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Unformatted text preview: ME375 Handouts Hydraulic (Fluid) Systems
• Fundamental Principles
– Pascal’s Law: Pressure applied to the
pp
fluid is transmitted equally in all directions
– Transmit forces (incompressible) • Applications
– High force
– Heavy loads
– Precise motion School of Mechanical Engineering
Purdue University ME375 Hydraulic  1 Hydraulic System Modeling
• Basic Modeling Elements
–
–
–
– Resistance
Capacitance
Inertance
Pressure and Flow Sources • Interconnection Relationships
– Compatibility Law
Law
– Continuity Law • Derive Input/Output Models
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Purdue University ME375 Hydraulic  2 1 ME375 Handouts Key Concepts
• q : volumetric flow rate [m3/sec]
[m
2]
[N/
[N/m
• p : pressure [N/m
• v : volume [m3]
[m (
(
( )
)
) The analogy between a hydraulic system and an electrical system will be used
often.
often. Just as in electrical systems, the flow rate (current) is defined to be the
time rate of change (derivative) of volume (charge):
(charge):
q= d
&
v=v
dt The pressure, p, used in this chapter is the absolute pressure. You need to be
careful in determining whether the pressure is the absolute pressure or gage
pressure, p*. Gage pressure is the difference between the absolute pressure and
the atmospheric pressure, i.e.
p * = p − patmospheric
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Purdue University ME375 Hydraulic  3 Basic Modeling Elements
• Fluid Resistance
• Fluid Capacitance
Describes any physical element with
Describes any physical element with the
the characteristic that the pressure
characteristic that the rate of change in
drop, Δp, across the element is
pressure p in the element is proportional
proportional to the flow rate q.
to the difference between the input flow
rate, qIN , and the output flow rate, qOUT .
p1 + Δp − p2
+ Δp −
p1 q R p2 R Δp = p1 − p2 = p12 = R ⋅ q
1
1
q = Δp = p12
R
R pC qIN ( – Orifices, valves, nozzles and
friction in pipes can be modeled
as fluid resistors.
resistors. + pCr − pref q C qOUT qIN − qOUT C ) d
&
C
pC − pref = C ⋅ pCr = q IN − qOUT
dt 14243
pCr – Hydraulic cylinder chambers, tanks,
and accumulators are examples of
fluid capacitors.
capacitors. School of Mechanical Engineering
Purdue University ME375 Hydraulic  4 2 ME375 Handouts Basic Modeling Elements
Ex: Consider an open tank with a constant
crosscrosssectional area, A: Ex: Will the effective capacitance change if in
the previous open tank example, a load
mass M is floating on top of the tank? pr pr
h M pC qIN qOUT h
pC qIN pC = qOUT qIN − qOUT =
&
pCr = 1
(qIN − qOUT ) =
C =
⇒ C=
School of Mechanical Engineering
Purdue University ME375 Hydraulic  5 Basic Modeling Elements
• Fluid Inertance (Inductance)
Ex: Consider a section of pipe with crosscrosssectional area A and length L, filled
Describes any physical element with the
with fluid whose density is ρ :
characteristic that the pressure drop, Δp ,
p1
+ Δp −
p2
across the element is proportional to the
rate of change (derivative) of the flow
q
L
A
rate, q.
p1 + Δp − p2 p1 q
I Δ p = p12 = ( p1 − p2 ) = I Start with force balance: F = ma
ma + Δp −
I p2
q d
&
q = I ⋅q
dt – Long pipes are examples of fluid
inertances.
inertances.
Q: What will happen if you suddenly shut off one end
of a long tube ?  (Water Hammer effect)
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Purdue University ⇒ I= ρL
A
ME375 Hydraulic  6 3 ME375 Handouts Basic Modeling Elements
• Pressure Source (Pump)
– An ideal pressure source of a
hydraulic system is capable of
maintaining the desired pressure,
regardless of the flow required for
what it is driving.
driving. p1 − pS + p2 pS q p21 = p2 − p1 = pS • Flow Source (Pump)
– An ideal flow source is capable of
delivering the desired flow rate,
regardless of the pressure
required to drive the load.
load. p1 p2 qS q q = qS
School of Mechanical Engineering
Purdue University ME375 Hydraulic  7 Interconnection Laws
• Continuity Law
• Compatibility Law
– The algebraic sum of the flow rates
– The sum of the pressure drops
at any junction in the loop is zero.
around a loop must be zero.
