; ME375 Dynamic System Modeling and Control
Learning Center
Plans & pricing Sign in
Sign Out
Your Federal Quarterly Tax Payments are due April 15th Get Help Now >>

ME375 Dynamic System Modeling and Control


  • pg 1
									       CHAPTER 6
        MESB 374
System Modeling and Analysis

  Hydraulic (Fluid) Systems
Hydraulic (Fluid) Systems
 • Basic Modeling Elements
   –   Resistance
   –   Capacitance
   –   Inertance
   –   Pressure and Flow Sources

 • Interconnection Relationships
   – Compatibility Law
   – Continuity Law

 • Derive Input/Output Models
• q : volumetric flow rate [m3/sec]                     (   current       )
• V : volume [m3]                                       (   charge        )
• p : pressure [N/m2]                                   (   voltage      )

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):
                               q       V V
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.
                                p*  p  patmospheric
Basic Modeling Elements
• Fluid Resistance
  Describes any physical element with Ex: The flow that goes through an orifice or a
  the characteristic that the pressure    valve and the turbulent flow that goes
  drop, Dp , across the element is        through a pipe is related to the pressure
  proportional to the flow rate, q.       drop by
                                                      q  k p12
    p1 + Dp    p2           + Dp 
                        p1                p2   Find the effective flow resistance of the
q                                              element at certain operating point ( q , p12).
          R                     R     q              q
       Dp  p1  p2  p12  R  q                    q
                1     1
           q     Dp  p12
                R     R
                                                   1   dq              k
    – Orifices, valves, nozzles and                                 
                                                   R d p12  q , p  2 p12
      friction in pipes can be modeled                            12

      as fluid resistors.                                     2 p12  2q
                                                         R          2
                                                                k    k
  Basic Modeling Elements
                                                  Ex: Consider an open tank with a constant
  • Fluid Capacitance                                 cross-sectional area, A:
      Describes any physical element
      with the characteristic that the rate                                 pr
      of change in pressure, p, in the
      element is proportional to the                                  gh         h
      difference between the input flow                                      pC
      rate, qIN , and the output flow rate,             qIN                           qOUT
      qOUT .
                 pref                   + pCr 
                                                         pC          gh  pr  pCr   gh
              pC         qOUT
qIN                                                                  d            d
                                qIN - qOUT
                                             C    qIN  qOUT           Volum    Ah   Ah
             C                                                       dt           dt
          d                                             pCr          gh
        C    pC  pref  C  pCr  q IN  qOUT
                             &                                       qIN  qOUT    Ah   A
              4 3
          dt 1 24                                              C                   
                   pCr                                                   pCr       gh  g
       – Hydraulic cylinder chambers,
         tanks, and accumulators are
         examples of fluid capacitors.
Fluid Capacitance Examples
Ex: Calculate the equivalent fluid capacitance Ex: Will the effective capacitance change if in
    for a hydraulic chamber with only an inlet     the previous open tank example, a load
    port.                                          mass M is floating on top of the tank?
          qIN                                                               M
                               chamber volume
                     C               V
   Recall the bulk modulus (b ) of a fluid is
   defined by:                                               qIN                          qOUT
                     dpCr 
                         dt 
       b V
                 V                                  pC   gh  pr  Mg  pCr   gh  Mg A
            dV        dV   
                          dt                   qIN  qOUT     d           d
                                                                  Volum    Ah   Ah
                                                             dt        dt
                                                    pCr   gh
                                                                qIN  qOUT    Ah   A
            V  d                                       C                    
          q      p                                               pCr       gh  g
             b  dt Cr
 Basic Modeling Elements
• Fluid Inertance (Inductance)                     Ex: Consider a section of pipe with cross-
 Describes any physical element with the               sectional area A and length L, filled
 characteristic that the pressure drop, Dp , across    with fluid whose density is  :
 the element is proportional to the rate of change
 (derivative) of the flow rate, q.                    F1  Ap1 p1 + Dp  p2 F2  Ap2
                                                                   F  Ap
                                                                        1       1

      p1   + Dp       p2          + Dp                    q
                            p1                  p2                                              A
 q                                                                        L
                                      I     q        Start with force balance: F = ma

     Dp  p12  ( p1  p2 )  I
                                     q  I q
                                            &         F  F  F    1       2    A p1  p2   Ap12
                                                      m   LA
                                                                  dv    d q
                                                      Ap12   AL   AL  
  – Long pipes are examples of fluid                              dt    dt  A 
                                                      F       m
    inertances.                                              L dq a
                                                      p12 
                                                             A dt
                                                                               I
Basic Modeling Elements
                                                           Voltage Source
• Pressure Source (Pump)
  – An ideal pressure source of a                pS +
                                          p1              p2
    hydraulic system is capable of
    maintaining the desired pressure,                           q
    regardless of the flow required for
    what it is driving.                    p21  p2  p1  pS

