3 Flowcharts Step-by-step Name of PowerPointCIMName of LessonFlowchartsA flowchart is a schematic representation of an algorithm or a process.A schematic is a diagram that uses special symbols in place of actual pictures. In a wiring schematic, for example, a squiggly line is used to represent a resistor.An algorithm is a series of steps that tell how to complete a task. For example, an algorithm for reducing a mathematical expression could be: First, perform the operations contained within the parentheses. Then calculate the exponent, if preset. Next, multiply and divide from left to right. Finally, add and subtract from left to right. Here we have a simple flowchart that might show the steps in feeding a dog.Step-by-stepProject Lead The Way, Inc.Copyright 2008
4 Parts of a Flowchart Start/end Input/output Process Arrows Display Name of PowerPointCIMName of LessonParts of a FlowchartStart/endInput/outputProcessArrowsDisplayDecisionProject Lead The Way, Inc.Copyright 2008
5 Control SystemsDesigned to provide consistent process control and reliabilityControl system protocols are an established set of commands or functions typically created in a computer programming languageStartLight onMotor 1 on CWfor 5 secMotor 1 offEnd
6 Control Systems Open loop: no feedback used in processes Closed loop: feedback used in the programming to make operational and process decisions (temp, time, analog value, digital value, etc.)Digital: signals have 2 states: 1 (closed) or 0 (open) (Switch, phototransistor)Analog: data represented continuously with variable quantities (photoresistor, NTC resistor, potentiometer)
7 When the program starts, lamp (M1) is turned ON, and the computer checks to see if the switch (I1, wired normally closed) is being pressed. The program will loop back until the switch is pressed. When the switch is pressed, the value of variable A will be incremented by 1. The computer will then wait 0.2 seconds before checking the value of variable A to see if it is greater than or equal to 13. If the value of variable A is less than or equal to 12, the program will loop back to the beginning. If variable A is greater than or equal to 13, the lamp will turn OFF and the program will end.
9 Fluid PowerA system that transmits and controls energy through the use of pressurized liquid or gasPneumatics - the media used is airHydraulics - the media used is oil or water
10 4 Basic Components of Fluid Systems 1. Reservoir (Tank): Storage device which holds the fluid2. Pump or Compressor: Device used to move fluids3. Valves: Regulate the direction of fluid flow4. Actuator (Cylinder): Mechanical device for moving or controlling a mechanism or system
11 Common Pneumatic System Components Transmission LinesRegulatorFilterDrainDirectional Control ValveReceiver TankCompressor: Compresses air into the receiver tank.Pressure Relief Valve: Limits the maximum pressure in the system. It is a safety device that will allow excess pressure to escape to prevent damaging components in the circuit.Receiver Tank: A device that holds the air in a pneumatic system.Transmission Lines: Used to transport fluid in a circuit.Directional Control Valve: Used to control which path a fluid takes in a circuit.Cylinder: Also called an actuator. Used to convert fluid power to linear mechanical power.Drain: Removes moisture from the system.Regulator: A valve used to control pressure in the branch of a circuit.Filter: Used to remove contamination from fluids. Filters are also often next to lubricators. Lubrication helps prevent wear on the components in the system.All systems also require seals and gaskets between components to improve the air-tight nature of the system.Pressure Relief ValveCylinderCompressorNational Fluid Power Association & Fluid Power Distributors Association
12 Properties of Compressed Air AvailabilityEasily stored in large volumesSimplicity in design and controlLow system cost due to low component costEnvironmentally friendly
