nota pneumatic and hydraulic system

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.5.3 Manual forward and return stroke control with double solenoid valve.

The piston rod of a cylinder is to be controlled by brief actuation of two
pushbuttons (S1 : advance S2 : retract)

a) b)

2 42

Y1 Y2 Y1 Y2
13 53
1

+24V 1 2 +24V

c) 3 3 d) 3 3 3 3

S1 S2 S1 S2 K1 K2

44 44 4 4

A1 A1

Y1 Y2 K1 K2 Y1 Y2
0V
0V A2 A2

Figure 6.12 Manual forward and return stroke control with signal storage by double
solenoid valve

The two pushbuttons act directly and indirectly on the coils of a double
solenoid valve (figure 6.12(c) and (d)).
When pushbutton S1 is pressed, solenoid coil Y1 is energized. The
double solenoid valve switches and the piston rod advances. If the
pushbutton is released during the advancing movement, the piston rod
continues extending to the forward end position because the valve retains
its switch position.When pushbutton S2 is pressed, solenoid coil Y2 is
energized. The double solenoid valve switches again, and the piston rod
returns. Releasing pushbutton S2 has no effect on the return movement.

BPLK 15/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.5.4 Semi automatic return stroke control with double solenoid valve

The aim for the piston rod of a double acting cylinder to be advanced is
when pushbutton S1 is actuated. When the forward end position is
reached, the piston rod is to return automatically.

a)
A1

42

Y1 Y2
53
1

b) +24V 1 2 c) +24V 33 3 3
A1 K1 K2
3 3 S1
S1 A1 44 4 4

4 4

A1 A1

Y1 Y2 K1 K2 Y1 Y2
0V
A2 A2
0V

Figure 6.13 Semi automatic return stroke control with signal storage by double solenoid
valve

BPLK 16/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.5.5 Automatic movement with double solenoid valve

The piston rod of a cylinder is to advance and retract automatically as
soon as control switch S1 is actuated. When the control switch is reset,
the piston rod is to occupy the retracted end position.

a) A0 A1

42

Y1 Y2
53
1

b) +24V c) +24V

3 3 3 3 3 3
K1 K2
S1 A1 S1 A1
4 4
44 44
1 1

A0 A0

2 2

A1 A1

Y1 Y2 K1 K2 Y1 Y2
0V
0V A2 A2

Figure 6.14 Automatic forward and return stroke control with signal storage by double
solenoid valve

BPLK 17/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.6 RELAY

A relay is an electromagnetically actuated switch. When a voltage is applied to
the solenoid coil, an electromagnet field is created. This causes the armature to
be attracted to the coil core. The armature actuates the relay contacts, either
closing or opening them, depending on the design. A return spring returns the
armature to its initial position when the current to the coil is interrupted.

Cover Armature 12 22
14 24
Return
spring A1

Coil

A2 11 21

Symbol

Contacts

Coil connections

Contact connections

Figure 6.15 Construction of relay

A relay coil can switch one or more contacts. In addition to the type of relay
described above, there are other types of electromagnetically actuated switch,
such as the retentive relay, the time relay, and the contactor.

6.6.1 Application of relays

In electro-pneumatic control system, relays are used for the following
functions:
 Signal multiplication
 Delaying and conversion of signals
 Association of information
 Isolation of control circuit from main circuit

In purely electrical controllers, the relay is also used for isolation of DC and
AC result.

BPLK 18/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.6.2 Retentive relay

The retentive relay responds to current pulses:
 The armature is energized when a positive pulse is applied.
 The armature is de-energized when a negative pulse is applied
 If no input signal is applied, the previously set switch position retained

(retention).

The behaviour of a retentive relay is analogous to that of a pneumatic
double pilot valve, which responds to pressure pulses.

