Lecture 25 HYDRAULIC CIRCUIT DESIGN AND ANALYSIS [CONTINUED]
|
|
- Augustus Adams
- 5 years ago
- Views:
Transcription
1 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
2 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 m /s, find the cylinder speed and load-carrying capacity for the Solution: (a) Extending stroke. (b) Retracting stroke. (a) We have Qp ve xt 0.6 m/s A 6510 r F load-extension N pa r 5 (b) We have Qp vret 0.6 m/s A A (10 6) 5 10 p r
3 F p A A load-extension ( ) p r (10 65) 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?
4 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
5 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
6 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 p 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
7 F F p 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 N / m (50 15) cm 10 m p1 65 cm 10 m p MPa 7
8 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
9 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 ( ) 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 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 kpa = kpa Example1.9 Design a suitable hydraulic circuit to raise and lower a load of magnitude 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 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 ( ) = 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
10 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 ( ) = kgf Of this, 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 ( ) = 17 kgf, together with the load force = kgf, tries to extend the cylinder. 10
11 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 = = 566 kgf Therefore p R = ( )/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 m /s. The following data are applicable: Kinematic viscosity of oil 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 (b)determine the heat generation rate. (c) Determine the extending and retracting speeds of cylinder. 11
12 Cylinder 1 Cylinder F 1 Tee k=1.8 F 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
13 Re Flow through path (Fig. 1.6) is given by VD VD VD / Q Flow through path 6 (Fig 1.6) is given by m /s ( ) m /s Q6 0.0 Similarly for paths 8 and 9 we can write Velocity calculation: Reynolds number calculation: Q 8 Q m /s ( m / s) v1 (0.0508) ( m ) 1. m/s ( m / s) v,.19 m/s (0.017) (m ) ( m /s) v.9 m/s (0.05) (m ) ( m / s) v m/s (0.05) ( m ) ( m / s) v8, 9.9 m/s (0.017) (m ) Re(1) Re, Re() Re(6) Re(8, 9) 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
14 Hence the losses are 6 f Re H L (1) H L () H L () H L () H L (6) H m Pa 560 Pa m Pa m Pa m 500 Pa m 0900 Pa HL m Pa L(8) (9) Total force can now be calculated as (0.0 ) F1 F [( ) ( )] ( ) ( ) F1 F [16000] [90] =1000 N (b)we have Heat generation rate (power loss in W) = Pressure Discharge ={( ) (0.005) ( ) ( ) ( )} = = 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
15 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 ) (0.0 ) m/s v Q (m /s) ret Aannulus (m ) m/s (0.0 ) (0.10 ) Example 1.11 Figure 1.6 shows a regenerative circuit in which an 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 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
16 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 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 kw pump kw Qpump 0.00 m /s 6897 kpa Fregen pblank end AP prod end Aannulus Q Q Q pump m /s From the derivation of regenerative circuits, we can write Q A p Qpump Ar (0.10) (0.0) / 0.00 / m /s 16
17 Q A A p r Qpump Ar [(0.0) (0.10) ]/ 0.00 (0.10) / m /s Velocity calculation 0.00 (m /s) 1 v 1.0 m/s ( ) (m ) 0.00 (m /s) v 1.56 m/s (0.05 ) (m ) (m / s) v 6. m/s (0.05 ) ( m) (m / s) v.69 m/s (0.05 ) ( m) Reynolds number calculation Re(1) Re() Re() Re() Assume that all flows are laminar; head losses can be calculated as follows: H L (1) m Pa 580 Pa H L () m 1600 Pa H L () m 8000 Pa
18 H L () m 5800 Pa The force is given by (0.0) F [(6897 kpa) ( )] [(0.0) (0.10) ] [6897 kpa ( )] Solving we get (b) Determination of heat generation rate Power loss = QΔp F =5 kn = Pipe 1 loss + pump loss+ pipe loss + pipe loss + pipe loss = ( ) = Power loss = Heat generation rate =.16 kw (c) Cylinder speed for each position of DCV Upper position of DCV Q Q Q pump m /s Spring-centered position of DCV v Qpump (m / s) 0.00 ext Ap (m ) 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 ) m/s (0.0 ) (0.10 ) 18
19 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 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
20 The flow rate is given by Q v A cyl p m 1 L 60 s s m 0.9 LPM 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 m 1000 N/m C V v cyl A piston p ( F / A PRV load piston SG LPM kpa ) 0
Hydraulic Circuit Design
26/10/05 Hydraulics & Pneumatics Hydraulic Circuit Design M. S. Sham Prasad Asst. Professor Dept. of I&P Engg. National Institute of Engg, Mysore Regenerative circuit : A Regenerative circuit is one in
More informationTest Which component has the highest Energy Density? A. Accumulator. B. Battery. C. Capacitor. D. Spring.
