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Facts and Theory of Air

Facts and Theory of Air. For industrial pneumatics. 《气动技术》双语电子教案 吴洪明 Chapter 1. WUHAN UNIVERSITY OF TECHNOLOGY LOGISTICS ENG. DEPT. 武汉理工大学 物流工程系. 《气动技术》双语电子教案 吴洪明 Chapter 1. Contents. Composition of air. Constant pressure. Atmospheric pressure. Constant volume.

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Facts and Theory of Air

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  1. Facts and Theory of Air For industrial pneumatics 《气动技术》双语电子教案 吴洪明 Chapter 1 WUHAN UNIVERSITY OF TECHNOLOGY LOGISTICS ENG. DEPT. 武汉理工大学 物流工程系

  2. 《气动技术》双语电子教案 吴洪明 Chapter 1 Contents • Composition of air • Constant pressure • Atmospheric pressure • Constant volume • Industrial compressed air • General gas law • Pressure • Adiabatic compression • Pressure units • Water in compressed air • Pressure and force • Low temperature drying • The gas laws • Flow of compressed air • Constant temperature • Air quality Click the section to advance directly to it

  3. 《气动技术》双语电子教案 吴洪明 Chapter 1 Composition of air Composition by Volume Nitrogen 78.09% N2 Oxygen 20.95% O2 Argon 0.93% Ar Others 0.03% • The air we breathe is springy, squashy and fluid in substance • We take it for granted that wherever there is space it will be filled with air • Air is composed mainly of nitrogen and oxygen

  4. 《气动技术》双语电子教案 吴洪明 Chapter 1 Atmospheric pressure • The atmospheric pressure is caused by the weight of air above us • It gets less as we climb a mountain, more as we descend into a mine • The pressure value is also influenced by changing weather conditions

  5. 1013.25 m bar 《气动技术》双语电子教案 吴洪明 Chapter 1 Standard Atmosphere • A standard atmosphere is defined by The International Civil Aviation Organisation. The pressure and temperature at sea level is 1013.25 milli bar absolute and 288 K (15OC)

  6. 《气动技术》双语电子教案 吴洪明 Chapter 1 ISO Atmospheres • ISO Recommendation R 554 • Standard Atmospheres for conditioning and/or testing of material, components or equipment • 20OC, 65% RH, 860 to 1060 mbar • 27OC, 65% RH, 860 to 1060 mbar • 23OC, 50% RH, 860 to 1060 mbar • Tolerances ± 2OC ± 5%RH • Reduced tolerances ± 1OC ± 2%RH • Standard Reference Atmosphere to which tests made at other atmospheres can be corrected • 20OC, 65% RH, 1013 mbarNo qualifying altitude is given as it is concerned only with the effect of temperature, humidity and pressure

  7. 1015 mb 1012 mb 1008 mb 1000 mb 996 mb LOW 《气动技术》双语电子教案 吴洪明 Chapter 1 Atmospheric pressure • We see values of atmospheric pressure on a weather map • The lines called isobars show contours of pressure in millibar • These help predict the wind direction and force

  8. 760 mm Hg 《气动技术》双语电子教案 吴洪明 Chapter 1 Mercury barometer • Atmospheric pressure can be measured as the height of a liquid column in a vacuum • 760 mm Hg = 1013.9 millibar approximately • A water barometer tube would be over 10 metres long. Hg = 13.6 times the density of H2O • For vacuum measurement 1 mm Hg = 1 Torr760 Torr = nil vacuum0 Torr = full vacuum

  9. 《气动技术》双语电子教案 吴洪明 Chapter 1 Atmosphere and vacuum • The power of atmospheric pressure is apparent in industry where pick and place suction cups and vacuum forming machines are used • Air is removed from one side allowing atmospheric pressure on the other to do the work

  10. 16 17 16 15 Extended Industrial range 14 15 14 13 13 12 12 11 11 10 Typical Industrial range 10 9 9 8 Absolute pressure bar a Gauge pressure bar g 8 7 7 6 Low range 6 5 5 4 3 4 3 2 1 2 Atmosphere 1 0 0 Full vacuum 《气动技术》双语电子教案 吴洪明 Chapter 1 Industrial compressed air • Pressures are in “bar g” gauge pressure ( the value above atmosphere) • Zero gauge pressure is atmospheric pressure • Absolute pressures are used for calculationsPa = Pg + atmosphere • For quick calculationsassume 1 atmosphere is 1000 mbar • For standard calculations 1 atmosphere is1013 mbar