+24V 1 2
STAR T 3
3
4 K1

4

K1 A1
0V Y1

A2

2

Figure 6.16 Example circuit by using a relay

BPLK 19/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.7 MEMORY BLOCKS

6.7.1 Relay circuit with latching

When the “ON” pushbutton is actuated in the circuit in figure 6.17(a), the
relay coil is energized. The relay is energized, and contact K1 closes.
After the “ON” pushbutton is released, current continues to flow via
contact K1 through the coil, and the relay remains in the actuated
position. The “ON” signal is stored. This is therefore a relay circuit with
latching function.

a) +24V +24V 3 3
K1 4
33 b)
4
ON K1 ON

44 OFF 1
1
K1 2
OFF 0V A1

2 A2

K1 A1
0V A2

Dominant ON Dominant OFF

Figure 6.17 Latching circuit

When the “OFF” pushbutton is pressed the flow of current is interrupted
and the relay becomes de-energized. If the “ON” and “OFF” pushbuttons
are both pressed at the same time, the relay coil is energized. This circuit
is referred to as a dominant ON latching circuit.

The circuit in figure 6.17(b) exhibits the same behaviour as the circuit in
figure 6.17(a) provided that either only the “ON” pushbutton or only the
“OFF” pushbutton is pressed. The behaviour is different when both
pushbuttons are pressed. The relay coil is not energized. This circuit is
referred to as a dominant OFF latching circuit.

BPLK 20/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

6.8 CHANGE OVER BLOCKS

6.9 REED SWITCHES

6.9.1 Reed switch

Reed switches are magnetically actuated proximity switches. They
consist of two contact reeds in a glass tube filled with inert gas. The field
of a magnet causes the two reeds to close, allowing current to flow. In
reed switches that act as normally closed contacts, the contact reeds are
closed by small magnets. This magnetic field is overcome by the
considerably stronger magnetic field of the switching magnets.

Figure 6.18 Reed switch (normally open contact)

BPLK 21/22 DMV 1152/3152/4152/5152

ELECTRICAL COMPONENTS AND CIRCUIT DEVELOPMENT

SUMMARY

In this chapter we have studied to :

REFERENCES

1. Fundamentals Of Pneumatic Control Engineering Text Book, Festo Didactic, 1989
2. Electro-Pneumatics Basic Level TP 201 Text Book, Festo, 1998
3. FluidSim Pneumatic Software Version 4.0

BPLK 22/22 DMV 1152/3152/4152/5152

SOLENOID VALVES

CHAPTER 7: SOLENOID VALVES

INTRODUCTION

This chapter will introduce the students

LEARNING OBJECTIVES

The objectives of this unit are to :
1.

BPLK 1/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.1 SOLENOID OPERATING PRINCIPLE

7.1.1 Definitions

A solenoid is a 3-dimensional coil. In physics, the term solenoid refers to a loop
of wire, often wrapped around a metallic core, which produces a magnetic field
when an electrical current is passed through it. Solenoids are important because
they can create controlled magnetic fields and can be used as electromagnets.
The term solenoid refers specifically to a magnet designed to produce a uniform
magnetic field in a volume of space (where some experiment might be carried
out).

In engineering, the term solenoid may also refer to a variety of transducer
devices that convert energy into linear motion. The term is also often used to
refer to a solenoid valve, which is an integrated device containing an
electromechanical solenoid which actuates either a pneumatic or hydraulic valve,
or a solenoid switch, which is a specific type of relay that internally uses an
electromechanical solenoid to operate an electrical switch; for example, an
automobile starter solenoid, or a linear solenoid, which is an electromechanical
solenoid.

BPLK Figure 7.1 Magnetic field create by a solenoid

2/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.1.2 Functions

An electro-pneumatic control system works with two forms of energy:
 Electrical energy in the signal control section
 Compressed air in the power section

Electrically actuated directional valves form the interface between the two parts
of an electro-pneumatic control. They are switched by the output signals of the
signal control section and open or close connections in the power section. The
most important tasks of electrically actuated directional control valves include:
 Switching supply air on or off
 Extension and retraction of cylinder drives

7.1.2.1 Actuation of a single-acting cylinder

Figure 7.2(a) shows an electrically actuated valve that controls the motion
of a single-acting cylinder. It has three ports and two switching positions:
 If no current is applied to the solenoid coil of the directional control

valve, the cylinder chamber above the directional control valve is
vented. The position rod is retracted.
 If current applied to the solenoid coil, the directional control valve
switches and the chamber is pressurized. The piston extends.
 When the current is interrupted, the valve switches back. The cylinder
chamber is vented and the piston rod retracts.