Test 1 1. Which statement is True? A. Pneumatic systems are more suitable than hydraulic systems to drive powerful machines. B. Mechanical systems transfer energy for longer distances than hydraulic systems.
More informationModule 6 Assignment Part A
Module 6 Assignment Part A TOTAL MARKS Part A = 192 TOTAL QUESTIONS Part A = 36 Question 1 [3 Marks] What does pressure refer to in relation to hydrostatics and what is it dependent on? Question 2 [14
More informationLECTURE 24 TO 26 - HYDRAULIC CIRCUIT DESIGN FREQUENTLY ASKED QUESTIONS
LECUE 24 O 26 - HYDAULIC CICUI DESIGN EQUENLY ASKED QUESIONS 1. List three important considerations to be taken into account while designing a hydraulic circuit here are 3 important considerations in designing
More informationLecture 4. Lab this week: Review: Pilot-Open-Check. Cartridge valves Flow divider Properties of Hydraulic Fluids. Course feedback (2mins)
109 Lab this week: Lab 8 Sequencing circuit Lab 9 Flow divider Lecture 4 Review: Pilot-Open-Check Area ratio and pressure divider Cartridge valves Flow divider Properties of Hydraulic Fluids Viscosity
More informationWhat does pressure refer to in relation to hydrostatics and what is it dependent on?
Question 1 [3 Marks] What does pressure refer to in relation to hydrostatics and what is it dependent on? Question 2 [14 Marks] Make a circuit diagram of a regular hydraulic plant that is used to control
More informationExperiments on Hydraulic and Pneumatic circuit trainers HYDRAULIC CIRCUITS:
Experiments on Hydraulic and Pneumatic circuit trainers HYDRAULIC CIRCUITS: Hydraulic circuits are used in high power and high load applications such as earth moving equipment. The pressure used in industrial
More informationLecture 11 HYDRAULIC MOTORS [CONTINUED]
Lecture 11 HYDRULIC MOORS [CONINUED] 1.12Performance of Hydraulic Motors he performance of hydraulic motors depends upon many factors such as precision of their parts, tolerances between the mating parts,
More informationModule 6. Actuators. Version 2 EE IIT, Kharagpur 1
Module 6 ctuators Version 2 II, Kharagpur 1 Lesson 28 Industrial Hydraulic ircuits Version 2 II, Kharagpur 2 Lesson Objectives fter learning the lesson students should be able to escribe typical industrial
More informationFLUID POWER FLUID POWER EQUIPMENT TUTORIAL HYDRAULIC AND PNEUMATIC CYLINDERS. This work covers part of outcome 2 of the Edexcel standard module:
FLUID POWER FLUID POWER EQUIPMENT TUTORIAL HYDRAULIC AND PNEUMATIC CYLINDERS This work covers part of outcome 2 of the Edexcel standard module: UNIT 21746P APPLIED PNEUMATICS AND HYDRAULICS The material
More informationthree different ways, so it is important to be aware of how flow is to be specified
Flow-control valves Flow-control valves include simple s to sophisticated closed-loop electrohydraulic valves that automatically adjust to variations in pressure and temperature. The purpose of flow control
More informationDetermination of power loss of combine harvester travel gear
Agronomy Research 13(1), 5 3, 015 Determination of power loss of combine harvester travel gear L. Beneš *, P. Heřmánek and P. Novák Czech University of Life Sciences Prague, Faculty of Engineering, Department
More informationApplied Fluid Mechanics
Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and
More informationFluidics. Hydraulic circuit. Course teacher Prof. Mahbubur Razzaque
COURSE NUMBER: ME 433 Fluidics Hydraulic circuit Course teacher Prof. Mahbubur Razzaque 1 FAIL-SAFE CIRCUITS Fail-safe circuits are those designed to prevent injury to the operator or damage to equipment.
