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Mechatronics Outreach Day

Mechatronics Outreach Day. Mohammad I. Kilani Associate Professor & Head Department of Mechatronics The University of Jordan, Amman, Jordan. Education.

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Mechatronics Outreach Day

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  1. Mechatronics Outreach Day Mohammad I. Kilani Associate Professor & Head Department of Mechatronics The University of Jordan, Amman, Jordan

  2. Education • Ph.D. Mechanical Engineering, Florida State University, Tallahassee, Florida, USA. Dissertation Title "Development of a surface micromachined spiral-channel viscous pump". • M.Eng. 1991, Mechanical Engineering, Carnegie Mellon University, Pittsburgh, Pennsylvania, USA. M.Eng Project: “Design for Assembly Analysis of Part Features and Interactions”. • M.S. 1988, Mechanical Engineering, (Computer Integrated Manufacturing) George Washington University, Washington, D.C. USA. MS Thesis: “Force Control of Robotic Manipulators”. • B.Eng. 1986, Mechanical Engineering, University of Jordan, Amman, Jordan

  3. Professional Careers • Associate Professor & Chair, Mechatronics Engineering Department, The University of Jordan, Amman, Jordan • Associate Professor, Mechanical Engineering Department, King Faisal University, Al-Ahsa, Saudi Arabia 2008 –2011 • Assistant Professor, Mechanical Engineering Department, The University of Jordan, Amman, Jordan • Visiting Research Scientist, Institute of Mictrotechnology, Braunschweig, Germany, through DFG research scholarship, 2006 • Training Leader and Facilitator: TUV Akademie Middle East • Engineering Consultant, Nayzak Dies and Moulds Mfg. Company

  4. Hydraulic Basic Principles Prove Pascal’s law 10 N Force • Hydraulics is the technology or study of liquid pressure and flow. Liquids are materials which pour and conform to the shape of their container. Example are oil and water at room temperature. • Stationary fluids provide no resistance to shear stresses. This allows fluids to take the shape of the container they are in, and also leads to Pascal’s law, which states that the pressure at any given point in a fluid is the same in all directions. • Pressure applied to a confined fluid is transmitted undiminished in all directions, and acts with equal force on equal areas, and at right angles to them. 1 square cm stopper 10 N per square cm Pressure of 100 kPa

  5. Multiplication of Force • Since liquid transmit the same amount of pressure in all directions. The force transmitted to the output piston is multiplied by a factor equal to the area ratio of the output piston to the input piston

  6. Multiplication of Force • Even though force is multiplied, the power or energy is not. In fact, the energy or power input is usually larger than the obtained output. • The transmission ratio is defined as the velocity at the input to that at the output in case of no leakage, and is a function of the geometry of the setup. The ratio of the output force to the input force is usually less than the transmission ratio due to frictional losses.

  7. Introduction to Fluid Power An Introductory Fluid Power System Lifting and Lowering a Load

  8. An Introductory Fluid Power System: Load Raising Directional Control Valve in Load Lifting Position

  9. An Introductory Fluid Power System: Load Lowering Directional Control Valve in Load Lowering Position

  10. An Introductory Fluid Power System: Locked Position Directional Control Valve in Load Locked Position

  11. An Introductory Fluid Power System:Directional Control Valve Four Way Three Position Manually (Lever) Operated Spring Centered, Directional Control Valve

  12. An Introductory Fluid Power System: Symbolic Presentation Fload

  13. Operation of a Basic Hydraulic Circuit • When the directional control valve lever is moved upward, the pumped oil flows through path P – B of the to the lower part of the cylinder. • Since the oil is under pressure, it pushes up the piston inside the cylinder, causing the piston rod to retract, and the load to be raised. The oil in the upper side of the piston is drained back to the reservoir through path A – T of the directional control valve.

  14. Operation of a Basic Hydraulic Circuit • When the directional control valve lever is centered, all four ports are blocked and oil can not escape from either side of the cylinder. This stops the movement of the piston and causes the oil to flow from the pump back to the reservoir through the pressure relief valve. • The pressure in the lower end of the cylinder will be at an intermediate level due to the presence of the load.

  15. Operation of a Basic Hydraulic Circuit • When the directional control valve lever is moved toward the valve body, the pumped oil flows through the path P – A of the directional control valve to the upper end of the cylinder. The oil pushes the piston downward, which lowers the attached load. • The oil in the lower end of the cylinder of the piston is drained back to the reservoir through path B – T of the directional control valve.

