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Hydraulic (Fluid) Systems• Basic Modeling Elements
– Resistance
– Capacitance
– Inertance
– Pressure and Flow Sources
• Interconnection Relationships– Compatibility Law
– Continuity Law
• Derive Input/Output Models
The analogy between the hydraulic system and the electrical system will be used often. Just as in electrical systems, the flow rate (current) is defined to be the time rate of change (derivative) of volume (charge):
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 gauge pressure, p*. Gauge pressure is the difference between the absolute pressure and the atmospheric pressure, i.e.
Variables• q : volumetric flow rate [m3/sec] ( )
p p patmospheric*
dq V V
dt
current
voltage• p : pressure [N/m2] ( )
charge• V : volume [m3] ( )
• Fluid ResistanceDescribes any physical element with the characteristic that the pressure drop,p ,across the element is proportional to the flow rate, q.
– Orifices, valves, nozzles and friction in pipes can be modeled as fluid resistors.
Basic Modeling Elements
p p p p R q
qR
pR
p
1 2 12
12
1 1
Ex: The flow that goes through an orifice or a valve and the turbulent flow that goes through a pipe is related to the pressure drop by
Find the effective flow resistance of the element at certain operating point ( ).
q p, 12
q k p 12
q
p12
12 12
122
12,
1
2
2 2
q p
d q k
R d p p
p qR
k k
qR
+ p p1 p2+ p
R
p1 p2
p12
Basic Modeling Elements
• Fluid CapacitanceDescribes any physical element with the characteristic that the rate of change in pressure, p, in the element is proportional to the difference between the input flow rate, qIN , and the output flow rate, qOUT .
– Hydraulic cylinder chambers, tanks, and accumulators are examples of fluid capacitors.
gh
Cd
dtp p C p q qC ref Cr IN OUT
pCr
qIN
pref
C
pC qOUT
CqIN - qOUT
+ pCr
Ex: Consider an open tank with a constant cross-sectional area, A:
qIN qOUT
pC
h
pr
C
IN OUT
Cr
p
q q
p
C
d dVolum Ah Ah
dt dt
rgh p Crp gh
gh IN OUT
Cr
q q Ah A
p ggh
Ex:Will the effective capacitance change if in the previous open tank example, a load mass M is floating on top of the tank?
Ex: Calculate the equivalent fluid capacitance for a hydraulic chamber with only an inlet port.
Recall the bulk modulus () of a fluid is defined by:
Fluid Capacitance Examples
qIN
pr
C
pC
chamber volume V
Cr
C
V dq p
dt
qIN qOUT
pC
h
pr
M
Cr
Cr
dpdtdp
V VdVdV
dt
C
IN OUT
Cr
p
q q
p
C
d dVolum Ah Ah
dt dt
rMggh p A Cr
Mgp gh A
gh IN OUT
Cr
q q Ah A
p ggh
Basic Modeling Elements• Fluid Inertance (Inductance)
Describes any physical element with the characteristic that the pressure drop, p , across the element is proportional to the rate of change (derivative) of the flow rate, q.
– Long pipes are examples of fluid inertances.
1 1F Ap
p p p p Id
dtq I q 12 1 2( )
Ex: Consider a section of pipe with cross-sectional area A and length L, filled with fluid whose density is :
Start with force balance: F = ma Iq
p1 p2+ p
q
+ p
I
p1 p2
L
q
p1 p2+ p
A
IL
A
2 2F Ap1 1F Ap
1 2 1 2 12F F F A p p Ap m LA
12
mFa
dv d qAp AL AL
dt dt A
12
I
L dqp
A dt
• 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.
• Flow Source (Pump)– An ideal flow source is capable
of delivering the desired flow rate, regardless of the pressure required to drive the load.
Basic Modeling Elements
pS +p1 p2
pS q
p p p pS21 2 1
p1 p2
qS q
q qS
Voltage Source
Current Source
Interconnection Laws• Compatibility Law
– The sum of the pressure drops around a loop must be zero.