7.1.2.2 Actuation of a double-acting cylinder

The double acting cylinder drive in figure 7.2(b) is actuated by a
directional control valve with five ports and two switching positions.
 If no current is applied to the solenoid coil, the left cylinder chamber is

vented, the right chamber pressurized. The piston rod is retracted.
 If current is applied to the solenoid coil, the directional control valve

switches. The left chamber is pressurized, the right chamber vented.
The piston rod extends.

BPLK 3/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

 When the current is interrupted, the valve switches back and the
piston rod retracts.

Figure 7.2 Actuation of a pneumatic cylinder a)Single acting b) Double-acting

BPLK 4/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.2 DC SOLENOID

The solenoid coils are chosen to match the signal control section of the electro-
pneumatic control system. If the signal control section operates with DC voltage of 24V,
for example, the corresponding type of coil should be chosen.

To ensure proper operation of the solenoid coil, the voltage supplied to it from the signal
control section must be within certain limits. For the 24V coil type, the limits are as
follows:

Minimum voltage: Vmin = 24V . (100% - 10%) = 24V . 0.9 = 21.6V
Maximum voltage: Vmax = 24V . (100% + 10%)= 24V . 1.1 = 26.4V

7.2.1 Protective circuit of a solenoid coil

The electric circuit is opened or closed by a contact in the signal control section
of the control system. When the contact is opened, the current through the
solenoid coil suddenly decays. As a result of the rapid change in current
intensity, in conjunction with the inductance of the coil, a very high voltage is
induced briefly in the coil. Arcing may occur at the opening contact. Even after
only a short operating time, this leads to destruction of the contact. A protective
circuit is therefore necessary.

Figure 7.3 shows the protective circuit for a DC coil. While the contact is closed,
current I1 flows through the solenoid and the diode is de-energized (figure 7.3(a)).
When the contact is opened, the flow of current in the main circuit is interrupted
(figure 7.3(b)). The circuit is now closed via the diode. In that way current can
continue flowing through the coil until the energy stored in the magnetic field is
dissipated.

As a result of the protective circuit, current IM is no longer subject to sudden
decay, instead it is continuously reduced over a certain length of time. The
induced voltage peak is considerably lower, ensuring that the contact and
solenoid coil are not damaged.

BPLK 5/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

Figure 7.3 Protective circuit of a solenoid coil

In addition to the protective circuit required for operation of the valve, further
auxiliary functions can be integrated in the cable connection, for example:
 Indicator lamp (lights up when the solenoid is actuated)
 Switching delay (to allow delayed actuation)

BPLK 6/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.3 ELECTRIC TO PNEUMATIC ENERGY CONVERSION

 Conversion systems are required if control systems are used, that operate with
compressed air and electricity as working media.

 Solenoid valves convert electrical signals into pneumatic signals.

Solenoid valve consist of:
 A pneumatic valve
 A coil that switches the valve

2

1M1
13

3 3
2 2

1 1

unactuated actuated

Figure 7.4 Electric to pneumatic energy conversion

BPLK 7/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.4 3/2 WAY SINGLE SOLENOID VALVE, NORMALLY
CLOSED/OPEN

Figure 7.5 shows two cross-sections of a directly controlled electrically actuated 3/2 way
valve.
 In its initial position, the working port 2 is linked to the exhaust port 3 by the slot in

the armature (figure 7.5(a)).
 If the solenoid is energized, the magnetic field forces the armature up against the

pressure of the spring (figure 7.5(b)). The lower sealing seat opens and the path is
free for flow from pressure port 1 to working port 2. The upper sealing seat closes,
shutting off the path between port 1 and port 3.
 If the solenoid coil is de-energized, the armature is retracted to its initial position by
the return spring (figure 7.5(a)). The path between port 2 and port 3 is opened and
the path between port 1 and port 2 closed. The compressed air is vented via the
armature tube at port 3.