  16. An Introductory Fluid Power System: Engineering Issues As an Engineer, what issues will you be concerned about in the presented system 1 minute to write on a piece of paper!

  17. An Introductory Fluid Power System: Basic Performance • How much load will the system be able to raise? • How fast does the load go up? • Is it possible to control lifting and lowering speeds? How?

  18. An Introductory Fluid Power System:Efficiency/Operating Cost Issues • How much power will I need for (i) raising the load, (ii) lowering the load, and (iii) stopping the load? • What is the needed pressure at the pump outlet? • What is the pressure at the pump inlet? • Can I make the system more efficient, particularly during lowering and stopping the load

  19. An Introductory Fluid Power System: Safety Issues • Is the system safe enough? • What happens in case of hydraulic line rupture, pump failure, or electrical power shutdown in the position shown?

  20. An Introductory Fluid Power System:Expansion Issues • Is it possible to give the cylinder a command position, and have it go to that position automatically, without operator intervention? • Can I have two or more cylinders work in tandem? • Can I have two or more cylinders work in series?

  21. Introduction to Fluid Power Work and Power How much power can a person produce? What experiments would you perform to estimate your own power production?

  22. Work and Power • A person is required to lift a 50 N box a distance of 2 m. The work done by this person is given by: • Work = Force x Displacement • Work = 50 N x 2 m = 100 N.m (Joule) • If the person is asked to raise the box in 2 second. How much power is he producing? Power is the time rate of doing work. • Power = Work/time • Power = 100 J/2 s = 50 J/s (Watt) • A person doing the same work in 1 seconds produces 100 W of power How much power can a person produce?

  23. Power Produced by a Weight Lifter • The Power produced by popular weight lifters during the lift process ranges between 2 – 2.5 kW * Estimated

  24. Power Produced By a Cyclist • Average power for the first 30 seconds is 300 W http://mapawatt.com/wp-content/uploads/2009/07/Hour-watts-power-output.jpg

  25. Power Produced By a Horse • 1 Horsepower = 745.6 W • Lifting a 70 kg man at 23 km/h up a 10 degree slope

  26. Power Produced By an Automobile Engine

  27. Introduction to Fluid Power Other Applications Requiring Work and Power

  28. Applications requiring work and power • Lift a load against gravity • Manufacturing processes: (metal forming, cutting, punching, deep drawing, cutting, turning, milling, etc. involves applying a force for a certain distance. • Changing the speed of an object • Moving objects against friction, air drag, water drag: e.g., car, ships, airplanes, etc. • Walking, speaking, etc. How much power can a person produce?

  29. Introduction to Fluid Power Power Conversion and Power Transmissions

  30. Forms of Power

  31. Power Converters – Electromechanical Electric Generator T x ω F x v V x I Mechanical to Electric Electric Motor Linear Electric Actuator T x ω F x v V x I Electric to Mechanical

  32. Power Converters – Hydromechanical Pump T x ω F x v P x Q Mechanical to Fluid Rotary Actuator Linear Actuator T x ω F x v P x Q Fluid to Mechanical

  33. Power Transmitters – Direct Power Transmission Power Transmitter To x ωo Fo x vo Ti x ωi Fi x vi Why would we perform power transmission? Give Some Examples on power transmission devices

  34. Reasons for Power Transmission Power Transmitter • Manipulation of the values of the potential – flow variable, e.g. when the potential variable at the load (torque or force) is not compatible with the source (too high or too low) • Load is placed at a location different from that of the source. Need to “Transport” power from the source location to the load location To x ωo Fo x vo Ti x ωi Fi x vi

  35. Direct Power Transmission Power Transmitter To x ωo Fo x vo Ti x ωi Fi x vi Direct (Mechanical) • Examples: • Mechanical Linkages: (Levers, mechanisms, etc.) • Pulleys and Ropes • Sprockets and Chains • Gear Boxes • Belt Drives

  36. Indirect Power Transmitters(Back – to – Back Converter) Power Converter Electric Power Converter To x ωo Fo x vo Ti x ωi Fi x vi Hydraulic Pneumatic

