/s A 6510 r F load-extension 105 10 65 10 6850 N pa r 5 (b) We have Qp 0.0016 vret 0.6 m/s A A (10 6) 5 10 p r
F p A A load-extension ( ) p r 10510 (10 65) 10 5 6850 N Example 1. What is wrong with the circuit diagram given in Fig. 1.17? Unloading Valve Figure 1.17 Solution: A check valve is needed in the hydraulic line just upstream from where the pilot line to the unloading valve is connected to the hydraulic line. Otherwise the unloading valve would behave like a pressure-relief valve and thus, valuable energy would be wasted. Example 1. What unique feature does the circuit of Figure 1.18 provide in the operation of the hydraulic cylinder?
Figure 1.18 Solution: 1. It provides mid-stroke stop and hold of the hydraulic cylinder (during both the extension and retraction strokes) by deactivation of the four-way, three-position DCV.. It provides two speeds of the hydraulic cylinder during the extension stroke: When the three-way, two-position DCV is unactuated in spring offset mode, extension speed is normal. When this DCV is actuated, extension speed increases by the regenerative capability of the circuit. Example 1. For the circuit of Fig. 1.19, give the sequence of operation of cylinders 1 and when the pump is turned ON. Assume that both cylinders are initially fully retracted. Figure 1.19
Solution: Cylinder 1 extends, cylinder extends. Cylinder 1 retracts, cylinder retracts. Example 1.5 What safety features does Fig. 1.0possess in addition to a pressure-relief valve. If the load on cylinder 1 is greater than the load on cylinder, how will the cylinder move when DCV is shifted into the extending or retracting mode? Explain your answer. Figure 1.0 Solution: Both solenoid-actuated DCVs must be actuated in order to extend or retract the hydraulic cylinder. Cylinder will extend through its complete stroke receiving full pump flow while cylinder 1 will not move. The moment cylinder will extend through its complete stroke, cylinder 1 will receive full pump flow and extend through its complete stroke. This is because the system pressure builds up until load resistance is overcome to move cylinder with the smaller load. Then the pressure continues to increase until the load on cylinder 1 is overcome. This then causes cylinder 1 to extend. In retraction mode, the cylinders move in the same sequence. Example 1.6 Assuming that the two double-rod cylinders of Fig. 1.1 are identical, what unique feature does the circuit in Fig. 1.1 possess. 5
Figure 1.1 Solution: Both cylinder strokes would be synchronized. Example 1.7 For the hydraulic system is shown in Fig. 1. (a) What is the pump pressure for forward stroke if the cylinder loads are 000 N each and cylinder 1 has the piston area of 65 cm and zero back pressure? (b) What is pump pressure for retraction stroke (loads pull to right), if the piston and rod areas of cylinder equal to 50 cm and 15 cm, respectively, and zero back pressure? (c) Solve using a back pressure p of 00 kpa instead of zero, the piston area and rod area of cylinder equal 50 and 15 cm, respectively. Solution: (a)pressure acting during forward stroke is F1 F 000 000 p1 6.77 MPa A 6510 (b)for cylinder we can write For cylinder 1, force balance gives p1 p ( A A ) p A F p r p But Ap Ap1 Ar1. So we can write p ( A A ) F p1 r1 1 p A and rod side pressure of second cylinder is given by F p 1 6
F F 000 000 p 1570000 Pa 1.57 MPa 1 Ap Ar 510 F 1 F A r1 A r p p A p1 p 1 A p p (c)for cylinder 1, we have Similarly for cylinder, we have Figure 1. p A p ( A A ) F 1 p1 p1 r1 1 p A p ( A A ) F p p r Adding both equations and noting that Ap Ap1 Ar1 yield p A p ( A A ) F F 1 p1 p r 1 F 1 F p ( Ap Ar ) p1 Ap1 000 N 000 N 00000 N / m (50 15) cm 10 m p1 65 cm 10 m p 1 6.9 MPa 7
