Facts and Theory of Air For industrial pneumatics.

Facts and Theory of Air For industrial pneumatics

Contents Composition of air Atmospheric pressure Industrial compressed air Pressure Pressure units Pressure and force The gas laws Constant temperature Constant pressure Constant volume General gas law Adiabatic compression Water in compressed air Low temperature drying Flow of compressed air Air quality Click the section to advance directly to it

Composition of air 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 Composition by Volume Nitrogen78.09% N 2 Oxygen20.95% O 2 Argon0.93% Ar Others0.03%

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

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 (15 O C) 1013.25 m bar

ISO Atmospheres ISO Recommendation R 554 Standard Atmospheres for conditioning and/or testing of material, components or equipment 20 O C, 65% RH, 860 to 1060 mbar 27 O C, 65% RH, 860 to 1060 mbar 23 O C, 50% RH, 860 to 1060 mbar Tolerances ± 2 O C ± 5%RH Reduced tolerances ± 1 O C ± 2%RH Standard Reference Atmosphere to which tests made at other atmospheres can be corrected 20 O C, 65% RH, 1013 mbar No qualifying altitude is given as it is concerned only with the effect of temperature, humidity and pressure

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 LOW 1015 mb 1012 mb 1008 mb 1000 mb 996 mb

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 H 2 O For vacuum measurement 1 mm Hg = 1 Torr 760 Torr = nil vacuum 0 Torr = full vacuum 760 mm Hg

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

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 calculations Pa = Pg + atmosphere For quick calculations assume 1 atmosphere is 1000 mbar For standard calculations 1 atmosphere is 1013 mbar Low range Typical Industrial range 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Absolute pressure bar aGauge pressure bar g Full vacuum Atmosphere Extended Industrial range

Pressure 1 bar = 100000 N/m 2 (Newtons per square metre) 1 bar = 10 N/cm 2 For measuring lower pressures the millibar (mbar) is used 1000 mbar = 1 bar For measurements in pounds per square inch (psi) 1 psi = 68.95mbar 14.5 psi = 1bar

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/m 2 1 mm Hg = 1.334 mbar approx. 1 mm H 2 O = 0.0979 mbar approx. 1 Torr = 1mmHg abs (for vacuum) More units of pressure

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.

Pressure and force The thrust developed by a piston due to air pressure is the effective area multiplied by the pressure Thrust= D2D2 40 P Newtons  D mm P bar Where D = The bore of a cylinder in mm P = The pressure in bar. We require an answer in Newtons 1bar = 100000 N/m 2 D 2 is therefore divided by 1000000 to bring it to m 2 and P is multiplied by 100000 to bring it to N/m 2. The result is a division by 10 shown in the product 40 above

Pressure and force The force contained by a cylinder barrel is the projected area multiplied by the pressure l D Force = D. l. P 10 Newtons Where D = the cylinder bore mm l = length of pressurised chamber mm P = the pressure in bar

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

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 P1P2

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 = constant V T P T

Constant Temperature

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. 0 2468 16 0 2 4 6 8 10 12 Volume V Pressure P bar absolute P1.V1 = P2.V2 = constant 101214 16

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. 0 2468 16 0 2 4 6 8 10 12 101214 16 Volume V Pressure P bar absolute P1.V1 = P2.V2 = constant

Constant Pressure

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 20 o C a change of 73.25 o C will produce a 25% change of volume. 0 o Celsius = 273K 0 0.250.50.75 12 -60 -40 -20 0 20 40 60 Volume Temperature Celsius 1.251.51.75 80 100 293K V1 V2 T1 (K) T2 (K) = c=

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 20 o C a change of 73.25 o C will produce a 25% change of volume. 0 o Celsius = 273K 0 0.250.50.75 12 -60 -40 -20 0 20 40 60 Volume Temperature Celsius 1.251.51.75 80 100 366.25K V1 V2 T1 (K) T2 (K) = c=

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 20 o C a change of 73.25 o C will produce a 25% change of volume. 0 o Celsius = 273K 0 0.250.50.75 12 -60 -40 -20 0 20 40 60 Volume Temperature Celsius 1.251.51.75 80 100 219.75K V1 V2 T1 (K) T2 (K) = c=

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 20 o C a change of 73.25 o C will produce a 25% change of volume. 0 o Celsius = 273K 0 0.250.50.75 12 -60 -40 -20 0 20 40 60 Volume Temperature Celsius 1.251.51.75 80 100 366.25K 219.75K 293K V1 V2 T1 (K) T2 (K) = c=

