Chapter two fluid power system lecture note

CHAPTER 2
REVIEW OF FLUID
MECHANICS
1
5/10/2021 By : - Hailemariam G.

2
2.1 Introduction
Contents
2.3 Energy transfer
2.4 Pascal’s law and application
2.5 Conservation of energy
2.6 Continuity equation
2.7 Bernoulli’s equation
2.8 Frictional losses in hydraulic pipelines and fittings
2.9 Losses in valves and fittings
2.2 Fluid properties

2.1 Introduction
A hydraulic fluid has four primary functions
• Transmit power
• Lubricate moving parts
• Seal clearances between moving parts
• Dissipate heat
To accomplish the above, they should have the
following properties: Good lubricity; ideal viscosity;
chemical stability; compatibility with system
materials; fire resistance; good heat transfer
capability; low density; and the like
3

Need to be changed periodically due to fluid
breakdown or contamination (viscosity and acidity
increase).
Now a days, a hydraulic fluid test kit that provide a
quick and easy method to test hydraulic system:
• viscosity
• water content
• foreign particle contamination level.
4

Gases
• Gases are fluids that are readily compressible.
• Their volume will vary to fill the vessel containing
them
Air is the only gas commonly used in fluid power:
• Inexpensive and readily available
• Fire resistant
• Not messy
• Can be exhausted back into the atmosphere
5

2.2 Fluid properties
Properties to be dealt with include density,
pressure, compressibility, viscosity, and viscosity
index.
Density: designated by ρ = m/V (kg/m3)
Specific weight: γ=W/V (N/m3) = ρg
Specific gravity: ratio-density comparison with
that of water SG=ρ/ρH2O.
Pressure: Force (normal) per unit area (N/m2=Pa,
larger pressures: kPa, MPa, GPa)
In the case of any liquid, pressure at the bottom of
the column: p=γH.
Liquid column height or head (H)=p/γ
6

Atmospheric Pressure: surface of earth is at the
bottom of sea of air with heights of hundred’s of km.
It will exert a pressure-the so-called atmospheric
pressure which varies with elevation. Standard
atmospheric pressure at sea level and 15oC is given
as 101.325kPa.
Gage and absolute pressure
• Gage pressure: measured relative to the surrounding
atmospheric pressure.
• Absolute pressure: Measured relative to perfect
vacuum.
pabs = pgage + patm
7

Fig.2.1. Schematic representation of gage and absolute
pressures
8

Bulk Modulus: a measure of incompressibility
Typical value for oil, 1.72x106 kPa
Viscosity: a measure of fluids resistance to flow.
Two types
• Dynamic (Absolute) viscosity
Often expressed in CGS system as dyn s/cm2 where
1N = 105 dyn
)
(
/
kPa
V
V
p


−
=

SI
in
s
Pa
m
s
N
dy
du
)
/
(
/
=
= 2


9

1 dyn s/cm2 = 1 poise
A more convenient unit is the centipoise (cP)
• Kinematic viscosity
ν = μ/ρ (m2/s in SI units)
In CGS units 1cm2/s is called a stoke. A smaller
value is the centistoke (cS).
Measurement of Viscosity:
There are several ways of determining the viscosity
of fluids in the laboratory. The more common one is
the Saybolt viscometer given in units of Saybolt
Universal Seconds (SUS)
10

A relationship between SUS and cS
SUS
t
t
t
cS
SUS
t
t
t
cS
100
135
220
0
100
195
226
0

−
=

−
=
,
.
)
(
,
.
)
(


11

Viscosity Index
Viscosity index (VI) is a relative measure of an
oil’s viscosity change with change in temperature. A
good property of oil is that which brings little change
of viscosity with temperature. These are said to have
high VI. The VI of a hydraulic oil can be found by
using fig-chp2Fig2.2a.pptx
where
L = Viscosity in SUS of 0 – VI oil at 100oF
U = Viscosity in SUS of unknown – VI oil at 100oF
H =Viscosity in SUS of 100 – VI oil at 100oF 12
100
x
H
L
U
L
VI
−
−
=

This requires the determination of the viscosities of
the reference oils (0 and 100 VI) and the unknown oil
from actual tests.
The VI values can be greater than 100. As relative
values are required, the common point for all is at
T=210oF fig-chp2Fig2.2b.pptx
At 100oF: oil A, L = 2115 SUS, oil B, U = 1551
SUS and oil C, H = 986 SUS.
Oils A and C are reference oils.
VI (oil B)
13
50
100
x
986
2115
1551
2115
100
x
H
L
U
L
=
−
−
=
−
−
=

2.3 ENERGY TRANSFER
Transfer of energy is key in the operation of
hydraulic systems. fig-chp2Fig2.3.pptx gives a block
diagram illustrating how energy is transferred
throughout the hydraulic system.
The prime mover (electricity or combustion
engine) delivers energy to a pump. Outlet from
pump-fluid with high pressure and velocity. Directed
to actuators through pipelines and valves to generate
linear or rotary motions to drive external loads.
During this process energy is lost due to friction and
shows up as heat.
14