– This is the consequence of the
– Similar to the Kirchhoff voltage
conservation of mass.
mass.
law.
law.
∑ Δp j = Closedpij = 0
∑
– Similar to the Kirchhoff current law.
Closed
Loop p1 ∑ Loop B A qj = 0 Any
Node p2
or
C ∑q IN = pr School of Mechanical Engineering
Purdue University OUT q2 q1 p r 1 + p1 2 + p 2 r = 0 ∑q q1 + q 2 = q o
qo
ME375 Hydraulic  8 4 ME375 Handouts Modeling Steps
• Understand System Function and Identify Input/Output
Variables
• Draw Simplified Schematics Using Basic Elements
Si
El
• Develop Mathematical Model
–
–
–
–
– Label Each Element and the Corresponding Pressures.
Label Each Node and the Corresponding Flow Rates.
Write Down the Element Equations for Each Element.
Apply Interconnection Laws.
Check that the Number of Unknown Variables equals the Number of
Equations.
Equations.
– Eliminate Intermediate Variables to Obtain Standard Forms:
• Laplace Transform
• Block Diagrams
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Purdue University ME375 Hydraulic  9 Example
Derive the input/output model for the following fluid system. The pump supplies a constant
system.
pressure pS to the system and we are interested in finding out the volumetric flow rate
through the nozzle at the end of the pipe.
pipe.
pr
Valve pr pS pr • Label the pressures at nodes and flow rates
• Write down element equations: School of Mechanical Engineering
Purdue University ME375 Hydraulic  10 5 ME375 Handouts Example
• No. of unknowns and equations: laws:
• Interconnection laws: • Eliminate intermediate variables and obtain I/O model: Q: Can you draw an equivalent electrical circuit of this hydraulic system ? Note that pressure is
analogous to voltage and flow rate is analogues to electric current.
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Purdue University ME375 Hydraulic  11 School of Mechanical Engineering
Purdue University ME375 Hydraulic  12 Example
Electrical Analogy: 6 ME375 Handouts Motion Control of Hydraulic Cylinders
Hydraulic actuation is attractive for applications
when large power is needed while maintaining a
reasonable weight. Not counting the weight of
weight.
the pump and reservoir, hydraulic actuation has
the edge in powertoweight ratio compared with
powertoother cost effective actuation sources. Earth
sources.
moving applications (wheel loaders, excavators,
mining equipment, ...) are typical examples
...)
where hydraulic actuators are used extensively.
extensively.
A typical motion application involves a
hydraulic cylinder connected to certain
mechanical linkages (inertia load). The motion
of the cylinder is regulated via a valve that is
used to regulate the flow rate to the cylinder. It
cylinder.
is well known that such systems chatter during
sudden stops and starts. Can you analyze the
starts.
cause and propose solutions? M RV RV
pS
pr School of Mechanical Engineering
Purdue University ME375 Hydraulic  13 Motion Control of Hydraulic Cylinders
Let’s look at a simplified problem:
The input in the system to the right is the
input flow rate qIN and the output is the
velocity of the mass, V.
A: Cylinder bore area
C: Cylinder chamber capacitance
B: Viscous friction coefficient between piston
head and cylinder wall.
wall.
• Derive the input/output model and transfer
function between qIN and V.
• Draw the block diagram of the system.
the block diagram of the system
• Can this model explain the vibration when we
suddenly close the valve? V A C M School of Mechanical Engineering
Purdue University pL pr
B qIN
RV
pS
pSr
pr ME375 Hydraulic  14 7 ME375 Handouts Motion Control of Hydraulic Cylinders
Element equations and interconnection equations: Take Laplace transforms: Block diagram representation: School of Mechanical Engineering
Purdue University ME375 Hydraulic  15 Motion Control of Hydraulic Cylinders
Transfer function between qIN and V: How would the velocity response look if we
suddenly open the valve to reach constant
input flow rate Q for some time T and
suddenly close the valve to stop the flow? Analyze the transfer function:
Natural Frequency Damping Ratio Steady State Gain School of Mechanical Engineering
Purdue University ME375 Hydraulic  16 8 ...
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This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue UniversityWest Lafayette.
 Fall '10
 Meckle

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