More informationBasic Hydraulics and Pneumatics
Basic Hydraulics and Pneumatics Module 2: Actuators and directional control valves PREPARED BY Academic Services August 2011 Applied Technology High Schools, 2011 ATM 1122 Basic Hydraulics Module 2: Actuators
More informationProjekthandbuch. Project Manual Industriehydraulik. Industrial Hydraulics RE 00846/ Trainee's manual. Schülerhandbuch
Electric Drives and Controls Hydraulics Linear Motion and Assembly Technologies Pneumatics Service Projekthandbuch Project Manual Industriehydraulik Industrial Hydraulics RE 00846/04.07 Schülerhandbuch
More informationDEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Sub. Code/Name: ME1305 Applied Hydraulics and Pneumatics Year/Sem: III/V
DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub. Code/Name: ME1305 Applied Hydraulics and Pneumatics Year/Sem: III/V UNIT-1 FLUID POWER SYSTEMS AND FUNDAMENTALS 1. Define fluid power? 2. List the
More informationHydraulic energy control, conductive part
Chapter 2 2 Hydraulic energy control, conductive part Chapter 2 Hydraulic energy control, conductive part To get the hydraulic energy generated by the hydraulic pump to the actuator, cylinder or hydraulic
More informationTUTORIAL QUESTIONS FOR COURSE TEP 4195
TUTORIL QUESTIONS FOR COURSE TEP 4195 Data: Hydraulic Oil Density 870 kg/m 3 bsolute viscosity 0.03 Ns/m 2 Spool valve discharge coefficient 0.62. 1) hydrostatic transmission has a variable displacement
More informationLecture 4. Lab 8 Check valve and pilot-operated check valves Lab 9 Flow divider. Update: Identifying lab objectives Review: Metering/Bleed-off
98 Lecture 4 Labs coming week: Lab 8 Check valve and pilot-operated check valves Lab 9 Flow divider Update: Identifying lab objectives Review: Metering/Bleed-off (More) efficient circuits? Check and Pilot-to-Open-Check
More informationBASIC HYDRAULICS PRINCIPLES OF HYDRAULIC PRESSURE AND FLOW LEARNING ACTIVITY PACKET BB831-XA03XEN
BASIC HYDRAULICS LEARNING ACTIVITY PACKET PRINCIPLES OF HYDRAULIC PRESSURE AND FLOW TM BB831-XA03XEN LEARNING ACTIVITY PACKET 3 PRINCIPLES OF HYDRAULIC PRESSURE AND FLOW INTRODUCTION Previous LAPs discussed
More informationFormulas and units
Formulas and units Hydraulic system and circuit design is limited only by the creativity of the application engineer. All basic circuit design begins with the ultimate actuator functions in mind however.
More informationFLUID FLOW. Introduction
FLUID FLOW Introduction Fluid flow is an important part of many processes, including transporting materials from one point to another, mixing of materials, and chemical reactions. In this experiment, you
More informationName common, basic subsystems that compose
Chapter 13 Applying Hydraulic Power Typical Circuits and Systems 1 Objectives Name common, basic subsystems that compose complex hydraulic systems and describe their function. Compare the design and operation
More informationINDIAN INSTITUTE OF TECHNOLOGY KHARAGPUR NPTEL ONLINE CERTIFICATION COURSE. On Industrial Automation and Control
INDIAN INSTITUTE OF TECHNOLOGY KHARAGPUR NPTEL ONLINE CERTIFICATION COURSE On Industrial Automation and Control By Prof. S. Mukhopadhyay Department of Electrical Engineering IIT Kharagpur Topic Lecture
More informationCH.4 Basic Components of Hydraulic and Pneumatic System/16 M HAP/17522/AE5G
Content : 4.1 Hydraulic and Pneumatic actuators. 10 Marks Hydraulic Actuators - Hydraulic cylinders (single, double acting and telescopic) construction and working, Hydraulic motors (gear and piston type)
More informationTUTORIAL QUESTIONS FOR THE INDUSTRIAL HYDRAULICS COURSE TEP 4205
TUTORIAL QUESTIONS FOR THE INDUSTRIAL HYDRAULICS COURSE TEP 4205 The book for the course is Principles of Hydraulic System Design, by Peter J Chapple. Published by Coxmoor Publishing Co., UK. Available
More informationModule 5: Valves. CDX Diesel Hydraulics. Terms and Definitions. Categories of Valves. Types of Pressure Control Valves
Terms and Definitions Categories of Valves Types of Pressure Control Valves Types and Operation of Pressure Relief Valves Operation of an Unloading Valve Operation of a Sequencing Valve Operation of a