Figure 7.5 3/2 way solenoid valve with manual override (normally closed)

BPLK 8/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

Figure 7.6 shows an electrically actuated 3/2 way valve, normally open. Figure 7.6(a)
shows the valve in its initial position, figure 7.6(b) actuated. Compared to the initial
position of the close valve (figure 7.5) the pressure and exhaust ports are reversed.

Figure 7.6 3/2 way valve with manual override (normally open)

BPLK 9/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.5 PILOT OPERATED SOLENOID VALVES

7.5.1 Pilot controlled directional control valve

In pilot controlled directional control valves, the valve piston is indirectly actuated.
 The armature of a solenoid opens or closes an air duct from port 1.
 If the armature is open, compressed air from port 1 actuates the valve piston.

Figure 7.7 explains the mode of operation of the pilot control.
 If the coil is de-energized, the armature is pressed against the lower sealing

seat by the spring. The chamber of the upper side of the piston is vented
(figure 7.7(a)).
 If the coil is energized, the solenoid pulls the armature down. The chamber
on the upper side of the piston is pressurized (figure 7.7(b)).

An electrical signal a) b)
is applied at the Armature Air duct
solenoid coil Valve piston

The solenoid coil
actuates the
pilot valve

The pilot control
actuates the valve

Figure 7.7 Pilot controlled directional control valve

BPLK 10/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.6 3/2, 4/2, 5/2 WAY SINGLE SOLENOID VALVE, PILOT OPERATED

7.6.1 3/2 way valve pilot controlled

Figure 7.8 shows two cross-sections of an electrically actuated pilot controlled
3/2 way valve.
 In its initial position, the piston surface is only subject to atmospheric

pressure, so the return spring pushes the piston up (figure 7.8(a)). Ports 2
and 3 are connected.
 If the solenoid coil is energized, the chamber below the valve piston is
connected to pressure port 1 (figure 7.8(b)). The force on the upper surface
of the valve piston increases, pressing the piston down. The connection
between ports 2 and 3 is closed, the connection between ports 1 and 2
opened. The valve remains in this position as long as the solenoid coil is
energized.
 If the solenoid coil id de-energized, the valve switches back to its initial
position.

A minimum supply pressure (control pressure) is required to actuate the pilot
controlled valve against the spring pressure. This pressure is given in the valve
specifications and lies – depending on type – in the range of about 2 to 3 bar.

2
1M1

13

a) b)

3 3
2 2

1 1

unactuated actuated

Figure 7.8 Pilot controlled 3/2 way solenoid valve (normally closed, with manually override)

BPLK 11/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.6.2 Pilot controlled 5/2 way valve

Figure 7.9 shows the two switching positions of an electrically actuated pilot
controlled 5/2 way valve.
 In its initial position, the piston is at the left stop (figure 7.9 (a)). Ports 1 and 2,

ports 4 and 5 are connected.
 If the solenoid coil is energized, the valve spool moves to the right stop

(figure 7.9(b)). In this position, ports 1 and 4, 2 and 3 are connected.
 If the solenoid is de-energized, the return spring returns the valve spool to its

initial position.

a)

1M1 42
(14)
53
84 1

1M1 (14) 84 5 4 1 2 3 42
b)

1M1 3
(14) 1

84 5

1M1 (14)

84 5 4 1 2 3

Figure 7.9 Pilot controlled 5/2 way solenoid valve.

BPLK 12/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.7 5/2 WAY DOUBLE SOLENOID VALVE, PILOT OPERATED

Figure 7.10 shows two cross-sections of a pilot controlled 5/2 way double solenoid valve.
 If the piston is at the right stop, ports 1 and 2 and 4 and 5 are connected (figure

7.10(a)).
 If the right solenoid coil is energized, the piston moves to the left stop and ports 1

and 4 and 2 and 3 are connected (figure 7.10(b)).
 If the valve is to be retracted to its initial position, it is not sufficient to de-energized

the right solenoid coil. Rather, the left solenoid coil must be energized.

