Pneumatics & Hydraulics Technology
Chapter 1
Pneumatics
Overview
Instructor: Phan Thi Thu Thuy
Higher Engineering Education Alliance Program
Contents
1
Overview of pneumatics
Definition of pneumatics
History of pneumatics
Applications of pneumatics
Advantages and Disadvantages
2
Fundamentals of pneumatics
Physical properties of air
Characteristics of air
Control theory
2
Objectives
1. Be able to write pneumatics branch after
attending a lecture within 3 minutes
2. Given lecture note, be able to list at least 5
applications of pneumatics within 3 minutes
3. Given lecture note, be able to discuss 5
advantages and 4 disadvantages of
pneumatics system within 10 minutes
4. Given lecture note, be able to define 3 gas
laws and 3 equations with 100% accuracy.
3
1 Overview of pneumatics
Definition of pneumatics
History of pneumatics
Applications of pneumatics
Advantages and Disadvantages
4
Definition of pneumatics
60 AD - Pneumatics (from the Greek
πνευματικός pneumatikos, coming from the
wind) is the use of pressurized gas to do
work in science and technology.
Pneumatics is a branch of technology,
which deals with the study and application
of use of pressurized gas to affect
mechanical motion
5
History of pneumatics
It was near the end of the 17th Century that
Torricelli, Mariotte and later, Bernoulli carried on
experiments to study the elements of pressure or
force in the discharge of water through orifices in
the sides of tanks and through short pipes.
In that same period the
French scientist, Blaise Pascal,
evolved the fundamental law for
the science of hydraulics.
6
History of pneumatics
Establishment of Industry Standards
 During the last forty years and particularly during the period
of World War II, the use of fluid power has grown by leaps
and bounds and among many fields of application a most
important one has been that of production machinery.
 In fact, by 1951 applications in this field had increased so
rapidly that a number of conferences were held in Detroit,
Michigan, for the purpose of establishing a set of standard
for industrial hydraulics and pneumatic equipment.
 These are now well known as the “Joint Industry Conference
Standards for Industrial Equipment”
7
History of pneumatics
Establishment of Industry Standards
Represented in these conferences:
- Hydraulic equipment manufactures,
- Pneumatic equipment manufactures,
- Machine tool builders,
- Packing and seal manufacturers,
- Press manufacturers,
- Resistance welding manufacturers,
- Tubing and fitting manufacturers
- And industrial users.
8
History of pneumatics
Establishment of Industry Standards
 In 1966, a new set of graphic standards was released
by the United States of America Standards Institute. The
Institute changed its name to the American National
Standards Institute, Inc., on Oct 6, 1969.
 Through their use, the maintenance man can readily trace
the flow of fluids through the controls without reference to
the control manufacturer’s catalogs. This is great
timesaver, especially on complicated layouts.
 Today, many machines containing fluid power systems
have their circuit diagrams attached.
9
Applications of pneumatics
Pneumatic components can perform the following
types of motion:
- Linear: moving in a straight line such as on a paper trimmer
- Rotary: turning round in a circle, such as a wheel turning, gear, roller
Pneumatics is used in carrying out machining and
working operations:
- Drilling
- Finishing
- Turning
- Forming
- Milling
- Quality control
- Sawing
11
Applications of pneumatics
 Some industrial applications employing pneumatics
are listed below:
General methods of material handling:
- Clamping
- Shifting
- Positioning
- Orienting
12
Applications of pneumatics
General applications:
- Packaging
- Feeding
- Metering
- Door or chute control
- Transfer of materials
- Turning and inverting of parts
- Sorting of parts
- Stacking of components
- Stamping and embossing of components
13
Applications of pneumatics
S1
S2
S4
Cylinder A
S3
Cylinder B
Transfer station
Parts are to be transferred from vertical
magazine to a processing station
14
Applications of pneumatics
 Mps transfer / FESTO MPS system
 Robot Station with Assembly
 Tripod EXPT - the parallel kinematic robots
Festo_Exohand
Festo_MultiChoice Gripper
15
Test question
 Machines spray markings onto roads
 Automatic dumpling manufacturing system
 Rubbish sytem
Name at least 4 applications of pneumatics in 3
video clips ?