Conservation of energy dictates as
Energyin – Energylost = Energyout (Nm = J)
Power is the rate of doing work or spending
energy (J/s = W)
The rate at which the prime mover adds energy to
the pump equals the power input to the hydraulic
system. Likewise, the rate at which the actuator
delivers energy to the external load equals the
power(P) out put of the hydraulic system
For linear motion P = Fv
For rotary motion P = Tω
kW and hp are the common units for power 1 hp
=0.746 kW
15

Efficiency defined by
The efficiency of any system or component is
always less than 100% and is calculated to determine
power losses.
In hydraulic systems these losses are due to fluid
leakage past close-fitting parts, fluid friction due to
fluid movement, and mechanical friction due to the
rubbing of mating parts.
power
input
power
output
=

16

2.4 Pascal’s law and application
Pressure applied to a confined fluid is transmitted
undiminished in all directions throughout the fluid
and acts perpendicular to the surfaces in contact with
the fluid.
This results in transmission and multiplication of
force. Looking at the simple hydraulic jack fig-
chp2fig2.4.pptx .
2
2
2
1
1
1
A
F
p
A
F
p =
=
=
17

A force multiplication factor is achieved as:
Next, the piston stroke ratio S2/S1 equals the
piston area ratio A1/A2 .
The equality of energyin and energyout will give :
F1S1=F2S2
This result occurs because the force multiplication
factor equals the motion reduction factor.
18
1
2
1
2
A
A
F
F
=
2
1
1
2
A
A
S
S
=

Applications
• Hand operated hydraulic jack
fig-chp2fig2.5.pptx shows a simple jack.
A downward force applied forces the oil into the
piston
( opening the right check valve and closing the left
check valve) with a force multiplied by the ratio of
the areas of the pistons.
19

• Air-to-Hydraulic Pressure Booster
Here pneumatic pressure (low) is raised
considerably through the usage of small piston on the
hydraulic side. Force is transmitted directly. fig-
chp2fig2.6.pptx shows one arrangement.
As shown in the figure, application where an air-
to-hydraulic pressure boaster is supplying high-
pressure oil to a hydraulic cylinder whose short
stroke piston is used to clamp a workpiece to a
machine tool table.
20

2.5 Conservation of energy
Three different forms of energy to be considered here:
Potential energy: Energy due to elevation-( mgZ)
Kinetic energy: Energy due to translation –(mV2/2)
Flow energy: Energy due to pressure –(mp/ρ)
All have units of Nm.
Conservation of energy requires the total energy of
a fluid at any section to be the same although there
may be a conversion of energy from one form into
another.
= constant
2
2
V
W
p
W
WZ
Et +
+
=

21

2.6 Continuity equation
For steady flow in a pipeline, the mass flow rate
at any section is the same.
or ρ1A1V1 = ρ2A2V2
For incompressible fluids the density stays the
same. Then the continuity equation can be expressed
in terms of volume flow rate as:
For circular pipes the above gives a relation
between velocity and diameter as:
2
1 m
m 
 =
22
)
s
/
m
(
Q
V
A
V
A
Q 3
2
2
2
1
1
1

 =
=
=
2
1
2
2
1
D
D
V
V








=

2.7 Hydraulic power
A hydraulic power is defined as the power
delivered by a hydraulic fluid to a load-driving
device. Consider fig-chp2fig2.7.pptx where a
piston is moving against a resisting force.
• How do we determine how large a piston diameter is
required for the cylinder?
For a known inside pressure p inside the cylinder
the diameter or area required to balance this resisting
force will be
From pA = Fload A = Fload/p
23

• What is the pump flow rate required to drive the
piston in a specified time?
The volume flow rate from the pump is the same
as the volumetric displacement rate of the cylinder
given by
= AV (m3/s) V = velocity of piston = stroke/time
• How much hydraulic power does the fluid deliver to
the cylinder?
Power= energy/time = (FS)/t = (pA)S/t = pAV =
(S=stroke, unit for power is kW if p is in kPa)
Commonly horsepower is used for power and the
equivalence is 1 hp=0.746 kW 24
Q

Q
p 

When all the powers in the hydraulic system are
involved, the picture will be as shown in fig-
chp2fig2.8.pptx with the corresponding powers.
example-chp2example2.1.docx
example-chp2example2.2.docx
2.7 Bernoulli’s equation
Elementary fluid mechanics shows that pressure at a
depth of H in a fluid is given by
p = γH (Pa)
This allows the expression of pressure as a height of
liquid column termed as head
H = p/γ = p/ρg (m)
25

Bernoulli equation is a conservation of energy
expression given by (from state 1 to state 2):
Division by the mass gives :
The above is expression of energy as a head term.
26
g
Wv
p
W
WZ
g
Wv
p
W
WZ
2
2
2
2
2
2
2
1
1
1 +
+
=
+
+


