Facts and Theory of Air
For industrial pneumatics
1
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
2
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
Nitrogen 78.09% N2
Oxygen
20.95% O2
Argon
0.93% Ar
Others
0.03%
3
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
4
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
5
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
O
 20 C, 65% RH, 1013 mbar
No qualifying altitude is given as it is concerned only with the effect
of temperature, humidity and pressure
6
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
1015 mb
1012 mb
1008 mb
1000 mb
996 mb
LOW
7
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 Torr
760 Torr = nil vacuum
0 Torr = full vacuum
760 mm Hg
8
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
9




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
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Gauge pressure bar g

Absolute pressure bar a
Industrial compressed air
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Extended
Industrial
range
Typical
Industrial
range
Low
range
Atmosphere
Full vacuum
10
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 = 68.95mbar
14.5 psi = 1bar
11
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
12
Pressure and force
13
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.
14
Pressure and force
D mm

The thrust developed by a
piston due to air pressure
is the effective area
multiplied by the pressure
P bar
D2 P
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
15
Pressure and force

The force contained by a
cylinder barrel is the
projected area multiplied
by the pressure
l
D
Force = D . l . P Newtons
10
Where
D = the cylinder bore mm
l = length of pressurised chamber mm
P = the pressure in bar
16
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
17
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
18
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
P2
P1
19
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
20
The gas laws
21
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

P.V = constant
V
= constant
T

Constant volume
P
= constant
T
22
Constant Temperature
23
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.
Pressure P
bar absolute
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10 12 14 16
Volume V
P1.V1 = P2.V2 = constant
24
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.
Pressure P
bar absolute
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10 12 14 16
Volume V
P1.V1 = P2.V2 = constant
25
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.
Pressure P
bar absolute
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10 12 14 16
Volume V
P1.V1 = P2.V2 = constant
26
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.
Pressure P
bar absolute
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10 12 14 16
Volume V
P1.V1 = P2.V2 = constant
27
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.
Pressure P
bar absolute
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10 12 14 16
Volume V
P1.V1 = P2.V2 = constant
28
Constant Pressure
29
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
Temperature
Celsius
100
80
60
40
293K
20
0
-20
-40
-60
0
0.25 0.5 0.75
1
1.25 1.5 1.75
2
Volume
V1
V2
=
=c
T1(K)
T2(K)
30
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
Temperature
Celsius
100
366.25K
80
60
40
20
0
-20
-40
-60
0
0.25 0.5 0.75
1
1.25 1.5 1.75
2
Volume
V1
V2
=
=c
T1(K)
T2(K)
31
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
Temperature
Celsius
100
80
60
40
20
0
-20
-40
-60
219.75K
0
0.25 0.5 0.75
1
1.25 1.5 1.75
2
Volume
V1
V2
=
=c
T1(K)
T2(K)
32
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
Temperature
Celsius
100
366.25K
80
60
40
293K
20
0
-20
-40
-60
219.75K
0
0.25 0.5 0.75
1
1.25 1.5 1.75
2
Volume
V1
V2
=
=c
T1(K)
T2(K)
33
Constant volume
34
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
Temperature
Celsius
100
80
60
40
20
8
6
0
-20
4
12
2
14
0
-40
-60
10
bar
16
bar absolute
0
5
10
15
20
P1
P2
=
=c
T1(K)
T2(K)
35
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
Temperature
Celsius
100
80
60
40
20
8
6
0
-20
4
12
2
14
0
-40
-60
10
bar
16
bar absolute
0
5
10
15
20
P1
P2
=
=c
T1(K)
T2(K)
36
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
Temperature
Celsius
100
80
60
40
20
8
6
0
-20
4
12
2
14
0
-40
-60
10
bar
16
bar absolute
0
5
10
15
20
P1
P2
=
=c
T1(K)
T2(K)
37
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
Temperature
Celsius
100
80
60
40
20
8
6
0
-20
4
12
2
14
0
-40
-60
10
bar
16
bar absolute
0
5
10
15
P1
P2
=
=c
T1(K)
T2(K)
38
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.
P1 .V1
P2 .V2
=
= constant
T1
T2
39
Adiabatic and polytropic
compression
For compressed air
40
Adiabatic compression

bar a
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)
16
PV 1. 4 = c
14
adiabatic
12
PV 1. 2 = c
10
polytropic
8
PV = c
6
isothermal
4