16
Advantages
1. Availability: Air is available practically
everywhere in unlimited quantities.
2. Transportation: Air can be easily transported
in pipelines, even over large distance. Not
necessary to return the compress air.
3. Storage: Air compressor need not be in
continuous operation. Compressed air can
be stored in a reservoir and removed as
required. In addition, the reservoir can be
transportable.
17
Advantages
4. Temperature: Compressed air is relatively
insensitive to temperature fluctuations. This
ensures reliable operation, even under
extreme conditions.
5. Explosion proof: No risk of explosion or
fire; expensive protection against explosion
is not required
18
Advantages
6. Cleanliness: Unlubricated exhaust air is
clean. Any unlubricated air which escapes
through leaking pipes or components does not
cause contamination. This is an important
point when considering the food, wood and
textile industries.
7. Components: The operating components
are of simple construction and are therefore
relatively inexpensive.
19
Advantages
8. Speed: Compressed air is a very very fast
working medium; cylinders have working
speed of 1- 2 m/s. This enables high working
speeds to be attained.
9. Adjustable: With compressed air components,
speeds and forces are infinitely variable.
10. Overload safe: Pneumatic tools and operating
components can be loaded to the point of
stopping and are therefore overload safe.
20
Disadvantages
1. Preparation: require to remove dirt and
humidity
2. Compressible: Not possible to achieve
uniform and constant piston speeds
3. Force requirement: low working loads; the
require output force is economical only up to
about 20 KN to 30 KN; commonly used
working pressure of 7 bar
4. Exhaust air: Noisy; the oil mist with
compressed air escapes with the exhaust to
the atmosphere
21
Dirt and Moisture are damaging
Moisture in compressed air used in a manufacturing plant causes
problems in the operation of pneumatic systems, solenoid valves and
air motors and can adversely affect the process or product being
manufactured. For many years, problems from moisture in
compressed air lines were simply tolerated as unavoidable. Moisture:
 Causes corrosion of air or gas operated instruments, giving false
readings, interrupting or shutting down plant processes.
 Causes rust and increased wear of moving parts in production
equipment as it washes away lubrication
 Can freeze in control lines in cold weather, which may cause
faulty operation of controls
 Can endanger process industries where many operations are
dependent upon the proper functioning of pneumatic controls. The
malfunctioning of these controls due to rust, scale, and
clogged orifices can result in damage to product or in costly
shutdowns
22
2
Fundamentals of pneumatics
Physical properties of air
Characteristics of air
Control theory
23
Physical properties of air
24
Physical properties of air
Composition of air
The surface of the
globe
is
entirely
covered by a mantle
of air. It is an
abundant gas mixture
with the following
composition.
Composition by Volume
Nitrogen 78.09% N2
Oxygen 20.95% O2
Argon
0.93% Ar
Others
0.03%
25
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
26
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
28
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
30
The Story of Otto von Guericke's 1654 demonstration of
the power of a vacuum at the bottom of a sea of air.
Clip 1
Clip 2
Clip 3
 In 1654 Otto von Guericke gave the citizens of Magdeburg a remarkable lesson in the
force of the atmosphere.
 He machined two hollow hemispheres, twenty inches in diameter, so they fit snuggly into
a sealed sphere. He pumped the air out of it. Then he put sixteen horses, eight on each
side, to the task of pulling the halves apart. The horses couldn't, of course. It would've
taken a force of over two tons to separate the halves. ... John Lienhard, at the University
of Houston, where we're interested in the way inventive minds work.
