LECTURE-18
HYDRAULIC SYSTEM COMPONENTS
Dr. Dhafer Manea Hachim AL-HASNAWI
Assist Proof
Al-Furat Al-Awsat Technical University
Engineering Technical College / Najaf
email:coj.dfr@atu.edu.iq
Learning Objectives
After the completion of this module, the
student will be able to:
Identify the fundamental parts of a hydraulic
system.
Observe how hydraulic components can be
connected together to construct a hydraulic circuit.
Identify the main components of the hydraulic work
station TP 501.
Explain the main parts of the hydraulic power pack.
Explain the importance of using standard hydraulic
symbols.
3 Hydraulic system components
All industrial hydraulic systems consist of the following basic components
Power input device:
The pump and motor together are called the power input device; the
pump provides power to the hydraulic system by pumping oil from the
reservoir/tank. The pump’s shaft is rotated by an external force which is
most often an electric motor as illustrated in Fig 1.5.
Control device
Power input device Tank
Valve
Pump Liquid
Motor
Pipes or tubes
Power output device
Actuator
3 Hydraulic system components
Control device: Valves control the direction, pressure,
and flow of the hydraulic fluid from the pump to the
actuator/cylinder.
Power output device: The hydraulic power is
converted to mechanical power inside the power
output device. The output device can be either a
cylinder which produces linear motion or a motor
which produces rotary motion.
Liquid: the liquid is the medium used in hydraulic
systems to transmit power. The liquid is typically oil,
and it is stored in a tank or reservoir.
Conductors: The conductors are the pipes or hoses
needed to transmit the oil between the hydraulic
components.
Tip: “Watch the hydraulic system video”
3.1 Hydraulic power pack
The hydraulic power pack combines the pump,
the motor, and the tank. The hydraulic power
pack unit provides the energy required for the
hydraulic system. The parts of the hydraulic
power pack unit are shown in Fig. 1.6.
.1.6: The main parts of the hydraulic power pack
3.2 Activity 1: Hydraulic station component
identification
In this activity, you will identify the
components of the Festo Hydraulic work
station in your lab:
Locate the power pack unit and identify its
parts.
Locate the out put device (actuators).
Locate the control devices (valves).
Locate the conductors (hoses).
3.3 Hydraulic symbols
The way hydraulic components direct
and control liquid around a circuit can be
complex.
This would cause difficulty for one (a) Electric motor
engineer explaining to another engineer
how the circuit works.
A common form of representing
components and circuits is used to more
easily explain what is happening.
(b) Hydraulic pump
This form of representation uses
common symbols to represent
components and the ways in which they
are connected to form circuits. Fig. 1.7
shows some of the components’ symbols (c) Tank or reservoir
used in hydraulics.
The symbols don’t show the component
construction, or size, however, it is a
standard form that is used by all
engineers to represent that specific (d)Pressure relief valve
component.
Fig.1.7: (a) Electric motor. (b) Hydraulic pump.
(c) Tank or reservoir. (d) Pressure relief valve.
Power Pack Symbols
The simplified and detailed symbols of the
hydraulic power pack are shown in Fig. 1.8.
(a) Simplified
(b) Detailed
Fig.1.8: (a) Simplified symbol of the hydraulic power pack.
4- Fundamental laws of Hydraulics
All hydraulic systems operate
following a defined relationship
between area, force and pressure.
Laws have been established to
explain the behavior of hydraulic
systems.
Hydraulic systems use the ability
of a fluid to distribute an applied
force to a desired location.
4- Fundamental laws of Hydraulics
4.1 Pressure
When a force (F) is applied on an
area (A) of an enclosed liquid, a
pressure (P) is produced as
shown in Fig.
Pressure is the distribution of a
given force over a certain area.
Pressure can be quoted in bar,
pounds per square inch (PSI) or
Pascal (Pa) .
4.1 Pressure
Where
Force is in newtons (N) and
Area is in square meters (m2).