If neither solenoid coil is energized, friction holds the piston in the last position selected.
This also applies if both solenoids coils are energized simultaneously, as they oppose
each other with equal force.

a) 4 2
1M1 1M2
(14) (12)
84 5 3 82
1

1M1 (14) 1M2 (12) 42

b) 84 5 4 1 2 3 82 1M1 1M2
(14) (12)
3
84 5 1 82

1M1 (14) 1M2 (12)

84 5 4 1 2 3 82

Figure 7.10 Pilot controlled 5/2 way double solenoid valve

BPLK 13/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.8 VALVE SWITCHING CHARACTERISTICS

7.8.1 Construction and mode operation

Electrically actuated directional control valve are switched with the aid of
solenoids. They can be divided into two groups:
 Spring return valves only remain in the actuated position as long as current

flows through the solenoid.
 Double solenoid valves retain the last switched position even no current flows

through the solenoid.

7.8.1.1 Initial position

In the initial position, all solenoids of an electrically actuated directional
control valve are de-energized and the solenoids are inactive. A double
solenoid valve has no clear initial position, as it does not have a return
spring.

7.8.1.2 Port designation

Directional control valves are also differentiated by the number of ports
and the number of switching positions. The valve designation results from
the number of ports and positions, for example:
 Spring return 3/2 way valve
 5/2 way double solenoid valve

BPLK 14/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

7.9 PNEUMATIC-ELECTRIC (PE) CONVERTER

14

14 14
unactuated actuated

Figure 7.11 Pneumatic-Electric Converter

7.9.1 Conversion of pneumatic signals into electric signals.
 The PE converter is actuated with compressed air.
 An electrical signal is generated if pressure reaches a preset value.
 The pressure of a pneumatic signal acts against an adjustable spring.
 A stem actuates an electrical switch contact if the pressure acting on a
diaphragm overcomes the spring force.
 The electrical switching element can be either a normally closed or normally
open contact or a changeover switch.

BPLK 15/16 DMV 1152/3152/4152/5152

SOLENOID VALVES

SUMMARY

In this chapter we have studied to :
1. E

REFERENCES

1. Fundamentals Of Pneumatic Control Engineering Text Book, Festo Didactic, 1989
2. Electro-Pneumatics Basic Level TP 201 Text Book, Festo, 1998
3. FluidSim Pneumatic Software Version 4.0

BPLK 16/16 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

CHAPTER 8: INTRODUCTION TO HYDRAULIC SYSTEM

INTRODUCTION

LEARNING OBJECTIVES

BPLK 1/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.1 IMPORTANT CHARACTERISTIC OF HYDRAULIC SYSTEM

8.1.1 Introduction to hydraulics
Over the past few decades, the field of hydraulics has expended into the
technical fields of mechanical engineering. The replacement of “water
hydraulics”, a principle known and used for centuries, by all hydraulics,
using mineral oils as an operational fluid, is something which grew out of
development of all-tight sealing elements on a plastic base. Hydraulic
systems often including electronics and pneumatics, can be constructed
to provide an extremely high performance and can be designed to be
adjustable and controllable. Automation in mechanical engineering has
recently come to rely on the interplay of hydraulically moved and
electrically and pneumatically controlled mechanical elements, based on
the principle.
Muscles from hydraulics, nerves from electricity.

8.1.2 Definitions of hydraulics
The science of the movements and conditions of balance of fluids is
designated as “hydraulics”. The science of the conditions of balance of
fluids under the effect of external forces (hydrostatics) and of flow laws
(hydrodynamics) combine as hydromechanics, forming the physical basis
for hydraulics. The word “hydraulics” includes all drive, regulating and
control devices, of which the movements and forces produced are
brought about with the aid of hydrostatic pressure. The hydraulic fluid is
the energy transfer agent; accordingly, in a hydraulic system, energy is
converted.

In other word, hydraulics means that, the generation of forces and motion
using hydraulic fluids. The hydraulic fluids represent the medium for
power transmission.

BPLK 2/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

The place held by hydraulics in (modern) automation technology illustrated the
wide range of applications for which it can be used.