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
0
0
2
4
6
8 10 12 14 16 Volume
41
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
42
Water in compressed air
43
Water in compressed air


fully
saturated
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
Condensate
Drain
44
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.
Temperature Celsius
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
0
-20
-40
0
10
20
30
40
50
60
70
80
Grams of water vapour / cubic metre of air g/m3
45
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
46
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
47
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
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
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
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
51
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
52
Low temperature drying
53
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
Humid air in
Dry air out
M
Drain
Refrigeration
plant
54
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
Temperature Celsius
25% RH
50% RH
100% RH
40
20
0
-20
-40
0
10
20
30
40
50
60
70
80
Grams of water vapour / cubic metre of air g/m3
55
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
Temperature Celsius
25% RH
50% RH
100% RH
40
20
0
-20
-40
0
10
20
30
40
50
60
70
80
Grams of water vapour / cubic metre of air g/m3
56
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
Temperature Celsius
25% RH
50% RH
100% RH
40
20
0
-20
-40
0
10
20
30
40
50
60
70
80
Grams of water vapour / cubic metre of air g/m3
57
Flow of compressed air
58
Flow units


Flow is measured as a
volume of free air per unit
of time
Popular units are :



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 litre or
cubic decimetre
Litres or cubic
decimetres per second
l/s or dm3/s
Cubic metres per minute
m3/m
Standard cubic feet per
minute (same as cubic
feet of free air) scfm


1 cubic foot
1 cubic metre
or 1000 dm3
59
Free air flow



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
Actual volume of 1 litre
of free air at pressure
0
1 litre
1bar a
1/
2
2bar a
1/
4
4bar a
1/
8
8bar a
1/
16
16bar a
60
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 1.894
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
9
8
7
6
5
P1 bar
4
absolute 3
2
1
0
0
9
8
7
6
P2 bar
absolute 5
4
3
2
1
0
0
P1 is 9 bar a
reservoir to
atmosphere
2P2
atm
5
10
time
15
20
1/ P
2 1
P1 is 9 bar a
source to
reservoir
atm
5
10
15
20
61
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 P1 the pressure
P1
P2
P2 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
62
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”
0.5
Critical pressure ratio b = 0.15
Conductance
C= 0.062 dm/s/bar a
For the horizontal part
of the curve only
Flow 0.4
dm3/s
free 0.3
air
0.2
P1 is the zero
flow point for
each curve
0.1
0
0
1
2
3
4
5
Downstream Pressure P2 bar gauge
6
7
63
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
P2
Q = C P1
1-
Where :
P1 = upstream pressure bar
P2 = downstream pressure bar
C = conductance dm3/s/bar a
b = critical pressure ratio
Q = flow dm3/s
P1
2
-b
1-b
64
Air Quality
65
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/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
66
Compressed air quality
ISO 8573-1
Class
1
2
3
4
5
6
7
Solids
Water
Oil
particle concentration
size max maximum
µm
mg/m3
Max Pressure
Dew point OC
concentration
mg/m3
– 70
– 40
– 20
0.01
0.1
1
5
25
-
0.1
1
5
15
40
-
0.1
1
5
8
10
-
+3
+7
+ 10
Not Specified
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
67
End
68
69
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)
70
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
71
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

72
Temperature conversion
393
373
240
220
120

100
200
353
180
80

160
333
140
60
120
313
100
40
80
293
273
60
40
20
253
0
20


0

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/ + 32
5
-20
-20
233
OK
-40
OF
-40
OC
73
Click the arrow to return
74