  11. 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure • For measuring lower pressures the millibar (mbar) is used • 1000 mbar = 1 bar • For measurements in pounds per square inch (psi)1 psi = 68.95mbar14.5 psi = 1bar • 1 bar = 100000 N/m2 (Newtons per square metre) • 1 bar = 10 N/cm2

  12. 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure units • There are many units of pressure measurement. Some of these and their equivalents are listed below. • 1 bar = 100000 N/m2 • 1 bar = 100 kPa • 1 bar = 14.50 psi • 1 bar =10197 kgf/m2 • 1 mm Hg = 1.334 mbar approx. • 1 mm H2O = 0.0979 mbar approx. • 1 Torr = 1mmHg abs (for vacuum) More units of pressure

  13. Pressure and force 《气动技术》双语电子教案 吴洪明 Chapter 1

  14. 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • Compressed air exerts a force of constant value to every internal contact surface of the pressure containing equipment. • Liquid in a vessel will be pressurised and transmit this force • For every bar of gauge pressure, 10 Newtons are exerted uniformly over each square centimetre.

  15. D mm P bar 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • The thrust developed by a piston due to air pressure is the effective area multiplied by the pressure p D2 P Thrust = Newtons 40 Where D = The bore of a cylinder in mm P = The pressure in bar. We require an answer in Newtons 1bar = 100000 N/m2 D2 is therefore divided by 1000000 to bring it to m2 and P is multiplied by 100000 to bring it to N/m2. The result is a division by 10 shown in the product 40 above

  16. l D D. l . P Force Newtons = 10 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • The force contained by a cylinder barrel is the projected area multiplied by the pressure Where D = the cylinder bore mm l = length of pressurised chamber mm P = the pressure in bar

  17. 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • If both ports of a double acting cylinder are connected to the same pressure source, the cylinder will move out due to the difference in areas either side of the piston • If a through rod cylinder is applied in this way it will be in balance and not move in either direction

  18. P1 P2 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • In a balanced spool valve the pressure acting at any port will not cause the spool to move because the areas to the left and right are equal and will produce equal and opposite forces • P1 and P2 are the supply and exhaust pressures

  19. P2 P1 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • In a balanced spool valve the pressure acting at any port will not cause the spool to move because the areas to the left and right are equal and will produce equal and opposite forces • P1 and P2 are the supply and exhaust pressures

  20. P1 P2 《气动技术》双语电子教案 吴洪明 Chapter 1 Pressure and force • In a balanced spool valve the pressure acting at any port will not cause the spool to move because the areas to the left and right are equal and will produce equal and opposite forces • P1 and P2 are the supply and exhaust pressures

  21. The gas laws 《气动技术》双语电子教案 吴洪明 Chapter 1

  22. 《气动技术》双语电子教案 吴洪明 Chapter 1 The gas laws • For any given mass of air the variable properties are pressure, volume and temperature. • By assuming one of the three variables to be held at a constant value, we will look at the relationship between the other two for each case • Constant temperature • Constant pressure • Constant volume P.V = constant V = constant T P = constant T

  23. Constant Temperature 《气动技术》双语电子教案 吴洪明 Chapter 1

  24. Pressure P bar absolute 16 14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 0 16 Volume V P1.V1 = P2.V2 = constant 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant temperature • Boyle’s law states: the product of absolute pressure and volume of a given mass of gas remains constant if the temperature of the gas remains constant. • This process is called isothermal (constant temperature). It must be slow enough for heat to flow out of and in to the air as it is compressed and expanded.

  25. Pressure P bar absolute 16 14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 0 16 Volume V P1.V1 = P2.V2 = constant 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant temperature • Boyle’s law states: the product of absolute pressure and volume of a given mass of gas remains constant if the temperature of the gas remains constant. • This process is called isothermal (constant temperature). It must be slow enough for heat to flow out of and in to the air as it is compressed and expanded.

  26. Pressure P bar absolute 16 14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 0 16 Volume V P1.V1 = P2.V2 = constant 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant temperature • Boyle’s law states: the product of absolute pressure and volume of a given mass of gas remains constant if the temperature of the gas remains constant. • This process is called isothermal (constant temperature). It must be slow enough for heat to flow out of and in to the air as it is compressed and expanded.