  37. Power Transmission Methods • Direct (Mechanical) • Gear trains and shafts • Levers and linkages • Ropes and Pulleys • Chains and Sprockets • Belt Drives • Indirect (Back to Back) • Electric • (Generators – Transformers – Motors) • Hydraulic • (Pump – Hydraulic motor or hydraulic actuator) • Pneumatic • (Compressor – Pneumatic cylinder) Power Transmitter Ti x ωi Fi x vi To x ωo Fo x vo

  38. Transmission Ratio • We define the speed magnification ratio, for a power transmitter as the ratio of the output speed to the input speed. • If the power transmission system is ideal, the output power is equal to the input power (no losses), then Rotational Power Transmitter Linear Power Transmitter Ti x ωi Fi x vi Fo x vo To x ωo

  39. Transmission Ratio Fo • In a number of power transmission systems, a definite relationship exists between the input motion and the output motion. This relationship is maintained by a set of constraints provided in the transmission system. • Example constraints in mechanical power transmission systems include the constant distance between any two points in a rigid body, the equal displacement of the two pitch points on the pitch circles of two gears in mesh, and the constant length of ropes and chain sprockets in rope-pulleys and chain-sprocket drives. • In hydraulic power transmission systems, input – output motion constraint is provided by the incompressibility of the hydraulic fluid in the system and the conservation of mass principle Fi O A B di do

  40. Transmission Ratio • If the arm of the lever mechanism is treated as a rigid body, the constant distance constraint means that the angular displacement (Δϴ) is the same for all lines in the body. This means that the angular speed (ω = dϴ/dt) is the same for all points in the lever. • We define the transmission ratio as the ratio of input speed to output speed. This ratio is sometimes called the speed reduction ratio. Fo Fi O A B di do

  41. Transmission Ratio • If no power losses exist (no friction), the output power produced by the lever is equal to the input power. Therefore, we have • The mechanical advantage or the force amplification ratio is the ratio of the ideal output force to the input force. In an ideal transmitter with no power loss, the mechanical advantage is equal to the speed reduction ratio. Fo Fi O A B di do

  42. Transmission Ratio Rotational Power Transmitter Linear Power Transmitter Ti x ωi Fi x vi Fo x vo To x ωo • For an ideal transmission system, the torque (force) amplification ratio is equal to the speed reduction ratio. This ratio is called the transmission ratio and it is completely defined by the geometry of the system.

  43. Transmission Ratio Linear Power Transmitter Fo x vo Fi x vi Fo Fi O di do

  44. Transmission Ratio RotationalPower Transmitter To x vo Ti x vi FC ro To ri A B Ti FC

  45. Overall Efficiency • The overall efficiency is defined is the ratio of the power produced by the system to the power delivered to the system • For an ideal power transmission systems with no frictional losses, Tiand To are related by To = rTi. If the system is also a non-slipping mechanical system or a non-leaking converter, the following relation hold. Power Transmitter Ti x ωi To x ωo

  46. Mechanical Efficiency • When frictional losses exist the torque produced by the transmitter is less than that of an ideal frictionless transmitter, (To < rTi). The mechanical efficiency is defined as the ratio of the output torque produced by the system to the torque produced by an ideal frictionless transmitter for the same torque input. • The mechanical efficiency is also the ratio between the ideal input torque needed by a frictionless system to the actual input torque needed by the system to produce the same torque output. Power Transmitter Ti x ωi To x ωo

  47. Volumetric Efficiency • For non-slipping mechanical transmitters (gear trains, levers, pulleys and chain-sprocket), the speed ratio relation, ωo = ωi /r, holds regardless of frictional loss. This relation is also valid for non-leaking back-to-back converters based on mass/current conservation. When the systems have slippage or leakage, the output speed is reduced. The volumetric Efficiency is defined as: Power Transmitter Ti x ωi To x ωo

  48. Efficiency Relationships • The overall efficiency may be written in terms of the mechanical efficiency and the volumetric efficiency by utilizing the relationships Power Transmitter Ti x ωi To x ωo

  49. Introduction to Fluid Power Power Transmission Comparison Factor Power Transmitter Ti x ωi To x ωo What factors will you consider when comparing the different methods of power transmission?

  50. Comparison Factors: Transmission Distance Power Transmitter • Effect on initial cost (capital) • Effect on running cost (transmission efficiency ) Ti x ωi To x ωo

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