13 HydraulicsPrinciples of EngineeringTMUnit 4 - Lesson 4.3 – Fluid SystemsHydraulicsAn area of engineering science that deals with liquid flow and pressureUse syringes to demonstrate the basic concept of hydraulic flowPressure actsequally everywhere!!!Project Lead The Way, Inc.Copyright 2007
14 A Hydraulic System Hydraulics Principles of EngineeringTM Unit 4 - Lesson 4.3 – Fluid SystemsA Hydraulic SystemProject Lead The Way, Inc.Copyright 2007
15 Hydraulic Fluids Liquid pumped through a hydraulic system Principles of EngineeringTMUnit 4 - Lesson 4.3 – Fluid SystemsHydraulic FluidsLiquid pumped through a hydraulic systemPetroleum-based or synthetic oilServe four major functions:1. Power transmission2. Lubrication of moving parts3. Sealing of spaces between moving parts4. Heat removalRelatively Incompressible!Four major functions:Power transmission:Pumps apply pressure to hydraulic fluids, which causes the fluid to move (since they can’t be compressed). This movement causes components connected to the hydraulic system to move. Thus, power is transmitted.Lubrication of moving parts:Hydraulic systems provide their own lubrication!Sealing of spaces between moving parts:Hydraulic pressure between moving parts (e.g. piston) makes a seal between those parts.Heat removal:Hydraulic fluids move passed hot components and then carry that heat away via conduction mechanisms.Project Lead The Way, Inc.Copyright 2007
16 What can Fluid Power Do ? Fluid Power Systems Transmit force over great distancesMultiply an input forceIncrease the distance an output will moveWhat can Fluid Power Do ?Operation of system valves for air, water or chemicalsOperation of heavy or hot doorsUnloading of hoppers in building, steel making, mining and chemical industriesRamming and tamping in concrete and asphalt laying
17 Properties of Gases Gases are affected by 3 variables Temperature (T)Pressure (p)Volume (V)Gases have no definite volumeGases are highly compressibleGases are lighter than liquids
18 Properties of Gases Absolute Pressure Gauge Pressure: Pressure on a gauge does not account for atmospheric pressure on all sides of the systemAbsolute Pressure: Atmospheric pressure plus gauge pressureNot on formula sheetGauge Pressure + Atmospheric Pressure = Absolute PressureAbsolute pressure is used to complete pneumatic calculations.Atmospheric pressure is also known as barometric pressure. It is the weight of the air molecules. An application of air pressure is when your ears pop. The pressure is balancing on the inside and outside of your ear.Atmospheric pressure equals 14.7 lb/in.2If a gauge reads 120 psi, what is the absolute pressure?120 lb/in lb/in.2 = lb/in.2
19 Properties of Gases Absolute Temperature 0°F does not represent a true 0°Absolute Zero = -460°FAbsolute Temperature is measured in degrees Rankine (°R)°R = °F + 460Not on formula sheetTrue 0° represents the lack of molecular movement. Molecular movement increases as temperature increases.Pneumatics equations should use absolute temperature.If the temperature of the air in a system is 65 °F, what is the absolute temperature?Answer:65 °F = 525 °R
20 Boyle’s LawThe volume of a gas at constant temperature varies inversely with the pressure exerted on it.NASAp1 (V1) = p2 (V2)SymbolDefinitionExample UnitVVolumein.3
21 Boyle’s Law ExampleA cylinder is filled with 40 in.3 of air at a pressure of 60 psi. The cylinder is compressed to 10 in.3. What is the resulting absolute pressure?p1 = 60 lb/in.2 V1 = 40 in.3p2 = ? V2 = 10 in.3Convert p1 to absolute pressure.p1 = 60 lb/in lb/in.2 = 74.7 lb/in.2
22 Charles’ LawVolume of gas increases or decreases as the temperature increases or decreases, provided the amount of gas and pressure remain constant.NASANote: T1 and T2 refer to absolute temperature.
23 Charles' Law ExampleAn expandable container is filled with 28 in.3 of air and is sitting in ice water that is 32°F. The container is removed from the icy water and is heated to 200°F. What is the resulting volume?V1 = 28in.3V2 = ?T1 = 32°FT2 = 200°FConvert T to absolute temperature.T1 = 32°F + 460°F =482°RT2 = 200°F + 460°F =660°RThe temperature readings must be converted to absolute temperature in order for the equation to work.