31
Physical properties of air
Base Quantities
Unit
Symbol
Units and Units symbols
Technical system
System SI
Length
L
Metre (m)
Metre (m)
Mass
m
Kp.s2/m
Kilogram
(kg)
Time
t
Second (s)
Second (s)
Temperature
T
Degree Celsius (0C)
Kelvin (K)
32
Physical properties of air
Derived Quantities
Unit
Symbol
Derived Units and Units symbols
Technical system
System SI
Force
F
Kilopond (kp)
Newton (N) =
1 kg.m/s2
Area
A
Square metre (m2)
Square metre
(m2)
Volume
V
Cubic metre (m3)
Cubic metre
(m3)
Flowrate
Q
(m3/s)
(m3/s)
Pressure
P
Atmosphere (at)
Pascal (Pa)
1 Pa = 1 N/m2
1 bar = 105 Pa
33
Pressure
1 bar = 105N/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
F
P
A
(N/m2)
34
Pressure units
There are many units of pressure measurement.
Some of these and their equivalents are listed
below.
1 bar = 105 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)
35
Pressure Unit Conversion Table
Pa
bar
Pa
1
1x10-5
bar
1x105
1
KG/cm2 9,80665x104 9.80665x10-1
KG/cm2
atm
mmH20
mmHg
psi
1,01972x10-5 9,86923x10-6 1,01972x10-1 7,50062x10-3 1,45038x10-8
9,86923x10-1 1,01972x104 7,50062x102
1,45048x10
1
9,67841x10-1
1x104
7,35559x102
1,42234x10
7,60x102
1,46960x10
1,01972
atm
1,01325x105
1,01325
1,03323
1
1,03323x104
mmH20
9,80665
9,80665x10-5
1x10-4
9,67841x10-5
1
7,35559x10-2 1,42234x10-3
mmHg
1,33322x102 1,33322x10-5 1,35951x10-3 1,31579x10-3 1,35951x10-3
1
1,93368x10-2
psi
6,89473x103 6,89473x10-2 7,03065x10-2 6,80457x10-2 7,03067x102
5,17147x10
1
36
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Gauge pressure bar g
 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
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
37
Characteristics of air
38
Characteristics of air
In common with all gases, air has no particular
shape. Its shape changes with the slightest
resistance, i.e. its assume the shape of its
surroundings.
Air can be compressed and it endeavours to
expand.
39
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
(Boyle’s law )
= constant
(Charles’ law)
= constant
(Boyle’s law and
Charles’ law =
combined gas law)
T
 Constant volume
P
T
40
Constant Temperature
41
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
42
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
43
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
44
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
45
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
46
Constant Pressure
47
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
293K
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)
48
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)
49
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.
 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)
50
Constant pressure
 Charles’ law states: for a Temperature
Celsius
given mass of gas at
100
366.25K
constant pressure the
80
volume is proportional to
60
the absolute temperature.
40
 Assuming no friction a
20
293K
volume will change to
0
maintain constant pressure. -20
-40
 From an ambient of 20oC a
219.75K
-60
0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 Volume
change of 73.25oC will
produce a 25% change of
volume.
V1
V2
=
=c
 0o Celsius = 273K
T1(K)
T2(K)
51
Constant volume
52
Constant volume
 From Boyle’s law and Charles’
Temperature
law we can also see that if the Celsius
volume of a given mass of air 100
80
were to be kept at a constant
60
value, the pressure will be
40
proportional to the absolute
20
temperature K.
0
-20
 For a volume at 20oC and 10
-40
bar absolute a change in
-60
5
0
o
temperature of 60 C will
produce a change in pressure
of 2.05 bar
 0oC = 273K
8
6
10
4
12
14
2
0
bar
16
bar absolute
10
15
20
P1
P2
=
=c
T1(K)
T2(K)
53
Constant volume
 From Boyle’s law and Charles’ Temperature
Celsius
law we can also see that if the 100
volume of a given mass of air
80
were to be kept at a constant
60
40
value, the pressure will be
20
proportional to the absolute
0
temperature K.