W
g
v
p
Z
g
v
p
Z
2
2
2
2
2
2
2
1
1
1 +
+
=
+
+



The above equation has to be modified to include
the frictional losses along the line considered. Some
of the input energy is lost as heat due to friction and
this will also be shown as head loss.
Consider the energy supplied by the hydraulic
pump (Hp) to the fluid entering section 1 and the fluid
expends energy (Hm) to drive the actuator and which
in return drives an external load at section 2. If the
loss on the line between sections 1 and 2 is HL, then
the energy equation becomes
g
V
p
Z
H
H
H
g
V
p
Z L
m
p
2
2
2
2
2
2
2
1
1
1 +
+
=
−
−
+
+
+


27

2.8 Frictional losses in hydraulic pipelines and
fittings
A typical hydraulic system is shown in fig-
chp2fig2.10.pptx.
The resistance to flow is essentially a measure of
the viscosity of the fluid. The greater the viscosity of
a fluid, the less readily it flows and the more energy
is required to move it. This energy is loss because it
is dissipated into heat and thus represents wasted
energy. Energy losses occur in valves and fittings.
Examples are bends, couplings, tees, elbows, filters
and strainers.
28

In order to keep all energy losses in a fluid power
system to a minimum level, the proper selection of
the sizes of the pipes, valves and fittings is essential.
In general, the smaller the pipe diameter as well as
valve and fitting size, the greater the losses.
However, using large diameter pipes, valves and
fittings results in greater cost and poor space
utilization.
Thus, the selection of component sizes represents
a compromise between energy losses and component
cost and space requirements.
29

The energy equation and the continuity equation can
be used to perform complete analysis of a fluid power
system. Like, the pressure drops, flow rates, and
power losses for all components of fluid power
system
Whether the flow is laminar or turbulent also affects
the pipeline loss.
Reynods No. is used to check whether the flow is
laminar or turbulent
Re = (DVρ)/μ =DV/ν
Laminar flow Re < 2000
Transition : 2000<Re<4000
Turbulent flow: Re > 4000
30

Head losses in pipes can be found using Darcy’s
equation
where f = the friction factor (dimensionless)
L= length of pipe (m)
D= pipe inside diameter
V= average fluid velocity (m/s)
g= acceleration of gravity (m/s2)
g
2
V
D
L
f
H
2
L =
31

Frictional losses in laminar flow:
Where f is the friction factor and for laminar flow
is given by
g
2
V
D
L
f
H
2
L =
Re
64
f =
32














=
g
v
D
L
N
H
R
L
2
64 2

For turbulent flow the friction factor is also a
function of the surface roughness (ε) of the pipe,
better referred as relative roughness given by
Relative roughness
With the knowledge of Re, and relative roughness,
Moody chart can be used to determine f. Typical
roughness heights are given in Table 2.1
33
D

=

Table 2.1
Type of pipe ε (mm)
Glass or plastic smooth
Drawn tubing 0.0015
Commercial steel or 0.046
wrought iron
Asphalted cast iron 0.12
Galvanized iron 0.15
Cast iron 0.26
Riveted steel 1.8
34

2.9 Losses in valves and fittings
Additional losses occur due to valves and fittings
on the line. Experiments have shown that these
losses are proportional to the kinetic energy head
given as:
The constant proportionality (K) is called the K
factor (also called loss coefficient) of the valve or
fitting. Common values for the K factor are given in
Table 2.2
35
g
V
K
HL
2
2
=

Table 2.2 K factors of common valves and fittings
36

The losses in valves and fittings can be determined
using Darcy’s relation by replacing the length with
the so-called equivalent length determined as
follows:
This will give
The total head loss on a line with valves and fittings
will be:
g
2
V
D
L
f
g
2
V
K
H
H
2
e
2
)
pipe
(
L
)
fitting
or
vlave
(
L
=
=
f
KD
Le =
37
( )

+
= e
2
L L
L
g
2
V
D
1
f
H

Hydraulic circuit analysis
The hydraulic circuit analysis will mainly involve
the determination of :
• head and power losses
• flow velocities
• discharges and
• diameters of pipes etc.
38

Fig. 2-2a Typical curves for a viscosity index test

Fig. 2-3 Block diagram of hydraulic system showing major
components along with energy input and output terms.

Fig. 2-4 Operation of a simple hydraulic jack

Fig.2.5 Principles of operation of a hydraulic jack

Fig.2.6 An air to hydraulic pressure booster system

Fig.2.7 Cylinder example for determining hydraulic power

Fig.2.8 Conversion of power from input electrical to mechanical
to hydraulic to output mechanical in a hydraulic system.

Fig. 2.10 Energy transfer into and out of a hydraulic system includes
energy loss (in the form of heat) due to frictional fluid flow
through pipes, valves, and fittings.

Chapter two fluid power system lecture note

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