– Similar to the Kirchhoff’s voltage law.
• Continuity Law– The algebraic sum of the flow rates
at any junction in the loop is zero.
– This is the consequence of the conservation of mass.
– Similar to the Kirchhoff’s current law.
p pj ij ClosedLoop
ClosedLoop
0
q
q q
j
IN OUT
0AnyNode
or
q1q2
qo
q q qo1 2
A
B
C
pr
p1 p2
p p pr r1 12 2 0
Modeling Steps• Understand System Function and Identify Input/Output Variables
• Draw Simplified Schematics Using Basic Elements
• 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.
– Eliminate Intermediate Variables to Obtain Standard Forms:
• Laplace Transform
• Block Diagrams
In Class ExerciseDerive the input/output model for the following fluid system. The pump supplies a constant 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.
• Label the pressures at nodes and flow rates
• Write down element equations:1
1 2
1 2
sc o
c cr
c p
r eq
p q R
dq C p
dtd
p I qdt
p q R
pSpprr
pprr
pprr
Valve
sp1q
cp
cq2q
2p3p1p
+
_
C
oR
pIpR
vR NR
iRo
C
Rv
pr
ps
1qpc
cq
Ip1p pR 2p
3p
RN
eqR
2q
Equivalent electrical circuit
In Class Exercise• No. of unknowns and equations:
• 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. (Please refer to the previous slide)
1 1 2 1, , , , , ,sc s cr c rp q q p p q p
Loop 1:
Loop 2:
Node 2:
0sc cr srp p p
1 1 0c r rcp p p
1 2cq q q
we are interested in it
1 1 0cr srR q p p
2 2 0p eq cr
dI q R q p
dt
1 2crdP
q C qdt
2 2cr p eq
dp I q R q
dt
1
1 2 2 2
cr
cr p eq sr
pq
d dR C p q I q R q p
dt dt
2
2 21 1 1 22p eq p eq sr
d q dqR CI R CR I R R q p
dt dt
Motion Control of Hydraulic CylindersHydraulic actuation is attractive for applications when large power is needed while maintaining a reasonable weight. Not counting the weight of the pump and reservoir, hydraulic actuation has the edge in power-to-weight ratio compared with other cost effective actuation sources. Earth moving applications (wheel loaders, excavators, mining equipment, ...) are typical examples where hydraulic actuators are used 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 is well known that such system chatters during sudden stop and start. Can you analyze the cause and propose solutions?
M
pS
pprr
RRVVRRVV
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.
• Derive the input/output model and transfer function between qIN and V.
• Draw the block diagram of the system.
• Can this model explain the vibration when we suddenly close the valve?
M
pSr
pprr
RRVV
ppSS
qqININ
AA
ppLLpprr
CCvv
BB
i
C
pr
qIN1qcq
Lp
M
B
vvfc
Motion Control of Hydraulic CylindersElement equations and interconnection equations:
Block diagram representation:
Take Laplace transforms:
Hydraulic system Mechanical systemHydraulic-Mechanical
1
c Lr
IN c
dq C p
dtq q q
1
c Lrf Ap
q Av
cMv f Bv
1
c Lr
IN c
Q s CsP s
Q s Q s Q s
1
c LrF s AP s
Q s AV s
cMsV s F s BV s
Hydraulic System Mechanical SystemH-M Coupling
A LrP s X sQIN(s) 1
Cs
V s
-
A
1
Ms B1
s
V s
cF s
1Q s
Motion Control of Hydraulic CylindersTransfer function between qIN and V:
Analyze the transfer function:Natural Frequency
Damping Ratio
Steady State Gain
How would the velocity response look like 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?
2
2 2
22
2 nn
QVIN
V s AG
Q s MCs BCs A
AMC
B As sM MC
2
n
AAMC MC
2
2 2n
B A B CM BM A MMC
2
1AMCK
A AMC
In reality, large M, small C
very small damping ratio
reasonable value of natural frequency
Oscillation cannot die out quickly
Chattering !!
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