-20
o
 For a volume at 20 C and 10
-40
-60
bar absolute a change in
5
0
temperature of 60oC will
produce a change in pressure
of 2.05 bar
 0oC = 273K
8
6
10
4
12
2
14
0
bar
16
bar absolute
10
15
20
P1
P2
=
=c
T1(K)
T2(K)
54
Constant volume
 From Boyle’s law and
Charles’ law we can also see Temperature
Celsius
that if the volume of a given
100
mass of air were to be kept at 80
a constant value, the
60
40
pressure will be proportional
20
to the absolute temperature
0
K.
-20
 For a volume at 20oC and 10
-40
bar absolute a change in
-60
5
0
o
temperature of 60 C will
produce a change in pressure
of 2.05 bar
 0oC = 273K
8
6
10
4
12
14
2
0
bar
16
bar absolute
10
15
20
P1
P2
=
=c
T1(K)
T2(K)
55
Constant volume
 From Boyle’s law and
Temperature
Celsius
Charles’ law we can also see 100
that if the volume of a given
80
mass of air were to be kept at 60
40
a constant value, the
20
pressure will be proportional
0
to the absolute temperature
-20
K.
-40
-60
 For a volume at 20oC and 10
5
0
bar absolute a change in
temperature of 60oC will
produce a change in pressure
of 2.05 bar
 0oC = 273K
8
6
10
4
12
14
2
0
bar
16
bar absolute
10
15
P1
P2
=
=c
T1(K)
T2(K)
56
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
=
T1
=
constant
T2
57
Test question
 Question 1: It says that, for an ideal gas at constant
pressure, the volume is proportional to the ________ (in
kelvins).
 Question 2: which gas law is stated as follows?
The equation shows that, as absolute temperature
increases, the volume of the gas also increases in
proportion.
 Question 3: An ideal gas is contained in a volume V at
temperature T. If the volume is doubled at constant
pressure, the temperature will be:
a. Unchanged
b. Halved
c. Doubled
d. Unable to be determined
without more information
58
Test question
 Question 4: An ideal gas is sealed in a container at
constant volume. If the temperature T is increased to 4T,
the pressure will be:
a. Unchanged
b. Increased to 4P
c. Decreased by 4P
d. Unable to be determined
without more information
 Question 5: A sample of an ideal gas is held at constant
temperature. If the pressure is decreased to 1/2 P, the
volume will be:
a. Increased to 2V
b. Decreased to 1/2 V
c. Unchanged
d. Unable to be determined
without more information
59
Pressure and force
60
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.
61
Pressure and force
The force contained by a cylinder barrel is the
projected area multiplied by the pressure
F  P.A
F
 .D .P
2
4
Dm
P bar
F : Force (N).
D : Diameter of piston (m).
P : Pressure(N/m2).
63
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
64
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
65
Flow of compressed air
66
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 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 litre or
cubic decimetre
 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
67
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
Volume (litre)
Actual volume of 1 litre
of free air at pressure
1.0
0.5
0.25
0.125
0.06325
0
Bar Pa
Bar Pg
1
0
2
1
4
3
8
7
16
15
68
Control theory
Control types according to DIN 19226
Form of information representation
Pneumatic control system
Signal flow and the control chain
69
Control theory
Differentiation between the control systems can
be made according to the following viewpoints.
The following illustrations respresent the control
types according to DIN 19226. There are three
main groups.
The categorisation of a control system to the
three main groups depends on the problem
definition. If it is a case of program control, the
project designer has the choice of the three
subgroups of program control .
70
Control types according to DIN 19226
71
Control types according to DIN 19226
Pilot control system
There is always a clear relationship between the
command or reference value and the output
value provided disturbance variables do not
cause any deviations. Pilot controls do not have a
memory function.
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Control types according to DIN 19226
Memory control system
When the command or reference value is
removed or cancelled, in particular after
completion of the input signal, the output
value achieved is retained (memorised).
A different command value or an opposing
input signal is required to return the output
value to an initial value. Memory control
systems always have a storage function.
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Control types according to DIN 19226
Program control
The three types of program control are:
 Step diagram control
 Sequence control system
 Time (schedule) control
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Program control
Step diagram control
In the case of step diagram control, the
reference variables are supplied by a program
generator (program memory), whose output
variables depend on the path travelled or the
position of a moving part of the controlled
system.