A basic distinction is made between:
 Stationary hydraulics
 Mobile hydraulics

Mobile hydraulic systems move on wheels or tracks, for example, unlike
stationary hydraulic systems which remain firmly fixed in one direction. A
characteristic feature of mobile hydraulics is that the valves are frequently
manually operated. In the case of stationary hydraulics, however, mainly solenoid
valves are used. Other areas include marine, mining and aircraft hydraulics.

The following application areas are important for stationary hydraulics:
 Production and assembly machines of all types
 Transfer lines
 Lifting and conveying devices
 Presses
 Injection moulding machines
 Rolling lines
 Lifts

Typical application fields for mobile hydraulics include:
 Construction machinery
 Tippers, excavators, elevating platforms
 Lifting and conveying devices
 Agricultural machinery

BPLK 3/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.1.3 Hydro-mechanics

Hydro-mechanics

Hydrostatics Hydrodynamics
AF

S
p

Force effect through pressure Force effect through mass
area acceleration

Figure 8.1 Hydromechanics

Hydrostatic pressure: Is the pressure which rises above a certain level
in a liquid owing to the weight of the liquid
mass.

ps = h.ρ.g

ps = hydrostatic pressure (gravitational pressure) [Pascal,bar,N/m2]
h = level of the column of liquid [m]
ρ = density of the liquid [kg/m3]
g = acceleration due to gravity [m/s2]

BPLK 4/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Example 1. (Column)

h

Given that :
h = 300m
ρ = 1000 kg/m3
g = 9.81 m/s2 ≈ 10m/s2
ps = h.ρ.g

= 300 (1000) 10 [m.kg/m3.m/s2]
= 3,000,000 [kg/m.s2]
= 3,000,000 N/m2
ps = 3,000,000 Pa (30 bar)

BPLK 5/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Example 2. (Reservoir)

h

Given :
h = 15 m
ρ = 1000 kg/m3
g = 9.81 m/s2 ≈ 10 m/s2

ps = h.ρ.g
= 15 (1000) 10 [m . kg/m3 . m/s2]
= 150,000 [kg/ms2]
= 150,000 N/m2

ps = 150,000 Pa (1.5 bar)

8.1.4 The pressure p exists when a force F is imposed on an enclosed fluid
with a surface A.

p = F/A [N/m2]

p = pressure (bar / Pa / N/m2)
F = force (N)
A = area (m2)

BPLK 6/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Unit : 1Pa = 1 N/m2
1 bar = 100000 N/m2 = 105 Pa

Example 3.

A cylinder is supplied with 100 bar pressure, its effective piston surface is equal to 7.85
cm2. Find the maximum force which can be attained.

Given that:

p = 100 bar = 1000 N/cm2
A = 7.85 cm2

F =p.A
= 1000(7.85) [N.cm2/cm2]

F = 7850 N

Example 4.

A lifting platform is to lift a load of 15000 N and is to have a system pressure of 75 bar.
How large does the piston surface A need to be?

Given that:

F = 15000 N
p = 75 bar = 75 x 105 N/m2

A = F/p
= 15000/75 x 105 [N/N.m2]
= 0.002 [N . m2/N]

A = 0.002 m2 = 20cm2

BPLK 7/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Given that, = 10 cm2
A1 = 10,000 N
F

p = F/A
= 10,000/0.001 [N/m2]
= 100 x 105 Pa (100 bar)

Given that,

p = 100 x 105 Pa
A = 1 cm2 = 0.0001m2

F =p.A
= 100 x 105 x 0.0001 [Pa.m2]
= 1000 [N.m2/m2]
= 1000 N

8.1.4 Pressure Transmission

If a force F1 acts via an area A1 on an enclosed liquid, a pressure p is
produced which extends throughout the whole of the liquid (Pascal’ Law).
The same pressure applies at every point of the closed system.(see
figure 8.2)

p = F/A

BPLK Figure 8.2 Pressure transmission

8/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.1.5 Power Transmission

The same pressure applies at every point in a closed system. For the
reason, the shape of the container has no significance.