1 Pascal (Pa) =1 N/m2.
1 bar= 100,000 Pa= 105 Pa.
10 bar= 1 MPa (mega Pascals)
4.1 Pressure
If the pressure is calculated using a force in
Newton, and area in square millimeters, the
pressure in bar can be calculated.
Example 1-1.
A cylinder is supplied with 100 bar pressure; its
effective piston surface is equal to 700 mm2.
Find the maximum force which can be
attained.
P= 100 bar = 100X100000 N/m2.
A= 700/1000000=0.0007 m2.
F= P.A= 100X100000X0.0007= 7,000 N
4.2 Pascal’s Law
Pascal’s law states that:
“The pressure in a confined
fluid is transmitted equally to
the whole surface of its
container”
When force F is exerted on
area A on an enclosed
liquid, pressure P is
produced. The same
pressure applies at every
point of the closed system
as shown in Fig. 1.10a. Fig.1.10: (a) Pascal’s law.
4.2 Pascal’s Law
Fig.1.10b shows that, if a downward force is
applied to piston A, it will be transmitted
through the system to piston B.
According to Pascal’s law, the pressure at
piston A (P1) equals the pressure at piston B
(P2)
Piston A
Piston B
P1 P2 Fig.1.10: (b)Power transmission
4.2 Pascal’s Law
P1 P2
Fluid pressure is measured in terms of the force exerted per
unit area.
F
P
A F1
P1
A1
F2
P2
A2
F1 F2
A1 A2
The values F1, A2 can be calculated using the following formula:
F1
A1 F2 A1 F2
, and A2
A2 F1
4.2 Pascal’s Law
Example 1-2.
In Fig.11, find the weight of
the car in N, if the area of
piston A is 0.0006m2, the
area of piston B is 0.0105 m2,
and the force applied on
piston A is 500 N. Piston A
Pisto
Solution:
P1 P2
F1 F2 F1 A2 500 0.0105
F2 F2
A1 A2 A1 0.0006
F2 8750 N 8.75 kN
4.2 Pascal’s Law
Piston A
Example 1-3. Piston B
In Fig 1.11, if the weight of the car is 10,000 N, the
diameter of piston A is 0.01 m, and the force applied
on piston A is 250 N. Calculate the area of piston B.
Solution:
1. Calculate the area of piston A, the piston shape is
circular as shown in Fig. 1.10a, accordingly the area
will be calculated using the following formula.
2 2
D (0.01)
A1 3.14 0.0000785 m 2
4 4
F1 250 N F2 10,000 N
4.2 Pascal’s Law
2. Apply Pascal’s law
F1 F
P1 P2 A1
2
A2
3. Use Pascal’s law to calculate the area of piston B
A1 F2
A2
F1
2
(D )
A2 2 0.003140m 2
4
0.0000785 10,000
A2 0.00314m 2
250
4.3 Liquid flow
4.3.1 Flow rate versus flow velocity
The flow rate is the volume of fluid that moves through the system in a
given period of time.
Flow rates determine the speed at which the output device (e.g., a
cylinder) will operate.
The flow velocity of a fluid is the distance the fluid travels in a given
period of time.
These two quantities are often confused, so care should be taken to
note the distinction. The following equation relates the flow rate and
flow velocity of a liquid to the size (area) of the conductors (pipe,
tube or hose) through which it flows.
Q =V x A
Where:
Q= flow rate ( m³ /s )
V= flow velocity (m / s )
A= area (m² )
4.3 Liquid flow
This is shown graphically in Fig. 1.11. Arrows
are used to represent the fluid flow. It is
important to note that the area of the pipe or
tube being used.
Q, V
A
Fig.1.11: Flow velocity and flow rate
4.3 Liquid flow
Example 1-4.
A fluid flows at a velocity of 2 m/s through a pipe
with a diameter of 0.2 m. Determine the flow
rate.
