4 Actuator Control
6.2 Steady State Modeling and Simulation
Physically, steady state simulation of a hydraulic system is characterized by:
• constant actuator speed
• constant pump speed
• incompressible fluid
• all mechanical parts in valves are stationary
The steady state equations for the basic components of a hydraulic system are gone through in the following tables. For each component a short descriptive name is given together with a symbol. Also, the governing equations are listed both in SI-units and typical fluid power units. The typical fluid power units (FLP-units) as defined in this note are listed in Table 6.1:
Table 6.1 SI-units and FLP-units
Symbol Description SI-units FLP-units
Q Flow m3 /s l/min
n Omdrejningstal o/s o/min
D Fortrængning m3/o cm3/o
M Moment Nm Nm
p Tryk N/m2 bar
F Kraft N N
v Hastighed m/s m/s
A Areal m 2 mm 2
This yields the following basic equations for some of the most common components in hydraulic systems. The pressure node is simply a volume of fluid without significant pressure variations.
Table 6.2 Steady state equations for basic components
Description Symbol Equations
SI-units
Equations Hyd-units
Pressure node
Qi
Q1
Qn
0 Q
1 i
∑ i
=
0 Q
1 i
∑ i
=
n = n =
Pump Q
n
D
D n Q= ⋅
Q=1000n⋅D
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Motor Q
In the above a number of simplifications have been made. As an example efficiencies are not included in the pump and actuator equations. Also, the spring of the check valve is ignored, the slope of the pressure relief valve characteristic is not included. The 2-way flow control valve is simplified to an orifice if the pressure drop is smaller than the closing pressure of the spring of the main spool and to a perfect flow controller if the pressure drop is above this value. All of these simplified models can, however, be adjusted/ if so desired. As an example, consider the pressure relief valve and imagine
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that the slope of the p-Q characteristics should be included. In that case the governing equations simply change into:
cr
α is the slope of the p-Q characteristic of the valve. If the slope for some reason is not constant but varies significantly this may also be introduced. It is simply a question of how detailed information is required at the current stage of the design evaluation. In Table 6.2 only a few valves are shown, however, similar equations may be set up for any type of valve.
The most important aspect of the governing equations for the valves in Table 6.2 are the fact that there are at least/typically two modes of operation. Hence, a governing equation exists for both modes and for each mode there is an inequality that must be fulfilled in order for the mode to be active. In the following the choice of mode of operation is referred to as the configuration parameter of the valve. For a pure steady state analysis the configuration parameter of each valve must be chosen beforehand, i.e., it is necessary to make a number of qualified guesses. Only af choosing/guessing the configuration parameter of each valve can the governing equations be formulated and solved. After solving the equations each choice of configuration parameter must be validated by checking wether the corresponding inequality is fulfilled. If not, the configuration parameter must be changed and the analysis redone. Potentially, this leads to possible different system configurations, where n is the number of valves with two modes of operation. In practice, however, the mode of operation of most valves are easily recognized for a given situation.
2n
A step-wise approach for steady-state analysis of any hydraulic system can now be set up:
1. Identify pressure nodes.
2. Identify the components that demarcate each pressure node.
3. Choose/guess configuration parameters for all components with more than one mode of operation.
4. Set up equations:
a) Flow continuity for each pressure node.
b) Flow continuity for each pump and actuator.
c) Static equilibrium for each actuator.
d) Flow through restrictions (orifices, filters etc.).
e) Equations associated with the choise of configuration parameters.
f) Static equilibrium for spool and poppet valves.
5. Solve equations numerically.
6. Are the computed variables physically meaningful?
• Yes: Analysis completed.
• No: Have all combinations of configuration parameters been examined?
• Yes: Analysis cannot be carried out.
• No: Go to 3) and choose/guess configuration parameters differently.
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As an example the hydraulic system shown in Figure 6.1 has been subjected to this type of analysis. From the figure it is seen that a total number of 9 equations has been established. In this case, there is a single component with two operation modes, namely the pressure relief valve, and the guess is that the valve is closed. For a given system with a known pump speed and a known output torque the 9 equations may be solved to yield the 9 unknowns: . Hence, pressure, flow and actuator speed are the classical output of steady state analysis, however, it is also possible to prescribe the motor speed and introduce the motor displacement as a variable thereby changing the equation solving from pure analysis to component sizing. After solving the equations it is necessary to examine that all pressures are non-negative and that the choice of configuration parameter is correct. In this case this is simply done by ensuring that is smaller than .
Figure 6.1. Hydraulic circuit and corresponding steady state equations.
Clearly, the number of equations might seem excessive, and the system of equations could easily be reduced substantially by substituting the different expressions into each other.
The set of equations are non-linear because of the orifice equations and therefore have to be solved numerically, typically by means of Newton-Raphson iteration. Often the numerical solver will have difficulties solving the equations simply because they are formulated in SI-units. If this is the case then one should simply reformulate the problem using FLP-units. Also, the numerical solver may encounter problems with the orifice equation because the sign of the pressure drop may become negative during iteration. This may be avoided by using the following formulation:
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⎪⎩
Also, numerical problems will be encountered if for some reason the correct solution includes zero flow through an orifice. In that case the orifice equation must be replaced with a restriction flow equation that reflects the laminar regime, i.e., Q∝Δp.