24 Charles' Law ExampleAn expandable container is filled with 28 in.3 of air and is sitting in ice water that is 32°F. The container is removed from the icy water and is heated to 200°F. What is the resulting volume?V1 = 28in.3V2 = ?T1 = 32°FT2 = 200°FConvert T to absolute temperatureT1 = 32°F + 460°F = 492°RT2 = 200°F + 460°F = 660°R
25 Gay-Lussac’s LawAbsolute pressure of a gas increases or decreases as the temperature increases or decreases, provided the amount of gas and the volume remain constant.Not on formula sheet)Note: T1 and T2 refer to absolute temperature.p1 and p2 refer to absolute pressure.
26 Gay-Lussac’s Law Example A 300 in.3 sealed air tank is sitting outside. In the morning the temperature inside the tank is 62°F, and the pressure gauge reads 120 lb/in.2. By afternoon the temperature inside the tank is expected to be close to 90°F. What will the absolute pressure be at that point?V = 300 in.3 T1 = 62°Fp1 = 120 lb/in.2 T2 = 90°Fp2 = ?Convert p to absolute pressure.p1= 120 lb/in lb/in.2= lb/in.2Convert T to absolute temperature.T1 = 62°F + 460°F = 522°RT2 = 90°F + 460°F = 550°R
27 Gay-Lussac’s Law Example A 300 in.3 sealed air tank is sitting outside. In the morning the temperature inside the tank is 62°F, and the pressure gauge reads 120 lb/in2. By afternoon the temperature inside the tank is expected to be closer to 90°F. What will the absolute pressure be at that point?If the absolute pressure is lb/in.2, what is the pressure reading at the gauge?141.9 lb/in.2 – 14.7 lb/in.2 = lb/in.2
28 In hydrostatic systems: P1 = P2 or F1/A1 = F2/A2 Pascal’s LawPressure exerted by a confined fluid acts undiminished equally in all directions.Pressure: The force per unit area exerted by a fluid against a surfaceSymbolDefinitionExample UnitpPressurelb/in.2FForcelbAAreain.2The area of a cylinder will be the surface area of the piston.In hydrostatic systems:P1 = P2 or F1/A1 = F2/A2Not on formulasheet
29 Pascal’s Law ExampleHow much pressure can be produced with a 3 in. diameter (d) cylinder and 50 lb of force?d = 3 in. p = ?F = 50 lb A = ?
30 Application of Pascal’s Law in a Simple Hydrostatic System How much force must you exert on piston A to lift a load on piston B of 500 lbs? What is the ideal mechanical advantage of this system?
31 Hydrostatic system: P1 = P2 P = F P1=P F1 = F2A A1 A2F1 = lbs1 in^2 500 in^2F1 = 1 lb
32 Additional ExamplesI have a car lift with a 12” radius. How heavy a car can I lift if a 3 Lb force is applied to a piston with a 1” radius?2. I have a gas with a pressure of 53 kPa at a temperature of 470 C. I heat the gas an additional 200 degrees. What will the new pressure be if the volume is constant?
33 I have a car lift with a 12” radius I have a car lift with a 12” radius. How heavy a car can I lift if a 3 Lb force is applied to a piston with a 1” radius? Hydrostatic systemP = F P1=P2AP1 = 3 lbpi*1”*1”P2 = Xpi*12”*12”X = 3 lb * pi *12” * 12” = 432 lbpi*1”*1”
34 2. I have a gas with a pressure of 53 kPa at a temperature of 470 C 2. I have a gas with a pressure of 53 kPa at a temperature of 470 C. I heat the gas an additional 200 degrees. What will the new pressure be if the volume is constant?P1 * V1 = P2 * V2T T2T1 = = 320T2 = = 52053 kPa = P2P2 = 520 * 53 kPa320Watch additional heating versus heating to a certain temperature. If the problem saidI heat the gas to 200 degrees, then T2 = = 473 degrees.