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 20 o C and 10 bar absolute a change in temperature of 60 o C will produce a change in pressure of 2.05 bar 0 o C = 273K 0 5 1020 -60 -40 -20 0 20 40 60 Temperature Celsius 15 80 100 0 2 4 6 8 bar 10 12 14 16 P1 P2 T1 (K) T2 (K) = c= bar absolute

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 20 o C and 10 bar absolute a change in temperature of 60 o C will produce a change in pressure of 2.05 bar 0 o C = 273K 0 5 1020 -60 -40 -20 0 20 40 60 Temperature Celsius 15 80 100 0 2 4 6 8 bar 10 12 14 16 P1 P2 T1 (K) T2 (K) = c= bar absolute

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 20 o C and 10 bar absolute a change in temperature of 60 o C will produce a change in pressure of 2.05 bar 0 o C = 273K 0 5 10 -60 -40 -20 0 20 40 60 bar absolute Temperature Celsius 15 80 100 0 2 4 6 8 bar 10 12 14 16 P1 P2 T1 (K) T2 (K) = c=

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. = constant P 1.V 1 T1T1 P 2.V 2 T2T2 =

Adiabatic and polytropic compression For compressed air

Adiabatic compression 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) 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 2468 0 2 4 6 8 10 12 bar a 101214 16 PV 1. 4 = c adiabatic PV 1. 2 = c polytropic PV = c isothermal Volume 0

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

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 Drain fully saturated air Condensate

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. -40 -20 01020304050 0 20 40 Grams of water vapour / cubic metre of air g/m 3 607080 Temperature Celsius 25% RH50% RH100% RH At 20 o Celsius 100% RH = 17.4 g/m 3 50% RH = 8.7 g/m 3 25% RH = 4.35 g/m 3

Water in compressed air The illustration shows four cubes each representing 1 cubic metre of atmospheric air at 20 o C. 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

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

Water in compressed air 4 cubic metres at 50%RH and 1000 mbar atmospheric pressure contained in the space of 1 cubic metre produce a pressure of 3 bar gauge 17.4 grams of water remain as a vapour producing 100% RH (relative humidity) and 17.4 grams condense to liquid water This is a continuous process, so once the gauge pressure is over 1 bar, every time a cubic metre of air is compressed, and added to the contained 1 cubic metre, a further 8.7 grams of water are condensed

Low temperature drying

Low temperature drier Humid air enters the first heat exchanger where it is cooled by the dry air going out The air enters the second heat exchanger where it is refrigerated The condensate is collected and drained away As the dry refrigerated air leaves it is warmed by the incoming humid air M Dry air out Humid air in Drain Refrigeration plant

Low temperature drying If 1 cubic metre of fully saturated compressed air ( 100 % RH ) is cooled to just above freezing point, approximately 75% of the vapour content will be condensed out. When it is warmed back to 20 O C it will be dried to nearly 25% RH -40 -20 01020304050 0 20 40 Grams of water vapour / cubic metre of air g/m 3 607080 Temperature Celsius 25% RH50% RH100% RH

Flow of compressed air

Flow units Flow is measured as a volume of free air per unit of time Popular units are : Litres or cubic decimetres per second l/s or dm 3 /s Cubic metres per minute m 3 /m Standard cubic feet per minute (same as cubic feet of free air) scfm 1 m 3 /m = 35.31 scfm 1 dm 3 /s = 2.1 scfm 1 scfm = 0.472 l/s 1 scfm = 0.0283 m 3 /min 1 cubic metre or 1000 dm 3 1 litre or cubic decimetre 1 cubic foot

Free air flow Actual volume of 1 litre of free air at pressure The space between the bars represents the actual volume in the pipe occupied by 1 litre of free air at the respective absolute pressures. Flow takes place as the result of a pressure differential, at 1bar absolute (0 bar gauge) there will be flow only to a vacuum pressure If the velocity were the same each case will flow twice the one above 0 1/81/8 1 / 16 1/41/4 1/21/2 1 litre 1bar a 2bar a 4bar a 8bar a 16bar a

