Four Hydraulic Cylinder Simulation

This example shows how to use Simulink® to create a model with four hydraulic cylinders. The model has a single pump and four actuators. The same pump pressure, p1, drives each cylinder assembly, and the sum of their flows loads the pump. Although each of the four control valves could be controlled independently, as in an active suspension system, all four receive the same commands in this model, a linear ramp in orifice area from zero to 0.002 sq.m.

Opening Model

To open the model, enter sldemo_hdryl4 at the command line. To run the model, on the Simulink Toolstrip, click Run.

The model logs relevant data to MATLAB workspace, in the Simulink.SimulationOutput object out. Signal logging data is stored within the out object, in a structure called sldemo_hydcyl4_output. Logged signals have a blue badge. For more information, see Visualize and Access Signal Logging Data.

Model Parameters

The pump flow begins at 0.005 m3/sec, then drops to 0.0025 m3/sec at t=0.05 sec. By assuming individual values for K, A, and beta, each one of the four cylinders exhibits a distinctive transient response. The table gives the characteristics of the four actuators.

----------------------------------------------------------------
Parameter       |  Actuator1   Actuator2   Actuator3   Actuator4
----------------|-----------------------------------------------
Spring Constant |  K           K/4         4K          K
Piston Area     |  Ac          Ac/4        4Ac         Ac
Bulk Modulus    |  Beta        Beta        Beta        Beta/1000
----------------------------------------------------------------
Beta = 7e8  Pa  [fluid bulk modulus]
K    = 5e4  N/m [spring constant]
Ac   = 1e-3 m^2 [cylinder cross-sectional area]

The ratio of area and spring constant is the same for all pistons, so the pistons should have the same steady-state output. The dominant time constant for each actuator subsystem is proportional to (result obtained from dimensional analysis), so you can expect the piston assembly 2 to be faster than assembly 1. Piston assembly 3 is expected to be slower than piston assembly 1 or 2. The piston assembly 4 has a significantly lower bulk modulus beta,as would be the case with air, thus you can expect piston assembly 4 to respond more sluggishly than piston assembly 1.

Results

These plots show the piston position and pump supply pressure during the simulation.

The four actuators receive the initial jolt of flow at t=0 as a pressure impulse. The pump pressure, p1, which is initially high, drops rapidly due to the high flow demand from the four loads. During the initial transient, about 4 msec, distinct responses identify the individual dynamic characteristics of each assembly unit.

As predicted by the parameter values, actuator 2 responds faster than actuator 1. The third and fourth pistons are much slower because they require more working fluid to move the same distance. For actuator 3, the piston displaces more volume due to its larger cross-sectional area. For actuator 4, although the displaced volume is the same as in actuator 1, the device requires more fluid because it is subsequently compressed.

As the pump pressure falls to the level within the cylinders, the distinctions in behavior blur. The individual responses blend into an overall system response that maintains the flow balance between the components. At t=0.05 sec, the pump flow drops to a level that is close to the equilibrium, and the actuator flows are nearly zero. The individual steady state piston positions are equal, as predicted by the design.