facts and theory of air for industrial pneumatics
TRANSCRIPT
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
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
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
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
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
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
10 12 14
14
16
Volume V
Pressure Pbar absolute
P1.V1 = P2.V2 = constant
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
10 12 14
14
16
Volume V
Pressure Pbar absolute
P1.V1 = P2.V2 = constant
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
10 12 14
14
16
Volume V
Pressure Pbar absolute
P1.V1 = P2.V2 = constant
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
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
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
100 366.25K
V1 V2T1(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 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
219.75K
V1 V2T1(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 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
100 366.25K
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 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 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
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
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
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
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
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
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
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
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
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 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
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
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
End
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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