Figure 8.3 Power transmission

The following equation was:
p1 = F1/A1 and p2 = F2/A2

The following equation applies when the system is in equilibrium:

p1 = p2
When the two equations are balanced, the following formula is produced:

F1/A1 = F2/A2

BPLK 9/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Example 5.

Given that,
Load, m = 1500kg
Force due to weight F2 = m.g
F2 = 1500(10) [kg.m/s2]
= 15,000N

Given that, = 40cm2
A1 = 1200cm2
A2

BPLK 10/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Example 6

It has been proved that the force F1 of 100N is too great for actuation by hand lever.
What must the size of the piston surface A2 be when only a piston force of F1 = 100N is
available?

BPLK 11/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.1.6 Displacement Transmission

If a load is to be lifted a distance a distance s2 in line with the principle
described above, the piston P1 must displace a specific quantity of liquid
which lifts the piston P2 by a distance s2.

Figure 8.4 Displacement Transmission

The necessary displacement volume is calculated as follows:
V1 = s1 x A1 and V2 = s2 x A2

Since the displacement volumes are identical (V1 = V2), the following equation is
valid:

s1 x A1 = s2 x A2

The displacement of the piston is in inverse ratio to its area. This law can be
used to calculate the values s1 and s2. For example, for s2

and

BPLK 12/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Figure 8.5 Displacement Transmission -example

Given that: 40 cm2 Given that: 1200 cm2
A1 = 1200 cm2 A2 = 30 cm
A2 = 15 cm s1 = 0.3 cm
s1 = s2 =

BPLK 13/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.1.7 Pressure Transfer

Figure 8.6 Pressure transfer
The hydrostatic pressure p1 exerts a force F1 on the area A1 which is transferred
via the piston rod onto the small piston. Thus, the force F2 acts on the area A2
and produces the hydrostatic pressure p2. Since piston area A2 is smaller than
piston area A1, the pressure p2 is greater than the pressure p1. Here too, the
following law applies:

F1 = p1 x A1 and F2 = p2 x A2
p1 x A1 = p2 x A2
and

BPLK 14/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

In the case of the double-acting cylinder, excessively high pressures may be
produced when the flow from the piston rod area is blocked:

Figure 8.7 Pressure transfer by double-acting cylinder

Given that: 10 x 105 Pa Given that: 20 x 105 Pa
p1 = 1200 cm2 p1 = 100 x 105 Pa
A1 = 150 cm2 p2 = 8 cm2 = 0.0008 m2
A2 = A1 =

= 80 bar = 0.00016m2 = 1.6cm2

BPLK 15/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Flow rate is the term used to describe the volume of liquid flowing through
a pipe in a specific period of time. For example, approximately one minute is
required to fill a 10 liter bucket from a tap. Thus, the flow rate amounts to 10
l/min.

Figure 8.8 Flow rate
In hydraulics, the flow rate is designated as Q. The following equation applies:

Q = Flow rate [m3/s]

V = volume [m3]

t = time [s]

The equation for the volume (V) and the time (t) can be derived from the formula
for the flow rate. The following equation is produced:

BPLK 16/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Given that: 4.2 l/min
Q= 10 s
t=

V=

V = 0.7 l

Result:
A flow rate of 4.2 litres per minute produces a volume of 0.7 litres in 10 seconds.

Given that: 105 l
V= 4.2 l/min
Q=

t=

=
t = 25 min

Result:
25 minutes are required to transport a volume of 105 litres at a flow rate of 4.2
litres per minute.

BPLK 17/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.2 DESIGN OF A HYDRAULIC SYSTEM

Energy Transmission

Figure 8.2 Hydraulic system structure

BPLK 18/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.2.1 Structure of the hydraulic system

From the general point of view, the operation of a hydraulic system
includes:
 energy conversion
 energy transmission
 energy control

Initially, the mechanical energy, powering up the system is transformed
into hydraulic energy. The latter is transmitted, controlled and finally
converted back into mechanical energy. This is done by hydraulic
components, placed in a pre-defined sequence in order to achieve a
desired function. Figure 8.2 shows the basic hydraulic system and the
components from which it is built up.