Solution:
1. Calculate the pipe area
2
D (0.2) 2
A 3.14 0.0314 m 2
4 4
2. Calculate the flow rate
m3
Q V A Q 2 0.0314 0.0628
Sec
4.3.2 The continuity equation
Hydraulic systems commonly have a pump that
produces a constant flow rate. If we assume that
the fluid is incompressible (oil), this situation is
referred to as steady flow. This simply means
that whatever volume of fluid flows through one
section of the system must also flow through any
other section. Fig. 1.12 shows a system where
flow is constant and the diameter varies
A1 V1
A2 V2
Q1 Q2
Fig.1.12: Continuity of flow.
4.3.2 The continuity equation
The following equation applies in this system:
Q1 Q2
Therefore,
V1 A1 V2 A2
The following example illustrates the significance
of the continuity equation shown above.
4.3.2 The continuity equation
Example 1-5.
A fluid flows at a velocity of 0.2 m/s at point 1 in
the system shown in Fig. 1.12. The diameter at
point 1 is 50mm and the diameter at point 2 is 30
mm. Determine the flow velocity at point 2. Also
determine the flow rate in m/s.
1. Calculate the areas
(50 10 3 ) 2
2
D
A1 1 3.14 * 1.963 10 -3 m 2
4 4
(30 10 3 ) 2
2
D2
A2 3.14 * 7.068 10 - 4 m 2
4 4
4.3.2 The continuity equation
2. Calculate the velocity at point 2
Q1 Q2
Therefore,
V1 A1 V2 A2
A1 1.963 10 -3
V2 V1 0.2 0.55m / s
A2 7.068 10 -4
3. Calculate the flow rate in m/s
Q1 V1 A1 0.2 1.963 10 -3 3.926 10 4 m 3 s
4.3.2 The continuity equation
The example shows that in a system with a
steady flow rate, a reduction in area (pipe size)
corresponds to an increase in flow velocity by
the same factor. If the pipe diameter increases,
the flow velocity is reduced by the same factor.
This is an important concept to understand
because in an actual hydraulic system, the pipe
size changes repeatedly as the fluid flows
through hoses, fittings, valves, and other
devices.
5 Reading the pressure gauge
The pressure gauge indicates the amount of
pressure in a system. Technicians read these
gauges to determine if a machine is operating
properly.
Most pressure gauges have a face plate that
is graduated either in US units (psi) or SI units
(Pascal or bar) note that;
1 bar=0.1 mega pascals as explained
5 Reading the pressure gauge
A pointer rotates on the
graduated scale as the
pressure changes to Face plate
indicate the pressure in
the system. The pressure Pointer
gauge used in the
hydraulic power pack is
shown in Fig. 1.13. The
outer black scale indicates
pressure units of bar, and psi
bar
the inner red scale
indicates pressure units in Minimum
reading
SI units US units Maximum
reading
psi
Fig. 1.13: A pressure gauge.
5 Reading the pressure gauge
Each scale is graduated with a series of numbers ranging
from 0 to a maximum number. In case of the gauge shown, it
is graduated from 0 to a maximum reading of 100 bar or a
maximum reading of 1450 psi. The maximum reading is
always called the range of the gauge.
To read the pressure gauge, you only need to read the inner
red scale or the outer red scale to which the pointer points. If
the pointer points to a position between the two numbers,
you read the gauge to the closest graduation.
In the bar scale there are 4 graduations between 0 and 20;
this means the value of each graduation is 20/4=5 bar. In the
psi scale there are 4 graduations between 0 and 200; this
means the value of each graduation is 200/4=50 psi.
5.1 Activity 2: Setting the hydraulic
pressure to 30 bar.
Procedures:
1- Switch on the electrical power supply
first and then the hydraulic power
pack.
2- Use the pressure relief valve to set the
pressure to 30 bars.
3- While you are adjusting the pressure
observe the pressure gauge.
4- When the pressure gauge indicates 30
bar, switch off the hydraulic power
pack first, and then the electrical
power supply
Fig. 1.13: The hydraulic
power pack.
For more information, refer to the movie
section
