teoria neumatica

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

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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 VolumeNitrogen 78.09% N2

Oxygen 20.95% O2

Argon 0.93% ArOthers 0.03%

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 (15OC)

1013.25 m bar

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

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 H2O

For vacuum measurement 1 mm Hg = 1 Torr760 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 calculationsPa = Pg + atmosphere

For quick calculationsassume 1 atmosphere is 1000 mbar

For standard calculations 1 atmosphere is1013 mbar

Lowrange

TypicalIndustrialrange

01234

5

67

8

910

111213

1415

1617

01234

5

67

8

910

111213

1415

16

Abs

olut

e pr

essu

re b

ar a

Gau

ge p

ress

ure

bar

g

Full vacuum

Atmosphere

ExtendedIndustrialrange

Pressure

1 bar = 100000 N/m2 (Newtons per square metre)

1 bar = 10 N/cm2

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

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

Pressure and force

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 = D2

40P Newtons

D mm

P bar

WhereD = The bore of a cylinder in mmP = The pressure in bar. We require an answer in Newtons 1bar = 100000 N/m2 D2 is therefore divided by 1000000 to bringit 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

Pressure and force

The force contained by a cylinder barrel is the projected area multiplied by the pressure

l

D

Force =D . l . P10

Newtons

WhereD = the cylinder bore mml = length of pressurised chamber mmP = 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

P1 P2

Pressure and force



P1P2

Pressure and force



P1 P2

The gas laws

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

= constantV

T

= constantP

T

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 2 4 6 8 160

2

4

6

8

10

12

Volume V

Pressure Pbar absolute

P1.V1 = P2.V2 = constant

10 12 14

14

16


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 2 4 6 8 160

2

4

6

8

10

12

10 12 14

14

16

Volume V

Pressure Pbar 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 20oC a change of 73.25oC will produce a 25% change of volume.

0o Celsius = 273K

0 0.25 0.5 0.75 1 2-60

-40

-20

0

20

40

60

Volume

TemperatureCelsius

1.25 1.5 1.75

80

100

293K

V1 V2T1(K) T2(K)

= c=

Constant pressure




0o Celsius = 273K

0 0.25 0.5 0.75 1 2-60

-40

-20

0

20

40

60

Volume

TemperatureCelsius

1.25 1.5 1.75

80

100 366.25K

V1 V2T1(K) T2(K)

= c=

Constant pressure




0o Celsius = 273K

0 0.25 0.5 0.75 1 2-60

-40

-20

0

20

40

60

Volume

TemperatureCelsius

1.25 1.5 1.75

80

100

219.75K

V1 V2T1(K) T2(K)

= c=

Constant pressure




0o Celsius = 273K

0 0.25 0.5 0.75 1 2-60

-40

-20

0

20

40

60

Volume

TemperatureCelsius

1.25 1.5 1.75

80

100 366.25K

219.75K

293K

V1 V2T1(K) T2(K)

= c=

Constant volume

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

0 5 10 20-60

-40

-20

0

20

40

60

TemperatureCelsius

15

80

100

0

2

4

68

bar

10

12

14

16

P1 P2T1(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 20oC and 10 bar absolute a change in temperature of 60oC will produce a change in pressure of 2.05 bar

0oC = 273K

0 5 10-60

-40

-20

0

20

40

60

bar absolute

TemperatureCelsius

15

80

100

0

2

4

68

bar

10

12

14

16

P1 P2T1(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.

= constantP1 .V1

T1

P2 .V2

T2

=

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

2 4 6 80

2

4

6

8

10

12

bar a

10 12 14

14

16

16

PV 1. 4 = cadiabatic

PV 1. 2 = cpolytropic

PV = cisothermal

Volume0

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

fullysaturated

air

Condensate


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

0 10 20 30 40 50

0

20

40

Grams of water vapour / cubic metre of air g/m3

60 70 80

Tem

per

atu

re C

els

ius

25% RH 50% RH 100% RH

At 20o Celsius100% RH = 17.4 g/m3

50% RH = 8.7 g/m3

25% RH = 4.35 g/m3


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


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


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 20OC it will be dried to nearly 25% RH

-40

-20

0 10 20 30 40 50

0

20

40

Grams of water vapour / cubic metre of air g/m3

60 70 80

Tem

per

atu

re C

els

ius

25% RH 50% RH 100% 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 secondl/s or dm3/s

Cubic metres per minutem3/m

Standard cubic feet per minute (same as cubic feet of free air) scfm

1 m3/m = 35.31 scfm 1 dm3/s = 2.1 scfm 1 scfm = 0.472 l/s 1 scfm = 0.0283 m3/min 1 cubic metre

or 1000 dm3

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

1/8

1/16

1/4

1/2

1 litre1bar 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, P1 must be approx. 2 times P2 or more

When exhausting air from a reservoir at high pressure to atmosphere the flow will be constant until P1 is less than 2 P2

When charging a reservoir the flow will be constant until P2 is 1/2 P1

1.894

P1 barabsolute

time

P1 is 9 bar a reservoir to atmosphere

2P2

0 5 10 200123456

15

789

atm

0 5 10 200123456

15

789

P2 barabsolute

P1 is 9 bar a source to reservoir

1/2P1

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 sourcepressures P1 the pressureP2 is plotted against theflow through the valve untilit reaches a maximum

The result is a set of curvesshowing the flow characteristicsof the valve

P1 P2

Flow through valves

From these curves the critical pressure ratio “b” can be found. “b” represents the ratio of P2 to P1 at which the flow velocity goes sonic. Also the conductance“C”at this point which represents the flow “dm³/ second / bar absolute”

Downstream Pressure P2 bar gauge

Critical pressure ratio b = 0.15

0 1 2 3 4 5 6 70

0.1

0.2

0.3

0.4

0.5ConductanceC= 0.062 dm/s/bar aFor the horizontal partof the curve only

Flowdm3/sfreeair

P1 is the zeroflow 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 P1 1 -1 - b

P2

P1

- b

2

Where : P1 = upstream pressure barP2 = downstream pressure barC = conductance dm3/s/bar ab = critical pressure ratioQ = flow dm3/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 maxand 0.1 mg/m 3 max

water not specified 0.01 mg/m3 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

particlesize max

µm

Solids

concentration

mg/m3

Water

Max Pressure Dew point OC

Oil

concentrationmg/m3

1 0.1 0.1 – 70 0.01

2 1 1 – 40 0.1

3 5 5 – 20 1

4 15 8 + 3 5

5 40 10 + 7 25

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 H2O = 0.0979 mbar approx.

1 kPa = 10.0 mbar 1 MPa = 10 bar 1 kgf/cm2 = 981 mbar 1 N/m2 = 0.01 mbar 1 Torr = 1mmHg abs (for vacuum)

Pressure units

1 bar = 100000 N/m2 1 bar = 1000000 dyn/cm2

1 bar = 10197 kgf/m2

1 bar = 100 kPa 1 bar = 14.50 psi 1 bar = 0.98690 standard atmospheres

Pressure units

1 dyn/cm2 = 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 H2O = 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 OK

On the Celsius scale 0OC and 100OC are the freezing and boiling points for water

OK = OC + 273.15 The Fahrenheit and

Celsius scales coincide at - 40O

OF = OC. 9/5 + 32

OK

-40

-20

0

20

40

60

80

100

120

140

160

180

200

220

240

OF OC

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