More informationFamiliarize yourself with the pressure loss phenomenon. The Discussion of this exercise covers the following point:
Exercise 3-2 Pressure Loss EXERCISE OBJECTIVE Familiarize yourself with the pressure loss phenomenon. DISCUSSION OUTLINE The Discussion of this exercise covers the following point: Pressure loss Major
More informationLESSON 2 BASIC CONSTRUCTION AND OPERATION OF HYDRAULIC ACTUATING DEVICES, FLOW CONTROL, AND DIRECTIONAL DEVICES. STP Tasks:
LESSON 2 BASIC CONSTRUCTION AND OPERATION OF HYDRAULIC ACTUATING DEVICES, FLOW CONTROL, AND DIRECTIONAL DEVICES STP Tasks: 552-758-1003 552-758-1071 OVERVIEW LESSON DESCRIPTION: In this lesson you will
More informationProject Manual Industrial Hydraulics
Electric Drives and Controls Hydraulics Linear Motion and Assembly Technologies Pneumatics Service Project Manual Industrial Hydraulics RE 00845/04.07 Trainer s manual Electric Drives and Controls Hydraulics
More informationAT 2303 AUTOMOTIVE POLLUTION AND CONTROL Automobile Engineering Question Bank
AT 2303 AUTOMOTIVE POLLUTION AND CONTROL Automobile Engineering Question Bank UNIT I INTRODUCTION 1. What are the design considerations of a vehicle?(jun 2013) 2..Classify the various types of vehicles.
More informationExperiment (4): Flow measurement
Introduction: The flow measuring apparatus is used to familiarize the students with typical methods of flow measurement of an incompressible fluid and, at the same time demonstrate applications of the
More informationCOMPRESSIBLE FLOW ANALYSIS IN A CLUTCH PISTON CHAMBER
COMPRESSIBLE FLOW ANALYSIS IN A CLUTCH PISTON CHAMBER Masaru SHIMADA*, Hideharu YAMAMOTO* * Hardware System Development Department, R&D Division JATCO Ltd 7-1, Imaizumi, Fuji City, Shizuoka, 417-8585 Japan
More informationHovercraft
1 Hovercraft 2017-2018 Names: Score: / 44 Show all equations and work. Point values are shown in parentheses at the end of the question. Assume g=9.8 m/s/s for all calculations. Include units in your answer.
More informationBasic Hydraulics. Module 2: Actuators and directional control valves. Curriculum Development Unit PREPARED BY. August 2013
Basic Hydraulics Module 2: Actuators and directional control valves PREPARED BY Curriculum Development Unit August 2013 Applied Technology High Schools, 2013 ATM-312 Basic Hydraulics Module 2: Actuators
More informationLECTURE-23: Basic concept of Hydro-Static Transmission (HST) Systems
MODULE-6 : HYDROSTATIC TRANSMISSION SYSTEMS LECTURE-23: Basic concept of Hydro-Static Transmission (HST) Systems 1. INTRODUCTION The need for large power transmissions in tight space and their control
More informationCode No: R Set No. 1
Code No: R05310304 Set No. 1 III B.Tech I Semester Regular Examinations, November 2007 KINEMATICS OF MACHINERY ( Common to Mechanical Engineering, Mechatronics, Production Engineering and Automobile Engineering)
More informationAir Cylinders Drive System Full Stroke Time & Stroke End Velocity. How to Read the Graph
1-1 Best Pneumatics Air Cylinders Drive System Full Time & End Velocity How to Read the Graph This graph shows the full stroke time and stroke end velocity when a cylinder drive system is composed of the
More informationUniversity of Jordan School of Engineering Mechatronics Engineering Department. Fluid Power Engineering Lab
University of Jordan School of Engineering Mechatronics Engineering Department 0908464 09 The University of Jordan School of Engineering MECHATRONICS ENGINEERING DEPARTMENT EXPERIMENT N0. 1 Introduction
More information2. Hydraulic Valves, Actuators and Accessories. 24 Marks
2. Hydraulic Valves, Actuators and Accessories 24 Marks Co related to chapter 602.2 Describe working principle of various components used in hydraulic & pneumatic systems. 602.3 Choose valves, actuators
More informationActuators and directional control valves
Actuators and directional control valves 1. Differentiate between the main types of directional control valves. 2. Demonstrate the function and uses of 3/2 way valve, push button actuated. 3. Demonstrate
More informationLecture 7. Coming week s lab: Integrative lab (your choice!)