Cross-sectional illustrate the functionality of the components, which is
given on the left part of the figure. The right part of the figure shows the
corresponding standard symbol of each component. A single thick line
denotes the pipe fitting of the components.

BPLK 19/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

8.2.2 Comparison of systems

There are other technologies besides hydraulics which can be used in the
context of control technology for generating forces, movements and
signals:
 Mechanics
 Electricity
 Pneumatics
It is important to remember here that each technology has its own
preferred application areas. To illustrate this, a table has been drawn up
on the next page which compares typical data for the three most
commonly used technologies – electricity, pneumatics and hydraulics.

This comparison reveals some important advantages of hydraulics:
 Transmission of large forces using small components, i.e great power

intensity
 Precise positioning
 Start up under heavy load
 Even movements independent of load, since liquids are scarcely

compress-slide and flow control valves can be used
 Smooth operation and reversal
 Good control and regulation
 Favourable heat dissipation.

Compared to other technologies, hydraulics has the following
disadvantages:
 Pollution of the environment by waste oil (danger of fire or accidents)
 Sensitivity to dirt
 Danger resulting from excessive pressures (severed lines)
 Temperature dependence (change in viscocity)
 Unfavourable efficiency factor

BPLK 20/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

Leakage Electricity Hydraulics Pneumatics
Environmental Contamination No disadvantages
influences Risk of explosion in apart from energy
certain areas, Sensitive in case of loss
Energy storage insensitive to temperature Explosion-proof,
Energy transmission temperature fluctuation, risk of insensitive to
fire in case of temperature
Operating speed Difficult, only in leakage
Power supply costs small quantities Limited, with the Easy
Linear motion using batteries help of gases
Unlimited with Up to 1000m flow
Rotary motion power loss Up to 100m flow rate v = 20-40 m/s,
Positioning rate v = 2.6m/s, signal speed 20-
accuracy Low signal speed up to 40m/s
Stability 0.25 1000m/s v = 1.5m/s
v = 0.5m/s Very high
Forces Difficult and High
expensive, small 2.5
forces, speed 1 Simple using
regulation only Simple using cylinders, limited
possible at great cylinders, good forces, speed highly
cost speed control, very load dependent
Simple and powerful large forces
Simple, inefficient,
Precision to ± 1µm Simple, high turning high speed
and easier to moment, low speed Without load
achieve Precision up to ± 1 alteration precision
µm can be achieved of 1/10 mm possible
Very good values depending on
can be achieved expenditure Low air is
using mechanical High since oil is compressible
links almost
incompressible in Protected against
Overload able addition, the overload, forces
ineffective since pressure level is limited by
mechanical linkage considerably higher pneumatic pressure
connected on the than for pneumatics and cylinder
load side, very high Protected against diameter F < 30kN
forces, can be overload, with high at 6 bar
realized system pressure of
up to 600 bar, very
large forces can be
generated F <
3000kN

BPLK 21/22 DMV 1152/3152/4152/5152

INTRODUCTION TO HYDRAULIC SYSTEM

SUMMARY

In this chapter we have studied to :

REFERENCES

1. Fundamentals Of Pneumatic Control Engineering Text Book, Festo Didactic, 1989
2. Electro-Pneumatics Basic Level TP 201 Text Book, Festo, 1998
3. FluidSim Pneumatic Software Version 4.0

BPLK 22/22 DMV 1152/3152/4152/5152

HYDRAULIC SYMBOLS

CHAPTER 9: HYDRAULIC SYMBOLS

INTRODUCTION

LEARNING OBJECTIVES

BPLK 1/15 DMV 2152/3152/4152/6152

HYDRAULIC SYMBOLS

9.1 BASIC SYMBOLS / FUNCTION SYMBOLS / OPERATION
MODEL

BPLK 2/15 DMV 2152/3152/4152/6152

HYDRAULIC SYMBOLS

BPLK 3/15 DMV 2152/3152/4152/6152

HYDRAULIC SYMBOLS

BPLK 4/15 DMV 2152/3152/4152/6152