Example 1.8 For the double-pump system in Fig. 1., what should be pressure setting of the unloading valve and pressure-relief valve under the following conditions: (a) Sheet metal punching operation requires a force of 8000 N. (b) A hydraulic cylinder has a.75 cm diameter piston and a 1.5 cm diameter rod. (c) During the rapid extension of the cylinder, a frictional pressure loss of 675 kpa occurs in the line from the high-flow pump to the blank end of the cylinder. During the same time, a 50 kpa pressure loss occurs in the return line from the rod end of the cylinder to the oil tank. Frictional pressure losses in these lines are negligibly small during the punching operation. (d) Assume that the unloading valve and relief-valve pressure setting (for their full pump flow requirements) should be 50% higher than the pressure required to overcome frictional pressure losses and the cylinder punching load, respectively. / DCV (solenoid operated) CV1 Pressure relief valve Highpressure line Pressure unloading valve Highpressure low-flow Low-pressure line Electric motor Low-pressure high-flow pump Figure 1. Solution: 8
Unloading valve: Back pressure force on the cylinder equals pressure loss in the return line times the effective area of the cylinder ( Ap Ar) : N π Fback pressure 50000 (0.075 0.015 ) m N m Pressure at the blank end of the cylinder required to overcome back pressure force equals the back pressure force divided by the area of the cylinder piston: N pcyl blank end 11 kpa π (0.075 )m Thus,the pressure setting of unloading valve equals 1.50 675 11 kpa 180 kpa Pressure relief valve: Pressure required to overcome the punching operation equals the punching load divided by the area of the cylinder piston: 8000 N ppunching 70 kpa π (0.075 )m Thus, the pressure setting of pressure-relief valve equals 1.50 70 kpa = 10860 kpa Example1.9 Design a suitable hydraulic circuit to raise and lower a load of magnitude 10000 kgf at a speed of 100 mm/s. The speed must be equal both during raising and lowering of the load. The load is essentially overrunning. The load must be lowered gradually onto the platform. Calculate the flow through the control valves and indicate the pressure gauge readings both at the cap end and at the rod end during raising and lowering. Explain your reasons for your choice of the hydraulic components. Neglect mechanical and hydraulic losses. Assume 100 mm bore for the cylinder and a rod diameter = 5 mm. Solution: In any double-acting cylinder the rod end area is smaller than the cap end area to the extent of the piston rod cross-sectional area and so the pressure required to raise/lower the load is derived from the rod end area. Accordingly, the pressure required to raise the load is 10000 160 kgf/cm 160 bar (10.5 ) The relief-valve setting pressure is 175 bar. The cap end area =.758(10) = 75.8 cm. If the cylinder has to extend and retract at the rate of 100 mm/s,the flow required at the cap end of the cylinder is (75.8 10) = 758. cm /s or 7 LPM Flow required at the rod end would be (6.6 10) = 66 cm /s or 7.5 LPM. Referring to the circuit in Fig. 1., a constant delivery pump with a pressure rating of 175 bar capable of delivering 50 LPM has been chosen. A variable delivery pump would not help because both velocity and pressure are constant throughout the cycle. It is required that the piston must travel both during extension 9
10000 kgf and retraction at the same speed. Thus, flow control valve (1) is used on the cap end of the cylinder because pump supply is constant but cap end and rod end areas differ. Since it is an overrunning load, a flow control valve became necessary () on the rod end (meter-out flow control). Because it is required that the load must be positioned on the platform gradually, flow control () and solenoid-operated DCV () become necessary. Toward the end of the stroke, the load makes contact with a limit switch. This energizes valve () to divert rod end flow through the flow control valve () so that the load is decelerated from 100 mm/s to 0 mm/s. In order to ensure accurate speed control pressure- and temperature-compensated flow, flow control valves were chosen. Flow through the flow control valve () during deceleration would be = 188 cm /s or 11 LPM. While raising the load, the required flow to the rod end of the cylinder is 7.5 LPM. But the pump is supplying 50 LPM. The excess flow must pass over the relief valve which is set at 175 bar. The reliefvalve setting pressure of 175 bar creates a retracting force on the rod end equaling (175 6.6) = 10955 kgf Of this, 10000 kgf is required just to balance the load. p C P R () (1) () (1) () () () Sensor Figure 1. The remaining 955 kgf acting in the retracting direction has to be balanced by the backpressure due to the flow control valve (1). Consequently, the pressure gauge P c at the cap end during retraction would read 955/6.6 = 15 bar. When the load is lowered at a speed of 100 mm/s, the cylinder extends at a velocity of 100 mm/s. The flow entering the cap end of the cylinder is 7 LPM, which is less than the pump delivery, which is 50 LPM. The pressure gauge P c would read 175 bar because the extra flow must be dumped over the relief valve. In this operating condition, the extension force (175 75.8) = 17 kgf, together with the load force = 10000 kgf, tries to extend the cylinder. 10
According to Newton s first law of motion the net force must be equal to zero for an object moving at a constant velocity, neglecting friction. So the balancing force at the rod end should be 7 kgf. Therefore, the pressure gauge P r at the rod would read 77 / 6.6 = 79 bar. At the end of high-speed extension solenoid valve () is energized to decelerate the load from 100 mm/s to 0 mm/s. The time duration around which this occurs depends on the valve response time that can be assumed as 0 ms or 0.00 s. The deceleration Vf Vi a t where final velocity Vf 0 mm/s, initial velocity Vi 100 mm/s, δtis the time element = 0.00 s during which time the change occurs. Substituting the relevant values, we obtain the value of a as 500 mm/s.therefore, the force required to decelerate the cylinder is F = ma Now m= 10000/9.81 = 1019 kg. So F = 1019.5 = 566 kgf Therefore p R = (7 + 566)/6.6 = 6 bar Example 1.10 For the fluid power system shown in Fig. 1.5, (a) Determine the external loads F 1 and F that each hydraulic cylinder can sustain while moving in an extending direction. Take frictional pressure losses into account. The pump produces a pressure of increase of 6.90 MPa from the inlet port to the discharge port and a flow rate of 0.005 m /s. The following data are applicable: Kinematic viscosity of oil 0.000090 Specific weight of oil 780 N/m Cylinder piston diameter 0.0 m Cylinder rod diameter 0.10 m All elbows are 90 with k factor 0.75 m /s Pipe lengths and diameters are given: Pipe Number Length (m) Diameter Pipe number Length(m) Diameter 1 1.8 0.0508 6.05 0.05 9.15 0.017 7.05 0.05 6.10 0.017 8 1. 0.017.05 0.05 9 1. 0.017 5.05 0.05 (b)determine the heat generation rate. (c) Determine the extending and retracting speeds of cylinder. 11
Cylinder 1 Cylinder F 1 Tee k=1.8 F 6 7 8 5 Tee k=1.8 DCV, k=5 PRV 9 Check valve k= Elbow 1 Elbow Figure 1.5 Solution: (a) Cylinders 1 and are identical and are connected by identical lines. Therefore, they receive equal flows and sustain equal loads (F 1 = F ). Velocity is calculated from discharge and area as Q (m /s) v A (m ) Head loss in the systems is given by Reynolds number is given by H L fl v 1 p K 1 D p g 1
Re Flow through path (Fig. 1.6) is given by VD VD VD / 0.005 Q Flow through path 6 (Fig 1.6) is given by 0.0016m /s (0.0 0.10 ) 0.0016 0.00095 m /s Q6 0.0 Similarly for paths 8 and 9 we can write Velocity calculation: Reynolds number calculation: Q 8 Q9 0.00095 0.00189 m /s 0.005 ( m / s) v1 (0.0508) ( m ) 1. m/s 0.005 ( m / s) v,.19 m/s (0.017) (m ) 0.0016 ( m /s) v.9 m/s (0.05) (m ) 0.00095 ( m / s) v6 1.86 m/s (0.05) ( m ) 0.00189 ( m / s) v8, 9.9 m/s (0.017) (m ) 1. 0.0508 Re(1) 677 0.00009.19 0.017 Re, 1087 0.00009.9 0.05 Re() 680 0.00009 1.86 0.05 Re(6) 508 0.00009.9 0.017 Re(8, 9) 815 0.00009 All flows are laminar; hence we can calculate the losses in each branch. The general formula is 1 fl p v HL K 1 D p g where 1
Hence the losses are 6 f Re H L (1) H L () H L () H L () H L (6) H 6 1.8 1. 7.5 0. m 780 0. Pa 560 Pa 677 0.0508 9.81 6 9.15.19 10.9 m 85500 Pa 1087 0.017 9.81 6 1..19 6.8 15. m 10000 Pa 1087 0.017 9.81 6.05.9 1.8.1 m 500 Pa 680 0.05 9.81 6.05 1.86 0.67 m 0900 Pa 508 0.05 9.81 6 6.11..9 HL 5.75 1.8 m 100500 Pa 815 0.017 9.81 L(8) (9) Total force can now be calculated as (0.0 ) F1 F [(6900000) (560 85500 10000 500)] (0.0 0.10 ) (0900 100500) F1 F [16000] [90] =1000 N (b)we have Heat generation rate (power loss in W) = Pressure Discharge ={(560 85500 10000) (0.005) ( 0900 0.00095) (500 0.0016) (100500 0.00189)} = 5 + 9.5 + 81.9 + 190 = 85W = 0.85 kw (c) Cylinder piston diameter = 0.0 m (0.0 ) Area of piston ( A p ) = m Cylinder rod diameter = 0.10 m (0.10 ) Area of rod = m (0.0 ) (0.10 ) Annulus area Aannulus = Now 1