Lecture 7 Coming week s lab: Integrative lab (your choice!) Today: Systems review exercise due end of class Your feedback Review: sequencing and asynchronous circuit analysis Hydraulic hybrid vehicles
More informationLecture 6. This week: Lab 13: Hydraulic Power Steering [ Lab 14: Integrated Lab (Hydraulic test bench) ]
133 Lecture 6 This week: Lab 13: Hydraulic Power Steering [ Lab 14: Integrated Lab (Hydraulic test bench) ] 4-way directional control valve; proportional valve; servo-valve Modeling / Analysis of a servo-valve
More informationSIDDHARTH INSTITUTE OF ENGINEERING & TECHNOLOGY :: PUTTUR (AUTONOMOUS) QUESTION BANK UNIT I I.C ENGINES
SIDDHARTH INSTITUTE OF ENGINEERING & TECHNOLOGY :: PUTTUR UNIT I I.C ENGINES 1 (a) Explain any six types of classification of Internal Combustion engines. (6M) (b) With a neat sketch explain any three
More informationPneumatic Valve 4V100 Series
tic Valves Pneumatic Valve 4V100 4V100 Low power consumption exhaust for the pilot and main valve, small size, large flow, good appearance, flexible, can be integrated installation, high wear resistant
More informationI) Clamping the work piece II) Drilling the work piece. III) Unclamping the work piece. 10
Seventh Semester B.E. III IA Test, 2014 USN 1 P E M E PES INSTITUTE OF TECHNOLOGY (Bangalore South Campus) (Hosur Road, 1KM before Electronic City, Bangalore-560 100) Department of Mechanical Engineering
More informationPNEUMATIC & HYDRAULIC SYSTEMS
PNEUMATIC & HYDRAULIC SYSTEMS CHAPTER SIX PNEUMATIC SYSTEM DESIGN AND DEVELOPMENT Dr. Ibrahim Naimi Symbols And Standards In Pneumatics The development of pneumatic systems is assisted by a uniform approach
More informationLECTURE 15 TO 17 DIRECTIONAL CONTROL VALVES FREQUENTLY ASKED QUESTIONS
LECURE 15 O 17 DIRECIONL CONROL VLVES FREQUENLY SKED QUESIONS 1. Explain briefly the function of directional control valves o o start, stop, accelerate, decelerate and change the direction of motion of
More informationChapter 13: Application of Proportional Flow Control
Chapter 13: Application of Proportional Flow Control Objectives The objectives for this chapter are as follows: Review the benefits of compensation. Learn about the cost to add compensation to a hydraulic
More informationInput, Control and Processing elements
PNEUMATIC & HYDRAULIC SYSTEMS CHAPTER FIVE Input, Control and Processing elements Dr. Ibrahim Naimi Valves The function of valves is to control the fluid path or the pressure or the flow rate. Depending
More informationFig. 1 Two stage helical gearbox
Lecture 17 DESIGN OF GEARBOX Contents 1. Commercial gearboxes 2. Gearbox design. COMMERICAL GEARBOX DESIGN Fig. 1 Two stage helical gearbox Fig. 2. A single stage bevel gearbox Fig. 4 Worm gearbox HELICAL
More informationModule 4: Actuators. CDX Diesel Hydraulics. Terms and Definitions. Cylinder Actuators
Terms and Definitions Cylinder Actuators Symbols for Actuators Terms and Definitions II Cylinders Providing Linear Motion Cylinders Providing Angular Motion Parts of Actuators Mounting of Actuators Seals
More informationAir Cylinders Drive System Full Stroke Time & Stroke End Velocity. How to Read the Graph
1 Best Pneumatics Air Cylinders Drive System Full Time & End Velocity How to Read the Graph This graph shows the full stroke time and stroke end velocity when a cylinder drive system is composed of the
More informationECH 4224L Unit Operations Lab I Fluid Flow FLUID FLOW. Introduction. General Description
FLUID FLOW Introduction Fluid flow is an important part of many processes, including transporting materials from one point to another, mixing of materials, and chemical reactions. In this experiment, you
More informationUNIT - 4 TESTING OF DC MACHINES
UNIT - 4 TESTING OF DC MACHINES Testing of DC machines can be broadly classified as i) Direct method of Testing ii) Indirect method of testing DIRECT METHOD OF TESTING: In this method, the DC machine is
More informationFLUID POWER TUTORIAL HYDRAULIC PUMPS APPLIED PNEUMATICS AND HYDRAULICS H1
FLUID POWER TUTORIAL HYDRAULIC PUMPS This work covers outcome 2 of the Edexcel standard module: APPLIED PNEUMATICS AND HYDRAULICS H1 The material needed for outcome 2 is very extensive so the tutorial
More informationElectric Motors and Drives