  27. Pressure P bar absolute 16 14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 0 16 Volume V P1.V1 = P2.V2 = constant 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant temperature • Boyle’s law states: the product of absolute pressure and volume of a given mass of gas remains constant if the temperature of the gas remains constant. • This process is called isothermal (constant temperature). It must be slow enough for heat to flow out of and in to the air as it is compressed and expanded.

  28. Pressure P bar absolute 16 14 12 10 8 6 4 2 0 2 4 6 8 10 12 14 0 16 Volume V P1.V1 = P2.V2 = constant 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant temperature • Boyle’s law states: the product of absolute pressure and volume of a given mass of gas remains constant if the temperature of the gas remains constant. • This process is called isothermal (constant temperature). It must be slow enough for heat to flow out of and in to the air as it is compressed and expanded.

  29. Constant Pressure 《气动技术》双语电子教案 吴洪明 Chapter 1

  30. Temperature Celsius 100 80 60 40 293K 20 0 -20 -40 -60 1 2 0 Volume 0.25 0.5 0.75 1.25 1.5 1.75 V1 V2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant pressure • Charles’ law states: for a given mass of gas at constant pressure the volume is proportional to the absolute temperature. • Assuming no friction a volume will change to maintain constant pressure. • From an ambient of 20oC a change of 73.25oC will produce a 25% change of volume. • 0o Celsius = 273K

  31. Temperature Celsius 100 366.25K 80 60 40 20 0 -20 -40 -60 1 2 0 Volume 0.25 0.5 0.75 1.25 1.5 1.75 V1 V2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant pressure • Charles’ law states: for a given mass of gas at constant pressure the volume is proportional to the absolute temperature. • Assuming no friction a volume will change to maintain constant pressure. • From an ambient of 20oC a change of 73.25oC will produce a 25% change of volume. • 0o Celsius = 273K

  32. Temperature Celsius 100 80 60 40 20 0 -20 -40 219.75K -60 1 2 0 Volume 0.25 0.5 0.75 1.25 1.5 1.75 V1 V2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant pressure • Charles’ law states: for a given mass of gas at constant pressure the volume is proportional to the absolute temperature. • Assuming no friction a volume will change to maintain constant pressure. • From an ambient of 20oC a change of 73.25oC will produce a 25% change of volume. • 0o Celsius = 273K

  33. Temperature Celsius 100 366.25K 80 60 40 293K 20 0 -20 -40 219.75K -60 1 2 0 Volume 0.25 0.5 0.75 1.25 1.5 1.75 V1 V2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant pressure • Charles’ law states: for a given mass of gas at constant pressure the volume is proportional to the absolute temperature. • Assuming no friction a volume will change to maintain constant pressure. • From an ambient of 20oC a change of 73.25oC will produce a 25% change of volume. • 0o Celsius = 273K

  34. Constant volume 《气动技术》双语电子教案 吴洪明 Chapter 1

  35. Temperature Celsius 100 80 60 40 20 8 6 10 0 4 12 -20 14 2 bar 16 0 -40 bar absolute -60 10 20 0 5 15 P1 P2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant volume • From Boyle’s law and Charles’ law we can also see that if the volume of a given mass of air were to be kept at a constant value, the pressure will be proportional to the absolute temperature K. • For a volume at 20oC and 10 bar absolute a change in temperature of 60oC will produce a change in pressure of 2.05 bar • 0oC = 273K

  36. Temperature Celsius 100 80 60 40 20 8 6 10 0 4 12 -20 14 2 bar 16 0 -40 bar absolute -60 10 20 0 5 15 P1 P2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant volume • From Boyle’s law and Charles’ law we can also see that if the volume of a given mass of air were to be kept at a constant value, the pressure will be proportional to the absolute temperature K. • For a volume at 20oC and 10 bar absolute a change in temperature of 60oC will produce a change in pressure of 2.05 bar • 0oC = 273K

  37. Temperature Celsius 100 80 60 40 20 8 6 10 0 4 12 -20 14 2 bar 16 0 -40 bar absolute -60 10 20 0 5 15 P1 P2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant volume • From Boyle’s law and Charles’ law we can also see that if the volume of a given mass of air were to be kept at a constant value, the pressure will be proportional to the absolute temperature K. • For a volume at 20oC and 10 bar absolute a change in temperature of 60oC will produce a change in pressure of 2.05 bar • 0oC = 273K