Q v cyl (m /s) A (m ) where each cylinder receives one half of pump flow because of the configuration of cylinder. Extension velocity is given by Retracting velocity is given by v Qblank end (m /s) ext Ap (m ) 0.0016 (0.0 ) 0.089 m/s v Q (m /s) ret Aannulus (m ) 0.005 0.051 m/s (0.0 ) (0.10 ) Example 1.11 Figure 1.6 shows a regenerative circuit in which an 18.65 kw electric motor drives a 90% efficient pump. The pump discharge pressure is 6897 kpa. Take frictional pressure losses into account. (a)determine the external load F that the hydraulic cylinder can sustain in the regenerative mode (springcentered position of DCV). (b) Determine the heat generation rate due to frictional pressures losses in the regenerative mode. (c) Determine the cylinder speed for each position of the DCV. The following data are applicable: Kinematic viscosity of oil 0.000090 m /s Specific weight of oil 7850 N/m Cylinder piston diameter 0.0 m Cylinder rod diameter 0.10 m All elbows are 90 with k factor 0.75 Pipe lengths and diameters are given Pipe number Length (m) Diameter 1 0.61 0.0508 6.10 0.05 9.15 0.05 9.15 0.05 5 6.10 0.05 15
Elbow Cylinder Elbow 1 5 Elbow Strainer in the tank K = 10 K = 5 Figure 1.6 Solution: (a) Determination of external load, considering all losses: Let us first calculate the flow rate at different branches as shown in Fig. 1.7. Before we calculate the losses, we calculate the pump power as The flow rate is given by We can write the force balance Now Pump power η P 0.9018.65 16.79 kw pump 16.79 kw Qpump 0.00 m /s 6897 kpa Fregen pblank end AP prod end Aannulus Q Q Q pump 1 0.00 m /s From the derivation of regenerative circuits, we can write Q A p Qpump Ar (0.10) (0.0) / 0.00 / 0.0097 m /s 16
Q A A p r Qpump Ar [(0.0) (0.10) ]/ 0.00 (0.10) / 0.0079 m /s Velocity calculation 0.00 (m /s) 1 v 1.0 m/s (0.0508 ) (m ) 0.00 (m /s) v 1.56 m/s (0.05 ) (m ) 0.0097 (m / s) v 6. m/s (0.05 ) ( m) 0.0079 (m / s) v.69 m/s (0.05 ) ( m) Reynolds number calculation 1.0 0.0508 Re(1) 655 0.00009 1.56 0.05 Re() 76 0.00009 6. 0.05 Re() 990 0.00009.69 0.05 Re() 0 0.00009 Assume that all flows are laminar; head losses can be calculated as follows: H L (1) 6 6.10 1.0 10 0.7 m 7850 0. Pa 580 Pa 655 0.0508 9.81 H L () 6 6.10 1.56 5.08 m 1600 Pa 76 0.05 9.81 H L () 6 9.15 6. 0.75 10. m 8000 Pa 990 0.05 9.81 17
H L () 6 9.15.69 0.75 7. m 5800 Pa 0 0.05 9.81 The force is given by (0.0) F [(6897 kpa) (5.8 16. 80.)] [(0.0) (0.10) ] [6897 kpa (5.8 16. 58.)] Solving we get (b) Determination of heat generation rate Power loss = QΔp F 0 168 =5 kn = Pipe 1 loss + pump loss+ pipe loss + pipe loss + pipe loss = 0.00 5.8 + (18.7 16.8) + 0.00 16. +0.0097 80. + 0.0079 58. = 0.01+1.9+0.0+0.78+0. Power loss = Heat generation rate =.16 kw (c) Cylinder speed for each position of DCV Upper position of DCV Q Q Q pump 1 0.00 m /s Spring-centered position of DCV v Qpump (m / s) 0.00 ext Ap (m ) 0.0751 m/s (0.0 ) Lower position of DCV v Qpump (m /s) 0.00 ext Arod (m ) 0.97 m/s (0.10 ) v Qpump (m / s) 0.00 ret Aannulus (m ) 0.100 m/s (0.0 ) (0.10 ) 18
Example 1.1 For the meter-in flow control valve system of Fig. 1.7, the following data are given: Desired cylinder speed 0.5 m/s Cylinder piston diameter 0.508 m Cylinder load 10 N Specific gravity of oil 0.9 Pressure-relief valve setting 6895 kpa Determine the required capacity coefficient of flow control valve. 10 N Figure 1.7 Solution: We have C V are LPM / C V V cyl A piston p ( F / A PRV load piston SG kpa.therefore, we have the following units for the terms in the above equation: ) (1.7) Q v A LPM, p kpa, ( F ) / ( A ) Pressur e kpa cyl p PRV load piston 19
The flow rate is given by Q v A cyl p m 1 L 60 s s 0.5 0.000m 0.9 LPM 0.001 m 1 min Now it is given that pprv 6895 kpa and F A load piston Substituting values in Eq. (1.7), we get 10 N 1 kpa 6570 kpa 0.000 m 1000 N/m C V v cyl A piston p ( F / A PRV load piston SG 0.9 6895 6570 0.9 0.9 LPM 1.6 19.0 kpa ) 0