EML 2322L MAE Design and Manufacturing Laboratory Electric Motors and Drives To calculate the peak power and torque produced by an electric motor, you will need to know the following: Motor supply voltage:
More informationUNIT - 3 Friction and Belt Drives
UNIT - 3 Friction and Belt Drives 1.State the laws of dynamic or kinetic friction (03 Marks) (June 2015) Laws of Kinetic or Dynamic Friction Following are the laws of kinetic or dynamic friction: 1. The
More informationLecture 6. Systems review exercise To be posted this afternoon Due in class (10/23/15)
153 Systems review exercise To be posted this afternoon Due in class (10/23/15) Lecture 6 Coming week: Lab 13: Hydraulic Power Steering Lab 14: Integrated Lab (Hydraulic test bench) Topics today: 2 min
More informationOil Hydraulics Basic Technology Textbook
Technical education document Oil Hydraulics Basic Technology Textbook DAIKIN INDUSTRIES, LTD. Training Dept. Chapter 1 Basis of Oil Hydraulics CONTENTS 1. Overview of oil hydraulics...1 2. Basic structure
More informationMarine Engineering Exam Resource Review of Hydraulics
1. What is Pascal s law? Pressure confined on a confined fluid will transmit the pressure in all directions and act with equal force on all areas at right angles. 2. How does the law pertain to hydraulics?
More informationThe University of Melbourne Engineering Mechanics
The University of Melbourne 436-291 Engineering Mechanics Tutorial Twelve General Plane Motion, Work and Energy Part A (Introductory) 1. (Problem 6/78 from Meriam and Kraige - Dynamics) Above the earth
More informationMaintenance and service manual for elevator control valve
Maintenance and service manual for elevator control valve EV 100 blain Biofial s.a. Pontou 52 street (behind B K.T.E.O.) P.C. 54628 Thessaloniki Greece Tel: +302310755032 Fax: +302310754929 www.biofial.gr
More informationFLUID FLOW Introduction General Description
FLUID FLOW Introduction Fluid flow is an important part of many processes, including transporting materials from one point to another, mixing of materials, and chemical reactions. In this experiment, you
More informationIntroduction. General Information. Systems Operation
Systems Operation Introduction Reference: For illustrated Specifications, refer to the Specifications For 416, 426, 428, 436, 438, & Series II Backhoe Loaders Transmission, Form No. SENR3131. If the specifications
More information210C-1 Selection materials
7 2C- Selection materials 8 Calculation of cylinder buckling ) Be sure to calculate the cylinder buckling. 2) In the case of using a hydraulic cylinder, the stress and buckling must be considered depending
More informationChapter 15. Inertia Forces in Reciprocating Parts
Chapter 15 Inertia Forces in Reciprocating Parts 2 Approximate Analytical Method for Velocity & Acceleration of the Piston n = Ratio of length of ConRod to radius of crank = l/r 3 Approximate Analytical
More informationSeven Features of the Koganei Vacuum Valve
Seven Features of the Koganei Valve Koganei Original Solenoid Construction No burning damage to solenoid No need to stock solenoids as spare parts. Starting and energizing current values are extremely
More informationSimple Gears and Transmission
Simple Gears and Transmission Simple Gears and Transmission page: of 4 How can transmissions be designed so that they provide the force, speed and direction required and how efficient will the design be?
More informationLecture- 6: Multi quadrant Operation. Multi quadrant Operation
Lecture- 6: Multi quadrant Operation Multi quadrant Operation For consideration of multi quadrant operation of drives, it is useful to establish suitable conventions about the signs of torque and speed.
More informationChapter 15. Inertia Forces in Reciprocating Parts
Chapter 15 Inertia Forces in Reciprocating Parts 2 Approximate Analytical Method for Velocity and Acceleration of the Piston n = Ratio of length of ConRod to radius of crank = l/r 3 Approximate Analytical
More informationDENISON HYDRAULICS open loop pump controls series P140 A-mod, P260 B-mod service information
DENISON HYDRAULICS open loop pump controls series P10 A-mod, P260 B-mod service information Publ. S1-AM02-A replaces S1-AM02 01-97 CONTENTS typical characteristics-------------------------------------------------------------------------------
More informationProportional Valve.