  38. Temperature Celsius 100 80 60 40 20 8 6 10 0 4 12 -20 14 2 bar 16 0 -40 bar absolute -60 10 0 5 15 P1 P2 = = c T1(K) T2(K) 《气动技术》双语电子教案 吴洪明 Chapter 1 Constant volume • From Boyle’s law and Charles’ law we can also see that if the volume of a given mass of air were to be kept at a constant value, the pressure will be proportional to the absolute temperature K. • For a volume at 20oC and 10 bar absolute a change in temperature of 60oC will produce a change in pressure of 2.05 bar • 0oC = 273K

  39. P1 .V1 P2 .V2 constant = = T1 T2 《气动技术》双语电子教案 吴洪明 Chapter 1 The general gas law • The general gas law is a combination of Boyle’s law and Charles’ law where pressure, volume and temperature may all vary between states of a given mass of gas but their relationship result in a constant value.

  40. Adiabatic and polytropic compression For compressed air 《气动技术》双语电子教案 吴洪明 Chapter 1

  41. PV 1. 4 = c adiabatic 16 14 PV 1. 2 = c polytropic 12 10 bar a PV = c isothermal 8 6 4 2 0 Volume 0 2 4 6 8 10 12 14 16 《气动技术》双语电子教案 吴洪明 Chapter 1 Adiabatic compression • For adiabatic compression and expansion P V n = c for air n = 1.4 • In the cylinder of an air compressor the process is fast but some heat will be lost through the cylinder walls therefore the value of n will be less 1.3 approximately for a high speed compressor • In theory, when a volume of air is compressed instantly, the process is adiabatic (there is no time to dissipate heat through the walls of the cylinder)

  42. 《气动技术》双语电子教案 吴洪明 Chapter 1 Polytropic compression • In practice such as in a shock absorbing application there will be some heat loss during compression • The compression characteristic will be somewhere between adiabatic and isothermal • The value of n will be less than 1.4 dependent on the rate of compression. Typically PV 1.2 = c can be used but is applicable only during the process

  43. Water in compressed air 《气动技术》双语电子教案 吴洪明 Chapter 1

  44. fully saturated air Condensate Drain 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • When large quantities of air are compressed, noticeable amounts of water are formed • The natural moisture vapour contained in the atmosphere is squeezed out like wringing out a damp sponge • The air will still be fully saturated (100% RH) within the receiver

  45. 25% RH 50% RH 100% RH 40 At 20o Celsius 100% RH = 17.4 g/m3 50% RH = 8.7 g/m3 25% RH = 4.35 g/m3 20 Temperature Celsius 0 -20 -40 0 10 20 30 40 50 60 70 80 Grams of water vapour / cubic metre of air g/m3 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • The amount of water vapour contained in a sample of the atmosphere is measured as relative humidity %RH. This percentage is the proportion of the maximum amount that can be held at the prevailing temperature.

  46. 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • The illustration shows four cubes each representing 1 cubic metre of atmospheric air at 20oC. Each of these volumes are at a relative humidity of 50% (50%RH). This means that they actually contain 8.7 grams of water vapour, half of the maximum possible 17.4 grams

  47. 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • When the compressor squashes these four cubic metres to form one cubic metre there will be 4 times 8.7 grams, but only two of them can be held as a vapour in the new 1 cubic metre space. The other two have to condense out as water droplets

  48. 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • When the compressor squashes these four cubic metres to form one cubic metre there will be 4 times 8.7 grams, but only two of them can be held as a vapour in the new 1 cubic metre space. The other two have to condense out as water droplets

  49. 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • When the compressor squashes these four cubic metres to form one cubic metre there will be 4 times 8.7 grams, but only two of them can be held as a vapour in the new 1 cubic metre space. The other two have to condense out as water droplets

  50. 《气动技术》双语电子教案 吴洪明 Chapter 1 Water in compressed air • When the compressor squashes these four cubic metres to form one cubic metre there will be 4 times 8.7 grams, but only two of them can be held as a vapour in the new 1 cubic metre space. The other two have to condense out as water droplets

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