Sonic flow The limiting speed at which air can flow is the speed of sound For sonic flow to exist, P 1 must be approx. 2 times P 2 or more When exhausting air from a reservoir at high pressure to atmosphere the flow will be constant until P 1 is less than 2 P 2 When charging a reservoir the flow will be constant until P 2 is 1 / 2 P 1 1.894 P 1 bar absolute time P1 is 9 bar a reservoir to atmosphere 2P 2 0 51020 0 1 2 3 4 5 6 15 7 8 9 atm 0 51020 0 1 2 3 4 5 6 15 7 8 9 P 2 bar absolute P1 is 9 bar a source to reservoir 1/2P11/2P1 atm

Flow through valves Valve flow performance is usually indicated by a flow factor of some kind, such as “C”, “b”, “Cv”, “Kv” and others. The most accurate way of determining the performance of a pneumatic valve is through its values of “C” (conductance) and “b” (critical pressure ratio). These figures are determined by testing the valve to ISO 6358 For a range of steady source pressures P 1 the pressure P 2 is plotted against the flow through the valve until it reaches a maximum The result is a set of curves showing the flow characteristics of the valve P1P1 P2P2

Flow through valves From these curves the critical pressure ratio “b” can be found. “b” represents the ratio of P 2 to P 1 at which the flow velocity goes sonic. Also the conductance“C”at this point which represents the flow “dm³/ second / bar absolute” Downstream Pressure P 2 bar gauge Critical pressure ratio b = 0.15 01234567 0 0.1 0.2 0.3 0.4 0.5 Conductance C= 0.062 dm/s/bar a For the horizontal part of the curve only Flow dm 3 /s free air P 1 is the zero flow point for each curve

Flow through valves If a set of curves are not available but the conductance and critical pressure ratio are known the value of flow for any pressure drop can be calculated using this formulae Q = C P 1 1 - 1 - b P2P2 P1P1 - b 2 Where : P 1 = upstream pressure bar P 2 = downstream pressure bar C = conductance dm 3 /s/bar a b = critical pressure ratio Q = flow dm 3 /s

Air Quality

Air filtration quality ISO 8573-1 Compressed air for general use Part 1 Contaminants and quality classes Allowable levels of contamination are given a quality class number Specified according to the levels of these contaminants: solid particles water oil An air quality class is stated as three air quality numbers e.g. 1.7.1 solids 0.1 µm max and 0.1 mg/m 3 max water not specified 0.01 mg/m 3 max This is the filtration class resulting from a Norgren Ultraire Filter To obtain pressure dew points that are low, also use an air drier

Compressed air quality Class particle size max µmµm Solids concentration mg/m 3 Water Max Pressure Dew point O C Oil concentration mg/m 3 10.1 – 70 0.01 211 – 40 0.1 355 – 20 1 4158+ 35 54010+ 725 6--+ 10- 7--Not Specified- maximum Pressure dew point is the temperature to which compressed air must be cooled before water vapour in the air starts to condense into water particles ISO 8573-1

Pressure units Standard Atmosphere = 1.01325 bar abs Technical Atmosphere = 0.98100 bar abs 1 mm Hg = 1.334mbar approx. 1 mm H 2 O = 0.0979 mbar approx. 1 kPa = 10.0 mbar 1 MPa = 10 bar 1 kgf/cm 2 = 981 mbar 1 N/m 2 = 0.01 mbar 1 Torr = 1mmHg abs (for vacuum)

Pressure units 1 bar = 100000 N/m2 1 bar = 1000000 dyn/cm 2 1 bar = 10197 kgf/m 2 1 bar = 100 kPa 1 bar = 14.50 psi 1 bar = 0.98690 standard atmospheres

Pressure units 1 dyn/cm 2 = 0.001mbar 1 psi = 68.95mbar Standard atmosphere = 14.7 psi approx. Standard atmosphere = 760 Torr approx. 1 inch Hg = 33.8 mbar approx. 1 inch H 2 O = 2.49mbar approx. 100 mbar is about as hard as the average person can blow

Temperature conversion -40 -20 0 20 40 60 80 100 120 233 253 273 293 313 333 353 373 393 The absolute temperature scale is measured in degrees Kelvin O K On the Celsius scale 0 O C and 100 O C are the freezing and boiling points for water O K = O C + 273.15 The Fahrenheit and Celsius scales coincide at - 40 O O F = O C. 9 / 5 + 32 OKOK -40 -20 0 20 40 60 80 100 120 140 160 180 200 220 240 OFOF OCOC

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