Proportional Valve Electro-Hydraulic Proportional Pilot Relief Valve Electro-Hydraulic Proportional Flow & Directional Control Valve Electro-Hydraulic Proportional Relief & Flow Control Valve Electronic
More informationLinear Actuator with Ball Screw Series OSP-E..S. Contents Description Overview Technical Data Dimensions 89
Linear Actuator with Ball Screw Series OSP-E..S Contents Description Page Overview 79-82 Technical Data 83-88 Dimensions 89 79 The System Concept ELECTRIC LINEAR ACTUATOR FOR HIGH ACCURACY APPLICATIONS
More informationSteering unit LAGZ. Data sheet. Series 2 x
Steering unit LAGZ Data sheet Nominal sizes 125 620 Series 2 x Maximum flow 50 l / min HE 11868 / 09.2017 2 LAGZ HE 11868 / 09.2017 Page Content 4 4 5 6 7 8 9 10 11 12 13 14 Features Ordering details Function,
More informationPressure Control Equipment
Control High-pressure Precision General Purpose Special Fluid/Deionized Water (Pure Water) Detection Control Control (General purpose, high-pressure, precision, vacuum, special fluid, deionized water (pure
More informationMr.Giridhar Bosch Rexroth (India) Limited A Drive & Control Company. Cylinder Presentation
Mr.Giridhar Bosch Rexroth (India) Limited A Drive & Control Company Cylinder Presentation Introduction Good Morning Ladies & Gentlemen, We spend the next hour in understanding the Hydraulic Actuator Cylinder.
More informationBuckling of Pump Barrel and Rod String Stability in Pumping Wells
This is a revised version of manuscript PO-1115-0011 "Stability of Pump Barrels and Rod String in Pumping Wells" (2015). This manuscript has been submitted to SPE Production & Operations. Manuscript has
More informationSeries: hydraulic-type control in pneumatic machinery
Adding Adding the the control, control, rigidity rigidity and and power power of of hydraulics hydraulics to to aa pneumatic pneumatic machine machine Control with Air-Oil Tanks Air-Oil tanks provide a
More informationt2_ il PioC Oper. led Relet Valve Symbol ~~~ CHAPTER 12 Pilot Operated Pressure Control Valves
CHAPTER 12 Pilot Operated Pressure Control Valves Unlike a simple or direct operated pressure control valve, where a spool is held biased by spring pressure only, a pilot operated valve has its spool biased
More informationPneumatic Systems. Module 3: Logic Operations in Electropneumatics. IAT Curriculum Unit PREPARED BY. August 2008
Pneumatic Systems Module : Logic Operations in Electropneumatics PREPARED BY IAT Curriculum Unit August 2008 Institute of Applied Technology, 2008 2 Module : Logic Operations in Electro-pneumatics Module
More informationEC380EHR, EC480EHR, EC750EHR. Volvo Excavators
EC380EHR, EC480EHR, EC750EHR Volvo Excavators Volvo EC380EHR, E480EHR, EC750EHR in detail Engine The latest generation, Volvo engine Stage V emissions certified diesel engine fully meets the demands of
More informationTheory of Machines. CH-1: Fundamentals and type of Mechanisms
CH-1: Fundamentals and type of Mechanisms 1. Define kinematic link and kinematic chain. 2. Enlist the types of constrained motion. Draw a label sketch of any one. 3. Define (1) Mechanism (2) Inversion
More informationV Series AXIAL PISTON PUMP ORDER CODE
V Series AXIAL PISTON PUMP ORDER CODE V 15 A 1 R B S - A 1 D X A 11 Customers demand Pump Series V Links Type (only V 15-18): none - Standard A - SAE A 2 bolts Ma. Displacement (cm 3 /n): 15, 18, 23, 25,
More information9/13/2017. Friction, Springs and Scales. Mid term exams. Summary. Investigating friction. Physics 1010: Dr. Eleanor Hodby
Day 6: Friction s Friction, s and Scales Physics 1010: Dr. Eleanor Hodby Reminders: Homework 3 due Monday, 10pm Regular office hours Th, Fri, Mon. Finish up/review lecture Tuesday Midterm 1 on Thursday
More informationGSM. Modular Design. Versions of the series. Type GSM. Gripper type P Z W R. Size. {} AS IS without O.D. clamping I.D. clamping
GSM s Modular Design Versions of the series Type GSM Gripper type P Z W R Size 32 40 50 64 30 38 45 16 20 25 32 40 16 20 25 32 40 Gripping force safety device {} AS IS without O.D. clamping I.D. clamping
More informationNotes on Hydraulic Systems
Notes on Hydraulic Systems Dr. P A Sastry Professor, Dept. of Mechanical Engg. MVSR Engineering College, Hyderabad. The accompanying material on Hydraulic Systems (HS) is intended to cover the syllabus
More informationSimplus
Simplus in Latin means Simple. We focus on making direct drive 1 actuators that are simple to use, plus the additional benefits of: small form factor higher performance better reliability 1 direct drive
More informationForklift Hydraulic System Design
Applied Mechanics and Materials Online: 0-0- ISSN: 66-748, Vols. 0-06, pp 748-75 doi:0.408/www.scientific.net/amm.0-06.748 0 Trans Tech Publications, Switzerland Forklift Hydraulic System Design Yu Chun,
More informationHours / 100 Marks Seat No.