• Flow Source (Pump)                                      Current Source
  – An ideal flow source is capable of     p1              p2
    delivering the desired flow rate,              qS
    regardless of the pressure
    required to drive the load.                  q  qS
 Interconnection Laws
• Compatibility Law                         • Continuity Law
  – The sum of the pressure drops             – The algebraic sum of the flow rates
    around a loop must be zero.                 at any junction in the loop is zero.
  – Similar to the Kirchhoff’s voltage        – This is the consequence of the
    law.                                        conservation of mass.
                                              – Similar to the Kirchhoff’s current
               Dp j     p      ij   0
                          Loop                                
                                                                     qj  0
         p1                            p2
                    B                                    or   q   IN      qOUT

         A                             C            q1                        q2

                                                                            q1  q2  qo
         pr1  p12  p2 r  0                                    qo
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
   – Eliminate Intermediate Variables to Obtain Standard Forms:
      • Laplace Transform
      • Block Diagrams
In Class Exercise
Derive 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.
             ps                C        qc             Ip
                      q1                     q2                        Rp
        pS                                                                                                pr
                          Ro       pc                             p1               p2 R          p3 R N

• Label the pressures at nodes and flow rates                q1                  Ip q2 p          Rp      p2
                                                                       pc               1
• Write down element equations:
                                                   p         Ro                      i
        psc  q1 Ro                               +s              qc                                Rv
                                                                                   Req                         p3
        qc  C      pcr                           _                     C
                dt                                                                                  RN
        pc1  I p    q2
                  dt                                                        pr
        p1r  q2 Req
                                                            Equivalent electrical circuit
In Class Exercise
• No. of unknowns and equations:
    psc , q1 , qs , pcr , pc1 , q2 , p1r

• Interconnection laws:                   we are interested in it

       Loop 1:        psc  pcr  psr  0  R1q1  pcr  psr  0
       Loop 2:        pc1  p1r  prc  0  I p                      q2  Req q2  pcr  0
       Node 2:        q1  qc  q2                     q1  C       q2
• Eliminate intermediate variables and obtain I/O model:
    pcr  I p      q2  Req q2
                                                                     d 2 q2
                                                               R1CI p 2   R1CReq  I p  2   R1  Req  q2  psr
                dt                                                                        dq
         d                  d                                       dt                   dt
   R1  C    pcr  q2   I p    q2  Req q2              psr
         dt                 dt
                 q1                       pcr

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)
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        M
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                  RV             RV
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?
Motion Control of Hydraulic Cylinders
                                                               v                       A                    C
Let’s look at a simplified problem:                  M
  The input in the system to the right is the                      pr             pL
  input flow rate qIN and the output is the                                 B
  velocity of the mass, V.
    A: Cylinder bore area                                                                          qIN
    C: Cylinder chamber capacitance                                                                        RV
    B: Viscous friction coefficient between                                                 pS
        piston head and cylinder wall.                                                           pSr
• Derive the input/output model and transfer
  function between qIN and V.
• Draw the block diagram of the system.                       pL
• Can this model explain the vibration when we                          i              fc
                                               qIN       qc
  suddenly close the valve?                                                  q1                        M

  Motion Control of Hydraulic Cylinders
Element equations and interconnection equations:
             Hydraulic system                 Hydraulic-Mechanical           Mechanical system
         qc  C            pLr                     f c  ApLr
                       dt                                                    Mv  f c  Bv
                                                   q1  Av
         qIN         q1  qc
Take Laplace transforms:
    Qc  s   CsPLr  s                          Fc  s   APLr  s 
                                                                               MsV  s   Fc  s   BV  s 
    QIN  s   Q1  s   Qc  s                 Q1  s   AV  s 

Block diagram representation:
    QIN(s)                   1
                                  PLr  s               Fc  s               V  s              X  s
                                                                      1                     1
              -              Cs                A                    Ms  B                  s
                  Q1  s 
                                               A         V  s

             Hydraulic System           H-M Coupling                     Mechanical System
Motion Control of Hydraulic Cylinders
Transfer function between qIN and V:                How would the velocity response look like
            V  s         A                        if we suddenly open the valve to reach
     GQV           
           QIN  s  MCs  BCs  A2
                        2                           constant input flow rate Q for some time T
                  A                                 and suddenly close the valve to stop the
                     MC                            flow?
         s2  B       s A
                  M            MC
               2n            n
                                                  In reality, large M, small C
Analyze the transfer function:
Natural Frequency                                 reasonable value of natural frequency
              2                        A
      n  A MC                                  very small damping ratio
Damping Ratio
         B                                        Oscillation cannot die out quickly
                                    2A   B    C
           M B                     
         2n           M            MC   2A   M   Chattering !!
Steady State Gain
     K          MC  1
               A      A

To top