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Program control
Sequence control system
The sequence program is stored in a program
generator which runs through the program stepby-step according to the status attained by the
system being controlled.
This program may either be permanently
installed or else read from punched cards,
magnetic tapes or other suitable memories.
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Program control
Time (schedule) control
 In a time (schedule) control system, the command
values are supplied by a time-dependent program
generator.
 Characteristics of a timing control system are, thus,
the existence of a program generator and a timedependent program sequence. Program generators
may be :
– Cam shafts
– Cams
– Punched cards
– Punched tape
– Programs in an electronic memory
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Form of information representation
 Differentiation between the control systems: in the
form of representation of information
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Pneumatic control system
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Signal flow and the control chain
PNEUMATICS/HYDRAULICS
- Cylinders
- Motors
- Indicators
Directional
Control Valves
- Directional
Control Valves
- Non-return valves
- Pressure valves
- Switches
- Pushbuttons
- Limit switches
- Program generators
- Proximity signalers
- Sensors
Compressed air
ELECTRICS/ ELECTRONICS
ACTUATING DEVICES
FINAL CONTROL
ELEMENTS
PROCESSING ELEMENTS
(CONTROL ELEMENTS)
INPUT ELEMENTS
- Electric motor
- Solenoids
- Linear motors
- Power contactors
- Power transistors
- Semiconductors
- Contactors
- Relays
- Electronic
Components
- Switches
- Pushbuttons
- Limit switches
- Program generators
- Proximity signalers
- Indicators
- Generators
Electrics/electronics energy
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Pneumatic Circuit Diagram
1.5
Actuator
A
1.4
Final control
element
S P
1.3
1.1
A
B
A
Signal
processor
A
1.2
R
Signal input
P
R
P
R
0.2
0.1
P
A
R
Compressed air
Energy supply
(source)
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Electro – pneumatic circuit diagram
S1
Cylinder A
S2
B
A
A
Y1
S4
B
Y2
S
R
S
P
+24V
S3
Cylinder B
1
2
R
P
3
4
5
6
7
8
Start
S1
0V
K1
S2
K3
K4
K1
K2
K2
S3
S4
K1
K2
K3
K4
Y1
Y2
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Test question: Fill in the blanks
Advantages of pneumatic system
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Test question: Fill in the blanks
Disadvantages of pneumatic system
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Homework
3 in
 A 2 inch diameter piston with a 3 inch stroke, compresses
air at atmospheric pressure into a 2 cubic inch space at the
top of the cylinder. Use Boyle’s law to determine the
pressure after the gas has been compressed ?
Compressed Air
Air
a. 39.9 psig
2 in
b. 54.6 psia
c. 69.3 psig
d. 84 psig
e.119.6 psia
(choose the right answer
and attach the solution)
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Reference Textbook
 Peter Croser, Frank Ebel, Pneumatics – Textbook Basic Level, Festo Didactic
1999 (any edition is useful).
 D. Waller, H. Werner – Pneumatics – Workbook Basic Level, Festo Didactic 2002
(any edition is useful).
 G. Prede, D. Scholz – Electropneumatics – Textbook Basic Level, Festo Didactic
2002 (any edition is useful).
 D. Waller, H. Werner – Electropneumatics – Workbook Basic Level, Festo
Didactic 2002 (any edition is useful).
 Michael J.Pinches, John G.Ashby, Power Hydraulics, Prentice-Hall 1989
 H. Exner, R. Freitag, Basic Principles and Components of Fluid Technology,
Rexroth 1991.
 http://www.festo-didactic.com/int-en/
 http://www.norgren.com/
 http://www.pneumatica.be/pneumatics/index.htm
 http://www.hydraulicspneumatics.com
 http://pneumatic-source.com
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HIGHER ENGINEERI NG EDUC ATION ALLI ANCE PROGRAM
Contact: thuyptt@hcmute.edu.vn