17412 16117 3 Hours / 100 Seat No. Instructions (1) All Questions are Compulsory. (2) Answer each next main Question on a new page. (3) Illustrate your answers with neat sketches wherever necessary. (4)
More informationPneumatic & Hydraulic SYSTEMS
Pneumatic & Hydraulic SYSTEMS CHAPTER EIGHT HYDRAULIC PUMPS AND ACTUATORS Dr. Ibrahim Naimi The higher the discharge pressure, the lower the volumetric efficiency because internal leakage
More informationSPEC. SHEET No. GR-800E /EU-01 DATE November, 2012 TADANO ROUGH TERRAIN CRANE M ODEL : GR-800EX (Left-hand steering) G E N E R A L D ATA CRANE
DATE November, 212 TADANO ROUGH TERRAIN CRANE M ODEL : GR-8EX (Left-hand steering) G E N E R A L D ATA CRANE CAPACITY 8, kg at 3. m BOOM 5-section, 12. m 47. m DIMENSION Overall length approx. 14,375 mm
More informationa) Calculate the overall aerodynamic coefficient for the same temperature at altitude of 1000 m.
Problem 3.1 The rolling resistance force is reduced on a slope by a cosine factor ( cos ). On the other hand, on a slope the gravitational force is added to the resistive forces. Assume a constant rolling
More informationDHANALAKSHMI COLLEGE OF ENGINEERING
DHANALAKSHMI COLLEGE OF ENGINEERING (Dr.VPR Nagar, Manimangalam, Tambaram) Chennai - 601 301 DEPARTMENT OF MECHANICAL ENGINEERING III YEAR MECHANICAL - VI SEMESTER ME 6601 DESIGN OF TRANSMISSION SYSTEMS
More informationChapter B-3. Chapter 3. Actuators and output devices. Festo Didactic TP101
155 Chapter 3 Actuators and output devices Festo Didactic TP101 156 An actuator is an output device for the conversion of supply energy into useful work. The output signal is controlled by the control
More informationThe product information and specifications within this catalogue should be viewed as a guide only and are subject to change without notice.
Wheel gears... 3 Axial piston pumps... 3 Axial piston motors... 3 Winches... 4 Radial piston pumps... 4 Radial piston motors... 4 Slewing rings... 5 Gear pumps... 5 Gear flow dividers... 5 Cylinders...
More informationSection 6.1. Implement Circuit - General System. General: TF Configuration TB Configurations Implement Control Valve:
Section 6.1 Implement Circuit - General System General: TF Configuration... 6.1.3 TB Configurations... 6.1.5 Implement Pump Breakdown... 6.1.6 Operational Description: General... 6.1.7 Compensator Control...
More informationJob Sheet 1 Introduction to Fluid Power
Job Sheet 1 Introduction to Fluid Power Fluid Power Basics Fluid power relies on a hydraulic system to transfer energy from a prime mover, or input power source, to an actuator, or output device (Figure
More informationVariable Axial Piston Eaton Vickers PVH Pump
Variable Axial Piston Eaton Vickers PVH Pump PVH High pressure axial piston pumps with low noise PVH high flow, high performance pumps are a family of variable displacement, inline piston units that incorporate
More information